# A Theory of Besov and Triebel-Lizorkin Spaces on Metric Measure Spaces Modeled on Carnot-Carathéodory Spaces

Authors: Yongsheng Han; Detlef Müller; Dachun Yang
Journal: Abstract and Applied Analysis
 (2008)

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## Abstract

We work on RD-spaces𝒳, namely, spaces of homogeneous type in the
sense of Coifman and Weiss with the additional property that a reverse doubling property holds in 𝒳. An important example is the Carnot-Carathéodory
space with doubling measure. By constructing an approximation of the identity with bounded support of Coifman type, we develop a theory of Besov
and Triebel-Lizorkin spaces on the underlying spaces. In particular, this
includes a theory of Hardy spaces Hp(𝒳) and local Hardy spaces hp(𝒳) on RD-spaces, which appears to be new in this setting. Among other things, we
give frame characterization of these function spaces, study interpolation of
such spaces by the real method, and determine their dual spaces when p≥1.
The relations among homogeneous Besov spaces and Triebel-Lizorkin spaces,
inhomogeneous Besov spaces and Triebel-Lizorkin spaces, Hardy spaces, and
BMO are also presented. Moreover, we prove boundedness results on these
Besov and Triebel-Lizorkin spaces for classes of singular integral operators,
which include non-isotropic smoothing operators of order zero in the sense of
Nagel and Stein that appear in estimates for solutions of the Kohn-Laplacian
on certain classes of model domains in ℂN. Our theory applies in a wide
range of settings.


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## Body

## 1. Introduction

The scales ofBesov spacesBp,qs and Triebel-Lizorkin spacesFp,qs on ℝn, respectively, domains in ℝn
for the full range of
parameters, s∈ℝ and 0<p, q≤∞,
were introduced between 1959 and 1975. They cover many well-known classical concrete
function spaces such as Hölder-Zygmund spaces, Sobolev spaces, fractional
Sobolev spaces (also often referred to as Bessel-potential spaces), local Hardy
spaces, and bmo,
which have their own history. A comprehensive treatment of these function
spaces and their history can be found in Triebel's monographs [1, 2]. For further developments,
including analogous theories of function spaces on fractals, we refer to
[3–6].Metric spaces play a prominent role in many fields of
mathematics. In particular, they constitute natural generalizations of
manifolds admitting all kinds of singularities and still providing rich
geometric structure; see [7–9]. Analysis on metric measure spaces has been studied
quite intensively in recent years; see, for example, Semmes's survey [10] for a more detailed
discussion and references. Of particular interest is the study of functional
inequalities, like Sobolev and Poincaré inequalities, on metric measure spaces;
see, for example, [11–16]. Also the theory of
function spaces on metric measure spaces has seen a rapid development in recent
years. Since Hajłasz in [17] introduced Sobolev spaces on any metric measure
spaces, a series of papers has been devoted to the construction and
investigation of Sobolev spaces of various types on metric measure spaces; see,
for example, [12, 13, 18–27].It is well known that Calderón-Zygmund operators are
in general not bounded onL1(ℝn), and the Hardy space H1(ℝn) is a good substitute for L1(ℝn). Coifman and Weiss [28]
introduced atomic Hardy spaces Hatp(𝒳) for p∈(0,1] when 𝒳 is a general space of homogeneous type in the
sense of Coifman and Weiss [29], that is, μ is known only to be doubling. Moreover, under
the assumption that the measure of any ball in 𝒳 is equivalent to its radius (i.e., if 𝒳 is essentially a so-called Ahlfors1-regular metric measure space), Coifman
and Weiss [28] further
established a molecular characterization of Hat1(𝒳),
and if p∈(1/2,1], Macías and Segovia [30] gave a maximal function characterization of Hatp(𝒳).
For p in this range, a Lusin-area characterization
for Hatp(𝒳) was given in [31], and Duong and Yan in
[32] characterized
these atomic Hardy spaces in terms of Lusin-area functions associated with
certain Poisson semigroups. However, the results in [28, 30, 32] require that 𝒳 is an Ahlfors 1-regular metric measure space, and the methods
do not extend to arbitrary spaces of homogeneous type, even though Macías and
Segovia [33] proved
that any space of homogeneous type is topologically equivalent to an Ahlfors 1-regular metric measure space. On the other
hand, via Littlewood-Paley theory, a theory of Besov and Triebel-Lizorkin
spaces on Ahlfors n-regular metric measure spaces was established
in [34], which was
further completed in [2, 31, 35–37], and it turns out that some ideas from [31, 34, 35] still work on general
metric measure spaces considered in this paper.In this paper, we work on RD-spaces𝒳, that is, spaces of homogeneous type in the sense of Coifman and Weiss with the
additional property that a reverse doubling property holds in 𝒳, or equivalently, that there exists a constant a0>1 such that for all x∈𝒳 and 0<r<diam(𝒳)/a0,
the annulus B(x,a0r)∖B(x,r) is nonempty, where and in what follows, diam(𝒳) denotes the diameter of the metric space (𝒳,d). An important class of RD-spaces is provided by Carnot-Carathéodory spaces with
a doubling measure, which have been the object of intensive studies for quite a
while. A Carnot-Carathéodory (or sub-Riemannian) space is a connected smooth
manifold endowed with a Hörmander system of vectors {X1,X2,…,Xk},
which span, together with their commutators of order ≤m, the tangent space at each point, and the distance function d is in this case given by the
Carnot-Carathéodory or control distance associated with the Hörmander system.
Carnot-Carathéodory spaces arise in many places in mathematics, including
control theory, the theory of hypoelliptic differential operators, and several
areas of harmonic analysis and complex analysis; see
[38–40] for a general discussion
and detailed references. Also, this type of metric measure space plays an
important role in connection with problems related to snowflaked transforms;
see [13, page 99] or
[2]. Examples of noncompact
Carnot-Carathéodory spaces naturally arise as boundaries of unbounded model
polynomial domains in ℂN appearing in the work of Nagel and Stein; see
[41–43].Our theory of function spaces on RD-spaces will use
Carnot-Carathéodory spaces as a basic model, in particular well-known estimates
of heat kernels on such spaces.By constructing an approximation of the identity with
bounded support of Coifman type on RD-spaces, we develop a theory of Besov and
Triebel-Lizorkin spaces on the underlying spaces. In particular, this includes
a theory of Hardy spacesHp(𝒳) and local Hardy spaces hp(𝒳) on RD-spaces, which appears to be new in this
setting. Among other things, we give frame characterization of these function
spaces, study interpolation of such spaces by the real method and determine
their dual spaces when p≥1. The relations among homogeneous Besov spaces and Triebel-Lizorkin spaces,
inhomogeneous Besov spaces and Triebel-Lizorkin spaces, Hardy spaces, and BMO are also presented. Moreover, we prove
boundedness results on these Besov and Triebel-Lizorkin spaces for classes of
singular integral operators in [44], which include nonisotropic smoothing operators of
order zero in the sense of Nagel and Stein that appear in estimates for
solutions of the Kohn-Laplacian on certain classes of model domains in ℂN; see [43, 45–47].We point out that a theory of Hardy spaces on
RD-spaces was also established in [48] by using spaces of test functions, the theory of
distributions, and the boundedness criterion of singular integrals on spaces of
test functions, which are all developed in the current paper, and by assuming that
there exists a suitable Calderón reproducing formula in L2(𝒳).
These Hardy spaces are proved to coincide with some of Triebel-Lizorkin spaces
in this paper. Also, some spaces of Lipschitz type on RD-spaces were recently
studied in [49]. Via
Calderón reproducing formulae developed in the current paper, the relations of
Besov and Triebel-Lizorkin spaces introduced in the current paper with those
spaces of Lipschitz type in [49] and with various known Sobolev spaces as mentioned
above were established in [49]. As an application, a difference characterization of
Besov and Triebel-Lizorkin on RD-spaces was also obtained in [49]. Moreover, it is possible
to establish smooth atomic and molecular characterizations and lifting property
of these spaces by using fractional integrals and derivatives, and it is also
possible to develop a corresponding product theory. However, to limit the
length of this paper, we will leave these topics to forthcoming papers.Our setting of RD-spaces includes spaces with a
“local” dimension strictly less than the global dimension, such as
certain classes of nilpotent Lie groups. In such situations, the Lipschitz
classes (in the sense of Coifman and Weiss [28]) that we consider in this paper do not compare with
the usual Hölder classes (which are particular Besov spaces), which is why some
of our results, for example, about duality of Besov and Triebel-Lizorkin
spaces, assume a different form compared to the corresponding “classical”
results on ℝn,
respectively, on Ahlfors n-regular spaces.


### 1.1. Underlying Spaces

We first recall the notion of a space of homogeneous
type in the sense of Coifman and Weiss [28, 29] and then introduce the so-called RD-spaces, which are
particular spaces of homogeneous type.Definition 1.1.
Let(𝒳,d) be a metric space with a regular Borel measureμ, which means that μ is a nonnegative countably subadditive set
function defined on all subsets of 𝒳, open sets are measurable, and every set is contained in a Borel set with the
same measure, such that all balls defined by d have finite and positive measure. For any x∈𝒳 and r>0, set
(1.1)B(x,r)={y∈𝒳:d(x,y)<r}.
(i)
The triple(𝒳,d,μ) is called a space of homogeneous type if there exists
a constant C0≥1 such that for all x∈𝒳 and r>0,(1.2)μ(B(x,2r))≤C0μ(B(x,r))(doublingproperty).(ii)
Let0<κ≤n. The triple (𝒳,d,μ) is called a (κ,n)‐space if there exist constants 0<C1≤1 and C2≥1 such that for all x∈𝒳, 0<r<diam(𝒳)/2, and 1≤λ<diam(𝒳)/(2r),
(1.3)C1λκμ(B(x,r))≤μ(B(x,λr))≤C2λnμ(B(x,r)),
where diam(𝒳)=supx,y∈𝒳d(x,y).
A space of homogeneous type will be called anRD-space if it is a (κ,n)-space for some 0<κ≤n, that is, if some “reverse” doubling condition holds.Clearly, anyAhlforsn-regular metric measure space(𝒳,d,μ), which means that there exists some n>0 such that μ(B(x,r))~rn for all x∈𝒳 and 0<r<diam(𝒳)/2, is an (n,n)-space.Remark 1.2.
(i) Obviously, any(k,n)-space is a space of homogeneous type with C0=C22n. Conversely, any space of homogeneous type satisfies the second inequality of
(1.3) with C2=C0 and n=log2C0. Comparing with spaces of homogeneous type, the only additional restriction in (k,n)-spaces is the first inequality of (1.3).
If𝒳 is a (κ,n)-space, the first inequality in (1.3) implies
that there exist constants a0>1/C11/κ≥1 and C˜1=C1a0κ>1 such that for all x∈𝒳 and 0<r<diam(𝒳)/a0,
(1.4)μ(B(x,a0r))≥C˜1μ(B(x,r))(reversedoublingproperty),
(if a0=2, this is the classical reverse
doubling condition), and therefore,
(1.5)B(x,a0r)∖B(x,r)≠∅.
Conversely, assume thatμ satisfies the second inequality of (1.3)
(i.e., μ is doubling) and (1.5) holds for some a0>1 and for all x∈𝒳 and 0<r<diam(𝒳)/a0. Then, 𝒳 is a (κ,n)-space for some κ>0.
To see this, by ideas in [50, pages 11-12] (see also [51, pages 269-270]), it
suffices to show that there exist constants a1>a0 and C˜1>1 such that for all x∈𝒳 and 0<r<diam(𝒳)/2a0,(1.6)μ(B(x,a1r))≥C˜1μ(B(x,r)),
which further implies that the
first inequality in (1.3) also holds for some κ>0, that is, 𝒳 is a (κ,n)-space. To this end, fix any σ∈(0,1]. Then, if 0<r<diam(𝒳)/2a0, we have (1+σ)r<diam(𝒳)/a0. Thus, by the assumption,
(1.7)B(x,a0(1+σ)r)∖B(x,(1+σ)r)≠∅.
Choose y∈B(x,a0(1+σ)r)∖B(x,(1+σ)r). It is easy to see that(1.8)B(y,σr)∩B(x,r)=∅,B(y,σr)⊂B(x,[σ+a0(1+σ)]r), and
(1.9)B(x,[σ+a0(1+σ)]r)⊂B(y,[σ+2a0(1+σ)]r),
which together with the second inequality in (1.3) imply
(1.10)μ(B(x,[σ+a0(1+σ)]r))≥μ(B(x,r))+μ(B(y,σr))≥μ(B(x,r))+C2−1(σσ+2a0(1+σ))nμ(B(y,[σ+2a0(1+σ)]r))≥μ(B(x,r))+C2−1(σσ+2a0(1+σ))nμ(B(x,[σ+a0(1+σ)]r)).
This implies (1.6) with a1≡σ+a0(1+σ)>a0 and
(1.11)C˜1≡[1−C2−1(σσ+2a0(1+σ))n]−1>1.Thus, 𝒳 is a (κ,n)-space.
Therefore,𝒳is an RD-space if and only if𝒳is a space of homogeneous type with the
additional property that there exists a constanta0>1such that for allx∈𝒳and0<r<diam(𝒳)/a0, B(x,a0r)∖B(x,r)≠∅.
(ii) From (i), it is obvious that an RD-space has no
isolated points.
(iii) It is proved in [7–9] that some curvature-dimension condition on metric
measure spaces implies the doubling property of the considered measure.Remark 1.3.
We recall that two metricsd and d˜ are said to be equivalent if d/d˜ is uniformly bounded and uniformly bounded
away from zero. In what follows, we always regard two (κ,n)-spaces or spaces of homogeneous type with
equivalent metrics as the  same space.Remark 1.4.
Letd be a quasimetric, which means that d is a nonnegative and symmetric function on 𝒳×𝒳, d(x,y)=0 if and only if x=y, and there exists a constant A0≥1 such that for all x,y,z∈𝒳,
(1.12)d(x,y)≤A0(d(x,z)+d(z,y)).
Macías and Segovia [33] proved that for any
quasimetric d, there exists an equivalent quasimetric d˜ such that all balls corresponding to d˜ are open  in the topology induced by d˜, and there exist constants A0′>0 and θ∈(0,1) such that for all x,y,z∈𝒳,
(1.13)|d˜(x,z)−d˜(y,z)|≤A0′d˜(x,y)θ[d˜(x,z)+d˜(y,z)]1−θ,
which means d˜ has some regularity; see [33, Theorem 2].
Notice that all results below are true ifd is a quasimetric and has some regularity, and
Remark 1.3 is also true for both such equivalent quasimetrics. From this and
the above result of Macías and Segovia, it follows that all results in this
paper are still true if d on 𝒳 is only known to be a quasimetric (especially,
if 𝒳 is a so-called d-space of Triebel; see [2, page 189]), which is
another advantage of the results in this paper compared to all known results so
far.
In what follows, for the simplicity of the
presentation, we always assume thatd is a metric, which means A0=1 and θ=1.Remark 1.5.
If𝒳 is a closed subset of ℝn and 𝒳 is a (κ,n)-space with the additional normalization
assumption that μ(B(x,1))~1 for all x∈𝒳, Jonsson in [52, 53] introduced certain Besov spaces for some special
indices on such sets with the aid of difference or local polynomial
approximation (equivalently by atoms) and obtained some trace theorems for
restrictions to 𝒳.
Moreover, Bylund and Gudayol [54] gave out some conditions
such that a compact pseudometric space becomes a (κ,n)-space with μ being a probability measure.Remark 1.6.
Under rather general
circumstances, regular Borel measures have two useful properties, which are
often calledinner regularity and outer regularity. More precisely,
if μ is a regular Borel measure, then for each
Borel set A of finite measure, μ(A) is the supremum of the numbers μ(C), where C runs through all closed subsets of A; moreover, if the metric balls have finite measure, and A is a Borel set, then μ(A) is the infimum of the numbers μ(U), where U runs through all open supersets of A; see [55, Theorem 2.2.2] or [13, page 3]. These properties are used in establishing Lebesgue's
differentiation theorem; see [13,
pages 4–7] for the details.We now recall the definition ofCarnot-Carathéodory spaces. Let 𝒳 be a connected smooth manifold and let {X1,…,Xk} be k given smooth real vector fields on 𝒳 satisfying Hörmander's condition of orderm, that is, these vector fields together with their commutators of order at most m span the tangent space to 𝒳 at each point. To develop a theory of Besov
and Triebel-Lizorkin spaces on 𝒳, we make use of the notion of control distances associated to the vector fields.
One possibility of defining a control
distance is as follows: for x,y∈𝒳 and δ>0, let AC(x,y,δ) denote the collection of absolutely continuous
mappings
φ: [0,1]→𝒳 with φ(0)=x and φ(1)=y such that for almost every t∈[0,1], φ'(t)=∑j=1kajXj(φ(t)), with |aj|≤δ. Then, the control metric d(x,y) from x to y is the infimum of the set of δ>0 such that AC(x,y,δ)≠∅. Hörmander's condition makes sure that d(x,y)<∞ for every x,y∈𝒳.The following three specific examples of
Carnot-Carathéodory spaces which are also(κ,n)-spaces naturally come from harmonic analysis
and several complex variables.
(a)
Compact
case. If 𝒳 is a compactn-dimensional Carnot-Carathéodory space and
is endowed with any fixed smooth measure μ with strictly positive density, by [56, Theorem 1] (or [42, Theorem 2.2.4]), we know
that 𝒳 is an (n,nm)-space; see also [44].(b)
Noncompact case. Let Ω={(z,w)∈ℂ2:ℑm[w]>P(z)}, where P is a real, subharmonic, nonharmonic polynomial
of degree m. Namely, Ω is an unbounded model domain of polynomial type in ℂ2. Then, 𝒳=∂Ω can be identified with ℂ×ℝ={(z,t):z∈ℂ,t∈ℝ}. The basic (0,1) Levi vector field is then Z¯=∂/∂z¯−i(∂P/∂z¯)(∂/∂t), and we write Z¯=X1+iX2. The real vector fields {X1,X2} and their commutators of orders ≤m span the tangent space at each point. If we
endow ℂ×ℝ with the Lebesgue measure, then by [43, Proposition 3.1.1] we know
that 𝒳=ℂ×ℝ is a (4,m+2)-space; see also [42, 44, 56].(c)
(Noncompact case)Lie groups of polynomial growth (see
[56–60]). Let G be a connected Lie group and fix a left
invariant Haar measure μ on G. We assume that G has polynomial volume growth, that is, if U is a compact neighborhood of the identity
element e of G, then there is a constant C>0 such that μ(Un)≲nC for all n∈ℕ (see [56]). Then G is unimodular. Furthermore, there is a
nonnegative integer n∞ such that μ(Un)~nn∞ as n→∞; see [61] and also
[57, 62]. Let X1,…,Xn be left invariant vector fields on G that satisfy Hörmander's condition, that is,
they together with their successive Lie brackets [Xi1,[Xi2,[…,Xik]…] span the tangent space of G at every point of G. Let d be the associated control metric. Then this
metric is left invariant and compatible with the topology on G; see [57, 58]. Moreover, by the results
in [56, 57], we know that there is n0∈ℕ, independent of x, such that μ(B(x,r))~rn0 when 0<r≤1, and μ(B(x,r))~rn∞ when r≥1. From this, it is easy to verify that G is a (min{n0,n∞},max{n0,n∞})-space.


### 1.2. Outline of Some Basic Methods

Let𝒳 be a Carnot-Carathéodory space as in Case (a)
with n≥3 or as in Case (b), and let {X1,…,Xk} be a family of real vector fields on 𝒳, which are of finite-type m. Consider the sub-Laplacianℒ on 𝒳 in self-adjoint form, given by
(1.14)ℒ=∑j=1kXj∗Xj.
Here (Xj∗φ,ψ)=(φ,Xjψ), where
(1.15)(φ,ψ)=∫𝒳φ(x)ψ(x)¯dμ(x)
and φ,ψ∈Cc∞(𝒳), the space of C∞ functions on
𝒳 with compact support. In general, Xj∗=−Xj+aj, where aj∈𝒞∞(𝒳). The solution of the following initial value problem for the
heat equation,(1.16)∂u∂s(x,s)+ℒxu(x,s)=0
with u(x,0)=f(x), is given by us(x,s)=Hs[f](x), where Hs is the operator given via the spectral theorem
by Hs=e−sℒ, and an appropriate self-adjoint extension of the nonnegative operator ℒ initially defined on Cc∞(𝒳). It was proved in [44]
that for f∈L2(𝒳),
(1.17)Hs[f](x)=∫𝒳H(s,x,y)f(y)dμ(y),
where x∈𝒳. Moreover, H(s,x,y) has some nice regularity properties (see
[44, Proposition
2.3.1] and [63]).By abstracting from the properties ofheat kernelsH(s,x,y), we will introduce notions of approximations of the identity and spaces of test
functions and their dual spaces on arbitrary spaces of homogeneous type. These
two spaces are, respectively, the substitutes of the space of Schwartz
functions and the space of tempered distributions; see, for example, [31, 34]. Following Coifman's idea
in [64], we then
construct an approximation of the identity with bounded support on metric
measure spaces. We will show that our spaces of test functions are invariant
under a large class of singular integral operators. The integral kernels
associated to these operators satisfy a certain
“second difference regularity condition,” which also turns out to be
necessary to this result; see [31, 35]. This theorem is a main tool for establishing a
Calderón reproducing formula. Let {Sk}k∈ℤ be such an approximation of the identity. Set
next Dk=Sk−Sk−1 for k∈ℤ. Based on Coifman's ideas (see [64] for the details), on Lp(𝒳) with p∈(1,∞), we can decompose the identity operator I in the strong sense as(1.18)I=∑k=−∞∞Dk=(∑k=−∞∞Dk)(∑j=−∞∞Dj)=∑|k−j|≤NDkDj+∑|k−j|>NDkDj=TN+RN.The error term RN will be small of order 2−δN in norm with δ>0. Using the above-mentioned theorem, we prove that if N is large enough, then RN is bounded on the space of test functions with
the operator norm less than 1. Therefore, if N is large enough and DkN=∑|j|≤NDk+j for k∈ℤ, we then obtain the following Calderón-type reproducing  formulae:(1.19)f=∑k=−∞∞TN−1DkNDk(f)=∑k=−∞∞DkDkNTN−1(f),where TN−1 is the inverse of TN and the series converge in Lp(𝒳), 1<p<∞, and in the space of test functions and its dual space. We also obtain a
corresponding discrete version. The Calderón reproducing formula is another
main tool of this paper. As soon as (1.19) is established, we can introduce
Besov and Triebel-Lizorkin spaces on 𝒳 via approximations of the identity and prove
that these spaces are independent of the choice of approximations of the
identity. The Calderón reproducing formula is also employed to establish the
atomic decomposition of these spaces and to prove boundedness results for
operators acting on these spaces.Remark 1.7.
Let{Sk}k∈ℤ be an approximation of the identity as
constructed in Theorem 2.6 below. An essential difference in the theory occurs
when the measure, μ(𝒳), of the underlying space 𝒳 is finite versus the case when the measure of
the underlying space is infinite. When μ(𝒳)<∞, it is not true that ∥Sk(f)∥Lp(𝒳)→0 as k→−∞ for all p∈(1,∞) and f∈Lp(𝒳); see the proof of Proposition 3.1(i) below. Thus, when μ(𝒳)<∞, in the Calderón-type reproducing formulae (1.19), we should replace D0 by S0 and Dk for k∈{−1,−2,…} by 0. This means that when μ(𝒳)<∞,
we always have an inhomogeneous term corresponding to S0, which needs some special care.


### 1.3. Notation

Finally, we introduce some notation and make some
conventions. Throughout the paper,A~B means that the ratio A/B is bounded and bounded away from zero by
constants that do not depend on the relevant variables in A and B; A≲B and B≳A mean that the ratio A/B is bounded by a constant independent of the
relevant variables. For any p∈[1,∞], we denote by p′ its conjugate index, namely, 1/p+1/p′=1. We also denote by C a positive constant which is independent of
the main parameters, but it may vary from line to line. Constants with
subscripts, such as C0, do not change in different occurrences. If E is a subset of a metric space (𝒳,d), we denote by χE the characteristic function of E and define(1.20)diamE=supx,y∈Ed(x,y).
We also set ℕ={1,2,…} and ℤ+=ℕ∪{0}. For any a,b∈ℝ, we denote min{a,b}, max{a,b}, and max{a,0} by a⋀b, a∨b, and a+, respectively.If(𝒳,d,μ) is a space of homogeneous type, we also
introduce the volume functionsVδ(x)=μ(B(x,δ)) and V(x,y)=μ(B(x,d(x,y))) for all x,y∈𝒳 and δ>0. By (1.2), it is easy to see that V(x,y)~V(y,x); see also [44]. We will use this fact without further mentioning.Throughout the whole paper, forϵ∈(0,1] and |s|<ϵ, we set(1.21)p(s,ϵ)≡max{nn+ϵ,nn+ϵ+s}.


## 1.1. Underlying Spaces

We first recall the notion of a space of homogeneous
type in the sense of Coifman and Weiss [28, 29] and then introduce the so-called RD-spaces, which are
particular spaces of homogeneous type.Definition 1.1.
Let(𝒳,d) be a metric space with a regular Borel measureμ, which means that μ is a nonnegative countably subadditive set
function defined on all subsets of 𝒳, open sets are measurable, and every set is contained in a Borel set with the
same measure, such that all balls defined by d have finite and positive measure. For any x∈𝒳 and r>0, set
(1.1)B(x,r)={y∈𝒳:d(x,y)<r}.
(i)
The triple(𝒳,d,μ) is called a space of homogeneous type if there exists
a constant C0≥1 such that for all x∈𝒳 and r>0,(1.2)μ(B(x,2r))≤C0μ(B(x,r))(doublingproperty).(ii)
Let0<κ≤n. The triple (𝒳,d,μ) is called a (κ,n)‐space if there exist constants 0<C1≤1 and C2≥1 such that for all x∈𝒳, 0<r<diam(𝒳)/2, and 1≤λ<diam(𝒳)/(2r),
(1.3)C1λκμ(B(x,r))≤μ(B(x,λr))≤C2λnμ(B(x,r)),
where diam(𝒳)=supx,y∈𝒳d(x,y).
A space of homogeneous type will be called anRD-space if it is a (κ,n)-space for some 0<κ≤n, that is, if some “reverse” doubling condition holds.Clearly, anyAhlforsn-regular metric measure space(𝒳,d,μ), which means that there exists some n>0 such that μ(B(x,r))~rn for all x∈𝒳 and 0<r<diam(𝒳)/2, is an (n,n)-space.Remark 1.2.
(i) Obviously, any(k,n)-space is a space of homogeneous type with C0=C22n. Conversely, any space of homogeneous type satisfies the second inequality of
(1.3) with C2=C0 and n=log2C0. Comparing with spaces of homogeneous type, the only additional restriction in (k,n)-spaces is the first inequality of (1.3).
If𝒳 is a (κ,n)-space, the first inequality in (1.3) implies
that there exist constants a0>1/C11/κ≥1 and C˜1=C1a0κ>1 such that for all x∈𝒳 and 0<r<diam(𝒳)/a0,
(1.4)μ(B(x,a0r))≥C˜1μ(B(x,r))(reversedoublingproperty),
(if a0=2, this is the classical reverse
doubling condition), and therefore,
(1.5)B(x,a0r)∖B(x,r)≠∅.
Conversely, assume thatμ satisfies the second inequality of (1.3)
(i.e., μ is doubling) and (1.5) holds for some a0>1 and for all x∈𝒳 and 0<r<diam(𝒳)/a0. Then, 𝒳 is a (κ,n)-space for some κ>0.
To see this, by ideas in [50, pages 11-12] (see also [51, pages 269-270]), it
suffices to show that there exist constants a1>a0 and C˜1>1 such that for all x∈𝒳 and 0<r<diam(𝒳)/2a0,(1.6)μ(B(x,a1r))≥C˜1μ(B(x,r)),
which further implies that the
first inequality in (1.3) also holds for some κ>0, that is, 𝒳 is a (κ,n)-space. To this end, fix any σ∈(0,1]. Then, if 0<r<diam(𝒳)/2a0, we have (1+σ)r<diam(𝒳)/a0. Thus, by the assumption,
(1.7)B(x,a0(1+σ)r)∖B(x,(1+σ)r)≠∅.
Choose y∈B(x,a0(1+σ)r)∖B(x,(1+σ)r). It is easy to see that(1.8)B(y,σr)∩B(x,r)=∅,B(y,σr)⊂B(x,[σ+a0(1+σ)]r), and
(1.9)B(x,[σ+a0(1+σ)]r)⊂B(y,[σ+2a0(1+σ)]r),
which together with the second inequality in (1.3) imply
(1.10)μ(B(x,[σ+a0(1+σ)]r))≥μ(B(x,r))+μ(B(y,σr))≥μ(B(x,r))+C2−1(σσ+2a0(1+σ))nμ(B(y,[σ+2a0(1+σ)]r))≥μ(B(x,r))+C2−1(σσ+2a0(1+σ))nμ(B(x,[σ+a0(1+σ)]r)).
This implies (1.6) with a1≡σ+a0(1+σ)>a0 and
(1.11)C˜1≡[1−C2−1(σσ+2a0(1+σ))n]−1>1.Thus, 𝒳 is a (κ,n)-space.
Therefore,𝒳is an RD-space if and only if𝒳is a space of homogeneous type with the
additional property that there exists a constanta0>1such that for allx∈𝒳and0<r<diam(𝒳)/a0, B(x,a0r)∖B(x,r)≠∅.
(ii) From (i), it is obvious that an RD-space has no
isolated points.
(iii) It is proved in [7–9] that some curvature-dimension condition on metric
measure spaces implies the doubling property of the considered measure.Remark 1.3.
We recall that two metricsd and d˜ are said to be equivalent if d/d˜ is uniformly bounded and uniformly bounded
away from zero. In what follows, we always regard two (κ,n)-spaces or spaces of homogeneous type with
equivalent metrics as the  same space.Remark 1.4.
Letd be a quasimetric, which means that d is a nonnegative and symmetric function on 𝒳×𝒳, d(x,y)=0 if and only if x=y, and there exists a constant A0≥1 such that for all x,y,z∈𝒳,
(1.12)d(x,y)≤A0(d(x,z)+d(z,y)).
Macías and Segovia [33] proved that for any
quasimetric d, there exists an equivalent quasimetric d˜ such that all balls corresponding to d˜ are open  in the topology induced by d˜, and there exist constants A0′>0 and θ∈(0,1) such that for all x,y,z∈𝒳,
(1.13)|d˜(x,z)−d˜(y,z)|≤A0′d˜(x,y)θ[d˜(x,z)+d˜(y,z)]1−θ,
which means d˜ has some regularity; see [33, Theorem 2].
Notice that all results below are true ifd is a quasimetric and has some regularity, and
Remark 1.3 is also true for both such equivalent quasimetrics. From this and
the above result of Macías and Segovia, it follows that all results in this
paper are still true if d on 𝒳 is only known to be a quasimetric (especially,
if 𝒳 is a so-called d-space of Triebel; see [2, page 189]), which is
another advantage of the results in this paper compared to all known results so
far.
In what follows, for the simplicity of the
presentation, we always assume thatd is a metric, which means A0=1 and θ=1.Remark 1.5.
If𝒳 is a closed subset of ℝn and 𝒳 is a (κ,n)-space with the additional normalization
assumption that μ(B(x,1))~1 for all x∈𝒳, Jonsson in [52, 53] introduced certain Besov spaces for some special
indices on such sets with the aid of difference or local polynomial
approximation (equivalently by atoms) and obtained some trace theorems for
restrictions to 𝒳.
Moreover, Bylund and Gudayol [54] gave out some conditions
such that a compact pseudometric space becomes a (κ,n)-space with μ being a probability measure.Remark 1.6.
Under rather general
circumstances, regular Borel measures have two useful properties, which are
often calledinner regularity and outer regularity. More precisely,
if μ is a regular Borel measure, then for each
Borel set A of finite measure, μ(A) is the supremum of the numbers μ(C), where C runs through all closed subsets of A; moreover, if the metric balls have finite measure, and A is a Borel set, then μ(A) is the infimum of the numbers μ(U), where U runs through all open supersets of A; see [55, Theorem 2.2.2] or [13, page 3]. These properties are used in establishing Lebesgue's
differentiation theorem; see [13,
pages 4–7] for the details.We now recall the definition ofCarnot-Carathéodory spaces. Let 𝒳 be a connected smooth manifold and let {X1,…,Xk} be k given smooth real vector fields on 𝒳 satisfying Hörmander's condition of orderm, that is, these vector fields together with their commutators of order at most m span the tangent space to 𝒳 at each point. To develop a theory of Besov
and Triebel-Lizorkin spaces on 𝒳, we make use of the notion of control distances associated to the vector fields.
One possibility of defining a control
distance is as follows: for x,y∈𝒳 and δ>0, let AC(x,y,δ) denote the collection of absolutely continuous
mappings
φ: [0,1]→𝒳 with φ(0)=x and φ(1)=y such that for almost every t∈[0,1], φ'(t)=∑j=1kajXj(φ(t)), with |aj|≤δ. Then, the control metric d(x,y) from x to y is the infimum of the set of δ>0 such that AC(x,y,δ)≠∅. Hörmander's condition makes sure that d(x,y)<∞ for every x,y∈𝒳.The following three specific examples of
Carnot-Carathéodory spaces which are also(κ,n)-spaces naturally come from harmonic analysis
and several complex variables.
(a)
Compact
case. If 𝒳 is a compactn-dimensional Carnot-Carathéodory space and
is endowed with any fixed smooth measure μ with strictly positive density, by [56, Theorem 1] (or [42, Theorem 2.2.4]), we know
that 𝒳 is an (n,nm)-space; see also [44].(b)
Noncompact case. Let Ω={(z,w)∈ℂ2:ℑm[w]>P(z)}, where P is a real, subharmonic, nonharmonic polynomial
of degree m. Namely, Ω is an unbounded model domain of polynomial type in ℂ2. Then, 𝒳=∂Ω can be identified with ℂ×ℝ={(z,t):z∈ℂ,t∈ℝ}. The basic (0,1) Levi vector field is then Z¯=∂/∂z¯−i(∂P/∂z¯)(∂/∂t), and we write Z¯=X1+iX2. The real vector fields {X1,X2} and their commutators of orders ≤m span the tangent space at each point. If we
endow ℂ×ℝ with the Lebesgue measure, then by [43, Proposition 3.1.1] we know
that 𝒳=ℂ×ℝ is a (4,m+2)-space; see also [42, 44, 56].(c)
(Noncompact case)Lie groups of polynomial growth (see
[56–60]). Let G be a connected Lie group and fix a left
invariant Haar measure μ on G. We assume that G has polynomial volume growth, that is, if U is a compact neighborhood of the identity
element e of G, then there is a constant C>0 such that μ(Un)≲nC for all n∈ℕ (see [56]). Then G is unimodular. Furthermore, there is a
nonnegative integer n∞ such that μ(Un)~nn∞ as n→∞; see [61] and also
[57, 62]. Let X1,…,Xn be left invariant vector fields on G that satisfy Hörmander's condition, that is,
they together with their successive Lie brackets [Xi1,[Xi2,[…,Xik]…] span the tangent space of G at every point of G. Let d be the associated control metric. Then this
metric is left invariant and compatible with the topology on G; see [57, 58]. Moreover, by the results
in [56, 57], we know that there is n0∈ℕ, independent of x, such that μ(B(x,r))~rn0 when 0<r≤1, and μ(B(x,r))~rn∞ when r≥1. From this, it is easy to verify that G is a (min{n0,n∞},max{n0,n∞})-space.


## 1.2. Outline of Some Basic Methods

Let𝒳 be a Carnot-Carathéodory space as in Case (a)
with n≥3 or as in Case (b), and let {X1,…,Xk} be a family of real vector fields on 𝒳, which are of finite-type m. Consider the sub-Laplacianℒ on 𝒳 in self-adjoint form, given by
(1.14)ℒ=∑j=1kXj∗Xj.
Here (Xj∗φ,ψ)=(φ,Xjψ), where
(1.15)(φ,ψ)=∫𝒳φ(x)ψ(x)¯dμ(x)
and φ,ψ∈Cc∞(𝒳), the space of C∞ functions on
𝒳 with compact support. In general, Xj∗=−Xj+aj, where aj∈𝒞∞(𝒳). The solution of the following initial value problem for the
heat equation,(1.16)∂u∂s(x,s)+ℒxu(x,s)=0
with u(x,0)=f(x), is given by us(x,s)=Hs[f](x), where Hs is the operator given via the spectral theorem
by Hs=e−sℒ, and an appropriate self-adjoint extension of the nonnegative operator ℒ initially defined on Cc∞(𝒳). It was proved in [44]
that for f∈L2(𝒳),
(1.17)Hs[f](x)=∫𝒳H(s,x,y)f(y)dμ(y),
where x∈𝒳. Moreover, H(s,x,y) has some nice regularity properties (see
[44, Proposition
2.3.1] and [63]).By abstracting from the properties ofheat kernelsH(s,x,y), we will introduce notions of approximations of the identity and spaces of test
functions and their dual spaces on arbitrary spaces of homogeneous type. These
two spaces are, respectively, the substitutes of the space of Schwartz
functions and the space of tempered distributions; see, for example, [31, 34]. Following Coifman's idea
in [64], we then
construct an approximation of the identity with bounded support on metric
measure spaces. We will show that our spaces of test functions are invariant
under a large class of singular integral operators. The integral kernels
associated to these operators satisfy a certain
“second difference regularity condition,” which also turns out to be
necessary to this result; see [31, 35]. This theorem is a main tool for establishing a
Calderón reproducing formula. Let {Sk}k∈ℤ be such an approximation of the identity. Set
next Dk=Sk−Sk−1 for k∈ℤ. Based on Coifman's ideas (see [64] for the details), on Lp(𝒳) with p∈(1,∞), we can decompose the identity operator I in the strong sense as(1.18)I=∑k=−∞∞Dk=(∑k=−∞∞Dk)(∑j=−∞∞Dj)=∑|k−j|≤NDkDj+∑|k−j|>NDkDj=TN+RN.The error term RN will be small of order 2−δN in norm with δ>0. Using the above-mentioned theorem, we prove that if N is large enough, then RN is bounded on the space of test functions with
the operator norm less than 1. Therefore, if N is large enough and DkN=∑|j|≤NDk+j for k∈ℤ, we then obtain the following Calderón-type reproducing  formulae:(1.19)f=∑k=−∞∞TN−1DkNDk(f)=∑k=−∞∞DkDkNTN−1(f),where TN−1 is the inverse of TN and the series converge in Lp(𝒳), 1<p<∞, and in the space of test functions and its dual space. We also obtain a
corresponding discrete version. The Calderón reproducing formula is another
main tool of this paper. As soon as (1.19) is established, we can introduce
Besov and Triebel-Lizorkin spaces on 𝒳 via approximations of the identity and prove
that these spaces are independent of the choice of approximations of the
identity. The Calderón reproducing formula is also employed to establish the
atomic decomposition of these spaces and to prove boundedness results for
operators acting on these spaces.Remark 1.7.
Let{Sk}k∈ℤ be an approximation of the identity as
constructed in Theorem 2.6 below. An essential difference in the theory occurs
when the measure, μ(𝒳), of the underlying space 𝒳 is finite versus the case when the measure of
the underlying space is infinite. When μ(𝒳)<∞, it is not true that ∥Sk(f)∥Lp(𝒳)→0 as k→−∞ for all p∈(1,∞) and f∈Lp(𝒳); see the proof of Proposition 3.1(i) below. Thus, when μ(𝒳)<∞, in the Calderón-type reproducing formulae (1.19), we should replace D0 by S0 and Dk for k∈{−1,−2,…} by 0. This means that when μ(𝒳)<∞,
we always have an inhomogeneous term corresponding to S0, which needs some special care.


## 1.3. Notation

Finally, we introduce some notation and make some
conventions. Throughout the paper,A~B means that the ratio A/B is bounded and bounded away from zero by
constants that do not depend on the relevant variables in A and B; A≲B and B≳A mean that the ratio A/B is bounded by a constant independent of the
relevant variables. For any p∈[1,∞], we denote by p′ its conjugate index, namely, 1/p+1/p′=1. We also denote by C a positive constant which is independent of
the main parameters, but it may vary from line to line. Constants with
subscripts, such as C0, do not change in different occurrences. If E is a subset of a metric space (𝒳,d), we denote by χE the characteristic function of E and define(1.20)diamE=supx,y∈Ed(x,y).
We also set ℕ={1,2,…} and ℤ+=ℕ∪{0}. For any a,b∈ℝ, we denote min{a,b}, max{a,b}, and max{a,0} by a⋀b, a∨b, and a+, respectively.If(𝒳,d,μ) is a space of homogeneous type, we also
introduce the volume functionsVδ(x)=μ(B(x,δ)) and V(x,y)=μ(B(x,d(x,y))) for all x,y∈𝒳 and δ>0. By (1.2), it is easy to see that V(x,y)~V(y,x); see also [44]. We will use this fact without further mentioning.Throughout the whole paper, forϵ∈(0,1] and |s|<ϵ, we set(1.21)p(s,ϵ)≡max{nn+ϵ,nn+ϵ+s}.


## 2. Approximations of the Identity and Spaces of Test Functions

In this section, we will work onspaces of homogeneous type with the
constant C0 as in Definition 1.1, (𝒳,μ,d),
where μ(𝒳) can be finite or infinite.  We first present
some basic estimates which will be used throughout the whole paper. We then
introduce the notion of an approximation of the identity. Following Coifman's
idea in [64], we then
construct an approximation of the identity with bounded support on 𝒳.
We also introduce spaces of test functions and establish the boundedness of
singular integrals on these spaces, which are key tools of the whole theory.


### 2.1. Approximations of the Identity

Throughout the whole paper, we denote byMf the Hardy-Littlewood maximal function on 𝒳 for any f∈Lloc1(𝒳).
It is well known that M is bounded on Lp(𝒳) with p∈(1,∞];
see [29, 44]. Some basic estimates used
throughout the whole paper are stated in the following lemma, whose main part
is included in [48, Lemma 2.1], and the remaining
statements are obvious.Lemma 2.1.
Letδ>0, a>0, r>0, and θ∈(0,1).
Then the following hold.
(i)
∫d(x,y)≤δ(d(x,y)a/V(x,y))dμ(y)≤Cδa and ∫d(x,y)≥δ(1/V(x,y))(δa/d(x,y)a)dμ(y)≤C uniformly in x∈𝒳 and δ>0.(ii)
∫𝒳(1/(Vδ(x)+V(x,y)))(δa/(δ+d(x,y))a)d(x,y)ηdμ(y)≤Cδη uniformly in x∈𝒳 and δ>0,
if a>η≥0.(iii)
Ifx,x′,x1∈𝒳 satisfy d(x,x′)≤θ(r+d(x,x1)),
then 1/(r+d(x′,x1))≤C(1/(r+d(x,x1))), 1/(Vr(x1)+V(x1,x′))≤C(1/(Vr(x1)+V(x1,x))), and(2.1)1Vr(x′)+V(x′,x1)≤C1Vr(x)+V(x,x1)uniformly in r>0 and x,x′,x1∈𝒳.(iv)
For allf∈Lloc1(𝒳) and all x∈𝒳, ∫d(x,y)>δ(1/V(x,y))(δa/d(x,y)a)|f(y)|dμ(y)≤CM(f)(x) uniformly in δ>0, f∈Lloc1(𝒳) and x∈𝒳.(v)
For anyϵ>0, ∫d(y,x)≥δ(1/(Vr(x)+V(y,x)))(1/(r+d(y,x))ϵ)dμ(y)≤C(1/(r+δ)ϵ) uniformly in x∈𝒳 and δ,r>0.(vi)
For any fixedα>0,
if d(x,y)≤αr,
then Vr(x)~Vαr(x)~Vαr(y)~Vr(y) uniformly in x,y∈𝒳 and r>0.(vii)
For allr>0 and all x,y∈𝒳, Vr(x)+V(x,y)~Vr(x)+Vr(y)+V(x,y)~Vr(y)+V(x,y).Motivated by the properties of the heat kernel defined
in (1.17) in the case of Carnot-Carathéodory spaces, we introduce the following
notion of an approximation of the identity on 𝒳.Definition 2.2.
Letϵ1∈(0,1], ϵ2>0, and ϵ3>0.
A sequence {Sk}k∈ℤ of bounded linear integral operators on L2(𝒳) is said to be an approximation of the identity of order(ϵ1,ϵ2,ϵ3) (for short, (ϵ1,ϵ2,ϵ3)-ATI) if there exists a constant C3>0 such that for all k∈ℤ and all x,x′,y, and y′∈𝒳,Sk(x,y), the integral kernel of Sk is a measurable function from 𝒳×𝒳 into ℂ satisfying
(i)
|Sk(x,y)|≤C3(1/(V2−k(x)+V2−k(y)+V(x,y)))(2−kϵ2/(2−k+d(x,y))ϵ2);(ii)
|Sk(x,y)−Sk(x′,y)|≤C3(d(x,x′)/(2−k+d(x,y)))ϵ1(1/(V2−k(x)+V2−k(y)+V(x,y)))×(2−kϵ2/(2−k+d(x,y))ϵ2) for d(x,x′)≤(2−k+d(x,y))/2;(iii)
property (ii) also holds withx and y interchanged;(iv)
|[Sk(x,y)−Sk(x,y′)]−[Sk(x′,y)−Sk(x′,y′)]|≤C3(d(x,x′)/(2−k+d(x,y)))ϵ1×(d(y,y′)/(2−k+d(x,y)))ϵ1(1/(V2−k(x)+V2−k(y)+V(x,y)))(2−kϵ3/(2−k+d(x,y))ϵ3) for d(x,x′)≤(2−k+d(x,y))/3 and d(y,y′)≤(2−k+d(x,y))/3;(v)
∫𝒳Sk(x,y)dμ(y)=1;(vi)
∫𝒳Sk(x,y)dμ(x)=1.In case the ATI has bounded support, in the sense thatSk(x,y)=0 when d(x,y)≳2−k,
then the conditions (i), (ii), (iii), and (iv) of Definition 2.2 simplify as
follows (see Proposition 2.5).Definition 2.3.
Letϵ1∈(0,1].
A sequence {Sk}k∈ℤ of bounded linear integral operators on L2(𝒳) is said to be an approximation of the identity of orderϵ1with bounded support (for short, ϵ1-ATI with bounded support) if there exist
constants C4,C5>0 such that for all k∈ℤ and all x,x′,y, and y′∈𝒳,Sk(x,y), the integral kernel of Sk is a measurable function from 𝒳×𝒳 into ℂ satisfying (v) and (vi) of Definition 2.2 as
above, and
(i)
Sk(x,y)=0 if d(x,y)≥C52−k and |Sk(x,y)|≤C4(1/(V2−k(x)+V2−k(y)));(ii)
|Sk(x,y)−Sk(x′,y)|≤C42kϵ1d(x,x′)ϵ1(1/(V2−k(x)+V2−k(y))) for d(x,x′)≤max{C5,1}21−k;(iii)
property (ii) also holds withx and y interchanged;(iv)
|[Sk(x,y)−Sk(x,y′)]−[Sk(x′,y)−Sk(x′,y′)]|≤C422kϵ1d(x,x′)ϵ1d(y,y′)ϵ1×(1/(V2−k(x)+V2−k(y))) for d(x,x′)≤max{C5,1}21−k and d(y,y′)≤max{C5,1}21−k.We callϵ1 the regularity of (ϵ1,ϵ2,ϵ3)-ATI  {Sk}k∈ℤ.Remark 2.4.
(i) Assume that𝒳 is as in Case (a) with n≥3 or as in Case (b) in the introduction, and let H(s,x,y) be the heat kernel defined in (1.17). Define Sk(x,y)=H(2−2k,x,y) for k∈ℤ and x,y∈𝒳.
Using [44, Proposition 2.3.1], we can verify that {Sk}k∈ℤ is an (ϵ1,ϵ2,ϵ3)-ATI for any ϵ1∈(0,1], ϵ2>0, and ϵ3>0.
Moreover, in this case, Sk(x,y)=Sk(y,x).
(ii) If𝒳 is any (compact or noncompact) Ahlfors n-regular metric measure space or any Lie group
with polynomial growth, then one can construct an ϵ1-ATI with bounded support for any ϵ1∈(0,1] by following Coifman's idea in [64].
(iii) We remark that when we consider the existence of ATIs, the condition (iv) of Definition2.2 is not
essential, in the sense that if there exist {Sk}k∈ℤ with k∈ℤ satisfying (i), (ii), (iii), (v), and (vi) of
Definition 2.2, then {Sk∘Sk}k∈ℤ satisfy (i) through (vi) of Definition 2.2.Proposition 2.5.
Suppose{Sk}k∈ℤ is a sequence of bounded linear integral
operators on L2(𝒳) such that Sk(x,y)=0 whenever d(x,y)≥C52−k.
Then {Sk}k∈ℤ is an ATI if and only if it is an ATI with bounded support.Proof.
Obviously, from the assumption thatSk(x,y)=0 whenever d(x,y)≥C52−k,
it easily follows that Sk satisfies Definition 2.2(i) if and only if(2.2)|Sk(x,y)|≲1V2−k(x)+V2−k(y),which appears in Definition 2.3(i).
We now establish the equivalence between Definitions2.2(ii) and 2.3(ii), with the proof for the equivalence between
(iii) of Definitions 2.2 and (iii) of Definition 2.3 being similar. Notice that Sk(x,y)−Sk(x′,y)≠0 implies that d(x,y)<C52−k or d(x′,y)<C52−k.
Thus, Sk(x,y)−Sk(x′,y)≠0 together with d(x,x′)≤2max{C5,1}2−k shows that d(x,y)<3max{C5,1}2−k,
and hence, 2−k+d(x,y)~2−k and V2−k(x)+V2−k(y)+V(x,y)~V2−k(x)+V2−k(y).
From these estimates, it immediately follows that Definition 2.2(ii) implies Definition 2.3(ii), and conversely Definition 2.3(ii) implies that Definition 2.2(ii) holds whenever d(x,x′)≤2max{C5,1}2−k.
We still need to prove that Definition 2.3(ii) also implies that Definition 2.2(ii) holds when 2max{C5,1}2−k<d(x,x′)≤(2−k+d(x,y))/2.
However, if 2max{C5,1}2−k<d(x,x′)≤(2−k+d(x,y))/2,
we have d(x,y)>3max{C5,1}2−k and d(x′,y)≥d(x,x′)−2−k>C52−k.
Thus, in this case, Sk(x,y)=0=Sk(x′,y),
and therefore, Definition 2.2(ii) automatically holds. This proves the
equivalence between Definitions 2.2(ii) and 2.3(ii).
We now establish the equivalence between Definitions2.2(iv) and 2.3(iv). Notice that [Sk(x,y)−Sk(x,y′)]−[Sk(x′,y)−Sk(x′,y′)]≠0 implies that d(x,y)<C52−k or d(x′,y)<C52−k or d(x,y′)<C52−k or d(x′,y′)<C52−k.
This together with d(x,x′)≤max{C5,1}21−k and d(y,y′)≤max{C5,1}21−k further shows that we have d(x,y)<5max{C5,1}21−k,
and hence, 2−k+d(x,y)~2−k and V2−k(x)+V2−k(y)+V(x,y)~V2−k(x)+V2−k(y).
From these estimates, we see that Definition 2.2(iv) implies Definition 2.3(iv), and conversely, Definition 2.3(iv) also implies that Definition 2.2(iv) holds when d(x,x′)≤2max{C5,1}2−k and d(y,y′)≤max{C5,1}21−k.
We still need to prove that Definition 2.3(iv) also implies that Definition 2.2(iv) holds in the three cases listed below. Recall that we always assume
that d(x,x′)≤(2−k+d(x,y))/3 and d(y,y′)≤(2−k+d(x,y))/3 in  Definition 2.2(iv).
Case 1.
d(x,x′)≤2max{C5,1}2−k and d(y,y′)>2max{C5,1}2−k.
In this case, d(x,y)≥3d(y,y′)−2−k≥C52−k, d(x,y′)≥2d(y,y′)−2−k>C52−k, d(x′,y)>2d(y,y′)−2−k>C52−k,
and d(x′,y′)>d(y,y′)−2−k>C52−k.
Thus Sk(x,y)=Sk(x,y′)=Sk(x′,y)=Sk(x′,y′)=0 and therefore Definition 2.2(iv) automatically
holds in this case.Case 2.
d(x,x′)>2max{C5,1}2−k and d(y,y′)≤2max{C5,1}2−k.
This case is similar to Case 1 by symmetry.Case 3.
d(x,x′)>2max{C5,1}2−k and d(y,y′)>2max{C5,1}2−k.
In this case, similar to Case 1, we have that d(x,y)≥3d(y,y′)−2−k>C52−k, d(x,y′)≥2d(y,y′)−2−k>C52−k, d(x′,y)≥2d(x,x′)−2−k>C52−k,
and d(x′,y′)≥d(x,y)−d(x,x′)−d(y,y′)≥(d(x,y)−21−k)/3>d(y,y′)−2−k>C52−k.
We also have Sk(x,y)=Sk(x,y′)=Sk(x′,y)=Sk(x′,y′)=0 and therefore (iv) automatically holds in this
case. This establishes the equivalence between (iv) and (iv)′, and hence
completes the proof of  Proposition 2.5.Following Coifman's idea in [64], for any ϵ1∈(0,1],
we can construct ϵ1-ATIs with bounded support on 𝒳 (cf. also [65, Theorem (1.13)]).Theorem 2.6.
Let(𝒳,d,μ) be a space of homogeneous type as in
Definition 1.1 and ϵ1∈(0,1].
Then there exists an approximation of the identity {Sk}k∈ℤ of order ϵ1 with bounded support on 𝒳,
with constant C5=4.
Moreover, for all k∈ℤ and x,y∈𝒳, Sk(x,y)=Sk(y,x).Proof.
Obviously, we only need to prove the theorem forϵ1=1.
Let h∈C1(ℝ), h(t)=1 if t∈[0,1], h(t)=0 if t≥2,
and 0≤h(t)≤1 for all t∈ℝ.
For any k∈ℤ, f∈Lloc1(𝒳) and u∈𝒳,
we then define(2.3)Tkf(u)=∫𝒳h(2kd(u,w))f(w)dμ(w).Obviously, we have V2−k(u)≤Tk1(u)≤V21−k(u). Fix any x∈𝒳.
By Lemma 2.1(vi), for all u∈B(x,25−k),
we further have(2.4)Tk1(u)~V2−k(x).Thus, if z∈B(x,24−k) and h(2kd(z,u))≠0,
then u∈B(x,25−k),
and by (2.4), we further have(2.5)Tk(1Tk1)(z)=∫𝒳h(2kd(z,u))1Tk1(u)dμ(u)~1.For all x,y∈𝒳,
we define(2.6)Sk(x,y)=1Tk1(x){∫𝒳h(2kd(x,z))1Tk(1/Tk1)(z)h(2kd(z,y))dμ(z)}1Tk1(y).We now prove that {Sk}k∈ℤ is a 1-ATI with bounded support and C5=4.
It is obvious that for all x,y∈𝒳, Sk(x,y)=Sk(y,x),
and that if d(x,y)≥22−k,
then Sk(x,y)=0.
Also, it is easy to show that ∫𝒳Sk(x,y)dμ(y)=1. Moreover, if d(x,y)<22−k,
by (2.4) and (2.5) together with Lemma 2.1(vi), we obtain(2.7)0≤Sk(x,y)≲1V2−k(x)~1V2−k(x)+V2−k(y).Thus Sk(x,y) satisfies (i)′ of Definition 2.2 with C5=4.
We now show thatSk(x,y) has the desired regularity in the first
variable when d(x,x′)≤23−k.
Notice that in this case, Sk(x,y)−Sk(x′,y)≠0 implies that d(x,y)<24−k,
and hence,(2.8)1V2−k(x)+V2−k(y)~1V2−k(x).By (2.6), we
have(2.9)Sk(x,y)−Sk(x′,y)=[1Tk1(x)−1Tk1(x′)]{∫𝒳h(2kd(x,z))1Tk(1/Tk1)(z)h(2kd(z,y))dμ(z)}1Tk1(y)+1Tk1(x′){∫𝒳[h(2kd(x,z))−h(2kd(x′,z))]1Tk(1/Tk1)(z)h(2kd(z,y))dμ(z)}1Tk1(y)≡Z1+Z2.To estimate Z1,
by the choice of h,
the mean value theorem, and (2.4) together with Lemma 2.1(vi), we first have(2.10)|1Tk1(x)−1Tk1(x′)|≲1[V2−k(x)]22kd(x,x′)μ(B(x,24−k))~2kd(x,x′)V2−k(x),which together with (2.4),
(2.5), and (2.8) further yields(2.11)|Z1|≲2kd(x,x′)V2−k(x)~2kd(x,x′)V2−k(x)+V2−k(y).
To estimateZ2,
notice that h(2kd(x,z))−h(2kd(x′,z))≠0 implies d(x,z)<24−k.
By this observation, (2.5), the mean value theorem together with Lemma 2.1(vi), we obtain(2.12)|∫𝒳[h(2kd(x,z))−h(2kd(x′,z))]1Tk(1/Tk1)(z)h(2kd(z,y))dμ(z)|≲2kd(x,x′)μ(B(x,24−k))≲2kd(x,x′)V2−k(x),which together with (2.4) and (2.8) yields(2.13)|Z2|≲2kd(x,x′)V2−k(x)~2kd(x,x′)V2−k(x)+V2−k(y).In combination with the estimate
for Z1,
this shows that Sk(x,y) has the desired regularity with respect to the
first variable when d(x,x′)≤23−k.
We finally prove thatSk(x,y) satisfies the desired second difference
condition, when d(x,x′)≤23−k and d(y,y′)≤23−k. Notice that in this case, [Sk(x,y)−Sk(x′,y)]−[Sk(x,y′)−Sk(x′,y′)]≠0 implies that d(x,y)<25−k.
Thus, (2.8) also holds in this case. By (2.6), we have(2.14)[Sk(x,y)−Sk(x′,y)]−[Sk(x,y′)−Sk(x′,y′)]=[1Tk1(x)−1Tk1(x′)]{∫𝒳h(2kd(x,z))1Tk(1/Tk1)(z)h(2kd(z,y))dμ(z)}×[1Tk1(y)−1Tk1(y′)]+[1Tk1(x)−1Tk1(x′)]1Tk1(y′)×{∫𝒳h(2kd(x,z))1Tk(1/Tk1)(z)[h(2kd(z,y))−h(2kd(z,y′))]dμ(z)}+1Tk1(x′){∫𝒳[h(2kd(x,z))−h(2kd(x′,z))]×1Tk(1/Tk1)(z)h(2kd(z,y))dμ(z)}[1Tk1(y)−1Tk1(y′)]+1Tk1(x′)1Tk1(y′){∫𝒳[h(2kd(x,z))−h(2kd(x′,z))]×1Tk(1/Tk1)(z)[h(2kd(z,y))−h(2kd(z,y′))]dμ(z)}≡Z3+Z4+Z5+Z6.The estimates (2.10), (2.5), and (2.8) together with Lemma 2.1(vi) show that(2.15)|Z3|≲2kd(x,x′)V2−k(x)μ(B(x,23−k))2kd(y,y′)V2−k(x)~22kd(x,x′)d(y,y′)1V2−k(x)+V2−k(y).The estimate for Z4 is similar to the one for Z5,
while the estimate for Z5 can be deduced immediately from the estimates
(2.12), (2.10), and (2.4) together with (2.8). Finally, to estimate Z6,
notice that h(2kd(x,z))−h(2kd(x′,z))≠0 and d(x,x′)≤23−k implies that z∈B(x,24−k).
This observation together with the mean value theorem, (2.5), and Lemma 2.1(vi) yields(2.16)|∫𝒳[h(2kd(x,z))−h(2kd(x′,z))]1Tk(1/Tk1)(z)[h(2kd(z,y))−h(2kd(z,y′))]dμ(z)|≲22kd(x,x′)d(y,y′)V2−k(x),which together with (2.4) and (2.8) further gives(2.17)|Z6|≲22kd(x,x′)d(y,y′)1V2−k(x)~22kd(x,x′)d(y,y′)1V2−k(x)+V2−k(y).This proves that Sk(x,y) has the desired second difference property,
and hence, completes the proof of Theorem 2.6.In the sequel, for anyf∈Lp(𝒳) with p∈[1,∞] and x∈𝒳,
we set(2.18)Skf(x)=∫𝒳Sk(x,y)f(y)dμ(y).We also let(2.19)Lb∞(𝒳)={f∈L∞(𝒳):fhasboundedsupport}.Some basic properties of ATIs are presented in the next proposition.Proposition 2.7.
Let{Sk}k∈ℤ be an (ϵ1,ϵ2,ϵ3)-ATI with ϵ1∈(0,1],ϵ2>0 and ϵ3>0,
and let Skt denote the adjoint operator to Sk,
whose integral kernel is given by Skt(x,y)=Sk(y,x). Then the following hold.
(i)
For anyk∈ℤ and any x,y∈𝒳, ∫𝒳|Sk(x,y)|dμ(y)≤C and ∫𝒳|Sk(x,y)|dμ(x)≤C.(ii)
For anyk∈ℤ and any f∈Lloc1(𝒳), |Skf(x)|≤CMf(x).(iii)
For1≤p≤∞,
there exists a constant Cp>0 such that for all f∈Lp(𝒳),(2.20)∥Skf∥Lp(𝒳)≤Cp∥f∥Lp(𝒳).(iv)
For1≤p<∞ and any f∈Lp(𝒳), ∥Skf−f∥Lp(𝒳)→0 when k→∞.(v)
Properties (ii) through (iv) also hold forSk replaced by Skt.Proof.
Definition2.2(i) together with Lemma 2.1(i) shows that for any k∈ℤ and any x∈𝒳,(2.21)∫𝒳|Sk(x,y)|dμ(y)=∫d(x,y)≤2−k|Sk(x,y)|dμ(y)+∫d(x,y)>2−k|Sk(x,y)|dμ(y)≲1+∫d(x,y)>2−k1V(x,y)2−kϵ2d(x,y)ϵ2dμ(y)≲1.A similar argument proves that ∫𝒳|Sk(x,y)|dμ(x)≲1.
To prove (ii), by Definition2.2(i) together with
Lemma 2.1(iv), we have(2.22)|Skf(x)|≲1V2−k(x)∫d(x,y)≤2−k|f(y)|dμ(y)+∫d(x,y)>2−k1V(x,y)2−kϵ2d(x,y)ϵ2|f(y)|dμ(y)≲Mf(x).
From (ii) and theLp(𝒳)-boundedness of M,
we obtain (iii) for p∈(1,∞].
For p=1,
we obtain (iii) from (i), by Fubini's theorem.
To prove (iv), we first recall that Lebesgue's
differentiation theorem holds in𝒳,
that is, if f is locally integrable, then for almost all x∈𝒳, x is a Lebesgue point which means(2.23)limr→01μ(B(x,r))∫B(x,r)|f(y)−f(x)|dμ(y)=0,since the measure μ is regular; see [13, page 4] and [66, page 11].
For any givenf∈Lp(𝒳) with p∈[1,∞],
assume that x0 is a Lebesgue point of f,
then Mf(x0)+|f(x0)|<∞.
By the conditions (v) and (i) of Sk in Definition 2.2, we have(2.24)|Skf(x0)−f(x0)|=|∫𝒳Sk(x0,y)[f(y)−f(x0)]dμ(y)|≤C3∫d(x0,y)<2−k1V2−k(x0)+V2−k(y)+V(x0,y)2−kϵ2(2−k+d(x0,y))ϵ2|f(y)−f(x0)|dμ(y)+C3∑l=0∞∫2l−12−k≤d(x0,y)<2l2−k⋯≤CC3∑l=0∞12lϵ21V2l2−k(x0)∫d(x0,y)<2l2−k|f(y)−f(x0)|dμ(y).For any δ>0,
we choose L0∈ℕ so that(2.25)CC3[Mf(x0)+|f(x0)|]∑l=L0+1∞12lϵ2<δ2.Since x0 is a Lebesgue point, by the definition, we
know that there exists K∈ℕ such that when k>K,(2.26)CC3∑l=0L012lϵ21V2l2−k(x0)∫d(x0,y)<2l2−k|f(y)−f(x0)|dμ(y)<δ2.Thus limk→∞Skf(x0)=f(x0),
and therefore for almost everywhere x∈𝒳,(2.27)limk→∞Skf(x)=f(x).This fact together with |Skf(x)|≲Mf(x) (see (ii) of this proposition), the Lp(𝒳)-boundedness of M with p∈(1,∞) again and Lebesgue's dominated convergence
theorem, gives that limk→∞∥Skf−f∥Lp(𝒳)=0 for all f∈Lp(𝒳).
Whenp=1,
we first consider f∈Lb∞(𝒳).
Assume that suppf⊂B(y0,r0) for some y0∈𝒳 and r0>0.
For some fixed L>2r0,
by Hölder's inequality and the conditions (i) and (v) of Sk in Definition 2.2 together with Lemma 2.1(i),
we have(2.28)∥Skf−f∥L1(𝒳)=∫d(x,y0)<L|Skf(x)−f(x)|dμ(x)+∫d(x,y0)≥L|Skf(x)−f(x)|dμ(x)≤μ(BL(y0))1/2∥Skf−f∥L2(𝒳)+C3∫d(y,y0)<r0|f(y)|{∫d(x,y)≥L/21V(x,y)2−kϵ2d(x,y)ϵ2dμ(x)}dμ(y)≤μ(BL(y0))1/2∥Skf−f∥L2(𝒳)+CC32−kϵ2L−ϵ2∥f∥L1(𝒳),which together with Lb∞(𝒳)⊂L2(𝒳) and the above proved conclusion for p∈(1,∞) implies that limk→∞∥Skf−f∥L1(𝒳)=0 for all f∈Lb∞(𝒳).
This and the density of Lb∞(𝒳) in L1(𝒳) further yield that limk→∞∥Skf−f∥L1(𝒳)=0 for all f∈L1(𝒳),
which verify (iv).
Property (v) can be deduced from (i) through (iv)
together with the symmetry, which completes the proof of Proposition2.7.We now introduce the space of test functions on𝒳.Definition 2.8.
Letx1∈𝒳, r>0, 0<β≤1, and γ>0.
A function f on 𝒳 is said to be a test function of type(x1,r,β,γ) if there exists a constant C≥0 such that
(i)
|f(x)|≤C(1/(Vr(x1)+V(x1,x)))(r/(r+d(x1,x)))γ for all x∈𝒳;(ii)
|f(x)−f(y)|≤C(d(x,y)/(r+d(x1,x)))β(1/(Vr(x1)+V(x1,x)))(r/(r+d(x1,x)))γ for all x,y∈𝒳 satisfying that d(x,y)≤(r+d(x1,x))/2.
Moreover, we
denote by 𝒢(x1,r,β,γ) the set of all test functions of type (x1,r,β,γ),
and if f∈𝒢(x1,r,β,γ),
we define its norm by(2.29)∥f∥𝒢(x1,r,β,γ)=inf{C:(i)and(ii)hold}.The space 𝒢(x1,r,β,γ) is called the space of test functions.This definition of a test function of type(x1,r,β,γ) gives a quantified meaning to the notation of
a sufficiently “smooth” function which is essentially supported in the
ball of radius r centered at x1 in the sense that it decays of sufficiently
high order (measured by γ) at infinity, and is Hölder continuous of
order β (at the right scale r).Obviously,Sk(x,⋅) for any fixed k∈ℤ and x∈𝒳 in Definition 2.2 is a test function of type (x,2−k,ϵ1,ϵ2),
and Sk(⋅,y) for any fixed k∈ℤ and y∈𝒳 in Definition 2.2 is a test function of type (y,2−k,ϵ1,ϵ2).
Moreover, for η∈(0,1],
let(2.30)∥f∥C˙η(𝒳)=supx,y∈𝒳x≠y|f(x)−f(y)|d(x,y)ηand define the homogeneous Hölder space(2.31)C˙η(𝒳)={f∈C(𝒳):∥f∥C˙η(𝒳)<∞};we also consider the inhomogeneous Hölder space(2.32)Cη(𝒳)={f∈C(𝒳):∥f∥Cη(𝒳)<∞},Cbη(𝒳)={f∈Cη(𝒳):fhasboundedsupport},where ∥f∥Cη(𝒳)=∥f∥L∞(𝒳)+∥f∥C˙η(𝒳).
Then any f∈Cbη(𝒳) is also a test  function of type (x1,r,η,γ) for any x1∈𝒳, r>0, and γ>0.Now fixx1∈𝒳 and let 𝒢(β,γ)=𝒢(x1,1,β,γ). It is easy to see that(2.33)𝒢(x0,r,β,γ)=𝒢(β,γ)with the equivalent norms for
all x0∈𝒳 and r>0. Furthermore, it is easy to check that 𝒢(β,γ) is a Banach space with respect to the norm in 𝒢(β,γ).It is well known that even when𝒳=ℝn, 𝒢(β1,γ) is not dense in 𝒢(β2,γ) if β1>β2,
which will bring us some inconvenience. To overcome this defect, in what
follows, for given ϵ∈(0,1],
let 𝒢0ϵ(β,γ) be the completion of the space 𝒢(ϵ,ϵ) in 𝒢(β,γ) when 0<β,γ≤ϵ.
Obviously, 𝒢0ϵ(ϵ,ϵ)=𝒢(ϵ,ϵ).
Moreover, f∈𝒢0ϵ(β,γ) if and only if f∈𝒢(β,γ) and there exist {fn}n∈ℕ⊂𝒢(ϵ,ϵ) such that ∥f−fn∥𝒢(β,γ)→0 as n→∞.
If f∈𝒢0ϵ(β,γ),
we then define ∥f∥𝒢0ϵ(β,γ)=∥f∥𝒢(β,γ).
Then, obviously, 𝒢0ϵ(β,γ) is a Banach space and we also have ∥f∥𝒢0ϵ(β,γ)=limn→∞∥fn∥𝒢(β,γ) for the above chosen {fn}n∈ℕ.We define the dual space(𝒢0ϵ(β,γ))′ to be the set of all linear functionals ℒ from 𝒢0ϵ(β,γ) to ℂ with the property that there exists C≥0 such that for all f∈𝒢0ϵ(β,γ),(2.34)|ℒ(f)|≤C∥f∥𝒢0ϵ(β,γ).We denote by 〈h,f〉 the natural pairing of elements h∈(𝒢0ϵ(β,γ))′ and f∈𝒢0ϵ(β,γ). Clearly, for all h∈(𝒢0ϵ(β,γ))′,〈h,f〉 is well defined for all f∈𝒢0ϵ(x1,r,β,γ) with  x1∈𝒳 and r>0.In the sequel, we define(2.35)𝒢°(x1,r,β,γ)={f∈𝒢(x1,r,β,γ):∫𝒳f(x)dμ(x)=0},which is called the space of test functions with mean zero.
The space 𝒢°0ϵ(β,γ) is defined to be the completion of the space 𝒢°(ϵ,ϵ) in 𝒢°(β,γ) when 0<β,γ<ϵ.
Moreover, if f∈𝒢°0ϵ(β,γ),
we then define ∥f∥𝒢°0ϵ(β,γ)=∥f∥𝒢(β,γ).By essentially following a procedure similar to that
of [44, Lemma 3.5.1],
we can establish some kind of decomposition for test functions with mean
zero.Proposition 2.9.
Letx1∈𝒳, r>0, 0<β≤1, and γ>0,
and let f∈𝒢°(x1,r,β,γ).
Then for any γ˜∈(0,γ] and all x∈𝒳,(2.36)f(x)=∑l=0∞2−lγ˜φl(x),where φl is an
adjusted bump function associated with the ball B(x1,2lr),
which means that there exists a constant C>0 independent of r and l such that
(i)
suppφl⊂B(x1,2lr);(ii)
|φl(x)|≤C(1/V2lr(x1)) for all x∈𝒳;(iii)
∥φl∥C˙η(𝒳)≤C(2lr)−η(1/V2lr(x1)) for all 0<η≤β;(iv)
∫𝒳φl(x)dμ(x)=0.
Moreover, the
series in (2.36) converges to f pointwise, as well as in Lp(𝒳) for p∈[1,∞],
and in 𝒢°0β⋀γ˜(β′,γ′) with 0<β′,γ′<(β⋀γ˜) and also in (𝒢°0β⋀γ˜(β′,γ′))′ with 0<β′,γ′≤(β⋀γ).Proof.
Fix a nonnegative functionα∈C1(ℝ) such that α(t)=1 if t≤1/2 and α(t)=0 if t≥1.
Let A0(x)=α(d(x,x1)/r)f(x), and for l∈ℕ,(2.37)Al(x)=[α(d(x,x1)2lr)−α(d(x,x1)2l−1r)]f(x).Then f(x)=∑l=0∞Al(x), and for all γ˜∈(0,γ],(2.38)|Al(x)|≲2−lγ˜1V2lr(x1).Define al=∫𝒳Al(x)dμ(x) and vl=∑j=0laj. Let η˜l(x)=α(d(x,x1)/2lr) and ηl(x)=η˜l(x)[∫𝒳η˜l(z)dμ(z)]−1. Finally, if we define(2.39)A˜l(x)=Al(x)−alηl(x)+vl(ηl(x)−ηl+1(x))and φl(x)=2lγ˜A˜l(x), then it is easy to verify that φl satisfies (i) through (iv) of the proposition
and (2.36) holds pointwise.
Obviously,φl∈𝒢°(β,γ) and therefore φl∈𝒢°0β⋀γ˜(β′,γ′) for 0<β′,γ′≤(β⋀γ˜).
Moreover, if l∈ℕ is large enough, then there exists a constant Cr,x1>0,
which is independent of x,
such that for all x∈B(x1,2lr), 1+d(x,x1)≤Cr,x12l and hence(2.40)∥φl∥𝒢°0β⋀γ˜(β′,γ′)≤Cr,x12lγ′.From this, it follows that if γ′<γ˜,(2.41)∥f−∑l=0L2−lγ˜φl∥𝒢°0β⋀γ˜(β′,γ′)≲Cr,x1∑l=L+1∞2−l(γ˜−γ′)~Cr,x12−L(γ˜−γ′)→0,as L→∞.
This shows that (2.36) holds in 𝒢°0β⋀γ˜(β′,γ′) with 0<β′,γ′<(β⋀γ˜).
By Fatou's lemma and Minkowski's inequality, we also obtain that for any p∈[1,∞],(2.42)∥f−∑l=0L2−lγ˜φl∥Lp(𝒳)≤∑l=L+1∞2−lγ˜∥φl∥Lp(𝒳)≲∑l=L+1∞2−lγ˜1V2lr(x1)1−1/p≲1Vr(x1)1−1/p2−Lγ˜→0,as L→∞.
That is, (2.36) holds in Lp(𝒳) with p∈[1,∞].
Finally, for anyψ∈𝒢°0β⋀γ(β′,γ′) with 0<β′,γ′≤(β⋀γ),
from Hölder's inequality, Fatou lemma and Minkowski's inequality, it follows
that(2.43)|〈f,ψ〉−∑l=0L2−lγ˜〈φl,ψ〉|≤∥∑l=L+1∞2−lγ˜φl∥L1(𝒳)∥ψ∥L∞(𝒳)≲(∑l=L+1∞2−lγ˜)∥ψ∥L∞(𝒳)≲2−Lγ˜∥ψ∥L∞(𝒳)→0,as L→∞.
Thus, (2.36) also holds in (𝒢°0β⋀γ(β′,γ′))′ with 0<β′,γ′≤(β⋀γ),
which completes the proof of  Proposition 2.9.The following proposition is a slight variant of Proposition2.9.Proposition 2.10.
Letx1∈𝒳, r>0, 0<β≤1, and γ>0,
and let f∈(x1,r,β,γ).
Then for any γ˜∈(0,γ] and all x∈𝒳,(2.44)f(x)=∑l=0∞2−lγ˜φl(x),where φl is an adjusted bump function associated with
the ball B(x1,2lr),
which means that there exists a constant C>0 independent of r and l such that
(i)
suppφl⊂B(x1,2lr);(ii)
|φl(x)|≤C(1/V2lr(x1)) for all x∈𝒳;(iii)
∥φl∥C˙η(𝒳)≤C(2lr)−η(1/V2lr(x1)) for all 0<η≤β.
Moreover, the
series in (2.44) converges to f pointwise, as well as in Lp(𝒳) for p∈[1,∞],
and in 𝒢0β⋀γ˜(β′,γ′) with 0<β′,γ′<(β⋀γ˜) and also in (𝒢0β⋀γ(β′,γ′))′ with 0<β′,γ′≤(β⋀γ).Proof.
Letα∈C1(ℝ) be as in the proof of Proposition 2.9. Let A0(x)=α(d(x,x1)/r)f(x), and for l∈ℕ,(2.45)Al(x)=[α(d(x,x1)2lr)−α(d(x,x1)2l−1r)]f(x).
Then for l∈ℤ+, setting φl(x)=2lγ˜Al(x), we can verify that the {φl}l∈ℤ+ have all the properties stated in the
proposition, which completes the proof of Proposition 2.10.In what follows, for anyβ∈(0,1], γ>0, and r>0,
we let(2.46)𝒢°b(x1,r,β,γ)={f∈𝒢°(x1,r,β,γ):fhasboundedsupport},𝒢b(x1,r,β,γ)={f∈𝒢(x1,r,β,γ):fhasboundedsupport}.Also, for η∈(0,1],
let(2.47)C°bη(𝒳)={f∈Cbη(𝒳):∫𝒳f(x)dμ(x)=0}.From Propositions2.7, 2.9, and 2.10, it is easy to deduce the following useful result; we omit the details.Corollary 2.11.
Letϵ1∈(0,1] be as in Definition 2.2, 0<β≤ϵ1 and γ>0.
Then,
(i)
bothC°bβ(𝒳) and 𝒢°b(x1,r,β,γ) for any fixed x1∈𝒳 and r>0 are dense in 𝒢°(x1,r,β,γ);(ii)
bothCbβ(𝒳) and 𝒢b(x1,r,β,γ) for any fixed x1∈𝒳 and r>0 are dense in both 𝒢(x1,r,β,γ) and Lp(𝒳) with p∈[1,∞).


### 2.2. Boundedness of Singular Integrals on Spaces of Test Functions with Mean Zero

Letβ∈(0,1].
In analogy with the topology of the space 𝒟(ℝn),
in what follows, we endow the spaces Cbβ(𝒳) (resp., C°bβ(𝒳)) with the strict inductive limit topology (see
[30, page 273] or
[67]) arising from
the decomposition Cbβ(𝒳)=⋃nCβ(Bn) (resp., C°bβ(𝒳)=⋃nC°β(Bn)), where {Bn}n is any increasing sequence of closed balls
with the same center and 𝒳=⋃nBn,
and the space Cβ(Bn) (resp., C°β(Bn)) means the set of all functions f∈Cβ(𝒳) (resp., f∈C°β(𝒳)) with suppf⊂Bn,
whose topology is given by the norm ∥⋅∥Cβ(𝒳).
It is well known that the topology of Cbβ(𝒳) (resp., C°bβ(𝒳)) is independent of the choice of closed balls {Bn}n;
see [30, page 273] or
[67]. Their dual
spaces (Cbβ(𝒳))′ and (C°bβ(𝒳))′ will be endowed with the weak*topology.We first have the following basic facts.Proposition 2.12.
Letβ∈(0,1] and let T be a continuous linear operator from Cbβ(𝒳) to (Cbβ(𝒳))′.
Assume that T has a
distributional kernel K,
which is locally integrable away from the diagonal of 𝒳×𝒳,
in the sense that for all φ,ψ∈Cbβ(𝒳) with disjoint supports,
(2.48)〈Tφ,ψ〉=∬𝒳×𝒳K(x,y)φ(y)ψ(x)dμ(x)dμ(y).Assume also that there exists a
constant CT>0 such that for all x,x′,y∈𝒳 with d(x,x′)≤d(x,y)/2 and x≠y,(2.49)|K(x,y)−K(x′,y)|≤CTd(x,x′)ϵV(x,y)d(x,y)ϵ.Then T can be extended to a continuous linear
operator from Cβ(𝒳) to (C°bβ(𝒳))′.Proof.
For anyf∈Cβ(𝒳) and g∈C°bβ(𝒳),
suppose suppg⊂B(x0,r) for some x0∈𝒳 and r>0.
Choose ψ∈Cbβ(𝒳) such that ψ(x)=1 when x∈B(x0,2r) and ψ(x)=0 when x∉B(x0,4r).
It is easy to see that ψf∈Cbβ(𝒳),
which shows that 〈T(ψf),g〉 is well defined. On the other hand, we define(2.50)〈T((1−ψ)f),g〉=∬𝒳×𝒳[K(x,y)−K(x0,y)](1−ψ(y))f(y)g(x)dμ(x)dμ(y).By (2.49) and Lemma 2.1(i), it
is easy to see that the right-hand side of the above equality is finite;
furthermore, if f has also bounded support, this equals with(2.51)∬𝒳×𝒳K(x,y)(1−ψ(y))f(y)g(x)dμ(x)dμ(y),which coincides with (2.48).
Moreover, it is easy to verify that(2.52)〈T(ψf),g〉+〈T((1−ψ)f),g〉is independent of choice of ψ.
Thus we can define Tf by letting(2.53)〈Tf,g〉=〈T(ψf),g〉+〈T((1−ψ)f),g〉.In this sense, we have Tf∈(C°bβ(𝒳))′,
which completes the proof of Proposition 2.12.We also need the weak boundedness property and the
strong weak boundedness property of operators; see [68] for the definition of the
weak boundedness property on ℝn and [34] for the definition of the strong weak boundedness
property on Ahlfors 1-regular metric measure spaces.In what follows, forβ∈(0,1],
we denote by Cbβ(𝒳×𝒳) the set of all functions f on 𝒳×𝒳 with bounded support, which satisfy that for
any x∈𝒳,
both f(x,⋅) and f(⋅,x) are in C˙β(𝒳).Definition 2.13.
Letβ∈(0,1] and let T be a continuous linear mapping from Cbβ(𝒳) to (Cbβ(𝒳))′.
The operator T is said to have the weak boundedness property of orderβ (for short, T∈WBP(β)) if there exists a constant C>0 such that for all ϕ,ψ∈Cbβ(𝒳) with suppϕ,suppψ⊂B(z,r) for some z∈𝒳 and r>0, ∥ϕ∥C˙β(𝒳)≤r−β and ∥ψ∥C˙β(𝒳)≤r−β,(2.54)|〈Tϕ,ψ〉|≤Cμ(B(z,r)).The minimal constant C as above is denoted by ∥T∥WBP(β).Remark 2.14.
(i) Letβ∈(0,1] and γ>0.
Let us also endow the space 𝒢b(β,γ) (resp., 𝒢°b(β,γ)) with the strict inductive limit topology in a similar way as the
space Cbβ(𝒳) (resp., C°bβ(𝒳)) and its dual space (𝒢b(β,γ))′ (resp., (𝒢°b(β,γ))′) with the weak* topology. Then, as topological spaces, Cbβ(𝒳)=𝒢b(β,γ) and (Cbβ(𝒳))′=(𝒢b(β,γ))′ (resp., C°bβ(𝒳)=𝒢°b(β,γ) and (C°bβ(𝒳))′=(𝒢°b(β,γ))′).
(ii) We remark that if𝒳 is a (κ,n)-space as in Definition 1.1, then there exists
a constant C>0 such that for all ϕ∈C˙bβ(𝒳) and all x∈𝒳,(2.55)|ϕ(x)|≤C[diam(suppϕ)]β∥ϕ∥C˙β(𝒳).To see this, assume that suppϕ⊂B(x0,r) for some x0∈𝒳 and r>0.
By Remark 1.2, we can find a y0∈𝒳 such that r≤d(y0,x0)≤2r.
Then ϕ(y0)=0 and for all x∈suppϕ,(2.56)|ϕ(x)|=|ϕ(x)−ϕ(y0)|≤d(x,y0)β∥ϕ∥C˙β(𝒳)≤(3r)β∥ϕ∥C˙β(𝒳),which is just (2.55). By this
observation, we see that the functions ϕ and ψ in Definition 2.13 also satisfy ∥ϕ∥L∞(𝒳)≲1 and ∥ψ∥L∞(𝒳)≲1.
(iii) From Hölder's inequality, it is easy to deduce
that ifT is bounded on Lp(𝒳) with p∈(1,∞),
then T∈WBP(β) for any β∈(0,1] and(2.57)∥T∥WBP(β)≤∥T∥Lp(𝒳)→Lp(𝒳),where and in what follows, ∥T∥Lp(𝒳)→Lp(𝒳) denotes the operator norm of T from Lp(𝒳) to Lp(𝒳).
(iv) Following [34], we can also introduce a slightly stronger property
than WBP,
called strong weak boundedness property in [34] as follows: let T be as in Definition 2.13. The operator T is said to have the strong weak boundedness property of orderβ (for short, T∈SWBP(β)) if T has a distributional kernel K∈(Cbβ(𝒳×𝒳))′ such that (2.48) holds and there exists a
constant C>0 such that for all f∈Cbβ(𝒳×𝒳) with suppf⊂B(x0,r)×B(x0,r) for some x0∈𝒳 and r>0, ∥f(⋅,y)∥C˙β(𝒳)≤r−β for all y∈𝒳 and ∥f(x,⋅)∥C˙β(𝒳)≤r−β for all x∈𝒳,(2.58)|〈K,f〉|≤Cμ(B(x0,r)).The minimal constant C as above is denoted by ∥T∥SWBP(β).
By the observation in (ii) of this remark, the
functionf also satisfies that ∥f∥L∞(𝒳×𝒳)≲1.
(v) Leti=1,2, βi∈(0,1] and let T be a continuous linear mapping from Cbβi(𝒳) to (Cbβi(𝒳))′ with a distributional kernel K as in (2.48) of Proposition 2.12. If K satisfies the size condition that for all x,y∈𝒳 with x≠y,(2.59)|K(x,y)|≤CT1V(x,y),then T∈WBP(β1) if and only if T∈WBP(β2).
This fact can be proved by an argument similar to that used in the proof of
Proposition 1 in [64] and we omit the details.In what follows, standard notions from distribution
theory carry over to continuous linear functionals onCbβ(𝒳).
For instance, if 𝒰⊂𝒳 is an open subset, the restriction of ω∈(Cbβ(𝒳))′ to 𝒰 is defined by 〈ω|𝒰,φ〉=〈ω,φ〉 for all φ∈Cbβ(𝒳) supported in 𝒰.Using some ideas from Meyer in [69], we can obtain the
following useful estimates which play a key role in establishing the
boundedness of singular integrals on spaces of test functions with mean zero.Lemma 2.15.
Fix a bump functionθ∈Cb1(ℝ) with 0≤θ(x)≤1 for all x∈suppθ⊂{x∈ℝ:|x|≤2} and θ(x)=1 on {x∈𝒳:|x|≤1}.
For any fixed z∈𝒳 and r>0,
let θz,r(y)=θ(d(z,y)/r) for all y∈𝒳 and put ωz,r=1−θz,r.
Let T be as in Proposition 2.12. If T(1)∈(C°bβ(𝒳))′ is 0, then the following
hold.
(i)
The restriction of the linear functionalT(θz,r)∈(Cbβ(𝒳))′ to the ball B(z,r/2) is a measurable function, and for a.e.  x∈B(z,r/2),(2.60)T(θz,r)(x)=Cz,r−∫𝒳[K(x,y)−K(z,y)]ωz,r(y)dμ(y),where Cz,r is a constant independent of x.(ii)
If further assuming thatT∈WBP(β),
then there exists a constant C>0 such that for a.e.  x∈B(z,r/2),(2.61)|T(θz,r)(x)|≤C(CT+∥T∥WBP(β)).Proof.
Assume thatT(1)=0 in (C°bβ(𝒳))′.
For any f∈C°bβ(𝒳) with suppf⊂B(z,r/2),
since ∫𝒳f(x)dμ(x)=0,
we have(2.62)0=〈T(1),f〉=〈T(θz,r)+∫𝒳[K(⋅,y)−K(z,y)]ωz,r(y)dμ(y),f〉.Let α∈C1(ℝ) be as in the proof of Proposition 2.9, and η˜z,r(y)=α(d(y,z)/(r/2)).
Set(2.63)ηz,r(y)=η˜z,r(y)[∫𝒳η˜z,r(w)dμ(w)]−1.Then ηz,r∈Cb1(𝒳), suppηz,r⊂B(z,r/2) and ∫𝒳ηz,r(y)dμ(y)=1.
For any f∈Cbβ(𝒳) with suppf⊂B(z,r/2),
we set f˜(y)=f(y)−ηz,r(y)∫𝒳f(y)dμ(y);
then f˜∈C°bβ(𝒳) with suppf˜⊂B(z,r/2).
Applying (2.62) to f˜ and using Corollary 2.11(ii) in the case p=1 show that the restriction of the linear
functional T(θz,r)∈(Cbβ(𝒳))′ to the ball B(z,r/2) is a measurable function, and for a.e.  x∈B(z,r/2),(2.64)〈T(θz,r)(x)+∫𝒳[K(x,y)−K(z,y)]ωz,r(y)dμ(y),f〉=∫𝒳〈T(θz,r)+∫𝒳[K(⋅,y)−K(z,y)]ωz,r(y)dμ(y),ηz,r〉f(x)dμ(x),which implies (i) with(2.65)Cz,r=〈T(θz,r)+∫𝒳[K(⋅,y)−K(z,y)]ωz,r(y)dμ(y),ηz,r〉,by noticing that Cz,r is a constant independent of x and the choice of ηz,r,
but, it may depend on z and r.
To verify (ii), by (2.62) together with the definition
of T(1), (2.49) of Proposition 2.12 and Lemma 2.1(i), we have(2.66)|〈T(θz,r),f〉|=|−〈T(ωz,r),f〉|=|∬𝒳×𝒳[K(x,y)−K(z,y)]ωz,r(y)f(x)dμ(y)dμ(x)|≤CT∫𝒳[∫2d(z,x)<d(z,y)d(x,z)ϵV(z,y)d(z,y)ϵdμ(y)]|f(x)|dμ(x)≲CT∥f∥L1(𝒳),which together with Corollary 2.11(ii) in the cases r=1 and p=1 again shows that for a.e.  x∈B(z,r/2),(2.67)T(θz,r)(x)=Ωz,r(x)+Cz,r,where ∥Ωz,r∥L∞(𝒳)≲CT and Cz,r is a constant independent of x.
We now estimate Cz,r by using that T∈WBP(β).
To this end, let g∈Cbβ(𝒳) with suppg⊂B(z,r), ∥g∥L∞(𝒳)≤1, ∥g∥C˙β(𝒳)≤r−β,
and ∫𝒳g(x)dμ(x)~μ(B(z,r)).
From (2.67) and T∈WBP(β),
it follows that(2.68)|Cz,r||∫𝒳g(x)dμ(x)|=|〈T(θz,r),g〉−∫𝒳Ωz,r(x)g(x)dμ(x)|≲(∥T∥WBP(β)+CT)μ(B(z,r)),which implies that |Cz,r|≲∥T∥WBP(β)+CT, and hence, completes the proof of Lemma
2.15.We recall the notion of the space of functions with
bounded mean oscillation,BMO(𝒳),
which was first introduced by John and Nirenberg in [70], and was proved to be the
dual space of Hat1(𝒳) in [28].Definition 2.16.
Let1≤q<∞.
The spaceBMOq(𝒳) is defined to be the set of
all f∈Llocq(𝒳) such that(2.69)∥f∥BMOq(𝒳)=supx∈𝒳,r>0{1μ(B(x,r))∫B(x,r)|f(y)−mB(x,r)(f)|qdμ(y)}1/q<∞.When q=1,
one denotes
BMO1(𝒳) simply by BMO(𝒳).It was proved in [28] that for any 1≤q1,q2<∞, BMOq1(𝒳) and BMOq2(𝒳) are equal as vector spaces and the seminorms ∥⋅∥BMOq1(𝒳) and ∥⋅∥BMOq2(𝒳) are equivalent. Moreover, if we let 𝒩≡ℂ be the subspace of all constant functions on𝒳,
then the quotient space BMO(𝒳)/𝒩 becomes a Banach space in a natural  way.Remark 2.17.
Letβ∈(0,ϵ] and γ>0.
It is easy to see that T(1)=0 in (C°bβ(𝒳))′ if and only if T(1)∈(C°bβ(𝒳))′ is constant; see also [71, page 22]. Moreover, from Corollary 2.11(i), Proposition 5.21, Theorems 5.19(i) and 6.11 below, it
is easy to see that T(1)∈(C°bβ(𝒳))′ is constant if and only if T(1)∈BMO(𝒳) is constant, which is also equivalent to T(1)=0 in BMO(𝒳).We now establish a basic boundedness result for
singular integrals on spaces of test functions with mean zero, which will be a
key tool for the whole paper.Theorem 2.18.
Letϵ∈(0,1], β∈(0,ϵ), and let T be as in Proposition 2.12 with the
distributional kernel K satisfying the following additional conditions
that
(i)
for allx,y∈𝒳 with x≠y, |K(x,y)|≤CT(1/V(x,y));(ii)
for allx,y,y′∈𝒳 with d(y,y′)≤d(x,y)/2 and x≠y,(2.70)|K(x,y)−K(x,y′)|≤CTd(y,y′)ϵV(x,y)d(x,y)ϵ;(iii)
for allx,x′,y,y′∈𝒳 with d(x,x′)≤d(x,y)/3,d(y,y′)≤d(x,y)/3 and x≠y,(2.71)|[K(x,y)−K(x′,y)]−[K(x,y′)−K(x′,y′)]|≤CTd(x,x′)ϵd(y,y′)ϵV(x,y)d(x,y)2ϵ.
If T∈WBP(β) and T(1)=0 in (C°bβ(𝒳))′,
then T extends to a bounded linear operator from 𝒢°(x1,r,β,γ) to 𝒢(x1,r,β,γ) for all x1∈𝒳, r>0, and γ∈(0,ϵ).
Moreover, there exists a constant Cβ,γ,C0>0 such that for all f∈𝒢°(x1,r,β,γ) with any x1∈𝒳,
any r>0,
and any γ∈(0,ϵ),(2.72)∥Tf∥𝒢(x1,r,β,γ)≤Cβ,γ,C0(CT+∥T∥WBP(β))∥f∥𝒢(x1,r,β,γ).Compared with the corresponding results in [31, 35], Theorem 2.18 has three
advantages. (1) Theorem 2.18 is true on spaces of homogeneous type, while the
corresponding results in [31, 35] are proved only for ℝn or for spaces of homogeneous type with μ(B(x,r))~r for all x∈𝒳 and 0<r≤diam𝒳,
respectively. (2) We do not assume that T∗(1)=0 in Theorem 2.18. Since T∗(1)=0 was also assumed in [31, 35], by the T(1)-theorem in those settings, one knows that T is bounded on L2(𝒳).
Thus, for any f∈𝒢°(x1,r,β,γ) for some x1∈𝒳, r>0, β∈(0,1] and γ>0, Tf∈L2(𝒳),
which makes the proof much easier. (3) We only require T∈WBP(β) instead of T∈SWBP(β) in [35].To prove Theorem2.18, we first recall the following
construction given by Christ in [72], which provides an analogue of the grid of Euclidean dyadic cubes on spaces of homogeneous type.Lemma 2.19.
Let𝒳 be a space of homogeneous type. Then there
exists a collection {Qαk⊂𝒳:k∈ℤ,α∈Ik} of open subsets, where Ik is some index set, and  constants δ∈(0,1) and C6,C7>0 such that
(i)
μ(𝒳∖⋃αQαk)=0 for each fixed k and Qαk∩Qβk=∅ if α≠β;(ii)
for anyα,β,k,l with l≥k, either Qβl⊂Qαk or Qβl∩Qαk=∅;(iii)
for each(k,α) and each l<k there is a unique β such that Qαk⊂Qβl;(iv)
diam(Qαk)≤C6δk;(v)
eachQαk contains some ball B(zαk,C7δk),
where zαk∈𝒳.In fact, we can think ofQαk as being a dyadic cube with diameter rough δk and centered at zαk. In what follows, to simplify our presentation,
we always suppose δ=1/2;
otherwise, we need to replace 2−k in the definition of ATIs by δk and some other changes are also necessary; see
[34, pages 96–98] for
more details.To prove Theorem2.18, we need another technical lemma,
where we need Lemma 2.19 and T∈WBP(β); see also [69, Lemma 3] and [34, Lemma (3.12)].In what follows, ifTf∈(Cbβ(𝒳))′ and g∈Cbβ(𝒳),
we sometimes will write(2.73)∫𝒳Tf(x)g(x)dμ(x)in place of 〈Tf,g〉,
in order to indicate more clearly the dependence on the variable x.Lemma 2.20.
Letϵ∈(0,1], β∈(0,ϵ), γ∈(0,ϵ] and let T be as in Proposition 2.12 with the
distributional kernel K satisfying the additional size condition
(2.59). Let θ be as in Lemma 2.15 and for any fixed x1∈𝒳 and r>0,
one defines
θ˜≡θx1,20r and ω˜≡ωx1,20r in the same way as in Lemma 2.15. If T∈WBP(β) and T(1)=0 in (C°bβ(𝒳))′,
then for any f∈𝒢b(x1,r,β,γ),
the restriction of the linear functional Tf∈(Cbβ(𝒳))′ to the ball B(x1,10r) is a measurable function, and for a.e.  x∈B(x1,10r),(2.74)Tf(x)=∫𝒳K(x,y)[f(y)−f(x)]θ˜(y)dμ(y)+∫𝒳K(x,y)f(y)ω˜(y)dμ(y)+f(x)T(θ˜)(x),where the first two integrals
are absolutely convergent.Proof.
We make use of some ideas used in the proofs of Lemma 3 in [69] and
Lemma (3.12) in [34]. For any fixed f∈𝒢b(x1,r,β,γ) and any ψ∈Cbβ(𝒳) with suppψ⊂B(x1,10r),
we have(2.75)〈Tf,ψ〉=〈T(fθ˜),ψ〉+〈T(fω˜),ψ〉=∫𝒳T(θ˜[f−f(x)])(x)ψ(x)dμ(x)+∫𝒳T(θ˜)(x)f(x)ψ(x)dμ(x)+∬𝒳×𝒳K(x,y)ψ(x)f(y)ω˜(y)dμ(x)dμ(y)≡Y1+Y2+Y3,where the first two integrals
have to be interpreted in the sense of “distributions,” whereas the third
one is an absolutely convergent integral.
Actually, by Lemma2.15(ii), T(θ˜) is a bounded function in B(x1,10r),
and we have(2.76)|Y2|≤(supx∈B(x1,10r)|T(θ˜)(x)|)∫𝒳|ψ(y)f(y)|dμ(y)≲1Vr(x1)∥ψ∥L1(𝒳),which is the desired estimate.
Notice that ifd(x,x1)<10r,
then x∉suppω˜.
Moreover, if we put(2.77)Y3,1(x)≡∫𝒳K(x,y)f(y)ω˜(y)dμ(y),then for x∈B(x1,10r),(2.78)|Y3,1(x)|≲1Vr(x1)+V(x1,x).In fact, we first notice that ω˜(y)≠0 implies that d(y,x1)≥20r≥2d(x,x1).
From this, it follows that V(x1,y)≥μ(B(x1,2d(x,x1)))≥V(x1,x) and d(y,x1)≤2d(x,y),
which together with (1.2) shows that(2.79)V(x1,y)~V(y,x1)=μ(B(y,d(y,x1)))≤μ(B(y,2d(x,y)))≲V(y,x)~V(x,y).These estimates together with
Lemma 2.1(i) and (2.59) imply that(2.80)|Y3,1(x)|≲∫d(y,x1)≥20r1V(x,y)1Vr(x1)+V(x1,y)rγ(r+d(y,x1))γdμ(y)≲1Vr(x1)+V(x1,x),which is just (2.78). Noticing
that suppψ∩supp(fω˜)=∅,
by the assumption (2.48) of Proposition 2.12 and (2.78), we have(2.81)|Y3|=|∫𝒳Y3,1(x)ψ(x)dμ(x)|≲1Vr(x1)∥ψ∥L1(𝒳),which is again the desired
estimate.
Letk∈ℕ
be large enough,
which will be determined later, and let {Sk}k∈ℤ be an ATI with bounded support as constructed in Theorem 2.6. Let η∈C2(ℝ) be radial and η(t)=1 when t∈[0,23], η(t)=0 when |t|>24 and 0≤η(t)≤1 for all t∈ℝ.
For any x,y∈ℝ,
we then define λk(x,y)=∫𝒳Sk(x,z)η(2kd(z,y))dμ(z). It is easy to see that λk(x,y)=0 if d(x,y)≥25−k,
and that(2.82)λk(x,y)=∫𝒳Sk(x,z)dμ(z)=1,ifd(x,y)<2−k.
For any fixedk∈ℕ,
choose J∈ℕ so large that C62−J≤23−k.
Put(2.83)NJ={i∈IJ:QiJ∩B(x,22−k)≠∅forsomex∈B(x1,10r)},and for all x∈B(x1,10r), NJx={i∈IJ:QiJ∩B(x,22−k)≠∅}, where QiJ is the dyadic cube as in Lemma 2.19. Let ♯NJ and ♯NJx denote the cardinality of NJ and NJx,
respectively. It is easy to see that if QiJ∩B(x,22−k)≠∅ for some x∈B(x1,10r),
then QiJ⊂B(x1,24−k+10r) and B(x1,24−k+10r)⊂B(zQiJ,2(24−k+10r)),
where zQiJ is the center of QiJ as in Lemma 2.19. From this, and Lemma 2.19(i)
and (v), it is easy to see that(2.84)♯NJ≲2nJ(r+2−k)n.Similarly, ♯NJx≲2(J−k)n.
We now claim that for any givenk∈ℕ and any given x∈B(x1,10r),(2.85)λk(x,⋅)=limJ→∞∑i∈NJSk(x,zQiJ)η(2kd(zQiJ,⋅))μ(QiJ),(2.86)1−λk(x,⋅)=limJ→∞∑i∈NJSk(x,zQiJ)[1−η(2kd(zQiJ,⋅))]μ(QiJ)hold in Cβ(𝒳).
We only prove (2.85), the proof of (2.86) being similar. To this end, it is easy to see that(2.87)λk(x,y)−∑i∈NJSk(x,zQiJ)η(2kd(zQiJ,y))μ(QiJ)=∑i∈NJx∫QiJ[Sk(x,z)−Sk(x,zQiJ)]η(2kd(z,y))dμ(z)+∑i∈NJx∫QiJSk(x,zQiJ)[η(2kd(z,y))−η(2kd(zQiJ,y))]dμ(z)≡Y4(y)+Y5(y).Since z∈QiJ,
by Lemma 2.19(iv), we have d(z,zQiJ)≤C62−J,
which together with the regularity of Sk yields(2.88)|Y4(y)|≲∑i∈NJx2k−JV2−k(x)μ(QiJ)≲2k−JV2−k(x)μ(B(x,24−k))≲2k−J→0,as J→∞.
Similarly, by the size condition of Sk and the mean value theorem, we have(2.89)|Y5(y)|≲2k−J1V2−k(x)∑i∈NJxμ(QiJ)≲2k−J→0,as J→∞.
Thus, as J→∞,(2.90)∥λk(x,⋅)−∑i∈NJSk(x,zQiJ)η(2kd(zQiJ,⋅))μ(QiJ)∥L∞(𝒳)→0.
For anyy,y′∈𝒳,
if d(y,y′)≥2k−J,
by (2.88), we then have(2.91)|Y4(y)−Y4(y′)|≲2(k−J)(1−β)d(y,y′)β.If d(y,y′)<2k−J,
by the regularity of Sk and the mean value theorem, we
have(2.92)|Y4(y)−Y4(y′)|=|∑i∈NJx∫QiJ[Sk(x,z)−Sk(x,zQiJ)][η(2kd(z,y))−η(2kd(z,y′))]dμ(z)|≲22k−JV2−k(x)d(y,y′)μ(B(x,24−k))≲2k(3−β)−J(2−β)d(y,y′)β.
Similarly, ifd(y,y′)≥2k−J,
by (2.89), we then have(2.93)|Y5(y)−Y5(y′)|≲2(k−J)(1−β)d(y,y′)β.To estimate Y5(y)−Y5(y′) when d(y,y′)<2k−J,
by the mean value theorem, we first see that(2.94)|[η(2kd(z,y))−η(2kd(zQiJ,y))]−[η(2kd(z,y′))−η(2kd(zQiJ,y′))]|≲2kd(z,zQiJ),|[η(2kd(z,y))−η(2kd(zQiJ,y))]−[η(2kd(z,y′))−η(2kd(zQiJ,y′))]|≲2kd(y,y′).Taking a suitable geometric mean
of these estimates, we find that(2.95)|[η(2kd(z,y))−η(2kd(zQiJ,y))]−[η(2kd(z,y′))−η(2kd(zQiJ,y′))]|≲2kd(y,y′)βd(z,zQiJ)1−β.Using this estimate and the size
condition of Sk,
we obtain(2.96)|Y5(y)−Y5(y′)|=|∑i∈NJx∫QiJSk(x,zQiJ){[η(2kd(z,y))−η(2kd(zQiJ,y))]−[η(2kd(z,y′))−η(2kd(zQiJ,y′))]}dμ(z)|≲2kd(y,y′)β2−J(1−β)1V2−k(x)∑i∈NJxμ(QiJ)≲2kd(y,y′)β2−J(1−β).Thus,(2.97)∥λk(x,⋅)−∑i∈NJSk(x,zQiJ)η(2kd(zQiJ,⋅))μ(QiJ)∥C˙β(𝒳)≲2k(3−β)2−J(1−β)→0,as J→∞.
This establishes (2.85), and hence also (2.86).
We now decomposeY1 into(2.98)Y1=∫𝒳T(θ˜[f−f(x)][1−λk(x,⋅)])(x)ψ(x)dμ(x)+∫𝒳T(θ˜[f−f(x)]λk(x,⋅))(x)ψ(x)dμ(x)≡Y1,1+Y1,2,where both integrals have to be
interpreted in the sense of “distributions.”
To estimateY1,1,
by (2.86) and (2.84), observe that(2.99)Y1,1=limJ→∞∑i∈NJ∫𝒳T(θ˜[f−f(x)][1−η(2kd(zQiJ,⋅))])(x)Sk(x,zQiJ)ψ(x)dμ(x)μ(QiJ)where the integral has to be
interpreted in the sense of “distributions.” However, noticing that suppθ˜(⋅)[1−η(2kd(zQiJ,⋅))]∩suppSk(⋅,zQiJ)=∅,
by (2.48), we then further have(2.100)Y1,1=limJ→∞∑i∈NJμ(QiJ)∬𝒳×𝒳K(x,y)θ˜(y)[f(y)−f(x)][1−η(2kd(zQiJ,y))]×Sk(x,zQiJ)ψ(x)dμ(x)dμ(y).Notice that(2.101)∑i∈NJμ(QiJ)∬𝒳×𝒳|K(x,y)θ˜(y)[f(y)−f(x)][1−η(2kd(zQiJ,y))]||Sk(x,zQiJ)ψ(x)|dμ(x)dμ(y)≲μ(B(x1,24−k+10r))×(∫𝒳|ψ(x)|{∫d(x,y)≤(r+d(x1,x))/2|θ˜(y)|1V(x,y)(d(x,y)r+d(x1,x))β×1Vr(x1)+V(x,x1)(rr+d(x1,x))γdμ(y)+∫d(x,y)<50rd(x,y)>(r+d(x,x1))/2|θ˜(y)|[1Vr(x1)+V(x1,y)(rr+d(x1,y))γ+1Vr(x1)+V(x1,x)(rr+d(x1,x))γ]×1V(x,y)dμ(y)}dμ(x))≲μ(B(x1,24−k+10r))1Vr(x1)∥ψ∥L1(𝒳). Thus, by Lebesgue's dominated
convergence theorem, we have(2.102)Y1,1=∬𝒳×𝒳K(x,y)θ˜(y)[f(y)−f(x)]ψ(x){∫𝒳[1−η(2kd(z,y))]Sk(x,z)dμ(z)}dμ(x)dμ(y)=∬𝒳×𝒳K(x,y)θ˜(y)[f(y)−f(x)]ψ(x)[1−λk(x,y)]dμ(x)dμ(y),where the last integral
converges absolutely; and moreover, by (2.82) and an argument similar to (2.101), we further  have |Y1,1|≲(1/Vr(x1))∥ψ∥L1(𝒳).
We next prove that(2.103)limk→∞Y1,2=0,making use of our assumption
that T∈WBP(β).
To this end, by (2.85) and (2.84), we have(2.104)Y1,2=limJ→∞∑i∈NJμ(QiJ)∫𝒳T(θ˜[f−f(x)]η(2kd(zQiJ,⋅)))(x)Sk(x,zQiJ)ψ(x)dμ(x)=limJ→∞∑i∈NJμ(QiJ){∫𝒳T(θ˜[f−f(zQiJ)]η(2kd(zQiJ,⋅)))(x)Sk(x,zQiJ)ψ(x)dμ(x)+∫𝒳T(θ˜η(2kd(zQiJ,⋅)))(x)[f(zQiJ)−f(x)]Sk(x,zQiJ)ψ(x)dμ(x)}.
Choosek∈ℕ such that 2−k≤r/25.
We claim that(2.105)|∫𝒳T(θ˜[f−f(zQiJ)]η(2kd(zQiJ,⋅)))(x)Sk(x,zQiJ)ψ(x)dμ(x)|≲2−kβrβ1Vr(x1)∥f∥𝒢(x1,r,β,γ)∥ψ∥Cβ(𝒳),(2.106)|∫𝒳T(θ˜η(2kd(zQiJ,⋅)))(x)[f(zQiJ)−f(x)]Sk(x,zQiJ)ψ(x)dμ(x)|≲2−kβrβ1Vr(x1)∥f∥𝒢(x1,r,β,γ)∥ψ∥Cβ(𝒳).
We only show (2.105), the proof  of (2.106) being similar.
To see (2.105), put ϕ(y)=θ˜(y)[f(y)−f(zQiJ)]η(2kd(zQiJ,y)) and(2.107)g(x)=Sk(x,zQiJ)ψ(x).Notice that if f,h∈Cβ(𝒳),
then(2.108)∥fh∥C˙β(𝒳)≲∥f∥L∞(𝒳)∥h∥C˙β(𝒳)+∥f∥C˙β(𝒳)∥h∥L∞(𝒳).From this, it follows that(2.109)∥ϕ∥C˙β(𝒳)≲∥f∥𝒢(x1,r,β,γ)1rβVr(x1),and that(2.110)∥g∥C˙β(𝒳)≲∥ψ∥Cβ(𝒳)1V2−k(zQiJ)2kβ.The estimates (2.109) and (2.110) together with T∈WBP(β) imply the claim (2.105).
By (2.105) and (2.106), we know that if 2−k<r/25,
then(2.111)|Y1,2|≲2−kβrβ1Vr(x1)∥f∥𝒢(x1,r,β,γ)∥ψ∥Cβ(𝒳)limJ→∞∑i∈NJμ(QiJ)≲2−kβrβ∥f∥𝒢(x1,r,β,γ)∥ψ∥Cβ(𝒳),which implies (2.103).
Thus, we have that for anyψ∈Cbβ(𝒳) with suppψ⊂B(x1,10r), |〈Tf,ψ〉|≲∥ψ∥L1(𝒳), so that Tf agrees with L∞(𝒳) function on B(x1,10r).
Moreover, by (2.102) and Lebesgue's dominated convergence theorem, we
have(2.112)〈Tf,ψ〉=∬𝒳×𝒳K(x,y)ψ(x)θ˜(y)[f(y)−f(x)]dμ(y)dμ(x)+∬𝒳×𝒳K(x,y)ψ(x)ω˜(y)f(y)dμ(y)dμ(x)+∫𝒳T(θ˜)(x)f(x)ψ(x)dμ(x),as was to be proved.Proof of Theorem2.18.
By Corollary2.11(i), we only need to prove
Theorem 2.18 for 𝒢°b(x1,r,β,γ).
Letf∈𝒢°b(x1,r,β,γ).
We first verify that Tf(x) satisfies (i) of Definition 2.8 for a.e. x∈𝒳.
To this end, we consider two cases.
Case 1  (d(x,x1)<10r).
In this case, letθ˜ and ω˜ be as in Lemma 2.20. By Lemma 2.20, for a.e. x∈B(x1,10r),
we have(2.113)Tf(x)=∫𝒳K(x,y)[f(y)−f(x)]θ˜(y)dμ(y)+∫𝒳K(x,y)f(y)ω˜(y)dμ(y)+f(x)T(θ˜)(x)≡Z1+Z2+Z3.
Lemma 2.15(ii) shows that for a.e. x∈B(x1,10r), |Z3|≲|f(x)|≲1/(Vr(x1)+V(x1,x)), which gives the desired estimate. By (2.78),
we have |Z2|≲1/(Vr(x1)+V(x1,x)). For Z1,
the facts that θ˜(y)≠0 and d(x,x1)<10r imply that d(x,y)<50r.
The size condition on K and the regularity of f yield that(2.114)|Z1|≤∫d(x,y)≤(r+d(x,x1))/2|K(x,y)[f(y)−f(x)]θ˜(y)|dμ(y)+∫d(x,y)>(r+d(x,x1))/2|K(x,y)f(x)θ˜(y)|dμ(y)+∫d(x,y)>(r+d(x,x1))/2|K(x,y)f(y)θ˜(y)|dμ(y)≲{1rβ1Vr(x1)+V(x1,x)∫d(x,y)≤50rd(x,y)βV(x,y)dμ(y)+1Vr(x1)+V(x1,x)∫50r≥d(x,y)>(r+d(x,x1))/2,d(x1,y)≤40r1V(x,y)dμ(y)}+∫50r≥d(x,y)>(r+d(x,x1))/2,d(x1,y)≤40r1V(x,y)1Vr(x1)+V(x1,y)dμ(y)≡Z1,1+Z1,2.By Lemma 2.1(i) and (1.2), we obtain(2.115)Z1,1≲1Vr(x1)+V(x1,x)[1+V(x,50r)V(x,r/2)]≲1Vr(x1)+V(x1,x),Z1,2≲min{1Vr(x1),∫r/2<d(x,y)≤50r,d(x,x1)/5≤r≤d(x1,y)1V(x,y)V(x1,y)dμ(y)+1Vr(x1)∫d(x,y)>d(x,x1)/2,r>d(x1,y)1V(x,y)dμ(y)}≲1Vr(x1)+V(x1,x).Combining the estimates for Z1,1 and Z1,2 yields the desired estimate for Z1,
which verifies that Tf(x) satisfies (i) of Definition 2.8 when d(x,x1)<10r.Case 2  (d(x,x1)≡R≥10r).
In this case, for anym∈ℕ,
let(2.116)Bm=B(x1,10(m+1)r)∖B(x1,10mr).For any fixed x0∈Bm,
we put Im(y)=θ(10d(x0,y)/37mr), Jm(y)=θ(4d(x1,y)/5mr) and define Lm(y) by Lm(y)=1−Im(y)−Jm(y).
Notice that if y∈suppIm∩suppJm,
then d(y,x0)<37mr/5 and d(y,x1)<5mr/2,
and hence d(x0,x1)≤d(x0,y)+d(y,x1)<(99/10)mr<10mr,
which contracts the choice of x0∈Bm.
Therefore, Lm(y)≥0.
We also define f1(y)=f(y)Im(y), f2(y)=f(y)Jm(y) and f3(y)=f(y)Lm(y).
We first establish some estimates on fi with i=1,2,3,
when x∈Bm.
Obviouslyf1(y)≠0 implies that d(y,x0)<37mr/5,
and therefore(2.117)d(x1,y)≥d(x1,x0)−d(x0,y)>325Rwhich together with (1.2) and
the size condition of f shows that(2.118)|f1(y)|≲1VR(x1)(rR)γ,∀y∈𝒳.
We now claim that(2.119)|f1(y)−f1(y′)|≲1VR(x1)d(y,y′)βRβ(rR)γ,∀y,y′∈𝒳.To prove (2.119), we consider two
cases.Case 1  (d(y,y′)≤(r+d(x1,y))/2).
In this case, we divide(2.120)f1(y)−f1(y′)=[f(y)−f(y′)]Im(y)χ0(y,y′)+f(y′)[Im(y)−Im(y′)]=Z4+Z5,where χ0(y,y′)=χ{y∈𝒳:d(x0,y)<37mr/5}(y)+χ{y′∈𝒳:d(x0,y′)<37mr/5}(y′).
The regularity off shows that(2.121)|Z4|≲(d(y,y′)r+d(x1,y))β1Vr(x1)+V(x1,y)(rr+d(x1,y))γχ0(y,y′).If χ0(y,y′)≠0,
then d(x0,y)<37mr/5 or d(x0,y′)<37mr/5.
Notice that d(y,y′)≤(r+d(x1,y))/2 implies that 3d(x1,y)/2≥d(x1,y′)−r/2.
If d(x0,y′)<37mr/5,
we then further have 3d(x1,y)/2≥d(x1,x0)−d(x0,y′)−r/2>21mr/10>R/10, and hence d(x1,y)>R/15.
This together with (2.117) implies that if χ0(y,y′)≠0,
then we have(2.122)d(y,x1)>R15.By (2.122) and (1.2), we obtain VR(x1)≲V(x1,y).
These estimates prove that |Z4|≲(1/VR(x1))(d(y,y′)/R)β(r/R)γ.
The size condition off implies that(2.123)|Z5|≲1Vr(x1)+V(x1,y′)(rr+d(x1,y′))γ|Im(y)−Im(y′)|χ0(y,y′).We now claim that(2.124)1Vr(x1)+V(x1,y′)χ0(y,y′)≲1VR(x1).To prove (2.124), we consider two
subcases.
Subcase 1.1  (r≥d(x1,y′)).
In this case,(1/(Vr(x1)+V(x1,y′)))χ0(y,y′)≲(1/Vr(x1))χ0(y,y′). If d(x0,y′)<37mr/5,
then d(x1,y′)≥d(x1,x0)−d(x0,y′)>13mr/5>2r, which contradicts the assumption that r≥d(x1,y′).
Thus, we further have(2.125)1Vr(x1)+V(x1,y′)χ0(y,y′)≲1Vr(x1)χ{y∈𝒳:d(x0,y)<37mr/5}(y).Since d(x0,y)<37mr/5 and x∈Bm,
by (2.117), we have d(y,x1)>3R/25. Moreover, from d(y,x1)≤d(y,y′)+d(y′,x1) and d(y,y′)≤(r+d(x1,y))/2,
it follows that d(y,x1)≤r+2d(y′,x1)≤3r.
Thus, in this case, R<25r and hence VR(x1)≤Vr(x1).
Thus, the claim (2.124) holds in this case.Subcase 1.2  (r<d(x1,y′)).
In this case, fromr<d(x1,y′) and d(y,y′)≤(r+d(x1,y))/2,
it follows that d(x1,y)≤r+2d(x1,y′)<3d(x1,y′),
which together with (2.122) implies that VR(x1)≲V(x1,y′).
Thus, (1/(Vr(x1)+V(x1,y′)))χ0(y,y′)≲1/VR(x1). Thus, (2.124) also holds in this case, which
completes the proof of claim (2.124).
Notice that ifd(x0,y)<37mr/5,
then d(y,x1)≤d(x1,x0)+d(x0,y)≲R, and if d(x0,y′)<37mr/5,
since d(y,y′)≤(r+d(y,x1))/2,
we then have d(y,x1)≤r+2d(y′,x1)≤r+2[d(y′,x0)+d(x0,x1)]≲R. These estimates together with d(y,y′)≤(r+d(y,x1))/2 again implies that if χ0(y,y′)≠0,
then d(y,y′)≲R and r+d(y,x1)≤2(r+d(x1,y′)),
which together with (2.122) further yields that r+d(x1,y′)≳R.
All these estimates, (2.124), and the mean value theorem show that for any β∈(0,1],(2.126)|Z5|≲1VR(x1)(rR)γ|d(y,x)−d(y′,x)|R≲1VR(x1)(d(y,y′)R)β(rR)γ.Case 2  (d(y,y′)≥(r+d(y,x1))/2).
In this case, notice again thatf1(y)−f1(y′)≠0 implies that d(x0,y)<37mr/5 or d(x0,y′)<37mr/5.
If d(x0,y′)<37mr/5,
by d(y,x1)<2d(y,y′),
we have 10mr≤d(x0,x1)≤d(x0,y′)+d(y′,y)+d(y,x1)<37mr/5+3d(y,y′). Thus, by x∈Bm,
we further have 3R/25<13mr/5<3d(y,y′), which together with (2.117) and d(y,x1)<2d(y,y′) again implies that if f1(y)−f1(y′)≠0 and d(y,y′)≥(r+d(y,x1))/2,
then R<17d(y,y′). This estimate together with (2.118) yields that for any β>0,(2.127)|f1(y)−f1(y′)|≤|f1(y)|+|f1(y′)|≲1VR(x1)(rR)γ≲1VR(x1)(d(y,y′)R)β(rR)γ.Thus, (2.119) holds.
As forf3,
first observe that obviously, for all y∈𝒳,(2.128)|f3(y)|≲1Vr(x1)+V(x1,y)(rr+d(y,x1))γχ{d(x1,y)>R/16}(y).
From (2.128) and Lemma 2.1(i), it follows that(2.129)∫𝒳|f3(y)|dμ(y)≲rγ∫d(y,x1)>R/161V(x1,y)1d(y,x1)γdμ(y)≲(rR)γ.
Notice thatsuppf1⊂B(x1,137R/50).
From this, the estimates (2.118) and (2.129) together with ∫𝒳f(x)dμ(x)=0 and (1.2), it follows that
(2.130)|∫𝒳f2(y)dμ(y)|=|∫𝒳f1(y)dμ(y)+∫𝒳f3(y)dμ(y)|≲1VR(x1)(rR)γV137R/50(x1)+(rR)γ≲(rR)γ.
Sincex∈Bm,
we have 1/(Vr(x1)+V(x1,x))~1/VR(x1). Now, for any m∈ℕ and y∈𝒳,
put Um,r(y)=θ(d(x1,y)/42(m+1)r).
Notice that suppf1∩supp(1−Um,r)=∅.
By Lemma 2.20, for a.e. x∈Bm,
we have(2.131)Tf1(x)=∫𝒳K(x,y)[f1(y)−f1(x)]Um,r(y)dμ(y)+f1(x)T(Um,r)(x)≡Z6(x)+Z7(x).From Lemma 2.15(ii) and (2.118),
it follows that for a.e. x∈Bm,(2.132)|Z7(x)|≲|f1(x)|≲1VR(x1)(rR)γ.As for Z6(x),
notice that x∈Bm together with Um,r(y)≠0 implies that d(x,y)<18R;
hence, by (2.119), the size condition on K and Lemma 2.1(i), we obtain(2.133)|Z6(x)|≲∫d(x,y)<18R1V(x,y)1VR(x1)d(x,y)βRβ(rR)γdμ(y)≲1VR(x1)(rR)γ.Combining the estimate for Z6(x) with Z7(x) gives us the desired estimate for Tf1(x) for a.e. x∈Bm.
Sincex∉suppf2,
we can write(2.134)Tf2(x)=∫𝒳[K(x,y)−K(x,x1)]f2(y)dμ(y)+K(x,x1)∫𝒳f2(y)dμ(y)≡Z8(x)+Z9(x).The assumption (ii) of the
theorem, the support condition of f2,
and the fact that γ<ϵ together with Lemma 2.1(i) yield that(2.135)|Z8(x)|≤∫d(x1,y)≤R/2|K(x,y)−K(x,x1)||f2(y)|dμ(y)≲1V(x,x1)d(x,x1)ϵ∫d(x1,y)≤R/2d(x1,y)ϵ1V(x1,y)rγd(x1,y)γdμ(y)≲1VR(x1)(rR)γ.From the size condition on K,
and (2.130), it also follows that(2.136)|Z9(x)|≲1V(x,x1)(rR)γ≲1VR(x1)(rR)γ,which together with the estimate
for Z8(x) gives the desired estimate for Tf2(x) for a.e. x∈Bm.
Notice thatf3(x)≠0 implies that d(x1,y)>R/16 and d(x0,y)>37mr/10.
If we now further assume that x∈Bm∩B(x0,mr/10),
then d(x,y)≥d(y,x0)−d(x0,x)>18mr/5>9R/50. From this, (1.17), and (1.2), it follows that VR(x1)~VR(x)≲V(x,y).
This together with (2.128), the size condition on K and Lemma 2.1(i) yields that(2.137)|Tf3(x)|=|∫d(x,y)>9R/50,d(x1,y)>R/16K(x,y)f3(y)dμ(y)|≲∫d(x,y)>9R/50,d(x1,y)>R/161V(x,y)1V(x1,y)rγd(y,x1)γdμ(y)≲1VR(x1)(rR)γ,which is the desired estimate.
Thus, we have verified thatTf(x) for a.e. x∈Bm∩B(x0,mr/10) satisfies the size condition (i) of Definition 2.8, with constants independent of m, r, and x1.
By the Besicovitch covering lemma (see, e.g., [13, Theorem 1.16, pages 8-9]), this implies that for a.e. x∈Bm, Tf(x) satisfies the size condition (i) of Definition 2.8, with constants independent of m, r, and x1.
In combination with Case 1, we thus see that for a.e. x∈𝒳, Tf(x) satisfies the size condition (i) of Definition 2.8.
We now turn to verify thatTf(x) for a.e. x∈𝒳 satisfies the regularity condition (ii) of
Definition 2.8. In what follows, we fix x′ near x and put δ≡d(x,x′).
We first remark that for a.e. x,x′∈𝒳 such that (1/160)(r+R)<δ≤(1/2)(r+R),
by the size condition of Tf(x) and Lemma 2.1(iii), we obtain(2.138)|Tf(x)−Tf(x′)|≲1Vr(x1)+V(x1,x)(rr+d(x,x1))γ+1Vr(x1)+V(x1,x′)(rr+d(x′,x1))γ≲(d(x,x′)r+d(x,x1))β1Vr(x1)+V(x1,x)(rr+d(x,x1))γ,which is the desired estimate.
In what follows, we only need to estimate |Tf(x)−Tf(x′)| for a.e. x,x′∈𝒳 such that δ≤(1/160)(r+R) by considering two cases. Recall that  R≡d(x,x1).
Case 1  (R<10r).
In this case, we divide the ballB(x1,10r) into the union of annuli Bm1={x∈𝒳:10r/(m1+1)≤d(x,x1)<10r/m1},
where m1∈ℕ.
For any fixed x0∈Bm1 and m2∈ℕ,
we put um1,m2(y)=θ(20m1m2d(x0,y)/(m1+10)r) and let wm1,m2(y)=1−θm1,m2(y) for all y∈𝒳. By Lemmas 2.15 and 2.20, for a.e. x∈B(x0,(m1+10)r/40m1m2),(2.139)Tf(x)=∫𝒳K(x,y)[f(y)−f(x)]um1,m2(y)dμ(y)+[∫𝒳K(x,y)f(y)wm1,m2(y)dμ(y)+f(x)T(um1,m2)(x)]≡Γ1(x)+Γ2(x),(2.140)|T(um1,m2)(x)|≲CT+∥T∥WBP(β),(2.141)T(um1,m2)(x)=Cm1,m2,r−∫𝒳[K(x,y)−K(x0,y)]wm1,m2(y)dμ(y),where Cm1,m2,r is a constant independent of x.
For any(2.142)x∈B(x0,(m1+11)r160(m1+1)(m2+1)),we then consider x′ in the annulus (r+R)/160(m2+1)<d(x,x′)≤(r+R)/160m2.
It is easy to check x′∈B(x0,(m1+10)r/40m1m2).
Assume that x,x′ both satisfy (2.139), (2.140), and (2.141). Notice that um1,m2(y)≠0 implies that d(y,x0)≤(m1+10)r/10m1m2,
and hence d(x,y)≤d(x,x0)+d(x0,y)<17(m1+11)r/160m1m2<68d(x,x′)≡68δ and  d(x′,y)≤69δ.
Therefore, the size condition on K,
the regularity of f,
and Lemma 2.1(i) show that(2.143)|Γ1(x)|=|∫𝒳K(x,y)[f(y)−f(x)]um1,m2(y)dμ(y)|≤∫d(x,y)≤68δ|K(x,y)||f(y)−f(x)|dμ(y)≲(d(x,x′)r+d(x,x1))β1Vr(x1)V(x1,x)(rr+d(x1,x))γ,and a similar estimate also
holds for Γ1(x′).
Thus, we clearly obtain the desired estimate for Γ1(x)−Γ1(x′).
Now by (2.141), we have(2.144)Γ2(x)−Γ2(x′)=∫𝒳[K(x,y)−K(x′,y)][f(y)−f(x)]wm1,m2(y)dμ(y)+f(x){T(um1,m2)(x)+∫𝒳[K(x,y)−K(x′,y)]wm1,m2(y)dμ(y)}−f(x′)T(um1,m2)(x′)=∫𝒳[K(x,y)−K(x′,y)][f(y)−f(x)]wm1,m2(y)dμ(y)+[f(x)−f(x′)]T(um1,m2)(x′)≡Γ2,1+Γ2,2.The estimate (2.140) and the
regularity of f then show that(2.145)|Γ2,2|≲|f(x)−f(x′)|≲(d(x,x′)r+d(x,x1))β1Vr(x1)+V(x1,x)(rr+d(x1,x))γ.Notice that wm1,m2(y)≠0 implies that d(x,y)>6δ.
The regularity (2.49) on K,
the regularity and the size condition of f,
and Lemma 2.1(i) together with the fact that β<ϵ give(2.146)|Γ2,1|≲∫d(x,y)>6δ,d(x,y)≤(r+d(x,x1))/2d(x,x′)ϵV(x,y)d(x,y)ϵ(d(x,y)r+d(x,x1))β1Vr(x1)+V(x1,x)(rr+d(x1,x))γdμ(y)+∫d(x,y)>6δ,d(x,y)>(r+d(x,x1))/2d(x,x′)ϵV(x,y)d(x,y)ϵ×[1Vr(x1)+V(x1,x)(rr+d(x1,x))γ+1Vr(x1)+V(x1,y)(rr+d(x1,y))γ]dμ(y)≲(d(x,x′)r+d(x,x1))β1Vr(x1)+V(x1,x)(rr+d(x1,x))γ,where, in the last inequality,
we used that r+d(x1,x)≤11r≲r+d(x1,y) and(2.147)1Vr(x1)+V(x1,y)≤1Vr(x1)≲1Vr(x1)+V(x1,x),which follows from V(x1,x)≤μ(B(x1,10r))≲Vr(x1), by (1.2). Thus, for any x0∈Bm1,
and a.e. x∈B(x0,(m1+11)r/160(m1+1)(m2+1)) and a.e. x′∈𝒳 satisfying (r+R)/160(m2+1)<d(x,x′)≤(r+R)/160m2, Tf(x)−Tf(x′) has the desired regularity with constants
independent of m1 and m2.
Again, by the Besicovitch covering lemma, we further see that for a.e. x∈Bm1 and a.e. x′∈𝒳 satisfying (r+R)/160(m2+1)<d(x,x′)≤(r+R)/160m2, Tf(x)−Tf(x′) has the desired regularity with constants
independent of m1 and m2,
which implies that there exists a measurable set 𝒳1 such that μ(𝒳1)=0 and for all x∈B(x1,10r)∖𝒳1 and all x′∈𝒳∖𝒳1 with d(x,x′)≤(1/160)(r+R), Tf(x)−Tf(x′) has the desired regularity.Case 2  (R≥10r).
In this case, we renameBm in Case 2 of the proof for the size condition
of Tf by B˜m1.
Namely, for m1∈ℕ,
we consider annuli(2.148)B˜m1=B(x1,10(m1+1)r)∖B(x1,10m1r).For any fixed x˜0∈B˜m1,
we define Im1, Jm1,
and Lm1 in the same way as in Case 1 with m and x0 replaced, respectively, by m1 and x˜0 here. Let f1(y)=f(y)Im1(y), f2(y)=f(y)Jm1(y) and f3(y)=f(y)Lm1(y) for all y∈𝒳. Then the estimates (2.119) and (2.130) still hold for f1, f2,
and f3 here, when x∈B˜m1.
Now, form2∈ℕ,
we put u˜m1,m2(y)=θ(20m2d(x˜0,y)/(10m1+11)r) and let w˜m1,m2(y)=1−θ˜m1,m2(y) for all y∈𝒳. By Lemmas 2.15 and 2.20, for a.e. x∈B(x˜0,(10m1+11)r/40m2),
(2.149)Tf1(x)=∫𝒳K(x,y)[f1(y)−f1(x)]u˜m1,m2(y)dμ(y)+[∫𝒳K(x,y)f1(y)w˜m1,m2(y)dμ(y)+f1(x)T(u˜m1,m2)(x)]≡Γ3(x)+Γ4(x),(2.150)|T(u˜m1,m2)(x)|≲CT+∥T∥WBP(β),(2.151)T(u˜m1,m2)(x)=C˜m1,m2,r−∫𝒳[K(x,y)−K(x˜0,y)]w˜m1,m2(y)dμ(y),where C˜m1,m2,r is a constant independent of x.
For any x∈B(x˜0,(10m1+1)r/160(m2+1)),
we then consider x′ in the annulus (r+R)/160(m2+1)<d(x,x′)≤(r+R)/160m2.
It is easy to check x′∈B(x˜0,(10m1+11)r/40m2).
We also restrict x∈B˜m1.
Assume that x,x′ both satisfy (2.149), (2.150), and (2.151).
Notice that u˜m1,m2(y)≠0 implies that d(y,x˜0)≤(10m1+11)r/10m2,
which implies d(x,y)≤d(x,x˜0)+d(x˜0,y)<177(10m1+1)r/160m2<354d(x,x′)≡354δ and d(x′,y)≤355δ.
Therefore, the size condition on K, (2.119), and Lemma 2.1(i)  show that(2.152)|Γ3(x)|≲∫d(x,y)≤354δ1V(x,y)1VR(x1)d(x,y)βRβ(rR)γdμ(y)≲1VR(x1)(d(x,x′)R)β(rR)γ,and a similar estimate also
holds for Γ4(x′).
Thus, we obtain the desired estimate for Γ3(x)−Γ4(x′).
Now by (2.151) and some computations similar to Γ2(x)−Γ2(x′),
we have(2.153)Γ4(x)−Γ4(x′)=∫𝒳[K(x,y)−K(x′,y)][f1(y)−f1(x)]w˜m1,m2(y)dμ(y)+[f1(x)−f1(x′)]T(u˜m1,m2)(x′)≡Γ4,1+Γ4,2.The estimate (2.150) together with (2.119) then yields(2.154)|Γ4,2|≲|f1(x)−f1(x′)|≲1VR(x1)(d(x,x′)R)β(rR)γ.Notice that w˜m1,m2(y)≠0 implies d(x,y)>7δ.
Then by the regularity (2.49) on K,
(2.119), and Lemma 2.1(i) together with β<ϵ,
we further have(2.155)|Γ4,1|≲∫d(x,y)>7δd(x,x′)ϵV(x,y)d(x,y)ϵ1VR(x1)(d(x,y)R)β(rR)γdμ(y)≲1VR(x1)(d(x,x′)R)β(rR)γ.Thus, for any x˜0∈B˜m1,
and a.e. x∈B(x˜0,(10m1+1)r/160(m2+1))∩B˜m1 and a.e. x′∈𝒳 satisfying (r+R)/160(m2+1)<d(x,x′)≤(r+R)/160m2, Tf1(x)−Tf1(x′) has the desired regularity with constants
independent of m1 and m2.
Notice thatf2(x)=0=f2(x′).
We write(2.156)|Tf2(x)−Tf2(x′)|=|∫𝒳[K(x,y)−K(x′,y)]f2(y)dμ(y)|≤|∫𝒳{[K(x,y)−K(x′,y)]−[K(x,x1)−K(x′,x1)]}f2(y)dμ(y)|+|K(x,x1)−K(x′,x1)||∫𝒳f2(y)dμ(y)|≡Γ5+Γ6.Since d(x,x′)≤(11/1600)R,
the regularity (2.49) on K,
(2.130), and β<ϵ prove(2.157)Γ6≲d(x,x′)ϵV(x1,x)d(x1,x)ϵ(rR)γ≲1VR(x1)(d(x,x′)R)β(rR)γ.Notice that f2(y)≠0 implies that d(x1,y)≤R/4.
Then the regularity (iii) on K together with the size condition of f,
the fact that γ<ϵ and Lemma 2.1(i) yields that(2.158)Γ5≲1VR(x1)∫d(x1,y)≤R/4d(x,x′)ϵd(x1,y)ϵd(x1,x)2ϵ1Vr(x1)+V(x1,y)(rr+d(x1,y))γdμ(y)≲1VR(x1)(d(x,x′)R)β(rR)γ.Thus, we have also obtained the
desired regularity for Tf2(x)−Tf2(x′) for all x∈B˜m1 and d(x,x′)≤(1/160)(r+R).
Notice thatf3(y)≠0 implies that d(x1,y)>R/16.
Moreover, if(2.159)x∈B(x˜0,(10m1+1)r160(m2+1))∩B˜m1,then f3(y)≠0 also implies that d(x,y)>7R/40.
Since δ≤(11/1600)R,
we then have δ≤(11/280)d(x,y).
Then the regularity (2.49) on K,
the size condition of f,
and Lemma 2.1(i) show that(2.160)|Tf3(x)−Tf3(x′)|≲∫d(x1,y)>R/16,d(x,y)>7R/40|K(x,y)−K(x′,y)||f3(y)|dμ(y)≲∫d(x1,y)>R/16,d(x,y)>7R/40d(x,x′)ϵV(x,y)d(x,y)ϵ1Vr(x1)+V(x1,y)rγd(x1,y)γdμ(y)≲1VR(x1)(d(x,x′)R)β(rR)γ,where in the last step, we used
the fact that β<ϵ and d(x,x′)≤(11/1600)R,
and in the third-to-last inequality, we used the fact that(2.161)VR(x1)~V(x,x1)≤μ(B(x,40d(x,y)7))≲V(x,y).Thus, for all x∈B(x˜0,(10m1+1)r/160(m2+1))∩B˜m1 and d(x,x′)≤(1/160)(r+R),
we obtain the desired regularity for Tf3(x)−Tf3(x′).
Summarizing the above estimates, we see that for anyx˜0∈B˜m1,
and a.e. x∈B˜m1∩B(x˜0,(10m1+1)r/160(m2+1)) and a.e. x′∈𝒳 satisfying (r+R)/160(m2+1)<d(x,x′)≤(r+R)/160m2,Tf(x)−Tf(x′) has the desired regularity with constants
independent of m1 and m2.
By the Besicovitch covering lemma again, similarly to Case 1, we obtain that
there exists a measurable set 𝒳2 such that μ(𝒳2)=0,
and for all x∈𝒳∖(B(x1,10r)∪𝒳2) and all x′∈𝒳∖𝒳2 with d(x,x′)≤(1/160)(r+R), Tf(x)−Tf(x′) has the desired regularity.
Thus, there exists a measurable set𝒳3 such that for all x,x′∈𝒳∖𝒳3 with d(x,x′)≤(1/2)(r+d(x,x1)), Tf(x)−Tf(x′) satisfies the regularity condition (ii) of Definition 2.8, which implies that for all x,x′∈𝒳∖𝒳3 with d(x,x′)≤r/2,(2.162)|Tf(x)−Tf(x′)|≲1Vr(x1)rβd(x,x′)β.Now, for any x∈𝒳,
we choose {xn}n∈ℕ⊂(𝒳∖𝒳3) such that d(xn,x)→0 as n→∞.
Then {Tf(xn)}n∈ℕ is a Cauchy sequence in ℂ.
We then define g(x)=limn→∞Tf(xn).
It is easy to show that g is a well-defined continuous function lying in 𝒢(x1,r,β,γ),
which agrees with Tf almost everywhere. We may thus choose g as a representative of Tf,
for which we then have(2.163)∥Tf∥𝒢(x1,r,β,γ)≲(CT+∥T∥WBP(β))∥f∥𝒢(x1,r,β,γ).
This completes the proof  of Theorem 2.18.Remark 2.21.
It was proved in [35] that the condition (iii) of
Theorem 2.18 is also necessary for an operator T with distributional kernel K to be bounded from 𝒢°(x1,r,β,γ) to 𝒢(x1,r,β,γ) for all x1∈𝒳 and r>0;
see [35, Theorem 4] or
[31].By Remark2.14(iii), we immediately obtain the
following conclusion, which is convenient in applications.Corollary 2.22.
Letϵ∈(0,1], β∈(0,ϵ), and let T be as in Proposition 2.12, with the
distributional kernel K satisfying (i), (ii), and (iii) of Theorem 2.18. If T is bounded on Lp(𝒳) for a certain p∈(1,∞) and T(1)=0 in (C°bβ(𝒳))′,
then T extends to a bounded linear operator from 𝒢°(x1,r,β,γ) to 𝒢(x1,r,β,γ) for all x1∈𝒳, r>0 and γ∈(0,ϵ).
Moreover, there exists a constant Cβ,γ,C0>0 such that for all f∈𝒢°(x1,r,β,γ) with any x1∈𝒳,
any r>0,
and any γ∈(0,ϵ),(2.164)∥Tf∥𝒢(x1,r,β,γ)≤Cβ,γ,C0(CT+∥T∥Lp(𝒳)→Lp(𝒳))∥f∥𝒢(x1,r,β,γ).By an argument similar to the proof of Lemma2.20, we
can easily obtain the following result, which is of independent interest; see also [69, Lemma 2], [71, Lemma 1, page 119] for ℝn, and [34, Lemma (3.12)] for Ahlfors 1-regular metric measure spaces. Moreover, instead of T∈SWBP(β) in [34, Lemma (3.12)], we only need that T∈WBP(β).Corollary 2.23.
LetT be as in Lemma 2.20. Then there exists a
constant C>0 such that for all f∈C˙bβ(𝒳),(2.165)∥Tf∥L∞(𝒳)≤C(CT+∥T∥WBP(β))[diam(suppf)]β∥f∥C˙β(𝒳).Proof.
Assume thatsuppf⊂B(x0,r) for some x0∈𝒳 and r>0.
Let θx0,10r be defined in the same way as in Lemma 2.15. An
argument similar to the proof of Lemma 2.20 yields that when d(x,x0)<5r,(2.166)Tf(x)=∫𝒳K(x,y)[f(y)−f(x)]θx0,10r(y)dμ(y)+f(x)∫𝒳K(x,y)θx0,10r(y)dμ(y).From this, Lemma 2.1(i), and
Lemma 2.15(ii) together with (2.55), it follows that(2.167)|Tf(x)|≲CT∥f∥C˙β(𝒳)∫d(x,y)<25r1V(x,y)d(x,y)βdμ(y)+∥f∥L∞(𝒳)(CT+∥T∥WBP(β))≲(CT+∥T∥WBP(β))rβ∥f∥C˙β(𝒳).If d(x,x0)≥5r and d(y,x0)<r,
then d(y,x)>4d(x,x0)/5,
and by Lemma 2.1(i) and (2.55), we also have(2.168)|Tf(x)|≤CT∫d(x,y)>4d(x,x0)/51V(x,y)|f(y)|dμ(y)≲CT1V(x0,x)∥f∥L∞(𝒳)Vr(x0)≲CT[diam(suppf)]β∥f∥C˙β(𝒳),which completes the proof of
Corollary 2.23.From Theorem2.18, it is easy to deduce the following
result which is convenient in applications.Corollary 2.24.
LetT be as in Theorem 2.18 or Corollary 2.22. If ϵ˜∈(0,ϵ),
then T is bounded from 𝒢°0ϵ˜(x1,r,β,γ) to 𝒢0ϵ˜(x1,r,β,γ) for all x1∈𝒳, r>0, and 0<β,γ<ϵ˜.
Moreover, there exists a constant C>0 such that for all f∈𝒢°0ϵ˜(x1,r,β,γ) with any x1∈𝒳 and any r>0,(2.169)∥Tf∥𝒢0ϵ˜(x1,r,β,γ)≤Cβ,γ,C0(CT+∥T∥WBP(β))∥f∥𝒢0ϵ˜(x1,r,β,γ)(or resp., ∥Tf∥𝒢0ϵ˜(x1,r,β,γ)≤Cβ,γ,C0(CT+∥T∥Lp(𝒳)→Lp(𝒳))∥f∥𝒢0ϵ˜(x1,r,β,γ)).Proof.
Letf∈𝒢°0ϵ˜(x1,r,β,γ).
By the definition of 𝒢°0ϵ˜(x1,r,β,γ), there exists(2.170){fn}n=1∞⊂𝒢°(x1,r,ϵ˜,ϵ˜)such that(2.171)limn→∞∥fn−f∥𝒢(x1,r,β,γ)=0.By Theorem 2.18 or Corollary 2.22,
we have Tfn∈𝒢(x1,r,ϵ˜,ϵ˜) and(2.172)∥Tfn−Tf∥𝒢(x1,r,β,γ)≲∥f−fn∥𝒢(x1,r,β,γ)→0,as n→∞.
Thus, Tf∈𝒢0ϵ˜(x1,r,β,γ) and(2.173)∥Tf∥𝒢0ϵ˜(x1,r,β,γ)=∥Tf∥𝒢(x1,r,β,γ)≲∥f∥𝒢(x1,r,β,γ)=∥f∥𝒢0ϵ˜(x1,r,β,γ),which completes the proof of
Corollary 2.24.


### 2.3. Boundedness of Singular Integrals on Spaces of Test Functions

In this subsection, we establish the boundedness on𝒢(x1,r,β,γ) of singular integrals. Since the functions in 𝒢(x1,r,β,γ) may have nonvanishing integral, this requires,
as usual, some extra size decay conditions on the integral kernels; see, for
example, [71, page 123].The following result is an inhomogeneous variant of
Proposition2.12.Proposition 2.25.
Letβ∈(0,1], σ>0, r0>0, and let T be as in Proposition 2.12 with the
distributional kernel K satisfying the following extra size condition
that for all x,y∈𝒳 with d(x,y)≥r0,(2.174)|K(x,y)|≤CT1V(x,y)r0σd(x,y)σ.Then T can be extended to a continuous linear operator
from Cβ(𝒳) to (Cbβ(𝒳))′.Proof.
Letf∈Cβ(𝒳) and g∈Cbβ(𝒳),
and suppose suppg⊂B(x0,r) for some x0∈𝒳 and r>0.
Choose ψ∈Cbβ(𝒳) such that ψ(x)=1 when x∈B(x0,2max{r,r0}) and ψ(x)=0 when x∉B(x0,4max{r,r0}).
It is easy to see that ψf∈Cbβ(𝒳),
which implies that 〈T(ψf),g〉 is well defined. On the other hand, we define(2.175)〈T((1−ψ)f),g〉=∬𝒳×𝒳K(x,y)(1−ψ(y))f(y)g(x)dμ(y)dμ(x).By the size condition (2.174) and
Lemma 2.1(i), it is easy to check that the right-hand side of the above
equality is finite; furthermore, if f has also bounded support, this coincides with
(2.48). Moreover, it is easy to verify that 〈T(ψf),g〉+〈T((1−ψ)f),g〉 is independent of choice of ψ.
Thus we can define Tf by(2.176)〈Tf,g〉=〈T(ψf),g〉+〈T((1−ψ)f),g〉,so that Tf∈(Cbβ(𝒳))′.
It is clear that then T is continuous, which completes the proof of
Proposition 2.25.We now establish an inhomogeneous variant of Lemma2.15. Here, in contrast to Lemma 2.15, we do not need T∈WBP(β).Lemma 2.26.
Letθ be as in Lemma 2.15. For any fixed z∈𝒳 and r>0,
let θz,max{r,r0} and ωz,max{r,r0} be defined as in Lemma 2.15. Let T be as in Proposition 2.25. If, for a certain β∈(0,ϵ], T(1)∈(Cbβ(𝒳))′ is a constant CT(1),
then there exists a constant C>0 such that
(i)
for allx∈B(z,max{r,r0}/2), |T(ωz,max{r,r0})(x)|≤CCT;(ii)
the restriction of the linear functionalT(θz,max{r,r0})+T(ωz,max{r,r0})∈(Cbβ(𝒳))′ to the ball B(z,max{r,r0}/2) is a constant; namely, for all x∈B(z,max{r,r0}/2),(2.177)T(θz,max{r,r0})(x)+T(ωz,max{r,r0})(x)=CT(1);(iii)
for allx∈B(z,max{r,r0}/2), |T(θz,max{r,r0})(x)|≤C(CT+|CT(1)|).Proof.
By the definitions ofθz,max{r,r0} and ωz,max{r,r0},
we know that(2.178)ωz,max{r,r0}(y)≠0implies that d(y,z)>max{r,r0},
which together with d(x,z)<max{r,r0}/2 shows that d(x,y)>max{r,r0}/2.
Therefore, by (2.174) and Lemma 2.1(i), for all x∈𝒳 with d(x,z)<max{r,r0}/2,
we have(2.179)|T(ωz,max{r,r0})(x)|=|∫d(x,y)>max{r,r0}/2K(x,y)ωz,max{r,r0}(y)dμ(y)|≤CT∫d(x,y)>max{r,r0}/21V(x,y)r0σd(x,y)σdμ(y)≤CCT,which is (i).
To see (ii), for anyf∈𝒢b(β,γ) with suppf⊂B(z,max{r,r0}/2),
by the definition of T(1)=CT(1) in (Cbβ(𝒳))′,
we have(2.180)CT(1)∫𝒳f(x)dμ(x)=〈T(θz,max{r,r0}),f〉+〈T(ωz,max{r,r0}),f〉,which together with Corollary 2.11(ii) gives the conclusion (ii) of this lemma.
The conclusion (iii) can be deduced from the
conclusion (i) and the conclusion (ii), which completes the proof of Lemma2.26.We recall the notion of the spacebmo(𝒳),
which, when 𝒳=ℝn and μ is the n-dimensional Lebesgue measure, was first
introduced by Goldberg [73]. A variant in the setting of spaces of homogeneous
type can be found in [74].Definition 2.27.
Let1≤q<∞.
The spacebmoq(𝒳) is defined to be the set of all f∈Llocq(𝒳) such that(2.181)∥f∥bmoq(𝒳)=supx∈𝒳,0<r<1{1μ(B(x,r))∫B(x,r)|f(y)−mB(x,r)(f)|qdμ(y)}1/q+supx∈𝒳,r≥1{1μ(B(x,r))∫B(x,r)|f(y)|qdμ(y)}1/qis finite. When q=1,
one denotes
bmo1(𝒳) simply by bmo(𝒳).It was proved in [74] that for any 1≤q1,q2<∞, bmoq1(𝒳)=bmoq2(𝒳) with equivalent norms.Remark 2.28.
From Corollary2.11(ii), Proposition 5.46, Theorems 5.44(i) and 6.28 below, it is easy to see that T(1)=CT(1) in (Cbβ(𝒳))′ if and only if T(1)=CT(1) in bmo(𝒳).We can now present the inhomogeneous variant of
Theorem2.18 as follows.Theorem 2.29.
Letϵ∈(0,1], σ>0, r0>0, and let T be as in Proposition 2.25 with the
distributional kernel K also satisfying the conditions (i) through
(iii) in Theorem 2.18 and the following additional regularity condition: for all x,x′,y∈𝒳 with d(x,x′)≤d(x,y)/2 and d(x,y)≥r0,(2.182)|K(x,y)−K(x′,y)|≤CT1V(x,y)(d(x,x′)d(x,y))ϵ(r0d(x,y))σ.If T∈WBP(β) for a certain β∈(0,ϵ) and T(1)=CT(1) in (Cbβ(𝒳))′,
then T extends to a bounded linear operator from 𝒢(x1,r0,β,γ) to itself for any x1∈𝒳 and any γ∈(0,σ]∩(0,ϵ).
Moreover, there is a constant Cβ,γ,σ,C0>0 such that for all f∈𝒢(x1,r0,β,γ) with any x1∈𝒳 and any γ∈(0,σ]∩(0,ϵ),(2.183)∥Tf∥𝒢(x1,r0,β,γ)≤Cβ,γ,σ,C0(CT+|CT(1)|)∥f∥𝒢(x1,r0,β,γ).Proof.
We prove Theorem2.29 by essentially following the same argument as the
proof of Theorem 2.18, and we only give an outline. We use the notations as in
the proof of Theorem 2.18. By Corollary 2.11(ii), we only need to prove Theorem
2.29 for 𝒢b(x1,r0,β,γ).
Letf∈𝒢b(x1,r0,β,γ).
We first verify that Tf(x) for a.e. x∈𝒳 satisfies Definition 2.8(i) by considering
two cases.
Case 1  (d(x,x1)<10r0).
In this case, letθx1,20r0 and ωx1,20r0 be defined in the same way as in Lemma 2.15.
Instead of Lemma 2.15(ii) by Lemma 2.26(iii), following the procedure of Lemma 2.20 (here we need T∈WBP(β)), for a.e. x0∈B(x110r0),
we have(2.184)Tf(x)=∫𝒳K(x,y)[f(y)−f(x)]θx1,20r0(y)dμ(y)+∫𝒳K(x,y)f(y)ωx1,20r0(y)dμ(y)+f(x)T(θx1,20r0)(x).Using Lemma 2.26(iii) to replace
Lemma 2.15(ii) and following the same proof as that of Theorem 2.18 give the
desired estimate of this case.Case 2  (d(x,x1)≡R≥10r0).
In this case, we use all the notation as in the proof of Theorem2.18, but with r replaced by r0. The estimates (2.118), (2.119), (2.128), and (2.129) with r replaced by r0 hold in the current setting. Instead of
(2.130), by Lemma 2.1(ii), we now have(2.185)|∫𝒳f2(y)dμ(y)|≲1.
Replacing Lemma2.15(ii) by Lemma 2.26(iii) and using T∈WBP(β),
by following the proof of Lemma 2.20, for a.e. x∈Bm,
we have(2.186)Tf1(x)=∫𝒳K(x,y)[f1(y)−f1(x)]Um,r0(y)dμ(y)+f1(x)T(Um,r0)(x).Then Lemma 2.26(iii) together
with an argument similar to the proof of Theorem 2.18 gives the desired estimate
for Tf1(x) in this case.
Sincex∉suppf2,
we can write(2.187)Tf2(x)=∫𝒳[K(x,y)−K(x,x1)]f2(y)dμ(y)+K(x,x1)∫𝒳f2(y)dμ(y)≡Z8(x)+Z9(x).The estimate for Z8(x) is as in the proof of Theorem 2.18. Since R≥10r0 and γ≤σ,
by (2.174) and (2.185), we have(2.188)|Z9(x)|≲1VR(x1)(r0R)σ≲1VR(x1)(r0R)γ,which is the desired estimate.
The estimate forTf3(x) in this case is also similar to the proof of
Theorem 2.18. Thus, T(f)(x) for a.e. x∈𝒳 satisfies Definition 2.2(i).
To verify thatT(f)(x) for a.e. x∈𝒳 satisfies Definition 2.2(ii), similarly to
the proof of Theorem 2.18, we can always assume that d(x,x′)≤(1/160)(r0+d(x,x1)),
and we only need to consider the following two cases.
Case 1  (R<10r0).
In this case, if we replace Lemma2.15(i) and (ii), respectively, by Lemma 2.26(ii) and (iii), the same proof as in the proof of Theorem 2.18 gives the desired
estimates.Case 2  (R≥10r0).
In this case, the same argument as in the proof of Theorem2.18 together with
Lemma 2.15 replaced by Lemma 2.26 yields the desired estimates for Tf1(x) and Tf3(x).
Noticing that f2(x)=0=f2(x′),
we now estimate Tf2 by writing(2.189)|Tf2(x)−Tf2(x′)|=|∫𝒳[K(x,y)−K(x′,y)]f2(y)dμ(y)|≤|∫𝒳{[K(x,y)−K(x′,y)]−[K(x,x1)−K(x′,x1)]}f2(y)dμ(y)|+|K(x,x1)−K(x′,x1)||∫𝒳f2(y)dμ(y)|≡Γ5+Γ6.The estimate for Γ5 is as in the proof of Theorem 2.18. To estimate Γ6,
by (2.182) and (2.185) together with γ≤σ,
we have(2.190)Γ6≲1V(x,x1)(d(x,x′)d(x,x1))ϵ(r0d(x,x1))σ≲1VR(x1)(d(x,x′)R)β(r0R)γ,which is the desired estimate.
Then an extension via the Besicovitch covering lemma as in the proof of Theorem
2.18 completes the proof of  Theorem 2.29.Remark 2.30.
A regularity condition similar to (2.182) also appears in [71, page 123]. This additional
regularity assumption is used only in the estimate of Γ6.
Instead of (2.182) by requiring γ∈(0,σ(ϵ−β)/ϵ],
we can obtain a similar conclusion. In fact, since d(x′,x1)≥d(x,x1)−d(x,x′)≥189r0/20≥r0 and d(x′,x1)≥189R/200≥R/2,
by (2.174) and (2.185), we have(2.191)Γ6≲1V(x,x1)(r0d(x,x1))σ+1V(x′,x1)(r0d(x′,x1))σ≲1V(x,x1)(r0d(x,x1))σ.On the other hand, by (2.49) and
(2.185), we also have Γ6≲(1/V(x,x1))(d(x,x′)/d(x,x1))ϵ. Combining both estimates yields that(2.192)Γ6≲1V(x,x1)(d(x,x′)d(x,x1))β(r0d(x,x1))σ(1−β/ϵ)≲1VR(x1)(d(x,x′)R)β(r0R)γ,which is the desired estimate.By Remark2.14(iii) again, we can obtain the following
conclusion which is convenient in  applications.Corollary 2.31.
Letϵ∈(0,1], σ>0, r0>0, and let T be as in Proposition 2.25 with the
distributional kernel K also satisfying Theorem 2.18(i) through (iii)
and (2.182). If T is bounded on Lp(𝒳) for a certain p∈(1,∞) and T(1)=CT(1) in (Cbβ(𝒳))′,
then T extends to a bounded linear operator from 𝒢(x1,r0,β,γ) to itself for any x1∈𝒳,
any β∈(0,ϵ), and any γ∈(0,σ]∩(0,ϵ).
Moreover, there is a constant Cβ,γ,σ,C0>0 such that for all f∈𝒢(x1,r0,β,γ) with any x1∈𝒳 and any γ∈(0,σ]∩(0,ϵ),(2.193)∥Tf∥𝒢(x1,r0,β,γ)≤Cβ,γ,σ,C0(CT+|CT(1)|)∥f∥𝒢(x1,r0,β,γ).The proof of the following corollary is similar to
that of Corollary2.24. We omit the details.Corollary 2.32.
LetT be as in Theorem 2.29 or Corollary 2.31. If ϵ˜∈(0,ϵ),
then T is bounded from 𝒢0ϵ˜(x1,r0,β,γ) to 𝒢0ϵ˜(x1,r0,β,γ) for all x1,β,γ as in Theorem 2.29. Moreover, there exists a
constant Cβ,γ,σ,C0>0 such that for all f∈𝒢0ϵ˜(x1,r0,β,γ)  with any x1∈𝒳,(2.194)∥Tf∥𝒢(x1,r0,β,γ)≤Cβ,γ,σ,C0(CT+|CT(1)|)∥f∥𝒢(x1,r0,β,γ).


## 2.1. Approximations of the Identity

Throughout the whole paper, we denote byMf the Hardy-Littlewood maximal function on 𝒳 for any f∈Lloc1(𝒳).
It is well known that M is bounded on Lp(𝒳) with p∈(1,∞];
see [29, 44]. Some basic estimates used
throughout the whole paper are stated in the following lemma, whose main part
is included in [48, Lemma 2.1], and the remaining
statements are obvious.Lemma 2.1.
Letδ>0, a>0, r>0, and θ∈(0,1).
Then the following hold.
(i)
∫d(x,y)≤δ(d(x,y)a/V(x,y))dμ(y)≤Cδa and ∫d(x,y)≥δ(1/V(x,y))(δa/d(x,y)a)dμ(y)≤C uniformly in x∈𝒳 and δ>0.(ii)
∫𝒳(1/(Vδ(x)+V(x,y)))(δa/(δ+d(x,y))a)d(x,y)ηdμ(y)≤Cδη uniformly in x∈𝒳 and δ>0,
if a>η≥0.(iii)
Ifx,x′,x1∈𝒳 satisfy d(x,x′)≤θ(r+d(x,x1)),
then 1/(r+d(x′,x1))≤C(1/(r+d(x,x1))), 1/(Vr(x1)+V(x1,x′))≤C(1/(Vr(x1)+V(x1,x))), and(2.1)1Vr(x′)+V(x′,x1)≤C1Vr(x)+V(x,x1)uniformly in r>0 and x,x′,x1∈𝒳.(iv)
For allf∈Lloc1(𝒳) and all x∈𝒳, ∫d(x,y)>δ(1/V(x,y))(δa/d(x,y)a)|f(y)|dμ(y)≤CM(f)(x) uniformly in δ>0, f∈Lloc1(𝒳) and x∈𝒳.(v)
For anyϵ>0, ∫d(y,x)≥δ(1/(Vr(x)+V(y,x)))(1/(r+d(y,x))ϵ)dμ(y)≤C(1/(r+δ)ϵ) uniformly in x∈𝒳 and δ,r>0.(vi)
For any fixedα>0,
if d(x,y)≤αr,
then Vr(x)~Vαr(x)~Vαr(y)~Vr(y) uniformly in x,y∈𝒳 and r>0.(vii)
For allr>0 and all x,y∈𝒳, Vr(x)+V(x,y)~Vr(x)+Vr(y)+V(x,y)~Vr(y)+V(x,y).Motivated by the properties of the heat kernel defined
in (1.17) in the case of Carnot-Carathéodory spaces, we introduce the following
notion of an approximation of the identity on 𝒳.Definition 2.2.
Letϵ1∈(0,1], ϵ2>0, and ϵ3>0.
A sequence {Sk}k∈ℤ of bounded linear integral operators on L2(𝒳) is said to be an approximation of the identity of order(ϵ1,ϵ2,ϵ3) (for short, (ϵ1,ϵ2,ϵ3)-ATI) if there exists a constant C3>0 such that for all k∈ℤ and all x,x′,y, and y′∈𝒳,Sk(x,y), the integral kernel of Sk is a measurable function from 𝒳×𝒳 into ℂ satisfying
(i)
|Sk(x,y)|≤C3(1/(V2−k(x)+V2−k(y)+V(x,y)))(2−kϵ2/(2−k+d(x,y))ϵ2);(ii)
|Sk(x,y)−Sk(x′,y)|≤C3(d(x,x′)/(2−k+d(x,y)))ϵ1(1/(V2−k(x)+V2−k(y)+V(x,y)))×(2−kϵ2/(2−k+d(x,y))ϵ2) for d(x,x′)≤(2−k+d(x,y))/2;(iii)
property (ii) also holds withx and y interchanged;(iv)
|[Sk(x,y)−Sk(x,y′)]−[Sk(x′,y)−Sk(x′,y′)]|≤C3(d(x,x′)/(2−k+d(x,y)))ϵ1×(d(y,y′)/(2−k+d(x,y)))ϵ1(1/(V2−k(x)+V2−k(y)+V(x,y)))(2−kϵ3/(2−k+d(x,y))ϵ3) for d(x,x′)≤(2−k+d(x,y))/3 and d(y,y′)≤(2−k+d(x,y))/3;(v)
∫𝒳Sk(x,y)dμ(y)=1;(vi)
∫𝒳Sk(x,y)dμ(x)=1.In case the ATI has bounded support, in the sense thatSk(x,y)=0 when d(x,y)≳2−k,
then the conditions (i), (ii), (iii), and (iv) of Definition 2.2 simplify as
follows (see Proposition 2.5).Definition 2.3.
Letϵ1∈(0,1].
A sequence {Sk}k∈ℤ of bounded linear integral operators on L2(𝒳) is said to be an approximation of the identity of orderϵ1with bounded support (for short, ϵ1-ATI with bounded support) if there exist
constants C4,C5>0 such that for all k∈ℤ and all x,x′,y, and y′∈𝒳,Sk(x,y), the integral kernel of Sk is a measurable function from 𝒳×𝒳 into ℂ satisfying (v) and (vi) of Definition 2.2 as
above, and
(i)
Sk(x,y)=0 if d(x,y)≥C52−k and |Sk(x,y)|≤C4(1/(V2−k(x)+V2−k(y)));(ii)
|Sk(x,y)−Sk(x′,y)|≤C42kϵ1d(x,x′)ϵ1(1/(V2−k(x)+V2−k(y))) for d(x,x′)≤max{C5,1}21−k;(iii)
property (ii) also holds withx and y interchanged;(iv)
|[Sk(x,y)−Sk(x,y′)]−[Sk(x′,y)−Sk(x′,y′)]|≤C422kϵ1d(x,x′)ϵ1d(y,y′)ϵ1×(1/(V2−k(x)+V2−k(y))) for d(x,x′)≤max{C5,1}21−k and d(y,y′)≤max{C5,1}21−k.We callϵ1 the regularity of (ϵ1,ϵ2,ϵ3)-ATI  {Sk}k∈ℤ.Remark 2.4.
(i) Assume that𝒳 is as in Case (a) with n≥3 or as in Case (b) in the introduction, and let H(s,x,y) be the heat kernel defined in (1.17). Define Sk(x,y)=H(2−2k,x,y) for k∈ℤ and x,y∈𝒳.
Using [44, Proposition 2.3.1], we can verify that {Sk}k∈ℤ is an (ϵ1,ϵ2,ϵ3)-ATI for any ϵ1∈(0,1], ϵ2>0, and ϵ3>0.
Moreover, in this case, Sk(x,y)=Sk(y,x).
(ii) If𝒳 is any (compact or noncompact) Ahlfors n-regular metric measure space or any Lie group
with polynomial growth, then one can construct an ϵ1-ATI with bounded support for any ϵ1∈(0,1] by following Coifman's idea in [64].
(iii) We remark that when we consider the existence of ATIs, the condition (iv) of Definition2.2 is not
essential, in the sense that if there exist {Sk}k∈ℤ with k∈ℤ satisfying (i), (ii), (iii), (v), and (vi) of
Definition 2.2, then {Sk∘Sk}k∈ℤ satisfy (i) through (vi) of Definition 2.2.Proposition 2.5.
Suppose{Sk}k∈ℤ is a sequence of bounded linear integral
operators on L2(𝒳) such that Sk(x,y)=0 whenever d(x,y)≥C52−k.
Then {Sk}k∈ℤ is an ATI if and only if it is an ATI with bounded support.Proof.
Obviously, from the assumption thatSk(x,y)=0 whenever d(x,y)≥C52−k,
it easily follows that Sk satisfies Definition 2.2(i) if and only if(2.2)|Sk(x,y)|≲1V2−k(x)+V2−k(y),which appears in Definition 2.3(i).
We now establish the equivalence between Definitions2.2(ii) and 2.3(ii), with the proof for the equivalence between
(iii) of Definitions 2.2 and (iii) of Definition 2.3 being similar. Notice that Sk(x,y)−Sk(x′,y)≠0 implies that d(x,y)<C52−k or d(x′,y)<C52−k.
Thus, Sk(x,y)−Sk(x′,y)≠0 together with d(x,x′)≤2max{C5,1}2−k shows that d(x,y)<3max{C5,1}2−k,
and hence, 2−k+d(x,y)~2−k and V2−k(x)+V2−k(y)+V(x,y)~V2−k(x)+V2−k(y).
From these estimates, it immediately follows that Definition 2.2(ii) implies Definition 2.3(ii), and conversely Definition 2.3(ii) implies that Definition 2.2(ii) holds whenever d(x,x′)≤2max{C5,1}2−k.
We still need to prove that Definition 2.3(ii) also implies that Definition 2.2(ii) holds when 2max{C5,1}2−k<d(x,x′)≤(2−k+d(x,y))/2.
However, if 2max{C5,1}2−k<d(x,x′)≤(2−k+d(x,y))/2,
we have d(x,y)>3max{C5,1}2−k and d(x′,y)≥d(x,x′)−2−k>C52−k.
Thus, in this case, Sk(x,y)=0=Sk(x′,y),
and therefore, Definition 2.2(ii) automatically holds. This proves the
equivalence between Definitions 2.2(ii) and 2.3(ii).
We now establish the equivalence between Definitions2.2(iv) and 2.3(iv). Notice that [Sk(x,y)−Sk(x,y′)]−[Sk(x′,y)−Sk(x′,y′)]≠0 implies that d(x,y)<C52−k or d(x′,y)<C52−k or d(x,y′)<C52−k or d(x′,y′)<C52−k.
This together with d(x,x′)≤max{C5,1}21−k and d(y,y′)≤max{C5,1}21−k further shows that we have d(x,y)<5max{C5,1}21−k,
and hence, 2−k+d(x,y)~2−k and V2−k(x)+V2−k(y)+V(x,y)~V2−k(x)+V2−k(y).
From these estimates, we see that Definition 2.2(iv) implies Definition 2.3(iv), and conversely, Definition 2.3(iv) also implies that Definition 2.2(iv) holds when d(x,x′)≤2max{C5,1}2−k and d(y,y′)≤max{C5,1}21−k.
We still need to prove that Definition 2.3(iv) also implies that Definition 2.2(iv) holds in the three cases listed below. Recall that we always assume
that d(x,x′)≤(2−k+d(x,y))/3 and d(y,y′)≤(2−k+d(x,y))/3 in  Definition 2.2(iv).
Case 1.
d(x,x′)≤2max{C5,1}2−k and d(y,y′)>2max{C5,1}2−k.
In this case, d(x,y)≥3d(y,y′)−2−k≥C52−k, d(x,y′)≥2d(y,y′)−2−k>C52−k, d(x′,y)>2d(y,y′)−2−k>C52−k,
and d(x′,y′)>d(y,y′)−2−k>C52−k.
Thus Sk(x,y)=Sk(x,y′)=Sk(x′,y)=Sk(x′,y′)=0 and therefore Definition 2.2(iv) automatically
holds in this case.Case 2.
d(x,x′)>2max{C5,1}2−k and d(y,y′)≤2max{C5,1}2−k.
This case is similar to Case 1 by symmetry.Case 3.
d(x,x′)>2max{C5,1}2−k and d(y,y′)>2max{C5,1}2−k.
In this case, similar to Case 1, we have that d(x,y)≥3d(y,y′)−2−k>C52−k, d(x,y′)≥2d(y,y′)−2−k>C52−k, d(x′,y)≥2d(x,x′)−2−k>C52−k,
and d(x′,y′)≥d(x,y)−d(x,x′)−d(y,y′)≥(d(x,y)−21−k)/3>d(y,y′)−2−k>C52−k.
We also have Sk(x,y)=Sk(x,y′)=Sk(x′,y)=Sk(x′,y′)=0 and therefore (iv) automatically holds in this
case. This establishes the equivalence between (iv) and (iv)′, and hence
completes the proof of  Proposition 2.5.Following Coifman's idea in [64], for any ϵ1∈(0,1],
we can construct ϵ1-ATIs with bounded support on 𝒳 (cf. also [65, Theorem (1.13)]).Theorem 2.6.
Let(𝒳,d,μ) be a space of homogeneous type as in
Definition 1.1 and ϵ1∈(0,1].
Then there exists an approximation of the identity {Sk}k∈ℤ of order ϵ1 with bounded support on 𝒳,
with constant C5=4.
Moreover, for all k∈ℤ and x,y∈𝒳, Sk(x,y)=Sk(y,x).Proof.
Obviously, we only need to prove the theorem forϵ1=1.
Let h∈C1(ℝ), h(t)=1 if t∈[0,1], h(t)=0 if t≥2,
and 0≤h(t)≤1 for all t∈ℝ.
For any k∈ℤ, f∈Lloc1(𝒳) and u∈𝒳,
we then define(2.3)Tkf(u)=∫𝒳h(2kd(u,w))f(w)dμ(w).Obviously, we have V2−k(u)≤Tk1(u)≤V21−k(u). Fix any x∈𝒳.
By Lemma 2.1(vi), for all u∈B(x,25−k),
we further have(2.4)Tk1(u)~V2−k(x).Thus, if z∈B(x,24−k) and h(2kd(z,u))≠0,
then u∈B(x,25−k),
and by (2.4), we further have(2.5)Tk(1Tk1)(z)=∫𝒳h(2kd(z,u))1Tk1(u)dμ(u)~1.For all x,y∈𝒳,
we define(2.6)Sk(x,y)=1Tk1(x){∫𝒳h(2kd(x,z))1Tk(1/Tk1)(z)h(2kd(z,y))dμ(z)}1Tk1(y).We now prove that {Sk}k∈ℤ is a 1-ATI with bounded support and C5=4.
It is obvious that for all x,y∈𝒳, Sk(x,y)=Sk(y,x),
and that if d(x,y)≥22−k,
then Sk(x,y)=0.
Also, it is easy to show that ∫𝒳Sk(x,y)dμ(y)=1. Moreover, if d(x,y)<22−k,
by (2.4) and (2.5) together with Lemma 2.1(vi), we obtain(2.7)0≤Sk(x,y)≲1V2−k(x)~1V2−k(x)+V2−k(y).Thus Sk(x,y) satisfies (i)′ of Definition 2.2 with C5=4.
We now show thatSk(x,y) has the desired regularity in the first
variable when d(x,x′)≤23−k.
Notice that in this case, Sk(x,y)−Sk(x′,y)≠0 implies that d(x,y)<24−k,
and hence,(2.8)1V2−k(x)+V2−k(y)~1V2−k(x).By (2.6), we
have(2.9)Sk(x,y)−Sk(x′,y)=[1Tk1(x)−1Tk1(x′)]{∫𝒳h(2kd(x,z))1Tk(1/Tk1)(z)h(2kd(z,y))dμ(z)}1Tk1(y)+1Tk1(x′){∫𝒳[h(2kd(x,z))−h(2kd(x′,z))]1Tk(1/Tk1)(z)h(2kd(z,y))dμ(z)}1Tk1(y)≡Z1+Z2.To estimate Z1,
by the choice of h,
the mean value theorem, and (2.4) together with Lemma 2.1(vi), we first have(2.10)|1Tk1(x)−1Tk1(x′)|≲1[V2−k(x)]22kd(x,x′)μ(B(x,24−k))~2kd(x,x′)V2−k(x),which together with (2.4),
(2.5), and (2.8) further yields(2.11)|Z1|≲2kd(x,x′)V2−k(x)~2kd(x,x′)V2−k(x)+V2−k(y).
To estimateZ2,
notice that h(2kd(x,z))−h(2kd(x′,z))≠0 implies d(x,z)<24−k.
By this observation, (2.5), the mean value theorem together with Lemma 2.1(vi), we obtain(2.12)|∫𝒳[h(2kd(x,z))−h(2kd(x′,z))]1Tk(1/Tk1)(z)h(2kd(z,y))dμ(z)|≲2kd(x,x′)μ(B(x,24−k))≲2kd(x,x′)V2−k(x),which together with (2.4) and (2.8) yields(2.13)|Z2|≲2kd(x,x′)V2−k(x)~2kd(x,x′)V2−k(x)+V2−k(y).In combination with the estimate
for Z1,
this shows that Sk(x,y) has the desired regularity with respect to the
first variable when d(x,x′)≤23−k.
We finally prove thatSk(x,y) satisfies the desired second difference
condition, when d(x,x′)≤23−k and d(y,y′)≤23−k. Notice that in this case, [Sk(x,y)−Sk(x′,y)]−[Sk(x,y′)−Sk(x′,y′)]≠0 implies that d(x,y)<25−k.
Thus, (2.8) also holds in this case. By (2.6), we have(2.14)[Sk(x,y)−Sk(x′,y)]−[Sk(x,y′)−Sk(x′,y′)]=[1Tk1(x)−1Tk1(x′)]{∫𝒳h(2kd(x,z))1Tk(1/Tk1)(z)h(2kd(z,y))dμ(z)}×[1Tk1(y)−1Tk1(y′)]+[1Tk1(x)−1Tk1(x′)]1Tk1(y′)×{∫𝒳h(2kd(x,z))1Tk(1/Tk1)(z)[h(2kd(z,y))−h(2kd(z,y′))]dμ(z)}+1Tk1(x′){∫𝒳[h(2kd(x,z))−h(2kd(x′,z))]×1Tk(1/Tk1)(z)h(2kd(z,y))dμ(z)}[1Tk1(y)−1Tk1(y′)]+1Tk1(x′)1Tk1(y′){∫𝒳[h(2kd(x,z))−h(2kd(x′,z))]×1Tk(1/Tk1)(z)[h(2kd(z,y))−h(2kd(z,y′))]dμ(z)}≡Z3+Z4+Z5+Z6.The estimates (2.10), (2.5), and (2.8) together with Lemma 2.1(vi) show that(2.15)|Z3|≲2kd(x,x′)V2−k(x)μ(B(x,23−k))2kd(y,y′)V2−k(x)~22kd(x,x′)d(y,y′)1V2−k(x)+V2−k(y).The estimate for Z4 is similar to the one for Z5,
while the estimate for Z5 can be deduced immediately from the estimates
(2.12), (2.10), and (2.4) together with (2.8). Finally, to estimate Z6,
notice that h(2kd(x,z))−h(2kd(x′,z))≠0 and d(x,x′)≤23−k implies that z∈B(x,24−k).
This observation together with the mean value theorem, (2.5), and Lemma 2.1(vi) yields(2.16)|∫𝒳[h(2kd(x,z))−h(2kd(x′,z))]1Tk(1/Tk1)(z)[h(2kd(z,y))−h(2kd(z,y′))]dμ(z)|≲22kd(x,x′)d(y,y′)V2−k(x),which together with (2.4) and (2.8) further gives(2.17)|Z6|≲22kd(x,x′)d(y,y′)1V2−k(x)~22kd(x,x′)d(y,y′)1V2−k(x)+V2−k(y).This proves that Sk(x,y) has the desired second difference property,
and hence, completes the proof of Theorem 2.6.In the sequel, for anyf∈Lp(𝒳) with p∈[1,∞] and x∈𝒳,
we set(2.18)Skf(x)=∫𝒳Sk(x,y)f(y)dμ(y).We also let(2.19)Lb∞(𝒳)={f∈L∞(𝒳):fhasboundedsupport}.Some basic properties of ATIs are presented in the next proposition.Proposition 2.7.
Let{Sk}k∈ℤ be an (ϵ1,ϵ2,ϵ3)-ATI with ϵ1∈(0,1],ϵ2>0 and ϵ3>0,
and let Skt denote the adjoint operator to Sk,
whose integral kernel is given by Skt(x,y)=Sk(y,x). Then the following hold.
(i)
For anyk∈ℤ and any x,y∈𝒳, ∫𝒳|Sk(x,y)|dμ(y)≤C and ∫𝒳|Sk(x,y)|dμ(x)≤C.(ii)
For anyk∈ℤ and any f∈Lloc1(𝒳), |Skf(x)|≤CMf(x).(iii)
For1≤p≤∞,
there exists a constant Cp>0 such that for all f∈Lp(𝒳),(2.20)∥Skf∥Lp(𝒳)≤Cp∥f∥Lp(𝒳).(iv)
For1≤p<∞ and any f∈Lp(𝒳), ∥Skf−f∥Lp(𝒳)→0 when k→∞.(v)
Properties (ii) through (iv) also hold forSk replaced by Skt.Proof.
Definition2.2(i) together with Lemma 2.1(i) shows that for any k∈ℤ and any x∈𝒳,(2.21)∫𝒳|Sk(x,y)|dμ(y)=∫d(x,y)≤2−k|Sk(x,y)|dμ(y)+∫d(x,y)>2−k|Sk(x,y)|dμ(y)≲1+∫d(x,y)>2−k1V(x,y)2−kϵ2d(x,y)ϵ2dμ(y)≲1.A similar argument proves that ∫𝒳|Sk(x,y)|dμ(x)≲1.
To prove (ii), by Definition2.2(i) together with
Lemma 2.1(iv), we have(2.22)|Skf(x)|≲1V2−k(x)∫d(x,y)≤2−k|f(y)|dμ(y)+∫d(x,y)>2−k1V(x,y)2−kϵ2d(x,y)ϵ2|f(y)|dμ(y)≲Mf(x).
From (ii) and theLp(𝒳)-boundedness of M,
we obtain (iii) for p∈(1,∞].
For p=1,
we obtain (iii) from (i), by Fubini's theorem.
To prove (iv), we first recall that Lebesgue's
differentiation theorem holds in𝒳,
that is, if f is locally integrable, then for almost all x∈𝒳, x is a Lebesgue point which means(2.23)limr→01μ(B(x,r))∫B(x,r)|f(y)−f(x)|dμ(y)=0,since the measure μ is regular; see [13, page 4] and [66, page 11].
For any givenf∈Lp(𝒳) with p∈[1,∞],
assume that x0 is a Lebesgue point of f,
then Mf(x0)+|f(x0)|<∞.
By the conditions (v) and (i) of Sk in Definition 2.2, we have(2.24)|Skf(x0)−f(x0)|=|∫𝒳Sk(x0,y)[f(y)−f(x0)]dμ(y)|≤C3∫d(x0,y)<2−k1V2−k(x0)+V2−k(y)+V(x0,y)2−kϵ2(2−k+d(x0,y))ϵ2|f(y)−f(x0)|dμ(y)+C3∑l=0∞∫2l−12−k≤d(x0,y)<2l2−k⋯≤CC3∑l=0∞12lϵ21V2l2−k(x0)∫d(x0,y)<2l2−k|f(y)−f(x0)|dμ(y).For any δ>0,
we choose L0∈ℕ so that(2.25)CC3[Mf(x0)+|f(x0)|]∑l=L0+1∞12lϵ2<δ2.Since x0 is a Lebesgue point, by the definition, we
know that there exists K∈ℕ such that when k>K,(2.26)CC3∑l=0L012lϵ21V2l2−k(x0)∫d(x0,y)<2l2−k|f(y)−f(x0)|dμ(y)<δ2.Thus limk→∞Skf(x0)=f(x0),
and therefore for almost everywhere x∈𝒳,(2.27)limk→∞Skf(x)=f(x).This fact together with |Skf(x)|≲Mf(x) (see (ii) of this proposition), the Lp(𝒳)-boundedness of M with p∈(1,∞) again and Lebesgue's dominated convergence
theorem, gives that limk→∞∥Skf−f∥Lp(𝒳)=0 for all f∈Lp(𝒳).
Whenp=1,
we first consider f∈Lb∞(𝒳).
Assume that suppf⊂B(y0,r0) for some y0∈𝒳 and r0>0.
For some fixed L>2r0,
by Hölder's inequality and the conditions (i) and (v) of Sk in Definition 2.2 together with Lemma 2.1(i),
we have(2.28)∥Skf−f∥L1(𝒳)=∫d(x,y0)<L|Skf(x)−f(x)|dμ(x)+∫d(x,y0)≥L|Skf(x)−f(x)|dμ(x)≤μ(BL(y0))1/2∥Skf−f∥L2(𝒳)+C3∫d(y,y0)<r0|f(y)|{∫d(x,y)≥L/21V(x,y)2−kϵ2d(x,y)ϵ2dμ(x)}dμ(y)≤μ(BL(y0))1/2∥Skf−f∥L2(𝒳)+CC32−kϵ2L−ϵ2∥f∥L1(𝒳),which together with Lb∞(𝒳)⊂L2(𝒳) and the above proved conclusion for p∈(1,∞) implies that limk→∞∥Skf−f∥L1(𝒳)=0 for all f∈Lb∞(𝒳).
This and the density of Lb∞(𝒳) in L1(𝒳) further yield that limk→∞∥Skf−f∥L1(𝒳)=0 for all f∈L1(𝒳),
which verify (iv).
Property (v) can be deduced from (i) through (iv)
together with the symmetry, which completes the proof of Proposition2.7.We now introduce the space of test functions on𝒳.Definition 2.8.
Letx1∈𝒳, r>0, 0<β≤1, and γ>0.
A function f on 𝒳 is said to be a test function of type(x1,r,β,γ) if there exists a constant C≥0 such that
(i)
|f(x)|≤C(1/(Vr(x1)+V(x1,x)))(r/(r+d(x1,x)))γ for all x∈𝒳;(ii)
|f(x)−f(y)|≤C(d(x,y)/(r+d(x1,x)))β(1/(Vr(x1)+V(x1,x)))(r/(r+d(x1,x)))γ for all x,y∈𝒳 satisfying that d(x,y)≤(r+d(x1,x))/2.
Moreover, we
denote by 𝒢(x1,r,β,γ) the set of all test functions of type (x1,r,β,γ),
and if f∈𝒢(x1,r,β,γ),
we define its norm by(2.29)∥f∥𝒢(x1,r,β,γ)=inf{C:(i)and(ii)hold}.The space 𝒢(x1,r,β,γ) is called the space of test functions.This definition of a test function of type(x1,r,β,γ) gives a quantified meaning to the notation of
a sufficiently “smooth” function which is essentially supported in the
ball of radius r centered at x1 in the sense that it decays of sufficiently
high order (measured by γ) at infinity, and is Hölder continuous of
order β (at the right scale r).Obviously,Sk(x,⋅) for any fixed k∈ℤ and x∈𝒳 in Definition 2.2 is a test function of type (x,2−k,ϵ1,ϵ2),
and Sk(⋅,y) for any fixed k∈ℤ and y∈𝒳 in Definition 2.2 is a test function of type (y,2−k,ϵ1,ϵ2).
Moreover, for η∈(0,1],
let(2.30)∥f∥C˙η(𝒳)=supx,y∈𝒳x≠y|f(x)−f(y)|d(x,y)ηand define the homogeneous Hölder space(2.31)C˙η(𝒳)={f∈C(𝒳):∥f∥C˙η(𝒳)<∞};we also consider the inhomogeneous Hölder space(2.32)Cη(𝒳)={f∈C(𝒳):∥f∥Cη(𝒳)<∞},Cbη(𝒳)={f∈Cη(𝒳):fhasboundedsupport},where ∥f∥Cη(𝒳)=∥f∥L∞(𝒳)+∥f∥C˙η(𝒳).
Then any f∈Cbη(𝒳) is also a test  function of type (x1,r,η,γ) for any x1∈𝒳, r>0, and γ>0.Now fixx1∈𝒳 and let 𝒢(β,γ)=𝒢(x1,1,β,γ). It is easy to see that(2.33)𝒢(x0,r,β,γ)=𝒢(β,γ)with the equivalent norms for
all x0∈𝒳 and r>0. Furthermore, it is easy to check that 𝒢(β,γ) is a Banach space with respect to the norm in 𝒢(β,γ).It is well known that even when𝒳=ℝn, 𝒢(β1,γ) is not dense in 𝒢(β2,γ) if β1>β2,
which will bring us some inconvenience. To overcome this defect, in what
follows, for given ϵ∈(0,1],
let 𝒢0ϵ(β,γ) be the completion of the space 𝒢(ϵ,ϵ) in 𝒢(β,γ) when 0<β,γ≤ϵ.
Obviously, 𝒢0ϵ(ϵ,ϵ)=𝒢(ϵ,ϵ).
Moreover, f∈𝒢0ϵ(β,γ) if and only if f∈𝒢(β,γ) and there exist {fn}n∈ℕ⊂𝒢(ϵ,ϵ) such that ∥f−fn∥𝒢(β,γ)→0 as n→∞.
If f∈𝒢0ϵ(β,γ),
we then define ∥f∥𝒢0ϵ(β,γ)=∥f∥𝒢(β,γ).
Then, obviously, 𝒢0ϵ(β,γ) is a Banach space and we also have ∥f∥𝒢0ϵ(β,γ)=limn→∞∥fn∥𝒢(β,γ) for the above chosen {fn}n∈ℕ.We define the dual space(𝒢0ϵ(β,γ))′ to be the set of all linear functionals ℒ from 𝒢0ϵ(β,γ) to ℂ with the property that there exists C≥0 such that for all f∈𝒢0ϵ(β,γ),(2.34)|ℒ(f)|≤C∥f∥𝒢0ϵ(β,γ).We denote by 〈h,f〉 the natural pairing of elements h∈(𝒢0ϵ(β,γ))′ and f∈𝒢0ϵ(β,γ). Clearly, for all h∈(𝒢0ϵ(β,γ))′,〈h,f〉 is well defined for all f∈𝒢0ϵ(x1,r,β,γ) with  x1∈𝒳 and r>0.In the sequel, we define(2.35)𝒢°(x1,r,β,γ)={f∈𝒢(x1,r,β,γ):∫𝒳f(x)dμ(x)=0},which is called the space of test functions with mean zero.
The space 𝒢°0ϵ(β,γ) is defined to be the completion of the space 𝒢°(ϵ,ϵ) in 𝒢°(β,γ) when 0<β,γ<ϵ.
Moreover, if f∈𝒢°0ϵ(β,γ),
we then define ∥f∥𝒢°0ϵ(β,γ)=∥f∥𝒢(β,γ).By essentially following a procedure similar to that
of [44, Lemma 3.5.1],
we can establish some kind of decomposition for test functions with mean
zero.Proposition 2.9.
Letx1∈𝒳, r>0, 0<β≤1, and γ>0,
and let f∈𝒢°(x1,r,β,γ).
Then for any γ˜∈(0,γ] and all x∈𝒳,(2.36)f(x)=∑l=0∞2−lγ˜φl(x),where φl is an
adjusted bump function associated with the ball B(x1,2lr),
which means that there exists a constant C>0 independent of r and l such that
(i)
suppφl⊂B(x1,2lr);(ii)
|φl(x)|≤C(1/V2lr(x1)) for all x∈𝒳;(iii)
∥φl∥C˙η(𝒳)≤C(2lr)−η(1/V2lr(x1)) for all 0<η≤β;(iv)
∫𝒳φl(x)dμ(x)=0.
Moreover, the
series in (2.36) converges to f pointwise, as well as in Lp(𝒳) for p∈[1,∞],
and in 𝒢°0β⋀γ˜(β′,γ′) with 0<β′,γ′<(β⋀γ˜) and also in (𝒢°0β⋀γ˜(β′,γ′))′ with 0<β′,γ′≤(β⋀γ).Proof.
Fix a nonnegative functionα∈C1(ℝ) such that α(t)=1 if t≤1/2 and α(t)=0 if t≥1.
Let A0(x)=α(d(x,x1)/r)f(x), and for l∈ℕ,(2.37)Al(x)=[α(d(x,x1)2lr)−α(d(x,x1)2l−1r)]f(x).Then f(x)=∑l=0∞Al(x), and for all γ˜∈(0,γ],(2.38)|Al(x)|≲2−lγ˜1V2lr(x1).Define al=∫𝒳Al(x)dμ(x) and vl=∑j=0laj. Let η˜l(x)=α(d(x,x1)/2lr) and ηl(x)=η˜l(x)[∫𝒳η˜l(z)dμ(z)]−1. Finally, if we define(2.39)A˜l(x)=Al(x)−alηl(x)+vl(ηl(x)−ηl+1(x))and φl(x)=2lγ˜A˜l(x), then it is easy to verify that φl satisfies (i) through (iv) of the proposition
and (2.36) holds pointwise.
Obviously,φl∈𝒢°(β,γ) and therefore φl∈𝒢°0β⋀γ˜(β′,γ′) for 0<β′,γ′≤(β⋀γ˜).
Moreover, if l∈ℕ is large enough, then there exists a constant Cr,x1>0,
which is independent of x,
such that for all x∈B(x1,2lr), 1+d(x,x1)≤Cr,x12l and hence(2.40)∥φl∥𝒢°0β⋀γ˜(β′,γ′)≤Cr,x12lγ′.From this, it follows that if γ′<γ˜,(2.41)∥f−∑l=0L2−lγ˜φl∥𝒢°0β⋀γ˜(β′,γ′)≲Cr,x1∑l=L+1∞2−l(γ˜−γ′)~Cr,x12−L(γ˜−γ′)→0,as L→∞.
This shows that (2.36) holds in 𝒢°0β⋀γ˜(β′,γ′) with 0<β′,γ′<(β⋀γ˜).
By Fatou's lemma and Minkowski's inequality, we also obtain that for any p∈[1,∞],(2.42)∥f−∑l=0L2−lγ˜φl∥Lp(𝒳)≤∑l=L+1∞2−lγ˜∥φl∥Lp(𝒳)≲∑l=L+1∞2−lγ˜1V2lr(x1)1−1/p≲1Vr(x1)1−1/p2−Lγ˜→0,as L→∞.
That is, (2.36) holds in Lp(𝒳) with p∈[1,∞].
Finally, for anyψ∈𝒢°0β⋀γ(β′,γ′) with 0<β′,γ′≤(β⋀γ),
from Hölder's inequality, Fatou lemma and Minkowski's inequality, it follows
that(2.43)|〈f,ψ〉−∑l=0L2−lγ˜〈φl,ψ〉|≤∥∑l=L+1∞2−lγ˜φl∥L1(𝒳)∥ψ∥L∞(𝒳)≲(∑l=L+1∞2−lγ˜)∥ψ∥L∞(𝒳)≲2−Lγ˜∥ψ∥L∞(𝒳)→0,as L→∞.
Thus, (2.36) also holds in (𝒢°0β⋀γ(β′,γ′))′ with 0<β′,γ′≤(β⋀γ),
which completes the proof of  Proposition 2.9.The following proposition is a slight variant of Proposition2.9.Proposition 2.10.
Letx1∈𝒳, r>0, 0<β≤1, and γ>0,
and let f∈(x1,r,β,γ).
Then for any γ˜∈(0,γ] and all x∈𝒳,(2.44)f(x)=∑l=0∞2−lγ˜φl(x),where φl is an adjusted bump function associated with
the ball B(x1,2lr),
which means that there exists a constant C>0 independent of r and l such that
(i)
suppφl⊂B(x1,2lr);(ii)
|φl(x)|≤C(1/V2lr(x1)) for all x∈𝒳;(iii)
∥φl∥C˙η(𝒳)≤C(2lr)−η(1/V2lr(x1)) for all 0<η≤β.
Moreover, the
series in (2.44) converges to f pointwise, as well as in Lp(𝒳) for p∈[1,∞],
and in 𝒢0β⋀γ˜(β′,γ′) with 0<β′,γ′<(β⋀γ˜) and also in (𝒢0β⋀γ(β′,γ′))′ with 0<β′,γ′≤(β⋀γ).Proof.
Letα∈C1(ℝ) be as in the proof of Proposition 2.9. Let A0(x)=α(d(x,x1)/r)f(x), and for l∈ℕ,(2.45)Al(x)=[α(d(x,x1)2lr)−α(d(x,x1)2l−1r)]f(x).
Then for l∈ℤ+, setting φl(x)=2lγ˜Al(x), we can verify that the {φl}l∈ℤ+ have all the properties stated in the
proposition, which completes the proof of Proposition 2.10.In what follows, for anyβ∈(0,1], γ>0, and r>0,
we let(2.46)𝒢°b(x1,r,β,γ)={f∈𝒢°(x1,r,β,γ):fhasboundedsupport},𝒢b(x1,r,β,γ)={f∈𝒢(x1,r,β,γ):fhasboundedsupport}.Also, for η∈(0,1],
let(2.47)C°bη(𝒳)={f∈Cbη(𝒳):∫𝒳f(x)dμ(x)=0}.From Propositions2.7, 2.9, and 2.10, it is easy to deduce the following useful result; we omit the details.Corollary 2.11.
Letϵ1∈(0,1] be as in Definition 2.2, 0<β≤ϵ1 and γ>0.
Then,
(i)
bothC°bβ(𝒳) and 𝒢°b(x1,r,β,γ) for any fixed x1∈𝒳 and r>0 are dense in 𝒢°(x1,r,β,γ);(ii)
bothCbβ(𝒳) and 𝒢b(x1,r,β,γ) for any fixed x1∈𝒳 and r>0 are dense in both 𝒢(x1,r,β,γ) and Lp(𝒳) with p∈[1,∞).


## 2.2. Boundedness of Singular Integrals on Spaces of Test Functions with Mean Zero

Letβ∈(0,1].
In analogy with the topology of the space 𝒟(ℝn),
in what follows, we endow the spaces Cbβ(𝒳) (resp., C°bβ(𝒳)) with the strict inductive limit topology (see
[30, page 273] or
[67]) arising from
the decomposition Cbβ(𝒳)=⋃nCβ(Bn) (resp., C°bβ(𝒳)=⋃nC°β(Bn)), where {Bn}n is any increasing sequence of closed balls
with the same center and 𝒳=⋃nBn,
and the space Cβ(Bn) (resp., C°β(Bn)) means the set of all functions f∈Cβ(𝒳) (resp., f∈C°β(𝒳)) with suppf⊂Bn,
whose topology is given by the norm ∥⋅∥Cβ(𝒳).
It is well known that the topology of Cbβ(𝒳) (resp., C°bβ(𝒳)) is independent of the choice of closed balls {Bn}n;
see [30, page 273] or
[67]. Their dual
spaces (Cbβ(𝒳))′ and (C°bβ(𝒳))′ will be endowed with the weak*topology.We first have the following basic facts.Proposition 2.12.
Letβ∈(0,1] and let T be a continuous linear operator from Cbβ(𝒳) to (Cbβ(𝒳))′.
Assume that T has a
distributional kernel K,
which is locally integrable away from the diagonal of 𝒳×𝒳,
in the sense that for all φ,ψ∈Cbβ(𝒳) with disjoint supports,
(2.48)〈Tφ,ψ〉=∬𝒳×𝒳K(x,y)φ(y)ψ(x)dμ(x)dμ(y).Assume also that there exists a
constant CT>0 such that for all x,x′,y∈𝒳 with d(x,x′)≤d(x,y)/2 and x≠y,(2.49)|K(x,y)−K(x′,y)|≤CTd(x,x′)ϵV(x,y)d(x,y)ϵ.Then T can be extended to a continuous linear
operator from Cβ(𝒳) to (C°bβ(𝒳))′.Proof.
For anyf∈Cβ(𝒳) and g∈C°bβ(𝒳),
suppose suppg⊂B(x0,r) for some x0∈𝒳 and r>0.
Choose ψ∈Cbβ(𝒳) such that ψ(x)=1 when x∈B(x0,2r) and ψ(x)=0 when x∉B(x0,4r).
It is easy to see that ψf∈Cbβ(𝒳),
which shows that 〈T(ψf),g〉 is well defined. On the other hand, we define(2.50)〈T((1−ψ)f),g〉=∬𝒳×𝒳[K(x,y)−K(x0,y)](1−ψ(y))f(y)g(x)dμ(x)dμ(y).By (2.49) and Lemma 2.1(i), it
is easy to see that the right-hand side of the above equality is finite;
furthermore, if f has also bounded support, this equals with(2.51)∬𝒳×𝒳K(x,y)(1−ψ(y))f(y)g(x)dμ(x)dμ(y),which coincides with (2.48).
Moreover, it is easy to verify that(2.52)〈T(ψf),g〉+〈T((1−ψ)f),g〉is independent of choice of ψ.
Thus we can define Tf by letting(2.53)〈Tf,g〉=〈T(ψf),g〉+〈T((1−ψ)f),g〉.In this sense, we have Tf∈(C°bβ(𝒳))′,
which completes the proof of Proposition 2.12.We also need the weak boundedness property and the
strong weak boundedness property of operators; see [68] for the definition of the
weak boundedness property on ℝn and [34] for the definition of the strong weak boundedness
property on Ahlfors 1-regular metric measure spaces.In what follows, forβ∈(0,1],
we denote by Cbβ(𝒳×𝒳) the set of all functions f on 𝒳×𝒳 with bounded support, which satisfy that for
any x∈𝒳,
both f(x,⋅) and f(⋅,x) are in C˙β(𝒳).Definition 2.13.
Letβ∈(0,1] and let T be a continuous linear mapping from Cbβ(𝒳) to (Cbβ(𝒳))′.
The operator T is said to have the weak boundedness property of orderβ (for short, T∈WBP(β)) if there exists a constant C>0 such that for all ϕ,ψ∈Cbβ(𝒳) with suppϕ,suppψ⊂B(z,r) for some z∈𝒳 and r>0, ∥ϕ∥C˙β(𝒳)≤r−β and ∥ψ∥C˙β(𝒳)≤r−β,(2.54)|〈Tϕ,ψ〉|≤Cμ(B(z,r)).The minimal constant C as above is denoted by ∥T∥WBP(β).Remark 2.14.
(i) Letβ∈(0,1] and γ>0.
Let us also endow the space 𝒢b(β,γ) (resp., 𝒢°b(β,γ)) with the strict inductive limit topology in a similar way as the
space Cbβ(𝒳) (resp., C°bβ(𝒳)) and its dual space (𝒢b(β,γ))′ (resp., (𝒢°b(β,γ))′) with the weak* topology. Then, as topological spaces, Cbβ(𝒳)=𝒢b(β,γ) and (Cbβ(𝒳))′=(𝒢b(β,γ))′ (resp., C°bβ(𝒳)=𝒢°b(β,γ) and (C°bβ(𝒳))′=(𝒢°b(β,γ))′).
(ii) We remark that if𝒳 is a (κ,n)-space as in Definition 1.1, then there exists
a constant C>0 such that for all ϕ∈C˙bβ(𝒳) and all x∈𝒳,(2.55)|ϕ(x)|≤C[diam(suppϕ)]β∥ϕ∥C˙β(𝒳).To see this, assume that suppϕ⊂B(x0,r) for some x0∈𝒳 and r>0.
By Remark 1.2, we can find a y0∈𝒳 such that r≤d(y0,x0)≤2r.
Then ϕ(y0)=0 and for all x∈suppϕ,(2.56)|ϕ(x)|=|ϕ(x)−ϕ(y0)|≤d(x,y0)β∥ϕ∥C˙β(𝒳)≤(3r)β∥ϕ∥C˙β(𝒳),which is just (2.55). By this
observation, we see that the functions ϕ and ψ in Definition 2.13 also satisfy ∥ϕ∥L∞(𝒳)≲1 and ∥ψ∥L∞(𝒳)≲1.
(iii) From Hölder's inequality, it is easy to deduce
that ifT is bounded on Lp(𝒳) with p∈(1,∞),
then T∈WBP(β) for any β∈(0,1] and(2.57)∥T∥WBP(β)≤∥T∥Lp(𝒳)→Lp(𝒳),where and in what follows, ∥T∥Lp(𝒳)→Lp(𝒳) denotes the operator norm of T from Lp(𝒳) to Lp(𝒳).
(iv) Following [34], we can also introduce a slightly stronger property
than WBP,
called strong weak boundedness property in [34] as follows: let T be as in Definition 2.13. The operator T is said to have the strong weak boundedness property of orderβ (for short, T∈SWBP(β)) if T has a distributional kernel K∈(Cbβ(𝒳×𝒳))′ such that (2.48) holds and there exists a
constant C>0 such that for all f∈Cbβ(𝒳×𝒳) with suppf⊂B(x0,r)×B(x0,r) for some x0∈𝒳 and r>0, ∥f(⋅,y)∥C˙β(𝒳)≤r−β for all y∈𝒳 and ∥f(x,⋅)∥C˙β(𝒳)≤r−β for all x∈𝒳,(2.58)|〈K,f〉|≤Cμ(B(x0,r)).The minimal constant C as above is denoted by ∥T∥SWBP(β).
By the observation in (ii) of this remark, the
functionf also satisfies that ∥f∥L∞(𝒳×𝒳)≲1.
(v) Leti=1,2, βi∈(0,1] and let T be a continuous linear mapping from Cbβi(𝒳) to (Cbβi(𝒳))′ with a distributional kernel K as in (2.48) of Proposition 2.12. If K satisfies the size condition that for all x,y∈𝒳 with x≠y,(2.59)|K(x,y)|≤CT1V(x,y),then T∈WBP(β1) if and only if T∈WBP(β2).
This fact can be proved by an argument similar to that used in the proof of
Proposition 1 in [64] and we omit the details.In what follows, standard notions from distribution
theory carry over to continuous linear functionals onCbβ(𝒳).
For instance, if 𝒰⊂𝒳 is an open subset, the restriction of ω∈(Cbβ(𝒳))′ to 𝒰 is defined by 〈ω|𝒰,φ〉=〈ω,φ〉 for all φ∈Cbβ(𝒳) supported in 𝒰.Using some ideas from Meyer in [69], we can obtain the
following useful estimates which play a key role in establishing the
boundedness of singular integrals on spaces of test functions with mean zero.Lemma 2.15.
Fix a bump functionθ∈Cb1(ℝ) with 0≤θ(x)≤1 for all x∈suppθ⊂{x∈ℝ:|x|≤2} and θ(x)=1 on {x∈𝒳:|x|≤1}.
For any fixed z∈𝒳 and r>0,
let θz,r(y)=θ(d(z,y)/r) for all y∈𝒳 and put ωz,r=1−θz,r.
Let T be as in Proposition 2.12. If T(1)∈(C°bβ(𝒳))′ is 0, then the following
hold.
(i)
The restriction of the linear functionalT(θz,r)∈(Cbβ(𝒳))′ to the ball B(z,r/2) is a measurable function, and for a.e.  x∈B(z,r/2),(2.60)T(θz,r)(x)=Cz,r−∫𝒳[K(x,y)−K(z,y)]ωz,r(y)dμ(y),where Cz,r is a constant independent of x.(ii)
If further assuming thatT∈WBP(β),
then there exists a constant C>0 such that for a.e.  x∈B(z,r/2),(2.61)|T(θz,r)(x)|≤C(CT+∥T∥WBP(β)).Proof.
Assume thatT(1)=0 in (C°bβ(𝒳))′.
For any f∈C°bβ(𝒳) with suppf⊂B(z,r/2),
since ∫𝒳f(x)dμ(x)=0,
we have(2.62)0=〈T(1),f〉=〈T(θz,r)+∫𝒳[K(⋅,y)−K(z,y)]ωz,r(y)dμ(y),f〉.Let α∈C1(ℝ) be as in the proof of Proposition 2.9, and η˜z,r(y)=α(d(y,z)/(r/2)).
Set(2.63)ηz,r(y)=η˜z,r(y)[∫𝒳η˜z,r(w)dμ(w)]−1.Then ηz,r∈Cb1(𝒳), suppηz,r⊂B(z,r/2) and ∫𝒳ηz,r(y)dμ(y)=1.
For any f∈Cbβ(𝒳) with suppf⊂B(z,r/2),
we set f˜(y)=f(y)−ηz,r(y)∫𝒳f(y)dμ(y);
then f˜∈C°bβ(𝒳) with suppf˜⊂B(z,r/2).
Applying (2.62) to f˜ and using Corollary 2.11(ii) in the case p=1 show that the restriction of the linear
functional T(θz,r)∈(Cbβ(𝒳))′ to the ball B(z,r/2) is a measurable function, and for a.e.  x∈B(z,r/2),(2.64)〈T(θz,r)(x)+∫𝒳[K(x,y)−K(z,y)]ωz,r(y)dμ(y),f〉=∫𝒳〈T(θz,r)+∫𝒳[K(⋅,y)−K(z,y)]ωz,r(y)dμ(y),ηz,r〉f(x)dμ(x),which implies (i) with(2.65)Cz,r=〈T(θz,r)+∫𝒳[K(⋅,y)−K(z,y)]ωz,r(y)dμ(y),ηz,r〉,by noticing that Cz,r is a constant independent of x and the choice of ηz,r,
but, it may depend on z and r.
To verify (ii), by (2.62) together with the definition
of T(1), (2.49) of Proposition 2.12 and Lemma 2.1(i), we have(2.66)|〈T(θz,r),f〉|=|−〈T(ωz,r),f〉|=|∬𝒳×𝒳[K(x,y)−K(z,y)]ωz,r(y)f(x)dμ(y)dμ(x)|≤CT∫𝒳[∫2d(z,x)<d(z,y)d(x,z)ϵV(z,y)d(z,y)ϵdμ(y)]|f(x)|dμ(x)≲CT∥f∥L1(𝒳),which together with Corollary 2.11(ii) in the cases r=1 and p=1 again shows that for a.e.  x∈B(z,r/2),(2.67)T(θz,r)(x)=Ωz,r(x)+Cz,r,where ∥Ωz,r∥L∞(𝒳)≲CT and Cz,r is a constant independent of x.
We now estimate Cz,r by using that T∈WBP(β).
To this end, let g∈Cbβ(𝒳) with suppg⊂B(z,r), ∥g∥L∞(𝒳)≤1, ∥g∥C˙β(𝒳)≤r−β,
and ∫𝒳g(x)dμ(x)~μ(B(z,r)).
From (2.67) and T∈WBP(β),
it follows that(2.68)|Cz,r||∫𝒳g(x)dμ(x)|=|〈T(θz,r),g〉−∫𝒳Ωz,r(x)g(x)dμ(x)|≲(∥T∥WBP(β)+CT)μ(B(z,r)),which implies that |Cz,r|≲∥T∥WBP(β)+CT, and hence, completes the proof of Lemma
2.15.We recall the notion of the space of functions with
bounded mean oscillation,BMO(𝒳),
which was first introduced by John and Nirenberg in [70], and was proved to be the
dual space of Hat1(𝒳) in [28].Definition 2.16.
Let1≤q<∞.
The spaceBMOq(𝒳) is defined to be the set of
all f∈Llocq(𝒳) such that(2.69)∥f∥BMOq(𝒳)=supx∈𝒳,r>0{1μ(B(x,r))∫B(x,r)|f(y)−mB(x,r)(f)|qdμ(y)}1/q<∞.When q=1,
one denotes
BMO1(𝒳) simply by BMO(𝒳).It was proved in [28] that for any 1≤q1,q2<∞, BMOq1(𝒳) and BMOq2(𝒳) are equal as vector spaces and the seminorms ∥⋅∥BMOq1(𝒳) and ∥⋅∥BMOq2(𝒳) are equivalent. Moreover, if we let 𝒩≡ℂ be the subspace of all constant functions on𝒳,
then the quotient space BMO(𝒳)/𝒩 becomes a Banach space in a natural  way.Remark 2.17.
Letβ∈(0,ϵ] and γ>0.
It is easy to see that T(1)=0 in (C°bβ(𝒳))′ if and only if T(1)∈(C°bβ(𝒳))′ is constant; see also [71, page 22]. Moreover, from Corollary 2.11(i), Proposition 5.21, Theorems 5.19(i) and 6.11 below, it
is easy to see that T(1)∈(C°bβ(𝒳))′ is constant if and only if T(1)∈BMO(𝒳) is constant, which is also equivalent to T(1)=0 in BMO(𝒳).We now establish a basic boundedness result for
singular integrals on spaces of test functions with mean zero, which will be a
key tool for the whole paper.Theorem 2.18.
Letϵ∈(0,1], β∈(0,ϵ), and let T be as in Proposition 2.12 with the
distributional kernel K satisfying the following additional conditions
that
(i)
for allx,y∈𝒳 with x≠y, |K(x,y)|≤CT(1/V(x,y));(ii)
for allx,y,y′∈𝒳 with d(y,y′)≤d(x,y)/2 and x≠y,(2.70)|K(x,y)−K(x,y′)|≤CTd(y,y′)ϵV(x,y)d(x,y)ϵ;(iii)
for allx,x′,y,y′∈𝒳 with d(x,x′)≤d(x,y)/3,d(y,y′)≤d(x,y)/3 and x≠y,(2.71)|[K(x,y)−K(x′,y)]−[K(x,y′)−K(x′,y′)]|≤CTd(x,x′)ϵd(y,y′)ϵV(x,y)d(x,y)2ϵ.
If T∈WBP(β) and T(1)=0 in (C°bβ(𝒳))′,
then T extends to a bounded linear operator from 𝒢°(x1,r,β,γ) to 𝒢(x1,r,β,γ) for all x1∈𝒳, r>0, and γ∈(0,ϵ).
Moreover, there exists a constant Cβ,γ,C0>0 such that for all f∈𝒢°(x1,r,β,γ) with any x1∈𝒳,
any r>0,
and any γ∈(0,ϵ),(2.72)∥Tf∥𝒢(x1,r,β,γ)≤Cβ,γ,C0(CT+∥T∥WBP(β))∥f∥𝒢(x1,r,β,γ).Compared with the corresponding results in [31, 35], Theorem 2.18 has three
advantages. (1) Theorem 2.18 is true on spaces of homogeneous type, while the
corresponding results in [31, 35] are proved only for ℝn or for spaces of homogeneous type with μ(B(x,r))~r for all x∈𝒳 and 0<r≤diam𝒳,
respectively. (2) We do not assume that T∗(1)=0 in Theorem 2.18. Since T∗(1)=0 was also assumed in [31, 35], by the T(1)-theorem in those settings, one knows that T is bounded on L2(𝒳).
Thus, for any f∈𝒢°(x1,r,β,γ) for some x1∈𝒳, r>0, β∈(0,1] and γ>0, Tf∈L2(𝒳),
which makes the proof much easier. (3) We only require T∈WBP(β) instead of T∈SWBP(β) in [35].To prove Theorem2.18, we first recall the following
construction given by Christ in [72], which provides an analogue of the grid of Euclidean dyadic cubes on spaces of homogeneous type.Lemma 2.19.
Let𝒳 be a space of homogeneous type. Then there
exists a collection {Qαk⊂𝒳:k∈ℤ,α∈Ik} of open subsets, where Ik is some index set, and  constants δ∈(0,1) and C6,C7>0 such that
(i)
μ(𝒳∖⋃αQαk)=0 for each fixed k and Qαk∩Qβk=∅ if α≠β;(ii)
for anyα,β,k,l with l≥k, either Qβl⊂Qαk or Qβl∩Qαk=∅;(iii)
for each(k,α) and each l<k there is a unique β such that Qαk⊂Qβl;(iv)
diam(Qαk)≤C6δk;(v)
eachQαk contains some ball B(zαk,C7δk),
where zαk∈𝒳.In fact, we can think ofQαk as being a dyadic cube with diameter rough δk and centered at zαk. In what follows, to simplify our presentation,
we always suppose δ=1/2;
otherwise, we need to replace 2−k in the definition of ATIs by δk and some other changes are also necessary; see
[34, pages 96–98] for
more details.To prove Theorem2.18, we need another technical lemma,
where we need Lemma 2.19 and T∈WBP(β); see also [69, Lemma 3] and [34, Lemma (3.12)].In what follows, ifTf∈(Cbβ(𝒳))′ and g∈Cbβ(𝒳),
we sometimes will write(2.73)∫𝒳Tf(x)g(x)dμ(x)in place of 〈Tf,g〉,
in order to indicate more clearly the dependence on the variable x.Lemma 2.20.
Letϵ∈(0,1], β∈(0,ϵ), γ∈(0,ϵ] and let T be as in Proposition 2.12 with the
distributional kernel K satisfying the additional size condition
(2.59). Let θ be as in Lemma 2.15 and for any fixed x1∈𝒳 and r>0,
one defines
θ˜≡θx1,20r and ω˜≡ωx1,20r in the same way as in Lemma 2.15. If T∈WBP(β) and T(1)=0 in (C°bβ(𝒳))′,
then for any f∈𝒢b(x1,r,β,γ),
the restriction of the linear functional Tf∈(Cbβ(𝒳))′ to the ball B(x1,10r) is a measurable function, and for a.e.  x∈B(x1,10r),(2.74)Tf(x)=∫𝒳K(x,y)[f(y)−f(x)]θ˜(y)dμ(y)+∫𝒳K(x,y)f(y)ω˜(y)dμ(y)+f(x)T(θ˜)(x),where the first two integrals
are absolutely convergent.Proof.
We make use of some ideas used in the proofs of Lemma 3 in [69] and
Lemma (3.12) in [34]. For any fixed f∈𝒢b(x1,r,β,γ) and any ψ∈Cbβ(𝒳) with suppψ⊂B(x1,10r),
we have(2.75)〈Tf,ψ〉=〈T(fθ˜),ψ〉+〈T(fω˜),ψ〉=∫𝒳T(θ˜[f−f(x)])(x)ψ(x)dμ(x)+∫𝒳T(θ˜)(x)f(x)ψ(x)dμ(x)+∬𝒳×𝒳K(x,y)ψ(x)f(y)ω˜(y)dμ(x)dμ(y)≡Y1+Y2+Y3,where the first two integrals
have to be interpreted in the sense of “distributions,” whereas the third
one is an absolutely convergent integral.
Actually, by Lemma2.15(ii), T(θ˜) is a bounded function in B(x1,10r),
and we have(2.76)|Y2|≤(supx∈B(x1,10r)|T(θ˜)(x)|)∫𝒳|ψ(y)f(y)|dμ(y)≲1Vr(x1)∥ψ∥L1(𝒳),which is the desired estimate.
Notice that ifd(x,x1)<10r,
then x∉suppω˜.
Moreover, if we put(2.77)Y3,1(x)≡∫𝒳K(x,y)f(y)ω˜(y)dμ(y),then for x∈B(x1,10r),(2.78)|Y3,1(x)|≲1Vr(x1)+V(x1,x).In fact, we first notice that ω˜(y)≠0 implies that d(y,x1)≥20r≥2d(x,x1).
From this, it follows that V(x1,y)≥μ(B(x1,2d(x,x1)))≥V(x1,x) and d(y,x1)≤2d(x,y),
which together with (1.2) shows that(2.79)V(x1,y)~V(y,x1)=μ(B(y,d(y,x1)))≤μ(B(y,2d(x,y)))≲V(y,x)~V(x,y).These estimates together with
Lemma 2.1(i) and (2.59) imply that(2.80)|Y3,1(x)|≲∫d(y,x1)≥20r1V(x,y)1Vr(x1)+V(x1,y)rγ(r+d(y,x1))γdμ(y)≲1Vr(x1)+V(x1,x),which is just (2.78). Noticing
that suppψ∩supp(fω˜)=∅,
by the assumption (2.48) of Proposition 2.12 and (2.78), we have(2.81)|Y3|=|∫𝒳Y3,1(x)ψ(x)dμ(x)|≲1Vr(x1)∥ψ∥L1(𝒳),which is again the desired
estimate.
Letk∈ℕ
be large enough,
which will be determined later, and let {Sk}k∈ℤ be an ATI with bounded support as constructed in Theorem 2.6. Let η∈C2(ℝ) be radial and η(t)=1 when t∈[0,23], η(t)=0 when |t|>24 and 0≤η(t)≤1 for all t∈ℝ.
For any x,y∈ℝ,
we then define λk(x,y)=∫𝒳Sk(x,z)η(2kd(z,y))dμ(z). It is easy to see that λk(x,y)=0 if d(x,y)≥25−k,
and that(2.82)λk(x,y)=∫𝒳Sk(x,z)dμ(z)=1,ifd(x,y)<2−k.
For any fixedk∈ℕ,
choose J∈ℕ so large that C62−J≤23−k.
Put(2.83)NJ={i∈IJ:QiJ∩B(x,22−k)≠∅forsomex∈B(x1,10r)},and for all x∈B(x1,10r), NJx={i∈IJ:QiJ∩B(x,22−k)≠∅}, where QiJ is the dyadic cube as in Lemma 2.19. Let ♯NJ and ♯NJx denote the cardinality of NJ and NJx,
respectively. It is easy to see that if QiJ∩B(x,22−k)≠∅ for some x∈B(x1,10r),
then QiJ⊂B(x1,24−k+10r) and B(x1,24−k+10r)⊂B(zQiJ,2(24−k+10r)),
where zQiJ is the center of QiJ as in Lemma 2.19. From this, and Lemma 2.19(i)
and (v), it is easy to see that(2.84)♯NJ≲2nJ(r+2−k)n.Similarly, ♯NJx≲2(J−k)n.
We now claim that for any givenk∈ℕ and any given x∈B(x1,10r),(2.85)λk(x,⋅)=limJ→∞∑i∈NJSk(x,zQiJ)η(2kd(zQiJ,⋅))μ(QiJ),(2.86)1−λk(x,⋅)=limJ→∞∑i∈NJSk(x,zQiJ)[1−η(2kd(zQiJ,⋅))]μ(QiJ)hold in Cβ(𝒳).
We only prove (2.85), the proof of (2.86) being similar. To this end, it is easy to see that(2.87)λk(x,y)−∑i∈NJSk(x,zQiJ)η(2kd(zQiJ,y))μ(QiJ)=∑i∈NJx∫QiJ[Sk(x,z)−Sk(x,zQiJ)]η(2kd(z,y))dμ(z)+∑i∈NJx∫QiJSk(x,zQiJ)[η(2kd(z,y))−η(2kd(zQiJ,y))]dμ(z)≡Y4(y)+Y5(y).Since z∈QiJ,
by Lemma 2.19(iv), we have d(z,zQiJ)≤C62−J,
which together with the regularity of Sk yields(2.88)|Y4(y)|≲∑i∈NJx2k−JV2−k(x)μ(QiJ)≲2k−JV2−k(x)μ(B(x,24−k))≲2k−J→0,as J→∞.
Similarly, by the size condition of Sk and the mean value theorem, we have(2.89)|Y5(y)|≲2k−J1V2−k(x)∑i∈NJxμ(QiJ)≲2k−J→0,as J→∞.
Thus, as J→∞,(2.90)∥λk(x,⋅)−∑i∈NJSk(x,zQiJ)η(2kd(zQiJ,⋅))μ(QiJ)∥L∞(𝒳)→0.
For anyy,y′∈𝒳,
if d(y,y′)≥2k−J,
by (2.88), we then have(2.91)|Y4(y)−Y4(y′)|≲2(k−J)(1−β)d(y,y′)β.If d(y,y′)<2k−J,
by the regularity of Sk and the mean value theorem, we
have(2.92)|Y4(y)−Y4(y′)|=|∑i∈NJx∫QiJ[Sk(x,z)−Sk(x,zQiJ)][η(2kd(z,y))−η(2kd(z,y′))]dμ(z)|≲22k−JV2−k(x)d(y,y′)μ(B(x,24−k))≲2k(3−β)−J(2−β)d(y,y′)β.
Similarly, ifd(y,y′)≥2k−J,
by (2.89), we then have(2.93)|Y5(y)−Y5(y′)|≲2(k−J)(1−β)d(y,y′)β.To estimate Y5(y)−Y5(y′) when d(y,y′)<2k−J,
by the mean value theorem, we first see that(2.94)|[η(2kd(z,y))−η(2kd(zQiJ,y))]−[η(2kd(z,y′))−η(2kd(zQiJ,y′))]|≲2kd(z,zQiJ),|[η(2kd(z,y))−η(2kd(zQiJ,y))]−[η(2kd(z,y′))−η(2kd(zQiJ,y′))]|≲2kd(y,y′).Taking a suitable geometric mean
of these estimates, we find that(2.95)|[η(2kd(z,y))−η(2kd(zQiJ,y))]−[η(2kd(z,y′))−η(2kd(zQiJ,y′))]|≲2kd(y,y′)βd(z,zQiJ)1−β.Using this estimate and the size
condition of Sk,
we obtain(2.96)|Y5(y)−Y5(y′)|=|∑i∈NJx∫QiJSk(x,zQiJ){[η(2kd(z,y))−η(2kd(zQiJ,y))]−[η(2kd(z,y′))−η(2kd(zQiJ,y′))]}dμ(z)|≲2kd(y,y′)β2−J(1−β)1V2−k(x)∑i∈NJxμ(QiJ)≲2kd(y,y′)β2−J(1−β).Thus,(2.97)∥λk(x,⋅)−∑i∈NJSk(x,zQiJ)η(2kd(zQiJ,⋅))μ(QiJ)∥C˙β(𝒳)≲2k(3−β)2−J(1−β)→0,as J→∞.
This establishes (2.85), and hence also (2.86).
We now decomposeY1 into(2.98)Y1=∫𝒳T(θ˜[f−f(x)][1−λk(x,⋅)])(x)ψ(x)dμ(x)+∫𝒳T(θ˜[f−f(x)]λk(x,⋅))(x)ψ(x)dμ(x)≡Y1,1+Y1,2,where both integrals have to be
interpreted in the sense of “distributions.”
To estimateY1,1,
by (2.86) and (2.84), observe that(2.99)Y1,1=limJ→∞∑i∈NJ∫𝒳T(θ˜[f−f(x)][1−η(2kd(zQiJ,⋅))])(x)Sk(x,zQiJ)ψ(x)dμ(x)μ(QiJ)where the integral has to be
interpreted in the sense of “distributions.” However, noticing that suppθ˜(⋅)[1−η(2kd(zQiJ,⋅))]∩suppSk(⋅,zQiJ)=∅,
by (2.48), we then further have(2.100)Y1,1=limJ→∞∑i∈NJμ(QiJ)∬𝒳×𝒳K(x,y)θ˜(y)[f(y)−f(x)][1−η(2kd(zQiJ,y))]×Sk(x,zQiJ)ψ(x)dμ(x)dμ(y).Notice that(2.101)∑i∈NJμ(QiJ)∬𝒳×𝒳|K(x,y)θ˜(y)[f(y)−f(x)][1−η(2kd(zQiJ,y))]||Sk(x,zQiJ)ψ(x)|dμ(x)dμ(y)≲μ(B(x1,24−k+10r))×(∫𝒳|ψ(x)|{∫d(x,y)≤(r+d(x1,x))/2|θ˜(y)|1V(x,y)(d(x,y)r+d(x1,x))β×1Vr(x1)+V(x,x1)(rr+d(x1,x))γdμ(y)+∫d(x,y)<50rd(x,y)>(r+d(x,x1))/2|θ˜(y)|[1Vr(x1)+V(x1,y)(rr+d(x1,y))γ+1Vr(x1)+V(x1,x)(rr+d(x1,x))γ]×1V(x,y)dμ(y)}dμ(x))≲μ(B(x1,24−k+10r))1Vr(x1)∥ψ∥L1(𝒳). Thus, by Lebesgue's dominated
convergence theorem, we have(2.102)Y1,1=∬𝒳×𝒳K(x,y)θ˜(y)[f(y)−f(x)]ψ(x){∫𝒳[1−η(2kd(z,y))]Sk(x,z)dμ(z)}dμ(x)dμ(y)=∬𝒳×𝒳K(x,y)θ˜(y)[f(y)−f(x)]ψ(x)[1−λk(x,y)]dμ(x)dμ(y),where the last integral
converges absolutely; and moreover, by (2.82) and an argument similar to (2.101), we further  have |Y1,1|≲(1/Vr(x1))∥ψ∥L1(𝒳).
We next prove that(2.103)limk→∞Y1,2=0,making use of our assumption
that T∈WBP(β).
To this end, by (2.85) and (2.84), we have(2.104)Y1,2=limJ→∞∑i∈NJμ(QiJ)∫𝒳T(θ˜[f−f(x)]η(2kd(zQiJ,⋅)))(x)Sk(x,zQiJ)ψ(x)dμ(x)=limJ→∞∑i∈NJμ(QiJ){∫𝒳T(θ˜[f−f(zQiJ)]η(2kd(zQiJ,⋅)))(x)Sk(x,zQiJ)ψ(x)dμ(x)+∫𝒳T(θ˜η(2kd(zQiJ,⋅)))(x)[f(zQiJ)−f(x)]Sk(x,zQiJ)ψ(x)dμ(x)}.
Choosek∈ℕ such that 2−k≤r/25.
We claim that(2.105)|∫𝒳T(θ˜[f−f(zQiJ)]η(2kd(zQiJ,⋅)))(x)Sk(x,zQiJ)ψ(x)dμ(x)|≲2−kβrβ1Vr(x1)∥f∥𝒢(x1,r,β,γ)∥ψ∥Cβ(𝒳),(2.106)|∫𝒳T(θ˜η(2kd(zQiJ,⋅)))(x)[f(zQiJ)−f(x)]Sk(x,zQiJ)ψ(x)dμ(x)|≲2−kβrβ1Vr(x1)∥f∥𝒢(x1,r,β,γ)∥ψ∥Cβ(𝒳).
We only show (2.105), the proof  of (2.106) being similar.
To see (2.105), put ϕ(y)=θ˜(y)[f(y)−f(zQiJ)]η(2kd(zQiJ,y)) and(2.107)g(x)=Sk(x,zQiJ)ψ(x).Notice that if f,h∈Cβ(𝒳),
then(2.108)∥fh∥C˙β(𝒳)≲∥f∥L∞(𝒳)∥h∥C˙β(𝒳)+∥f∥C˙β(𝒳)∥h∥L∞(𝒳).From this, it follows that(2.109)∥ϕ∥C˙β(𝒳)≲∥f∥𝒢(x1,r,β,γ)1rβVr(x1),and that(2.110)∥g∥C˙β(𝒳)≲∥ψ∥Cβ(𝒳)1V2−k(zQiJ)2kβ.The estimates (2.109) and (2.110) together with T∈WBP(β) imply the claim (2.105).
By (2.105) and (2.106), we know that if 2−k<r/25,
then(2.111)|Y1,2|≲2−kβrβ1Vr(x1)∥f∥𝒢(x1,r,β,γ)∥ψ∥Cβ(𝒳)limJ→∞∑i∈NJμ(QiJ)≲2−kβrβ∥f∥𝒢(x1,r,β,γ)∥ψ∥Cβ(𝒳),which implies (2.103).
Thus, we have that for anyψ∈Cbβ(𝒳) with suppψ⊂B(x1,10r), |〈Tf,ψ〉|≲∥ψ∥L1(𝒳), so that Tf agrees with L∞(𝒳) function on B(x1,10r).
Moreover, by (2.102) and Lebesgue's dominated convergence theorem, we
have(2.112)〈Tf,ψ〉=∬𝒳×𝒳K(x,y)ψ(x)θ˜(y)[f(y)−f(x)]dμ(y)dμ(x)+∬𝒳×𝒳K(x,y)ψ(x)ω˜(y)f(y)dμ(y)dμ(x)+∫𝒳T(θ˜)(x)f(x)ψ(x)dμ(x),as was to be proved.Proof of Theorem2.18.
By Corollary2.11(i), we only need to prove
Theorem 2.18 for 𝒢°b(x1,r,β,γ).
Letf∈𝒢°b(x1,r,β,γ).
We first verify that Tf(x) satisfies (i) of Definition 2.8 for a.e. x∈𝒳.
To this end, we consider two cases.
Case 1  (d(x,x1)<10r).
In this case, letθ˜ and ω˜ be as in Lemma 2.20. By Lemma 2.20, for a.e. x∈B(x1,10r),
we have(2.113)Tf(x)=∫𝒳K(x,y)[f(y)−f(x)]θ˜(y)dμ(y)+∫𝒳K(x,y)f(y)ω˜(y)dμ(y)+f(x)T(θ˜)(x)≡Z1+Z2+Z3.
Lemma 2.15(ii) shows that for a.e. x∈B(x1,10r), |Z3|≲|f(x)|≲1/(Vr(x1)+V(x1,x)), which gives the desired estimate. By (2.78),
we have |Z2|≲1/(Vr(x1)+V(x1,x)). For Z1,
the facts that θ˜(y)≠0 and d(x,x1)<10r imply that d(x,y)<50r.
The size condition on K and the regularity of f yield that(2.114)|Z1|≤∫d(x,y)≤(r+d(x,x1))/2|K(x,y)[f(y)−f(x)]θ˜(y)|dμ(y)+∫d(x,y)>(r+d(x,x1))/2|K(x,y)f(x)θ˜(y)|dμ(y)+∫d(x,y)>(r+d(x,x1))/2|K(x,y)f(y)θ˜(y)|dμ(y)≲{1rβ1Vr(x1)+V(x1,x)∫d(x,y)≤50rd(x,y)βV(x,y)dμ(y)+1Vr(x1)+V(x1,x)∫50r≥d(x,y)>(r+d(x,x1))/2,d(x1,y)≤40r1V(x,y)dμ(y)}+∫50r≥d(x,y)>(r+d(x,x1))/2,d(x1,y)≤40r1V(x,y)1Vr(x1)+V(x1,y)dμ(y)≡Z1,1+Z1,2.By Lemma 2.1(i) and (1.2), we obtain(2.115)Z1,1≲1Vr(x1)+V(x1,x)[1+V(x,50r)V(x,r/2)]≲1Vr(x1)+V(x1,x),Z1,2≲min{1Vr(x1),∫r/2<d(x,y)≤50r,d(x,x1)/5≤r≤d(x1,y)1V(x,y)V(x1,y)dμ(y)+1Vr(x1)∫d(x,y)>d(x,x1)/2,r>d(x1,y)1V(x,y)dμ(y)}≲1Vr(x1)+V(x1,x).Combining the estimates for Z1,1 and Z1,2 yields the desired estimate for Z1,
which verifies that Tf(x) satisfies (i) of Definition 2.8 when d(x,x1)<10r.Case 2  (d(x,x1)≡R≥10r).
In this case, for anym∈ℕ,
let(2.116)Bm=B(x1,10(m+1)r)∖B(x1,10mr).For any fixed x0∈Bm,
we put Im(y)=θ(10d(x0,y)/37mr), Jm(y)=θ(4d(x1,y)/5mr) and define Lm(y) by Lm(y)=1−Im(y)−Jm(y).
Notice that if y∈suppIm∩suppJm,
then d(y,x0)<37mr/5 and d(y,x1)<5mr/2,
and hence d(x0,x1)≤d(x0,y)+d(y,x1)<(99/10)mr<10mr,
which contracts the choice of x0∈Bm.
Therefore, Lm(y)≥0.
We also define f1(y)=f(y)Im(y), f2(y)=f(y)Jm(y) and f3(y)=f(y)Lm(y).
We first establish some estimates on fi with i=1,2,3,
when x∈Bm.
Obviouslyf1(y)≠0 implies that d(y,x0)<37mr/5,
and therefore(2.117)d(x1,y)≥d(x1,x0)−d(x0,y)>325Rwhich together with (1.2) and
the size condition of f shows that(2.118)|f1(y)|≲1VR(x1)(rR)γ,∀y∈𝒳.
We now claim that(2.119)|f1(y)−f1(y′)|≲1VR(x1)d(y,y′)βRβ(rR)γ,∀y,y′∈𝒳.To prove (2.119), we consider two
cases.Case 1  (d(y,y′)≤(r+d(x1,y))/2).
In this case, we divide(2.120)f1(y)−f1(y′)=[f(y)−f(y′)]Im(y)χ0(y,y′)+f(y′)[Im(y)−Im(y′)]=Z4+Z5,where χ0(y,y′)=χ{y∈𝒳:d(x0,y)<37mr/5}(y)+χ{y′∈𝒳:d(x0,y′)<37mr/5}(y′).
The regularity off shows that(2.121)|Z4|≲(d(y,y′)r+d(x1,y))β1Vr(x1)+V(x1,y)(rr+d(x1,y))γχ0(y,y′).If χ0(y,y′)≠0,
then d(x0,y)<37mr/5 or d(x0,y′)<37mr/5.
Notice that d(y,y′)≤(r+d(x1,y))/2 implies that 3d(x1,y)/2≥d(x1,y′)−r/2.
If d(x0,y′)<37mr/5,
we then further have 3d(x1,y)/2≥d(x1,x0)−d(x0,y′)−r/2>21mr/10>R/10, and hence d(x1,y)>R/15.
This together with (2.117) implies that if χ0(y,y′)≠0,
then we have(2.122)d(y,x1)>R15.By (2.122) and (1.2), we obtain VR(x1)≲V(x1,y).
These estimates prove that |Z4|≲(1/VR(x1))(d(y,y′)/R)β(r/R)γ.
The size condition off implies that(2.123)|Z5|≲1Vr(x1)+V(x1,y′)(rr+d(x1,y′))γ|Im(y)−Im(y′)|χ0(y,y′).We now claim that(2.124)1Vr(x1)+V(x1,y′)χ0(y,y′)≲1VR(x1).To prove (2.124), we consider two
subcases.
Subcase 1.1  (r≥d(x1,y′)).
In this case,(1/(Vr(x1)+V(x1,y′)))χ0(y,y′)≲(1/Vr(x1))χ0(y,y′). If d(x0,y′)<37mr/5,
then d(x1,y′)≥d(x1,x0)−d(x0,y′)>13mr/5>2r, which contradicts the assumption that r≥d(x1,y′).
Thus, we further have(2.125)1Vr(x1)+V(x1,y′)χ0(y,y′)≲1Vr(x1)χ{y∈𝒳:d(x0,y)<37mr/5}(y).Since d(x0,y)<37mr/5 and x∈Bm,
by (2.117), we have d(y,x1)>3R/25. Moreover, from d(y,x1)≤d(y,y′)+d(y′,x1) and d(y,y′)≤(r+d(x1,y))/2,
it follows that d(y,x1)≤r+2d(y′,x1)≤3r.
Thus, in this case, R<25r and hence VR(x1)≤Vr(x1).
Thus, the claim (2.124) holds in this case.Subcase 1.2  (r<d(x1,y′)).
In this case, fromr<d(x1,y′) and d(y,y′)≤(r+d(x1,y))/2,
it follows that d(x1,y)≤r+2d(x1,y′)<3d(x1,y′),
which together with (2.122) implies that VR(x1)≲V(x1,y′).
Thus, (1/(Vr(x1)+V(x1,y′)))χ0(y,y′)≲1/VR(x1). Thus, (2.124) also holds in this case, which
completes the proof of claim (2.124).
Notice that ifd(x0,y)<37mr/5,
then d(y,x1)≤d(x1,x0)+d(x0,y)≲R, and if d(x0,y′)<37mr/5,
since d(y,y′)≤(r+d(y,x1))/2,
we then have d(y,x1)≤r+2d(y′,x1)≤r+2[d(y′,x0)+d(x0,x1)]≲R. These estimates together with d(y,y′)≤(r+d(y,x1))/2 again implies that if χ0(y,y′)≠0,
then d(y,y′)≲R and r+d(y,x1)≤2(r+d(x1,y′)),
which together with (2.122) further yields that r+d(x1,y′)≳R.
All these estimates, (2.124), and the mean value theorem show that for any β∈(0,1],(2.126)|Z5|≲1VR(x1)(rR)γ|d(y,x)−d(y′,x)|R≲1VR(x1)(d(y,y′)R)β(rR)γ.Case 2  (d(y,y′)≥(r+d(y,x1))/2).
In this case, notice again thatf1(y)−f1(y′)≠0 implies that d(x0,y)<37mr/5 or d(x0,y′)<37mr/5.
If d(x0,y′)<37mr/5,
by d(y,x1)<2d(y,y′),
we have 10mr≤d(x0,x1)≤d(x0,y′)+d(y′,y)+d(y,x1)<37mr/5+3d(y,y′). Thus, by x∈Bm,
we further have 3R/25<13mr/5<3d(y,y′), which together with (2.117) and d(y,x1)<2d(y,y′) again implies that if f1(y)−f1(y′)≠0 and d(y,y′)≥(r+d(y,x1))/2,
then R<17d(y,y′). This estimate together with (2.118) yields that for any β>0,(2.127)|f1(y)−f1(y′)|≤|f1(y)|+|f1(y′)|≲1VR(x1)(rR)γ≲1VR(x1)(d(y,y′)R)β(rR)γ.Thus, (2.119) holds.
As forf3,
first observe that obviously, for all y∈𝒳,(2.128)|f3(y)|≲1Vr(x1)+V(x1,y)(rr+d(y,x1))γχ{d(x1,y)>R/16}(y).
From (2.128) and Lemma 2.1(i), it follows that(2.129)∫𝒳|f3(y)|dμ(y)≲rγ∫d(y,x1)>R/161V(x1,y)1d(y,x1)γdμ(y)≲(rR)γ.
Notice thatsuppf1⊂B(x1,137R/50).
From this, the estimates (2.118) and (2.129) together with ∫𝒳f(x)dμ(x)=0 and (1.2), it follows that
(2.130)|∫𝒳f2(y)dμ(y)|=|∫𝒳f1(y)dμ(y)+∫𝒳f3(y)dμ(y)|≲1VR(x1)(rR)γV137R/50(x1)+(rR)γ≲(rR)γ.
Sincex∈Bm,
we have 1/(Vr(x1)+V(x1,x))~1/VR(x1). Now, for any m∈ℕ and y∈𝒳,
put Um,r(y)=θ(d(x1,y)/42(m+1)r).
Notice that suppf1∩supp(1−Um,r)=∅.
By Lemma 2.20, for a.e. x∈Bm,
we have(2.131)Tf1(x)=∫𝒳K(x,y)[f1(y)−f1(x)]Um,r(y)dμ(y)+f1(x)T(Um,r)(x)≡Z6(x)+Z7(x).From Lemma 2.15(ii) and (2.118),
it follows that for a.e. x∈Bm,(2.132)|Z7(x)|≲|f1(x)|≲1VR(x1)(rR)γ.As for Z6(x),
notice that x∈Bm together with Um,r(y)≠0 implies that d(x,y)<18R;
hence, by (2.119), the size condition on K and Lemma 2.1(i), we obtain(2.133)|Z6(x)|≲∫d(x,y)<18R1V(x,y)1VR(x1)d(x,y)βRβ(rR)γdμ(y)≲1VR(x1)(rR)γ.Combining the estimate for Z6(x) with Z7(x) gives us the desired estimate for Tf1(x) for a.e. x∈Bm.
Sincex∉suppf2,
we can write(2.134)Tf2(x)=∫𝒳[K(x,y)−K(x,x1)]f2(y)dμ(y)+K(x,x1)∫𝒳f2(y)dμ(y)≡Z8(x)+Z9(x).The assumption (ii) of the
theorem, the support condition of f2,
and the fact that γ<ϵ together with Lemma 2.1(i) yield that(2.135)|Z8(x)|≤∫d(x1,y)≤R/2|K(x,y)−K(x,x1)||f2(y)|dμ(y)≲1V(x,x1)d(x,x1)ϵ∫d(x1,y)≤R/2d(x1,y)ϵ1V(x1,y)rγd(x1,y)γdμ(y)≲1VR(x1)(rR)γ.From the size condition on K,
and (2.130), it also follows that(2.136)|Z9(x)|≲1V(x,x1)(rR)γ≲1VR(x1)(rR)γ,which together with the estimate
for Z8(x) gives the desired estimate for Tf2(x) for a.e. x∈Bm.
Notice thatf3(x)≠0 implies that d(x1,y)>R/16 and d(x0,y)>37mr/10.
If we now further assume that x∈Bm∩B(x0,mr/10),
then d(x,y)≥d(y,x0)−d(x0,x)>18mr/5>9R/50. From this, (1.17), and (1.2), it follows that VR(x1)~VR(x)≲V(x,y).
This together with (2.128), the size condition on K and Lemma 2.1(i) yields that(2.137)|Tf3(x)|=|∫d(x,y)>9R/50,d(x1,y)>R/16K(x,y)f3(y)dμ(y)|≲∫d(x,y)>9R/50,d(x1,y)>R/161V(x,y)1V(x1,y)rγd(y,x1)γdμ(y)≲1VR(x1)(rR)γ,which is the desired estimate.
Thus, we have verified thatTf(x) for a.e. x∈Bm∩B(x0,mr/10) satisfies the size condition (i) of Definition 2.8, with constants independent of m, r, and x1.
By the Besicovitch covering lemma (see, e.g., [13, Theorem 1.16, pages 8-9]), this implies that for a.e. x∈Bm, Tf(x) satisfies the size condition (i) of Definition 2.8, with constants independent of m, r, and x1.
In combination with Case 1, we thus see that for a.e. x∈𝒳, Tf(x) satisfies the size condition (i) of Definition 2.8.
We now turn to verify thatTf(x) for a.e. x∈𝒳 satisfies the regularity condition (ii) of
Definition 2.8. In what follows, we fix x′ near x and put δ≡d(x,x′).
We first remark that for a.e. x,x′∈𝒳 such that (1/160)(r+R)<δ≤(1/2)(r+R),
by the size condition of Tf(x) and Lemma 2.1(iii), we obtain(2.138)|Tf(x)−Tf(x′)|≲1Vr(x1)+V(x1,x)(rr+d(x,x1))γ+1Vr(x1)+V(x1,x′)(rr+d(x′,x1))γ≲(d(x,x′)r+d(x,x1))β1Vr(x1)+V(x1,x)(rr+d(x,x1))γ,which is the desired estimate.
In what follows, we only need to estimate |Tf(x)−Tf(x′)| for a.e. x,x′∈𝒳 such that δ≤(1/160)(r+R) by considering two cases. Recall that  R≡d(x,x1).
Case 1  (R<10r).
In this case, we divide the ballB(x1,10r) into the union of annuli Bm1={x∈𝒳:10r/(m1+1)≤d(x,x1)<10r/m1},
where m1∈ℕ.
For any fixed x0∈Bm1 and m2∈ℕ,
we put um1,m2(y)=θ(20m1m2d(x0,y)/(m1+10)r) and let wm1,m2(y)=1−θm1,m2(y) for all y∈𝒳. By Lemmas 2.15 and 2.20, for a.e. x∈B(x0,(m1+10)r/40m1m2),(2.139)Tf(x)=∫𝒳K(x,y)[f(y)−f(x)]um1,m2(y)dμ(y)+[∫𝒳K(x,y)f(y)wm1,m2(y)dμ(y)+f(x)T(um1,m2)(x)]≡Γ1(x)+Γ2(x),(2.140)|T(um1,m2)(x)|≲CT+∥T∥WBP(β),(2.141)T(um1,m2)(x)=Cm1,m2,r−∫𝒳[K(x,y)−K(x0,y)]wm1,m2(y)dμ(y),where Cm1,m2,r is a constant independent of x.
For any(2.142)x∈B(x0,(m1+11)r160(m1+1)(m2+1)),we then consider x′ in the annulus (r+R)/160(m2+1)<d(x,x′)≤(r+R)/160m2.
It is easy to check x′∈B(x0,(m1+10)r/40m1m2).
Assume that x,x′ both satisfy (2.139), (2.140), and (2.141). Notice that um1,m2(y)≠0 implies that d(y,x0)≤(m1+10)r/10m1m2,
and hence d(x,y)≤d(x,x0)+d(x0,y)<17(m1+11)r/160m1m2<68d(x,x′)≡68δ and  d(x′,y)≤69δ.
Therefore, the size condition on K,
the regularity of f,
and Lemma 2.1(i) show that(2.143)|Γ1(x)|=|∫𝒳K(x,y)[f(y)−f(x)]um1,m2(y)dμ(y)|≤∫d(x,y)≤68δ|K(x,y)||f(y)−f(x)|dμ(y)≲(d(x,x′)r+d(x,x1))β1Vr(x1)V(x1,x)(rr+d(x1,x))γ,and a similar estimate also
holds for Γ1(x′).
Thus, we clearly obtain the desired estimate for Γ1(x)−Γ1(x′).
Now by (2.141), we have(2.144)Γ2(x)−Γ2(x′)=∫𝒳[K(x,y)−K(x′,y)][f(y)−f(x)]wm1,m2(y)dμ(y)+f(x){T(um1,m2)(x)+∫𝒳[K(x,y)−K(x′,y)]wm1,m2(y)dμ(y)}−f(x′)T(um1,m2)(x′)=∫𝒳[K(x,y)−K(x′,y)][f(y)−f(x)]wm1,m2(y)dμ(y)+[f(x)−f(x′)]T(um1,m2)(x′)≡Γ2,1+Γ2,2.The estimate (2.140) and the
regularity of f then show that(2.145)|Γ2,2|≲|f(x)−f(x′)|≲(d(x,x′)r+d(x,x1))β1Vr(x1)+V(x1,x)(rr+d(x1,x))γ.Notice that wm1,m2(y)≠0 implies that d(x,y)>6δ.
The regularity (2.49) on K,
the regularity and the size condition of f,
and Lemma 2.1(i) together with the fact that β<ϵ give(2.146)|Γ2,1|≲∫d(x,y)>6δ,d(x,y)≤(r+d(x,x1))/2d(x,x′)ϵV(x,y)d(x,y)ϵ(d(x,y)r+d(x,x1))β1Vr(x1)+V(x1,x)(rr+d(x1,x))γdμ(y)+∫d(x,y)>6δ,d(x,y)>(r+d(x,x1))/2d(x,x′)ϵV(x,y)d(x,y)ϵ×[1Vr(x1)+V(x1,x)(rr+d(x1,x))γ+1Vr(x1)+V(x1,y)(rr+d(x1,y))γ]dμ(y)≲(d(x,x′)r+d(x,x1))β1Vr(x1)+V(x1,x)(rr+d(x1,x))γ,where, in the last inequality,
we used that r+d(x1,x)≤11r≲r+d(x1,y) and(2.147)1Vr(x1)+V(x1,y)≤1Vr(x1)≲1Vr(x1)+V(x1,x),which follows from V(x1,x)≤μ(B(x1,10r))≲Vr(x1), by (1.2). Thus, for any x0∈Bm1,
and a.e. x∈B(x0,(m1+11)r/160(m1+1)(m2+1)) and a.e. x′∈𝒳 satisfying (r+R)/160(m2+1)<d(x,x′)≤(r+R)/160m2, Tf(x)−Tf(x′) has the desired regularity with constants
independent of m1 and m2.
Again, by the Besicovitch covering lemma, we further see that for a.e. x∈Bm1 and a.e. x′∈𝒳 satisfying (r+R)/160(m2+1)<d(x,x′)≤(r+R)/160m2, Tf(x)−Tf(x′) has the desired regularity with constants
independent of m1 and m2,
which implies that there exists a measurable set 𝒳1 such that μ(𝒳1)=0 and for all x∈B(x1,10r)∖𝒳1 and all x′∈𝒳∖𝒳1 with d(x,x′)≤(1/160)(r+R), Tf(x)−Tf(x′) has the desired regularity.Case 2  (R≥10r).
In this case, we renameBm in Case 2 of the proof for the size condition
of Tf by B˜m1.
Namely, for m1∈ℕ,
we consider annuli(2.148)B˜m1=B(x1,10(m1+1)r)∖B(x1,10m1r).For any fixed x˜0∈B˜m1,
we define Im1, Jm1,
and Lm1 in the same way as in Case 1 with m and x0 replaced, respectively, by m1 and x˜0 here. Let f1(y)=f(y)Im1(y), f2(y)=f(y)Jm1(y) and f3(y)=f(y)Lm1(y) for all y∈𝒳. Then the estimates (2.119) and (2.130) still hold for f1, f2,
and f3 here, when x∈B˜m1.
Now, form2∈ℕ,
we put u˜m1,m2(y)=θ(20m2d(x˜0,y)/(10m1+11)r) and let w˜m1,m2(y)=1−θ˜m1,m2(y) for all y∈𝒳. By Lemmas 2.15 and 2.20, for a.e. x∈B(x˜0,(10m1+11)r/40m2),
(2.149)Tf1(x)=∫𝒳K(x,y)[f1(y)−f1(x)]u˜m1,m2(y)dμ(y)+[∫𝒳K(x,y)f1(y)w˜m1,m2(y)dμ(y)+f1(x)T(u˜m1,m2)(x)]≡Γ3(x)+Γ4(x),(2.150)|T(u˜m1,m2)(x)|≲CT+∥T∥WBP(β),(2.151)T(u˜m1,m2)(x)=C˜m1,m2,r−∫𝒳[K(x,y)−K(x˜0,y)]w˜m1,m2(y)dμ(y),where C˜m1,m2,r is a constant independent of x.
For any x∈B(x˜0,(10m1+1)r/160(m2+1)),
we then consider x′ in the annulus (r+R)/160(m2+1)<d(x,x′)≤(r+R)/160m2.
It is easy to check x′∈B(x˜0,(10m1+11)r/40m2).
We also restrict x∈B˜m1.
Assume that x,x′ both satisfy (2.149), (2.150), and (2.151).
Notice that u˜m1,m2(y)≠0 implies that d(y,x˜0)≤(10m1+11)r/10m2,
which implies d(x,y)≤d(x,x˜0)+d(x˜0,y)<177(10m1+1)r/160m2<354d(x,x′)≡354δ and d(x′,y)≤355δ.
Therefore, the size condition on K, (2.119), and Lemma 2.1(i)  show that(2.152)|Γ3(x)|≲∫d(x,y)≤354δ1V(x,y)1VR(x1)d(x,y)βRβ(rR)γdμ(y)≲1VR(x1)(d(x,x′)R)β(rR)γ,and a similar estimate also
holds for Γ4(x′).
Thus, we obtain the desired estimate for Γ3(x)−Γ4(x′).
Now by (2.151) and some computations similar to Γ2(x)−Γ2(x′),
we have(2.153)Γ4(x)−Γ4(x′)=∫𝒳[K(x,y)−K(x′,y)][f1(y)−f1(x)]w˜m1,m2(y)dμ(y)+[f1(x)−f1(x′)]T(u˜m1,m2)(x′)≡Γ4,1+Γ4,2.The estimate (2.150) together with (2.119) then yields(2.154)|Γ4,2|≲|f1(x)−f1(x′)|≲1VR(x1)(d(x,x′)R)β(rR)γ.Notice that w˜m1,m2(y)≠0 implies d(x,y)>7δ.
Then by the regularity (2.49) on K,
(2.119), and Lemma 2.1(i) together with β<ϵ,
we further have(2.155)|Γ4,1|≲∫d(x,y)>7δd(x,x′)ϵV(x,y)d(x,y)ϵ1VR(x1)(d(x,y)R)β(rR)γdμ(y)≲1VR(x1)(d(x,x′)R)β(rR)γ.Thus, for any x˜0∈B˜m1,
and a.e. x∈B(x˜0,(10m1+1)r/160(m2+1))∩B˜m1 and a.e. x′∈𝒳 satisfying (r+R)/160(m2+1)<d(x,x′)≤(r+R)/160m2, Tf1(x)−Tf1(x′) has the desired regularity with constants
independent of m1 and m2.
Notice thatf2(x)=0=f2(x′).
We write(2.156)|Tf2(x)−Tf2(x′)|=|∫𝒳[K(x,y)−K(x′,y)]f2(y)dμ(y)|≤|∫𝒳{[K(x,y)−K(x′,y)]−[K(x,x1)−K(x′,x1)]}f2(y)dμ(y)|+|K(x,x1)−K(x′,x1)||∫𝒳f2(y)dμ(y)|≡Γ5+Γ6.Since d(x,x′)≤(11/1600)R,
the regularity (2.49) on K,
(2.130), and β<ϵ prove(2.157)Γ6≲d(x,x′)ϵV(x1,x)d(x1,x)ϵ(rR)γ≲1VR(x1)(d(x,x′)R)β(rR)γ.Notice that f2(y)≠0 implies that d(x1,y)≤R/4.
Then the regularity (iii) on K together with the size condition of f,
the fact that γ<ϵ and Lemma 2.1(i) yields that(2.158)Γ5≲1VR(x1)∫d(x1,y)≤R/4d(x,x′)ϵd(x1,y)ϵd(x1,x)2ϵ1Vr(x1)+V(x1,y)(rr+d(x1,y))γdμ(y)≲1VR(x1)(d(x,x′)R)β(rR)γ.Thus, we have also obtained the
desired regularity for Tf2(x)−Tf2(x′) for all x∈B˜m1 and d(x,x′)≤(1/160)(r+R).
Notice thatf3(y)≠0 implies that d(x1,y)>R/16.
Moreover, if(2.159)x∈B(x˜0,(10m1+1)r160(m2+1))∩B˜m1,then f3(y)≠0 also implies that d(x,y)>7R/40.
Since δ≤(11/1600)R,
we then have δ≤(11/280)d(x,y).
Then the regularity (2.49) on K,
the size condition of f,
and Lemma 2.1(i) show that(2.160)|Tf3(x)−Tf3(x′)|≲∫d(x1,y)>R/16,d(x,y)>7R/40|K(x,y)−K(x′,y)||f3(y)|dμ(y)≲∫d(x1,y)>R/16,d(x,y)>7R/40d(x,x′)ϵV(x,y)d(x,y)ϵ1Vr(x1)+V(x1,y)rγd(x1,y)γdμ(y)≲1VR(x1)(d(x,x′)R)β(rR)γ,where in the last step, we used
the fact that β<ϵ and d(x,x′)≤(11/1600)R,
and in the third-to-last inequality, we used the fact that(2.161)VR(x1)~V(x,x1)≤μ(B(x,40d(x,y)7))≲V(x,y).Thus, for all x∈B(x˜0,(10m1+1)r/160(m2+1))∩B˜m1 and d(x,x′)≤(1/160)(r+R),
we obtain the desired regularity for Tf3(x)−Tf3(x′).
Summarizing the above estimates, we see that for anyx˜0∈B˜m1,
and a.e. x∈B˜m1∩B(x˜0,(10m1+1)r/160(m2+1)) and a.e. x′∈𝒳 satisfying (r+R)/160(m2+1)<d(x,x′)≤(r+R)/160m2,Tf(x)−Tf(x′) has the desired regularity with constants
independent of m1 and m2.
By the Besicovitch covering lemma again, similarly to Case 1, we obtain that
there exists a measurable set 𝒳2 such that μ(𝒳2)=0,
and for all x∈𝒳∖(B(x1,10r)∪𝒳2) and all x′∈𝒳∖𝒳2 with d(x,x′)≤(1/160)(r+R), Tf(x)−Tf(x′) has the desired regularity.
Thus, there exists a measurable set𝒳3 such that for all x,x′∈𝒳∖𝒳3 with d(x,x′)≤(1/2)(r+d(x,x1)), Tf(x)−Tf(x′) satisfies the regularity condition (ii) of Definition 2.8, which implies that for all x,x′∈𝒳∖𝒳3 with d(x,x′)≤r/2,(2.162)|Tf(x)−Tf(x′)|≲1Vr(x1)rβd(x,x′)β.Now, for any x∈𝒳,
we choose {xn}n∈ℕ⊂(𝒳∖𝒳3) such that d(xn,x)→0 as n→∞.
Then {Tf(xn)}n∈ℕ is a Cauchy sequence in ℂ.
We then define g(x)=limn→∞Tf(xn).
It is easy to show that g is a well-defined continuous function lying in 𝒢(x1,r,β,γ),
which agrees with Tf almost everywhere. We may thus choose g as a representative of Tf,
for which we then have(2.163)∥Tf∥𝒢(x1,r,β,γ)≲(CT+∥T∥WBP(β))∥f∥𝒢(x1,r,β,γ).
This completes the proof  of Theorem 2.18.Remark 2.21.
It was proved in [35] that the condition (iii) of
Theorem 2.18 is also necessary for an operator T with distributional kernel K to be bounded from 𝒢°(x1,r,β,γ) to 𝒢(x1,r,β,γ) for all x1∈𝒳 and r>0;
see [35, Theorem 4] or
[31].By Remark2.14(iii), we immediately obtain the
following conclusion, which is convenient in applications.Corollary 2.22.
Letϵ∈(0,1], β∈(0,ϵ), and let T be as in Proposition 2.12, with the
distributional kernel K satisfying (i), (ii), and (iii) of Theorem 2.18. If T is bounded on Lp(𝒳) for a certain p∈(1,∞) and T(1)=0 in (C°bβ(𝒳))′,
then T extends to a bounded linear operator from 𝒢°(x1,r,β,γ) to 𝒢(x1,r,β,γ) for all x1∈𝒳, r>0 and γ∈(0,ϵ).
Moreover, there exists a constant Cβ,γ,C0>0 such that for all f∈𝒢°(x1,r,β,γ) with any x1∈𝒳,
any r>0,
and any γ∈(0,ϵ),(2.164)∥Tf∥𝒢(x1,r,β,γ)≤Cβ,γ,C0(CT+∥T∥Lp(𝒳)→Lp(𝒳))∥f∥𝒢(x1,r,β,γ).By an argument similar to the proof of Lemma2.20, we
can easily obtain the following result, which is of independent interest; see also [69, Lemma 2], [71, Lemma 1, page 119] for ℝn, and [34, Lemma (3.12)] for Ahlfors 1-regular metric measure spaces. Moreover, instead of T∈SWBP(β) in [34, Lemma (3.12)], we only need that T∈WBP(β).Corollary 2.23.
LetT be as in Lemma 2.20. Then there exists a
constant C>0 such that for all f∈C˙bβ(𝒳),(2.165)∥Tf∥L∞(𝒳)≤C(CT+∥T∥WBP(β))[diam(suppf)]β∥f∥C˙β(𝒳).Proof.
Assume thatsuppf⊂B(x0,r) for some x0∈𝒳 and r>0.
Let θx0,10r be defined in the same way as in Lemma 2.15. An
argument similar to the proof of Lemma 2.20 yields that when d(x,x0)<5r,(2.166)Tf(x)=∫𝒳K(x,y)[f(y)−f(x)]θx0,10r(y)dμ(y)+f(x)∫𝒳K(x,y)θx0,10r(y)dμ(y).From this, Lemma 2.1(i), and
Lemma 2.15(ii) together with (2.55), it follows that(2.167)|Tf(x)|≲CT∥f∥C˙β(𝒳)∫d(x,y)<25r1V(x,y)d(x,y)βdμ(y)+∥f∥L∞(𝒳)(CT+∥T∥WBP(β))≲(CT+∥T∥WBP(β))rβ∥f∥C˙β(𝒳).If d(x,x0)≥5r and d(y,x0)<r,
then d(y,x)>4d(x,x0)/5,
and by Lemma 2.1(i) and (2.55), we also have(2.168)|Tf(x)|≤CT∫d(x,y)>4d(x,x0)/51V(x,y)|f(y)|dμ(y)≲CT1V(x0,x)∥f∥L∞(𝒳)Vr(x0)≲CT[diam(suppf)]β∥f∥C˙β(𝒳),which completes the proof of
Corollary 2.23.From Theorem2.18, it is easy to deduce the following
result which is convenient in applications.Corollary 2.24.
LetT be as in Theorem 2.18 or Corollary 2.22. If ϵ˜∈(0,ϵ),
then T is bounded from 𝒢°0ϵ˜(x1,r,β,γ) to 𝒢0ϵ˜(x1,r,β,γ) for all x1∈𝒳, r>0, and 0<β,γ<ϵ˜.
Moreover, there exists a constant C>0 such that for all f∈𝒢°0ϵ˜(x1,r,β,γ) with any x1∈𝒳 and any r>0,(2.169)∥Tf∥𝒢0ϵ˜(x1,r,β,γ)≤Cβ,γ,C0(CT+∥T∥WBP(β))∥f∥𝒢0ϵ˜(x1,r,β,γ)(or resp., ∥Tf∥𝒢0ϵ˜(x1,r,β,γ)≤Cβ,γ,C0(CT+∥T∥Lp(𝒳)→Lp(𝒳))∥f∥𝒢0ϵ˜(x1,r,β,γ)).Proof.
Letf∈𝒢°0ϵ˜(x1,r,β,γ).
By the definition of 𝒢°0ϵ˜(x1,r,β,γ), there exists(2.170){fn}n=1∞⊂𝒢°(x1,r,ϵ˜,ϵ˜)such that(2.171)limn→∞∥fn−f∥𝒢(x1,r,β,γ)=0.By Theorem 2.18 or Corollary 2.22,
we have Tfn∈𝒢(x1,r,ϵ˜,ϵ˜) and(2.172)∥Tfn−Tf∥𝒢(x1,r,β,γ)≲∥f−fn∥𝒢(x1,r,β,γ)→0,as n→∞.
Thus, Tf∈𝒢0ϵ˜(x1,r,β,γ) and(2.173)∥Tf∥𝒢0ϵ˜(x1,r,β,γ)=∥Tf∥𝒢(x1,r,β,γ)≲∥f∥𝒢(x1,r,β,γ)=∥f∥𝒢0ϵ˜(x1,r,β,γ),which completes the proof of
Corollary 2.24.


## 2.3. Boundedness of Singular Integrals on Spaces of Test Functions

In this subsection, we establish the boundedness on𝒢(x1,r,β,γ) of singular integrals. Since the functions in 𝒢(x1,r,β,γ) may have nonvanishing integral, this requires,
as usual, some extra size decay conditions on the integral kernels; see, for
example, [71, page 123].The following result is an inhomogeneous variant of
Proposition2.12.Proposition 2.25.
Letβ∈(0,1], σ>0, r0>0, and let T be as in Proposition 2.12 with the
distributional kernel K satisfying the following extra size condition
that for all x,y∈𝒳 with d(x,y)≥r0,(2.174)|K(x,y)|≤CT1V(x,y)r0σd(x,y)σ.Then T can be extended to a continuous linear operator
from Cβ(𝒳) to (Cbβ(𝒳))′.Proof.
Letf∈Cβ(𝒳) and g∈Cbβ(𝒳),
and suppose suppg⊂B(x0,r) for some x0∈𝒳 and r>0.
Choose ψ∈Cbβ(𝒳) such that ψ(x)=1 when x∈B(x0,2max{r,r0}) and ψ(x)=0 when x∉B(x0,4max{r,r0}).
It is easy to see that ψf∈Cbβ(𝒳),
which implies that 〈T(ψf),g〉 is well defined. On the other hand, we define(2.175)〈T((1−ψ)f),g〉=∬𝒳×𝒳K(x,y)(1−ψ(y))f(y)g(x)dμ(y)dμ(x).By the size condition (2.174) and
Lemma 2.1(i), it is easy to check that the right-hand side of the above
equality is finite; furthermore, if f has also bounded support, this coincides with
(2.48). Moreover, it is easy to verify that 〈T(ψf),g〉+〈T((1−ψ)f),g〉 is independent of choice of ψ.
Thus we can define Tf by(2.176)〈Tf,g〉=〈T(ψf),g〉+〈T((1−ψ)f),g〉,so that Tf∈(Cbβ(𝒳))′.
It is clear that then T is continuous, which completes the proof of
Proposition 2.25.We now establish an inhomogeneous variant of Lemma2.15. Here, in contrast to Lemma 2.15, we do not need T∈WBP(β).Lemma 2.26.
Letθ be as in Lemma 2.15. For any fixed z∈𝒳 and r>0,
let θz,max{r,r0} and ωz,max{r,r0} be defined as in Lemma 2.15. Let T be as in Proposition 2.25. If, for a certain β∈(0,ϵ], T(1)∈(Cbβ(𝒳))′ is a constant CT(1),
then there exists a constant C>0 such that
(i)
for allx∈B(z,max{r,r0}/2), |T(ωz,max{r,r0})(x)|≤CCT;(ii)
the restriction of the linear functionalT(θz,max{r,r0})+T(ωz,max{r,r0})∈(Cbβ(𝒳))′ to the ball B(z,max{r,r0}/2) is a constant; namely, for all x∈B(z,max{r,r0}/2),(2.177)T(θz,max{r,r0})(x)+T(ωz,max{r,r0})(x)=CT(1);(iii)
for allx∈B(z,max{r,r0}/2), |T(θz,max{r,r0})(x)|≤C(CT+|CT(1)|).Proof.
By the definitions ofθz,max{r,r0} and ωz,max{r,r0},
we know that(2.178)ωz,max{r,r0}(y)≠0implies that d(y,z)>max{r,r0},
which together with d(x,z)<max{r,r0}/2 shows that d(x,y)>max{r,r0}/2.
Therefore, by (2.174) and Lemma 2.1(i), for all x∈𝒳 with d(x,z)<max{r,r0}/2,
we have(2.179)|T(ωz,max{r,r0})(x)|=|∫d(x,y)>max{r,r0}/2K(x,y)ωz,max{r,r0}(y)dμ(y)|≤CT∫d(x,y)>max{r,r0}/21V(x,y)r0σd(x,y)σdμ(y)≤CCT,which is (i).
To see (ii), for anyf∈𝒢b(β,γ) with suppf⊂B(z,max{r,r0}/2),
by the definition of T(1)=CT(1) in (Cbβ(𝒳))′,
we have(2.180)CT(1)∫𝒳f(x)dμ(x)=〈T(θz,max{r,r0}),f〉+〈T(ωz,max{r,r0}),f〉,which together with Corollary 2.11(ii) gives the conclusion (ii) of this lemma.
The conclusion (iii) can be deduced from the
conclusion (i) and the conclusion (ii), which completes the proof of Lemma2.26.We recall the notion of the spacebmo(𝒳),
which, when 𝒳=ℝn and μ is the n-dimensional Lebesgue measure, was first
introduced by Goldberg [73]. A variant in the setting of spaces of homogeneous
type can be found in [74].Definition 2.27.
Let1≤q<∞.
The spacebmoq(𝒳) is defined to be the set of all f∈Llocq(𝒳) such that(2.181)∥f∥bmoq(𝒳)=supx∈𝒳,0<r<1{1μ(B(x,r))∫B(x,r)|f(y)−mB(x,r)(f)|qdμ(y)}1/q+supx∈𝒳,r≥1{1μ(B(x,r))∫B(x,r)|f(y)|qdμ(y)}1/qis finite. When q=1,
one denotes
bmo1(𝒳) simply by bmo(𝒳).It was proved in [74] that for any 1≤q1,q2<∞, bmoq1(𝒳)=bmoq2(𝒳) with equivalent norms.Remark 2.28.
From Corollary2.11(ii), Proposition 5.46, Theorems 5.44(i) and 6.28 below, it is easy to see that T(1)=CT(1) in (Cbβ(𝒳))′ if and only if T(1)=CT(1) in bmo(𝒳).We can now present the inhomogeneous variant of
Theorem2.18 as follows.Theorem 2.29.
Letϵ∈(0,1], σ>0, r0>0, and let T be as in Proposition 2.25 with the
distributional kernel K also satisfying the conditions (i) through
(iii) in Theorem 2.18 and the following additional regularity condition: for all x,x′,y∈𝒳 with d(x,x′)≤d(x,y)/2 and d(x,y)≥r0,(2.182)|K(x,y)−K(x′,y)|≤CT1V(x,y)(d(x,x′)d(x,y))ϵ(r0d(x,y))σ.If T∈WBP(β) for a certain β∈(0,ϵ) and T(1)=CT(1) in (Cbβ(𝒳))′,
then T extends to a bounded linear operator from 𝒢(x1,r0,β,γ) to itself for any x1∈𝒳 and any γ∈(0,σ]∩(0,ϵ).
Moreover, there is a constant Cβ,γ,σ,C0>0 such that for all f∈𝒢(x1,r0,β,γ) with any x1∈𝒳 and any γ∈(0,σ]∩(0,ϵ),(2.183)∥Tf∥𝒢(x1,r0,β,γ)≤Cβ,γ,σ,C0(CT+|CT(1)|)∥f∥𝒢(x1,r0,β,γ).Proof.
We prove Theorem2.29 by essentially following the same argument as the
proof of Theorem 2.18, and we only give an outline. We use the notations as in
the proof of Theorem 2.18. By Corollary 2.11(ii), we only need to prove Theorem
2.29 for 𝒢b(x1,r0,β,γ).
Letf∈𝒢b(x1,r0,β,γ).
We first verify that Tf(x) for a.e. x∈𝒳 satisfies Definition 2.8(i) by considering
two cases.
Case 1  (d(x,x1)<10r0).
In this case, letθx1,20r0 and ωx1,20r0 be defined in the same way as in Lemma 2.15.
Instead of Lemma 2.15(ii) by Lemma 2.26(iii), following the procedure of Lemma 2.20 (here we need T∈WBP(β)), for a.e. x0∈B(x110r0),
we have(2.184)Tf(x)=∫𝒳K(x,y)[f(y)−f(x)]θx1,20r0(y)dμ(y)+∫𝒳K(x,y)f(y)ωx1,20r0(y)dμ(y)+f(x)T(θx1,20r0)(x).Using Lemma 2.26(iii) to replace
Lemma 2.15(ii) and following the same proof as that of Theorem 2.18 give the
desired estimate of this case.Case 2  (d(x,x1)≡R≥10r0).
In this case, we use all the notation as in the proof of Theorem2.18, but with r replaced by r0. The estimates (2.118), (2.119), (2.128), and (2.129) with r replaced by r0 hold in the current setting. Instead of
(2.130), by Lemma 2.1(ii), we now have(2.185)|∫𝒳f2(y)dμ(y)|≲1.
Replacing Lemma2.15(ii) by Lemma 2.26(iii) and using T∈WBP(β),
by following the proof of Lemma 2.20, for a.e. x∈Bm,
we have(2.186)Tf1(x)=∫𝒳K(x,y)[f1(y)−f1(x)]Um,r0(y)dμ(y)+f1(x)T(Um,r0)(x).Then Lemma 2.26(iii) together
with an argument similar to the proof of Theorem 2.18 gives the desired estimate
for Tf1(x) in this case.
Sincex∉suppf2,
we can write(2.187)Tf2(x)=∫𝒳[K(x,y)−K(x,x1)]f2(y)dμ(y)+K(x,x1)∫𝒳f2(y)dμ(y)≡Z8(x)+Z9(x).The estimate for Z8(x) is as in the proof of Theorem 2.18. Since R≥10r0 and γ≤σ,
by (2.174) and (2.185), we have(2.188)|Z9(x)|≲1VR(x1)(r0R)σ≲1VR(x1)(r0R)γ,which is the desired estimate.
The estimate forTf3(x) in this case is also similar to the proof of
Theorem 2.18. Thus, T(f)(x) for a.e. x∈𝒳 satisfies Definition 2.2(i).
To verify thatT(f)(x) for a.e. x∈𝒳 satisfies Definition 2.2(ii), similarly to
the proof of Theorem 2.18, we can always assume that d(x,x′)≤(1/160)(r0+d(x,x1)),
and we only need to consider the following two cases.
Case 1  (R<10r0).
In this case, if we replace Lemma2.15(i) and (ii), respectively, by Lemma 2.26(ii) and (iii), the same proof as in the proof of Theorem 2.18 gives the desired
estimates.Case 2  (R≥10r0).
In this case, the same argument as in the proof of Theorem2.18 together with
Lemma 2.15 replaced by Lemma 2.26 yields the desired estimates for Tf1(x) and Tf3(x).
Noticing that f2(x)=0=f2(x′),
we now estimate Tf2 by writing(2.189)|Tf2(x)−Tf2(x′)|=|∫𝒳[K(x,y)−K(x′,y)]f2(y)dμ(y)|≤|∫𝒳{[K(x,y)−K(x′,y)]−[K(x,x1)−K(x′,x1)]}f2(y)dμ(y)|+|K(x,x1)−K(x′,x1)||∫𝒳f2(y)dμ(y)|≡Γ5+Γ6.The estimate for Γ5 is as in the proof of Theorem 2.18. To estimate Γ6,
by (2.182) and (2.185) together with γ≤σ,
we have(2.190)Γ6≲1V(x,x1)(d(x,x′)d(x,x1))ϵ(r0d(x,x1))σ≲1VR(x1)(d(x,x′)R)β(r0R)γ,which is the desired estimate.
Then an extension via the Besicovitch covering lemma as in the proof of Theorem
2.18 completes the proof of  Theorem 2.29.Remark 2.30.
A regularity condition similar to (2.182) also appears in [71, page 123]. This additional
regularity assumption is used only in the estimate of Γ6.
Instead of (2.182) by requiring γ∈(0,σ(ϵ−β)/ϵ],
we can obtain a similar conclusion. In fact, since d(x′,x1)≥d(x,x1)−d(x,x′)≥189r0/20≥r0 and d(x′,x1)≥189R/200≥R/2,
by (2.174) and (2.185), we have(2.191)Γ6≲1V(x,x1)(r0d(x,x1))σ+1V(x′,x1)(r0d(x′,x1))σ≲1V(x,x1)(r0d(x,x1))σ.On the other hand, by (2.49) and
(2.185), we also have Γ6≲(1/V(x,x1))(d(x,x′)/d(x,x1))ϵ. Combining both estimates yields that(2.192)Γ6≲1V(x,x1)(d(x,x′)d(x,x1))β(r0d(x,x1))σ(1−β/ϵ)≲1VR(x1)(d(x,x′)R)β(r0R)γ,which is the desired estimate.By Remark2.14(iii) again, we can obtain the following
conclusion which is convenient in  applications.Corollary 2.31.
Letϵ∈(0,1], σ>0, r0>0, and let T be as in Proposition 2.25 with the
distributional kernel K also satisfying Theorem 2.18(i) through (iii)
and (2.182). If T is bounded on Lp(𝒳) for a certain p∈(1,∞) and T(1)=CT(1) in (Cbβ(𝒳))′,
then T extends to a bounded linear operator from 𝒢(x1,r0,β,γ) to itself for any x1∈𝒳,
any β∈(0,ϵ), and any γ∈(0,σ]∩(0,ϵ).
Moreover, there is a constant Cβ,γ,σ,C0>0 such that for all f∈𝒢(x1,r0,β,γ) with any x1∈𝒳 and any γ∈(0,σ]∩(0,ϵ),(2.193)∥Tf∥𝒢(x1,r0,β,γ)≤Cβ,γ,σ,C0(CT+|CT(1)|)∥f∥𝒢(x1,r0,β,γ).The proof of the following corollary is similar to
that of Corollary2.24. We omit the details.Corollary 2.32.
LetT be as in Theorem 2.29 or Corollary 2.31. If ϵ˜∈(0,ϵ),
then T is bounded from 𝒢0ϵ˜(x1,r0,β,γ) to 𝒢0ϵ˜(x1,r0,β,γ) for all x1,β,γ as in Theorem 2.29. Moreover, there exists a
constant Cβ,γ,σ,C0>0 such that for all f∈𝒢0ϵ˜(x1,r0,β,γ)  with any x1∈𝒳,(2.194)∥Tf∥𝒢(x1,r0,β,γ)≤Cβ,γ,σ,C0(CT+|CT(1)|)∥f∥𝒢(x1,r0,β,γ).


## 3. Continuous Calderón Reproducing Formulae

From now on till the end of this paper, we will always
assume that𝒳 is an RD-space. In this section, using
Corollaries 2.22 and 2.31, we establish homogeneous and inhomogeneous
Calderón reproducing formulae.


### 3.1. Homogeneous Continuous Calderón Reproducing Formulae

In this subsection, we always assume thatdiam(𝒳)=∞.Proposition 3.1.
Let{Sk}k∈ℤ be an (ϵ1,ϵ2,ϵ3)-ATI with ϵ1∈(0,1],ϵ2>0, and ϵ3>0 and let Skt be the adjoint operator of Sk for any k∈ℤ.
Then the following hold.
(i)
Forp∈(1,∞) and any f∈Lp(𝒳), ∥Skf∥Lp(𝒳)→0 when k→−∞.(ii)
LetDk=Sk−Sk−1 for k∈ℤ.
Then ∑k=−∞∞Dk=I in Lp(𝒳) for p∈(1,∞),
where I is the identity on Lp(𝒳).(iii)
Properties (i) and (ii) also hold whenSk is replaced by Skt.Proof.
Let1/p+1/q=1.
Then Definition 2.2(i), Hölder's inequality, (1.2) and (1.3) together with μ(𝒳)=∞ yield that
(3.1)|Skf(x)|≲1V2−k(x)∫d(x,y)<2−k|f(y)|dμ(y)+∫d(x,y)≥2−k1V(x,y)|f(y)|dμ(y)≲{1V2−k(x)1/p+[∑l=0∞∫2l2−k≤d(x,y)<2l+12−k1V(x,y)qdμ(y)]1/q}∥f∥Lp(𝒳)≲1V2−k(x)1/p∥f∥Lp(𝒳)→0,
as k→−∞.
This together with Proposition 2.7(ii) and Lebesgue's dominated convergence
theorem implies that ∥Skf∥Lp(𝒳)→0 when k→−∞.
Property (ii) is a simple corollary of Property (i)
together with Proposition2.7(iv). Property (iii) follows by symmetry, which
completes the proof of Proposition 3.1.Before we establish the continuous Calderón reproducing
formulae, we need a technical lemma. We recall that we denotemin{a,b} by a⋀b for any a,b∈ℝ.Lemma 3.2.
Letϵ1∈(0,1], ϵ2>0, ϵ3>0, {Sk}k∈ℤ,
and let {Ek}k∈ℤ be two (ϵ1,ϵ2,ϵ3)-ATIs. Let Pk=Sk−Sk−1 and Qk=Ek−Ek−1 for k∈ℤ.
Then for any ϵ1′∈(0,ϵ1⋀ϵ2),
there exist positive constants C (depending on ϵ1′, ϵ1, ϵ2,
and ϵ3), δ (depending on ϵ1′, ϵ1, and ϵ2), and σ (depending on ϵ1′, ϵ1, ϵ2,
and ϵ3) such that PlQk(x,y),
the kernel of PlQk,
satisfies the following estimates that for all x,y∈𝒳 and all k,l∈ℤ,
(3.2)|PlQk(x,y)|≤C2−|k−l|ϵ1′1V2−(k⋀l)(x)+V2−(k⋀l)(y)+V(x,y)2−(k⋀l)ϵ2(2−(k⋀l)+d(x,y))ϵ2;
for d(y,y′)≤(1/4)d(x,y) and all k,l∈ℤ,
(3.3)|PlQk(x,y)−PlQk(x,y′)|≤C2−|k−l|δ(d(y,y′)2−(l⋀k)+d(x,y))ϵ1′1V2−(k⋀l)(x)+V2−(k⋀l)(y)+V(x,y)2−(k⋀l)ϵ2(2−(k⋀l)+d(x,y))ϵ2;
for d(x,x′)≤(1/4)d(x,y) and all k,l∈ℤ,
(3.4)|PlQk(x,y)−PlQk(x′,y)|≤C2−|k−l|δ(d(x,x′)2−(l⋀k)+d(x,y))ϵ1′1V2−(k⋀l)(x)+V2−(k⋀l)(y)+V(x,y)2−(k⋀l)ϵ2(2−(k⋀l)+d(x,y))ϵ2;
and for d(x,x′)≤(1/8)d(x,y), d(y,y′)≤(1/8)d(x,y), and all k,l∈ℤ,
(3.5)|[PlQk(x,y)−PlQk(x′,y)]−[PlQk(x,y′)−PlQk(x′,y′)]|≤C2−|k−l|δ(d(x,x′)2−(l⋀k)+d(x,y))ϵ1′(d(y,y′)2−(l⋀k)+d(x,y))ϵ1′×1V2−(k⋀l)(x)+V2−(k⋀l)(y)+V(x,y)2−(k⋀l)σ(2−(k⋀l)+d(x,y))σ.Proof.
By symmetry, we may assume thatk≥l.
Noticing that(3.6)∫𝒳Qk(x,y)dμ(x)=0,
we can write
(3.7)|PlQk(x,y)|=|∫𝒳[Pl(x,z)−Pl(x,y)]Qk(z,y)dμ(z)|≤∫d(z,y)≤(2−l+d(x,y))/2|Pl(x,z)−Pl(x,y)||Qk(z,y)|dμ(z)+∫d(z,y)>(2−l+d(x,y))/2|Pl(x,z)||Qk(z,y)|dμ(z)+|Pl(x,y)|∫d(z,y)>(2−l+d(x,y))/2|Qk(z,y)|dμ(z)≡Z1+Z2+Z3.
The regularity of Pl and the size condition of Qk together with Lemma 2.1(i) show that for any ϵ1′∈(0,ϵ1⋀ϵ2),
(3.8)Z1≲2−kϵ1′1V2−l(x)+V2−l(y)+V(x,y)2−lϵ2(2−l+d(x,y))ϵ1+ϵ2×∫d(z,y)≤(2−l+d(x,y))/2d(z,y)ϵ1−ϵ1′1V(z,y)dμ(y)≲2−(k−l)ϵ1′1V2−l(x)+V2−l(y)+V(x,y)2−lϵ2(2−l+d(x,y))ϵ2.
The size conditions for Pl and Qk together with Lemma 2.1(i) prove that
(3.9)Z2≲1V2−l(x)2−kϵ2(2−l+d(x,y))ϵ1′∫d(z,y)>(2−l+d(x,y))/21V(z,y)1d(z,y)ϵ2−ϵ1′dμ(z)≲2−(k−l)ϵ21V2−l(x)2−lϵ2(2−l+d(x,y))ϵ2,Z2≲2−kϵ2(2−l+d(x,y))ϵ2∫d(z,y)>(2−l+d(x,y))/2|Pl(x,z)|1V(z,y)dμ(z)≲2−(k−l)ϵ21V2−l(y)+V(x,y)2−lϵ2(2−l+d(x,y))ϵ2,
where in the last step, we used
Proposition 2.7(i) and the fact that 1/V(z,y)≲1/(V2−l(y)+V(x,y)), by d(z,y)>(2−l+d(x,y))/2 together with (1.2). Thus,
(3.10)Z2≲2−(k−l)ϵ1′1V2−l(x)+V2−l(y)+V(x,y)2−lϵ2(2−l+d(x,y))ϵ2.
Similarly, by Lemma 2.1(v),(3.11)Z3≲1V2−l(x)+V2−l(y)+V(x,y)∫d(z,y)>(2−l+d(x,y))/21V2−k(y)+V(z,y)2−kϵ2(2−k+d(z,y))ϵ2dμ(z)≲2−(k−l)ϵ21V2−l(x)+V2−l(y)+V(x,y)2−lϵ2(2−l+d(x,y))ϵ2,
which completes the proof of
(3.2).
The proofs for (3.3) and (3.4) are similar and we only
verify (3.3). To this end, it suffices to prove
(3.12)|PlQk(x,y)−PlQk(x,y′)|≲(d(y,y′)2−l+d(x,y))ϵ11V2−l(x)+V2−l(y)+V(x,y)2−lϵ2(2−l+d(x,y))ϵ2
for d(y,y′)≤(1/4)d(x,y).
To see this, by (3.2), for any ϵ1′∈(0,ϵ1⋀ϵ2),
(3.13)|PlQk(x,y)−PlQk(x,y′)|≲2−(k−l)ϵ1′[1V2−l(x)+V2−l(y)+V(x,y)2−lϵ2(2−l+d(x,y))ϵ2+1V2−l(x)+V2−l(y′)+V(x,y′)2−lϵ2(2−l+d(x,y′))ϵ2].
The assumption that d(y,y′)≤(1/4)d(x,y) together with Lemma 2.1(iii) further yields that
(3.14)|PlQk(x,y)−PlQk(x,y′)|≲2−(k−l)ϵ1′1V2−l(x)+V2−l(y)+V(x,y)2−lϵ2(2−l+d(x,y))ϵ2.
Let σ∈(0,1).
Then the geometric mean between (3.12) and (3.14) gives
(3.15)|PlQk(x,y)−PlQk(x,y′)|=|PlQk(x,y)−PlQk(x,y′)|σ|PlQk(x,y)−PlQk(x,y′)|1−σ≲2−(k−l)ϵ1′(1−σ)(d(y,y′)2−l+d(x,y))σϵ11V2−l(x)+V2−l(y)+V(x,y)2−lϵ2(2−l+d(x,y))ϵ2,
which is (3.3).
We now verify (3.12). By (3.6), we can write
(3.16)|PlQk(x,y)−PlQk(x,y′)|=|∫𝒳[Pl(x,z)−Pl(x,y)][Qk(z,y)−Qk(z,y′)]dμ(z)|≤∑i=13∫Wi|Pl(x,z)−Pl(x,y)||Qk(z,y)−Qk(z,y′)|dμ(z)≡∑i=46Zi,
where
(3.17)W1={z∈𝒳:d(y,y′)≤2−l+d(x,y)4≤2−k+d(z,y)2},W2={z∈𝒳:d(y,y′)≤2−k+d(z,y)2≤2−l+d(x,y)4},W3={z∈𝒳:d(y,y′)≥2−k+d(z,y)2}.
The regularity forQk and the size condition for Pl together with the assumption k≥l implies that
(3.18)Z4≲∫W1[|Pl(x,z)|+1V2−l(x)+V2−l(y)+V(x,y)2−lϵ2(2−l+d(x,y))ϵ2]×d(y,y′)ϵ1(2−k+d(z,y))ϵ11V2−k(z)+V2−k(y)+V(z,y)2−kϵ2(2−k+d(z,y))ϵ2dμ(z)≡Z4,1+Z4,2.
We now claim that if k≥l and 2−l+d(x,y)≲2−k+d(z,y),
then for all x,y,z∈𝒳,
(3.19)1V2−k(y)+V(z,y)≲1V2−l(y)+V(x,y).
In fact, if 2−k≥d(z,y),
then 2−l+d(x,y)≲2−k+d(z,y)≲2−k,
and therefore V2−l(y)≲V2−k(y) and V(x,y)~V(y,x)≲V2−k(y).
Thus, (3.19) holds in this case. If 2−k<d(z,y),
then 2−l+d(x,y)≲d(z,y) and hence, V2−l(y)≲V(y,z)~V(z,y) and V(x,y)~V(y,x)≲V(z,y),
which verifies (3.19). Proposition 2.7(i) and the estimate (3.19) together with 2−l+d(x,y)≤2(2−k+d(z,y)) and k≥l imply that
(3.20)Z4,1≲d(y,y′)ϵ1(2−l+d(x,y))ϵ11V2−l(y)+V(x,y)2−lϵ2(2−l+d(x,y))ϵ2∫𝒳|Pl(x,z)|dμ(z)≲d(y,y′)ϵ1(2−l+d(x,y))ϵ11V2−l(y)+V(x,y)2−lϵ2(2−l+d(x,y))ϵ2,
and the size condition of Pl together with Lemma 2.1(i) yields that
(3.21)Z4,1≲d(y,y′)ϵ1(2−l+d(x,y))ϵ11V2−l(x){∫W1,d(z,y)≤2−k+∫W1,d(z,y)>2−k}1V2−k(y)+V(z,y)2−kϵ2(2−k+d(z,y))ϵ2dμ(z)≲d(y,y′)ϵ1(2−l+d(x,y))ϵ11V2−l(x)2−lϵ2(2−l+d(x,y))ϵ2.
Similarly, using Lemma 2.1(ii), we have
(3.22)Z4,2≲d(y,y′)ϵ1(2−l+d(x,y))ϵ11V2−l(x)+V2−l(y)+V(x,y)2−lϵ2(2−l+d(x,y))ϵ2×∫W11V2−k(y)+V(z,y)2−kϵ2(2−k+d(z,y))ϵ2dμ(z)≲d(y,y′)ϵ1(2−l+d(x,y))ϵ11V2−l(x)+V2−l(y)+V(x,y)2−lϵ2(2−l+d(x,y))ϵ2,
which completes the estimate for Z4.
The regularities ofPl and Qk and Lemma 2.1(ii) show that
(3.23)Z5≲∫W2d(y,z)ϵ1(2−l+d(x,y))ϵ11V2−l(x)+V2−l(y)+V(x,y)2−lϵ2(2−l+d(x,y))ϵ2×d(y,y′)ϵ1(2−k+d(y,z))ϵ11V2−k(y)+V2−k(z)+V(z,y)2−kϵ2(2−k+d(z,y))ϵ2dμ(z)≲d(y,y′)ϵ1(2−l+d(x,y))ϵ11V2−l(x)+V2−l(y)+V(x,y)2−lϵ2(2−l+d(x,y))ϵ2.From (1/4)d(x,y)≥d(y,y′)≥(1/2)(2−k+d(z,y)),
it follows that d(z,y)≤(1/2)d(x,y)≤(1/2)(2−l+d(x,y)). Then the regularity of Pl and the size condition of Qk together with Proposition 2.7(i) and d(y,z)≤2d(y,y′) prove that
(3.24)Z6≲1V2−l(x)+V2−l(y)+V(x,y)2−lϵ2(2−l+d(x,y))ϵ2×∫W3d(y,z)ϵ1(2−l+d(x,y))ϵ1[|Qk(z,y)|+|Qk(z,y′)|]dμ(z)≲d(y,y′)ϵ1(2−l+d(x,y))ϵ11V2−l(x)+V2−l(y)+V(x,y)2−lϵ2(2−l+d(x,y))ϵ2.
Thus, (3.12) holds and this finishes the proof of (3.3).
We finally prove (3.5) in the case k≥l.
To this end, it suffices to verify that for any ϵ1′∈(0,ϵ1⋀ϵ2), d(x,x′)≤(1/8)d(x,y) and d(y,y′)≤(1/8)d(x,y),
(3.25)|[PlQk(x,y)−PlQk(x′,y)]−[PlQk(x,y′)−PlQk(x′,y′)]|≲d(x,x′)ϵ1′(2−l+d(x,y))ϵ1′d(y,y′)ϵ1′(2−l+d(x,y))ϵ1′1V2−l(x)+V2−l(y)+V(x,y)2−lσ(2−l+d(x,y))σ,
where σ=(ϵ2−ϵ1′)⋀ϵ3>0. To see this, the estimate (3.4) implies that for any ϵ1′∈(0,ϵ1⋀ϵ2), d(x,x′)≤(1/8)d(x,y), and d(y,y′)≤(1/8)d(x,y),
(3.26)|[PlQk(x,y)−PlQk(x′,y)]−[PlQk(x,y′)−PlQk(x′,y′)]|≲2−(k−l)δd(x,x′)ϵ1′(2−l+d(x,y))ϵ1′1V2−l(x)+V2−l(y)+V(x,y)2−lϵ2(2−l+d(x,y))ϵ2+2−(k−l)δd(x,x′)ϵ1′(2−l+d(x,y′))ϵ1′1V2−l(x)+V2−l(y′)+V(x,y′)2−lϵ2(2−l+d(x,y′))ϵ2,
where δ>0 depends only on ϵ1′,
ϵ1,
and ϵ2. The assumption d(y,y′)≤(1/8)d(x,y) together with Lemma 2.1(iii) further shows
that for any ϵ1′∈(0,ϵ1⋀ϵ2), d(x,x′)≤(1/8)d(x,y) and d(y,y′)≤(1/8)d(x,y),
(3.27)|[PlQk(x,y)−PlQk(x′,y)]−[PlQk(x,y′)−PlQk(x′,y′)]|≲2−(k−l)δd(x,x′)ϵ1′(2−l+d(x,y))ϵ1′1V2−l(x)+V2−l(y)+V(x,y)2−lϵ2(2−l+d(x,y))ϵ2.
By the estimate (3.3) and
symmetry, we further obtain that for any ϵ1′∈(0,ϵ1⋀ϵ2), d(x,x′)≤(1/8)d(x,y) and d(y,y′)≤(1/8)d(x,y),
(3.28)|[PlQk(x,y)−PlQk(x′,y)]−[PlQk(x,y′)−PlQk(x′,y′)]|≲2−(k−l)δd(y,y′)ϵ1′(2−l+d(x,y))ϵ1′1V2−l(x)+V2−l(y)+V(x,y)2−lϵ2(2−l+d(x,y))ϵ2,
where δ>0 depends only on ϵ1′, ϵ1,
and ϵ2. Then the geometric mean among (3.25), (3.27), and (3.28) gives the desired estimate (3.5).
Using (3.6), we now prove (3.25) by writing that
(3.29)|[PlQk(x,y)−PlQk(x′,y)]−[PlQk(x,y′)−PlQk(x′,y′)]|=|∫𝒳{[Pl(x,z)−Pl(x′,z)]−[Pl(x,y)−Pl(x′,y)]}[Qk(z,y)−Qk(z,y′)]dμ(z)|≤∑i=14∫Wi|[Pl(x,z)−Pl(x′,z)]−[Pl(x,y)−Pl(x′,y)]||Qk(z,y)−Qk(z,y′)|dμ(z)≡∑i=14Yi,
where
(3.30)W1={z∈𝒳:d(y,y′)≤2−k+d(z,y)2≤2−l+d(x,y)8},W2={z∈𝒳:d(y,y′)≤2−l+d(x,y)8≤2−k+d(z,y)2,andd(x,x′)≤2−l+d(z,x)2},W3={z∈𝒳:d(y,y′)≤2−l+d(x,y)8≤2−k+d(z,y)2,andd(x,x′)>2−l+d(z,x)2},W4={z∈𝒳:d(y,y′)>2−k+d(z,y)2}.
Ifz∈W1,
then d(z,y)≤(1/4)(2−l+d(x,y)), d(x,x′)≤(1/8)d(x,y)≤(1/8)(2−l+d(x,y)), and d(y,y′)≤(1/2)(2−k+d(z,y)),
which together with the second difference condition of Pl,
the regularity of Qk and Lemma 2.1(ii) shows
(3.31)Y1≲d(x,x′)ϵ1(2−l+d(x,y))ϵ11V2−l(x)+V2−l(y)+V(x,y)2−lϵ3(2−l+d(x,y))ϵ3×∫W1d(y,z)ϵ1(2−l+d(x,y))ϵ1d(y,y′)ϵ1(2−k+d(z,y))ϵ1×1V2−k(z)+V2−k(y)+V(z,y)2−kϵ2(2−k+d(z,y))ϵ2dμ(z)≲d(x,x′)ϵ1(2−l+d(x,y))ϵ1d(y,y′)ϵ1(2−l+d(x,y))ϵ11V2−l(x)+V2−l(y)+V(x,y)2−lϵ3(2−l+d(x,y))ϵ3.
Ifz∈W4,
then (1/2)(2−k+d(z,y))<d(y,y′)≤(1/8)d(x,y),
and therefore, d(z,y)≤2d(y,y′)≤(1/4)d(x,y)≤(1/4)(2−l+d(x,y)) and d(x,x′)≤(1/8)d(x,y)≤(1/8)(2−l+d(x,y)).
From this, the regularity of Pl and Proposition 2.7(i), it follows that
(3.32)Y4≲d(x,x′)ϵ1(2−l+d(x,y))ϵ1d(y,y′)ϵ1(2−l+d(x,y))ϵ11V2−l(x)+V2−l(y)+V(x,y)×2−lϵ3(2−l+d(x,y))ϵ3∫𝒳[|Qk(z,y)|+|Qk(z,y′)|]dμ(z)≲d(x,x′)ϵ1(2−l+d(x,y))ϵ1d(y,y′)ϵ1(2−l+d(x,y))ϵ11V2−l(x)+V2−l(y)+V(x,y)2−lϵ3(2−l+d(x,y))ϵ3.
Ifz∈W2,
then d(x,x′)≤(1/2)(2−l+d(z,x)), d(y,y′)≤(1/2)(2−k+d(z,y)), and d(x,x′)≤(1/8)d(x,y)≤(1/8)(2−l+d(x,y)),
which together with the regularity of Pl and Qk implies that
(3.33)Y2≲∫W2[d(x,x′)ϵ1(2−l+d(x,z))ϵ11V2−l(x)+V2−l(z)+V(x,z)2−lϵ2(2−l+d(x,z))ϵ2+d(x,x′)ϵ1(2−l+d(x,y))ϵ11V2−l(x)+V2−l(y)+V(x,y)2−lϵ2(2−l+d(x,y))ϵ2]×d(y,y′)ϵ1(2−k+d(y,z))ϵ11V2−k(z)+V2−k(y)+V(z,y)2−kϵ2(2−k+d(z,y))ϵ2dμ(z)≡Y2,1+Y2,2.
The fact that z∈W2,
which implies that 2−k+d(z,y)≥(1/4)(2−l+d(x,y)),
together with Lemma 2.1(ii) yields that
(3.34)Y2,2≲d(x,x′)ϵ1(2−l+d(x,y))ϵ1d(y,y′)ϵ1(2−l+d(x,y))ϵ11V2−l(x)+V2−l(y)+V(x,y)×2−lϵ2(2−l+d(x,y))ϵ2∫𝒳1V2−k(y)+V(z,y)2−kϵ2(2−k+d(z,y))ϵ2dμ(z)≲d(x,x′)ϵ1(2−l+d(x,y))ϵ1d(y,y′)ϵ1(2−l+d(x,y))ϵ11V2−l(x)+V2−l(y)+V(x,y)2−lϵ2(2−l+d(x,y))ϵ2.
Similarly, the fact that 2−k+d(z,y)≥(1/4)(2−l+d(x,y)) implies that
(3.35)Y2,1≲d(y,y′)ϵ1(2−l+d(x,y))ϵ1∫W2d(x,x′)ϵ1(2−l+d(x,z))ϵ11V2−l(x)+V2−l(z)+V(x,z)×2−lϵ2(2−l+d(x,z))ϵ21V2−k(z)+V2−k(y)+V(z,y)2−kϵ2(2−k+d(z,y))ϵ2dμ(z).
The estimate (3.19) together with
the facts that k≥l and 2−k+d(z,y)≥(1/4)(2−l+d(x,y)) and Lemma 2.1(ii) yields that for any ϵ1′∈(0,ϵ1⋀ϵ2),
(3.36)Y2,1≲d(y,y′)ϵ1(2−l+d(x,y))ϵ11V2−l(y)+V(x,y)d(x,x′)ϵ1′2−lϵ2(2−l+d(x,y))ϵ2×∫𝒳1V2−l(x)+V(x,z)2−lϵ2(2−l+d(x,z))ϵ2+ϵ1′dμ(z)≲d(y,y′)ϵ1(2−l+d(x,y))ϵ11V2−l(y)+V(x,y)d(x,x′)ϵ1′(2−l+d(x,y))ϵ1′2−l(ϵ2−ϵ1′)(2−l+d(x,y))ϵ2−ϵ1′,
and similarly, Lemma 2.1(i) implies that
(3.37)Y2,1≲d(y,y′)ϵ1(2−l+d(x,y))ϵ11V2−l(x)d(x,x′)ϵ1′2lϵ1′∫W21V2−k(y)+V(z,y)2−kϵ2(2−k+d(z,y))ϵ2dμ(z)≲d(y,y′)ϵ1(2−l+d(x,y))ϵ11V2−l(x)d(x,x′)ϵ1′2lϵ1′×{2−lϵ2(2−l+d(x,y))ϵ2+2−lϵ2∫d(z,y)≥(1/8)(2−l+d(x,y)),d(z,y)≥2−k1V(z,y)1d(z,y)ϵ2dμ(z)}≲d(y,y′)ϵ1(2−l+d(x,y))ϵ11V2−l(x)d(x,x′)ϵ1′(2−l+d(x,y))ϵ1′2−l(ϵ2−ϵ1′)(2−l+d(x,y))ϵ2−ϵ1′,
which completes the estimate for Y2.
Finally, to estimateY3,
since d(y,y′)≤(1/2)(2−k+d(z,y)) and
(3.38)d(x,x′)≤d(x,y)8≤2−l+d(x,y)8,
by the regularities of Pl and Qk, we then have
(3.39)Y3≲∫W3[|Pl(x,z)|+|Pl(x′,z)|+d(x,x′)ϵ1(2−l+d(x,y))ϵ11V2−l(x)+V2−l(y)+V(x,y)2−lϵ2(2−l+d(x,y))ϵ2]×d(y,y′)ϵ1(2−k+d(z,y))ϵ11V2−k(z)+V2−k(y)+V(z,y)2−kϵ2(2−k+d(z,y))ϵ2dμ(z)≡∑i=13Y3,i.
The fact that 2−k+d(z,y)≥(1/4)(2−l+d(x,y)) and Lemma 2.1(ii) imply that
(3.40)Y3,3≲d(x,x′)ϵ1(2−l+d(x,y))ϵ1d(y,y′)ϵ1(2−l+d(x,y))ϵ11V2−l(x)+V2−l(y)+V(x,y)×2−lϵ2(2−l+d(x,y))ϵ2∫𝒳1V2−k(y)+V(z,y)2−kϵ2(2−k+d(z,y))ϵ2dμ(z)≲d(x,x′)ϵ1(2−l+d(x,y))ϵ1d(y,y′)ϵ1(2−l+d(x,y))ϵ11V2−l(x)+V2−l(y)+V(x,y)2−lϵ2(2−l+d(x,y))ϵ2.
The facts that k≥l, 2−k+d(z,y)≥(1/4)(2−l+d(x,y)),
(3.19), and Proposition 2.7(i) yield that for any ϵ1′∈(0,ϵ1⋀ϵ2),
(3.41)Y3,1≲d(y,y′)ϵ1(2−l+d(x,y))ϵ1∫W3|Pl(x,z)|1V2−k(z)+V2−k(y)+V(z,y)2−kϵ2(2−k+d(z,y))ϵ2dμ(z)≲d(x,x′)ϵ1′(2−l+d(x,y))ϵ1′d(y,y′)ϵ1(2−l+d(x,y))ϵ11V2−l(y)+V(x,y)2−l(ϵ2−ϵ1′)(2−l+d(x,y))ϵ2−ϵ1′,
where, in the last step, we used
the fact that 2−l<2d(x,x′) in this case. Similarly, by the size condition
of Pl and Lemma 2.1(i), we further have
(3.42)Y3,1≲d(y,y′)ϵ1(2−l+d(x,y))ϵ11V2−l(x)∫W31V2−k(y)+V(z,y)2−kϵ2(2−k+d(z,y))ϵ2dμ(z)≲d(y,y′)ϵ1(2−l+d(x,y))ϵ11V2−l(x)×{2−lϵ2(2−l+d(x,y))ϵ2+2−lϵ2∫d(z,y)≥(1/8)(2−l+d(x,y)),d(z,y)≥2−k1V(z,y)1d(z,y)ϵ2dμ(z)}≲d(x,x′)ϵ1′(2−l+d(x,y))ϵ1′d(y,y′)ϵ1(2−l+d(x,y))ϵ11V2−l(x)2−l(ϵ2−ϵ1′)(2−l+d(x,y))ϵ2−ϵ1′.
Notice that in this case, 2−l≤2d(x,x′)≤(1/4)d(x,y) and therefore V2−l(x)≤V(x,y).
Thus, by (3.19), we have that for z∈W3, 1/(V2−k(y)+V(z,y))≲1/(V2−l(y)+V(x,y))≲1/(V2−l(x)+V2−l(y)+V(x,y)), which together with Proposition 2.7(i) shows
(3.43)Y3,2≲d(y,y′)ϵ1(2−l+d(x,y))ϵ12−lϵ2(2−l+d(x,y))ϵ2∫W3|Pl(x′,z)|1V2−k(z)+V2−k(y)+V(z,y)dμ(z)≲d(x,x′)ϵ1′(2−l+d(x,y))ϵ1′d(y,y′)ϵ1(2−l+d(x,y))ϵ11V2−l(x)+V2−l(y)+V(x,y)2−l(ϵ2−ϵ1′)(2−l+d(x,y))ϵ2−ϵ1′.
This completes the proof of
(3.25), and hence, the proof of Lemma 3.2.Remark 3.3.
From the proof of (3.2) in Lemma 3.2, it is easy to see that (3.2) still
holds if Pl has the required regularity only in the second
variable, and Qk the first variable. This observation is useful
in applications.In what follows, let{Sk}k∈ℤ be an (ϵ1,ϵ2,ϵ3)-ATI with ϵ1∈(0,1],ϵ2>0, and ϵ3>0 as in Definition 2.2. Set Dk=Sk−Sk−1 for k∈ℤ.
To establish continuous Calderón reproducing formulae, by following Coifman's
idea (see [64]) and
Proposition 3.1(ii), we write for any N∈ℕ,
(3.44)I=(∑k=−∞∞Dk)(∑j=−∞∞Dj)=∑|l|>N∑k=−∞∞Dk+lDk+∑k=−∞∞DkNDk=RN+TN
in Lp(𝒳) with p∈(1,∞),
where DkN=∑|l|≤NDk+l. To verify that TN−1 exists and is bounded on any space of test
functions, we first prove that RN is bounded on L2(𝒳) with a small operator norm.Lemma 3.4.
LetN∈ℕ and let RN be as in (3.44). Then there exist constants C>0 and δ>0,
independent of N,
such that for all f∈L2(𝒳),
(3.45)∥RN(f)∥L2(𝒳)≤C2−Nδ∥f∥L2(𝒳).Proof.
To prove the lemma, by applying the Cotlar-Stein lemma (see [75, page 280] or [64]), we see that it suffices
to verify that for any σ∈(0,1), ϵ1′∈(0,ϵ1⋀ϵ2) with i=1,2,3,
and all k1,l1,k2,l2∈ℤ,
(3.46)∥Dk1+l1Dk1(Dk2+l2Dk2)t∥L2(𝒳)→L2(𝒳)≲2−|l1|ϵ1′σ2−|l2|ϵ2′σ2−|k1−k2|ϵ3′(1−σ),∥(Dk2+l2Dk2)tDk1+l1Dk1∥L2(𝒳)→L2(𝒳)≲2−|l1|ϵ1′σ2−|l2|ϵ2′σ2−|k1−k2|ϵ3′(1−σ).
We only prove (3.46), the proof for the last inequality being similar. In fact, Lemma 3.2 and Proposition 2.7 yield that
(3.47)∥Dk1+l1Dk1(Dk2+l2Dk2)t∥L2(𝒳)→L2(𝒳)≲2−|l1|ϵ1′2−|l2|ϵ2′,∥Dk1+l1Dk1Dk2tDk2+l2t∥L2(𝒳)→L2(𝒳)≲2−|k1−k2|ϵ3′.
Then the geometric mean of the above two estimates gives the estimate (3.46), which completes the proof of Lemma 3.4.We now establish some estimates for the kernel,RN(x,y),
of the operator RN.
To this end, we first give a technical lemma.Lemma 3.5.
For anyσ≥s>0 and x,y∈𝒳 with x≠y,
(3.48)∑k=−∞∞1V2−k(x)+V(x,y)2−ks(2−k+d(x,y))σ≲1V(x,y)1d(x,y)σ−s.Proof.
For anyx,y∈𝒳,
choose k0∈ℤ such that 2−k0<d(x,y)≤2−k0+1.
Then if k≤k0,
by (1.3), we have
(3.49)V2−k(x)=μ(B(x,2−(k−k0)2−k0))≥2−κ(k−k0)μ(B(x,2−k0))≳2−κ(k−k0)V(x,y).
Therefore,
(3.50)∑k=−∞∞1V2−k(x)+V(x,y)2−ks(2−k+d(x,y))σ≤1d(x,y)σ−s∑k=−∞k01V2−k(x)+1V(x,y)d(x,y)σ∑k=k0+1∞2−ks≲1V(x,y)1d(x,y)σ−s{∑k=−∞k02−κ(k−k0)+1}≲1V(x,y)1d(x,y)σ−s,
which completes the proof of Lemma 3.5.Lemma 3.6.
LetN∈ℕ,
let RN be as in (3.44), and let RN(x,y) be its kernel. Then for any ϵ1′∈(0,ϵ1⋀ϵ2), there exists a constant δ>0,
independent  of N,
such that RN satisfies all the conditions of Corollary 2.22
with ϵ replaced  by ϵ1′ and CRN≲2−δN.
Moreover, RN∗(1)=0.Proof.
Letϵ1′∈(0,ϵ1⋀ϵ2). Write
(3.51)RN(x,y)=∑l=N+1∞∑k=−∞∞(Dk+lDk)(x,y)+∑l=−∞−N−1∑k=−∞∞(Dk+lDk)(x,y)=RN1(x,y)+RN2(x,y).
By (3.2) and Lemma 3.5, we have
that for any x,y∈𝒳 with x≠y,
(3.52)|RN(x,y)|≤|RN1(x,y)|+|RN2(x,y)|≲∑l=N+1∞2−lϵ1′∑k=−∞∞1V2−k(x)+V(x,y)2−kϵ2(2−k+d(x,y))ϵ2+∑l=−∞−N−12lϵ1′∑k=−∞∞1V2−(k+l)(x)+V(x,y)2−(k+l)ϵ2(2−(k+l)+d(x,y))ϵ2≲2−Nϵ1′1V(x,y).
Thus, RN satisfies (i) of Theorem 2.18.
The estimate (3.3) and Lemma 3.5 show that for any x,y,y′∈𝒳 with x≠y and d(y,y′)≤(1/4)d(x,y),
(3.53)|RN(x,y)−RN(x,y′)|≤|RN1(x,y)−RN1(x,y′)|+|RN2(x,y)−RN2(x,y′)|≲d(y,y′)ϵ1′d(x,y)ϵ1′{∑l=N+1∞2−lδ∑k=−∞∞1V2−k(x)+V(x,y)2−kϵ2(2−k+d(x,y))ϵ2+∑l=−∞−N−12lδ∑k=−∞∞1V2−(k+l)(x)+V(x,y)2−(k+l)ϵ2(2−(k+l)+d(x,y))ϵ2}≲2−Nδd(y,y′)ϵ1′d(x,y)ϵ1′1V(x,y),
and the estimate (3.4) and the
symmetry also yield that
(3.54)|RN(y,x)−RN(y′,x)|≲2−Nδd(y,y′)ϵ1′d(x,y)ϵ1′1V(x,y),
which shows that RN satisfies (2.49) and (ii) of Theorem 2.18.
Using (3.5) and Lemma 3.5 then gives that for x,x′,y,y′∈𝒳 with x≠y and d(x,x′)≤(1/8)d(x,y) and d(y,y′)≤(1/8)d(x,y),
(3.55)|[RN(x,y)−RN(x′,y)]−[RN(x,y′)−RN(x′,y′)]|≤|[RN1(x,y)−RN1(x′,y)]−[RN1(x,y′)−RN1(x′,y′)]|+|[RN2(x,y)−RN2(x′,y)]−[RN2(x,y′)−RN2(x′,y′)]|≲d(x,x′)ϵ1′d(x,y)ϵ1′d(y,y′)ϵ1′d(x,y)ϵ1′{∑l=N+1∞2−lδ∑k=−∞∞1V2−k(x)+V(x,y)2−kσ(2−k+d(x,y))σ+∑l=−∞−N−12lδ∑k=−∞∞1V2−(k+l)(x)+V(x,y)2−(k+l)σ(2−(k+l)+d(x,y))σ}≲2−Nδd(x,x′)ϵ1′d(x,y)ϵ1′d(y,y′)ϵ1′d(x,y)ϵ1′1V(x,y),
which shows that RN satisfies (iii) of Theorem 2.18.
Moreover, by the vanishing moments ofDk,
we obviously have RN(1)=0=RN∗(1).
Thus, RN satisfies all the conditions of Corollary 2.22
with ϵ replaced by ϵ1′ and CRN≲2−δN,
which completes the proof of Lemma 3.6.From Lemma3.6 and Corollary 2.22, it is easy to deduce
the boundedness of TN on spaces of test functions when N is large enough.Proposition 3.7.
Letϵ1∈(0,1], ϵ2>0, ϵ3>0, and let {Sk}k∈ℤ be an (ϵ1,ϵ2,ϵ3)-ATI. For N∈ℕ,
let RN and TN be as in (3.44). Then there exist constants C8>0 and δ>0,
which are independent of N,
such that for all f∈𝒢°(x1,r,β,γ) with x1∈𝒳, r>0 and 0<β,γ<(ϵ1⋀ϵ2),
(3.56)∥RN(f)∥𝒢(x1,r,β,γ)≤C82−Nδ∥f∥𝒢(x1,r,β,γ).
Moreover, if N is so large that
(3.57)C82−Nδ<1,
then TN−1 exists and maps any space of test functions to
itself. More precisely, there exists a constant C>0 such that for all f∈𝒢°(x1,r,β,γ) with x1∈𝒳, r>0, and 0<β,γ<(ϵ1⋀ϵ2),
(3.58)∥TN−1(f)∥𝒢(x1,r,β,γ)≤C∥f∥𝒢(x1,r,β,γ).Proof.
Applying Corollary2.22 together with Lemmas 3.4 and 3.6 gives
(3.56). Moreover, if we choose N∈ℕ such that (3.57) holds, then by (3.56), we
have that for all f∈𝒢°(x1,r,β,γ),
(3.59)∥TN−1(f)∥𝒢(x1,r,β,γ)=∥(I−RN)−1(f)∥𝒢(x1,r,β,γ)=∥∑l=0∞(RN)l(f)∥𝒢(x1,r,β,γ)≤∑l=0∞(C82−Nδ)l∥f∥𝒢(x1,r,β,γ)≲∥f∥𝒢(x1,r,β,γ),
which completes the proof of Proposition 3.7.Proposition 3.8.
Letp∈(1,∞), ϵ1∈(0,1], ϵ2>0, ϵ3>0 and let {Sk}k∈ℤ be an (ϵ1,ϵ2,ϵ3)-ATI.
For N∈ℕ,
let RN and TN be as in (3.44). Then there exist constants C9>0 and δ>0,
which are independent of N,
such that for all f∈Lp(𝒳)(3.60)∥RN(f)∥Lp(𝒳)≤C92−Nδ∥f∥Lp(𝒳).
Moreover, if N is so large that
(3.61)C92−Nδ<1,
then TN−1 exists and is bounded on Lp(𝒳),
and there exists a constant C>0 such that for all f∈Lp(𝒳),
(3.62)∥TN−1(f)∥Lp(𝒳)≤C∥f∥Lp(𝒳).Proof.
We use the same notation as in Lemma3.6. From Lemma 3.6 together with
the proposition in [75, page 29], it is easy to see that RN is a singular integral satisfying the
condition (10) in [75, page 19] with A≲2−Nδ.
This fact and the corollary in [75, page 19] together with a duality argument and Lemma
3.4 prove that RN is also bounded on Lp(𝒳) for p∈(1,∞) and ∥RN∥Lp(𝒳)→Lp(𝒳)≲2−Nδ.
That is, (3.60) holds. If we choose N∈ℕ so large that (3.61) holds, by an argument
similar to the proof of Proposition 3.7, we can prove that TN−1 is bounded on Lp(𝒳) for p∈(1,∞).Letϵ1∈(0,1], ϵ2>0, ϵ3>0, and let {Sk}k=0∞ be an (ϵ1,ϵ2,ϵ3)-ATI.
Set Dk=Sk−Sk−1 for k∈ℤ.
For any f∈Lp(𝒳) with p∈(1,∞) and x∈𝒳,
the Littlewood-Paleyg-functiong˙(f) is defined by
(3.63)g˙(f)(x)={∑k=−∞∞|Dk(f)(x)|2}1/2.Lemma 3.9.
Letϵ1∈(0,1], ϵ2>0, ϵ3>0, and let {Sk}k∈ℤ be an (ϵ1,ϵ2,ϵ3)-ATI.
Let Dk=Sk−Sk−1 for k∈ℤ and let g˙(f) for f∈Lp(𝒳) with p∈(1,∞) be as in (3.63). Then there exists a constant Cp>0 such that for all f∈Lp(𝒳),
(3.64)∥g˙(f)∥Lp(𝒳)≤Cp∥f∥Lp(𝒳).Proof.
By Khinchin's inequality (see [76, page 165]) and Minkowski's inequality, we first have
that for any N∈ℕ,
(3.65)∥{∑k=−NN|Dk(f)|2}1/2∥Lp(𝒳)≲∥122N∑σ−N⋯∑σN∑k=−NNσkDk(f)∥Lp(𝒳)≲122N∑σ−N⋯∑σN∥∑k=−NNσkDk(f)∥Lp(𝒳),
where σk=1 or −1 for k∈{−N,…,N}.
For any fixed σ={σk}k=−NN,
we set TNσ=∑k=−NNσkDk and denote its kernel by KNσ.
Similarly to the proof of Lemma 3.6, it is easy to verify that KNσ and (KNσ)∗ are standard Calderón-Zygmund kernels, with
constants independent of N and σ.
Then if we can verify that TNσ is bounded on L2(𝒳) with ∥TNσ∥L2(𝒳)→L2(𝒳)≲1,
then by the corollary in [75, page 22] together with a duality argument, we obtain
that for p∈(1,∞), ∥TNσ∥Lp(𝒳)→Lp(𝒳)≲1. Therefore, for all N∈ℕ and all f∈Lp(𝒳),
(3.66)∥{∑k=−NN|Dk(f)|2}1/2∥Lp(𝒳)≲∥f∥Lp(𝒳).
Then Fatou's lemma further shows
that ∥g˙(f)∥Lp(𝒳)≲∥f∥Lp(𝒳). To finish the proof of Lemma 3.9, it remains
to verify that TNσ is bounded on L2(𝒳).
By the Cotlar-Stein lemma, it suffices to verify that for any ϵ1′∈(0,ϵ1⋀ϵ2) and all j,k∈ℤ,
(3.67)∥σkDk(σjDj)t∥L2(𝒳)→L2(𝒳)≲2−|k−j|ϵ1′,∥(σjDj)tσkDk∥L2(𝒳)→L2(𝒳)≲2−|k−j|ϵ1′.
However, these two estimates are
a simple corollary of (3.2) in Lemma 3.2. This completes the proof of Lemma
3.9.We can now establish acontinuous Calderón reproducing formula.Theorem 3.10.
Letϵ1∈(0,1], ϵ2>0, ϵ3>0, ϵ∈(0,ϵ1⋀ϵ2) and  let {Sk}k∈ℤ be an (ϵ1,ϵ2,ϵ3)-ATI.
Set Dk=Sk−Sk−1 for k∈ℤ.
Then there exists a family of linear operators {D˜k}k∈ℤ such that for all f∈𝒢°0ϵ(β,γ) with 0<β,γ<ϵ,
(3.68)f=∑k=−∞∞D˜kDk(f),
where the series converges in
both the norm of 𝒢°0ϵ(β,γ) and the norm of Lp(𝒳) for p∈(1,∞).
Moreover, the kernels of the operators D˜k satisfy the conditions (i) and (ii) of
Definition 2.2 with ϵ1 and ϵ2 replaced by ϵ′∈(ϵ,ϵ1⋀ϵ2), and ∫𝒳D˜k(x,y)dμ(y)=∫𝒳D˜k(x,y)dμ(x)=0.Proof.
Fix a large integerN such that (3.57) and (3.61) hold and,
therefore, Propositions 3.7 and 3.8 hold. Let DkN for k∈ℤ be as in (3.44). It is easy to check that DkN(⋅,y)∈𝒢°(y,2−j,ϵ1,ϵ2) for all k∈ℤ.
Define D˜k(x,y)=TN−1(DkN(⋅,y))(x) for k∈ℤ,
where TN−1 is defined as in Proposition 3.7. Then
Proposition 3.7 shows that D˜k for k∈ℤ satisfies all the conclusions of the theorem,
and formally, we also have (3.68). We still need to verify that the series in
(3.68) converges in both the norm of 𝒢°0ϵ(β,γ) and the norm of Lp(𝒳) with
p∈(1,∞).
Letϵ˜=ϵ1⋀ϵ2. We first prove that the series in (3.68) converges in the norm of 𝒢°0ϵ(β,γ) with 0<β,γ<ϵ.
Let f∈𝒢°(β′,γ′) with β<β′<ϵ˜ and γ<γ′<ϵ˜.
Then, for L∈ℕ,
we  write
(3.69)∑|k|≤LD˜kDk(f)=TN−1(∑|k|≤LDkNDk)(f)=TN−1(TN−∑|k|≥L+1DkNDk)(f)=TN−1TN(f)−TN−1(∑|k|≥L+1DkNDk)(f)=f−limj→∞(RN)j(f)−TN−1(∑|k|≥L+1DkNDk)(f).
We now verify that
(3.70)limL→∞∥f−∑|k|≤LD˜kDk(f)∥𝒢(β,γ)=0.
To see this, we write
(3.71)∥∑|k|≤LD˜kDk(f)−f∥𝒢(β,γ)≤limj→∞∥(RN)j(f)∥𝒢(β,γ)+∥TN−1(∑|k|≥L+1DkNDk)(f)∥𝒢(β,γ).
Notice that 𝒢°(β′,γ′)⊂𝒢°(β,γ).
By (3.56) and (3.57), we have
(3.72)limj→∞∥(RN)j(f)∥𝒢(β,γ)≤limj→∞(C82−Nδ)j∥f∥𝒢(β,γ)=0,
we remark that this is also true
even when β=β′ and γ=γ′.
We now prove that
(3.73)limL→∞∥TN−1(∑|k|≥L+1DkNDk)(f)∥𝒢(β,γ)=0.
To this end, by Proposition 3.7,
it suffices to verify that there exists some σ>0 such that for all 0<β<β′<ϵ˜ and 0<γ<γ′<ϵ˜ and all f∈𝒢°(β′,γ′),
(3.74)∥∑|k|≥L+1DkNDk(f)∥𝒢(β,γ)≲2−σL∥f∥𝒢(β′,γ′).
Similarly to the proof of (3.3),
by Lemma 2.1(iii) and the geometric mean, we can reduce the proof of (3.74) to
verifying that there exists some σ>0 such that for all f∈𝒢°(β′,γ′) and all x∈𝒳,
(3.75)|∑|k|≥L+1DkNDk(f)(x)|≲2−σL∥f∥𝒢(β′,γ′)1V1(x1)+V(x1,x)1(1+d(x,x1))γ,
and for all x,x′∈𝒳 with d(x,x′)≤(1/2)(1+d(x,x1)),
(3.76)|∑|k|≥L+1DkNDk(f)(x)−∑|k|≥L+1DkNDk(f)(x′)|≲∥f∥𝒢(β′,γ′)d(x,x′)β′(1+d(x,x1))β′1V1(x1)+V(x1,x)1(1+d(x,x1))γ′.
Similarly to the proofs of Lemmas3.4 and 3.6 and
using some estimates similar to (3.2) and (3.12), we easily obtain that for any L∈ℕ,
the operator TL=∑|k|≥L+1DkNDk satisfies all the conditions of Corollary 2.22
with ϵ replaced by ϵ˜, CTL≲1 and
(3.77)∥TL∥L2(𝒳)→L2(𝒳)≲1.
Corollary 2.22 then shows that TL is bounded on 𝒢°(β′,γ′) for any 0<β′,γ′<ϵ˜.
In particular, we see that (3.76)  holds.
To verify (3.75), we simply denote DkNDk by Ek.
By Lemma 3.2 and its proof, it is easy to see that Ek(x,y),
the kernel of Ek,
still satisfies (i) to (iv) of Definition 2.2 with a constant depending on N;
see (3.2) and (3.12). Moreover, Ek(1)=0.
For f∈𝒢°(β′,γ′),
(3.78)|∑k=L+1∞DkNDk(f)(x)|=|∑k=L+1∞∫𝒳Ek(x,y)[f(y)−f(x)]dμ(y)|≤∑k=L+1∞∫d(x,y)≤(1+d(x1,x))/2|Ek(x,y)||f(y)−f(x)|dμ(y)+∑k=L+1∞∫d(x,y)>(1+d(x1,x))/2|Ek(x,y)||f(y)|dμ(y)+∑k=L+1∞∫d(x,y)>(1+d(x1,x))/2|Ek(x,y)||f(x)|dμ(y)≡∑i=13Zi.
The size estimates for Ek and the regularity of f imply together with Lemma 2.1(ii) that
(3.79)Z1≲∑k=L+1∞∫d(x,y)≤(1+d(x1,x))/21V2−k(x)+V2−k(y)+V(x,y)2−kϵ2(2−k+d(x,y))ϵ2×d(x,y)β′(1+d(x1,x))β′1V1(x1)+V(x1,x)1(1+d(x1,x))γ′dμ(y)≲2−Lβ′1V1(x1)+V(x1,x)1(1+d(x1,x))γ′.
Similarly, Lemma 2.1(i) yields that
(3.80)Z3≲∑k=L+1∞∫d(x,y)>(1+d(x1,x))/21V2−k(x)+V(x,y)2−kϵ2(2−k+d(x,y))ϵ2dμ(y)×1V1(x1)+V(x1,x)1(1+d(x1,x))γ′≲2−Lϵ21V1(x1)+V(x1,x)1(1+d(x1,x))γ′.
Since d(x,y)>(1+d(x1,x))/2,
we have V(x,x1)≲V(x,y).
This estimate together with Lemma 2.1 shows that
(3.81)Z2≲∑k=L+1∞∫d(x,y)>(1+d(x1,x))/21V2−k(x)+V2−k(y)+V(x,y)2−kϵ2(2−k+d(x,y))ϵ2×1V1(x1)+V(x1,y)1(1+d(x1,y))γ′dμ(y)≲2−Lϵ2min{1V1(x1),1V(x,x1)}1(1+d(x1,x))ϵ2≲2−Lϵ21V1(x1)+V(x1,x)1(1+d(x1,x))γ′,
which completes the proof of
(3.75) for the operator ∑k=L+1∞DkNDk.
Since∫𝒳f(y)dμ(y)=0, we can write
(3.82)|∑k=−∞−L−1DkNDk(f)(x)|=|∑k=−∞−L−1∫𝒳[Ek(x,y)−Ek(x,x1)]f(y)dμ(y)|≤∑k=−∞−L−1∫d(x1,y)≤(2−k+d(x1,x))/2|Ek(x,y)−Ek(x,x1)||f(y)|dμ(y)+∑k=−∞−L−1∫d(x1,y)>(2−k+d(x1,x))/2|Ek(x,y)||f(y)|dμ(y)+∑k=−∞−L−1∫d(x1,y)>(2−k+d(x1,x))/2|Ek(x,x1)||f(y)|dμ(y)≡∑i=13Yi.
The regularity of Ek in the second variable and the size condition
of f together with Lemma 2.1(i) yield that
(3.83)Y1≲∑k=−∞−L−1∫d(x1,y)≤(2−k+d(x1,x))/2d(y,x1)ϵ1(2−k+d(x,x1))ϵ12−kϵ2(2−k+d(x,x1))ϵ21V2−k(x)+V2−k(x1)+V(x,x1)×1V1(x1)+V(x1,y)1(1+d(x1,y))γ′dμ(y)≲2−L(γ′−γ)1V1(x1)+V(x,x1)1(1+d(x,x1))γ.
Similarly, Lemma 2.1(ii) shows that
(3.84)Y3≲∑k=−∞−L−1∫d(x1,y)>(2−k+d(x1,x))/21V2−k(x)+V2−k(x1)+V(x,x1)×2−kϵ2(2−k+d(x,x1))ϵ21V1(x1)+V(x1,y)1(1+d(x1,y))γ′dμ(y)≲2−L(γ′−γ)/21V1(x1)+V(x,x1)1(1+d(x1,x))γ.
Since k<0 and d(x1,y)>(2−k+d(x1,x))/2>(1+d(x1,x))/2,
then V(x1,x)≲V(x1,y).
From this and Lemma 2.1(ii), it follows that
(3.85)Y2≲∑k=−∞−L−1∫d(x1,y)>(2−k+d(x1,x))/21V2−k(x)+V2−k(y)+V(x,y)×2−kϵ2(2−k+d(x,y))ϵ21V1(x1)+V(x1,y)1(1+d(x1,y))γ′dμ(y)≲2−L(γ′−γ)1V1(x1)+V(x,x1)1(1+d(x1,x))γ,
which completes the proof of (3.75). Hence, we obtain (3.74). Therefore for f∈𝒢°(β′,γ′) with β<β′<ϵ˜ and γ<γ′<ϵ˜,
(3.73) holds. Combining the estimate (3.73) with (3.72) shows (3.70).
Let nowf∈𝒢°0ϵ(β,γ) with 0<β,γ<ϵ.
By definition, there exists a sequence {fn}n=1∞⊂𝒢°(ϵ,ϵ) such that limn→∞∥f−fn∥𝒢(β,γ)=0. For any given L∈ℕ,
using some estimates similar to (3.2) and (3.12) (see also Remark 3.16),
similarly to the proof of Lemma 3.6, we can easily verify that for any L∈ℕ,
the operator T˜L=∑|k|≤LDkNDk satisfies all the conditions of Corollary 2.22
with ϵ replaced by ϵ˜, CT˜L≲1, and ∥T˜L∥L2(𝒳)→L2(𝒳)≲1. Thus T˜L is bounded on 𝒢°(β,γ) and T˜L∗(1)=0 by the vanishing moment of DkN,
which together with the boundedness of TN−1 on 𝒢°(β,γ) yields
(3.86)∥f−∑|k|≤LD˜kDk(f)∥𝒢(β,γ)≲∥f−fn∥𝒢(β,γ)+∥fn−TN−1(T˜L(fn))∥𝒢(β,γ)+∥TN−1(T˜L(fn))−TN−1(T˜L(f))∥𝒢(β,γ)≲∥f−fn∥𝒢(β,γ)+∥fn−TN−1(T˜L(fn))∥𝒢(β,γ).For any given δ>0,
fix some n∈ℕ such that ∥f−fn∥𝒢(β,γ)≤δ/4. By (3.70), for this chosen n,
there exists some L0∈ℕ such that if L>L0,
then
(3.87)∥fn−TN−1(T˜L(fn))∥𝒢(β,γ)<δ2.
Thus, when L>L0,
(3.88)∥f−∑|k|≤LD˜kDk(f)∥𝒢(β,γ)<δ.
Since T˜L∗(1)=0 and RN∗(1)=0 and since T˜L and TN−1 are bounded from 𝒢°(x,r,β,γ) to 𝒢(x,r,β,γ) for any x∈𝒳, r>0, and 0<β,γ<ϵ˜,
it follows that for
ϵ∈(0,ϵ˜), ∑|k|≤LD˜kDk(fn)=TN−1T˜L(fn)∈𝒢°(ϵ,ϵ). Moreover,
(3.89)∥{f−∑|k|≤LD˜kDk(f)}−{fn−∑|k|≤LD˜kDk(fn)}∥𝒢(β,γ)≤∥f−fn∥𝒢(β,γ)+∥TN−1T˜L(f−fn)∥𝒢(β,γ)≲∥f−fn∥𝒢(β,γ)→0,
as n→∞.
Thus, f−∑|k|≤LD˜kDk(f)∈𝒢°0ϵ(β,γ).
By (3.88), we further have that when L>L0,
(3.90)∥f−∑|k|≤LD˜kDk(f)∥𝒢°0ϵ(β,γ)<δ,which implies that (3.68) holds
in the norm of 𝒢°0ϵ(β,γ).
We now verify that (3.68) also holds in Lp(𝒳) for p∈(1,∞).
By Proposition 3.8 and the choice of N,
we have
(3.91)limj→∞∥(RN)j(f)∥Lp(𝒳)≤limj→∞(C92−Nδ)j∥f∥Lp(𝒳)=0,∥TN−1(∑|k|≥L+1DkNDk)(f)∥Lp(𝒳)≲∥(∑|k|≥L+1DkNDk)(f)∥Lp(𝒳).
Moreover, by Lemma 3.9 and
Hölder's inequality, we obtain
(3.92)∥(∑|k|≥L+1DkNDk)(f)∥Lp(𝒳)=sup∥h∥Lp′(𝒳)≤1|〈∑|k|≥L+1DkNDk(f),h〉|≤sup∥h∥Lp′(𝒳)≤1∥(∑|k|≥L+1|Dk(f)|2)1/2∥Lp(𝒳)∥(∑|k|≥L+1|(DkN)t(h)|2)1/2∥Lp′(𝒳)≲∥(∑|k|≥L+1|Dk(f)|2)1/2∥Lp(𝒳)≲∥f∥Lp(𝒳).
Thus,
(3.93)limL→∞∥(∑|k|≥L+1DkNDk)(f)∥Lp(𝒳)≲limL→∞∥(∑|k|≥L+1|Dk(f)|2)1/2∥Lp(𝒳)=0.
From these estimates, we finally
deduce that (3.68) holds in Lp(𝒳) with p∈(1,∞),
which completes the proof of Theorem 3.10.By an argument similar to the proof of Theorem3.10, we
obtain another continuous Calderón
reproducing formula (we omit the details).Theorem 3.11.
Letϵ1∈(0,1], ϵ2>0, ϵ3>0, ϵ∈(0,ϵ1⋀ϵ2), and let
{Sk}k∈ℤ be an (ϵ1,ϵ2,ϵ3)-ATI.
Set Dk=Sk−Sk−1 for k∈ℤ.
Then there exists a family of linear operators {D¯k}k∈ℤ such that for all f∈𝒢°0ϵ(β,γ) with 0<β,γ<ϵ,
(3.94)f=∑k=−∞∞DkD¯k(f),
where the series converges in
both the norm of 𝒢°0ϵ(β,γ) and the norm of Lp(𝒳) for p∈(1,∞).
Moreover, the kernels of the operators D¯k satisfy the conditions (i) and (iii) of
Definition 2.2 with ϵ1 and ϵ2 replaced by ϵ′∈(ϵ,ϵ1⋀ϵ2), and ∫𝒳D¯k(x,y)dμ(y)=∫𝒳D¯k(x,y)dμ(x)=0.To establish some Calderón reproducing formulae in
spaces of distributions, we first need to understand the action of the
operatorsDk on spaces of distributions.To this end, for allx,y∈𝒳,
let φ(x,y) satisfy the conditions (i) through (iii) with k=0 of Definition 2.2, and ∫𝒳φ(x,y)dμ(y)=0=∫𝒳φ(x,y)dμ(x).
Let ϵ be as in  Theorem 3.10, 0<β,γ<ϵ, and f∈𝒢°0ϵ(β,γ). We then define
(3.95)Ψ(f)(x)=∫𝒳φ(x,y)f(y)dμ(y).Let u∈(𝒢°0ϵ(β,γ))′.
In analogy with the theory of distributions on ℝn,
there exist two ways to define Ψ(u)∈(𝒢°0ϵ(β,γ))′.
One way is to define Ψ(u)∈(𝒢°0ϵ(β,γ))′ by duality, that is, for all f∈𝒢°0ϵ(β,γ), we put
(3.96)〈Ψ(u),f〉=〈u,Ψt(f)〉,
where Ψt denotes the integral operator with the kernel φt(x,y)=φ(y,x) for all x,y∈𝒳.
Alternatively, we define pointwise
(3.97)Ψ˜(u)(x)=〈u,φ(x,⋅)〉.
We now show that both definitions actually coincide.Lemma 3.12.
Letϵ be as in Theorem 3.10 and let 0<β,γ<ϵ.
Let u∈(𝒢°0ϵ(β,γ))′ and let Ψ(u) and Ψ˜(u) be defined, respectively, as in (3.96) and
(3.97). Then Ψ(u)=Ψ˜(u) in (𝒢°0ϵ(β,γ))′.Proof.
To establish this lemma, it suffices to show that for allf∈𝒢°(ϵ,ϵ),
(3.98)∫𝒳Ψ˜(u)(x)f(x)dμ(x)=〈u,Ψt(f)〉.
To this end, for L∈ℕ large enough, we define
(3.99)TL(f)(x)=∫B(x1,L)φ(y,x)f(y)dμ(y).
Let β,γ be as in the lemma. Using Theorem 2.18 and some
routine computations, we have TL(f)∈𝒢°(ϵ,ϵ) and limL→∞∥Ψt(f)−TL(f)∥𝒢(β,γ)=0.
Thus, 〈u,Ψt(f)〉=limL→∞〈u,TL(f)〉.
Now for fixedL∈ℕ large enough, and for any J∈ℕ,
let NJ={i∈IJ:QiJ∩B(x1,L)≠∅}, where {QiJ}J∈ℕ,i∈IJ are dyadic cubes of 𝒳 as in Lemma 2.19. If QiJ∩B(x1,L)≠∅ and L is large enough, then B(x1,L)⊂B(ziJ,3L) and QiJ⊂B(x1,2L),
where ziJ is the center of QiJ as in Lemma 2.19. These facts imply that ♯NJ≲(L2j)n.
By Lemma 2.19, we write
(3.100)TL(f)(x)=∑i∈NJ∫QiJ∩B(x1,L)[φ(y,x)−φ(yQiJ,x)]f(y)dμ(y)+∑i∈NJφ(yQiJ,x)∫QiJ∩B(x1,L)f(y)dμ(y)≡TL,J1(f)(x)+TL,J2(f)(x),
where yQiJ is any point in QiJ∩B(x1,L).
For fixed L∈ℕ large enough, using Theorem 2.18 and some
routine computations again, we have TL,J1(f)∈𝒢°(ϵ,ϵ) and
(3.101)limJ→∞∥TL,J1(f)∥𝒢(β,γ)=0.
Thus,
(3.102)〈u,Ψt(f)〉=limL→∞limJ→∞〈u,TL,J2(f)〉=limL→∞limJ→∞∑i∈NJΨ˜(u)(yQiJ)∫QiJ∩B(x1,L)f(y)dμ(y)=∫𝒳Ψ˜(u)(y)f(y)dμ(y)+limL→∞limJ→∞∑i∈NJ∫QiJ∩B(x1,L)[Ψ˜(u)(yQiJ)−Ψ˜(u)(y)]f(y)dμ(y).
It is not so difficult to verify that
(3.103)∑i∈NJ[φ(⋅,yQiJ)−φ(⋅,y)]χQiJ∩B(x1,L)(y)∈𝒢°(ϵ,ϵ),∥∑i∈NJ[φ(⋅,yQiJ)−φ(⋅,y)]χQiJ∩B(x1,L)(y)∥𝒢(β,γ)≤C2−Jϵ1
uniformly in y∈𝒳.
From this, it follows that |Ψ˜(u)(yQiJ)−Ψ˜(u)(y)|≤Cu2−Jϵ1 uniformly in y∈𝒳,
which along with Lebesgue's dominated convergence theorem shows that
(3.104)limJ→∞∑i∈NJ∫QiJ∩B(x1,L)[Ψ˜(u)(yQiJ)−Ψ˜(u)(y)]f(y)dμ(y)=0.
Thus, (3.98) is true and this
completes the proof of Lemma 3.12.Theorems3.10 and 3.11 in combination with a
duality argument and Lemma 3.12 show that continuous Calderón reproducing formulae also hold in spaces of distributions.Theorem 3.13.
Let all the notation be as in Theorems3.10 and 3.11. Then for all f∈(𝒢°0ϵ(β,γ))′ with 0<β,γ<ϵ,
(3.68) and (3.94) hold in (𝒢°0ϵ(β,γ))′.Finally, let us end this subsection by establishing aLittlewood-Paley theorem associated to ATIs via Theorem 3.10, which is a generalization
of [44, Proposition 2.5.1]. However, the method used here is different from that
in [44].To this end, we need the following Fefferman-Stein
vector-valued maximal function inequality in [77]; see also [44, Equation (2.11)] and  [75, Chapter II, Section 1].Lemma 3.14.
Let1<p<∞, 1<q≤∞, and let M be the Hardy-Littlewood maximal operator on 𝒳.
Let {fk}k∈ℤ⊂Lp(𝒳) be a sequence of measurable functions on 𝒳. Then
(3.105)∥{∑k=−∞∞|M(fk)|q}1/q∥Lp(𝒳)≤C∥{∑k=−∞∞|fk|q}1/q∥Lp(𝒳),
where C is independent of {fk}k∈ℤ.Proposition 3.15.
Letϵ1∈(0,1], ϵ2>0, ϵ3>0, and let {Sk}k∈ℤ be an (ϵ1,ϵ2,ϵ3)-ATI. Let Dk=Sk−Sk−1 for k∈ℤ and let g˙(f) for f∈Lp(𝒳) with p∈(1,∞) be as in (3.63). Then there exists a constant Cp>0 such that for all f∈Lp(𝒳),
(3.106)Cp−1∥f∥Lp(𝒳)≤∥g˙(f)∥Lp(𝒳)≤Cp∥f∥Lp(𝒳).Proof.
By Lemma3.9, we only need to verify the first inequality. To this end,
for any f∈Lp(𝒳),
Theorem 3.10 shows that there exist operators {D˜k}k∈ℤ as in Theorem 3.10 such that
(3.107)f=∑k=−∞∞D˜kDk(f)
in Lp(𝒳).
For 1<p<∞,
let 1/p+1/p′=1.
We first claim that for any h∈Lp′(𝒳),
(3.108)∥{∑k=−∞∞|D˜kt(h)|2}1/2∥Lp′(𝒳)≲∥h∥Lp′(𝒳).
Let ϵ∈(0,ϵ1⋀ϵ2).
In fact, by (3.2) of Lemma 3.2 and Remark 3.3, we have that for any ϵ1′∈(0,ϵ1⋀ϵ2),(3.109)|D˜ktD˜l(x,y)|≲2−|k−l|ϵ′1V2−(k⋀l)(x)+V2−(k⋀l)(y)+V(x,y)2−(k⋀l)ϵ(2−(k⋀l)+d(x,y))ϵ.From this and Lemma 2.1(iv), it
follows that for x∈𝒳,(3.110)|DktD˜l(h)(x)|≲2−|k−l|ϵ′∫𝒳1V2−(k⋀l)(x)+V2−(k⋀l)(y)+V(x,y)2−(k⋀l)ϵ(2−(k⋀l)+d(x,y))ϵ|h(y)|dμ(y)≲2−|k−l|ϵ′M(h)(x),which together with (3.107),
Hölder's inequality, Lemma 3.14, and Lemma 3.9 yields that
(3.111)∥{∑k=−∞∞|D˜kt(h)|2}1/2∥Lp′(𝒳)≲∥{∑k=−∞∞[∑l=−∞∞2−|k−l|ϵ′M(Dl(h))]2}1/2∥Lp′(𝒳)≲∥{∑l=−∞∞[M(Dl(h))]2}1/2∥Lp′(𝒳)≲∥h∥Lp′(𝒳).
Thus, (3.108) holds.
Using (3.108) and (3.107) together with a duality
argument and Hölder's inequality gives
(3.112)∥f∥Lp(𝒳)=sup∥h∥Lp′(𝒳)≤1|〈∑k=−∞∞D˜kDk(f),h〉|≤sup∥h∥Lp′(𝒳)≤1∥g˙(f)∥Lp(𝒳)∥{∑k=−∞∞|D˜kt(h)|2}1/2∥Lp′(𝒳)≲∥g˙(f)∥Lp(𝒳),
which completes the proof of Proposition 3.15.Remark 3.16.
From the proof of Proposition3.15, it is easy to see that Lemma 3.9 is
still true if Dk there is replaced by D˜kt for k∈ℤ,
which has regularity only in the second variable; see (3.108).


### 3.2. Inhomogeneous Continuous Calderón Reproducing Formulae

In this subsection, we have no restriction ondiam(𝒳),
which means diam(𝒳)<∞ or diam(𝒳)=∞.
We first introduce the following inhomogeneous approximation of the identity on 𝒳.Definition 3.17.
Letϵ1∈(0,1], ϵ2>0,
and ϵ3>0. A sequence {Sk}k∈ℤ+ of linear operators is said to be an inhomogeneous approximation of the identity of
order(ϵ1,ϵ2,ϵ3) (for short, (ϵ1,ϵ2,ϵ3)-IATI) if Sk for k∈ℤ+ satisfies Definition 2.2.
A sequence{Sk}k∈ℤ+ of linear operators is said to be an inhomogeneous approximation of the identity of
orderϵ1 with bounded support (for short, ϵ1-IATI with bounded support) if Sk for k∈ℤ+ satisfies Definition 2.3.The following proposition is a simple corollary of
Proposition2.7(iv) and (v).Proposition 3.18.
Letϵ1∈(0,1], ϵ2>0, ϵ3>0, let {Sk}k∈ℤ+ be an (ϵ1,ϵ2,ϵ3)-IATI, and let Skt be the adjoint operator of Sk for any k∈ℤ+.
Let Dk=Sk−Sk−1 for k∈ℕ and D0=S0.
Then I=∑k=0∞Dk in Lp(𝒳) for p∈[1,∞).
The same is true for {Skt}k∈ℤ+.To establish the continuous inhomogeneous Calderón reproducing formulae, we need a technical lemma, which is a variant of Lemma3.2.Lemma 3.19.
Letϵ1∈(0,1], ϵ2>0,
and ϵ3>0, {Sk}k∈ℤ+,
and let {Ek}k∈ℤ+ be two (ϵ1,ϵ2,ϵ3)-IATIs. Let Pk=Sk−Sk−1 and Qk=Ek−Ek−1 for k∈ℕ, P0=S0,
and Q0=E0.
Then for any ϵ1′∈(0,ϵ1⋀ϵ2), there exist constants C>0, δ>0, and σ>0 as in Lemma 3.2 such that the estimates (3.2)
to (3.5) are still true for these {Pl}k∈ℤ+ and {Qk}k∈ℤ+.Proof.
The proof of Lemma3.19 is essentially as in that of Lemma 3.2. The only
different situations are the cases when l=0 or k=0.
Let us prove (3.2) for l=0=k to show the difference. In this case, by the
size condition of P0 and Q0,
we have
(3.113)|P0Q0(x,y)|=|∫𝒳S0(x,z)Q0(z,y)dμ(z)|≲∫d(x,z)≤(1/2)d(x,y)1V1(x)+V1(z)+V(x,z)1(1+d(x,z))ϵ2×1V1(z)+V1(y)+V(z,y)1(1+d(z,y))ϵ2dμ(z)+∫d(x,z)>(1/2)d(x,y)⋯.Since d(x,z)>(1/2)d(x,y) implies that d(y,z)≤d(x,y)/2,
by symmetry, the estimates of the first and the second terms are similar and we
only estimate the first term. To this end, since d(x,z)≤d(x,y)/2≤(1+d(x,y))/2,
by Lemma 2.1(iii), we have 1/(1+d(z,y))≲1/(1+d(x,y)) and 1/(V1(z)+V(z,y))≲1/(V1(x)+V(x,y)), which further implies that 1/(V1(z)+V1(y)+V(z,y))≲1/(V1(x)+V1(y)+V(x,y)). These estimates together with Lemma 2.1(ii)
yield that
(3.114)|P0Q0(x,y)|≲1V1(x)+V1(y)+V(x,y)1(1+d(x,y))ϵ2,which is the desired estimate
and hence completes the proof of Lemma 3.19.Now, letϵ1∈(0,1], ϵ2>0, ϵ3>0, and {Sk}k∈ℤ+ be an (ϵ1,ϵ2,ϵ3)-IATI as in Definition 3.17. Throughout this subsection,
we always assume that Dk=Sk−Sk−1 for k∈ℕ, D0=S0 and Dk=0 if k∈{−1,−2,…}.
Similar to (3.44), by Proposition 3.18, for any N∈ℕ, we write
(3.115)I=(∑k=0∞Dk)(∑j=0∞Dj)=∑|l|>N∑k=0∞Dk+lDk+∑k=0∞DkNDk=RN+TN
in Lp(𝒳) with p∈(1,∞),
where DkN=∑|l|≤NDk+l.Repeating the proof of Lemma3.4, we
obtain the following.Lemma 3.20.
LetN∈ℕ and let RN be as in (3.115). Then there exist constants C>0 and δ>0,
independent of N,
such that for all f∈L2(𝒳),
(3.116)∥RN(f)∥L2(𝒳)≤C2−Nδ∥f∥L2(𝒳).We now establish some estimates for the kernel,RN(x,y),
of the operator RN.
To this end, we first give a technical lemma, which is a simple corollary of
Lemma 3.5.Lemma 3.21.
For anyσ≥s>0 and x,y∈𝒳 with x≠y,
(3.117)∑k=0∞1V2−k(x)+V(x,y)2−ks(2−k+d(x,y))σ≲1V(x,y)1d(x,y)σ−s.Lemma 3.22.
LetN∈ℕ,
let RN be as in (3.115), and let RN(x,y) be its kernel. Then for any ϵ∈(0,ϵ1⋀ϵ2), there exists a constant δ>0,
independent of N,
such that RN satisfies all the conditions of Corollary 2.22
and CRN≲2−δN.
Moreover, RN∗(1)=0.Replacing Lemmas3.2 and 3.5, respectively, by
Lemmas 3.19 and 3.21, and repeating the proof of Lemma 3.6, we obtain Lemma
3.22. In combination with Lemma 3.20, this leads to the following variant of Proposition
3.7. We omit  the details.Proposition 3.23.
Letϵ1∈(0,1], ϵ2>0, ϵ3>0, and let {Sk}k∈ℤ+ be an (ϵ1,ϵ2,ϵ3)-IATI.
For N∈ℕ,
let RN and TN be as in (3.115). Then there exist constants C8>0 and δ>0,
which are independent of N,
such that RN is bounded on any space of test functions, 𝒢°(x1,r,β,γ) with x1∈𝒳, r>0, and 0<β,γ<(ϵ1⋀ϵ2), and its operator norm is bounded by C82−Nδ.
Moreover, if N is so large that (3.57) holds, then TN−1 exists and is bounded on 𝒢°(x1,r,β,γ),
if x1∈𝒳, r>0 and 0<β,γ<(ϵ1⋀ϵ2).Via Lemmas3.20 and 3.22, similar to the proof of Proposition 3.8, we can obtain the following version of Proposition 3.8.Proposition 3.24.
Letp∈(1,∞), ϵ1∈(0,1], ϵ2>0, ϵ3>0, and let {Sk}k∈ℤ+ be an (ϵ1,ϵ2,ϵ3)-IATI.
For N∈ℕ,
let RN and TN be as in (3.115). Then there exist constants C9>0 and δ>0,
which are independent of N,
such that RN is bounded on Lp(𝒳) with the operator norm bounded by C92−Nδ.
Moreover, if N is so large that (3.61) holds, then TN−1 exists and is bounded on Lp(𝒳).Letϵ1∈(0,1], ϵ2>0, ϵ3>0, and let {Sk}k∈ℤ+ be an (ϵ1,ϵ2,ϵ3)-IATI.
Set Dk=Sk−Sk−1 for k∈ℕ and D0=S0.
For any f∈Lp(𝒳) with p∈(1,∞) and x∈𝒳,
the inhomogeneous Littlewood-Paleyg-functiong(f) is defined by
(3.118)g(f)(x)={∑k=0∞|Dk(f)(x)|2}1/2.
Applying the Cotlar-Stein lemma
and using a procedure similar to the proof of Lemma 3.9, we obtain the
boundedness on Lp(𝒳) with p∈(1,∞) for the Littlewood-Paley g-function as  below.Lemma 3.25.
Letϵ1∈(0,1], ϵ2>0, ϵ3>0, and let {Sk}k∈ℤ+ be an (ϵ1,ϵ2,ϵ3)-IATI.
Let Dk=Sk−Sk−1 for k∈ℕ, D0=S0, and g(f) for f∈Lp(𝒳) with p∈(1,∞) being as in (3.118). Then there exists a
constant Cp>0 such that for all f∈Lp(𝒳),
(3.119)∥g(f)∥Lp(𝒳)≤Cp∥f∥Lp(𝒳).We can now establish aninhomogeneous continuous Calderón reproducing formula.Theorem 3.26.
Letϵ1∈(0,1], ϵ2>0, ϵ3>0, ϵ∈(0,ϵ1⋀ϵ2), and let {Sk}k∈ℤ+ be an (ϵ1,ϵ2,ϵ3)-IATI. Set Dk=Sk−Sk−1 for k∈ℤ and D0=S0.
Then there exist N∈ℕ and a family of linear operators {D˜k}k∈ℤ+ such that for all f∈𝒢0ϵ(β,γ) with 0<β,γ<ϵ,
(3.120)f=∑k=0∞D˜kDk(f),
where the series converges both
in the norm of 𝒢0ϵ(β,γ) and the norm of Lp(𝒳) for p∈(1,∞).
Moreover, the kernels of the operators D˜k satisfy the conditions (i) and (ii) of
Definition 2.2 with ϵ1 and ϵ2 replaced by ϵ′∈(ϵ,ϵ1⋀ϵ2), and ∫𝒳D˜k(x,y)dμ(y)=∫𝒳D˜k(x,y)dμ(x)=1 when 0≤k≤N; =0 when k>N.Proof.
We prove this theorem by an argument similar to that of Theorem3.10. Fix
a large integer N∈ℕ such that (3.57) and (3.61) hold and,
therefore, Propositions 3.23 and 3.24 hold. Let DkN be as in (3.115). Then,
(3.121)DkN=∑|j|≤NDk+j={∑j=0k+NDj,for0≤k≤N;∑j=k−Nk+NDj,fork>N.
Thus, for k>N, DkN(⋅,y)∈𝒢°(y,2−k,ϵ1,ϵ2),
and we then define(3.122)D˜k(x,y)=TN−1(DkN(⋅,y))(x).By Proposition 3.23, we know that
for
k>N and any 0<β,γ<(ϵ1⋀ϵ2), D˜k(⋅,y)∈𝒢(y,2−k,β,γ).
Moreover, (TN−1)∗(1)=∑i=0∞(RN∗)i(1)=1,
which implies that
(3.123)∫𝒳D˜k(x,y)dμ(x)=∫𝒳DkN(x,y)dμ(x)=0.
Obviously, we have
(3.124)∫𝒳D˜k(x,y)dμ(y)=TN−1{∫𝒳DkN(⋅,y)dμ(y)}(x)=0.Thus, D˜k for k>N satisfies all the conditions of Theorem 3.26.
If k∈{0,…,N}, then
(3.125)∫𝒳DkN(x,y)dμ(x)=1=∫𝒳DkN(x,y)dμ(y).
Thus, in this case, we cannot
directly apply Proposition 3.23. Notice that
(3.126)TN−1=∑i=0∞(RN)i.
Using Lemma 3.19, we easily show
that for ϵ1′∈(0,ϵ1⋀ϵ2), there exist positive constants C˜ and δ depending on ϵ1′ such that RN(x,y) satisfies (2.174) and (2.185) with r0=1 and CT≲C˜2−Nδ.
Thus, by Corollary 2.31, we know that RN(S0(⋅,y))∈𝒢(y,1,ϵ1′,ϵ2) and(3.127)∥RN(S0(⋅,y))∥𝒢°(y,1,ϵ1′,ϵ2)≤C˜2−Nδ.
Notice that RN∗(1)=0.
Applying Proposition 3.23 yields that there exist two positive constants C8 and δ,
which are independent of N,
such that for all j∈ℕ and all y∈𝒳, RNj(S0(⋅,y))∈𝒢°(y,1,β,γ) and
(3.128)∥RNj(S0(⋅,y))∥𝒢°(y,1,β,γ)≤(C82−Nδ)j,
where β,γ∈(0,ϵ1⋀ϵ2).
Thus, if we choose N so large that (3.57) holds, then by (3.126), we
know that D˜k for k∈{0,1,…,N} satisfies the conditions (i) and (ii) of
Definition 2.2 with ϵ1 and ϵ2 replaced by ϵ∈(0,ϵ1⋀ϵ2). Moreover, from (3.126) together with RN(1)=0=RN∗(1), ∫𝒳S0(x,y)dμ(y)=1=∫𝒳S0(x,y)dμ(x) and for k∈ℕ, ∫𝒳Dk(x,y)dμ(y)=0=∫𝒳Dk(x,y)dμ(x), it follows that for k∈{0,1,…,N}, ∫𝒳D˜k(x,y)dμ(y)=1=∫𝒳D˜k(x,y)dμ(x).
Now it remains to prove that the series in (3.120)
converges in both the norm of 𝒢°0ϵ(β,γ) with 0<β,γ<ϵ and the norm of Lp(𝒳) with p∈(1,γ).
To verify (3.120) converges in 𝒢0ϵ(β,γ) with β,γ∈(0,ϵ),
similarly to (3.69), for f∈𝒢0ϵ(β′,γ′) with β′∈(β,ϵ˜) and γ′∈(γ,ϵ˜),
and L∈ℕ with L>N+1,
we have
(3.129)∑k=0LD˜kDk(f)=f−limj→∞(RN)j(f)−TN−1(∑k=L+1∞DkNDk)(f).
Thus,
(3.130)∥∑k=0LD˜kDk(f)−f∥𝒢(β,γ)≤limj→∞∥(RN)j(f)∥𝒢(β,γ)+∥TN−1(∑k=L+1∞DkNDk)(f)∥𝒢(β,γ).
Applying Corollary 2.31 again, we
obtain that there exists δ>0 such that for all x∈𝒳,
(3.131)|RN(f)(x)|≲2−Nδ1V1(x1)+V(x1,x)1(1+d(x1,x))γ′,and for any β∈(0,β′),
and x,x′∈𝒳 with d(x,x′)≤(1+d(x1,x))/2,
(3.132)|RN(f)(x)−RN(f)(x′)|≲2−Nδ(d(x,x′)1+d(x1,x))β1V1(x1)+V(x1,x)1(1+d(x1,x))γ′.
Notice that RN∗(1)=0,
which together with (3.131) and (3.132) shows that RN(f)∈𝒢°(β,γ).
From this and Proposition 3.23, it follows that
(3.133)limj→∞∥(RN)j(f)∥𝒢(β,γ)≤limj→∞(C82−Nδ)j∥f∥𝒢(β,γ)=0.
To prove that the second term on the right-hand side
of (3.130) tends to 0 as L→∞,
by (DkN)*(1)=0 when L≥N (see (3.121)), and Proposition 3.24, we only
need to verify that for a certain σ>0,
(3.134)∥∑k=L+1∞DkNDk(f)∥𝒢(β,γ)≲2−σL∥f∥𝒢(β′,γ′).
In fact, (3.134) can be deduced
from the following two estimates that for all x∈𝒳,
(3.135)|∑k=L+1∞DkNDk(f)(x)|≲2−(β′⋀γ′)L∥f∥𝒢(β′,γ′)1V1(x1)+V(x1,x)1(1+d(x1,x))γ′,and for x,x′∈𝒳 with d(x,x′)≤(1+d(x1,x))/2,(3.136)|∑k=L+1∞DkNDk(f)(x)−∑k=L+1∞DkNDk(f)(x′)|≲2−(1−θ)(β′⋀γ′)L∥f∥𝒢(β′,γ′)(d(x,x′)1+d(x1,x))θβ′1V1(x1)+V(x1,x)1(1+d(x1,x))γ′,
where θ∈(0,1).
LetEk=DkNDk.
It is easy to check that Ek for k∈ℕ satisfies all estimates of Definition 2.2 with ϵ replaced by  ϵ˜ and a coefficient depending on N.
Moreover, for k∈ℕ, Ek(1)=0=Ek∗(1).
Thus, by Lemma 2.1(ii),
(3.137)|DkNDk(f)(x)|≤∫d(x,y)≤(1+d(x1,x))/2|Ek(x,y)||f(y)−f(x)|dμ(y)+∫d(x,y)>(1+d(x1,x))/2|Ek(x,y)|[|f(y)|+|f(x)|]dμ(y)≤∥f∥𝒢(β′,γ′){∫d(x,y)≤(1+d(x1,x))/2|Ek(x,y)|(d(x,y)1+d(x1,x))β′×1V1(x1)+V(x1,x)1(1+d(x1,x))γ′dμ(y)+∫d(x,y)>(1+d(x1,x))/2|Ek(x,y)|[1V1(x1)+V(x1,y)1(1+d(x1,y))γ′+1V1(x1)+V(x1,x)1(1+d(x1,x))γ′]dμ(y)}≤∥f∥𝒢(β′,γ′){2−kβ′1V1(x1)+V(x1,x)1(1+d(x1,x))γ′+Z},
and by Lemma 2.1(i) and (ii), we further have
(3.138)Z≲∫d(x,y)>(1+d(x1,x))/21V2−k(x)+V(x,y)2−kϵ2(2−k+d(x,y))ϵ21V1(x1)+V(x1,y)1(1+d(x1,y))γ′dμ(y)≲min{2−kγ′V1(x1)∫d(x,y)>(1+d(x1,x))/21V(x,y)1d(x,y)γ′dμ(y),2−kγ′V(x1,x)1(1+d(x1,x))γ′∫𝒳1V1(x1)+V(x1,y)1(1+d(x1,y))γ′dμ(y)}≲2−kγ′1V1(x1)+V(x1,x)1(1+d(x1,x))γ′.
From this, it is easy to deduce that (3.135) holds.
On the other hand, similarly to the proof of (3.12), we
can verify that for x,x′∈𝒳 with d(x,x′)≤(1+d(x,x1))/4,
(3.139)|DkNDk(f)(x)−DkNDk(f)(x′)|≲∥f∥𝒢(β′,γ′)(d(x,x′)1+d(x1,x))β′1V1(x1)+V(x1,x)1(1+d(x1,x))γ′,
and by (3.137) together with
Lemma 2.1(iii),
(3.140)|DkNDk(f)(x)−DkNDk(f)(x′)|≲2−k(β′⋀γ′)∥f∥𝒢(β′,γ′)1V1(x1)+V(x1,x)1(1+d(x1,x))γ′.
These estimates together with
the geometric means yield (3.136). Thus,
(3.141)limL→∞∥∑k=0LD˜kDk(f)−f∥𝒢(β,γ)=0.
Then repeating the proof of
Theorem 3.10 further shows that (3.120) holds in the norm of 𝒢0ϵ(β,γ).
Its convergence in Lp(𝒳) for p∈(1,∞) can be proved in a way similar to Theorem 3.10
if instead of Lemma 3.9 by Lemma 3.25, which completes the proof of Theorem
3.26.Remark 3.27.
From the proof of Theorem3.26, it is easy to see that N in Theorem 3.26 can be chosen so that both
(3.57) and (3.61) are satisfied. In order to defray our notation and simplify
our presentation, we will assume in the sequel that N=0.A similar argument as for the proof of Theorem3.26
leads to the following variant of the
inhomogeneous continuous Calderón reproducing formula (we omit the
details).Theorem 3.28.
Letϵ1∈(0,1], ϵ2>0, ϵ3>0, ϵ∈(0,ϵ1⋀ϵ2), and let {Sk}k∈ℤ+ be an (ϵ1,ϵ2,ϵ3)-IATI.
Set Dk=Sk−Sk−1 for k∈ℤ and D0=S0.
Then there exists a family of linear operators {D¯k}k∈ℤ+ such that for all f∈𝒢0ϵ(β,γ) with 0<β,γ<ϵ,
(3.142)f=∑k=0∞DkD¯k(f),
where the series converges in
both the norm of 𝒢0ϵ(β,γ) and the norm of Lp(𝒳) for p∈(1,∞).
Moreover, the kernels of the operators D¯k satisfy the conditions (i) and (iii) of
Definition 2.2 with ϵ1 and ϵ2 replaced by ϵ′∈(ϵ,ϵ1⋀ϵ2), and ∫𝒳D¯k(x,y)dμ(y)=∫𝒳D¯k(x,y)dμ(x)=1 when k=0; =0 when k∈ℕ.Theorems3.26 and 3.28 in combination with a
duality argument and Lemma 3.12 show that the inhomogeneous Calderón reproducing formulae also hold in spaces of distributions.Theorem 3.29.
Let all the notation be as in Theorems3.26 and
3.28. Then for all f∈(𝒢0ϵ(β,γ))′ with 0<β,γ<ϵ,
(3.120) and (3.142) hold in (𝒢0ϵ(β,γ))′.Finally, we have the following analogue of Proposition3.15 for the inhomogeneous Littlewood-Paleyg-function, based on Theorem 3.26 (we omit the proof).Proposition 3.30.
Letϵ1∈(0,1], ϵ2>0, ϵ3>0, and let {Sk}k∈ℤ+ be an (ϵ1,ϵ2,ϵ3)-IATI.
Set Dk=Sk−Sk−1 for k∈ℤ, D0=S0 and g(f) for f∈Lp(𝒳) with p∈(1,∞) be as in (3.118). Then there exists a constant Cp>0 such that for all f∈Lp(𝒳),
(3.143)Cp∥f∥Lp(𝒳)≤∥g(f)∥Lp(𝒳)≤Cp∥f∥Lp(𝒳).Remark 3.31.
Similarly as in Remark3.16, it
is easy to see that Lemma 3.25 is still true if Dk therein is replaced by the kernel D˜kt for k∈ℤ+,
which has regularity only in the second variable.


## 3.1. Homogeneous Continuous Calderón Reproducing Formulae

In this subsection, we always assume thatdiam(𝒳)=∞.Proposition 3.1.
Let{Sk}k∈ℤ be an (ϵ1,ϵ2,ϵ3)-ATI with ϵ1∈(0,1],ϵ2>0, and ϵ3>0 and let Skt be the adjoint operator of Sk for any k∈ℤ.
Then the following hold.
(i)
Forp∈(1,∞) and any f∈Lp(𝒳), ∥Skf∥Lp(𝒳)→0 when k→−∞.(ii)
LetDk=Sk−Sk−1 for k∈ℤ.
Then ∑k=−∞∞Dk=I in Lp(𝒳) for p∈(1,∞),
where I is the identity on Lp(𝒳).(iii)
Properties (i) and (ii) also hold whenSk is replaced by Skt.Proof.
Let1/p+1/q=1.
Then Definition 2.2(i), Hölder's inequality, (1.2) and (1.3) together with μ(𝒳)=∞ yield that
(3.1)|Skf(x)|≲1V2−k(x)∫d(x,y)<2−k|f(y)|dμ(y)+∫d(x,y)≥2−k1V(x,y)|f(y)|dμ(y)≲{1V2−k(x)1/p+[∑l=0∞∫2l2−k≤d(x,y)<2l+12−k1V(x,y)qdμ(y)]1/q}∥f∥Lp(𝒳)≲1V2−k(x)1/p∥f∥Lp(𝒳)→0,
as k→−∞.
This together with Proposition 2.7(ii) and Lebesgue's dominated convergence
theorem implies that ∥Skf∥Lp(𝒳)→0 when k→−∞.
Property (ii) is a simple corollary of Property (i)
together with Proposition2.7(iv). Property (iii) follows by symmetry, which
completes the proof of Proposition 3.1.Before we establish the continuous Calderón reproducing
formulae, we need a technical lemma. We recall that we denotemin{a,b} by a⋀b for any a,b∈ℝ.Lemma 3.2.
Letϵ1∈(0,1], ϵ2>0, ϵ3>0, {Sk}k∈ℤ,
and let {Ek}k∈ℤ be two (ϵ1,ϵ2,ϵ3)-ATIs. Let Pk=Sk−Sk−1 and Qk=Ek−Ek−1 for k∈ℤ.
Then for any ϵ1′∈(0,ϵ1⋀ϵ2),
there exist positive constants C (depending on ϵ1′, ϵ1, ϵ2,
and ϵ3), δ (depending on ϵ1′, ϵ1, and ϵ2), and σ (depending on ϵ1′, ϵ1, ϵ2,
and ϵ3) such that PlQk(x,y),
the kernel of PlQk,
satisfies the following estimates that for all x,y∈𝒳 and all k,l∈ℤ,
(3.2)|PlQk(x,y)|≤C2−|k−l|ϵ1′1V2−(k⋀l)(x)+V2−(k⋀l)(y)+V(x,y)2−(k⋀l)ϵ2(2−(k⋀l)+d(x,y))ϵ2;
for d(y,y′)≤(1/4)d(x,y) and all k,l∈ℤ,
(3.3)|PlQk(x,y)−PlQk(x,y′)|≤C2−|k−l|δ(d(y,y′)2−(l⋀k)+d(x,y))ϵ1′1V2−(k⋀l)(x)+V2−(k⋀l)(y)+V(x,y)2−(k⋀l)ϵ2(2−(k⋀l)+d(x,y))ϵ2;
for d(x,x′)≤(1/4)d(x,y) and all k,l∈ℤ,
(3.4)|PlQk(x,y)−PlQk(x′,y)|≤C2−|k−l|δ(d(x,x′)2−(l⋀k)+d(x,y))ϵ1′1V2−(k⋀l)(x)+V2−(k⋀l)(y)+V(x,y)2−(k⋀l)ϵ2(2−(k⋀l)+d(x,y))ϵ2;
and for d(x,x′)≤(1/8)d(x,y), d(y,y′)≤(1/8)d(x,y), and all k,l∈ℤ,
(3.5)|[PlQk(x,y)−PlQk(x′,y)]−[PlQk(x,y′)−PlQk(x′,y′)]|≤C2−|k−l|δ(d(x,x′)2−(l⋀k)+d(x,y))ϵ1′(d(y,y′)2−(l⋀k)+d(x,y))ϵ1′×1V2−(k⋀l)(x)+V2−(k⋀l)(y)+V(x,y)2−(k⋀l)σ(2−(k⋀l)+d(x,y))σ.Proof.
By symmetry, we may assume thatk≥l.
Noticing that(3.6)∫𝒳Qk(x,y)dμ(x)=0,
we can write
(3.7)|PlQk(x,y)|=|∫𝒳[Pl(x,z)−Pl(x,y)]Qk(z,y)dμ(z)|≤∫d(z,y)≤(2−l+d(x,y))/2|Pl(x,z)−Pl(x,y)||Qk(z,y)|dμ(z)+∫d(z,y)>(2−l+d(x,y))/2|Pl(x,z)||Qk(z,y)|dμ(z)+|Pl(x,y)|∫d(z,y)>(2−l+d(x,y))/2|Qk(z,y)|dμ(z)≡Z1+Z2+Z3.
The regularity of Pl and the size condition of Qk together with Lemma 2.1(i) show that for any ϵ1′∈(0,ϵ1⋀ϵ2),
(3.8)Z1≲2−kϵ1′1V2−l(x)+V2−l(y)+V(x,y)2−lϵ2(2−l+d(x,y))ϵ1+ϵ2×∫d(z,y)≤(2−l+d(x,y))/2d(z,y)ϵ1−ϵ1′1V(z,y)dμ(y)≲2−(k−l)ϵ1′1V2−l(x)+V2−l(y)+V(x,y)2−lϵ2(2−l+d(x,y))ϵ2.
The size conditions for Pl and Qk together with Lemma 2.1(i) prove that
(3.9)Z2≲1V2−l(x)2−kϵ2(2−l+d(x,y))ϵ1′∫d(z,y)>(2−l+d(x,y))/21V(z,y)1d(z,y)ϵ2−ϵ1′dμ(z)≲2−(k−l)ϵ21V2−l(x)2−lϵ2(2−l+d(x,y))ϵ2,Z2≲2−kϵ2(2−l+d(x,y))ϵ2∫d(z,y)>(2−l+d(x,y))/2|Pl(x,z)|1V(z,y)dμ(z)≲2−(k−l)ϵ21V2−l(y)+V(x,y)2−lϵ2(2−l+d(x,y))ϵ2,
where in the last step, we used
Proposition 2.7(i) and the fact that 1/V(z,y)≲1/(V2−l(y)+V(x,y)), by d(z,y)>(2−l+d(x,y))/2 together with (1.2). Thus,
(3.10)Z2≲2−(k−l)ϵ1′1V2−l(x)+V2−l(y)+V(x,y)2−lϵ2(2−l+d(x,y))ϵ2.
Similarly, by Lemma 2.1(v),(3.11)Z3≲1V2−l(x)+V2−l(y)+V(x,y)∫d(z,y)>(2−l+d(x,y))/21V2−k(y)+V(z,y)2−kϵ2(2−k+d(z,y))ϵ2dμ(z)≲2−(k−l)ϵ21V2−l(x)+V2−l(y)+V(x,y)2−lϵ2(2−l+d(x,y))ϵ2,
which completes the proof of
(3.2).
The proofs for (3.3) and (3.4) are similar and we only
verify (3.3). To this end, it suffices to prove
(3.12)|PlQk(x,y)−PlQk(x,y′)|≲(d(y,y′)2−l+d(x,y))ϵ11V2−l(x)+V2−l(y)+V(x,y)2−lϵ2(2−l+d(x,y))ϵ2
for d(y,y′)≤(1/4)d(x,y).
To see this, by (3.2), for any ϵ1′∈(0,ϵ1⋀ϵ2),
(3.13)|PlQk(x,y)−PlQk(x,y′)|≲2−(k−l)ϵ1′[1V2−l(x)+V2−l(y)+V(x,y)2−lϵ2(2−l+d(x,y))ϵ2+1V2−l(x)+V2−l(y′)+V(x,y′)2−lϵ2(2−l+d(x,y′))ϵ2].
The assumption that d(y,y′)≤(1/4)d(x,y) together with Lemma 2.1(iii) further yields that
(3.14)|PlQk(x,y)−PlQk(x,y′)|≲2−(k−l)ϵ1′1V2−l(x)+V2−l(y)+V(x,y)2−lϵ2(2−l+d(x,y))ϵ2.
Let σ∈(0,1).
Then the geometric mean between (3.12) and (3.14) gives
(3.15)|PlQk(x,y)−PlQk(x,y′)|=|PlQk(x,y)−PlQk(x,y′)|σ|PlQk(x,y)−PlQk(x,y′)|1−σ≲2−(k−l)ϵ1′(1−σ)(d(y,y′)2−l+d(x,y))σϵ11V2−l(x)+V2−l(y)+V(x,y)2−lϵ2(2−l+d(x,y))ϵ2,
which is (3.3).
We now verify (3.12). By (3.6), we can write
(3.16)|PlQk(x,y)−PlQk(x,y′)|=|∫𝒳[Pl(x,z)−Pl(x,y)][Qk(z,y)−Qk(z,y′)]dμ(z)|≤∑i=13∫Wi|Pl(x,z)−Pl(x,y)||Qk(z,y)−Qk(z,y′)|dμ(z)≡∑i=46Zi,
where
(3.17)W1={z∈𝒳:d(y,y′)≤2−l+d(x,y)4≤2−k+d(z,y)2},W2={z∈𝒳:d(y,y′)≤2−k+d(z,y)2≤2−l+d(x,y)4},W3={z∈𝒳:d(y,y′)≥2−k+d(z,y)2}.
The regularity forQk and the size condition for Pl together with the assumption k≥l implies that
(3.18)Z4≲∫W1[|Pl(x,z)|+1V2−l(x)+V2−l(y)+V(x,y)2−lϵ2(2−l+d(x,y))ϵ2]×d(y,y′)ϵ1(2−k+d(z,y))ϵ11V2−k(z)+V2−k(y)+V(z,y)2−kϵ2(2−k+d(z,y))ϵ2dμ(z)≡Z4,1+Z4,2.
We now claim that if k≥l and 2−l+d(x,y)≲2−k+d(z,y),
then for all x,y,z∈𝒳,
(3.19)1V2−k(y)+V(z,y)≲1V2−l(y)+V(x,y).
In fact, if 2−k≥d(z,y),
then 2−l+d(x,y)≲2−k+d(z,y)≲2−k,
and therefore V2−l(y)≲V2−k(y) and V(x,y)~V(y,x)≲V2−k(y).
Thus, (3.19) holds in this case. If 2−k<d(z,y),
then 2−l+d(x,y)≲d(z,y) and hence, V2−l(y)≲V(y,z)~V(z,y) and V(x,y)~V(y,x)≲V(z,y),
which verifies (3.19). Proposition 2.7(i) and the estimate (3.19) together with 2−l+d(x,y)≤2(2−k+d(z,y)) and k≥l imply that
(3.20)Z4,1≲d(y,y′)ϵ1(2−l+d(x,y))ϵ11V2−l(y)+V(x,y)2−lϵ2(2−l+d(x,y))ϵ2∫𝒳|Pl(x,z)|dμ(z)≲d(y,y′)ϵ1(2−l+d(x,y))ϵ11V2−l(y)+V(x,y)2−lϵ2(2−l+d(x,y))ϵ2,
and the size condition of Pl together with Lemma 2.1(i) yields that
(3.21)Z4,1≲d(y,y′)ϵ1(2−l+d(x,y))ϵ11V2−l(x){∫W1,d(z,y)≤2−k+∫W1,d(z,y)>2−k}1V2−k(y)+V(z,y)2−kϵ2(2−k+d(z,y))ϵ2dμ(z)≲d(y,y′)ϵ1(2−l+d(x,y))ϵ11V2−l(x)2−lϵ2(2−l+d(x,y))ϵ2.
Similarly, using Lemma 2.1(ii), we have
(3.22)Z4,2≲d(y,y′)ϵ1(2−l+d(x,y))ϵ11V2−l(x)+V2−l(y)+V(x,y)2−lϵ2(2−l+d(x,y))ϵ2×∫W11V2−k(y)+V(z,y)2−kϵ2(2−k+d(z,y))ϵ2dμ(z)≲d(y,y′)ϵ1(2−l+d(x,y))ϵ11V2−l(x)+V2−l(y)+V(x,y)2−lϵ2(2−l+d(x,y))ϵ2,
which completes the estimate for Z4.
The regularities ofPl and Qk and Lemma 2.1(ii) show that
(3.23)Z5≲∫W2d(y,z)ϵ1(2−l+d(x,y))ϵ11V2−l(x)+V2−l(y)+V(x,y)2−lϵ2(2−l+d(x,y))ϵ2×d(y,y′)ϵ1(2−k+d(y,z))ϵ11V2−k(y)+V2−k(z)+V(z,y)2−kϵ2(2−k+d(z,y))ϵ2dμ(z)≲d(y,y′)ϵ1(2−l+d(x,y))ϵ11V2−l(x)+V2−l(y)+V(x,y)2−lϵ2(2−l+d(x,y))ϵ2.From (1/4)d(x,y)≥d(y,y′)≥(1/2)(2−k+d(z,y)),
it follows that d(z,y)≤(1/2)d(x,y)≤(1/2)(2−l+d(x,y)). Then the regularity of Pl and the size condition of Qk together with Proposition 2.7(i) and d(y,z)≤2d(y,y′) prove that
(3.24)Z6≲1V2−l(x)+V2−l(y)+V(x,y)2−lϵ2(2−l+d(x,y))ϵ2×∫W3d(y,z)ϵ1(2−l+d(x,y))ϵ1[|Qk(z,y)|+|Qk(z,y′)|]dμ(z)≲d(y,y′)ϵ1(2−l+d(x,y))ϵ11V2−l(x)+V2−l(y)+V(x,y)2−lϵ2(2−l+d(x,y))ϵ2.
Thus, (3.12) holds and this finishes the proof of (3.3).
We finally prove (3.5) in the case k≥l.
To this end, it suffices to verify that for any ϵ1′∈(0,ϵ1⋀ϵ2), d(x,x′)≤(1/8)d(x,y) and d(y,y′)≤(1/8)d(x,y),
(3.25)|[PlQk(x,y)−PlQk(x′,y)]−[PlQk(x,y′)−PlQk(x′,y′)]|≲d(x,x′)ϵ1′(2−l+d(x,y))ϵ1′d(y,y′)ϵ1′(2−l+d(x,y))ϵ1′1V2−l(x)+V2−l(y)+V(x,y)2−lσ(2−l+d(x,y))σ,
where σ=(ϵ2−ϵ1′)⋀ϵ3>0. To see this, the estimate (3.4) implies that for any ϵ1′∈(0,ϵ1⋀ϵ2), d(x,x′)≤(1/8)d(x,y), and d(y,y′)≤(1/8)d(x,y),
(3.26)|[PlQk(x,y)−PlQk(x′,y)]−[PlQk(x,y′)−PlQk(x′,y′)]|≲2−(k−l)δd(x,x′)ϵ1′(2−l+d(x,y))ϵ1′1V2−l(x)+V2−l(y)+V(x,y)2−lϵ2(2−l+d(x,y))ϵ2+2−(k−l)δd(x,x′)ϵ1′(2−l+d(x,y′))ϵ1′1V2−l(x)+V2−l(y′)+V(x,y′)2−lϵ2(2−l+d(x,y′))ϵ2,
where δ>0 depends only on ϵ1′,
ϵ1,
and ϵ2. The assumption d(y,y′)≤(1/8)d(x,y) together with Lemma 2.1(iii) further shows
that for any ϵ1′∈(0,ϵ1⋀ϵ2), d(x,x′)≤(1/8)d(x,y) and d(y,y′)≤(1/8)d(x,y),
(3.27)|[PlQk(x,y)−PlQk(x′,y)]−[PlQk(x,y′)−PlQk(x′,y′)]|≲2−(k−l)δd(x,x′)ϵ1′(2−l+d(x,y))ϵ1′1V2−l(x)+V2−l(y)+V(x,y)2−lϵ2(2−l+d(x,y))ϵ2.
By the estimate (3.3) and
symmetry, we further obtain that for any ϵ1′∈(0,ϵ1⋀ϵ2), d(x,x′)≤(1/8)d(x,y) and d(y,y′)≤(1/8)d(x,y),
(3.28)|[PlQk(x,y)−PlQk(x′,y)]−[PlQk(x,y′)−PlQk(x′,y′)]|≲2−(k−l)δd(y,y′)ϵ1′(2−l+d(x,y))ϵ1′1V2−l(x)+V2−l(y)+V(x,y)2−lϵ2(2−l+d(x,y))ϵ2,
where δ>0 depends only on ϵ1′, ϵ1,
and ϵ2. Then the geometric mean among (3.25), (3.27), and (3.28) gives the desired estimate (3.5).
Using (3.6), we now prove (3.25) by writing that
(3.29)|[PlQk(x,y)−PlQk(x′,y)]−[PlQk(x,y′)−PlQk(x′,y′)]|=|∫𝒳{[Pl(x,z)−Pl(x′,z)]−[Pl(x,y)−Pl(x′,y)]}[Qk(z,y)−Qk(z,y′)]dμ(z)|≤∑i=14∫Wi|[Pl(x,z)−Pl(x′,z)]−[Pl(x,y)−Pl(x′,y)]||Qk(z,y)−Qk(z,y′)|dμ(z)≡∑i=14Yi,
where
(3.30)W1={z∈𝒳:d(y,y′)≤2−k+d(z,y)2≤2−l+d(x,y)8},W2={z∈𝒳:d(y,y′)≤2−l+d(x,y)8≤2−k+d(z,y)2,andd(x,x′)≤2−l+d(z,x)2},W3={z∈𝒳:d(y,y′)≤2−l+d(x,y)8≤2−k+d(z,y)2,andd(x,x′)>2−l+d(z,x)2},W4={z∈𝒳:d(y,y′)>2−k+d(z,y)2}.
Ifz∈W1,
then d(z,y)≤(1/4)(2−l+d(x,y)), d(x,x′)≤(1/8)d(x,y)≤(1/8)(2−l+d(x,y)), and d(y,y′)≤(1/2)(2−k+d(z,y)),
which together with the second difference condition of Pl,
the regularity of Qk and Lemma 2.1(ii) shows
(3.31)Y1≲d(x,x′)ϵ1(2−l+d(x,y))ϵ11V2−l(x)+V2−l(y)+V(x,y)2−lϵ3(2−l+d(x,y))ϵ3×∫W1d(y,z)ϵ1(2−l+d(x,y))ϵ1d(y,y′)ϵ1(2−k+d(z,y))ϵ1×1V2−k(z)+V2−k(y)+V(z,y)2−kϵ2(2−k+d(z,y))ϵ2dμ(z)≲d(x,x′)ϵ1(2−l+d(x,y))ϵ1d(y,y′)ϵ1(2−l+d(x,y))ϵ11V2−l(x)+V2−l(y)+V(x,y)2−lϵ3(2−l+d(x,y))ϵ3.
Ifz∈W4,
then (1/2)(2−k+d(z,y))<d(y,y′)≤(1/8)d(x,y),
and therefore, d(z,y)≤2d(y,y′)≤(1/4)d(x,y)≤(1/4)(2−l+d(x,y)) and d(x,x′)≤(1/8)d(x,y)≤(1/8)(2−l+d(x,y)).
From this, the regularity of Pl and Proposition 2.7(i), it follows that
(3.32)Y4≲d(x,x′)ϵ1(2−l+d(x,y))ϵ1d(y,y′)ϵ1(2−l+d(x,y))ϵ11V2−l(x)+V2−l(y)+V(x,y)×2−lϵ3(2−l+d(x,y))ϵ3∫𝒳[|Qk(z,y)|+|Qk(z,y′)|]dμ(z)≲d(x,x′)ϵ1(2−l+d(x,y))ϵ1d(y,y′)ϵ1(2−l+d(x,y))ϵ11V2−l(x)+V2−l(y)+V(x,y)2−lϵ3(2−l+d(x,y))ϵ3.
Ifz∈W2,
then d(x,x′)≤(1/2)(2−l+d(z,x)), d(y,y′)≤(1/2)(2−k+d(z,y)), and d(x,x′)≤(1/8)d(x,y)≤(1/8)(2−l+d(x,y)),
which together with the regularity of Pl and Qk implies that
(3.33)Y2≲∫W2[d(x,x′)ϵ1(2−l+d(x,z))ϵ11V2−l(x)+V2−l(z)+V(x,z)2−lϵ2(2−l+d(x,z))ϵ2+d(x,x′)ϵ1(2−l+d(x,y))ϵ11V2−l(x)+V2−l(y)+V(x,y)2−lϵ2(2−l+d(x,y))ϵ2]×d(y,y′)ϵ1(2−k+d(y,z))ϵ11V2−k(z)+V2−k(y)+V(z,y)2−kϵ2(2−k+d(z,y))ϵ2dμ(z)≡Y2,1+Y2,2.
The fact that z∈W2,
which implies that 2−k+d(z,y)≥(1/4)(2−l+d(x,y)),
together with Lemma 2.1(ii) yields that
(3.34)Y2,2≲d(x,x′)ϵ1(2−l+d(x,y))ϵ1d(y,y′)ϵ1(2−l+d(x,y))ϵ11V2−l(x)+V2−l(y)+V(x,y)×2−lϵ2(2−l+d(x,y))ϵ2∫𝒳1V2−k(y)+V(z,y)2−kϵ2(2−k+d(z,y))ϵ2dμ(z)≲d(x,x′)ϵ1(2−l+d(x,y))ϵ1d(y,y′)ϵ1(2−l+d(x,y))ϵ11V2−l(x)+V2−l(y)+V(x,y)2−lϵ2(2−l+d(x,y))ϵ2.
Similarly, the fact that 2−k+d(z,y)≥(1/4)(2−l+d(x,y)) implies that
(3.35)Y2,1≲d(y,y′)ϵ1(2−l+d(x,y))ϵ1∫W2d(x,x′)ϵ1(2−l+d(x,z))ϵ11V2−l(x)+V2−l(z)+V(x,z)×2−lϵ2(2−l+d(x,z))ϵ21V2−k(z)+V2−k(y)+V(z,y)2−kϵ2(2−k+d(z,y))ϵ2dμ(z).
The estimate (3.19) together with
the facts that k≥l and 2−k+d(z,y)≥(1/4)(2−l+d(x,y)) and Lemma 2.1(ii) yields that for any ϵ1′∈(0,ϵ1⋀ϵ2),
(3.36)Y2,1≲d(y,y′)ϵ1(2−l+d(x,y))ϵ11V2−l(y)+V(x,y)d(x,x′)ϵ1′2−lϵ2(2−l+d(x,y))ϵ2×∫𝒳1V2−l(x)+V(x,z)2−lϵ2(2−l+d(x,z))ϵ2+ϵ1′dμ(z)≲d(y,y′)ϵ1(2−l+d(x,y))ϵ11V2−l(y)+V(x,y)d(x,x′)ϵ1′(2−l+d(x,y))ϵ1′2−l(ϵ2−ϵ1′)(2−l+d(x,y))ϵ2−ϵ1′,
and similarly, Lemma 2.1(i) implies that
(3.37)Y2,1≲d(y,y′)ϵ1(2−l+d(x,y))ϵ11V2−l(x)d(x,x′)ϵ1′2lϵ1′∫W21V2−k(y)+V(z,y)2−kϵ2(2−k+d(z,y))ϵ2dμ(z)≲d(y,y′)ϵ1(2−l+d(x,y))ϵ11V2−l(x)d(x,x′)ϵ1′2lϵ1′×{2−lϵ2(2−l+d(x,y))ϵ2+2−lϵ2∫d(z,y)≥(1/8)(2−l+d(x,y)),d(z,y)≥2−k1V(z,y)1d(z,y)ϵ2dμ(z)}≲d(y,y′)ϵ1(2−l+d(x,y))ϵ11V2−l(x)d(x,x′)ϵ1′(2−l+d(x,y))ϵ1′2−l(ϵ2−ϵ1′)(2−l+d(x,y))ϵ2−ϵ1′,
which completes the estimate for Y2.
Finally, to estimateY3,
since d(y,y′)≤(1/2)(2−k+d(z,y)) and
(3.38)d(x,x′)≤d(x,y)8≤2−l+d(x,y)8,
by the regularities of Pl and Qk, we then have
(3.39)Y3≲∫W3[|Pl(x,z)|+|Pl(x′,z)|+d(x,x′)ϵ1(2−l+d(x,y))ϵ11V2−l(x)+V2−l(y)+V(x,y)2−lϵ2(2−l+d(x,y))ϵ2]×d(y,y′)ϵ1(2−k+d(z,y))ϵ11V2−k(z)+V2−k(y)+V(z,y)2−kϵ2(2−k+d(z,y))ϵ2dμ(z)≡∑i=13Y3,i.
The fact that 2−k+d(z,y)≥(1/4)(2−l+d(x,y)) and Lemma 2.1(ii) imply that
(3.40)Y3,3≲d(x,x′)ϵ1(2−l+d(x,y))ϵ1d(y,y′)ϵ1(2−l+d(x,y))ϵ11V2−l(x)+V2−l(y)+V(x,y)×2−lϵ2(2−l+d(x,y))ϵ2∫𝒳1V2−k(y)+V(z,y)2−kϵ2(2−k+d(z,y))ϵ2dμ(z)≲d(x,x′)ϵ1(2−l+d(x,y))ϵ1d(y,y′)ϵ1(2−l+d(x,y))ϵ11V2−l(x)+V2−l(y)+V(x,y)2−lϵ2(2−l+d(x,y))ϵ2.
The facts that k≥l, 2−k+d(z,y)≥(1/4)(2−l+d(x,y)),
(3.19), and Proposition 2.7(i) yield that for any ϵ1′∈(0,ϵ1⋀ϵ2),
(3.41)Y3,1≲d(y,y′)ϵ1(2−l+d(x,y))ϵ1∫W3|Pl(x,z)|1V2−k(z)+V2−k(y)+V(z,y)2−kϵ2(2−k+d(z,y))ϵ2dμ(z)≲d(x,x′)ϵ1′(2−l+d(x,y))ϵ1′d(y,y′)ϵ1(2−l+d(x,y))ϵ11V2−l(y)+V(x,y)2−l(ϵ2−ϵ1′)(2−l+d(x,y))ϵ2−ϵ1′,
where, in the last step, we used
the fact that 2−l<2d(x,x′) in this case. Similarly, by the size condition
of Pl and Lemma 2.1(i), we further have
(3.42)Y3,1≲d(y,y′)ϵ1(2−l+d(x,y))ϵ11V2−l(x)∫W31V2−k(y)+V(z,y)2−kϵ2(2−k+d(z,y))ϵ2dμ(z)≲d(y,y′)ϵ1(2−l+d(x,y))ϵ11V2−l(x)×{2−lϵ2(2−l+d(x,y))ϵ2+2−lϵ2∫d(z,y)≥(1/8)(2−l+d(x,y)),d(z,y)≥2−k1V(z,y)1d(z,y)ϵ2dμ(z)}≲d(x,x′)ϵ1′(2−l+d(x,y))ϵ1′d(y,y′)ϵ1(2−l+d(x,y))ϵ11V2−l(x)2−l(ϵ2−ϵ1′)(2−l+d(x,y))ϵ2−ϵ1′.
Notice that in this case, 2−l≤2d(x,x′)≤(1/4)d(x,y) and therefore V2−l(x)≤V(x,y).
Thus, by (3.19), we have that for z∈W3, 1/(V2−k(y)+V(z,y))≲1/(V2−l(y)+V(x,y))≲1/(V2−l(x)+V2−l(y)+V(x,y)), which together with Proposition 2.7(i) shows
(3.43)Y3,2≲d(y,y′)ϵ1(2−l+d(x,y))ϵ12−lϵ2(2−l+d(x,y))ϵ2∫W3|Pl(x′,z)|1V2−k(z)+V2−k(y)+V(z,y)dμ(z)≲d(x,x′)ϵ1′(2−l+d(x,y))ϵ1′d(y,y′)ϵ1(2−l+d(x,y))ϵ11V2−l(x)+V2−l(y)+V(x,y)2−l(ϵ2−ϵ1′)(2−l+d(x,y))ϵ2−ϵ1′.
This completes the proof of
(3.25), and hence, the proof of Lemma 3.2.Remark 3.3.
From the proof of (3.2) in Lemma 3.2, it is easy to see that (3.2) still
holds if Pl has the required regularity only in the second
variable, and Qk the first variable. This observation is useful
in applications.In what follows, let{Sk}k∈ℤ be an (ϵ1,ϵ2,ϵ3)-ATI with ϵ1∈(0,1],ϵ2>0, and ϵ3>0 as in Definition 2.2. Set Dk=Sk−Sk−1 for k∈ℤ.
To establish continuous Calderón reproducing formulae, by following Coifman's
idea (see [64]) and
Proposition 3.1(ii), we write for any N∈ℕ,
(3.44)I=(∑k=−∞∞Dk)(∑j=−∞∞Dj)=∑|l|>N∑k=−∞∞Dk+lDk+∑k=−∞∞DkNDk=RN+TN
in Lp(𝒳) with p∈(1,∞),
where DkN=∑|l|≤NDk+l. To verify that TN−1 exists and is bounded on any space of test
functions, we first prove that RN is bounded on L2(𝒳) with a small operator norm.Lemma 3.4.
LetN∈ℕ and let RN be as in (3.44). Then there exist constants C>0 and δ>0,
independent of N,
such that for all f∈L2(𝒳),
(3.45)∥RN(f)∥L2(𝒳)≤C2−Nδ∥f∥L2(𝒳).Proof.
To prove the lemma, by applying the Cotlar-Stein lemma (see [75, page 280] or [64]), we see that it suffices
to verify that for any σ∈(0,1), ϵ1′∈(0,ϵ1⋀ϵ2) with i=1,2,3,
and all k1,l1,k2,l2∈ℤ,
(3.46)∥Dk1+l1Dk1(Dk2+l2Dk2)t∥L2(𝒳)→L2(𝒳)≲2−|l1|ϵ1′σ2−|l2|ϵ2′σ2−|k1−k2|ϵ3′(1−σ),∥(Dk2+l2Dk2)tDk1+l1Dk1∥L2(𝒳)→L2(𝒳)≲2−|l1|ϵ1′σ2−|l2|ϵ2′σ2−|k1−k2|ϵ3′(1−σ).
We only prove (3.46), the proof for the last inequality being similar. In fact, Lemma 3.2 and Proposition 2.7 yield that
(3.47)∥Dk1+l1Dk1(Dk2+l2Dk2)t∥L2(𝒳)→L2(𝒳)≲2−|l1|ϵ1′2−|l2|ϵ2′,∥Dk1+l1Dk1Dk2tDk2+l2t∥L2(𝒳)→L2(𝒳)≲2−|k1−k2|ϵ3′.
Then the geometric mean of the above two estimates gives the estimate (3.46), which completes the proof of Lemma 3.4.We now establish some estimates for the kernel,RN(x,y),
of the operator RN.
To this end, we first give a technical lemma.Lemma 3.5.
For anyσ≥s>0 and x,y∈𝒳 with x≠y,
(3.48)∑k=−∞∞1V2−k(x)+V(x,y)2−ks(2−k+d(x,y))σ≲1V(x,y)1d(x,y)σ−s.Proof.
For anyx,y∈𝒳,
choose k0∈ℤ such that 2−k0<d(x,y)≤2−k0+1.
Then if k≤k0,
by (1.3), we have
(3.49)V2−k(x)=μ(B(x,2−(k−k0)2−k0))≥2−κ(k−k0)μ(B(x,2−k0))≳2−κ(k−k0)V(x,y).
Therefore,
(3.50)∑k=−∞∞1V2−k(x)+V(x,y)2−ks(2−k+d(x,y))σ≤1d(x,y)σ−s∑k=−∞k01V2−k(x)+1V(x,y)d(x,y)σ∑k=k0+1∞2−ks≲1V(x,y)1d(x,y)σ−s{∑k=−∞k02−κ(k−k0)+1}≲1V(x,y)1d(x,y)σ−s,
which completes the proof of Lemma 3.5.Lemma 3.6.
LetN∈ℕ,
let RN be as in (3.44), and let RN(x,y) be its kernel. Then for any ϵ1′∈(0,ϵ1⋀ϵ2), there exists a constant δ>0,
independent  of N,
such that RN satisfies all the conditions of Corollary 2.22
with ϵ replaced  by ϵ1′ and CRN≲2−δN.
Moreover, RN∗(1)=0.Proof.
Letϵ1′∈(0,ϵ1⋀ϵ2). Write
(3.51)RN(x,y)=∑l=N+1∞∑k=−∞∞(Dk+lDk)(x,y)+∑l=−∞−N−1∑k=−∞∞(Dk+lDk)(x,y)=RN1(x,y)+RN2(x,y).
By (3.2) and Lemma 3.5, we have
that for any x,y∈𝒳 with x≠y,
(3.52)|RN(x,y)|≤|RN1(x,y)|+|RN2(x,y)|≲∑l=N+1∞2−lϵ1′∑k=−∞∞1V2−k(x)+V(x,y)2−kϵ2(2−k+d(x,y))ϵ2+∑l=−∞−N−12lϵ1′∑k=−∞∞1V2−(k+l)(x)+V(x,y)2−(k+l)ϵ2(2−(k+l)+d(x,y))ϵ2≲2−Nϵ1′1V(x,y).
Thus, RN satisfies (i) of Theorem 2.18.
The estimate (3.3) and Lemma 3.5 show that for any x,y,y′∈𝒳 with x≠y and d(y,y′)≤(1/4)d(x,y),
(3.53)|RN(x,y)−RN(x,y′)|≤|RN1(x,y)−RN1(x,y′)|+|RN2(x,y)−RN2(x,y′)|≲d(y,y′)ϵ1′d(x,y)ϵ1′{∑l=N+1∞2−lδ∑k=−∞∞1V2−k(x)+V(x,y)2−kϵ2(2−k+d(x,y))ϵ2+∑l=−∞−N−12lδ∑k=−∞∞1V2−(k+l)(x)+V(x,y)2−(k+l)ϵ2(2−(k+l)+d(x,y))ϵ2}≲2−Nδd(y,y′)ϵ1′d(x,y)ϵ1′1V(x,y),
and the estimate (3.4) and the
symmetry also yield that
(3.54)|RN(y,x)−RN(y′,x)|≲2−Nδd(y,y′)ϵ1′d(x,y)ϵ1′1V(x,y),
which shows that RN satisfies (2.49) and (ii) of Theorem 2.18.
Using (3.5) and Lemma 3.5 then gives that for x,x′,y,y′∈𝒳 with x≠y and d(x,x′)≤(1/8)d(x,y) and d(y,y′)≤(1/8)d(x,y),
(3.55)|[RN(x,y)−RN(x′,y)]−[RN(x,y′)−RN(x′,y′)]|≤|[RN1(x,y)−RN1(x′,y)]−[RN1(x,y′)−RN1(x′,y′)]|+|[RN2(x,y)−RN2(x′,y)]−[RN2(x,y′)−RN2(x′,y′)]|≲d(x,x′)ϵ1′d(x,y)ϵ1′d(y,y′)ϵ1′d(x,y)ϵ1′{∑l=N+1∞2−lδ∑k=−∞∞1V2−k(x)+V(x,y)2−kσ(2−k+d(x,y))σ+∑l=−∞−N−12lδ∑k=−∞∞1V2−(k+l)(x)+V(x,y)2−(k+l)σ(2−(k+l)+d(x,y))σ}≲2−Nδd(x,x′)ϵ1′d(x,y)ϵ1′d(y,y′)ϵ1′d(x,y)ϵ1′1V(x,y),
which shows that RN satisfies (iii) of Theorem 2.18.
Moreover, by the vanishing moments ofDk,
we obviously have RN(1)=0=RN∗(1).
Thus, RN satisfies all the conditions of Corollary 2.22
with ϵ replaced by ϵ1′ and CRN≲2−δN,
which completes the proof of Lemma 3.6.From Lemma3.6 and Corollary 2.22, it is easy to deduce
the boundedness of TN on spaces of test functions when N is large enough.Proposition 3.7.
Letϵ1∈(0,1], ϵ2>0, ϵ3>0, and let {Sk}k∈ℤ be an (ϵ1,ϵ2,ϵ3)-ATI. For N∈ℕ,
let RN and TN be as in (3.44). Then there exist constants C8>0 and δ>0,
which are independent of N,
such that for all f∈𝒢°(x1,r,β,γ) with x1∈𝒳, r>0 and 0<β,γ<(ϵ1⋀ϵ2),
(3.56)∥RN(f)∥𝒢(x1,r,β,γ)≤C82−Nδ∥f∥𝒢(x1,r,β,γ).
Moreover, if N is so large that
(3.57)C82−Nδ<1,
then TN−1 exists and maps any space of test functions to
itself. More precisely, there exists a constant C>0 such that for all f∈𝒢°(x1,r,β,γ) with x1∈𝒳, r>0, and 0<β,γ<(ϵ1⋀ϵ2),
(3.58)∥TN−1(f)∥𝒢(x1,r,β,γ)≤C∥f∥𝒢(x1,r,β,γ).Proof.
Applying Corollary2.22 together with Lemmas 3.4 and 3.6 gives
(3.56). Moreover, if we choose N∈ℕ such that (3.57) holds, then by (3.56), we
have that for all f∈𝒢°(x1,r,β,γ),
(3.59)∥TN−1(f)∥𝒢(x1,r,β,γ)=∥(I−RN)−1(f)∥𝒢(x1,r,β,γ)=∥∑l=0∞(RN)l(f)∥𝒢(x1,r,β,γ)≤∑l=0∞(C82−Nδ)l∥f∥𝒢(x1,r,β,γ)≲∥f∥𝒢(x1,r,β,γ),
which completes the proof of Proposition 3.7.Proposition 3.8.
Letp∈(1,∞), ϵ1∈(0,1], ϵ2>0, ϵ3>0 and let {Sk}k∈ℤ be an (ϵ1,ϵ2,ϵ3)-ATI.
For N∈ℕ,
let RN and TN be as in (3.44). Then there exist constants C9>0 and δ>0,
which are independent of N,
such that for all f∈Lp(𝒳)(3.60)∥RN(f)∥Lp(𝒳)≤C92−Nδ∥f∥Lp(𝒳).
Moreover, if N is so large that
(3.61)C92−Nδ<1,
then TN−1 exists and is bounded on Lp(𝒳),
and there exists a constant C>0 such that for all f∈Lp(𝒳),
(3.62)∥TN−1(f)∥Lp(𝒳)≤C∥f∥Lp(𝒳).Proof.
We use the same notation as in Lemma3.6. From Lemma 3.6 together with
the proposition in [75, page 29], it is easy to see that RN is a singular integral satisfying the
condition (10) in [75, page 19] with A≲2−Nδ.
This fact and the corollary in [75, page 19] together with a duality argument and Lemma
3.4 prove that RN is also bounded on Lp(𝒳) for p∈(1,∞) and ∥RN∥Lp(𝒳)→Lp(𝒳)≲2−Nδ.
That is, (3.60) holds. If we choose N∈ℕ so large that (3.61) holds, by an argument
similar to the proof of Proposition 3.7, we can prove that TN−1 is bounded on Lp(𝒳) for p∈(1,∞).Letϵ1∈(0,1], ϵ2>0, ϵ3>0, and let {Sk}k=0∞ be an (ϵ1,ϵ2,ϵ3)-ATI.
Set Dk=Sk−Sk−1 for k∈ℤ.
For any f∈Lp(𝒳) with p∈(1,∞) and x∈𝒳,
the Littlewood-Paleyg-functiong˙(f) is defined by
(3.63)g˙(f)(x)={∑k=−∞∞|Dk(f)(x)|2}1/2.Lemma 3.9.
Letϵ1∈(0,1], ϵ2>0, ϵ3>0, and let {Sk}k∈ℤ be an (ϵ1,ϵ2,ϵ3)-ATI.
Let Dk=Sk−Sk−1 for k∈ℤ and let g˙(f) for f∈Lp(𝒳) with p∈(1,∞) be as in (3.63). Then there exists a constant Cp>0 such that for all f∈Lp(𝒳),
(3.64)∥g˙(f)∥Lp(𝒳)≤Cp∥f∥Lp(𝒳).Proof.
By Khinchin's inequality (see [76, page 165]) and Minkowski's inequality, we first have
that for any N∈ℕ,
(3.65)∥{∑k=−NN|Dk(f)|2}1/2∥Lp(𝒳)≲∥122N∑σ−N⋯∑σN∑k=−NNσkDk(f)∥Lp(𝒳)≲122N∑σ−N⋯∑σN∥∑k=−NNσkDk(f)∥Lp(𝒳),
where σk=1 or −1 for k∈{−N,…,N}.
For any fixed σ={σk}k=−NN,
we set TNσ=∑k=−NNσkDk and denote its kernel by KNσ.
Similarly to the proof of Lemma 3.6, it is easy to verify that KNσ and (KNσ)∗ are standard Calderón-Zygmund kernels, with
constants independent of N and σ.
Then if we can verify that TNσ is bounded on L2(𝒳) with ∥TNσ∥L2(𝒳)→L2(𝒳)≲1,
then by the corollary in [75, page 22] together with a duality argument, we obtain
that for p∈(1,∞), ∥TNσ∥Lp(𝒳)→Lp(𝒳)≲1. Therefore, for all N∈ℕ and all f∈Lp(𝒳),
(3.66)∥{∑k=−NN|Dk(f)|2}1/2∥Lp(𝒳)≲∥f∥Lp(𝒳).
Then Fatou's lemma further shows
that ∥g˙(f)∥Lp(𝒳)≲∥f∥Lp(𝒳). To finish the proof of Lemma 3.9, it remains
to verify that TNσ is bounded on L2(𝒳).
By the Cotlar-Stein lemma, it suffices to verify that for any ϵ1′∈(0,ϵ1⋀ϵ2) and all j,k∈ℤ,
(3.67)∥σkDk(σjDj)t∥L2(𝒳)→L2(𝒳)≲2−|k−j|ϵ1′,∥(σjDj)tσkDk∥L2(𝒳)→L2(𝒳)≲2−|k−j|ϵ1′.
However, these two estimates are
a simple corollary of (3.2) in Lemma 3.2. This completes the proof of Lemma
3.9.We can now establish acontinuous Calderón reproducing formula.Theorem 3.10.
Letϵ1∈(0,1], ϵ2>0, ϵ3>0, ϵ∈(0,ϵ1⋀ϵ2) and  let {Sk}k∈ℤ be an (ϵ1,ϵ2,ϵ3)-ATI.
Set Dk=Sk−Sk−1 for k∈ℤ.
Then there exists a family of linear operators {D˜k}k∈ℤ such that for all f∈𝒢°0ϵ(β,γ) with 0<β,γ<ϵ,
(3.68)f=∑k=−∞∞D˜kDk(f),
where the series converges in
both the norm of 𝒢°0ϵ(β,γ) and the norm of Lp(𝒳) for p∈(1,∞).
Moreover, the kernels of the operators D˜k satisfy the conditions (i) and (ii) of
Definition 2.2 with ϵ1 and ϵ2 replaced by ϵ′∈(ϵ,ϵ1⋀ϵ2), and ∫𝒳D˜k(x,y)dμ(y)=∫𝒳D˜k(x,y)dμ(x)=0.Proof.
Fix a large integerN such that (3.57) and (3.61) hold and,
therefore, Propositions 3.7 and 3.8 hold. Let DkN for k∈ℤ be as in (3.44). It is easy to check that DkN(⋅,y)∈𝒢°(y,2−j,ϵ1,ϵ2) for all k∈ℤ.
Define D˜k(x,y)=TN−1(DkN(⋅,y))(x) for k∈ℤ,
where TN−1 is defined as in Proposition 3.7. Then
Proposition 3.7 shows that D˜k for k∈ℤ satisfies all the conclusions of the theorem,
and formally, we also have (3.68). We still need to verify that the series in
(3.68) converges in both the norm of 𝒢°0ϵ(β,γ) and the norm of Lp(𝒳) with
p∈(1,∞).
Letϵ˜=ϵ1⋀ϵ2. We first prove that the series in (3.68) converges in the norm of 𝒢°0ϵ(β,γ) with 0<β,γ<ϵ.
Let f∈𝒢°(β′,γ′) with β<β′<ϵ˜ and γ<γ′<ϵ˜.
Then, for L∈ℕ,
we  write
(3.69)∑|k|≤LD˜kDk(f)=TN−1(∑|k|≤LDkNDk)(f)=TN−1(TN−∑|k|≥L+1DkNDk)(f)=TN−1TN(f)−TN−1(∑|k|≥L+1DkNDk)(f)=f−limj→∞(RN)j(f)−TN−1(∑|k|≥L+1DkNDk)(f).
We now verify that
(3.70)limL→∞∥f−∑|k|≤LD˜kDk(f)∥𝒢(β,γ)=0.
To see this, we write
(3.71)∥∑|k|≤LD˜kDk(f)−f∥𝒢(β,γ)≤limj→∞∥(RN)j(f)∥𝒢(β,γ)+∥TN−1(∑|k|≥L+1DkNDk)(f)∥𝒢(β,γ).
Notice that 𝒢°(β′,γ′)⊂𝒢°(β,γ).
By (3.56) and (3.57), we have
(3.72)limj→∞∥(RN)j(f)∥𝒢(β,γ)≤limj→∞(C82−Nδ)j∥f∥𝒢(β,γ)=0,
we remark that this is also true
even when β=β′ and γ=γ′.
We now prove that
(3.73)limL→∞∥TN−1(∑|k|≥L+1DkNDk)(f)∥𝒢(β,γ)=0.
To this end, by Proposition 3.7,
it suffices to verify that there exists some σ>0 such that for all 0<β<β′<ϵ˜ and 0<γ<γ′<ϵ˜ and all f∈𝒢°(β′,γ′),
(3.74)∥∑|k|≥L+1DkNDk(f)∥𝒢(β,γ)≲2−σL∥f∥𝒢(β′,γ′).
Similarly to the proof of (3.3),
by Lemma 2.1(iii) and the geometric mean, we can reduce the proof of (3.74) to
verifying that there exists some σ>0 such that for all f∈𝒢°(β′,γ′) and all x∈𝒳,
(3.75)|∑|k|≥L+1DkNDk(f)(x)|≲2−σL∥f∥𝒢(β′,γ′)1V1(x1)+V(x1,x)1(1+d(x,x1))γ,
and for all x,x′∈𝒳 with d(x,x′)≤(1/2)(1+d(x,x1)),
(3.76)|∑|k|≥L+1DkNDk(f)(x)−∑|k|≥L+1DkNDk(f)(x′)|≲∥f∥𝒢(β′,γ′)d(x,x′)β′(1+d(x,x1))β′1V1(x1)+V(x1,x)1(1+d(x,x1))γ′.
Similarly to the proofs of Lemmas3.4 and 3.6 and
using some estimates similar to (3.2) and (3.12), we easily obtain that for any L∈ℕ,
the operator TL=∑|k|≥L+1DkNDk satisfies all the conditions of Corollary 2.22
with ϵ replaced by ϵ˜, CTL≲1 and
(3.77)∥TL∥L2(𝒳)→L2(𝒳)≲1.
Corollary 2.22 then shows that TL is bounded on 𝒢°(β′,γ′) for any 0<β′,γ′<ϵ˜.
In particular, we see that (3.76)  holds.
To verify (3.75), we simply denote DkNDk by Ek.
By Lemma 3.2 and its proof, it is easy to see that Ek(x,y),
the kernel of Ek,
still satisfies (i) to (iv) of Definition 2.2 with a constant depending on N;
see (3.2) and (3.12). Moreover, Ek(1)=0.
For f∈𝒢°(β′,γ′),
(3.78)|∑k=L+1∞DkNDk(f)(x)|=|∑k=L+1∞∫𝒳Ek(x,y)[f(y)−f(x)]dμ(y)|≤∑k=L+1∞∫d(x,y)≤(1+d(x1,x))/2|Ek(x,y)||f(y)−f(x)|dμ(y)+∑k=L+1∞∫d(x,y)>(1+d(x1,x))/2|Ek(x,y)||f(y)|dμ(y)+∑k=L+1∞∫d(x,y)>(1+d(x1,x))/2|Ek(x,y)||f(x)|dμ(y)≡∑i=13Zi.
The size estimates for Ek and the regularity of f imply together with Lemma 2.1(ii) that
(3.79)Z1≲∑k=L+1∞∫d(x,y)≤(1+d(x1,x))/21V2−k(x)+V2−k(y)+V(x,y)2−kϵ2(2−k+d(x,y))ϵ2×d(x,y)β′(1+d(x1,x))β′1V1(x1)+V(x1,x)1(1+d(x1,x))γ′dμ(y)≲2−Lβ′1V1(x1)+V(x1,x)1(1+d(x1,x))γ′.
Similarly, Lemma 2.1(i) yields that
(3.80)Z3≲∑k=L+1∞∫d(x,y)>(1+d(x1,x))/21V2−k(x)+V(x,y)2−kϵ2(2−k+d(x,y))ϵ2dμ(y)×1V1(x1)+V(x1,x)1(1+d(x1,x))γ′≲2−Lϵ21V1(x1)+V(x1,x)1(1+d(x1,x))γ′.
Since d(x,y)>(1+d(x1,x))/2,
we have V(x,x1)≲V(x,y).
This estimate together with Lemma 2.1 shows that
(3.81)Z2≲∑k=L+1∞∫d(x,y)>(1+d(x1,x))/21V2−k(x)+V2−k(y)+V(x,y)2−kϵ2(2−k+d(x,y))ϵ2×1V1(x1)+V(x1,y)1(1+d(x1,y))γ′dμ(y)≲2−Lϵ2min{1V1(x1),1V(x,x1)}1(1+d(x1,x))ϵ2≲2−Lϵ21V1(x1)+V(x1,x)1(1+d(x1,x))γ′,
which completes the proof of
(3.75) for the operator ∑k=L+1∞DkNDk.
Since∫𝒳f(y)dμ(y)=0, we can write
(3.82)|∑k=−∞−L−1DkNDk(f)(x)|=|∑k=−∞−L−1∫𝒳[Ek(x,y)−Ek(x,x1)]f(y)dμ(y)|≤∑k=−∞−L−1∫d(x1,y)≤(2−k+d(x1,x))/2|Ek(x,y)−Ek(x,x1)||f(y)|dμ(y)+∑k=−∞−L−1∫d(x1,y)>(2−k+d(x1,x))/2|Ek(x,y)||f(y)|dμ(y)+∑k=−∞−L−1∫d(x1,y)>(2−k+d(x1,x))/2|Ek(x,x1)||f(y)|dμ(y)≡∑i=13Yi.
The regularity of Ek in the second variable and the size condition
of f together with Lemma 2.1(i) yield that
(3.83)Y1≲∑k=−∞−L−1∫d(x1,y)≤(2−k+d(x1,x))/2d(y,x1)ϵ1(2−k+d(x,x1))ϵ12−kϵ2(2−k+d(x,x1))ϵ21V2−k(x)+V2−k(x1)+V(x,x1)×1V1(x1)+V(x1,y)1(1+d(x1,y))γ′dμ(y)≲2−L(γ′−γ)1V1(x1)+V(x,x1)1(1+d(x,x1))γ.
Similarly, Lemma 2.1(ii) shows that
(3.84)Y3≲∑k=−∞−L−1∫d(x1,y)>(2−k+d(x1,x))/21V2−k(x)+V2−k(x1)+V(x,x1)×2−kϵ2(2−k+d(x,x1))ϵ21V1(x1)+V(x1,y)1(1+d(x1,y))γ′dμ(y)≲2−L(γ′−γ)/21V1(x1)+V(x,x1)1(1+d(x1,x))γ.
Since k<0 and d(x1,y)>(2−k+d(x1,x))/2>(1+d(x1,x))/2,
then V(x1,x)≲V(x1,y).
From this and Lemma 2.1(ii), it follows that
(3.85)Y2≲∑k=−∞−L−1∫d(x1,y)>(2−k+d(x1,x))/21V2−k(x)+V2−k(y)+V(x,y)×2−kϵ2(2−k+d(x,y))ϵ21V1(x1)+V(x1,y)1(1+d(x1,y))γ′dμ(y)≲2−L(γ′−γ)1V1(x1)+V(x,x1)1(1+d(x1,x))γ,
which completes the proof of (3.75). Hence, we obtain (3.74). Therefore for f∈𝒢°(β′,γ′) with β<β′<ϵ˜ and γ<γ′<ϵ˜,
(3.73) holds. Combining the estimate (3.73) with (3.72) shows (3.70).
Let nowf∈𝒢°0ϵ(β,γ) with 0<β,γ<ϵ.
By definition, there exists a sequence {fn}n=1∞⊂𝒢°(ϵ,ϵ) such that limn→∞∥f−fn∥𝒢(β,γ)=0. For any given L∈ℕ,
using some estimates similar to (3.2) and (3.12) (see also Remark 3.16),
similarly to the proof of Lemma 3.6, we can easily verify that for any L∈ℕ,
the operator T˜L=∑|k|≤LDkNDk satisfies all the conditions of Corollary 2.22
with ϵ replaced by ϵ˜, CT˜L≲1, and ∥T˜L∥L2(𝒳)→L2(𝒳)≲1. Thus T˜L is bounded on 𝒢°(β,γ) and T˜L∗(1)=0 by the vanishing moment of DkN,
which together with the boundedness of TN−1 on 𝒢°(β,γ) yields
(3.86)∥f−∑|k|≤LD˜kDk(f)∥𝒢(β,γ)≲∥f−fn∥𝒢(β,γ)+∥fn−TN−1(T˜L(fn))∥𝒢(β,γ)+∥TN−1(T˜L(fn))−TN−1(T˜L(f))∥𝒢(β,γ)≲∥f−fn∥𝒢(β,γ)+∥fn−TN−1(T˜L(fn))∥𝒢(β,γ).For any given δ>0,
fix some n∈ℕ such that ∥f−fn∥𝒢(β,γ)≤δ/4. By (3.70), for this chosen n,
there exists some L0∈ℕ such that if L>L0,
then
(3.87)∥fn−TN−1(T˜L(fn))∥𝒢(β,γ)<δ2.
Thus, when L>L0,
(3.88)∥f−∑|k|≤LD˜kDk(f)∥𝒢(β,γ)<δ.
Since T˜L∗(1)=0 and RN∗(1)=0 and since T˜L and TN−1 are bounded from 𝒢°(x,r,β,γ) to 𝒢(x,r,β,γ) for any x∈𝒳, r>0, and 0<β,γ<ϵ˜,
it follows that for
ϵ∈(0,ϵ˜), ∑|k|≤LD˜kDk(fn)=TN−1T˜L(fn)∈𝒢°(ϵ,ϵ). Moreover,
(3.89)∥{f−∑|k|≤LD˜kDk(f)}−{fn−∑|k|≤LD˜kDk(fn)}∥𝒢(β,γ)≤∥f−fn∥𝒢(β,γ)+∥TN−1T˜L(f−fn)∥𝒢(β,γ)≲∥f−fn∥𝒢(β,γ)→0,
as n→∞.
Thus, f−∑|k|≤LD˜kDk(f)∈𝒢°0ϵ(β,γ).
By (3.88), we further have that when L>L0,
(3.90)∥f−∑|k|≤LD˜kDk(f)∥𝒢°0ϵ(β,γ)<δ,which implies that (3.68) holds
in the norm of 𝒢°0ϵ(β,γ).
We now verify that (3.68) also holds in Lp(𝒳) for p∈(1,∞).
By Proposition 3.8 and the choice of N,
we have
(3.91)limj→∞∥(RN)j(f)∥Lp(𝒳)≤limj→∞(C92−Nδ)j∥f∥Lp(𝒳)=0,∥TN−1(∑|k|≥L+1DkNDk)(f)∥Lp(𝒳)≲∥(∑|k|≥L+1DkNDk)(f)∥Lp(𝒳).
Moreover, by Lemma 3.9 and
Hölder's inequality, we obtain
(3.92)∥(∑|k|≥L+1DkNDk)(f)∥Lp(𝒳)=sup∥h∥Lp′(𝒳)≤1|〈∑|k|≥L+1DkNDk(f),h〉|≤sup∥h∥Lp′(𝒳)≤1∥(∑|k|≥L+1|Dk(f)|2)1/2∥Lp(𝒳)∥(∑|k|≥L+1|(DkN)t(h)|2)1/2∥Lp′(𝒳)≲∥(∑|k|≥L+1|Dk(f)|2)1/2∥Lp(𝒳)≲∥f∥Lp(𝒳).
Thus,
(3.93)limL→∞∥(∑|k|≥L+1DkNDk)(f)∥Lp(𝒳)≲limL→∞∥(∑|k|≥L+1|Dk(f)|2)1/2∥Lp(𝒳)=0.
From these estimates, we finally
deduce that (3.68) holds in Lp(𝒳) with p∈(1,∞),
which completes the proof of Theorem 3.10.By an argument similar to the proof of Theorem3.10, we
obtain another continuous Calderón
reproducing formula (we omit the details).Theorem 3.11.
Letϵ1∈(0,1], ϵ2>0, ϵ3>0, ϵ∈(0,ϵ1⋀ϵ2), and let
{Sk}k∈ℤ be an (ϵ1,ϵ2,ϵ3)-ATI.
Set Dk=Sk−Sk−1 for k∈ℤ.
Then there exists a family of linear operators {D¯k}k∈ℤ such that for all f∈𝒢°0ϵ(β,γ) with 0<β,γ<ϵ,
(3.94)f=∑k=−∞∞DkD¯k(f),
where the series converges in
both the norm of 𝒢°0ϵ(β,γ) and the norm of Lp(𝒳) for p∈(1,∞).
Moreover, the kernels of the operators D¯k satisfy the conditions (i) and (iii) of
Definition 2.2 with ϵ1 and ϵ2 replaced by ϵ′∈(ϵ,ϵ1⋀ϵ2), and ∫𝒳D¯k(x,y)dμ(y)=∫𝒳D¯k(x,y)dμ(x)=0.To establish some Calderón reproducing formulae in
spaces of distributions, we first need to understand the action of the
operatorsDk on spaces of distributions.To this end, for allx,y∈𝒳,
let φ(x,y) satisfy the conditions (i) through (iii) with k=0 of Definition 2.2, and ∫𝒳φ(x,y)dμ(y)=0=∫𝒳φ(x,y)dμ(x).
Let ϵ be as in  Theorem 3.10, 0<β,γ<ϵ, and f∈𝒢°0ϵ(β,γ). We then define
(3.95)Ψ(f)(x)=∫𝒳φ(x,y)f(y)dμ(y).Let u∈(𝒢°0ϵ(β,γ))′.
In analogy with the theory of distributions on ℝn,
there exist two ways to define Ψ(u)∈(𝒢°0ϵ(β,γ))′.
One way is to define Ψ(u)∈(𝒢°0ϵ(β,γ))′ by duality, that is, for all f∈𝒢°0ϵ(β,γ), we put
(3.96)〈Ψ(u),f〉=〈u,Ψt(f)〉,
where Ψt denotes the integral operator with the kernel φt(x,y)=φ(y,x) for all x,y∈𝒳.
Alternatively, we define pointwise
(3.97)Ψ˜(u)(x)=〈u,φ(x,⋅)〉.
We now show that both definitions actually coincide.Lemma 3.12.
Letϵ be as in Theorem 3.10 and let 0<β,γ<ϵ.
Let u∈(𝒢°0ϵ(β,γ))′ and let Ψ(u) and Ψ˜(u) be defined, respectively, as in (3.96) and
(3.97). Then Ψ(u)=Ψ˜(u) in (𝒢°0ϵ(β,γ))′.Proof.
To establish this lemma, it suffices to show that for allf∈𝒢°(ϵ,ϵ),
(3.98)∫𝒳Ψ˜(u)(x)f(x)dμ(x)=〈u,Ψt(f)〉.
To this end, for L∈ℕ large enough, we define
(3.99)TL(f)(x)=∫B(x1,L)φ(y,x)f(y)dμ(y).
Let β,γ be as in the lemma. Using Theorem 2.18 and some
routine computations, we have TL(f)∈𝒢°(ϵ,ϵ) and limL→∞∥Ψt(f)−TL(f)∥𝒢(β,γ)=0.
Thus, 〈u,Ψt(f)〉=limL→∞〈u,TL(f)〉.
Now for fixedL∈ℕ large enough, and for any J∈ℕ,
let NJ={i∈IJ:QiJ∩B(x1,L)≠∅}, where {QiJ}J∈ℕ,i∈IJ are dyadic cubes of 𝒳 as in Lemma 2.19. If QiJ∩B(x1,L)≠∅ and L is large enough, then B(x1,L)⊂B(ziJ,3L) and QiJ⊂B(x1,2L),
where ziJ is the center of QiJ as in Lemma 2.19. These facts imply that ♯NJ≲(L2j)n.
By Lemma 2.19, we write
(3.100)TL(f)(x)=∑i∈NJ∫QiJ∩B(x1,L)[φ(y,x)−φ(yQiJ,x)]f(y)dμ(y)+∑i∈NJφ(yQiJ,x)∫QiJ∩B(x1,L)f(y)dμ(y)≡TL,J1(f)(x)+TL,J2(f)(x),
where yQiJ is any point in QiJ∩B(x1,L).
For fixed L∈ℕ large enough, using Theorem 2.18 and some
routine computations again, we have TL,J1(f)∈𝒢°(ϵ,ϵ) and
(3.101)limJ→∞∥TL,J1(f)∥𝒢(β,γ)=0.
Thus,
(3.102)〈u,Ψt(f)〉=limL→∞limJ→∞〈u,TL,J2(f)〉=limL→∞limJ→∞∑i∈NJΨ˜(u)(yQiJ)∫QiJ∩B(x1,L)f(y)dμ(y)=∫𝒳Ψ˜(u)(y)f(y)dμ(y)+limL→∞limJ→∞∑i∈NJ∫QiJ∩B(x1,L)[Ψ˜(u)(yQiJ)−Ψ˜(u)(y)]f(y)dμ(y).
It is not so difficult to verify that
(3.103)∑i∈NJ[φ(⋅,yQiJ)−φ(⋅,y)]χQiJ∩B(x1,L)(y)∈𝒢°(ϵ,ϵ),∥∑i∈NJ[φ(⋅,yQiJ)−φ(⋅,y)]χQiJ∩B(x1,L)(y)∥𝒢(β,γ)≤C2−Jϵ1
uniformly in y∈𝒳.
From this, it follows that |Ψ˜(u)(yQiJ)−Ψ˜(u)(y)|≤Cu2−Jϵ1 uniformly in y∈𝒳,
which along with Lebesgue's dominated convergence theorem shows that
(3.104)limJ→∞∑i∈NJ∫QiJ∩B(x1,L)[Ψ˜(u)(yQiJ)−Ψ˜(u)(y)]f(y)dμ(y)=0.
Thus, (3.98) is true and this
completes the proof of Lemma 3.12.Theorems3.10 and 3.11 in combination with a
duality argument and Lemma 3.12 show that continuous Calderón reproducing formulae also hold in spaces of distributions.Theorem 3.13.
Let all the notation be as in Theorems3.10 and 3.11. Then for all f∈(𝒢°0ϵ(β,γ))′ with 0<β,γ<ϵ,
(3.68) and (3.94) hold in (𝒢°0ϵ(β,γ))′.Finally, let us end this subsection by establishing aLittlewood-Paley theorem associated to ATIs via Theorem 3.10, which is a generalization
of [44, Proposition 2.5.1]. However, the method used here is different from that
in [44].To this end, we need the following Fefferman-Stein
vector-valued maximal function inequality in [77]; see also [44, Equation (2.11)] and  [75, Chapter II, Section 1].Lemma 3.14.
Let1<p<∞, 1<q≤∞, and let M be the Hardy-Littlewood maximal operator on 𝒳.
Let {fk}k∈ℤ⊂Lp(𝒳) be a sequence of measurable functions on 𝒳. Then
(3.105)∥{∑k=−∞∞|M(fk)|q}1/q∥Lp(𝒳)≤C∥{∑k=−∞∞|fk|q}1/q∥Lp(𝒳),
where C is independent of {fk}k∈ℤ.Proposition 3.15.
Letϵ1∈(0,1], ϵ2>0, ϵ3>0, and let {Sk}k∈ℤ be an (ϵ1,ϵ2,ϵ3)-ATI. Let Dk=Sk−Sk−1 for k∈ℤ and let g˙(f) for f∈Lp(𝒳) with p∈(1,∞) be as in (3.63). Then there exists a constant Cp>0 such that for all f∈Lp(𝒳),
(3.106)Cp−1∥f∥Lp(𝒳)≤∥g˙(f)∥Lp(𝒳)≤Cp∥f∥Lp(𝒳).Proof.
By Lemma3.9, we only need to verify the first inequality. To this end,
for any f∈Lp(𝒳),
Theorem 3.10 shows that there exist operators {D˜k}k∈ℤ as in Theorem 3.10 such that
(3.107)f=∑k=−∞∞D˜kDk(f)
in Lp(𝒳).
For 1<p<∞,
let 1/p+1/p′=1.
We first claim that for any h∈Lp′(𝒳),
(3.108)∥{∑k=−∞∞|D˜kt(h)|2}1/2∥Lp′(𝒳)≲∥h∥Lp′(𝒳).
Let ϵ∈(0,ϵ1⋀ϵ2).
In fact, by (3.2) of Lemma 3.2 and Remark 3.3, we have that for any ϵ1′∈(0,ϵ1⋀ϵ2),(3.109)|D˜ktD˜l(x,y)|≲2−|k−l|ϵ′1V2−(k⋀l)(x)+V2−(k⋀l)(y)+V(x,y)2−(k⋀l)ϵ(2−(k⋀l)+d(x,y))ϵ.From this and Lemma 2.1(iv), it
follows that for x∈𝒳,(3.110)|DktD˜l(h)(x)|≲2−|k−l|ϵ′∫𝒳1V2−(k⋀l)(x)+V2−(k⋀l)(y)+V(x,y)2−(k⋀l)ϵ(2−(k⋀l)+d(x,y))ϵ|h(y)|dμ(y)≲2−|k−l|ϵ′M(h)(x),which together with (3.107),
Hölder's inequality, Lemma 3.14, and Lemma 3.9 yields that
(3.111)∥{∑k=−∞∞|D˜kt(h)|2}1/2∥Lp′(𝒳)≲∥{∑k=−∞∞[∑l=−∞∞2−|k−l|ϵ′M(Dl(h))]2}1/2∥Lp′(𝒳)≲∥{∑l=−∞∞[M(Dl(h))]2}1/2∥Lp′(𝒳)≲∥h∥Lp′(𝒳).
Thus, (3.108) holds.
Using (3.108) and (3.107) together with a duality
argument and Hölder's inequality gives
(3.112)∥f∥Lp(𝒳)=sup∥h∥Lp′(𝒳)≤1|〈∑k=−∞∞D˜kDk(f),h〉|≤sup∥h∥Lp′(𝒳)≤1∥g˙(f)∥Lp(𝒳)∥{∑k=−∞∞|D˜kt(h)|2}1/2∥Lp′(𝒳)≲∥g˙(f)∥Lp(𝒳),
which completes the proof of Proposition 3.15.Remark 3.16.
From the proof of Proposition3.15, it is easy to see that Lemma 3.9 is
still true if Dk there is replaced by D˜kt for k∈ℤ,
which has regularity only in the second variable; see (3.108).


## 3.2. Inhomogeneous Continuous Calderón Reproducing Formulae

In this subsection, we have no restriction ondiam(𝒳),
which means diam(𝒳)<∞ or diam(𝒳)=∞.
We first introduce the following inhomogeneous approximation of the identity on 𝒳.Definition 3.17.
Letϵ1∈(0,1], ϵ2>0,
and ϵ3>0. A sequence {Sk}k∈ℤ+ of linear operators is said to be an inhomogeneous approximation of the identity of
order(ϵ1,ϵ2,ϵ3) (for short, (ϵ1,ϵ2,ϵ3)-IATI) if Sk for k∈ℤ+ satisfies Definition 2.2.
A sequence{Sk}k∈ℤ+ of linear operators is said to be an inhomogeneous approximation of the identity of
orderϵ1 with bounded support (for short, ϵ1-IATI with bounded support) if Sk for k∈ℤ+ satisfies Definition 2.3.The following proposition is a simple corollary of
Proposition2.7(iv) and (v).Proposition 3.18.
Letϵ1∈(0,1], ϵ2>0, ϵ3>0, let {Sk}k∈ℤ+ be an (ϵ1,ϵ2,ϵ3)-IATI, and let Skt be the adjoint operator of Sk for any k∈ℤ+.
Let Dk=Sk−Sk−1 for k∈ℕ and D0=S0.
Then I=∑k=0∞Dk in Lp(𝒳) for p∈[1,∞).
The same is true for {Skt}k∈ℤ+.To establish the continuous inhomogeneous Calderón reproducing formulae, we need a technical lemma, which is a variant of Lemma3.2.Lemma 3.19.
Letϵ1∈(0,1], ϵ2>0,
and ϵ3>0, {Sk}k∈ℤ+,
and let {Ek}k∈ℤ+ be two (ϵ1,ϵ2,ϵ3)-IATIs. Let Pk=Sk−Sk−1 and Qk=Ek−Ek−1 for k∈ℕ, P0=S0,
and Q0=E0.
Then for any ϵ1′∈(0,ϵ1⋀ϵ2), there exist constants C>0, δ>0, and σ>0 as in Lemma 3.2 such that the estimates (3.2)
to (3.5) are still true for these {Pl}k∈ℤ+ and {Qk}k∈ℤ+.Proof.
The proof of Lemma3.19 is essentially as in that of Lemma 3.2. The only
different situations are the cases when l=0 or k=0.
Let us prove (3.2) for l=0=k to show the difference. In this case, by the
size condition of P0 and Q0,
we have
(3.113)|P0Q0(x,y)|=|∫𝒳S0(x,z)Q0(z,y)dμ(z)|≲∫d(x,z)≤(1/2)d(x,y)1V1(x)+V1(z)+V(x,z)1(1+d(x,z))ϵ2×1V1(z)+V1(y)+V(z,y)1(1+d(z,y))ϵ2dμ(z)+∫d(x,z)>(1/2)d(x,y)⋯.Since d(x,z)>(1/2)d(x,y) implies that d(y,z)≤d(x,y)/2,
by symmetry, the estimates of the first and the second terms are similar and we
only estimate the first term. To this end, since d(x,z)≤d(x,y)/2≤(1+d(x,y))/2,
by Lemma 2.1(iii), we have 1/(1+d(z,y))≲1/(1+d(x,y)) and 1/(V1(z)+V(z,y))≲1/(V1(x)+V(x,y)), which further implies that 1/(V1(z)+V1(y)+V(z,y))≲1/(V1(x)+V1(y)+V(x,y)). These estimates together with Lemma 2.1(ii)
yield that
(3.114)|P0Q0(x,y)|≲1V1(x)+V1(y)+V(x,y)1(1+d(x,y))ϵ2,which is the desired estimate
and hence completes the proof of Lemma 3.19.Now, letϵ1∈(0,1], ϵ2>0, ϵ3>0, and {Sk}k∈ℤ+ be an (ϵ1,ϵ2,ϵ3)-IATI as in Definition 3.17. Throughout this subsection,
we always assume that Dk=Sk−Sk−1 for k∈ℕ, D0=S0 and Dk=0 if k∈{−1,−2,…}.
Similar to (3.44), by Proposition 3.18, for any N∈ℕ, we write
(3.115)I=(∑k=0∞Dk)(∑j=0∞Dj)=∑|l|>N∑k=0∞Dk+lDk+∑k=0∞DkNDk=RN+TN
in Lp(𝒳) with p∈(1,∞),
where DkN=∑|l|≤NDk+l.Repeating the proof of Lemma3.4, we
obtain the following.Lemma 3.20.
LetN∈ℕ and let RN be as in (3.115). Then there exist constants C>0 and δ>0,
independent of N,
such that for all f∈L2(𝒳),
(3.116)∥RN(f)∥L2(𝒳)≤C2−Nδ∥f∥L2(𝒳).We now establish some estimates for the kernel,RN(x,y),
of the operator RN.
To this end, we first give a technical lemma, which is a simple corollary of
Lemma 3.5.Lemma 3.21.
For anyσ≥s>0 and x,y∈𝒳 with x≠y,
(3.117)∑k=0∞1V2−k(x)+V(x,y)2−ks(2−k+d(x,y))σ≲1V(x,y)1d(x,y)σ−s.Lemma 3.22.
LetN∈ℕ,
let RN be as in (3.115), and let RN(x,y) be its kernel. Then for any ϵ∈(0,ϵ1⋀ϵ2), there exists a constant δ>0,
independent of N,
such that RN satisfies all the conditions of Corollary 2.22
and CRN≲2−δN.
Moreover, RN∗(1)=0.Replacing Lemmas3.2 and 3.5, respectively, by
Lemmas 3.19 and 3.21, and repeating the proof of Lemma 3.6, we obtain Lemma
3.22. In combination with Lemma 3.20, this leads to the following variant of Proposition
3.7. We omit  the details.Proposition 3.23.
Letϵ1∈(0,1], ϵ2>0, ϵ3>0, and let {Sk}k∈ℤ+ be an (ϵ1,ϵ2,ϵ3)-IATI.
For N∈ℕ,
let RN and TN be as in (3.115). Then there exist constants C8>0 and δ>0,
which are independent of N,
such that RN is bounded on any space of test functions, 𝒢°(x1,r,β,γ) with x1∈𝒳, r>0, and 0<β,γ<(ϵ1⋀ϵ2), and its operator norm is bounded by C82−Nδ.
Moreover, if N is so large that (3.57) holds, then TN−1 exists and is bounded on 𝒢°(x1,r,β,γ),
if x1∈𝒳, r>0 and 0<β,γ<(ϵ1⋀ϵ2).Via Lemmas3.20 and 3.22, similar to the proof of Proposition 3.8, we can obtain the following version of Proposition 3.8.Proposition 3.24.
Letp∈(1,∞), ϵ1∈(0,1], ϵ2>0, ϵ3>0, and let {Sk}k∈ℤ+ be an (ϵ1,ϵ2,ϵ3)-IATI.
For N∈ℕ,
let RN and TN be as in (3.115). Then there exist constants C9>0 and δ>0,
which are independent of N,
such that RN is bounded on Lp(𝒳) with the operator norm bounded by C92−Nδ.
Moreover, if N is so large that (3.61) holds, then TN−1 exists and is bounded on Lp(𝒳).Letϵ1∈(0,1], ϵ2>0, ϵ3>0, and let {Sk}k∈ℤ+ be an (ϵ1,ϵ2,ϵ3)-IATI.
Set Dk=Sk−Sk−1 for k∈ℕ and D0=S0.
For any f∈Lp(𝒳) with p∈(1,∞) and x∈𝒳,
the inhomogeneous Littlewood-Paleyg-functiong(f) is defined by
(3.118)g(f)(x)={∑k=0∞|Dk(f)(x)|2}1/2.
Applying the Cotlar-Stein lemma
and using a procedure similar to the proof of Lemma 3.9, we obtain the
boundedness on Lp(𝒳) with p∈(1,∞) for the Littlewood-Paley g-function as  below.Lemma 3.25.
Letϵ1∈(0,1], ϵ2>0, ϵ3>0, and let {Sk}k∈ℤ+ be an (ϵ1,ϵ2,ϵ3)-IATI.
Let Dk=Sk−Sk−1 for k∈ℕ, D0=S0, and g(f) for f∈Lp(𝒳) with p∈(1,∞) being as in (3.118). Then there exists a
constant Cp>0 such that for all f∈Lp(𝒳),
(3.119)∥g(f)∥Lp(𝒳)≤Cp∥f∥Lp(𝒳).We can now establish aninhomogeneous continuous Calderón reproducing formula.Theorem 3.26.
Letϵ1∈(0,1], ϵ2>0, ϵ3>0, ϵ∈(0,ϵ1⋀ϵ2), and let {Sk}k∈ℤ+ be an (ϵ1,ϵ2,ϵ3)-IATI. Set Dk=Sk−Sk−1 for k∈ℤ and D0=S0.
Then there exist N∈ℕ and a family of linear operators {D˜k}k∈ℤ+ such that for all f∈𝒢0ϵ(β,γ) with 0<β,γ<ϵ,
(3.120)f=∑k=0∞D˜kDk(f),
where the series converges both
in the norm of 𝒢0ϵ(β,γ) and the norm of Lp(𝒳) for p∈(1,∞).
Moreover, the kernels of the operators D˜k satisfy the conditions (i) and (ii) of
Definition 2.2 with ϵ1 and ϵ2 replaced by ϵ′∈(ϵ,ϵ1⋀ϵ2), and ∫𝒳D˜k(x,y)dμ(y)=∫𝒳D˜k(x,y)dμ(x)=1 when 0≤k≤N; =0 when k>N.Proof.
We prove this theorem by an argument similar to that of Theorem3.10. Fix
a large integer N∈ℕ such that (3.57) and (3.61) hold and,
therefore, Propositions 3.23 and 3.24 hold. Let DkN be as in (3.115). Then,
(3.121)DkN=∑|j|≤NDk+j={∑j=0k+NDj,for0≤k≤N;∑j=k−Nk+NDj,fork>N.
Thus, for k>N, DkN(⋅,y)∈𝒢°(y,2−k,ϵ1,ϵ2),
and we then define(3.122)D˜k(x,y)=TN−1(DkN(⋅,y))(x).By Proposition 3.23, we know that
for
k>N and any 0<β,γ<(ϵ1⋀ϵ2), D˜k(⋅,y)∈𝒢(y,2−k,β,γ).
Moreover, (TN−1)∗(1)=∑i=0∞(RN∗)i(1)=1,
which implies that
(3.123)∫𝒳D˜k(x,y)dμ(x)=∫𝒳DkN(x,y)dμ(x)=0.
Obviously, we have
(3.124)∫𝒳D˜k(x,y)dμ(y)=TN−1{∫𝒳DkN(⋅,y)dμ(y)}(x)=0.Thus, D˜k for k>N satisfies all the conditions of Theorem 3.26.
If k∈{0,…,N}, then
(3.125)∫𝒳DkN(x,y)dμ(x)=1=∫𝒳DkN(x,y)dμ(y).
Thus, in this case, we cannot
directly apply Proposition 3.23. Notice that
(3.126)TN−1=∑i=0∞(RN)i.
Using Lemma 3.19, we easily show
that for ϵ1′∈(0,ϵ1⋀ϵ2), there exist positive constants C˜ and δ depending on ϵ1′ such that RN(x,y) satisfies (2.174) and (2.185) with r0=1 and CT≲C˜2−Nδ.
Thus, by Corollary 2.31, we know that RN(S0(⋅,y))∈𝒢(y,1,ϵ1′,ϵ2) and(3.127)∥RN(S0(⋅,y))∥𝒢°(y,1,ϵ1′,ϵ2)≤C˜2−Nδ.
Notice that RN∗(1)=0.
Applying Proposition 3.23 yields that there exist two positive constants C8 and δ,
which are independent of N,
such that for all j∈ℕ and all y∈𝒳, RNj(S0(⋅,y))∈𝒢°(y,1,β,γ) and
(3.128)∥RNj(S0(⋅,y))∥𝒢°(y,1,β,γ)≤(C82−Nδ)j,
where β,γ∈(0,ϵ1⋀ϵ2).
Thus, if we choose N so large that (3.57) holds, then by (3.126), we
know that D˜k for k∈{0,1,…,N} satisfies the conditions (i) and (ii) of
Definition 2.2 with ϵ1 and ϵ2 replaced by ϵ∈(0,ϵ1⋀ϵ2). Moreover, from (3.126) together with RN(1)=0=RN∗(1), ∫𝒳S0(x,y)dμ(y)=1=∫𝒳S0(x,y)dμ(x) and for k∈ℕ, ∫𝒳Dk(x,y)dμ(y)=0=∫𝒳Dk(x,y)dμ(x), it follows that for k∈{0,1,…,N}, ∫𝒳D˜k(x,y)dμ(y)=1=∫𝒳D˜k(x,y)dμ(x).
Now it remains to prove that the series in (3.120)
converges in both the norm of 𝒢°0ϵ(β,γ) with 0<β,γ<ϵ and the norm of Lp(𝒳) with p∈(1,γ).
To verify (3.120) converges in 𝒢0ϵ(β,γ) with β,γ∈(0,ϵ),
similarly to (3.69), for f∈𝒢0ϵ(β′,γ′) with β′∈(β,ϵ˜) and γ′∈(γ,ϵ˜),
and L∈ℕ with L>N+1,
we have
(3.129)∑k=0LD˜kDk(f)=f−limj→∞(RN)j(f)−TN−1(∑k=L+1∞DkNDk)(f).
Thus,
(3.130)∥∑k=0LD˜kDk(f)−f∥𝒢(β,γ)≤limj→∞∥(RN)j(f)∥𝒢(β,γ)+∥TN−1(∑k=L+1∞DkNDk)(f)∥𝒢(β,γ).
Applying Corollary 2.31 again, we
obtain that there exists δ>0 such that for all x∈𝒳,
(3.131)|RN(f)(x)|≲2−Nδ1V1(x1)+V(x1,x)1(1+d(x1,x))γ′,and for any β∈(0,β′),
and x,x′∈𝒳 with d(x,x′)≤(1+d(x1,x))/2,
(3.132)|RN(f)(x)−RN(f)(x′)|≲2−Nδ(d(x,x′)1+d(x1,x))β1V1(x1)+V(x1,x)1(1+d(x1,x))γ′.
Notice that RN∗(1)=0,
which together with (3.131) and (3.132) shows that RN(f)∈𝒢°(β,γ).
From this and Proposition 3.23, it follows that
(3.133)limj→∞∥(RN)j(f)∥𝒢(β,γ)≤limj→∞(C82−Nδ)j∥f∥𝒢(β,γ)=0.
To prove that the second term on the right-hand side
of (3.130) tends to 0 as L→∞,
by (DkN)*(1)=0 when L≥N (see (3.121)), and Proposition 3.24, we only
need to verify that for a certain σ>0,
(3.134)∥∑k=L+1∞DkNDk(f)∥𝒢(β,γ)≲2−σL∥f∥𝒢(β′,γ′).
In fact, (3.134) can be deduced
from the following two estimates that for all x∈𝒳,
(3.135)|∑k=L+1∞DkNDk(f)(x)|≲2−(β′⋀γ′)L∥f∥𝒢(β′,γ′)1V1(x1)+V(x1,x)1(1+d(x1,x))γ′,and for x,x′∈𝒳 with d(x,x′)≤(1+d(x1,x))/2,(3.136)|∑k=L+1∞DkNDk(f)(x)−∑k=L+1∞DkNDk(f)(x′)|≲2−(1−θ)(β′⋀γ′)L∥f∥𝒢(β′,γ′)(d(x,x′)1+d(x1,x))θβ′1V1(x1)+V(x1,x)1(1+d(x1,x))γ′,
where θ∈(0,1).
LetEk=DkNDk.
It is easy to check that Ek for k∈ℕ satisfies all estimates of Definition 2.2 with ϵ replaced by  ϵ˜ and a coefficient depending on N.
Moreover, for k∈ℕ, Ek(1)=0=Ek∗(1).
Thus, by Lemma 2.1(ii),
(3.137)|DkNDk(f)(x)|≤∫d(x,y)≤(1+d(x1,x))/2|Ek(x,y)||f(y)−f(x)|dμ(y)+∫d(x,y)>(1+d(x1,x))/2|Ek(x,y)|[|f(y)|+|f(x)|]dμ(y)≤∥f∥𝒢(β′,γ′){∫d(x,y)≤(1+d(x1,x))/2|Ek(x,y)|(d(x,y)1+d(x1,x))β′×1V1(x1)+V(x1,x)1(1+d(x1,x))γ′dμ(y)+∫d(x,y)>(1+d(x1,x))/2|Ek(x,y)|[1V1(x1)+V(x1,y)1(1+d(x1,y))γ′+1V1(x1)+V(x1,x)1(1+d(x1,x))γ′]dμ(y)}≤∥f∥𝒢(β′,γ′){2−kβ′1V1(x1)+V(x1,x)1(1+d(x1,x))γ′+Z},
and by Lemma 2.1(i) and (ii), we further have
(3.138)Z≲∫d(x,y)>(1+d(x1,x))/21V2−k(x)+V(x,y)2−kϵ2(2−k+d(x,y))ϵ21V1(x1)+V(x1,y)1(1+d(x1,y))γ′dμ(y)≲min{2−kγ′V1(x1)∫d(x,y)>(1+d(x1,x))/21V(x,y)1d(x,y)γ′dμ(y),2−kγ′V(x1,x)1(1+d(x1,x))γ′∫𝒳1V1(x1)+V(x1,y)1(1+d(x1,y))γ′dμ(y)}≲2−kγ′1V1(x1)+V(x1,x)1(1+d(x1,x))γ′.
From this, it is easy to deduce that (3.135) holds.
On the other hand, similarly to the proof of (3.12), we
can verify that for x,x′∈𝒳 with d(x,x′)≤(1+d(x,x1))/4,
(3.139)|DkNDk(f)(x)−DkNDk(f)(x′)|≲∥f∥𝒢(β′,γ′)(d(x,x′)1+d(x1,x))β′1V1(x1)+V(x1,x)1(1+d(x1,x))γ′,
and by (3.137) together with
Lemma 2.1(iii),
(3.140)|DkNDk(f)(x)−DkNDk(f)(x′)|≲2−k(β′⋀γ′)∥f∥𝒢(β′,γ′)1V1(x1)+V(x1,x)1(1+d(x1,x))γ′.
These estimates together with
the geometric means yield (3.136). Thus,
(3.141)limL→∞∥∑k=0LD˜kDk(f)−f∥𝒢(β,γ)=0.
Then repeating the proof of
Theorem 3.10 further shows that (3.120) holds in the norm of 𝒢0ϵ(β,γ).
Its convergence in Lp(𝒳) for p∈(1,∞) can be proved in a way similar to Theorem 3.10
if instead of Lemma 3.9 by Lemma 3.25, which completes the proof of Theorem
3.26.Remark 3.27.
From the proof of Theorem3.26, it is easy to see that N in Theorem 3.26 can be chosen so that both
(3.57) and (3.61) are satisfied. In order to defray our notation and simplify
our presentation, we will assume in the sequel that N=0.A similar argument as for the proof of Theorem3.26
leads to the following variant of the
inhomogeneous continuous Calderón reproducing formula (we omit the
details).Theorem 3.28.
Letϵ1∈(0,1], ϵ2>0, ϵ3>0, ϵ∈(0,ϵ1⋀ϵ2), and let {Sk}k∈ℤ+ be an (ϵ1,ϵ2,ϵ3)-IATI.
Set Dk=Sk−Sk−1 for k∈ℤ and D0=S0.
Then there exists a family of linear operators {D¯k}k∈ℤ+ such that for all f∈𝒢0ϵ(β,γ) with 0<β,γ<ϵ,
(3.142)f=∑k=0∞DkD¯k(f),
where the series converges in
both the norm of 𝒢0ϵ(β,γ) and the norm of Lp(𝒳) for p∈(1,∞).
Moreover, the kernels of the operators D¯k satisfy the conditions (i) and (iii) of
Definition 2.2 with ϵ1 and ϵ2 replaced by ϵ′∈(ϵ,ϵ1⋀ϵ2), and ∫𝒳D¯k(x,y)dμ(y)=∫𝒳D¯k(x,y)dμ(x)=1 when k=0; =0 when k∈ℕ.Theorems3.26 and 3.28 in combination with a
duality argument and Lemma 3.12 show that the inhomogeneous Calderón reproducing formulae also hold in spaces of distributions.Theorem 3.29.
Let all the notation be as in Theorems3.26 and
3.28. Then for all f∈(𝒢0ϵ(β,γ))′ with 0<β,γ<ϵ,
(3.120) and (3.142) hold in (𝒢0ϵ(β,γ))′.Finally, we have the following analogue of Proposition3.15 for the inhomogeneous Littlewood-Paleyg-function, based on Theorem 3.26 (we omit the proof).Proposition 3.30.
Letϵ1∈(0,1], ϵ2>0, ϵ3>0, and let {Sk}k∈ℤ+ be an (ϵ1,ϵ2,ϵ3)-IATI.
Set Dk=Sk−Sk−1 for k∈ℤ, D0=S0 and g(f) for f∈Lp(𝒳) with p∈(1,∞) be as in (3.118). Then there exists a constant Cp>0 such that for all f∈Lp(𝒳),
(3.143)Cp∥f∥Lp(𝒳)≤∥g(f)∥Lp(𝒳)≤Cp∥f∥Lp(𝒳).Remark 3.31.
Similarly as in Remark3.16, it
is easy to see that Lemma 3.25 is still true if Dk therein is replaced by the kernel D˜kt for k∈ℤ+,
which has regularity only in the second variable.


## 4. Discrete Calderón Reproducing Formulae

In this section, we will establish some discrete
Calderón reproducing formulae which play a key role in the theory of function
spaces, especially in obtaining a frame characterization. To obtain these
discrete Calderón reproducing formulae, we mainly use Corollaries2.22 and 2.31 again.In the following, fork∈ℤ and τ∈Ik,
we denote by Qτk,ν,ν=1,2,…,N(k,τ), the set of all cubes Qτ′k+j⊂Qτk, where Qτk is the dyadic cube as in Lemma 2.19 and j is a positive large integer such that (4.1)2−jC6<13. Denote by zτk,ν the “center” of Qτk,ν as in Lemma 2.19 and by yτk,ν a point in Qτk,ν.


### 4.1. Homogeneous Discrete Calderón Reproducing Formulae

In this subsection, we always assume thatdiam(𝒳)=∞,ϵ1∈(0,1], ϵ2>0, ϵ3>0, and {Sk}k∈ℤ is an (ϵ1,ϵ2,ϵ3)-ATI.
Set Dk=Sk−Sk−1 for k∈ℤ.Let all the notation be as in (3.44). We now introduce
the following discrete Riemann sum
operator on 𝒳, (4.2)S(f)(x)=∑k=−∞∞∑τ∈Ik∑ν=1N(k,τ)∫Qτk,νDkN(x,y)dμ(y)Dk(f)(yτk,ν).We first verify thatS is well defined and bounded on L2(𝒳) via the Littlewood-Paley theorem for the
homogeneous g-function as in (3.63), Proposition 3.15. To do
so, let us first establish the following estimate by using  Proposition
3.15.Lemma 4.1.
There exists a constantC>0 such that for all yτk,ν∈Qτk,ν and all f∈L2(𝒳), (4.3)∑k=−∞∞∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)|Dk(f)(yτk,ν)|2≤C∥f∥L2(𝒳)2.Proof.
By Theorem3.10, there exists a family of linear operators {D˜k}k=−∞∞ as in Theorem 3.10 such that for all f∈L2(𝒳), f=∑l=−∞∞D˜lDl(f). By Lemma 3.2 together with Remark 3.3, we have
that for any ϵ1′∈(0,ϵ1⋀ϵ2) and any ϵ1′′∈(0,ϵ1′),
all x,z∈𝒳 and all k,l∈ℤ, (4.4)|DkD˜l(z,x)|≲2−|k−l|ϵ1′′1V2−(k⋀l)(x)+V2−(k⋀l)(z)+V(z,x)2−(k⋀l)ϵ1′(2−(k⋀l)+d(z,x))ϵ1′. Notice that for all x∈𝒳 and any
z,y∈Qτk,ν, by Lemma 2.19(iv), we have that d(y,z)≤C62−j2−k≤C62−j2−(k⋀l)≤C62−j(2−(k⋀l)+d(x,y)), where j∈ℕ satisfies (4.1). Thus, for all x∈𝒳,
any y,z∈Qτk,ν and all k,l∈ℤ, Lemma 2.1(iii) shows that (4.5)|DkD˜l(z,x)|χQτk,ν(z)≲2−|k−l|ϵ1′′1V2−(k⋀l)(x)+V2−(k⋀l)(y)+V(y,x)2−(k⋀l)ϵ1′(2−(k⋀l)+d(y,x))ϵ1′χQτk,ν(y). From this and Lemma 2.1(iv), it
follows that for k∈ℤ, (4.6)|Dk(f)(yτk,ν)|≤∑l=−∞∞∫𝒳|(DkD˜l)(yτk,ν,x)||Dl(f)(x)|dμ(x)≲∑l=−∞∞2−|k−l|ϵ1′′M(Dl(f))(y)χQτk,ν(y), where M is the Hardy-Littlewood maximal function on 𝒳.
By (4.6), the construction of Qτk,ν (see Lemma 2.19), Lemmas 3.14 and 3.9, we
obtain (4.7)∑k=−∞∞∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)|Dk(f)(yτk,ν)|2≲∑k=−∞∞∫𝒳[∑l=−∞∞2−|k−l|ϵ1′′M(Dl(f))(y)]2dμ(y)≲∑l=−∞∞∥M(Dl(f))∥L2(𝒳)2≲∥f∥L2(𝒳)2,which proves Lemma 4.1.The next lemma can be proved in a way similar as in
the proof of Theorem (1.14) in [78, page 12]. The main idea is to combine Theorem 3.10,
Lemma 2.19, and Hölder's inequality with a duality argument. We omit the details
here; see also [36].Lemma 4.2.
Suppose that a sequence{aτk,ν:k∈ℤ,τ∈Ik,ν=1,…,N(k,τ)} of numbers satisfies ∑k=−∞∞∑τ∈Ik∑ν=1N(k,τ)|aτk,ν|2<∞. Then the function defined by (4.8)f(x)=∑k=−∞∞∑τ∈Ik∑ν=1N(k,τ)[μ(Qτk,ν)]−1/2aτk,ν∫Qτk,νDkN(x,y)dμ(y) is in L2(𝒳).
Moreover, (4.9)∥f∥L2(𝒳)2≤C∑k=−∞∞∑τ∈Ik∑ν=1N(k,τ)|aτk,ν|2.Lemmas4.1 and 4.2 yield the boundedness of the
discrete Riemann sum operator S on L2(𝒳).Proposition 4.3.
Let
the notation be the same as above withj satisfying (4.1). Then the discrete Riemann
sum operator S in (4.2) is bounded on L2(𝒳).
That is, there is a constant C>0,
only depending on N,
such that for all f∈L2(𝒳), (4.10)∥S(f)∥L2(𝒳)≤C∥f∥L2(𝒳).Next we prove that the discrete Riemann sum operatorS is invertible and S−1 maps 𝒢°(x1,r,β,γ) into itself. To do this, we define R=I−S and first establish some estimates on the
kernel, R(x,y),
of the operator R.
To this end, by (3.44), we write (4.11)R(f)(x)=(I−S)(f)(x)=∑k=−∞∞∑τ∈Ik∑ν=1N(k,τ)∫Qτk,νDkN(x,y)[Dk(f)(y)−Dk(f)(yτk,ν)]dμ(y)+∑|l|>N∑k=−∞∞Dk+lDk(f)(x)≡∑k=−∞∞Gk(f)(x)+RN(f)(x)≡G(f)(x)+RN(f)(x). Let Gk(x,y) be the kernel of Gk for k∈ℤ.
We now verify that Gk(x,y),
and hence G(x,y),
satisfies all the desired estimates. Clearly, (4.12)G(x,y)=∑k=−∞∞∑τ∈Ik∑ν=1N(k,τ)∫Qτk,νDkN(x,z)[Dk(z,y)−Dk(yτk,ν,y)]dμ(z)=∑k=−∞∞Gk(x,y). We need the following two
technical lemmas.Lemma 4.4.
Let0≤λ,μ≤1, ϵ1∈(0,1], μϵ1<ν1, λϵ1<ν2,
and k∈ℤ.
Then there exists a constant C>0 such that for all x,y∈𝒳 and all k∈ℤ, (4.13)∫𝒳d(x,x′)λϵ1(2−k+d(x,z))λϵ11V2−k(x)+V2−k(z)+V(x,z)2−kν1(2−k+d(x,z))ν1×d(y,y′)μϵ1(2−k+d(z,y))μϵ11V2−k(z)+V2−k(y)+V(z,y)2−kν2(2−k+d(z,y))ν2dμ(z)≤Cd(x,x′)λϵ1(2−k+d(x,y))λϵ1d(y,y′)μϵ1(2−k+d(x,y))μϵ11V2−k(x)+V2−k(y)+V(x,y)×{2−k(ν2−λϵ1)(2−k+d(x,y))ν2−λϵ1+2−k(ν1−μϵ1)(2−k+d(x,y))ν1−μϵ1}.Proof.
Write(4.14)∫𝒳d(x,x′)λϵ1(2−k+d(x,z))λϵ11V2−k(x)+V2−k(z)+V(x,z)2−kν1(2−k+d(x,z))ν1×d(y,y′)μϵ1(2−k+d(z,y))μϵ11V2−k(z)+V2−k(y)+V(z,y)2−kν2(2−k+d(z,y))ν2dμ(z)=∫d(x,z)≥d(x,y)/2d(x,x′)λϵ1(2−k+d(x,z))λϵ11V2−k(x)+V2−k(z)+V(x,z)×2−kν1(2−k+d(x,z))ν1d(y,y′)μϵ1(2−k+d(z,y))μϵ12−kν2(2−k+d(z,y))ν2×1V2−k(z)+V2−k(y)+V(z,y)dμ(z)+∫d(x,z)<d(x,y)/2⋯≡Z1+Z2. For Z1,
the fact d(x,z)≥d(x,y)/2 implies that V(x,y)≤|B(x,2d(x,z))|≲V(x,z),
which together with Lemma 2.1(ii) shows that (4.15)Z1≲d(x,x′)λϵ1(2−k+d(x,y))λϵ11V2−k(x)+V(x,y)2−kν1(2−k+d(x,y))ν1×d(y,y′)μϵ1∫𝒳1V2−k(y)+V(z,y)2−kν2(2−k+d(z,y))ν2+μϵ1dμ(z)≲d(x,x′)λϵ1(2−k+d(x,y))λϵ1d(y,y′)μϵ1(2−k+d(x,y))μϵ11V2−k(x)+V(x,y)2−k(ν1−μϵ1)(2−k+d(x,y))ν1−μϵ1. For Z2,
the fact d(x,z)<d(x,y)/2 implies that d(z,y)>d(x,y)/2 and therefore, V(x,y)~V(y,x)≲V(y,z)~V(z,y).
From these facts and Lemma 2.1(ii), it follows that (4.16)Z2≲d(x,x′)λϵ1d(y,y′)μϵ1(2−k+d(x,y))μϵ12−kν2(2−k+d(x,y))ν2×min{1V(x,y)∫𝒳1V2−k(x)+V(x,z)2−kν1(2−k+d(x,z))ν1+λϵ1dμ(z),1V2−k(x)1V(x,y)2kλϵ1μ(B(x,d(x,y)2))}≲d(x,x′)λϵ1(2−k+d(x,y))λϵ1d(y,y′)μϵ1(2−k+d(x,y))μϵ11V2−k(x)+V(x,y)2−k(ν2−λϵ1)(2−k+d(x,y))ν2−λϵ1.
Moreover, by symmetry, we also
have (4.17)∫𝒳d(x,x′)λϵ1(2−k+d(x,z))λϵ11V2−k(x)+V2−k(z)+V(x,z)2−kν1(2−k+d(x,z))ν1×d(y,y′)μϵ1(2−k+d(z,y))μϵ11V2−k(z)+V2−k(y)+V(z,y)2−kν2(2−k+d(z,y))ν2dμ(z)≲d(x,x′)λϵ1(2−k+d(x,y))λϵ1d(y,y′)μϵ1(2−k+d(x,y))μϵ11V2−k(y)+V(x,y)×{2−k(ν2−λϵ1)(2−k+d(x,y))ν2−λϵ1+2−k(ν1−μϵ1)(2−k+d(x,y))ν1−μϵ1}. Combining these estimates
completes the proof of Lemma 4.4.Lemma 4.5.
LetGk(x,y) for k∈ℤ be as above. Then for any λ,μ∈(0,1) satisfying that λϵ1<ϵ2,
there exists a constant CN>0,
independent of j,
such that
(i)
|Gk(x,y)|≤CN2−jϵ1(1/(V2−k(x)+V2−k(y)+V(x,y)))(2−kϵ2/(2−k+d(x,y))ϵ2);(ii)
fory,y′∈𝒳 with d(y,y′)≤(2−k+d(x,y))/2,(4.18)|Gk(x,y)−Gk(x,y′)|≤CN2−jϵ1d(y,y′)μϵ1(2−k+d(x,y))μϵ11V2−k(x)+V2−k(y)+V(x,y)×{2−kϵ3(2−k+d(x,y))ϵ3+2−k(ϵ2−μϵ1)(2−k+d(x,y))ϵ2−μϵ1};(iii)
|Gk(x,y)−Gk(x′,y)|≤CN2−jϵ1(d(x,x′)λϵ1/(2−k+d(x,y))λϵ1)(1/(V2−k(x)+V2−k(y)+V(x,y)))(2−k(ϵ2−λϵ1)/(2−k+d(x,y))ϵ2−λϵ1) for x,x′∈𝒳 with d(x,x′)≤(2−k+d(x,y))/2;(iv)
∫𝒳Gk(x,y)dμ(y)=0=∫𝒳Gk(x,y)dμ(x).Proof.
Since∫𝒳Dk(x,y)dμ(y)=0=∫𝒳DkN(x,y)dμ(x), by the definition of Gk in (4.12), we easily see that (iv) holds.
By the construction of dyadic cubes in Lemma2.19, we
also easily see that for any z∈Qτk,ν, (4.19)d(z,yτk,ν)≤C62−(k+j)=C62−j2−k≤C62−j(2−k+d(y,z)). We recall that j always satisfies (4.1). Then, the regularity
of Dk and the size estimates of DkN together with Lemma 4.4 in the case λ=μ=0 yield that (4.20)|Gk(x,y)|=|∑τ∈Ik∑ν=1N(k,τ)∫Qτk,νDkN(x,z)[Dk(z,y)−Dk(yτk,ν,y)]dμ(z)|≤CN2−jϵ11V2−k(x)+V2−k(y)+V(x,y)2−kϵ2(2−k+d(x,y))ϵ2, which shows Gk(x,y) satisfies (i).
Write(4.21)Gk(x,y)−Gk(x,y′)=∑τ∈Ik∑ν=1N(k,τ)∫Qτk,νDkN(x,z){[Dk(z,y)−Dk(yτk,ν,y)]−[Dk(z,y′)−Dk(yτk,ν,y′)]}dμ(z). We now verify that Gk(x,y) satisfies (ii) by considering the following
two cases.
Case 1  (d(y,y′)≤(2−k+d(z,y))/3).
In this case, from (4.19) with j satisfying (4.1), the size estimates of DkN and the second difference regularity of Dk and Lemma 4.4 with λ=0, μ∈(0,1), ν1=ϵ2,
and ν2=ϵ3,
it follows that (4.22)|Gk(x,y)−Gk(x,y′)|≲2−jϵ1∫𝒳|DkN(x,z)|d(y,y′)μϵ1(2−k+d(z,y))μϵ11V2−k(z)+V2−k(y)+V(z,y)2−kϵ3(2−k+d(z,y))ϵ3dμ(z)≤CN2−jϵ1d(y,y′)μϵ1(2−k+d(x,y))μϵ11V2−k(x)+V2−k(y)+V(x,y){2−kϵ3(2−k+d(x,y))ϵ3+2−k(ϵ2−μϵ1)(2−k+d(x,y))ϵ2−μϵ1}.
Case 2  ((2−k+d(z,y))/3<d(y,y′)≤(2−k+d(x,y))/2).
In this case, the estimate (i), Lemma2.1(iii) together with the fact that 2−k<3d(y,y′) show that for any μ∈(0,1), (4.23)|Gk(x,y)−Gk(x,y′)|≤|Gk(x,y)|+|Gk(x,y′)|≤CN2−jϵ1{1V2−k(x)+V2−k(y)+V(x,y)2−kϵ2(2−k+d(x,y))ϵ2+1V2−k(x)+V2−k(y′)+V(x,y′)2−kϵ2(2−k+d(x,y′))ϵ2}≤CN2−jϵ1d(y,y′)μϵ1(2−k+d(x,y))μϵ11V2−k(x)+V2−k(y)+V(x,y)2−k(ϵ2−μϵ1)(2−k+d(x,y))ϵ2−μϵ1, which completes the proof of
(ii).
We verify (iii) by writing(4.24)Gk(x,y)−Gk(x′,y)=∑τ∈Ik∑ν=1N(k,τ)∫Qτk,ν[DkN(x,z)−DkN(x′,z)][Dk(z,y)−Dk(yτk,ν,y)]dμ(z). If d(x,x′)≤(2−k+d(x,z))/2,
the estimate (4.19) with j satisfying (4.1), the regularity of DkN and Dk,
Lemma 4.4 with λ∈(0,1), μ=0,
and ν1=ν2=ϵ2 show that (4.25)|Gk(x,y)−Gk(x′,y)|≤CN2−jϵ1∫𝒳d(x,x′)λϵ1(2−k+d(x,z))λϵ11V2−k(x)+V2−k(z)+V(x,z)×2−kϵ2(2−k+d(x,z))ϵ21V2−k(z)+V2−k(y)+V(z,y)2−kϵ2(2−k+d(z,y))ϵ2dμ(z)≤CN2−jϵ1d(x,x′)λϵ1(2−k+d(x,y))λϵ11V2−k(x)+V2−k(y)+V(x,y)2−k(ϵ2−λϵ1)(2−k+d(x,y))ϵ2−λϵ1.
If(2−k+d(x,z))/2<d(x,x′)≤(2−k+d(x,y))/2,
the estimate (i), Lemma 2.1(iii) together with the fact 2−k<2d(x,x′) yield that (4.26)|Gk(x,y)−Gk(x′,y)|≤CN2−jϵ1{1V2−k(x)+V2−k(y)+V(x,y)2−kϵ2(2−k+d(x,y))ϵ2+1V2−k(x′)+V2−k(y)+V(x′,y)2−kϵ2(2−k+d(x′,y))ϵ2}≤CN2−jϵ1d(x,x′)λϵ1(2−k+d(x,y))λϵ11V2−k(x)+V2−k(y)+V(x,y)2−k(ϵ2−λϵ1)(2−k+d(x,y))ϵ2−λϵ1, which completes the proof of
(iii), and hence the proof of Lemma 4.5.Lemma 4.6.
LetN∈ℕ and G be as in (4.12). Then there exist a constant CN>0 independent of j such that (4.27)∥G∥L2(𝒳)→L2(𝒳)≤CN2−jϵ1.Proof.
Using Lemma4.5 and repeating the proof of (3.2) in Lemma 3.2 yield that
for any ϵ1′∈(0,ϵ1⋀ϵ2) and all k,j∈ℤ,(4.28)∥GkGjt∥L2(𝒳)→L2(𝒳)≤CN2−jϵ12−|k−j|ϵ1′,∥GjtGk∥L2(𝒳)→L2(𝒳)≤CN2−jϵ12−|k−j|ϵ1′, where CN>0 is a constant depending on ϵ1′,
but, independent of j and k.
These estimates together with the Cotlar-Stein lemma and G=∑k=−∞∞Gk yields the conclusion of Lemma 4.6.The following lemma is a key lemma to establish the
discrete Calderón reproducing formula.Lemma 4.7.
LetS be as in (4.2) and R=I−S.
Then R is a Calderón-Zygmund singular integral
operator, R(1)=0=R∗(1).
Moreover, for any ϵ1′∈(0,ϵ1⋀ϵ2),
there exist positive constants C10, CN,
and δ,
depending on ϵ1′,
such that the kernel, R(x,y),
of R satisfies all the conditions of Corollary 2.22
with ϵ replaced by ϵ1′ and (4.29)CR+∥R∥L2(𝒳)→L2(𝒳)≤C102−δN+CN2−jϵ1, where CR is the Calderón-Zygmund constant as in Theorem
2.18, C10 and δ are independent of N and j,
and CN is independent  of j.Proof.
From (4.11), it is easy to see that R(1)=0=R∗(1).
Moreover, Lemma 3.6 shows that RN(x,y),
the kernel of RN,
satisfies all the conditions of Corollary 2.22 with ϵ replaced by ϵ1′ and CRN+∥RN∥L2(𝒳)→L2(𝒳)≤C102−δN,
where C10 and δ are independent of
N and j.
This combining the formula (4.11) implies that to prove Lemma 4.7, it suffices
to verify that G(x,y),
the kernel of G,
satisfies (2.49) and Theorem 2.18(i), (ii), and (iii) with ϵ replaced by ϵ1′ and CG≤CN2−jϵ1,
where CN is independent of N.
In fact, combining Lemma 4.5(i) with Lemma 3.5 and (4.12) yields that for all x,y∈𝒳 with x≠y, (4.30)|G(x,y)|≤∑k=−∞∞|Gk(x,y)|≤CN2−jϵ1∑k=−∞∞1V2−k(x)+V(x,y)2−kϵ2(2−k+d(x,y))ϵ2≤CN2−jϵ11V(x,y), which shows G(x,y) satisfies Theorem 2.18(i).
Lemma4.5(ii) and (iii) and Lemma 3.5 together with
(4.12), respectively, show that for any μ∈(0,1),
all x,y,y′∈𝒳 with x≠y and d(y,y′)≤d(x,y)/2, (4.31)|G(x,y)−G(x,y′)|≤∑k=−∞∞|Gk(x,y)−Gk(x,y′)|≤CN2−jϵ1d(y,y′)μϵ1d(x,y)μϵ1∑k=−∞∞1V2−k(x)+V(x,y)×{2−kϵ3(2−k+d(x,y))ϵ3+2−k(ϵ2−μϵ1)(2−k+d(x,y))ϵ2−μϵ1}≤CN2−jϵ1d(y,y′)μϵ1d(x,y)μϵ11V(x,y), and for any λ∈(0,1),
all x,x′,y∈𝒳 with x≠y and d(x,x′)≤d(x,y)/2, (4.32)|G(x,y)−G(x′,y)|≤∑k=−∞∞|Gk(x,y)−Gk(x′,y)|≤CN2−jϵ1d(x,x′)λϵ1d(x,y)λϵ1∑k=−∞∞1V2−k(x)+V(x,y)2−k(ϵ2−λϵ1)(2−k+d(x,y))ϵ2−λϵ1≤CN2−jϵ1d(x,x′)λϵ1d(x,y)λϵ11V(x,y), which proves that G(x,y) satisfies Theorem 2.18(ii) and (2.49).
By (2.49) and Theorem 2.18(ii), it is easy to see that
it suffices to establish Theorem 2.18(iii) only for d(x,x′)≤d(x,y)/6 and d(y,y′)≤d(x,y)/6.
To this end, we write (4.33)[G(x,y)−G(x′,y)]−[G(x,y′)−G(x′,y′)]=∑k=−∞∞∑τ∈Ik∑ν=1N(k,τ)∫Qτk,ν[DkN(x,z)−DkN(x′,z)]×{[Dk(z,y)−Dk(yτk,ν,y)]−[Dk(z,y′)−Dk(yτk,ν,y′)]}dμ(z).
Ifd(x,x′)>(2−k+d(x,z))/2 and d(y,y′)>(2−k+d(y,z))/3,
then(4.34)d(x,x′)+d(y,y′)≥(21−k+d(x,z)+d(z,y))3>d(x,y)3, which contradicts the
assumptions that d(x,x′)≤d(x,y)/6 and d(y,y′)≤d(x,y)/6.
Thus, we still have the following three cases:
(i)
d(x,x′)≤(2−k+d(x,z))/2 and d(y,y′)≤(2−k+d(y,z))/3;(ii)
d(x,x′)≤(2−k+d(x,z))/2 and d(y,y′)>(2−k+d(y,z))/3;(iii)
d(x,x′)>(2−k+d(x,z))/2 and d(y,y′)≤(2−k+d(y,z))/3. In the case
(i), by (4.19) with j satisfying (4.1), the second difference
regularity of DkN and Dk,
Lemma 4.4 with λ,μ∈(0,1), ν1=ϵ2,
and ν2=ϵ1+ϵ3,
and Lemma 3.5, we obtain (4.35)|[G(x,y)−G(x′,y)]−[G(x,y′)−G(x′,y′)]|≤CN2−jϵ1∑k=−∞∞∫𝒳d(x,x′)λϵ1(2−k+d(x,z))λϵ11V2−k(x)+V2−k(z)+V(x,z)2−kϵ2(2−k+d(x,z))ϵ2d(y,y′)μϵ1(2−k+d(z,y))μϵ1×1V2−k(z)+V2−k(y)+V(z,y)2−k(ϵ1+ϵ3)(2−k+d(z,y))ϵ1+ϵ3dμ(z)≤CN2−jϵ1d(x,x′)λϵ1d(x,y)λϵ1d(y,y′)μϵ1d(x,y)μϵ11V(x,y). In the case (ii), we also have (2−k+d(z,y′))/2≤(2−k+d(z,y))/2+d(y,y′)/2≤2d(y,y′). From this, (4.19) with j satisfying (4.1), the regularity of DkN and Dk,
Lemma 4.4 with λ,μ∈(0,1) and ν1=ν2=ϵ2,
Lemmas 2.1(iii) and 3.5, it follows that (4.36)|[G(x,y)−G(x′,y)]−[G(x,y′)−G(x′,y′)]|≤CN2−jϵ1∑k=−∞∞∫𝒳d(x,x′)λϵ1(2−k+d(x,z))λϵ11V2−k(x)+V2−k(z)+V(x,z)2−kϵ2(2−k+d(x,z))ϵ2×[d(y,y′)μϵ1(2−k+d(z,y))μϵ11V2−k(z)+V2−k(y)+V(z,y)2−kϵ2(2−k+d(z,y))ϵ2+d(y,y′)μϵ1(2−k+d(z,y′))μϵ11V2−k(z)+V2−k(y′)+V(z,y′)2−kϵ2(2−k+d(z,y′))ϵ2]dμ(z)≤CN2−jϵ1d(x,x′)λϵ1d(x,y)λϵ1d(y,y′)μϵ1d(x,y)μϵ11V(x,y). For the last case (iii), we have (2−k+d(x′,z))/2≤(2−k+d(x,z))/2+d(x,x′)/2≤3d(x,x′)/2. This estimate, (4.19) with j satisfying (4.1), the size condition of DkN,
the second difference regularity of Dk,
Lemma 4.4 with λ,μ∈(0,1), ν1=ϵ2, and ν2=ϵ1+ϵ3,
Lemmas 2.1(iii) and 3.5 show that (4.37)|[G(x,y)−G(x′,y)]−[G(x,y′)−G(x′,y′)]|≤CN2−jϵ1∑k=−∞∞∫𝒳[|DkN(x,z)|+|DkN(x′,z)|]×d(y,y′)μϵ1(2−k+d(z,y))μϵ11V2−k(z)+V2−k(y)+V(z,y)2−k(ϵ1+ϵ3)(2−k+d(z,y))ϵ1+ϵ3dμ(z)≤CN2−jϵ1d(x,x′)λϵ1d(x,y)λϵ1d(y,y′)μϵ1d(x,y)μϵ11V(x,y). Thus, G(x,y) satisfies Theorem 2.18(iii). This finishes the
proof of Lemma 4.7.By Lemmas4.6 and 4.7, we obtain the boundedness
of R on 𝒢°(x1,r,β,γ) for any x1∈𝒳, r>0 and 0<β,γ<(ϵ1⋀ϵ2),
and in Lp(𝒳) for p∈(1,∞).
We omit the details.Proposition 4.8.
LetR and the notation be as in Lemma 4.7. Then R is bounded on Lp(𝒳) for p∈(1,∞) and on 𝒢°(x1,r,β,γ) for any x1∈𝒳, r>0 and 0<β,γ<(ϵ1⋀ϵ2).
That is, there exists a constant C11>0,
only depending on p, β, and γ such that for all f∈Lp(𝒳), (4.38)∥R(f)∥Lp(𝒳)≤C11(C102−δN+CN2−jϵ1)∥f∥Lp(𝒳),∥R(f)∥𝒢(x1,r,β,γ)≤C11(C102−δN+CN2−jϵ1)∥f∥𝒢(x1,r,β,γ).By using Proposition4.8 and repeating the proof of
Propositions 3.7 and 3.8, we can obtain the boundedness of S−1 on 𝒢°(x1,r,β,γ) for any x1∈𝒳, r>0 and 0<β,γ<ϵ,
and in Lp(𝒳) for p∈(1,∞).
We also omit the details.Corollary 4.9.
LetS be as in (4.2). Let N,j∈ℕ such that (4.1) and (4.39)C11(C102−δN+CN2−jϵ1)<1 hold. Then S has a bounded inverse in 𝒢°(x1,r,β,γ) for any x1∈𝒳, r>0, and 0<β,γ<(ϵ1⋀ϵ2),
and in Lp(𝒳) for p∈(1,∞).
Namely, there exists a constant C>0 depending only on β,γ, and p such that for all f∈𝒢°(x1,r,β,γ), (4.40)∥S−1(f)∥𝒢(x1,r,β,γ)≤C∥f∥𝒢(x1,r,β,γ) and for all f∈Lp(𝒳), (4.41)∥S−1(f)∥Lp(𝒳)≤C∥f∥Lp(𝒳).To establish the discrete Calderón reproducing
formulae, we still need the following technical lemma.Lemma 4.10.
Letj satisfy (4.1). For k∈ℤ,
any fixed yτk,ν∈Qτk,ν with τ∈Ik and ν∈{1,…,N(k,τ)},
and any x∈𝒳,
let (4.42)Ek(x,y)=∑τ∈Ik∑ν=1N(k,τ)∫Qτk,νDkN(x,z)dμ(z)Dk(yτk,ν,y). Then for any λ,μ∈(0,1),
there exists a constant CN>0 depending only on N, λ and μ such that
(i)
for allk∈ℤ and all x,y∈𝒳, (4.43)|Ek(x,y)|≤CN1V2−k(x)+V2−k(y)+V(x,y)2−kϵ2(2−k+d(x,y))ϵ2;(ii)
for allk∈ℤ and all x,y,y′∈𝒳 with d(y,y′)≤(2−k+d(x,y))/2, (4.44)|Ek(x,y)−Ek(x,y′)|≤CNd(y,y′)μϵ1(2−k+d(x,y))μϵ11V2−k(x)+V2−k(y)+V(x,y)2−k(ϵ2−μϵ1)(2−k+d(x,y))ϵ2−μϵ1;(iii)
property (ii) also holds withx and y interchanged and μ replaced by λ;(iv)
for allk∈ℤ and all x,x′,y,y′∈𝒳 with d(x,x′)≤(2−k+d(x,y))/4 and d(y,y′)≤(2−k+d(x,y))/4, (4.45)|[Ek(x,y)−Ek(x′,y)]−[Ek(x,y′)−Ek(x′,y′)]|≤CNd(x,x′)λϵ1(2−k+d(x,y))λϵ1d(y,y′)μϵ1(2−k+d(x,y))μϵ11V2−k(x)+V2−k(y)+V(x,y)×[2−k(ϵ2−λϵ1)(2−k+d(x,y))ϵ2−λϵ1+2−k(ϵ2−μϵ1)(2−k+d(x,y))ϵ2−μϵ1].Proof.
The main idea for the proof of this lemma is
to combine the techniques used in the proof of Lemma3.2 with Lemma 2.1(iii).
By (4.19) and Lemma 2.1(iii), for any z∈Qτk,ν and y∈𝒳, (4.46)12−k+d(yτk,ν,y)~12−k+d(z,y),(4.47)1V2−k(yτk,ν)+V2−k(y)+V(yτk,ν,y)~1V2−k(z)+V2−k(y)+V(z,y). To see (i), by the size
condition of Dk together with (4.46) and (4.47), and Lemma 4.4
with λ=μ=0 and ν1=ν2=ϵ2,
we have (4.48)|Ek(x,y)|≲∫𝒳|DkN(x,z)|1V2−k(z)+V2−k(y)+V(z,y)2−kϵ2(2−k+d(z,y))ϵ2dμ(z)≲1V2−k(x)+V2−k(y)+V(x,y)2−kϵ2(2−k+d(x,y))ϵ2, which verifies (i).
To verify (ii), we write(4.49)|Ek(x,y)−Ek(x,y′)|≤∑τ∈Ik∑ν=1N(k,τ)|∫Qτk,νDkN(x,z)dμ(z)[Dk(yτk,ν,y)−Dk(yτk,ν,y′)]|×[χ{d(y,y′)≤(2−k+d(yτk,ν,y))/2}(y,y′)+χ{d(y,y′)>(2−k+d(yτk,ν,y))/2}(y,y′)]≡Y1+Y2. For Y1,
by the regularity of Dk and the size condition of DkN together with (4.46), (4.47) and Lemma 4.4 with λ=0,μ∈(0,1), and ν1=ν2=ϵ2,
we obtain (4.50)Y1≲∫𝒳1V2−k(x)+V2−k(z)+V(x,z)2−kϵ2(2−k+d(x,z))ϵ2d(y,y′)μϵ1(2−k+d(z,y))μϵ1×1V2−k(z)+V2−k(y)+V(z,y)2−kϵ2(2−k+d(z,y))ϵ2dμ(z)≲d(y,y′)μϵ1(2−k+d(x,y))μϵ11V2−k(x)+V2−k(y)+V(x,y)2−k(ϵ2−μϵ1)(2−k+d(x,y))ϵ2−μϵ1. The property (i), Lemma 2.1(iii), and the fact that 2−k≤2d(y,y′) yield that (4.51)Y2≲1V2−k(x)+V2−k(y)+V(x,y)2−kϵ2(2−k+d(x,y))ϵ2+1V2−k(x)+V2−k(y′)+V(x,y′)2−kϵ2(2−k+d(x,y′))ϵ2≲d(y,y′)μϵ1(2−k+d(x,y))μϵ11V2−k(x)+V2−k(y)+V(x,y)2−k(ϵ2−μϵ1)(2−k+d(x,y))ϵ2−μϵ1, which completes the proof of
(ii).
By symmetry, we can deduce (iii) from (ii).
To prove (iv), we write(4.52)[Ek(x,y)−Ek(x′,y)]−[Ek(x,y′)−Ek(x′,y′)]=∑τ∈Ik∑ν=1N(k,τ)∫Qτk,ν[DkN(x,z)−DkN(x′,z)][Dk(yτk,ν,y)−Dk(yτk,ν,y′)]dμ(z). Notice that if d(x,x′)>(2−k+d(x,z))/2 with z∈Qτk,ν and d(y,y′)>(2−k+d(yτk,ν,y))/2, then d(x,x′)+d(y,y′)>(21−k+d(x,z)+d(yτk,ν,y))/2, which contradicts the assumptions that d(x,x′)≤(2−k+d(x,y))/4 and d(y,y′)≤(2−k+d(x,y))/4, since these estimates together with (4.1) and
Lemma 2.19 prove that for z∈Qτk,ν, d(x,x′)+d(y,y′)<(21−k+d(x,z)+d(yτk,ν,y))/2. Thus, if we let (4.53)W1={z∈Qτk,ν:d(x,x′)≤2−k+d(x,z)2andd(y,y′)≤2−k+d(yτk,ν,y)2},W2={z∈Qτk,ν:d(x,x′)≤2−k+d(x,z)2andd(y,y′)>2−k+d(yτk,ν,y)2},W3={z∈Qτk,ν:d(x,x′)>2−k+d(x,z)2andd(y,y′)≤2−k+d(yτk,ν,y)2}, we then have (4.54)[Ek(x,y)−Ek(x′,y)]−[Ek(x,y′)−Ek(x′,y′)]=∑i=13∑τ∈Ik∑ν=1N(k,τ)∫Qτk,ν∩Wi[DkN(x,z)−DkN(x′,z)][Dk(yτk,ν,y)−Dk(yτk,ν,y′)]dμ(z)=∑i=13Zi. For Z1,
the regularity of DkN and Dk together with the estimates (4.46) and (4.47)
and Lemma 4.4 yield that for any λ,μ∈(0,1), (4.55)|Z1|≲∫𝒳d(x,x′)λϵ1(2−k+d(x,z))λϵ11V2−k(x)+V2−k(z)+V(x,z)2−kϵ2(2−k+d(x,z))ϵ2×d(y,y′)μϵ1(2−k+d(z,y))μϵ11V2−k(z)+V2−k(y)+V(z,y)2−kϵ2(2−k+d(z,y))ϵ2dμ(z)≲d(x,x′)λϵ1(2−k+d(x,y))λϵ1d(y,y′)μϵ1(2−k+d(x,y))μϵ11V2−k(x)+V2−k(y)+V(x,y)×{2−k(ϵ2−λϵ1)(2−k+d(x,y))ϵ2−λϵ1+2−k(ϵ2−μϵ1)(2−k+d(x,y))ϵ2−μϵ1}. Similarly, for Z2,
from d(y,y′)>(2−k+d(yτk,ν,y))/2,
(4.46) and Lemma 2.1(iii) with d(y,y′)≤(2−k+d(x,y))/4,
it follows that for z∈Qτk,ν,1/2μϵ1<d(y,y′)μϵ1/(2−k+d(yτk,ν,y))μϵ1~d(y,y′)μϵ1/(2−k+d(z,y))μϵ1, which together with the regularity of DkN and the size condition of Dk,
(4.46), (4.47), Lemmas 4.4, and 2.1(iii) shows for any λ,μ∈(0,1), (4.56)|Z2|≲∫𝒳d(x,x′)λϵ1(2−k+d(x,z))λϵ11V2−k(x)+V2−k(z)+V(x,z)2−kϵ2(2−k+d(x,z))ϵ2×[d(y,y′)μϵ1(2−k+d(z,y))μϵ11V2−k(z)+V2−k(y)+V(z,y)2−kϵ2(2−k+d(z,y))ϵ2+1V2−k(z)+V2−k(y′)+V(z,y′)2−kϵ2(2−k+d(z,y′))ϵ2]dμ(z)≲d(x,x′)λϵ1(2−k+d(x,y))λϵ1d(y,y′)μϵ1(2−k+d(x,y))μϵ11V2−k(x)+V2−k(y)+V(x,y)×{2−k(ϵ2−λϵ1)(2−k+d(x,y))ϵ2−λϵ1+2−k(ϵ2−μϵ1)(2−k+d(x,y))ϵ2−μϵ1}. From Z2 and the symmetry, we can deduce the desired
estimate for Z3,
which completes the proof of Lemma 4.10.We can now establishdiscrete Calderón reproducing formulae.Theorem 4.11.
Letϵ1∈(0,1],ϵ2>0,ϵ3>0,ϵ∈(0,ϵ1⋀ϵ2), and let {Sk}k=−∞∞ be an (ϵ1,ϵ2,ϵ3)−ATI.
Set Dk=Sk−Sk−1 for any k∈ℤ.
Then for any fixed j∈ℕ as in Corollary 4.9, there exists a family of
linear operators {D˜k}k∈ℤ such that for any fixed yτk,ν∈Qτk,ν with k∈ℤ,τ∈Ik, and ν=1,…,N(k,τ),
and all f∈𝒢°0ϵ(β,γ) with 0<β,γ<ϵ, (4.57)f(x)=∑k=−∞∞∑τ∈Ik∑ν=1N(k,τ)∫Qτk,νD˜k(x,y)dμ(y)Dk(f)(yτk,ν)=∑k=−∞∞∑τ∈Ik∑ν=1N(k,τ)D˜k(x,yτk,ν)∫Qτk,νDk(f)(y)dμ(y)=∑k=−∞∞∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)D˜k(x,yτk,ν)Dk(f)(yτk,ν), where the series converges in
both the norm of 𝒢°0ϵ(β,γ) and the norm of Lp(𝒳) with p∈(1,∞).
Moreover, D˜k satisfies the conditions as in Theorem 3.10.Proof.
We only prove the first formula in (4.57), the proof of the second
formula in (4.57) being similar. Fix N,j∈ℕ such that (4.1) and (4.39) hold. Thus, for such N and j,
Corollary 4.9 holds. Let DkN for k∈ℤ be as in (3.44). For k∈ℤ,
let D˜k(x,y)=S−1[DkN(⋅,y)](x).
By (4.2) and Corollary 4.9, similarly to the proof of Theorem 3.10, it is easy
to see that we obtain all the conclusions of the theorem except for the
convergence of the series in the first formula in (4.57). To prove this, we
need to verify that all the series in the first summation and the second
summation of the first formula in (4.57) converges in the desired ways. To
simplify the presentation, by similarity, we prove this only for the series in
the first summation of the first formula in (4.57).
Similarly to the proof of Theorem3.10, for L∈ℕ,
we write (4.58)∑|k|≤L∑τ∈Ik∑ν=1N(k,τ)∫Qτk,νD˜k(x,y)dμ(y)Dk(f)(yτk,ν)=S−1[∑|k|≤L∑τ∈Ik∑ν=1N(k,τ)∫Qτk,νDkN(⋅,y)dμ(y)Dk(f)(yτk,ν)](x)=S−1{S(f)(⋅)−∑|k|>L∑τ∈Ik∑ν=1N(k,τ)∫Qτk,νDkN(⋅,y)dμ(y)Dk(f)(yτk,ν)}(x)=f(x)−limm→∞Rm(f)(x)−S−1{∑|k|>L∑τ∈Ik∑ν=1N(k,τ)∫Qτk,νDkN(⋅,y)dμ(y)Dk(f)(yτk,ν)}(x). Corollary 2.24 together with
Lemmas 4.6 and 4.7 shows that for all f∈𝒢°0ϵ(β,γ) with 0<β,γ<ϵ,(4.59)limm→∞∥Rm(f)∥𝒢°0ϵ(β,γ)≤limm→∞C11m(C102−δN+CN2−jϵ)m∥f∥𝒢°0ϵ(β,γ)=0, and for all f∈Lp(𝒳) with p∈(1,∞), (4.60)limm→∞∥Rm(f)∥Lp(𝒳)≤limm→∞C11m(C102−δN+CN2−jϵ)m∥f∥Lp(𝒳)=0. To finish the proof of the
theorem, we still need to verify that for all f∈𝒢°0ϵ(β,γ) with 0<β,γ<ϵ, (4.61)limL→∞∥S−1{∑|k|>L∑τ∈Ik∑ν=1N(k,τ)∫Qτk,νDkN(⋅,y)dμ(y)Dk(f)(yτk,ν)}∥𝒢°0ϵ(β,γ)=0, and for all f∈Lp(𝒳) with p∈(1,∞), (4.62)limL→∞∥S−1{∑|k|>L∑τ∈Ik∑ν=1N(k,τ)∫Qτk,νDkN(⋅,y)dμ(y)Dk(f)(yτk,ν)}∥Lp(𝒳)=0. We first verify (4.61). To this
end, letting ϵ˜=ϵ1⋀ϵ2,
similarly to the proof of Theorem 3.10, by Corollary 4.9, it suffices to prove
that there exists some σ>0 such that for all 0<β<β′<ϵ˜,0<γ<γ′<ϵ˜, all L∈ℕ,
and all f∈𝒢°(β′,γ′), (4.63)∥∑|k|>L∑τ∈Ik∑ν=1N(k,τ)∫Qτk,νDkN(⋅,y)dμ(y)Dk(f)(yτk,ν)∥𝒢(β,γ)≤C2−σL∥f∥𝒢(β′,γ′), where C>0 is independent of L and f.
An argument similar to the proof of (3.3), via Lemma 2.1(iii) and geometric
mean, reduces the proof of this estimate to verifying the following two
estimates that there exists some σ>0 such that for  all f∈𝒢°(β′,γ′) and  all x∈𝒳, (4.64)|∑|k|>L∑τ∈Ik∑ν=1N(k,τ)∫Qτk,νDkN(x,y)dμ(y)Dk(f)(yτk,ν)|≲2−σL∥f∥𝒢(β′,γ′)1V1(x1)+V(x1,x)1(1+d(x,x1))γ, and for all x,x′∈𝒳 with d(x,x′)≤(1/2)(1+d(x,x1)), (4.65)|∑|k|>L∑τ∈Ik∑ν=1N(k,τ)∫Qτk,νDkN(x,y)dμ(y)Dk(f)(yτk,ν)−∑|k|>L∑τ∈Ik∑ν=1N(k,τ)∫Qτk,νDkN(x′,y)dμ(y)Dk(f)(yτk,ν)|≲∥f∥𝒢(β′,γ′)d(x,x′)β′(1+d(x,x1))β′1V1(x1)+V(x1,x)1(1+d(x,x1))γ′. For L∈ℕ,
let TL be the operator associated with the kernel (4.66)KL(x,y)=∑|k|>L∑τ∈Ik∑ν=1N(k,τ)∫Qτk,νDkN(x,z)dμ(z)Dk(yτk,ν,y) with x,y∈𝒳.
By an argument similar to the proof of Lemma 4.7 together with Lemmas 4.10 and 3.5, we know TL satisfies all the conditions of Corollary 2.22
with ϵ replaced by ϵ˜ and CTL+∥TL∥L2(𝒳)→L2(𝒳)≲1.
Thus, Corollary 2.22 then shows that TL is bounded on 𝒢°(β′,γ′) for any 0<β′,γ′<ϵ˜.
In particular, (4.65) holds. Using Lemma 4.10 and an argument similar to the
proof of (3.75) also gives (4.64). Thus, (4.61) holds.
We now prove (4.62). To this end, Corollary 4.9 shows
that it suffices to verify that for all f∈Lp(𝒳) with p∈(1,∞), limL→∞∥TL(f)∥Lp(𝒳)=0. By Theorem 3.10, for f∈Lp(𝒳) and h∈Lp′(𝒳),f=∑l=−∞∞D˜lDl(f) and h=∑l=−∞∞D˜lDl(h), respectively, in Lp(𝒳) and Lp′(𝒳),
where D˜k for k∈ℤ are as in Theorem 3.10. From Remark 3.3, it is
easy to deduce that there exists ϵ1′∈(0,ϵ1⋀ϵ2) such that for all y∈Qτk,ν and z∈𝒳, (4.67)|DkD˜l(yτk,ν,z)|≲2−|k−l|ϵ1′1V2−k(y)+V2−k(z)+V(y,z)2−(k⋀l)ϵ1′(2−(k⋀l)+d(y,z))ϵ1′,|(DkN)tD˜l(y,z)|≲2−|k−l|ϵ1′1V2−k(y)+V2−k(z)+V(y,z)2−(k⋀l)ϵ1′(2−(k⋀l)+d(y,z))ϵ1′. These estimates and Lemma 2.1(iv) together with Hölder's inequality, Lemma 3.14, and Proposition 3.15 yield
that (4.68)∥TL(f)∥Lp(𝒳)=sup∥h∥Lp′(𝒳)≤1|〈TL(f),h〉|=sup∥h∥Lp′(𝒳)≤1|∑|k|>L∑τ∈Ik∑ν=1N(k,τ)∫Qτk,νDk(f)(yτk,ν)(DkN)t(h)(y)dμ(y)|≲sup∥h∥Lp′(𝒳)≤1∥{∑|k|>L[∑l=−∞∞2−|k−l|ϵ1′M(|Dl(f)|)]2}1/2∥Lp(𝒳)×∥{∑|k|>L[∑l=−∞∞2−|k−l|ϵ1′M(|(DkN)t(h)|)]2}1/2∥Lp′(𝒳)≲2−ϵ1′L/2∥{∑|l|<L/2[M(|Dl(f)|)]2}1/2∥Lp(𝒳)+∥{∑|l|≥L/2[M(|Dl(f)|)]2}1/2∥Lp(𝒳)→0, as L→∞.
That is, (4.62) also holds, which completes the proof of Theorem 4.11.By an argument similar to the proof of Theorem4.11, we
can establish the following variants
of the discrete Calderón reproducing formulae (we omit the details).Theorem 4.12.
Letϵ1∈(0,1],ϵ2>0,ϵ3>0,ϵ∈(0,ϵ1⋀ϵ2), and let {Sk}k=−∞∞ be an (ϵ1,ϵ2,ϵ3)−ATI.
Set Dk=Sk−Sk−1 for any k∈ℤ.
Then for any fixed j∈ℕ as in Corollary 4.9, there exists a family of
linear operators {D¯k}k∈ℤ such that for any fixed yτk,ν∈Qτk,ν with k∈ℤ,τ∈Ik, and ν=1,…,N(k,τ),
and all f∈𝒢°0ϵ(β,γ) with 0<β,γ<ϵ, (4.69)f(x)=∑k=−∞∞∑τ∈Ik∑ν=1N(k,τ)∫Qτk,νDk(x,y)dμ(y)D¯k(f)(yτk,ν)=∑k=−∞∞∑τ∈Ik∑ν=1N(k,τ)Dk(x,yτk,ν)∫Qτk,νD¯k(f)(y)dμ(y)=∑k=−∞∞∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)Dk(x,yτk,ν)D¯k(f)(yτk,ν), where the series converges in
both the norm of 𝒢°0ϵ(β,γ) and the norm of Lp(𝒳) with p∈(1,∞).
Moreover, D¯k satisfies the conditions as in Theorem 3.11.Theorems4.11 and 4.12 in combination with a
duality argument show that discrete
Calderón reproducing formulae on spaces of distributions.Theorem 4.13.
Let all the notation be as in
Theorems4.11 and 4.12. Then for all f∈(𝒢°0ϵ(β,γ))′ with 0<β,γ<ϵ,
(4.57) and (4.69) hold in (𝒢°0ϵ(β,γ))′.


### 4.2. Inhomogeneous Discrete Calderón Reproducing Formulae

Similarly to Subsection3.2, we can establish the
following inhomogeneous discrete
Calderón reproducing formulae (we omit the details). Here, again, we have
no restriction on diam(𝒳).Theorem 4.14.
Letϵ1∈(0,1],ϵ2>0,ϵ3>0,ϵ∈(0,ϵ1⋀ϵ2), and let {Sk}k∈ℤ+ be an (ϵ1,ϵ2,ϵ3)−IATI.
Set D0=S0 and Dk=Sk−Sk−1 for k∈ℕ.
Then for any fixed j,N∈ℕ such that (4.1) and (4.39) hold, there exists a
family of functions {D˜k(x,y)}k∈ℤ+ such that for any fixed yτk,ν∈Qτk,ν with k∈ℕ, τ∈Ik and ν=1,…,N(k,τ) and all f∈𝒢0ϵ(β,γ) with 0<β,γ<ϵ, (4.70)f(x)=∑k=0N∑τ∈Ik∑ν=1N(k,τ)∫Qτk,νD˜k(x,y)dμ(y)Dτ,1k,ν(f)+∑k=N+1∞∑τ∈Ik∑ν=1N(k,τ)∫Qτk,νD˜k(x,y)dμ(y)Dk(f)(yτk,ν)=∑τ∈I0∑ν=1N(0,τ)∫Qτ0,νD˜0(x,y)dμ(y)Dτ,10,ν(f)+∑k=1∞∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)D˜k(x,yτk,ν)Dτ,1k,ν(f)=∑τ∈I0∑ν=1N(0,τ)∫Qτ0,νD˜0(x,y)dμ(y)Dτ,10,ν(f)+∑k=1N∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)D˜k(x,yτk,ν)Dτ,1k,ν(f)+∑k=N+1∞∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)D˜k(x,yτk,ν)Dk(f)(yτk,ν), where the series
converges in both the norm of 𝒢0ϵ(β,γ) and the norm of Lp(𝒳) with p∈(1,∞),
and Dτ,1k,ν for
k∈ℤ+,τ∈Ik, and ν=1,…,N(k,τ) is the corresponding integral operator with
the kernel Dτ,1k,ν(z)=(1/μ(Qτk,ν))∫Qτk,νDk(u,z)dμ(u).
Moreover, D˜k for k≥N+1 satisfies the conditions (i) and (ii) of
Definition 2.2 with ϵ1 and ϵ2 replaced by ϵ′∈(ϵ,ϵ1⋀ϵ2);
and there exists a constant C>0 depending on ϵ′ such that the function D˜k(x,y) for k=0,1,…,N satisfies that
(i)
for allx,y∈𝒳,|D˜k(x,y)|≤C((1/(V1(x)+V1(y)+V(x,y)))(1/(1+d(x,y))ϵ′)),(ii)
for allx,x′,y∈𝒳 with d(x,x′)≤(1+d(x,y))/2, (4.71)|D˜k(x,y)−D˜k(x′,y)|≤C(d(x,x′)1+d(x,y))ϵ′1V1(x)+V1(y)+V(x,y)1(1+d(x,y))ϵ′; and ∫𝒳D˜k(x,y)dμ(x)=∫𝒳D˜k(x,y)dμ(y)=1 when 0≤k≤N; =0 when k>N.Theorem 4.15.
Letϵ1∈(0,1],ϵ2>0,ϵ3>0,ϵ∈(0,ϵ1⋀ϵ2), and let {Sk}k∈ℤ+ be an (ϵ1,ϵ2,ϵ3)−IATI.
Set D0=S0 and Dk=Sk−Sk−1 for k∈ℕ.
Then for any fixed j,N∈ℕ such that (4.1) and (4.39) hold, there exists a
family of functions {D¯k(x,y)}k∈ℤ+ such that for any fixed yτk,ν∈Qτk,ν with k∈ℕ,τ∈Ik, and ν=1,…,N(k,τ) and all f∈𝒢0ϵ(β,γ) with 0<β,γ<ϵ, (4.72)f(x)=∑k=0N∑τ∈Ik∑ν=1N(k,τ)∫Qτk,νDk(x,y)dμ(y)Dτ,1k,ν¯(f)+∑k=N+1∞∑τ∈Ik∑ν=1N(k,τ)∫Qτk,νDk(x,y)dμ(y)D¯k(f)(yτk,ν)=∑τ∈I0∑ν=1N(0,τ)∫Qτo,νD0(x,y)dμ(y)Dτ,10,ν¯(f)+∑k=1∞∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)Dk(x,yτk,ν)Dτ,1k,ν¯(f)=∑τ∈I0∑ν=1N(0,τ)∫Qτo,νD0(x,y)dμ(y)Dτ,10,ν¯(f)+∑k=1N∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)Dk(x,yτk,ν)Dτ,1k,ν¯(f)+∑k=N+1∞∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)Dk(x,yτk,ν)D¯k(f)(yτk,ν),  where the series
converges in both the norm of 𝒢0ϵ(β,γ) and the norm of Lp(𝒳) with p∈(1,∞),
and Dτ,1k,ν¯ for k∈ℤ+,τ∈Ik, and ν=1,…,N(k,τ) is the corresponding integral operator with
the kernel Dτ,1k,ν¯(z) as in (4.70) with Dk replaced by D¯k.
Moreover, D¯k for k≥N+1 satisfies the conditions (i) and (iii) of
Definition 2.2 with ϵ1 and ϵ2 replaced by ϵ′∈(ϵ,ϵ1⋀ϵ2);
and there exists a constant C>0 depending on ϵ′ such that the function D¯k(x,y) for k=0,1,…,N satisfies  that
(i)
for allx,y∈𝒳, |D¯k(x,y)|≤C((1/(V1(x)+V1(y)+V(x,y)))(1/(1+d(x,y))ϵ′)),(ii)
for allx,x′,y∈𝒳 with d(x,x′)≤(1+d(x,y))/2, (4.73)|D¯k(x,y)−D¯k(x′,y)|≤C(d(x,x′)1+d(x,y))ϵ′1V1(x)+V1(y)+V(x,y)1(1+d(x,y))ϵ′; and ∫𝒳D¯k(x,y)dμ(x)=∫𝒳D¯k(x,y)dμ(y)=1 when 0≤k≤N; =0 when k>N.Theorem 4.16.
Use
the same notation as in Theorems4.14 and 4.15. Then for all f∈(𝒢0ϵ(β,γ))′ with 0<β,γ<ϵ,
(4.70) and (4.72) hold in (𝒢0ϵ(β,γ))′.Remark 4.17.
Similarly to Remark3.27, in the sequel, to simplify the representation
of the results, we will always assume that N=0 in Theorems 4.14, 4.15,  and 4.16.


## 4.1. Homogeneous Discrete Calderón Reproducing Formulae

In this subsection, we always assume thatdiam(𝒳)=∞,ϵ1∈(0,1], ϵ2>0, ϵ3>0, and {Sk}k∈ℤ is an (ϵ1,ϵ2,ϵ3)-ATI.
Set Dk=Sk−Sk−1 for k∈ℤ.Let all the notation be as in (3.44). We now introduce
the following discrete Riemann sum
operator on 𝒳, (4.2)S(f)(x)=∑k=−∞∞∑τ∈Ik∑ν=1N(k,τ)∫Qτk,νDkN(x,y)dμ(y)Dk(f)(yτk,ν).We first verify thatS is well defined and bounded on L2(𝒳) via the Littlewood-Paley theorem for the
homogeneous g-function as in (3.63), Proposition 3.15. To do
so, let us first establish the following estimate by using  Proposition
3.15.Lemma 4.1.
There exists a constantC>0 such that for all yτk,ν∈Qτk,ν and all f∈L2(𝒳), (4.3)∑k=−∞∞∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)|Dk(f)(yτk,ν)|2≤C∥f∥L2(𝒳)2.Proof.
By Theorem3.10, there exists a family of linear operators {D˜k}k=−∞∞ as in Theorem 3.10 such that for all f∈L2(𝒳), f=∑l=−∞∞D˜lDl(f). By Lemma 3.2 together with Remark 3.3, we have
that for any ϵ1′∈(0,ϵ1⋀ϵ2) and any ϵ1′′∈(0,ϵ1′),
all x,z∈𝒳 and all k,l∈ℤ, (4.4)|DkD˜l(z,x)|≲2−|k−l|ϵ1′′1V2−(k⋀l)(x)+V2−(k⋀l)(z)+V(z,x)2−(k⋀l)ϵ1′(2−(k⋀l)+d(z,x))ϵ1′. Notice that for all x∈𝒳 and any
z,y∈Qτk,ν, by Lemma 2.19(iv), we have that d(y,z)≤C62−j2−k≤C62−j2−(k⋀l)≤C62−j(2−(k⋀l)+d(x,y)), where j∈ℕ satisfies (4.1). Thus, for all x∈𝒳,
any y,z∈Qτk,ν and all k,l∈ℤ, Lemma 2.1(iii) shows that (4.5)|DkD˜l(z,x)|χQτk,ν(z)≲2−|k−l|ϵ1′′1V2−(k⋀l)(x)+V2−(k⋀l)(y)+V(y,x)2−(k⋀l)ϵ1′(2−(k⋀l)+d(y,x))ϵ1′χQτk,ν(y). From this and Lemma 2.1(iv), it
follows that for k∈ℤ, (4.6)|Dk(f)(yτk,ν)|≤∑l=−∞∞∫𝒳|(DkD˜l)(yτk,ν,x)||Dl(f)(x)|dμ(x)≲∑l=−∞∞2−|k−l|ϵ1′′M(Dl(f))(y)χQτk,ν(y), where M is the Hardy-Littlewood maximal function on 𝒳.
By (4.6), the construction of Qτk,ν (see Lemma 2.19), Lemmas 3.14 and 3.9, we
obtain (4.7)∑k=−∞∞∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)|Dk(f)(yτk,ν)|2≲∑k=−∞∞∫𝒳[∑l=−∞∞2−|k−l|ϵ1′′M(Dl(f))(y)]2dμ(y)≲∑l=−∞∞∥M(Dl(f))∥L2(𝒳)2≲∥f∥L2(𝒳)2,which proves Lemma 4.1.The next lemma can be proved in a way similar as in
the proof of Theorem (1.14) in [78, page 12]. The main idea is to combine Theorem 3.10,
Lemma 2.19, and Hölder's inequality with a duality argument. We omit the details
here; see also [36].Lemma 4.2.
Suppose that a sequence{aτk,ν:k∈ℤ,τ∈Ik,ν=1,…,N(k,τ)} of numbers satisfies ∑k=−∞∞∑τ∈Ik∑ν=1N(k,τ)|aτk,ν|2<∞. Then the function defined by (4.8)f(x)=∑k=−∞∞∑τ∈Ik∑ν=1N(k,τ)[μ(Qτk,ν)]−1/2aτk,ν∫Qτk,νDkN(x,y)dμ(y) is in L2(𝒳).
Moreover, (4.9)∥f∥L2(𝒳)2≤C∑k=−∞∞∑τ∈Ik∑ν=1N(k,τ)|aτk,ν|2.Lemmas4.1 and 4.2 yield the boundedness of the
discrete Riemann sum operator S on L2(𝒳).Proposition 4.3.
Let
the notation be the same as above withj satisfying (4.1). Then the discrete Riemann
sum operator S in (4.2) is bounded on L2(𝒳).
That is, there is a constant C>0,
only depending on N,
such that for all f∈L2(𝒳), (4.10)∥S(f)∥L2(𝒳)≤C∥f∥L2(𝒳).Next we prove that the discrete Riemann sum operatorS is invertible and S−1 maps 𝒢°(x1,r,β,γ) into itself. To do this, we define R=I−S and first establish some estimates on the
kernel, R(x,y),
of the operator R.
To this end, by (3.44), we write (4.11)R(f)(x)=(I−S)(f)(x)=∑k=−∞∞∑τ∈Ik∑ν=1N(k,τ)∫Qτk,νDkN(x,y)[Dk(f)(y)−Dk(f)(yτk,ν)]dμ(y)+∑|l|>N∑k=−∞∞Dk+lDk(f)(x)≡∑k=−∞∞Gk(f)(x)+RN(f)(x)≡G(f)(x)+RN(f)(x). Let Gk(x,y) be the kernel of Gk for k∈ℤ.
We now verify that Gk(x,y),
and hence G(x,y),
satisfies all the desired estimates. Clearly, (4.12)G(x,y)=∑k=−∞∞∑τ∈Ik∑ν=1N(k,τ)∫Qτk,νDkN(x,z)[Dk(z,y)−Dk(yτk,ν,y)]dμ(z)=∑k=−∞∞Gk(x,y). We need the following two
technical lemmas.Lemma 4.4.
Let0≤λ,μ≤1, ϵ1∈(0,1], μϵ1<ν1, λϵ1<ν2,
and k∈ℤ.
Then there exists a constant C>0 such that for all x,y∈𝒳 and all k∈ℤ, (4.13)∫𝒳d(x,x′)λϵ1(2−k+d(x,z))λϵ11V2−k(x)+V2−k(z)+V(x,z)2−kν1(2−k+d(x,z))ν1×d(y,y′)μϵ1(2−k+d(z,y))μϵ11V2−k(z)+V2−k(y)+V(z,y)2−kν2(2−k+d(z,y))ν2dμ(z)≤Cd(x,x′)λϵ1(2−k+d(x,y))λϵ1d(y,y′)μϵ1(2−k+d(x,y))μϵ11V2−k(x)+V2−k(y)+V(x,y)×{2−k(ν2−λϵ1)(2−k+d(x,y))ν2−λϵ1+2−k(ν1−μϵ1)(2−k+d(x,y))ν1−μϵ1}.Proof.
Write(4.14)∫𝒳d(x,x′)λϵ1(2−k+d(x,z))λϵ11V2−k(x)+V2−k(z)+V(x,z)2−kν1(2−k+d(x,z))ν1×d(y,y′)μϵ1(2−k+d(z,y))μϵ11V2−k(z)+V2−k(y)+V(z,y)2−kν2(2−k+d(z,y))ν2dμ(z)=∫d(x,z)≥d(x,y)/2d(x,x′)λϵ1(2−k+d(x,z))λϵ11V2−k(x)+V2−k(z)+V(x,z)×2−kν1(2−k+d(x,z))ν1d(y,y′)μϵ1(2−k+d(z,y))μϵ12−kν2(2−k+d(z,y))ν2×1V2−k(z)+V2−k(y)+V(z,y)dμ(z)+∫d(x,z)<d(x,y)/2⋯≡Z1+Z2. For Z1,
the fact d(x,z)≥d(x,y)/2 implies that V(x,y)≤|B(x,2d(x,z))|≲V(x,z),
which together with Lemma 2.1(ii) shows that (4.15)Z1≲d(x,x′)λϵ1(2−k+d(x,y))λϵ11V2−k(x)+V(x,y)2−kν1(2−k+d(x,y))ν1×d(y,y′)μϵ1∫𝒳1V2−k(y)+V(z,y)2−kν2(2−k+d(z,y))ν2+μϵ1dμ(z)≲d(x,x′)λϵ1(2−k+d(x,y))λϵ1d(y,y′)μϵ1(2−k+d(x,y))μϵ11V2−k(x)+V(x,y)2−k(ν1−μϵ1)(2−k+d(x,y))ν1−μϵ1. For Z2,
the fact d(x,z)<d(x,y)/2 implies that d(z,y)>d(x,y)/2 and therefore, V(x,y)~V(y,x)≲V(y,z)~V(z,y).
From these facts and Lemma 2.1(ii), it follows that (4.16)Z2≲d(x,x′)λϵ1d(y,y′)μϵ1(2−k+d(x,y))μϵ12−kν2(2−k+d(x,y))ν2×min{1V(x,y)∫𝒳1V2−k(x)+V(x,z)2−kν1(2−k+d(x,z))ν1+λϵ1dμ(z),1V2−k(x)1V(x,y)2kλϵ1μ(B(x,d(x,y)2))}≲d(x,x′)λϵ1(2−k+d(x,y))λϵ1d(y,y′)μϵ1(2−k+d(x,y))μϵ11V2−k(x)+V(x,y)2−k(ν2−λϵ1)(2−k+d(x,y))ν2−λϵ1.
Moreover, by symmetry, we also
have (4.17)∫𝒳d(x,x′)λϵ1(2−k+d(x,z))λϵ11V2−k(x)+V2−k(z)+V(x,z)2−kν1(2−k+d(x,z))ν1×d(y,y′)μϵ1(2−k+d(z,y))μϵ11V2−k(z)+V2−k(y)+V(z,y)2−kν2(2−k+d(z,y))ν2dμ(z)≲d(x,x′)λϵ1(2−k+d(x,y))λϵ1d(y,y′)μϵ1(2−k+d(x,y))μϵ11V2−k(y)+V(x,y)×{2−k(ν2−λϵ1)(2−k+d(x,y))ν2−λϵ1+2−k(ν1−μϵ1)(2−k+d(x,y))ν1−μϵ1}. Combining these estimates
completes the proof of Lemma 4.4.Lemma 4.5.
LetGk(x,y) for k∈ℤ be as above. Then for any λ,μ∈(0,1) satisfying that λϵ1<ϵ2,
there exists a constant CN>0,
independent of j,
such that
(i)
|Gk(x,y)|≤CN2−jϵ1(1/(V2−k(x)+V2−k(y)+V(x,y)))(2−kϵ2/(2−k+d(x,y))ϵ2);(ii)
fory,y′∈𝒳 with d(y,y′)≤(2−k+d(x,y))/2,(4.18)|Gk(x,y)−Gk(x,y′)|≤CN2−jϵ1d(y,y′)μϵ1(2−k+d(x,y))μϵ11V2−k(x)+V2−k(y)+V(x,y)×{2−kϵ3(2−k+d(x,y))ϵ3+2−k(ϵ2−μϵ1)(2−k+d(x,y))ϵ2−μϵ1};(iii)
|Gk(x,y)−Gk(x′,y)|≤CN2−jϵ1(d(x,x′)λϵ1/(2−k+d(x,y))λϵ1)(1/(V2−k(x)+V2−k(y)+V(x,y)))(2−k(ϵ2−λϵ1)/(2−k+d(x,y))ϵ2−λϵ1) for x,x′∈𝒳 with d(x,x′)≤(2−k+d(x,y))/2;(iv)
∫𝒳Gk(x,y)dμ(y)=0=∫𝒳Gk(x,y)dμ(x).Proof.
Since∫𝒳Dk(x,y)dμ(y)=0=∫𝒳DkN(x,y)dμ(x), by the definition of Gk in (4.12), we easily see that (iv) holds.
By the construction of dyadic cubes in Lemma2.19, we
also easily see that for any z∈Qτk,ν, (4.19)d(z,yτk,ν)≤C62−(k+j)=C62−j2−k≤C62−j(2−k+d(y,z)). We recall that j always satisfies (4.1). Then, the regularity
of Dk and the size estimates of DkN together with Lemma 4.4 in the case λ=μ=0 yield that (4.20)|Gk(x,y)|=|∑τ∈Ik∑ν=1N(k,τ)∫Qτk,νDkN(x,z)[Dk(z,y)−Dk(yτk,ν,y)]dμ(z)|≤CN2−jϵ11V2−k(x)+V2−k(y)+V(x,y)2−kϵ2(2−k+d(x,y))ϵ2, which shows Gk(x,y) satisfies (i).
Write(4.21)Gk(x,y)−Gk(x,y′)=∑τ∈Ik∑ν=1N(k,τ)∫Qτk,νDkN(x,z){[Dk(z,y)−Dk(yτk,ν,y)]−[Dk(z,y′)−Dk(yτk,ν,y′)]}dμ(z). We now verify that Gk(x,y) satisfies (ii) by considering the following
two cases.
Case 1  (d(y,y′)≤(2−k+d(z,y))/3).
In this case, from (4.19) with j satisfying (4.1), the size estimates of DkN and the second difference regularity of Dk and Lemma 4.4 with λ=0, μ∈(0,1), ν1=ϵ2,
and ν2=ϵ3,
it follows that (4.22)|Gk(x,y)−Gk(x,y′)|≲2−jϵ1∫𝒳|DkN(x,z)|d(y,y′)μϵ1(2−k+d(z,y))μϵ11V2−k(z)+V2−k(y)+V(z,y)2−kϵ3(2−k+d(z,y))ϵ3dμ(z)≤CN2−jϵ1d(y,y′)μϵ1(2−k+d(x,y))μϵ11V2−k(x)+V2−k(y)+V(x,y){2−kϵ3(2−k+d(x,y))ϵ3+2−k(ϵ2−μϵ1)(2−k+d(x,y))ϵ2−μϵ1}.
Case 2  ((2−k+d(z,y))/3<d(y,y′)≤(2−k+d(x,y))/2).
In this case, the estimate (i), Lemma2.1(iii) together with the fact that 2−k<3d(y,y′) show that for any μ∈(0,1), (4.23)|Gk(x,y)−Gk(x,y′)|≤|Gk(x,y)|+|Gk(x,y′)|≤CN2−jϵ1{1V2−k(x)+V2−k(y)+V(x,y)2−kϵ2(2−k+d(x,y))ϵ2+1V2−k(x)+V2−k(y′)+V(x,y′)2−kϵ2(2−k+d(x,y′))ϵ2}≤CN2−jϵ1d(y,y′)μϵ1(2−k+d(x,y))μϵ11V2−k(x)+V2−k(y)+V(x,y)2−k(ϵ2−μϵ1)(2−k+d(x,y))ϵ2−μϵ1, which completes the proof of
(ii).
We verify (iii) by writing(4.24)Gk(x,y)−Gk(x′,y)=∑τ∈Ik∑ν=1N(k,τ)∫Qτk,ν[DkN(x,z)−DkN(x′,z)][Dk(z,y)−Dk(yτk,ν,y)]dμ(z). If d(x,x′)≤(2−k+d(x,z))/2,
the estimate (4.19) with j satisfying (4.1), the regularity of DkN and Dk,
Lemma 4.4 with λ∈(0,1), μ=0,
and ν1=ν2=ϵ2 show that (4.25)|Gk(x,y)−Gk(x′,y)|≤CN2−jϵ1∫𝒳d(x,x′)λϵ1(2−k+d(x,z))λϵ11V2−k(x)+V2−k(z)+V(x,z)×2−kϵ2(2−k+d(x,z))ϵ21V2−k(z)+V2−k(y)+V(z,y)2−kϵ2(2−k+d(z,y))ϵ2dμ(z)≤CN2−jϵ1d(x,x′)λϵ1(2−k+d(x,y))λϵ11V2−k(x)+V2−k(y)+V(x,y)2−k(ϵ2−λϵ1)(2−k+d(x,y))ϵ2−λϵ1.
If(2−k+d(x,z))/2<d(x,x′)≤(2−k+d(x,y))/2,
the estimate (i), Lemma 2.1(iii) together with the fact 2−k<2d(x,x′) yield that (4.26)|Gk(x,y)−Gk(x′,y)|≤CN2−jϵ1{1V2−k(x)+V2−k(y)+V(x,y)2−kϵ2(2−k+d(x,y))ϵ2+1V2−k(x′)+V2−k(y)+V(x′,y)2−kϵ2(2−k+d(x′,y))ϵ2}≤CN2−jϵ1d(x,x′)λϵ1(2−k+d(x,y))λϵ11V2−k(x)+V2−k(y)+V(x,y)2−k(ϵ2−λϵ1)(2−k+d(x,y))ϵ2−λϵ1, which completes the proof of
(iii), and hence the proof of Lemma 4.5.Lemma 4.6.
LetN∈ℕ and G be as in (4.12). Then there exist a constant CN>0 independent of j such that (4.27)∥G∥L2(𝒳)→L2(𝒳)≤CN2−jϵ1.Proof.
Using Lemma4.5 and repeating the proof of (3.2) in Lemma 3.2 yield that
for any ϵ1′∈(0,ϵ1⋀ϵ2) and all k,j∈ℤ,(4.28)∥GkGjt∥L2(𝒳)→L2(𝒳)≤CN2−jϵ12−|k−j|ϵ1′,∥GjtGk∥L2(𝒳)→L2(𝒳)≤CN2−jϵ12−|k−j|ϵ1′, where CN>0 is a constant depending on ϵ1′,
but, independent of j and k.
These estimates together with the Cotlar-Stein lemma and G=∑k=−∞∞Gk yields the conclusion of Lemma 4.6.The following lemma is a key lemma to establish the
discrete Calderón reproducing formula.Lemma 4.7.
LetS be as in (4.2) and R=I−S.
Then R is a Calderón-Zygmund singular integral
operator, R(1)=0=R∗(1).
Moreover, for any ϵ1′∈(0,ϵ1⋀ϵ2),
there exist positive constants C10, CN,
and δ,
depending on ϵ1′,
such that the kernel, R(x,y),
of R satisfies all the conditions of Corollary 2.22
with ϵ replaced by ϵ1′ and (4.29)CR+∥R∥L2(𝒳)→L2(𝒳)≤C102−δN+CN2−jϵ1, where CR is the Calderón-Zygmund constant as in Theorem
2.18, C10 and δ are independent of N and j,
and CN is independent  of j.Proof.
From (4.11), it is easy to see that R(1)=0=R∗(1).
Moreover, Lemma 3.6 shows that RN(x,y),
the kernel of RN,
satisfies all the conditions of Corollary 2.22 with ϵ replaced by ϵ1′ and CRN+∥RN∥L2(𝒳)→L2(𝒳)≤C102−δN,
where C10 and δ are independent of
N and j.
This combining the formula (4.11) implies that to prove Lemma 4.7, it suffices
to verify that G(x,y),
the kernel of G,
satisfies (2.49) and Theorem 2.18(i), (ii), and (iii) with ϵ replaced by ϵ1′ and CG≤CN2−jϵ1,
where CN is independent of N.
In fact, combining Lemma 4.5(i) with Lemma 3.5 and (4.12) yields that for all x,y∈𝒳 with x≠y, (4.30)|G(x,y)|≤∑k=−∞∞|Gk(x,y)|≤CN2−jϵ1∑k=−∞∞1V2−k(x)+V(x,y)2−kϵ2(2−k+d(x,y))ϵ2≤CN2−jϵ11V(x,y), which shows G(x,y) satisfies Theorem 2.18(i).
Lemma4.5(ii) and (iii) and Lemma 3.5 together with
(4.12), respectively, show that for any μ∈(0,1),
all x,y,y′∈𝒳 with x≠y and d(y,y′)≤d(x,y)/2, (4.31)|G(x,y)−G(x,y′)|≤∑k=−∞∞|Gk(x,y)−Gk(x,y′)|≤CN2−jϵ1d(y,y′)μϵ1d(x,y)μϵ1∑k=−∞∞1V2−k(x)+V(x,y)×{2−kϵ3(2−k+d(x,y))ϵ3+2−k(ϵ2−μϵ1)(2−k+d(x,y))ϵ2−μϵ1}≤CN2−jϵ1d(y,y′)μϵ1d(x,y)μϵ11V(x,y), and for any λ∈(0,1),
all x,x′,y∈𝒳 with x≠y and d(x,x′)≤d(x,y)/2, (4.32)|G(x,y)−G(x′,y)|≤∑k=−∞∞|Gk(x,y)−Gk(x′,y)|≤CN2−jϵ1d(x,x′)λϵ1d(x,y)λϵ1∑k=−∞∞1V2−k(x)+V(x,y)2−k(ϵ2−λϵ1)(2−k+d(x,y))ϵ2−λϵ1≤CN2−jϵ1d(x,x′)λϵ1d(x,y)λϵ11V(x,y), which proves that G(x,y) satisfies Theorem 2.18(ii) and (2.49).
By (2.49) and Theorem 2.18(ii), it is easy to see that
it suffices to establish Theorem 2.18(iii) only for d(x,x′)≤d(x,y)/6 and d(y,y′)≤d(x,y)/6.
To this end, we write (4.33)[G(x,y)−G(x′,y)]−[G(x,y′)−G(x′,y′)]=∑k=−∞∞∑τ∈Ik∑ν=1N(k,τ)∫Qτk,ν[DkN(x,z)−DkN(x′,z)]×{[Dk(z,y)−Dk(yτk,ν,y)]−[Dk(z,y′)−Dk(yτk,ν,y′)]}dμ(z).
Ifd(x,x′)>(2−k+d(x,z))/2 and d(y,y′)>(2−k+d(y,z))/3,
then(4.34)d(x,x′)+d(y,y′)≥(21−k+d(x,z)+d(z,y))3>d(x,y)3, which contradicts the
assumptions that d(x,x′)≤d(x,y)/6 and d(y,y′)≤d(x,y)/6.
Thus, we still have the following three cases:
(i)
d(x,x′)≤(2−k+d(x,z))/2 and d(y,y′)≤(2−k+d(y,z))/3;(ii)
d(x,x′)≤(2−k+d(x,z))/2 and d(y,y′)>(2−k+d(y,z))/3;(iii)
d(x,x′)>(2−k+d(x,z))/2 and d(y,y′)≤(2−k+d(y,z))/3. In the case
(i), by (4.19) with j satisfying (4.1), the second difference
regularity of DkN and Dk,
Lemma 4.4 with λ,μ∈(0,1), ν1=ϵ2,
and ν2=ϵ1+ϵ3,
and Lemma 3.5, we obtain (4.35)|[G(x,y)−G(x′,y)]−[G(x,y′)−G(x′,y′)]|≤CN2−jϵ1∑k=−∞∞∫𝒳d(x,x′)λϵ1(2−k+d(x,z))λϵ11V2−k(x)+V2−k(z)+V(x,z)2−kϵ2(2−k+d(x,z))ϵ2d(y,y′)μϵ1(2−k+d(z,y))μϵ1×1V2−k(z)+V2−k(y)+V(z,y)2−k(ϵ1+ϵ3)(2−k+d(z,y))ϵ1+ϵ3dμ(z)≤CN2−jϵ1d(x,x′)λϵ1d(x,y)λϵ1d(y,y′)μϵ1d(x,y)μϵ11V(x,y). In the case (ii), we also have (2−k+d(z,y′))/2≤(2−k+d(z,y))/2+d(y,y′)/2≤2d(y,y′). From this, (4.19) with j satisfying (4.1), the regularity of DkN and Dk,
Lemma 4.4 with λ,μ∈(0,1) and ν1=ν2=ϵ2,
Lemmas 2.1(iii) and 3.5, it follows that (4.36)|[G(x,y)−G(x′,y)]−[G(x,y′)−G(x′,y′)]|≤CN2−jϵ1∑k=−∞∞∫𝒳d(x,x′)λϵ1(2−k+d(x,z))λϵ11V2−k(x)+V2−k(z)+V(x,z)2−kϵ2(2−k+d(x,z))ϵ2×[d(y,y′)μϵ1(2−k+d(z,y))μϵ11V2−k(z)+V2−k(y)+V(z,y)2−kϵ2(2−k+d(z,y))ϵ2+d(y,y′)μϵ1(2−k+d(z,y′))μϵ11V2−k(z)+V2−k(y′)+V(z,y′)2−kϵ2(2−k+d(z,y′))ϵ2]dμ(z)≤CN2−jϵ1d(x,x′)λϵ1d(x,y)λϵ1d(y,y′)μϵ1d(x,y)μϵ11V(x,y). For the last case (iii), we have (2−k+d(x′,z))/2≤(2−k+d(x,z))/2+d(x,x′)/2≤3d(x,x′)/2. This estimate, (4.19) with j satisfying (4.1), the size condition of DkN,
the second difference regularity of Dk,
Lemma 4.4 with λ,μ∈(0,1), ν1=ϵ2, and ν2=ϵ1+ϵ3,
Lemmas 2.1(iii) and 3.5 show that (4.37)|[G(x,y)−G(x′,y)]−[G(x,y′)−G(x′,y′)]|≤CN2−jϵ1∑k=−∞∞∫𝒳[|DkN(x,z)|+|DkN(x′,z)|]×d(y,y′)μϵ1(2−k+d(z,y))μϵ11V2−k(z)+V2−k(y)+V(z,y)2−k(ϵ1+ϵ3)(2−k+d(z,y))ϵ1+ϵ3dμ(z)≤CN2−jϵ1d(x,x′)λϵ1d(x,y)λϵ1d(y,y′)μϵ1d(x,y)μϵ11V(x,y). Thus, G(x,y) satisfies Theorem 2.18(iii). This finishes the
proof of Lemma 4.7.By Lemmas4.6 and 4.7, we obtain the boundedness
of R on 𝒢°(x1,r,β,γ) for any x1∈𝒳, r>0 and 0<β,γ<(ϵ1⋀ϵ2),
and in Lp(𝒳) for p∈(1,∞).
We omit the details.Proposition 4.8.
LetR and the notation be as in Lemma 4.7. Then R is bounded on Lp(𝒳) for p∈(1,∞) and on 𝒢°(x1,r,β,γ) for any x1∈𝒳, r>0 and 0<β,γ<(ϵ1⋀ϵ2).
That is, there exists a constant C11>0,
only depending on p, β, and γ such that for all f∈Lp(𝒳), (4.38)∥R(f)∥Lp(𝒳)≤C11(C102−δN+CN2−jϵ1)∥f∥Lp(𝒳),∥R(f)∥𝒢(x1,r,β,γ)≤C11(C102−δN+CN2−jϵ1)∥f∥𝒢(x1,r,β,γ).By using Proposition4.8 and repeating the proof of
Propositions 3.7 and 3.8, we can obtain the boundedness of S−1 on 𝒢°(x1,r,β,γ) for any x1∈𝒳, r>0 and 0<β,γ<ϵ,
and in Lp(𝒳) for p∈(1,∞).
We also omit the details.Corollary 4.9.
LetS be as in (4.2). Let N,j∈ℕ such that (4.1) and (4.39)C11(C102−δN+CN2−jϵ1)<1 hold. Then S has a bounded inverse in 𝒢°(x1,r,β,γ) for any x1∈𝒳, r>0, and 0<β,γ<(ϵ1⋀ϵ2),
and in Lp(𝒳) for p∈(1,∞).
Namely, there exists a constant C>0 depending only on β,γ, and p such that for all f∈𝒢°(x1,r,β,γ), (4.40)∥S−1(f)∥𝒢(x1,r,β,γ)≤C∥f∥𝒢(x1,r,β,γ) and for all f∈Lp(𝒳), (4.41)∥S−1(f)∥Lp(𝒳)≤C∥f∥Lp(𝒳).To establish the discrete Calderón reproducing
formulae, we still need the following technical lemma.Lemma 4.10.
Letj satisfy (4.1). For k∈ℤ,
any fixed yτk,ν∈Qτk,ν with τ∈Ik and ν∈{1,…,N(k,τ)},
and any x∈𝒳,
let (4.42)Ek(x,y)=∑τ∈Ik∑ν=1N(k,τ)∫Qτk,νDkN(x,z)dμ(z)Dk(yτk,ν,y). Then for any λ,μ∈(0,1),
there exists a constant CN>0 depending only on N, λ and μ such that
(i)
for allk∈ℤ and all x,y∈𝒳, (4.43)|Ek(x,y)|≤CN1V2−k(x)+V2−k(y)+V(x,y)2−kϵ2(2−k+d(x,y))ϵ2;(ii)
for allk∈ℤ and all x,y,y′∈𝒳 with d(y,y′)≤(2−k+d(x,y))/2, (4.44)|Ek(x,y)−Ek(x,y′)|≤CNd(y,y′)μϵ1(2−k+d(x,y))μϵ11V2−k(x)+V2−k(y)+V(x,y)2−k(ϵ2−μϵ1)(2−k+d(x,y))ϵ2−μϵ1;(iii)
property (ii) also holds withx and y interchanged and μ replaced by λ;(iv)
for allk∈ℤ and all x,x′,y,y′∈𝒳 with d(x,x′)≤(2−k+d(x,y))/4 and d(y,y′)≤(2−k+d(x,y))/4, (4.45)|[Ek(x,y)−Ek(x′,y)]−[Ek(x,y′)−Ek(x′,y′)]|≤CNd(x,x′)λϵ1(2−k+d(x,y))λϵ1d(y,y′)μϵ1(2−k+d(x,y))μϵ11V2−k(x)+V2−k(y)+V(x,y)×[2−k(ϵ2−λϵ1)(2−k+d(x,y))ϵ2−λϵ1+2−k(ϵ2−μϵ1)(2−k+d(x,y))ϵ2−μϵ1].Proof.
The main idea for the proof of this lemma is
to combine the techniques used in the proof of Lemma3.2 with Lemma 2.1(iii).
By (4.19) and Lemma 2.1(iii), for any z∈Qτk,ν and y∈𝒳, (4.46)12−k+d(yτk,ν,y)~12−k+d(z,y),(4.47)1V2−k(yτk,ν)+V2−k(y)+V(yτk,ν,y)~1V2−k(z)+V2−k(y)+V(z,y). To see (i), by the size
condition of Dk together with (4.46) and (4.47), and Lemma 4.4
with λ=μ=0 and ν1=ν2=ϵ2,
we have (4.48)|Ek(x,y)|≲∫𝒳|DkN(x,z)|1V2−k(z)+V2−k(y)+V(z,y)2−kϵ2(2−k+d(z,y))ϵ2dμ(z)≲1V2−k(x)+V2−k(y)+V(x,y)2−kϵ2(2−k+d(x,y))ϵ2, which verifies (i).
To verify (ii), we write(4.49)|Ek(x,y)−Ek(x,y′)|≤∑τ∈Ik∑ν=1N(k,τ)|∫Qτk,νDkN(x,z)dμ(z)[Dk(yτk,ν,y)−Dk(yτk,ν,y′)]|×[χ{d(y,y′)≤(2−k+d(yτk,ν,y))/2}(y,y′)+χ{d(y,y′)>(2−k+d(yτk,ν,y))/2}(y,y′)]≡Y1+Y2. For Y1,
by the regularity of Dk and the size condition of DkN together with (4.46), (4.47) and Lemma 4.4 with λ=0,μ∈(0,1), and ν1=ν2=ϵ2,
we obtain (4.50)Y1≲∫𝒳1V2−k(x)+V2−k(z)+V(x,z)2−kϵ2(2−k+d(x,z))ϵ2d(y,y′)μϵ1(2−k+d(z,y))μϵ1×1V2−k(z)+V2−k(y)+V(z,y)2−kϵ2(2−k+d(z,y))ϵ2dμ(z)≲d(y,y′)μϵ1(2−k+d(x,y))μϵ11V2−k(x)+V2−k(y)+V(x,y)2−k(ϵ2−μϵ1)(2−k+d(x,y))ϵ2−μϵ1. The property (i), Lemma 2.1(iii), and the fact that 2−k≤2d(y,y′) yield that (4.51)Y2≲1V2−k(x)+V2−k(y)+V(x,y)2−kϵ2(2−k+d(x,y))ϵ2+1V2−k(x)+V2−k(y′)+V(x,y′)2−kϵ2(2−k+d(x,y′))ϵ2≲d(y,y′)μϵ1(2−k+d(x,y))μϵ11V2−k(x)+V2−k(y)+V(x,y)2−k(ϵ2−μϵ1)(2−k+d(x,y))ϵ2−μϵ1, which completes the proof of
(ii).
By symmetry, we can deduce (iii) from (ii).
To prove (iv), we write(4.52)[Ek(x,y)−Ek(x′,y)]−[Ek(x,y′)−Ek(x′,y′)]=∑τ∈Ik∑ν=1N(k,τ)∫Qτk,ν[DkN(x,z)−DkN(x′,z)][Dk(yτk,ν,y)−Dk(yτk,ν,y′)]dμ(z). Notice that if d(x,x′)>(2−k+d(x,z))/2 with z∈Qτk,ν and d(y,y′)>(2−k+d(yτk,ν,y))/2, then d(x,x′)+d(y,y′)>(21−k+d(x,z)+d(yτk,ν,y))/2, which contradicts the assumptions that d(x,x′)≤(2−k+d(x,y))/4 and d(y,y′)≤(2−k+d(x,y))/4, since these estimates together with (4.1) and
Lemma 2.19 prove that for z∈Qτk,ν, d(x,x′)+d(y,y′)<(21−k+d(x,z)+d(yτk,ν,y))/2. Thus, if we let (4.53)W1={z∈Qτk,ν:d(x,x′)≤2−k+d(x,z)2andd(y,y′)≤2−k+d(yτk,ν,y)2},W2={z∈Qτk,ν:d(x,x′)≤2−k+d(x,z)2andd(y,y′)>2−k+d(yτk,ν,y)2},W3={z∈Qτk,ν:d(x,x′)>2−k+d(x,z)2andd(y,y′)≤2−k+d(yτk,ν,y)2}, we then have (4.54)[Ek(x,y)−Ek(x′,y)]−[Ek(x,y′)−Ek(x′,y′)]=∑i=13∑τ∈Ik∑ν=1N(k,τ)∫Qτk,ν∩Wi[DkN(x,z)−DkN(x′,z)][Dk(yτk,ν,y)−Dk(yτk,ν,y′)]dμ(z)=∑i=13Zi. For Z1,
the regularity of DkN and Dk together with the estimates (4.46) and (4.47)
and Lemma 4.4 yield that for any λ,μ∈(0,1), (4.55)|Z1|≲∫𝒳d(x,x′)λϵ1(2−k+d(x,z))λϵ11V2−k(x)+V2−k(z)+V(x,z)2−kϵ2(2−k+d(x,z))ϵ2×d(y,y′)μϵ1(2−k+d(z,y))μϵ11V2−k(z)+V2−k(y)+V(z,y)2−kϵ2(2−k+d(z,y))ϵ2dμ(z)≲d(x,x′)λϵ1(2−k+d(x,y))λϵ1d(y,y′)μϵ1(2−k+d(x,y))μϵ11V2−k(x)+V2−k(y)+V(x,y)×{2−k(ϵ2−λϵ1)(2−k+d(x,y))ϵ2−λϵ1+2−k(ϵ2−μϵ1)(2−k+d(x,y))ϵ2−μϵ1}. Similarly, for Z2,
from d(y,y′)>(2−k+d(yτk,ν,y))/2,
(4.46) and Lemma 2.1(iii) with d(y,y′)≤(2−k+d(x,y))/4,
it follows that for z∈Qτk,ν,1/2μϵ1<d(y,y′)μϵ1/(2−k+d(yτk,ν,y))μϵ1~d(y,y′)μϵ1/(2−k+d(z,y))μϵ1, which together with the regularity of DkN and the size condition of Dk,
(4.46), (4.47), Lemmas 4.4, and 2.1(iii) shows for any λ,μ∈(0,1), (4.56)|Z2|≲∫𝒳d(x,x′)λϵ1(2−k+d(x,z))λϵ11V2−k(x)+V2−k(z)+V(x,z)2−kϵ2(2−k+d(x,z))ϵ2×[d(y,y′)μϵ1(2−k+d(z,y))μϵ11V2−k(z)+V2−k(y)+V(z,y)2−kϵ2(2−k+d(z,y))ϵ2+1V2−k(z)+V2−k(y′)+V(z,y′)2−kϵ2(2−k+d(z,y′))ϵ2]dμ(z)≲d(x,x′)λϵ1(2−k+d(x,y))λϵ1d(y,y′)μϵ1(2−k+d(x,y))μϵ11V2−k(x)+V2−k(y)+V(x,y)×{2−k(ϵ2−λϵ1)(2−k+d(x,y))ϵ2−λϵ1+2−k(ϵ2−μϵ1)(2−k+d(x,y))ϵ2−μϵ1}. From Z2 and the symmetry, we can deduce the desired
estimate for Z3,
which completes the proof of Lemma 4.10.We can now establishdiscrete Calderón reproducing formulae.Theorem 4.11.
Letϵ1∈(0,1],ϵ2>0,ϵ3>0,ϵ∈(0,ϵ1⋀ϵ2), and let {Sk}k=−∞∞ be an (ϵ1,ϵ2,ϵ3)−ATI.
Set Dk=Sk−Sk−1 for any k∈ℤ.
Then for any fixed j∈ℕ as in Corollary 4.9, there exists a family of
linear operators {D˜k}k∈ℤ such that for any fixed yτk,ν∈Qτk,ν with k∈ℤ,τ∈Ik, and ν=1,…,N(k,τ),
and all f∈𝒢°0ϵ(β,γ) with 0<β,γ<ϵ, (4.57)f(x)=∑k=−∞∞∑τ∈Ik∑ν=1N(k,τ)∫Qτk,νD˜k(x,y)dμ(y)Dk(f)(yτk,ν)=∑k=−∞∞∑τ∈Ik∑ν=1N(k,τ)D˜k(x,yτk,ν)∫Qτk,νDk(f)(y)dμ(y)=∑k=−∞∞∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)D˜k(x,yτk,ν)Dk(f)(yτk,ν), where the series converges in
both the norm of 𝒢°0ϵ(β,γ) and the norm of Lp(𝒳) with p∈(1,∞).
Moreover, D˜k satisfies the conditions as in Theorem 3.10.Proof.
We only prove the first formula in (4.57), the proof of the second
formula in (4.57) being similar. Fix N,j∈ℕ such that (4.1) and (4.39) hold. Thus, for such N and j,
Corollary 4.9 holds. Let DkN for k∈ℤ be as in (3.44). For k∈ℤ,
let D˜k(x,y)=S−1[DkN(⋅,y)](x).
By (4.2) and Corollary 4.9, similarly to the proof of Theorem 3.10, it is easy
to see that we obtain all the conclusions of the theorem except for the
convergence of the series in the first formula in (4.57). To prove this, we
need to verify that all the series in the first summation and the second
summation of the first formula in (4.57) converges in the desired ways. To
simplify the presentation, by similarity, we prove this only for the series in
the first summation of the first formula in (4.57).
Similarly to the proof of Theorem3.10, for L∈ℕ,
we write (4.58)∑|k|≤L∑τ∈Ik∑ν=1N(k,τ)∫Qτk,νD˜k(x,y)dμ(y)Dk(f)(yτk,ν)=S−1[∑|k|≤L∑τ∈Ik∑ν=1N(k,τ)∫Qτk,νDkN(⋅,y)dμ(y)Dk(f)(yτk,ν)](x)=S−1{S(f)(⋅)−∑|k|>L∑τ∈Ik∑ν=1N(k,τ)∫Qτk,νDkN(⋅,y)dμ(y)Dk(f)(yτk,ν)}(x)=f(x)−limm→∞Rm(f)(x)−S−1{∑|k|>L∑τ∈Ik∑ν=1N(k,τ)∫Qτk,νDkN(⋅,y)dμ(y)Dk(f)(yτk,ν)}(x). Corollary 2.24 together with
Lemmas 4.6 and 4.7 shows that for all f∈𝒢°0ϵ(β,γ) with 0<β,γ<ϵ,(4.59)limm→∞∥Rm(f)∥𝒢°0ϵ(β,γ)≤limm→∞C11m(C102−δN+CN2−jϵ)m∥f∥𝒢°0ϵ(β,γ)=0, and for all f∈Lp(𝒳) with p∈(1,∞), (4.60)limm→∞∥Rm(f)∥Lp(𝒳)≤limm→∞C11m(C102−δN+CN2−jϵ)m∥f∥Lp(𝒳)=0. To finish the proof of the
theorem, we still need to verify that for all f∈𝒢°0ϵ(β,γ) with 0<β,γ<ϵ, (4.61)limL→∞∥S−1{∑|k|>L∑τ∈Ik∑ν=1N(k,τ)∫Qτk,νDkN(⋅,y)dμ(y)Dk(f)(yτk,ν)}∥𝒢°0ϵ(β,γ)=0, and for all f∈Lp(𝒳) with p∈(1,∞), (4.62)limL→∞∥S−1{∑|k|>L∑τ∈Ik∑ν=1N(k,τ)∫Qτk,νDkN(⋅,y)dμ(y)Dk(f)(yτk,ν)}∥Lp(𝒳)=0. We first verify (4.61). To this
end, letting ϵ˜=ϵ1⋀ϵ2,
similarly to the proof of Theorem 3.10, by Corollary 4.9, it suffices to prove
that there exists some σ>0 such that for all 0<β<β′<ϵ˜,0<γ<γ′<ϵ˜, all L∈ℕ,
and all f∈𝒢°(β′,γ′), (4.63)∥∑|k|>L∑τ∈Ik∑ν=1N(k,τ)∫Qτk,νDkN(⋅,y)dμ(y)Dk(f)(yτk,ν)∥𝒢(β,γ)≤C2−σL∥f∥𝒢(β′,γ′), where C>0 is independent of L and f.
An argument similar to the proof of (3.3), via Lemma 2.1(iii) and geometric
mean, reduces the proof of this estimate to verifying the following two
estimates that there exists some σ>0 such that for  all f∈𝒢°(β′,γ′) and  all x∈𝒳, (4.64)|∑|k|>L∑τ∈Ik∑ν=1N(k,τ)∫Qτk,νDkN(x,y)dμ(y)Dk(f)(yτk,ν)|≲2−σL∥f∥𝒢(β′,γ′)1V1(x1)+V(x1,x)1(1+d(x,x1))γ, and for all x,x′∈𝒳 with d(x,x′)≤(1/2)(1+d(x,x1)), (4.65)|∑|k|>L∑τ∈Ik∑ν=1N(k,τ)∫Qτk,νDkN(x,y)dμ(y)Dk(f)(yτk,ν)−∑|k|>L∑τ∈Ik∑ν=1N(k,τ)∫Qτk,νDkN(x′,y)dμ(y)Dk(f)(yτk,ν)|≲∥f∥𝒢(β′,γ′)d(x,x′)β′(1+d(x,x1))β′1V1(x1)+V(x1,x)1(1+d(x,x1))γ′. For L∈ℕ,
let TL be the operator associated with the kernel (4.66)KL(x,y)=∑|k|>L∑τ∈Ik∑ν=1N(k,τ)∫Qτk,νDkN(x,z)dμ(z)Dk(yτk,ν,y) with x,y∈𝒳.
By an argument similar to the proof of Lemma 4.7 together with Lemmas 4.10 and 3.5, we know TL satisfies all the conditions of Corollary 2.22
with ϵ replaced by ϵ˜ and CTL+∥TL∥L2(𝒳)→L2(𝒳)≲1.
Thus, Corollary 2.22 then shows that TL is bounded on 𝒢°(β′,γ′) for any 0<β′,γ′<ϵ˜.
In particular, (4.65) holds. Using Lemma 4.10 and an argument similar to the
proof of (3.75) also gives (4.64). Thus, (4.61) holds.
We now prove (4.62). To this end, Corollary 4.9 shows
that it suffices to verify that for all f∈Lp(𝒳) with p∈(1,∞), limL→∞∥TL(f)∥Lp(𝒳)=0. By Theorem 3.10, for f∈Lp(𝒳) and h∈Lp′(𝒳),f=∑l=−∞∞D˜lDl(f) and h=∑l=−∞∞D˜lDl(h), respectively, in Lp(𝒳) and Lp′(𝒳),
where D˜k for k∈ℤ are as in Theorem 3.10. From Remark 3.3, it is
easy to deduce that there exists ϵ1′∈(0,ϵ1⋀ϵ2) such that for all y∈Qτk,ν and z∈𝒳, (4.67)|DkD˜l(yτk,ν,z)|≲2−|k−l|ϵ1′1V2−k(y)+V2−k(z)+V(y,z)2−(k⋀l)ϵ1′(2−(k⋀l)+d(y,z))ϵ1′,|(DkN)tD˜l(y,z)|≲2−|k−l|ϵ1′1V2−k(y)+V2−k(z)+V(y,z)2−(k⋀l)ϵ1′(2−(k⋀l)+d(y,z))ϵ1′. These estimates and Lemma 2.1(iv) together with Hölder's inequality, Lemma 3.14, and Proposition 3.15 yield
that (4.68)∥TL(f)∥Lp(𝒳)=sup∥h∥Lp′(𝒳)≤1|〈TL(f),h〉|=sup∥h∥Lp′(𝒳)≤1|∑|k|>L∑τ∈Ik∑ν=1N(k,τ)∫Qτk,νDk(f)(yτk,ν)(DkN)t(h)(y)dμ(y)|≲sup∥h∥Lp′(𝒳)≤1∥{∑|k|>L[∑l=−∞∞2−|k−l|ϵ1′M(|Dl(f)|)]2}1/2∥Lp(𝒳)×∥{∑|k|>L[∑l=−∞∞2−|k−l|ϵ1′M(|(DkN)t(h)|)]2}1/2∥Lp′(𝒳)≲2−ϵ1′L/2∥{∑|l|<L/2[M(|Dl(f)|)]2}1/2∥Lp(𝒳)+∥{∑|l|≥L/2[M(|Dl(f)|)]2}1/2∥Lp(𝒳)→0, as L→∞.
That is, (4.62) also holds, which completes the proof of Theorem 4.11.By an argument similar to the proof of Theorem4.11, we
can establish the following variants
of the discrete Calderón reproducing formulae (we omit the details).Theorem 4.12.
Letϵ1∈(0,1],ϵ2>0,ϵ3>0,ϵ∈(0,ϵ1⋀ϵ2), and let {Sk}k=−∞∞ be an (ϵ1,ϵ2,ϵ3)−ATI.
Set Dk=Sk−Sk−1 for any k∈ℤ.
Then for any fixed j∈ℕ as in Corollary 4.9, there exists a family of
linear operators {D¯k}k∈ℤ such that for any fixed yτk,ν∈Qτk,ν with k∈ℤ,τ∈Ik, and ν=1,…,N(k,τ),
and all f∈𝒢°0ϵ(β,γ) with 0<β,γ<ϵ, (4.69)f(x)=∑k=−∞∞∑τ∈Ik∑ν=1N(k,τ)∫Qτk,νDk(x,y)dμ(y)D¯k(f)(yτk,ν)=∑k=−∞∞∑τ∈Ik∑ν=1N(k,τ)Dk(x,yτk,ν)∫Qτk,νD¯k(f)(y)dμ(y)=∑k=−∞∞∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)Dk(x,yτk,ν)D¯k(f)(yτk,ν), where the series converges in
both the norm of 𝒢°0ϵ(β,γ) and the norm of Lp(𝒳) with p∈(1,∞).
Moreover, D¯k satisfies the conditions as in Theorem 3.11.Theorems4.11 and 4.12 in combination with a
duality argument show that discrete
Calderón reproducing formulae on spaces of distributions.Theorem 4.13.
Let all the notation be as in
Theorems4.11 and 4.12. Then for all f∈(𝒢°0ϵ(β,γ))′ with 0<β,γ<ϵ,
(4.57) and (4.69) hold in (𝒢°0ϵ(β,γ))′.


## 4.2. Inhomogeneous Discrete Calderón Reproducing Formulae

Similarly to Subsection3.2, we can establish the
following inhomogeneous discrete
Calderón reproducing formulae (we omit the details). Here, again, we have
no restriction on diam(𝒳).Theorem 4.14.
Letϵ1∈(0,1],ϵ2>0,ϵ3>0,ϵ∈(0,ϵ1⋀ϵ2), and let {Sk}k∈ℤ+ be an (ϵ1,ϵ2,ϵ3)−IATI.
Set D0=S0 and Dk=Sk−Sk−1 for k∈ℕ.
Then for any fixed j,N∈ℕ such that (4.1) and (4.39) hold, there exists a
family of functions {D˜k(x,y)}k∈ℤ+ such that for any fixed yτk,ν∈Qτk,ν with k∈ℕ, τ∈Ik and ν=1,…,N(k,τ) and all f∈𝒢0ϵ(β,γ) with 0<β,γ<ϵ, (4.70)f(x)=∑k=0N∑τ∈Ik∑ν=1N(k,τ)∫Qτk,νD˜k(x,y)dμ(y)Dτ,1k,ν(f)+∑k=N+1∞∑τ∈Ik∑ν=1N(k,τ)∫Qτk,νD˜k(x,y)dμ(y)Dk(f)(yτk,ν)=∑τ∈I0∑ν=1N(0,τ)∫Qτ0,νD˜0(x,y)dμ(y)Dτ,10,ν(f)+∑k=1∞∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)D˜k(x,yτk,ν)Dτ,1k,ν(f)=∑τ∈I0∑ν=1N(0,τ)∫Qτ0,νD˜0(x,y)dμ(y)Dτ,10,ν(f)+∑k=1N∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)D˜k(x,yτk,ν)Dτ,1k,ν(f)+∑k=N+1∞∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)D˜k(x,yτk,ν)Dk(f)(yτk,ν), where the series
converges in both the norm of 𝒢0ϵ(β,γ) and the norm of Lp(𝒳) with p∈(1,∞),
and Dτ,1k,ν for
k∈ℤ+,τ∈Ik, and ν=1,…,N(k,τ) is the corresponding integral operator with
the kernel Dτ,1k,ν(z)=(1/μ(Qτk,ν))∫Qτk,νDk(u,z)dμ(u).
Moreover, D˜k for k≥N+1 satisfies the conditions (i) and (ii) of
Definition 2.2 with ϵ1 and ϵ2 replaced by ϵ′∈(ϵ,ϵ1⋀ϵ2);
and there exists a constant C>0 depending on ϵ′ such that the function D˜k(x,y) for k=0,1,…,N satisfies that
(i)
for allx,y∈𝒳,|D˜k(x,y)|≤C((1/(V1(x)+V1(y)+V(x,y)))(1/(1+d(x,y))ϵ′)),(ii)
for allx,x′,y∈𝒳 with d(x,x′)≤(1+d(x,y))/2, (4.71)|D˜k(x,y)−D˜k(x′,y)|≤C(d(x,x′)1+d(x,y))ϵ′1V1(x)+V1(y)+V(x,y)1(1+d(x,y))ϵ′; and ∫𝒳D˜k(x,y)dμ(x)=∫𝒳D˜k(x,y)dμ(y)=1 when 0≤k≤N; =0 when k>N.Theorem 4.15.
Letϵ1∈(0,1],ϵ2>0,ϵ3>0,ϵ∈(0,ϵ1⋀ϵ2), and let {Sk}k∈ℤ+ be an (ϵ1,ϵ2,ϵ3)−IATI.
Set D0=S0 and Dk=Sk−Sk−1 for k∈ℕ.
Then for any fixed j,N∈ℕ such that (4.1) and (4.39) hold, there exists a
family of functions {D¯k(x,y)}k∈ℤ+ such that for any fixed yτk,ν∈Qτk,ν with k∈ℕ,τ∈Ik, and ν=1,…,N(k,τ) and all f∈𝒢0ϵ(β,γ) with 0<β,γ<ϵ, (4.72)f(x)=∑k=0N∑τ∈Ik∑ν=1N(k,τ)∫Qτk,νDk(x,y)dμ(y)Dτ,1k,ν¯(f)+∑k=N+1∞∑τ∈Ik∑ν=1N(k,τ)∫Qτk,νDk(x,y)dμ(y)D¯k(f)(yτk,ν)=∑τ∈I0∑ν=1N(0,τ)∫Qτo,νD0(x,y)dμ(y)Dτ,10,ν¯(f)+∑k=1∞∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)Dk(x,yτk,ν)Dτ,1k,ν¯(f)=∑τ∈I0∑ν=1N(0,τ)∫Qτo,νD0(x,y)dμ(y)Dτ,10,ν¯(f)+∑k=1N∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)Dk(x,yτk,ν)Dτ,1k,ν¯(f)+∑k=N+1∞∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)Dk(x,yτk,ν)D¯k(f)(yτk,ν),  where the series
converges in both the norm of 𝒢0ϵ(β,γ) and the norm of Lp(𝒳) with p∈(1,∞),
and Dτ,1k,ν¯ for k∈ℤ+,τ∈Ik, and ν=1,…,N(k,τ) is the corresponding integral operator with
the kernel Dτ,1k,ν¯(z) as in (4.70) with Dk replaced by D¯k.
Moreover, D¯k for k≥N+1 satisfies the conditions (i) and (iii) of
Definition 2.2 with ϵ1 and ϵ2 replaced by ϵ′∈(ϵ,ϵ1⋀ϵ2);
and there exists a constant C>0 depending on ϵ′ such that the function D¯k(x,y) for k=0,1,…,N satisfies  that
(i)
for allx,y∈𝒳, |D¯k(x,y)|≤C((1/(V1(x)+V1(y)+V(x,y)))(1/(1+d(x,y))ϵ′)),(ii)
for allx,x′,y∈𝒳 with d(x,x′)≤(1+d(x,y))/2, (4.73)|D¯k(x,y)−D¯k(x′,y)|≤C(d(x,x′)1+d(x,y))ϵ′1V1(x)+V1(y)+V(x,y)1(1+d(x,y))ϵ′; and ∫𝒳D¯k(x,y)dμ(x)=∫𝒳D¯k(x,y)dμ(y)=1 when 0≤k≤N; =0 when k>N.Theorem 4.16.
Use
the same notation as in Theorems4.14 and 4.15. Then for all f∈(𝒢0ϵ(β,γ))′ with 0<β,γ<ϵ,
(4.70) and (4.72) hold in (𝒢0ϵ(β,γ))′.Remark 4.17.
Similarly to Remark3.27, in the sequel, to simplify the representation
of the results, we will always assume that N=0 in Theorems 4.14, 4.15,  and 4.16.


## 5. Besov Spaces and Triebel-Lizorkin Spaces

In this section, we consider Besov and
Triebel-Lizorkin spaces on RD-spaces and study their relations. As
applications, we obtain boundedness results on these spaces for singular
integrals considered by Nagel and Stein [44]. Finally, we establish a variant of the T(1)-theorem of David and Journé in these
settings, and a variant of the T(1)-theorem of Stein in [75] is also  presented.To develop a theory of these function spaces, we need
two basic tools: the Calderón reproducing formulae from Sections3 and 4
and Plancherel-Pôlya inequalities, which will be established in this section.


### 5.1. Plancherel-Pôlya Inequalities and Definition ofB˙p,qs(𝒳) and F˙p,qs(𝒳)

Throughout this and the next subsection, we will always
assume thatμ(𝒳)=∞.
We first introduce the norms in B˙p,qs(𝒳) and F˙p,qs(𝒳) via certain ATIs. We then prove that they are independent of
the choices of ATIs and spaces of distributions. To this end, we
need to establish homogeneous Plancherel-Pôlya inequalities; see also [79].Adapting Triebel's approach to homogeneous Besov and
Triebel-Lizorkin spaces onℝn in [3], we make the following.Definition 5.1.
Letϵ1∈(0,1],ϵ2>0,ϵ3>0,ϵ∈(0,ϵ1⋀ϵ2), and let {Sk}k∈ℤ be an (ϵ1,ϵ2,ϵ3)-ATI.
For k∈ℤ,
set Dk=Sk−Sk−1.
(i)
For allf∈(𝒢°0ϵ(β,γ))′ with 0<β,γ<ϵ,|s|<ϵ,p(s,ϵ)<p≤∞, and 0<q≤∞,
one sets ∥f∥B˙p,qs(𝒳)≡{∑k∈ℤ2ksq∥Dk(f)∥Lp(𝒳)q}1/q with the usual modification made when p=∞ or q=∞.(ii)
For allf∈(𝒢°0ϵ(β,γ))′ with 0<β,γ<ϵ,|s|<ϵ,p(s,ϵ)<p<∞, and p(s,ϵ)<q≤∞,
one defines
∥f∥F˙p,qs(𝒳)≡∥{∑k∈ℤ2ksq|Dk(f)|q}1/q∥Lp(𝒳)  with the usual modification made when q=∞.To verify that the definitions of∥⋅∥B˙p,qs(𝒳) and ∥⋅∥F˙p,qs(𝒳) are independent of the choice of ATIs, we need two technical lemmas, which have independent interest.Lemma 5.2.
Letϵ>0,k′,k∈ℤ,
and yτk,ν be any point in Qτk,ν for τ∈Ik and ν=1,…,N(k,τ).
If n/(n+ϵ)<p≤∞,
then for any x∈𝒳,(5.1)∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)[1V2−(k′⋀k)(x)+V(x,yτk,ν)](p⋀1)[2−(k⋀k′)2−(k⋀k′)+d(x,yτk,ν)]ϵ(p⋀1)≤C[V2−(k′⋀k)(x)]1−(p⋀1), where C>0 is independent of x∈𝒳,k,k′,τ,  and ν.Proof.
Notice that for anyz∈Qτk,ν,
by Lemma 2.1(iii), we have(5.2)V2−(k′⋀k)(yτk,ν)+V(x,yτk,ν)~V2−(k′⋀k)(z)+V(x,z),2−(k⋀k′)+d(x,yτk,ν)~2−(k⋀k′)+d(x,z). These estimates together with
Lemma 2.19 and the second inequality in (1.3) yield that the left-hand side of
(5.1) is, up to a bounded multiplicative constant, controlled
by (5.3)∫𝒳[1V2−(k′⋀k)(x)+V(x,z)](p⋀1)[2−(k⋀k′)2−(k⋀k′)+d(x,z)]ϵ(p⋀1)dμ(z)≲[V2−(k′⋀k)(x)]1−(p⋀1)∑l=0∞2l[n−n(p⋀1)−ϵ(p⋀1)]≲[V2−(k′⋀k)(x)]1−(p⋀1), which completes the proof of
Lemma 5.2.Lemma 5.3.
Letϵ>0,k′,k∈ℤ,
and yτk,ν be any point in Qτk,ν for τ∈Ik and ν=1,…,N(k,τ).
If n/(n+ϵ)<r≤1,
then there exists a constant C>0 depending on r such that for all aτk,ν∈ℂ and all x∈𝒳, (5.4)∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)1V2−(k′⋀k)(x)+V(x,yτk,ν)2−(k⋀k′)ϵ(2−(k⋀k′)+d(x,yτk,ν))ϵ|aτk,ν|≤C2[(k⋀k′)−k]n(1−1/r){M(∑τ∈Ik∑ν=1N(k,τ)|aτk,ν|rχQτk,ν)(x)}1/r, where C>0 is also independent of k,k′,τ, and ν.Proof.
We first recall the following well-known inequality that for allr∈(0,1] and all aj∈ℂ with j in some countable set of indices,(5.5)(∑j|aj|)r≤∑j|aj|r. From this inequality, the fact
that for all τ∈Ik and ν=1,…,N(k,τ),
and all z∈𝒳, μ(Qτk,ν)χQτk,ν(z)~V2−k(z)χQτk,ν(z), and the fact that for all z∈𝒳 and all k,k′∈ℤ,
by (1.3), V2−(k⋀k′)(z)≲2[k−(k⋀k′)]nV2−k(z), together with (5.2), Lemma 2.1(vi),
and the second inequality of (1.3), it follows that the left-hand  side of (5.4)
is, up to a bounded multiplicative constant, controlled by(5.6){∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)r(1V2−(k′⋀k)(x)+V(x,yτk,ν))r2−(k⋀k′)rϵ(2−(k⋀k′)+d(x,yτk,ν))rϵ|aτk,ν|r}1/r≲2[(k⋀k′)−k]n(1−1/r){1V2−(k′⋀k)(x)∫d(x,z)<2−(k′⋀k)(∑τ∈Ik∑ν=1N(k,τ)|aτk,ν|rχQτk,ν(z))dμ(z)+∑l=0∞12l[rϵ+n(r−1)]1V2l+12−(k′⋀k)(x)×∫d(x,z)<2l+12−(k′⋀k)(∑τ∈Ik∑ν=1N(k,τ)|aτk,ν|rχQτk,ν(z))dμ(z)}1/r, which implies the desired
conclusion.Using these technical lemmas and the discrete Calderón
reproducing formulae, we can now establish thePlancherel-Pôlya inequality.Proposition 5.4.
Letϵ1∈(0,1],ϵ2>0,ϵ3>0,ϵ∈(0,ϵ1⋀ϵ2), and let {Sk}k∈ℤ and {Pk}k∈ℤ be two (ϵ1,ϵ2,ϵ3)-ATIs. For k∈ℤ,
set Dk=Sk−Sk−1 and Qk=Pk−Pk−1.
(i)
For allf∈(𝒢°0ϵ(β,γ))′ with 0<β,γ<ϵ, |s|<ϵ, p(s,ϵ)<p≤∞, and 0<q≤∞, (5.7){∑k∈ℤ2ksq(∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)[supz∈Qτk,ν|Dk(f)(z)|]p)q/p}1/q~{∑k∈ℤ2ksq(∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)[infz∈Qτk,ν|Qk(f)(z)|]p)q/p}1/q.(ii)
For allf∈(𝒢°0ϵ(β,γ))′ with 0<β,γ<ϵ,|s|<ϵ,p(s,ϵ)<p<∞, and p(s,ϵ)<q≤∞,(5.8)∥{∑k∈ℤ∑τ∈Ik∑ν=1N(k,τ)2ksq[supz∈Qτk,ν|Dk(f)(z)|]qχQτk,ν}1/q∥Lp(𝒳)~∥{∑k∈ℤ∑τ∈Ik∑ν=1N(k,τ)2ksq[infz∈Qτk,ν|Qk(f)(z)|]qχQτk,ν}1/q∥Lp(𝒳).Proof.
We first verify (5.7). By Theorem 4.13, there exist functions {D˜k}k∈ℤ such that for all f∈(𝒢°0ϵ(β,γ))′ with 0<β,γ<ϵ and all z∈𝒳, (5.9)f(z)=∑k′∈ℤ∑τ′∈Ik′∑ν′=1N(k′,τ′)μ(Qτ′k′,ν′)D˜k′(z,yτ′k′,ν′)Qk′(f)(yτ′k′,ν′)holds in (𝒢°0ϵ(β,γ))′,
where D˜k′ satisfies the same conditions as D˜k in Theorem 4.13 and yτ′k′,ν′ is any point in Qτ′k′,ν′.
For any ϵ′∈(0,ϵ),
by Lemma 3.2 (see also Remark 3.3), for all z∈𝒳,(5.10)|(DkD˜k′)(z,yτ′k′,ν′)|≲2−|k−k′|ϵ′1V2−(k⋀k′)(z)+V2−(k⋀k′)(yτ′k′,ν′)+V(z,yτ′k′,ν′)2−(k⋀k′)ϵ(2−(k⋀k′)+d(z,yτ′k′,ν′))ϵ. If p(s,ϵ)<p≤1,
by applying Dk to (5.9), and making use of (5.10), (5.5),
and (5.2) together with Lemma 5.2, we obtain(5.11)∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)[supz∈Qτk,ν|Dk(f)(z)|]p≲∑k′∈ℤ∑τ′∈Ik′∑ν′=1N(k′,τ′)2−|k−k′|ϵ′pμ(Qτ′k′,ν′)p[V2−(k⋀k′)(yτ′k′,ν′)]1−p|Qk′(f)(yτ′k′,ν′)|p. From this and the fact that (5.12)V2−(k⋀k′)(yτ′k′,ν′)≲2[k′−(k⋀k′)]nV2−k′(yτ′k′,ν′)~2[k′−(k⋀k′)]nμ(Qτ′k′,ν′) together with Hölder's
inequality when q/p≥1 and (5.5) when q/p<1,
it follows that if we choose ϵ′∈(0,ϵ) such that max{n/(n+ϵ′),n/(n+ϵ′+s)}<p≤1 and |s|<ϵ′,
the left-hand side of (5.7) is, up to a bounded multiplicative constant,
dominated by (5.13){∑k∈ℤ[∑k′∈ℤ∑τ′∈Ik′∑ν′=1N(k′,τ′)2−|k−k′|ϵ′p2(k−k′)spμ(Qτ′k′,ν′)p[V2−(k⋀k′)(yτ′k′,ν′)]1−p2k′sp|Qk′(f)(yτ′k′,ν′)|p]q/p}1/q≲{∑k′∈ℤ2k′sq(∑τ′∈Ik′∑ν′=1N(k′,τ′)μ(Qτ′k′,ν′)|Qk′(f)(yτ′k′,ν′)|p)q/p}1/q, which together with the
arbitrary choice of yτ′k′,ν′ yields (5.14){∑k∈ℤ2ksq(∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)[supz∈Qτk,ν|Dk(f)(z)|]p)q/p}1/q≲{∑k∈ℤ2ksq(∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)[infz∈Qτk,ν|Qk(f)(z)|]p)q/p}1/q. By symmetry, we then obtain
(5.7) when p(s,ϵ)<p≤1.
Ifp∈(1,∞] and if we choose ϵ′∈(0,ϵ) such that |s|<ϵ′,
by Hölder's inequality and Lemma 5.2, we obtain (5.15)∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)[supz∈Qτk,ν|Dk(f)(z)|]p≲∑k′∈ℤ∑τ′∈Ik′∑ν′=1N(k′,τ′)2−|k−k′|ϵ′2(k′−k)s(p−1)μ(Qτ′k′,ν′)|Qk′(f)(yτ′k′,ν′)|p, which together with Hölder's
inequality when q/p≥1 and (5.5) when q/p<1 yields that the left-hand side of (5.7) is, up
to a bounded multiplicative constant, controlled by (5.16){∑k∈ℤ[∑k′∈ℤ∑τ′∈Ik′∑ν′=1N(k′,τ′)2−|k−k′|ϵ′2(k−k′)sμ(Qτ′k′,ν′)2k′sp|Qk′(f)(yτ′k′,ν′)|p]q/p}1/q≲{∑k′∈ℤ2k′sq(∑τ′∈Ik′∑ν′=1N(k′,τ′)μ(Qτ′k′,ν′)|Qk′(f)(yτ′k′,ν′)|p)q/p}1/q. Then, since yτ′k′,ν′ was an arbitrary point in Qτ′k′,ν′,
we see that the estimate (5.14) also holds when p∈(1,∞],
which by symmetry then completes the proof of (5.7).
We now verify (5.8). By applying Dk to (5.9), and making use of (5.10) together
with (5.2), and Lemma 5.3, we obtain that for ϵ′>|s| and r>max{p,p(s,ϵ′),q}, (5.17){∑k∈ℤ∑τ∈Ik∑ν=1N(k,τ)2ksq[supz∈Qτk,ν|Dk(f)(z)|]qχQτk,ν(x)}1/q≲{∑k∈ℤ[∑k′∈ℤ2(k−k′)s−|k−k′|ϵ′2[(k⋀k′)−k′]n(1−1/r)×{M(∑τ′∈Ik′∑ν′=1N(k′,τ′)2k′sr|Qk′(f)(yτ′k′,ν′)|rχQτ′k′,ν′)(x)}1/r]q}1/q, which together with Hölder's
inequality when q∈(1,∞] or (5.5) when q≤1,
and Lemma 3.14 further implies that the left-hand side of (5.8) is, up to a bounded
multiplicative constant, dominated by (5.18)∥{∑k′∈ℤ[M(∑τ′∈Ik′∑ν′=1N(k′,τ′)2k′sr|Qk′(f)(yτ′k′,ν′)|rχQτ′k′,ν′)]q/r}r/q∥Lp/r(𝒳)1/r≲∥{∑k′∈ℤ∑τ′∈Ik′∑ν′=1N(k′,τ′)2k′sq|Qk′(f)(yτ′k′,ν′)|qχQτ′k′,ν′}1/q∥Lp(𝒳). Then, by symmetry and the fact that yτ′k′,ν′ was an arbitrary point in Qτ′k′,ν′,
we obtain (5.8), which completes the proof of Proposition 5.4.The following remark is useful in
applications.Remark 5.5.
Let all the notation be as in Proposition5.4, except that Sk (and therefore Dk) for k∈ℤ has regularity only in the second variable.
Then, there exists a constant C>0 such that
(i)
for allf∈(𝒢°0ϵ(β,γ))′ with 0<β,γ<ϵ,|s|<ϵ,p(s,ϵ)<p≤∞, and 0<q≤∞, (5.19){∑k∈ℤ2ksq(∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)[supz∈Qτk,ν|Dk(f)(z)|]p)q/p}1/q≤C{∑k∈ℤ2ksq(∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)[infz∈Qτk,ν|Qk(f)(z)|]p)q/p}1/q;(ii)
for allf∈(𝒢°0ϵ(β,γ))′ with 0<β,γ<ϵ,|s|<ϵ,p(s,ϵ)<p<∞, and p(s,ϵ)<q≤∞, (5.20)∥{∑k∈ℤ∑τ∈Ik∑ν=1N(k,τ)2ksq[supz∈Qτk,ν|Dk(f)(z)|]qχQτk,ν}1/q∥Lp(𝒳)≤C∥{∑k∈ℤ∑τ∈Ik∑ν=1N(k,τ)2ksq[infz∈Qτk,ν|Qk(f)(z)|]qχQτk,ν}1/q∥Lp(𝒳).Using Proposition5.4, we can easily verify that the
definitions of ∥⋅∥B˙p,qs(𝒳) and ∥⋅∥F˙p,qs(𝒳) are independent of the choices of ATIs. We omit the details.Proposition 5.6.
Let all the notation be as in Proposition5.4.
(i)
For allf∈(𝒢°0ϵ(β,γ))′ with 0<β,γ<ϵ,|s|<ϵ,p(s,ϵ)<p≤∞, and 0<q≤∞, (5.21){∑k∈ℤ2ksq∥Dk(f)∥Lp(𝒳)q}1/q~{∑k∈ℤ2ksq∥Qk(f)∥Lp(𝒳)q}1/q.(ii)
For allf∈(𝒢°0ϵ(β,γ))′ with 0<β,γ<ϵ,|s|<ϵ,p(s,ϵ)<p<∞, and p(s,ϵ)<q≤∞, (5.22)∥{∑k∈ℤ2ksq|Dk(f)|q}1/q∥Lp(𝒳)~∥{∑k∈ℤ2ksq|Qk(f)|q}1/q∥Lp(𝒳).We now verify that the definition of the norm∥⋅∥B˙p,qs(𝒳) and the norm ∥⋅∥F˙p,qs(𝒳) is independent of the choice of the underlying
space of distributions. We recall that a+=max{a,0} for any a∈ℝ.Proposition 5.7.
Let all the notation be as in Definition5.1.
(i)
Let|s|<ϵ,p(s,ϵ)<p≤∞, and 0<q≤∞.
If f∈(𝒢°0ϵ(β1,γ1))′ with(5.23)max{0,−s+n(1p−1)+}<β1<ϵ,max{n(1p−1)+,s−κp}<γ1<ϵ, and if ∥f∥B˙p,qs(𝒳)<∞,
then f∈(𝒢°0ϵ(β2,γ2))′ for every β2,γ2 satisfying (5.23).(ii)
Let|s|<ϵ,p(s,ϵ)<p<∞, and p(s,ϵ)<q≤∞.
If f∈(𝒢°0ϵ(β1,γ1))′ with β1,γ1 as in (5.23), and if ∥f∥F˙p,qs(𝒳)<∞,
then f∈(𝒢°0ϵ(β2,γ2))′ for every β2,γ2 satisfying (5.23).Proof.
Letψ∈𝒢°(ϵ,ϵ).
Adopting the notation from Theorem 4.11, we first claim that for k∈ℤ+, (5.24)|〈D˜k(⋅,y),ψ〉|≲2−kβ2∥ψ∥𝒢(β2,γ2)1V1(x1)+V(x1,y)1(1+d(x1,y))γ2, and that for k=−1,−2,…,(5.25)|〈D˜k(⋅,y),ψ〉|≲2kγ2′∥ψ∥𝒢(β2,γ2)1V2−k(x1)+V(x1,y)2−kγ2(2−k+d(x1,y))γ2, where γ2′ can be any positive number in (0,γ2).
In fact, to verify (5.24), by the vanishing moment of D˜k,
Lemma 2.1(ii), we have  (5.26)|〈D˜k(⋅,y),ψ〉|=|∫𝒳D˜k(z,y)[ψ(z)−ψ(y)]dμ(z)|≲∥ψ∥𝒢(β2,γ2){∫d(z,y)≤(1+d(y,x1))/2|D˜k(z,y)|1V1(x1)+V(x1,y)d(z,y)β2(1+d(x1,y))β2+γ2dμ(z)+∫d(z,y)>(1+d(y,x1))/2|D˜k(z,y)|[1V1(x1)+V(x1,z)1(1+d(x1,z))γ2+1V1(x1)+V(x1,y)1(1+d(x1,y))γ2]dμ(z)}≲2−kβ2∥ψ∥𝒢(β2,γ2)1V1(x1)+V(x1,y)1(1+d(x1,y))γ2×{1+∫𝒳1V1(x1)+V(x1,z)1(1+d(x1,z))γ2dμ(z)+∫𝒳1V2−k(y)+V(z,y)2−k(ϵ−β2)(2−k+d(z,y))ϵ−β2dμ(z)}, where the last quantity is, up
to a bounded multiplicative constant, controlled by the right-hand side of
(5.24), and in the last inequality, we used the  fact that for d(z,y)>(1+d(y,x1))/2,V(z,y)≳V1(y)+V(x1,y)~V1(x1)+V(x1,y);
see  also Lemma 2.1(vii). Thus, (5.24)  holds.
To see  (5.25), by ∫𝒳ψ(z)dμ(z)=0,
(i) and (ii) of Lemma 2.1, and the fact that for d(z,x1)>(2−k+d(x1,y))/2, V(x1,z)≳V2−k(x1)+V(x1,y), we obtain that for k=−1,−2,…, (5.27)|〈D˜k(⋅,y),ψ〉|=|∫𝒳[D˜k(z,y)−D˜k(x1,y)]ψ(z)dμ(z)|≲∥ψ∥𝒢(β2,γ2){∫d(z,x1)≤(2−k+d(y,x1))/2d(z,x1)ϵ(2−k+d(x1,y))ϵ2−kϵ(2−k+d(x1,y))ϵ×1V2−k(x1)+V2−k(y)+V(x1,y)×1V1(x1)+V(x1,z)1(1+d(x1,z))γ2dμ(z)+∫d(z,x1)>(2−k+d(y,x1))/2[1V2−k(z)+V2−k(y)+V(z,y)2−kϵ(2−k+d(z,y))ϵ+1V2−k(x1)+V2−k(y)+V(x1,y)2−kϵ(2−k+d(x1,y))ϵ]×1V1(x1)+V(x1,z)1(1+d(x1,z))γ2dμ(z)}≲1V2−k(x1)+V(x1,y)×{1(2−k+d(x1,y))γ2+2−kϵ(2−k+d(x1,y))ϵ1(2−k+d(x1,y))γ2′}, where the last quantity is, up
to a bounded multiplicative constant, dominated by the right-hand side of
(5.25); namely, (5.25)  holds.
Thus, Theorem4.13 together with (5.24), (5.25),  and
Lemma 2.1(iii) yield that
(5.28)|〈f,ψ〉|=|∑k=−∞∞∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)Dk(f)(yτk,ν)〈D˜k(⋅,yτk,ν),ψ〉|≲∥ψ∥𝒢(β2,γ2){∑k=0∞∑τ∈Ik∑ν=1N(k,τ)2−kβ2μ(Qτk,ν)|Dk(f)(yτk,ν)|×1V1(x1)+V(x1,yτk,ν)1(1+d(x1,yτk,ν))γ2+∑k=−∞−1∑τ∈Ik∑ν=1N(k,τ)2kγ2′μ(Qτk,ν)|Dk(f)(yτk,ν)|×1V2−k(x1)+V(x1,yτk,ν)2−kγ2(2−k+d(x1,yτk,ν))γ2}.
If p≤1,
by (5.5), (5.29)|〈f,ψ〉|≲∥ψ∥𝒢(β2,γ2){∑k=0∞2−k(β2+s)[∑τ∈Ik∑ν=1N(k,τ)2kspμ(Qτk,ν)p|Dk(f)(yτk,ν)|p×1(V1(x1)+V(x1,yτk,ν))p1(1+d(x1,yτk,ν))γ2p]1/p+∑k=−∞−12k(γ2′−s)[∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)p2ksp|Dk(f)(yτk,ν)|p×1(V2−k(x1)+V(x1,yτk,ν))p2−kγ2p(2−k+d(x1,yτk,ν))γ2p]1/p}. Notice that when p≤1,
by Lemma 2.1(vii) and γ2>n(1/p−1),
for k∈ℤ+, τ∈Ik and ν=1,…,N(k,τ)
, (5.30)μ(Qτk,ν)p−1(V1(x1)+V(x1,yτk,ν))p1(1+d(x1,yτk,ν))γ2p≲1V1(x1)[V1(yτk,ν)+V(x1,yτk,ν)μ(Qτk,ν)]1−p1(1+d(x1,yτk,ν))γ2p≲2kn(1−p)1V1(x1); and similarly, by the first
inequality of (1.3),  for k=−1,−2,…,τ∈Ik and ν=1,…,N(k,τ), (5.31)μ(Qτk,ν)p−1(V2−k(x1)+V(x1,yτk,ν))p2−kγ2p(2−k+d(x1,yτk,ν))γ2p≲2kκ1V1(x1)[V2−k(yτk,ν)+V(x1,yτk,ν)μ(Qτk,ν)]1−p2−kγ2p(2−k+d(x1,yτk,ν))γ2p≲2kκ1V1(x1). Thus, if p≤1,
then by Hölder's inequality when q≥1 or (5.5) when q<1 together with Proposition 5.4, (5.32)|〈f,ψ〉|≲1V1(x1)1/p∥ψ∥𝒢(β2,γ2){∑k=0∞2k[n(1/p−1)−(β2+s)]+∑k=−∞−12k[κ/p+γ2′−s]}×[∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)2ksp|Dk(f)(yτk,ν)|p]1/p≲1V1(x1)1/p∥ψ∥𝒢(β2,γ2)∥f∥B˙p,qs(𝒳), where we used the assumption
that β2>−s+n(1/p−1) and we chose γ2′∈((s−κ/p)+,γ2).
Similarly, ifp>1,
by Lemma 2.1(ii), we have  (5.33)|〈f,ψ〉|≲1V1(x1)1/p∥ψ∥𝒢(β2,γ2)×{∑k=0∞2−k(β2+s)[∫𝒳1V1(x1)+V(x1,y)1(1+d(x1,y))γ2p′dμ(y)]1/p′+∑k=0∞2k(κ/p−γ2′−s)[∫𝒳1V2−k(x1)+V(x1,y)2−kγ2p2′(2−k+d(x1,y))γ2p′dμ(y)]1/p′}×[∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)2ksp|Dk(f)(yτk,ν)|p]1/p≲1V1(x1)1/p∥ψ∥𝒢(β2,γ2)∥f∥B˙p,qs(𝒳), where we used β2>−s and chose γ2′>max{0,s−κ/p}.
Let nowh∈𝒢°0ϵ(β2,γ2).
Then there exists {hn}n=1∞⊂𝒢°(ϵ,ϵ) such that as n→∞,∥h−hn∥𝒢(β2,γ2)→0.
By (5.32) and (5.33), we obtain(5.34)|〈f,hn−hm〉|≲∥f∥B˙p,qs(𝒳)∥hn−hm∥𝒢(β2,γ2), which shows that limn→∞〈f,hn〉 exists and the limit is independent of the
choice of {hn}.
Thus, if we define 〈f,h〉=limn→∞〈f,hn〉, by (5.32) and (5.33), we have |〈f,h〉|≲∥f∥B˙p,qs(𝒳)∥h∥𝒢°0ϵ(β2,γ2). That is, f∈(𝒢°0ϵ(β2,γ2))′,
which completes the  proof of (i).
The conclusion (ii) can be deduced from (i) and the
fact that∥f∥B˙p,max(p,q)s(𝒳)≲∥f∥F˙p,qs(𝒳); see [3, Proposition 2.3.2/2] or Proposition 5.10(ii)
below.Now we can introduce the Besov spaces,B˙p,qs(𝒳),
and the Triebel-Lizorkin spaces, F˙p,qs(𝒳).Definition 5.8.
Letϵ1∈(0,1],ϵ2>0,ϵ3>0,ϵ∈(0,ϵ1⋀ϵ2), and let {Sk}k∈ℤ be an (ϵ1,ϵ2,ϵ3)-ATI.
For k∈ℤ,
set Dk=Sk−Sk−1.
(i)
Let|s|<ϵ,p(s,ϵ)<p≤∞, and 0<q≤∞.
The spaceB˙p,qs(𝒳) is defined to be the set of all f∈(𝒢°0ϵ(β,γ))′,
for some β,γ satisfying(5.35)max{s,0,−s+n(1p−1)+}<β<ϵ,max{s−κp,n(1p−1)+,−s+n(1p−1)+−κ(1−1p)+}<γ<ϵsuch that ∥f∥B˙p,qs(𝒳)={∑k=−∞∞2ksq∥Dk(f)∥Lp(𝒳)q}1/q<∞ with the usual modifications made when p=∞ or q=∞.(ii)
Let|s|<ϵ,p(s,ϵ)<p<∞, and p(s,ϵ)<q≤∞. The spaceF˙p,qs(𝒳) is defined to be the set of all f∈(𝒢°0ϵ(β,γ))′ for some β,γ satisfying (5.35) such that ∥f∥F˙p,qs(𝒳)=∥{∑k=−∞∞2ksq|Dk(f)|q}1/q∥Lp(𝒳)<∞ with the usual modification made when q=∞.Propositions5.6 and 5.7 show that the
definitions of the spaces B˙p,qs(𝒳) and F˙p,qs(𝒳) are independent of the choice of the
approximations of the identity and the distribution space, (𝒢°0ϵ(β,γ))′,
with β,γ as in (5.35).Remark 5.9.
To guarantee that the definitions of the spacesB˙p,qs(𝒳) and F˙p,qs(𝒳) are independent of the choice of the
distribution space (𝒢°0ϵ(β,γ))′,
we only need the restriction that β,γ satisfy (5.23); see Proposition 5.7. Moreover,
if we assume that max{s,0}<β<ϵ and max{n(1/p−1)+,−s+n(1/p−1)+−κ(1−1/p)+}<γ<ϵ,
we can verify that the space of test functions 𝒢°(β,γ) is contained in B˙p,qs(𝒳) and F˙p,qs(𝒳);
see Proposition 5.10 below. Thus, the spaces B˙p,qs(𝒳) and F˙p,qs(𝒳) are nonempty if we restrict β,γ as in (5.35).


### 5.2. Properties ofB˙p,qs(𝒳) and F˙p,qs(𝒳) and Boundedness of Singular Integrals

In this subsection, we first present some basic properties
ofB˙p,qs(𝒳) and F˙p,qs(𝒳).
Then we establish a Lusin-area characterization of F˙p,qs(𝒳),
and as an application, we discuss the relation between the spaces F˙p,qs(𝒳) and the atomic Hardy spaces Hatp(𝒳) of Coifman and Weiss in [28] and the Hardy spaces Hp(𝒳) in [48]. Finally, we show that the singular integrals
considered by Nagel and Stein in [44] act boundedly on B˙p,qs(𝒳) and F˙p,qs(𝒳).Proposition 5.10.
Letϵ1∈(0,1],ϵ2>0,ϵ∈(0,ϵ1⋀ϵ2), and |s|<ϵ.
(i)
Forp(s,ϵ)<p≤∞ and 0<q0≤q1≤∞,B˙p,q0s(𝒳)⊂B˙p,q1s(𝒳); and for p(s,ϵ)<p<∞ and p(s,ϵ)<q0≤q1≤∞,F˙p,q0s(𝒳)⊂F˙p,q1s(𝒳).(ii)
Ifp(s,ϵ)<p<∞ and p(s,ϵ)<q≤∞,
then B˙p,min(p,q)s(𝒳)⊂F˙p,qs(𝒳)⊂B˙p,max(p,q)s(𝒳).(iii)
Ifβ,γ as in (5.23), then B˙p,qs(𝒳)⊂(𝒢°0ϵ(β,γ))′ when p(s,ϵ)<p≤∞ and 0<q≤∞,
and F˙p,qs(𝒳)⊂(𝒢°0ϵ(β,γ))′ when p(s,ϵ)<p<∞ and p(s,ϵ)<q≤∞.(iv)
Ifmax{s,0}<β<ϵ and max{n(1/p−1)+,−s+n(1/p−1)+−κ(1−1/p)+}<γ<ϵ,
then 𝒢°(β,γ)⊂B˙p,qs(𝒳) when p(s,ϵ)<p≤∞ and 0<q≤∞,
and 𝒢°(β,γ)⊂F˙p,qs(𝒳) when p(s,ϵ)<p<∞ and p(s,ϵ)<q≤∞.(v)
If1<p<∞,
then F˙p,20(𝒳)=Lp(𝒳) with equivalent norms.(vi)
The spacesB˙p,qs(𝒳)/𝒩 with p(s,ϵ)<p≤∞ and 0<q≤∞ and the spaces F˙p,qs(𝒳)/𝒩 with p(s,ϵ)<p<∞ and p(s,ϵ)<q≤∞ are complete.Proof.
Property (i) is a simple corollary of (5.5). Property (ii) can be
deduced from Minkowski's inequality, (5.5), and the following generalized
Minkowski inequality that for all 1≤q≤∞, (5.36){∑k=−∞∞[∫𝒳|ak(x)|dμ(x)]q}1/q≲∫𝒳{∑k=−∞∞|ak(x)|q}1/qdμ(x); see also [3, Proposition 2.3.2/2] and [6, Proposition 2.3].
Property (iii) is implied by the proof of Proposition5.7 and Property (vi) can be easily deduced from Property (iii).
To verify (iv), similarly to the proofs of (5.24) and
(5.25), for f∈𝒢°(β,γ),
we have that for k∈ℤ+, (5.37)|Dk(f)(x)|≲2−kβ∥f∥𝒢(β,γ)1V1(x1)+V(x1,x)1(1+d(x1,x))γ, and that for k=−1,−2,…,(5.38)|Dk(f)(x)|≲2kγ'∥f∥𝒢(β,γ)1V2−k(x1)+V(x1,x)2−kγ(2−k+d(x1,x))γ, where γ′ can be any positive number in (0,γ).
Moreover, since γ>n(1/p−1)+, (5.39){∫𝒳1(V1(x1)+V(x1,x))p1(1+d(x1,x))γpdμ(x)}1/p≲{1V1(x1)p−1+∑l=0∞1V2lr(x1)p−112lγp}1/p≲1V1(x1)1−1/p, and for k=−1,−2,…,(5.40){∫𝒳1(V2−k(x1)+V(x1,x))p2−kγp(2−k+d(x1,x))γpdμ(x)}1/p≲{1V2−k(x1)p−1+∑l=0∞1V2l2−k(x1)p−112lγp}1/p≲2kκ(1−1/p)+−kn(1/p−1)+V1(x1)1/p−1. Choose γ′∈(0,γ) such that γ′>n(1/p−1)+−s−κ(1−1/p)+.
From the above estimates, it follows that (5.41)∥f∥B˙p,qs(𝒳)≲∥f∥𝒢(β,γ)1V1(x1)1−1/p{∑k=0∞2k(s−β)q+∑k=−∞−12k[s+γ'−n(1/p−1)++κ(1−1/p)+]q}1/q≲∥f∥𝒢(β,γ). Thus,
𝒢°(β,γ)⊂B˙p,qs(𝒳),
which together with (ii) implies that 𝒢°(β,γ)⊂F˙p,qs(𝒳).
Property (v) is a simple conclusion of Proposition3.15, which completes the proof of Proposition 5.10.We next give a Lusin-area characterization for
Triebel-Lizorkin spacesF˙p,qs(𝒳),
which will allow to establish some relations between the Triebel-Lizorkin
spaces F˙p,qs(𝒳),
the atomic Hardy spaces of Coifman and Weiss in [28], and the Hardy spaces Hp(𝒳) in [48].Definition 5.11.
Lets∈ℝ,a>0,q∈(0,∞], and let ϵ1∈(0,1],ϵ2>0,ϵ3>0,ϵ∈(0,ϵ1⋀ϵ2]. Let {Sk}k∈ℤ be an (ϵ1,ϵ2,ϵ3)-ATI.
For k∈ℤ,
set Dk=Sk−Sk−1.
The Lusin-area function (also
called the Littlewood-PaleyS-function) S˙q,as(f) for any f∈(𝒢°0ϵ(β,γ))′ with 0<β,γ≤ϵ and x∈𝒳 is given by (5.42)S˙q,as(f)(x)={∑k∈ℤ∫d(x,y)<a2−k2ksq|Dk(f)(y)|qdμ(y)Va2−k(x)}1/q, where the usual modification is
made when q=∞.Remark 5.12.
(i) By Lemma2.1(vi), if we replace Va2−k(x),
respectively, by Va2−k(y),Va2−k(y)+Va2−k(x),V2−k(x),V2−k(y),
or V2−k(x)+V2−k(y) in the definition of S˙q,as(f),
then the corresponding Littlewood-Paley S-functions S˙q,as(f) are pointwise equivalent. This is often useful
in applications.
(ii) It is easy to see thatS˙2,a0(f) is a discrete version of the corresponding
Littlewood-Paley S-function in [48].Theorem 5.13.
Letϵ1∈(0,1],ϵ2>0,ϵ3>0,ϵ∈(0,ϵ1⋀ϵ2),
and let {Sk}k∈ℤ be an (ϵ1,ϵ2,ϵ3)-ATI.
For k∈ℤ,
set Dk=Sk−Sk−1.
Let a>0,|s|<ϵ,p(s,ϵ)<p<∞,p(s,ϵ)<q≤∞, and let S˙q,as(f) be as in Definition 5.11 for any f∈(𝒢°0ϵ(β,γ))′ with β,γ as in (5.35). Then f∈F˙p,qs(𝒳) if and only if f∈(𝒢°0ϵ(β,γ))′ for some β,γ as in (5.35), and S˙q,as(f)∈Lp(𝒳).
Moreover, in this case, ∥f∥F˙p,qs(𝒳)~∥S˙q,as(f)∥Lp(𝒳).Proof.
We use the notation as in the proof of
Proposition5.4. It is easy to check that for any fixed constant C>0, (5.43)supz∈B(zτk,ν,C2−k)|(DkD˜k′)(z,yτ′k′,ν′)|≲2−|k−k′|ϵ′1V2−(k⋀k′)(zτk,ν)+V2−(k⋀k′)(yτ′k′,ν′)+V(zτk,ν,yτ′k′,ν′)2−(k⋀k′)ϵ(2−(k⋀k′)+d(zτk,ν,yτ′k′,ν′))ϵ. Using this to replace (5.10) and
then repeating the proof of (5.8)
in Proposition 5.4
yield (5.44)∥{∑k∈ℤ∑τ∈Ik∑ν=1N(k,τ)2ksq[supz∈B(zτk,ν,C2−k)|Dk(f)(z)|]qχQτk,ν}1/q∥Lp(𝒳)~∥{∑k∈ℤ∑τ∈Ik∑ν=1N(k,τ)2ksq[infz∈Qτk,ν|Dk(f)(z)|]qχQτk,ν}1/q∥Lp(𝒳)~∥f∥F˙p,qs(𝒳),which shows that(5.45)∥S˙q,as(f)∥Lp(𝒳)≤∥{∑k∈ℤ∑τ∈Ik∑ν=1N(k,τ)2ksq[supz∈B(zτk,ν,C2−k)|Dk(f)(z)|]qχQτk,ν}1/q∥Lp(𝒳)~∥f∥F˙p,qs(𝒳).
On the other hand, noticing that ifx∈B(zτk,ν,min{a,1}C72−(k+j)),
then (5.46)B(zτk,ν,min{a,1}C72−(k+j))⊂{y∈𝒳:d(x,y)<a2−k}, therefore, by Lemma
2.19, (5.47)S˙q,as(f)(x)={∑k∈ℤ∑τ∈Ik∑ν=1N(k,τ)∫d(x,y)<a2−k2ksq|Dk(f)(y)|qχQτk,ν(x)dμ(y)Va2−k(x)}1/q≥{∑k∈ℤ∑τ∈Ik∑ν=1N(k,τ)2ksq[infy∈B(zτk,ν,min{a,1}C72−(k+j))|Dk(f)(y)|]qχB(zτk,ν,min{a,1}C72−(k+j))(x)}1/q, which together with the
following estimate (5.48)∥{∑k∈ℤ2ksq|Dk(f)|q}1/q∥Lp(𝒳)≲∥{∑k∈ℤ∑τ∈Ik∑ν=1N(k,τ)2ksq[infy∈B(zτk,ν,min{a,1}C72−(k+j))|Dk(f)(y)|]qχB(zτk,ν,min{a,1}C72−(k+j))}1/q∥Lp(𝒳) yields that
∥S˙q,as(f)∥Lp(𝒳)≳∥f∥F˙p,qs(𝒳). Using the notation as in Lemma 5.3, we can
verify the estimate (5.48) by repeating the proof of (5.8) in Proposition 5.4
and replacing  Lemma 5.3 by the following estimate that for any fixed constant C>0, (5.49)∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)1V2−(k′⋀k)(x)+V2−(k′⋀k)(yτk,ν)+V(x,yτk,ν)2−(k⋀k′)ϵ(2−(k⋀k′)+d(x,yτk,ν))ϵ|aτk,ν|≲2[(k⋀k′)−k]n(1−1/r){M(∑τ∈Ik∑ν=1N(k,τ)|aτk,ν|rχB(zτk,ν,C2−k))(x)}1/r, which can be proved by the same
way as the proof of Lemma 5.3 via the following facts μ(B(zτk,ν,C2−k))~μ(Qτk,ν) and (5.50)μ(B(zτk,ν,C2−k))χB(zτk,ν,C2−k)(z)~V2−k(z)χB(zτk,ν,C2−k)(z).This finishes the proof of
Theorem 5.13.Definition 5.14.
Let all the notation be as in
Definition5.8. The Hardy spaceHp(𝒳) when n/(n+ϵ)<p≤1 is defined to be the Triebel-Lizorkin spaces F˙p,20(𝒳),
with norm ∥f∥Hp(𝒳)≡∥f∥F˙p,20(𝒳).We now recall the definition of atoms on spaces of
homogeneous type in [28].Definition 5.15.
A functiona on 𝒳 is called an Hp(𝒳)-atom if
(i)
suppa⊂B(x0,r) for some x0∈𝒳 and some r>0;(ii)
∥a∥L2(𝒳)≤[μ(B(x0,r))]1/2−1/p;(iii)
∫𝒳a(x)dμ(x)=0.If (i) to (iii) apply, we also say thata is an Hp(𝒳)-atom supported on B(x0,r).Theorem 5.16.
Assume thatϵ and β,γ are as in Definition 5.8. If n/(n+ϵ)<p≤1,
then f∈Hp(𝒳) if and only if there exist a sequence of
numbers {λk}k=0∞⊂ℂ with ∑k=0∞|λk|p<∞ and a sequence of Hp(𝒳)-atoms {ak}k=0∞ such that f=∑k=0∞λkak in (𝒢°0ϵ(β,γ))′.
Moreover, in this case, ∥f∥Hp(𝒳)~inf{(∑k=0∞|λk|p)1/p}, where the infimum is taken over all the above
decompositions  of f.This theorem can be proved by a literal repetition of Theorem 2.21 in [48],
with Lemma 2.22 therein replaced by Theorem 2.6 in this paper. We omit the details.Remark 5.17.
Theorem5.16 shows that the Hardy spaces
defined here are the same as those in [48]. Moreover, by [48, Remark 2.27], we know that the Hardy space H1(𝒳) also coincides with the atomic Hardy space Hat1(𝒳) of Coifman and Weiss in [28]. Moreover, when n/(n+ϵ)<p<1 and 𝒳 is an Ahlfors n-regular metric measure space, Hp(𝒳) also coincides with the atomic Hardy space Hatp(𝒳) of Coifman and Weiss in [28]. However, if 𝒳 is a general space of homogeneous type, it is
still unclear so far to us if Hp(𝒳)=Hatp(𝒳) when p<1;
see [48, Remark 2.30].We now recall the definition of the Lipschitz spaceL˙ips(𝒳) with s>0;
see [28].Definition 5.18.
Lets>0.
The Lipschitz (or Hölder) spaceL˙ips(𝒳) is defined to be the set of all functions f on X such that (5.51)∥f∥L˙ips(𝒳)=supx≠y|f(x)−f(y)|V(x,y)s<∞.Observe that these classes are rather Lipschitz (or
Hölder) classes with respect to the measure distanceρ(x,y)≡inf{μ(B):x,y∈B,Baball},
not the distance d.Following Coifman and Weiss [28], we immediately obtain from
Theorem 5.16 in this paper and [48, Remark 2.27] the duality between H1(𝒳) and BMO(𝒳) and between Hatp(𝒳) and L˙ip1/p−1(𝒳) when p<1,
which is [48, Corollary 2.29].Theorem 5.19.
(i) The spaceBMO(𝒳)/𝒩 is the dual space of H1(𝒳), in the following sense: if f=∑k=0∞λkak∈H1(𝒳) is as in Theorem 5.16, then for each g∈BMO(𝒳), limN→∞∑k=0Nλk∫𝒳ak(x)g(x)dμ(x) is a well-defined continuous linear functional ℒg:f↦〈f,g〉 with norm ≲∥g∥BMO(𝒳).
Conversely, each continuous linear functionalℒ on H1(𝒳)
has the form
ℒ=ℒg for some g∈BMO(𝒳) with ∥g∥BMO(𝒳)≲∥ℒ∥.
Moreover,ℒg=0 if and only if g∈𝒩.
(ii) Assume thatϵ is as in Definition 5.8, n/(n+ϵ)<p<1, and s=1/p−1.
Then L˙ips(𝒳) (more precisely, L˙ips(𝒳)/𝒩) is the dual space of Hp(𝒳) in the sense of (i).Remark 5.20.
It is easy to see that if𝒳=ℝn and μ is the n-dimensional Lebesgue measure, then C˙ns(𝒳)=L˙ips(𝒳) with equivalent norms. Thus,(5.52)B˙∞,∞ns(𝒳)=F˙∞,∞ns(𝒳)=L˙ips(𝒳) with  equivalent norms.
Therefore, the dual space of Hp(𝒳) with n/(n+ϵ)<p<1 is the space B˙∞,∞n(1/p−1)(𝒳);
see also [3, Theorem
2.11.3(ii)]. However, for a general space 𝒳 of homogeneous type, it seems that C˙ns(𝒳)≠L˙ips(𝒳),
unless μ(B(x,r))~rn for all x∈𝒳 and r>0,
namely, 𝒳 is an Ahlfors n-regular metric measure space. Thus, one
cannot expect that [3, Theorem 2.11.3] still holds when 𝒳 is a general space of homogeneous type
considered in this paper, which demonstrates an essential difference between
function spaces on general spaces of homogeneous type considered in this paper
and those on Ahlfors regular metric measure space.Using Proposition5.10 and Theorem 2.6, we obtain the
following density property of B˙p,qs(𝒳) and F˙p,qs(𝒳).Proposition 5.21.
Letϵ1, ϵ2, ϵ, and let |s|<ϵ be as in Definition 5.8. Then 𝒢°b(ϵ1,ϵ2) is dense in B˙p,qs(𝒳) when p(s,ϵ)<p<∞ and 0<q<∞,
and in F˙p,qs(𝒳) when p(s,ϵ)<p,q<∞.Proof.
By similarity, we only verify the conclusion onB˙p,qs(𝒳).
Let f∈B˙p,qs(𝒳) with s,p,q as in the proposition. By Proposition 5.10(iii), we know f∈(𝒢°0ϵ(β,γ))′ with β,γ as in (5.23). Let {Dk}k∈ℤ be an ATI with bounded support as constructed in Theorem
2.6. By Theorem 4.13 and using the notation as in Theorem 4.12, we have (5.53)f(x)=∑k′=−∞∞∑τ′∈Ik′∑ν′=1N(k′,τ′)μ(Qτ′k′,ν′)Dk′(x,yτ′k′,ν′)D¯k′(f)(yτ′k′,ν′) in (𝒢°0ϵ(β,γ))′.
It is easy to check that for any fixed k′∈ℤ and τ′∈Ik′,N(k′,τ′) is a finite set of indices. For any fixed k′∈ℤ,
choose {Ik′N}N∈ℕ such that Ik′N⊂Ik′N+1,Ik′N is a finite set of indices, and Ik′N→Ik′ as N→∞.
If for any N∈ℕ,
setting (5.54)fN(x)=∑|k′|≤N∑τ′∈Ik′N∑ν′=1N(k′,τ′)μ(Qτ′k′,ν′)Dk′(x,yτ′k′,ν′)D¯k′(f)(yτ′k′,ν′), then it is easy to check fN∈𝒢°b(ϵ1,ϵ2).
From the proof of Proposition 5.4 together with Remark 5.5, it is easy to see
that ∥f−fN∥B˙p,qs(𝒳)→0, as N→∞ (here we need p,q<∞). Thus, 𝒢°b(ϵ1,ϵ2) is dense in B˙p,qs(𝒳) with s,p,q as in the proposition, which completes the
proof of Proposition 5.21.We now discuss the boundedness onB˙p,qs(𝒳) and F˙p,qs(𝒳) of the singular integral operators introduced
by Nagel and Stein in [44]. To state their definition, we need first to recall
the notion of a normalized bump function on 𝒳.
A function φ on 𝒳 is said to be an ϵ-bump function associated to a ball B(x0,δ) for some x0∈𝒳 and δ>0,
if it is supported in that ball, and if there exists a constant C>0 independent of φ such that ∥φ∥C˙η(𝒳)≤Cδ−η for all η∈(0,ϵ].
Notice that then ∥φ∥L∞(𝒳)≤C (see Remark 2.14(ii)). If C=1 in this definition, then such a bump function
is called a normalizedϵ-bump function for the ball B(x0,δ).Letϵ∈(0,1].
A linear operator T,
which is initially assumed to be continuous from Cbη(𝒳) to (Cbη(𝒳))′ for all η∈(0,ϵ),
is called a singular integral of
orderϵ if T has a distributional kernel K(x,y) which is locally integrable away from the
diagonal of 𝒳×𝒳,
and satisfies the following conditions.(I-1)
Ifφ,ψ∈Cbη(𝒳) have disjoint supports, then (5.55)〈Tφ,ψ〉=∬𝒳×𝒳K(x,y)φ(y)ψ(x)dμ(x)dμ(y).(I-2)
Ifφ is a normalized ϵ-bump function associated to a ball of radius r,
then ∥Tφ∥C˙η(𝒳)≤Cr−η for all η∈(0,ϵ),
where C>0 is independent of φ.
More precisely, for each η∈(0,ϵ),
there is another η˜∈(0,ϵ) and a constant Cη,η˜>0 so that whenever φ is a Cbη˜(𝒳) function supported in a ball B(x0,r),
then rη∥Tφ∥C˙η(𝒳)≤Cη,η˜supη0≤η˜rη0∥φ∥C˙η0(𝒳).(I-3)
There exists a constantC>0 such that
(I-3)1
for
allx,y∈𝒳 with x≠y, |K(x,y)|≤C(1/V(x,y));(I-3)2
for
allx,x′,y∈𝒳 with d(x,x′)≤d(x,y)/2 and x≠y, (5.56)|K(x,y)−K(x′,y)|+|K(y,x)−K(y,x′)|≤Cd(x,x′)ϵV(x,y)d(x,y)ϵ.(I-4)
Properties (I-1) through (I-3) also hold withx and y interchanged. That is, these properties also
hold for the adjoint operator Tt defined by 〈Ttφ,ψ〉=〈φ,Tψ〉.Remark 5.22.
We remark that ifT is a singular integral operator of order ϵ,
then T extends to a continuous linear operator from Cη(𝒳) to (𝒢°b(η,γ))′ for all η∈(0,ϵ] and all γ>0 by Proposition 2.12.
We also claim that forf∈𝒢°b(η,δ) with η∈(0,ϵ] and δ>0, Tf can be defined as a distribution in (𝒢°0ϵ(β,γ))′ with 0<β,γ≤ϵ.
We first define Tf as a distribution in (𝒢°(β,γ))′ with 0<β≤ϵ and γ>0.
In fact, for any given β∈(0,ϵ],
noticing that 𝒢°b(η1,δ)⊂𝒢°b(η2,δ) when η1≥η2,
without loss of generality, we may assume that η≤β.
Assume that suppf⊂B(x0,r) for some x0∈𝒳 and r>0.
Let ψ∈Cbη(𝒳) such that ψ(x)=1 when x∈B(x0,2r) and ψ(x)=0 when x∉B(x0,4r).
Choose any g∈𝒢°(β,γ) with 0<β≤ϵ and γ>0.
It is easy to check that ψg∈Cbη(𝒳).
Notice that 𝒢°b(η,δ)⊂Cbη(𝒳). Both facts show that 〈Tf,ψg〉 is well defined. On the other hand, we define 〈Tf,(1−ψ)g〉 by (5.57)〈Tf,(1−ψ)g〉=∬𝒳×𝒳[K(x,y)−K(x,x0)]f(y)(1−ψ(x))g(x)dμ(x)dμ(y). By (I-3)2 and Lemma 2.1(i), it is easy to check that(5.58)|〈Tf,(1−ψ)g〉|≲∥f∥L1(𝒳)∥g∥L∞(𝒳)≲∥f∥𝒢(η,δ)∥g∥𝒢(β,γ). Moreover, if g∈𝒢°b(β,γ),
since ∫𝒳f(y)dμ(y)=0,
we then obtain (5.59)〈Tf,(1−ψ)g〉=∬𝒳×𝒳K(x,y)f(y)(1−ψ(x))g(x)dμ(x)dμ(y), which coincides with (I-1).
Furthermore, it is easy to verify that 〈Tf,ψg〉+〈Tf,(1−ψ)g〉 is independent of the choice of ψ.
Thus, we can define Tf by (5.60)〈Tf,g〉=〈Tf,ψg〉+〈Tf,(1−ψ)g〉. In this sense, we have Tf∈(𝒢°(β,γ))′ with 0<β≤ϵ and γ>0.
Now for any g∈𝒢°0ϵ(β,γ) with 0<β,γ≤ϵ,
let {gn}n∈ℕ⊂𝒢°(ϵ,ϵ) such that ∥gn−g∥𝒢(β,γ)→0 as n→∞.
We then define Tf∈(𝒢°0ϵ(β,γ))′ with 0<β,γ≤ϵ by 〈Tf,g〉=limn→∞〈Tf,gn〉. It is easy to check that 〈Tf,g〉 is independent of the choice of {gn}n∈ℕ⊂𝒢°(ϵ,ϵ).
In this sense, we have Tf∈(𝒢°0ϵ(β,γ))′ with 0<β,γ≤ϵ.In what follows, for anyϵ∈(0,1] and 0<β,γ≤ϵ,
we set (5.61)𝒢°0,bϵ(β,γ)={f∈𝒢°0ϵ(β,γ):fhasboundedsupport}.Theorem 5.23.
Letϵ1,ϵ2,ϵ, and |s|<ϵ be as in Definition 5.8. Let T be a singular integral of order ϵ satisfying (I-1) through (I-4). Then T is bounded on B˙p,qs(𝒳) when p(s,ϵ)<p<∞ and 0<q<∞ and bounded from B˙p,qs(𝒳)∩𝒢°b(ϵ1,ϵ2) to B˙p,qs(𝒳) when max{p,q}=∞,
and T is also bounded on F˙p,qs(𝒳) when p(s,ϵ)<p,q<∞, and bounded from F˙p,qs(𝒳)∩𝒢°b(ϵ1,ϵ2) to F˙p,qs(𝒳) when p(s,ϵ)<p<∞ and q=∞.Proof.
By Propositions5.21 and 5.10(vi) together with a density
argument, to prove the theorem, it suffices to verify the conclusions of the
theorem only for all f∈𝒢°0,bϵ(β,γ) with 0<β,γ<ϵ as in (5.35). Let {Sk}k∈ℤ be an ATI with bounded support as constructed in Theorem
2.6. Put Dk=Sk−Sk−1 for k∈ℤ.
We then interpret DkT=Tt(Dkt),
where Dkt(x,y)=Dk(y,x) for all x,y∈𝒳 (cf. Lemma 3.12). By Remark 5.22, we have DkT∈(𝒢°0ϵ(β,γ))′.
On the other hand, let f∈𝒢°0,bϵ(β,γ).
By Theorem 4.12, we have (5.62)f(z)=∑k′∈ℤ∑τ′∈Ik′∑ν′=1N(k′,τ′)μ(Qτ′k′,ν′)Dk′(z,yτ′k′,ν′)D¯k′(f)(yτ′k′,ν′), where the series converges in 𝒢°0ϵ(β,γ) and all the notation is as in Theorem 4.12. Thus, for all x∈𝒳,
we have (5.63)DkTf(x)=∑k′∈ℤ∑τ′∈Ik′∑ν′=1N(k′,τ′)μ(Qτ′k′,ν′)(DkTDk′(⋅,yτ′k′,ν′))(x)D¯k′(f)(yτ′k′,ν′). For k,k′∈ℤ,τ′∈Ik′,ν′=1,…,N(k′,τ′),yτ′k′,ν′∈Qτ′k′,ν′,
and x∈𝒳,
let (5.64)J(x,yτ′k′,ν′)≡(DkTDk′(⋅,yτ′k′,ν′))(x). We now claim that for any fixed η∈(0,ϵ), (5.65)|J(x,yτ′k′,ν′)|≲2−|k−k′|η(2−(k⋀k′)2−(k⋀k′)+d(x,yτ′k′,ν′))η1V2−(k⋀k′)(x)+V2−(k⋀k′)(yτ′k′,ν′)+V(x,yτ′k′,ν′).
To verify (5.65), by symmetry, we only need to verify
that if k≤k′,
then (5.66)|J(x,yτ′k′,ν′)|≲2(k−k′)η1V2−k(x), and moreover, if d(x,yτ′k′,ν′)≥25−k,
then (5.67)|J(x,yτ′k′,ν′)|≲(2−k′d(x,yτ′k′,ν′))η1V(x,yτ′k′,ν′). To see this, for y∈𝒳,
set (5.68)Ψ(y)=∫𝒳K(z,y)Dk(x,z)dμ(z). Notice that for any fixed x∈𝒳, Dk(x,z) is an adjusted bump function in z associated with the ball B(x,22−k).
Conditions (I-2) and (I-4) show that for any fixed η∈(0,ϵ),(5.69)2−kη∥Ψ∥C˙η(𝒳)≲1V2−k(x), which further implies that for
all y∈𝒳, (5.70)|Ψ(y)−Ψ(yτ′k′,ν′)|≲d(y,yτ′k′,ν′)η2−kη1V2−k(x). Thus, (5.71)|J(x,yτ′k′,ν′)|=|∫𝒳Ψ(y)Dk′(y,yτ′k′,ν′)dμ(y)|=|∫𝒳[Ψ(y)−Ψ(yτ′k′,ν′)]Dk′(y,yτ′k′,ν′)dμ(y)|≲2(k−k′)η1V2−k(x). That is, (5.66) holds.
On the other hand, whenk≤k′, z∈B(x,22−k), y∈B(yτ′k′,ν′,22−k′), and d(x,yτ′k′,ν′)≥25−k,
then by Condition (I-3), we have (5.72)|K(z,y)−K(z,yτ′k′,ν′)|≲2−k′ϵd(x,yτ′k′,ν′)ϵV(x,yτ′k′,ν′), which yields
that (5.73)|J(x,yτ′k′,ν′)|=|∬𝒳×𝒳[K(z,y)−K(z,yτ′k′,ν′)]Dk(z,x)Dk′(y,yτ′k′,ν′)dμ(y)dμ(z)|≲(2−k′d(x,yτ′k′,ν′))η1V(x,yτ′k′,ν′), namely, (5.67) holds. Thus,
(5.65) holds.
Notice that the estimate (5.65) is analogous to the
estimate (5.10) in the proof of Proposition 5.4. We can further proceed from
here on exactly as in the proof of Proposition 5.4 in order to obtain the
boundedness of T on B˙p,qs(𝒳) and F˙p,qs(𝒳).
This finishes the proof of Theorem 5.16.


### 5.3. Inhomogeneous Plancherel-Pôlya Inequalities and Definition ofBp,qs(𝒳) and Fp,qs(𝒳)

Throughout this and the next subsection,μ(𝒳) can be finite or infinite. We first introduce
the norms in Bp,qs(𝒳) and Fp,qs(𝒳)
via certain IATI and then verify that these norms are
independent of the choices of IATIs and spaces of distributions; see [37]. To this end, we need to
establish inhomogeneous Plancherel-Pôlya inequalities; see [80]. In what follows, for any
dyadic cube Q,
we set mQ(f)=(1/μ(Q))∫Qf(x)dμ(x).Definition 5.24.
Letϵ1∈(0,1],ϵ2>0,ϵ3>0,ϵ∈(0,ϵ1⋀ϵ2), and let {Sk}k∈ℤ+ be an (ϵ1,ϵ2,ϵ3)-IATI.
Set D0=S0 and Dk=Sk−Sk−1 for k∈ℕ.
Let {Qτ0,ν:τ∈I0,ν=1,…,N(0,τ)} with a fixed large j∈ℕ be dyadic cubes as in Section
4.
(i)
For allf∈(𝒢0ϵ(β,γ))′ with 0<β,γ<ϵ,|s|<ϵ,p(s,ϵ)<p≤∞, and 0<q≤∞,
we define(5.74)∥f∥Bp,qs(𝒳)={∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)[mQτ0,ν(|D0(f)|)]p}1/p+{∑k=1∞2ksq∥Dk(f)∥Lp(𝒳)q}1/qwith the usual
modification made when p=∞ or q=∞.(ii)
For allf∈(𝒢0ϵ(β,γ))′ with 0<β,γ<ϵ,|s|<ϵ,p(s,ϵ)<p<∞, and p(s,ϵ)<q≤∞,
we define (5.75)∥f∥Fp,qs(𝒳)={∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)[mQτ0,ν(|D0(f)|)]p}1/p+∥{∑k=1∞2ksq|Dk(f)|q}1/q∥Lp(𝒳) with the usual
modification made when q=∞.The following theorem is theinhomogeneous Plancherel-Pôlya inequalities.Proposition 5.25.
Letϵ1∈(0,1],ϵ2>0,ϵ3>0,ϵ∈(0,ϵ1⋀ϵ2), and let {Sk}k∈ℤ+ and {Pk}k∈ℤ+ be two (ϵ1,ϵ2,ϵ3)-IATIs. Set D0=S0 and Q0=P0,
and Dk=Sk−Sk−1 and Qk=Pk−Pk−1 for k∈ℕ.
Let {Qτ0,ν:τ∈I0,ν=1,…,N(0,τ)} with a fixed large j∈ℕ be dyadic cubes as in Section 4.
(i)
For allf∈(𝒢0ϵ(β,γ))′ with 0<β,γ<ϵ, |s|<ϵ, p(s,ϵ)<p≤∞, and 0<q≤∞,
{∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)[mQτ0,ν(|D0(f)|)]p}1/p+{∑k=1∞2ksq(∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)[supz∈Qτk,ν|Dk(f)(z)|]p)q/p}1/q(5.76)~{∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)[mQτ0,ν(|Q0(f)|)]p}1/p+{∑k=1∞2ksq(∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)[infz∈Qτk,ν|Qk(f)(z)|]p)q/p}1/q
with the usual modification made
when p=∞ or q=∞.(ii)
For allf∈(𝒢0ϵ(β,γ))′ with 0<β,γ<ϵ, |s|<ϵ, p(s,ϵ)<p<∞, and p(s,ϵ)<q≤∞,(5.77){∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)[mQτ0,ν(|D0(f)|)]p}1/p+∥{∑k=1∞∑τ∈Ik∑ν=1N(k,τ)2ksq[supz∈Qτk,ν|Dk(f)(z)|]qχQτk,ν}1/q∥Lp(𝒳)~{∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)[mQτ0,ν(|Q0(f)|)]p}1/p+∥{∑k=1∞∑τ∈Ik∑ν=1N(k,τ)2ksq[infz∈Qτk,ν|Qk(f)(z)|]qχQτk,ν}1/q∥Lp(𝒳)with the usual modification made
when q=∞.Proof.
We first verify (5.76). By Theorem 4.16 together with Remark 4.17, there
exists a family of functions {D˜k′(x,y)}k′∈ℤ+ such that for all f∈(𝒢0ϵ(β,γ))′ with 0<β,γ<ϵ and all z∈𝒳, (5.78)f(z)=∑τ′∈I0∑ν′=1N(0,τ′)∫Qτ′0,ν′D˜0(x,y)dμ(y)Qτ′,10,ν′(f)+∑k′=1∞∑τ′∈Ik′∑ν′=1N(k′,τ′)μ(Qτ′k′,ν′)D˜k′(z,yτ′k′,ν′)Qk′(f)(yτ′k′,ν′), where Qτ′,10,ν′ denotes the integral operator with kernel (5.79)Qτ′,10,ν′(z)=1μ(Qτ′0,ν′)∫Qτ′0,ν′Q0(u,z)dμ(u) and {D˜k′}k′∈ℤ+ satisfy the same conditions as {D˜k}k∈ℤ+ in
Theorem 4.14. From (5.78) together with (5.80)|Qτ′,10,ν′(f)|=|∫𝒳Qτ′,10,ν′(z)f(z)dμ(z)|=|1μ(Qτ′0,ν′)∫Qτ′0,ν′Q0(f)(u)dμ(u)|≤mQτ′0,ν′(|Q0(f)|), it follows that (5.81)mQτ0,ν(|D0(f)|)=1μ(Qτ0,ν)∫Qτ0,ν|D0(f)(z)|dμ(z)≤∑τ′∈I0∑ν′=1N(0,τ′)mQτ′0,ν′(|Q0(f)|)1μ(Qτ0,ν)∫Qτ0,ν∫Qτ′0,ν′|(D0D˜0)(z,y)|dμ(y)dμ(z)+∑k′=1∞∑τ′∈Ik′∑ν′=1N(k′,τ′)|Qk′(f)(yτ′k′,ν′)|μ(Qτ′k′,ν′)1μ(Qτ0,ν)∫Qτ0,ν|(D0D˜k′)(z,yτ′k′,ν′)|dμ(z)≡Z1+Z2. By Lemma 3.19 together with
Remark 3.3, we have that for any ϵ∈(0,ϵ1⋀ϵ2),
all k′∈ℤ+ and all z,y∈𝒳,(5.82)|(D0D˜k′)(z,y)|≲2−k′ϵ1V1(z)+V1(y)+V(z,y)1(1+d(z,y))ϵ. From this together with Lemma
2.1(iii), it follows that (5.83)supz∈Qτ0,νsupy∈Qτ′0,ν′|(D0D˜0)(z,y)|≲infz∈Qτ0,νinfy∈Qτ′0,ν′1V1(z)+V1(y)+V(z,y)1(1+d(z,y))ϵ. Thus, (5.84)Z1≲∑τ′∈I0∑ν′=1N(0,τ′)mQτ′0,ν′(|Q0(f)|)μ(Qτ′0,ν′)infz∈Qτ0,νinfy∈Qτ′0,ν′1V1(z)+V1(y)+V(z,y)1(1+d(z,y))ϵ. Therefore, if n/(n+ϵ)<p≤1,
by (5.5) and Lemma 5.2, we then have(5.85){∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)[Z1]p}1/p≲{∑τ′∈I0∑ν′=1N(0,τ′)[mQτ′0,ν′(|Q0(f)|)]p[μ(Qτ′0,ν′)]p[V1(zτ′0,ν′)]1−p}1/p≲{∑τ′∈I0∑ν′=1N(0,τ′)μ(Qτ′0,ν′)[mQτ′0,ν′(|Q0(f)|)]p}1/p, where in the last step, we used
the fact that for any yτ′0,ν′∈Qτ′0,ν′, (5.86)V1(yτ′0,ν′)~μ(Qτ′0,ν′). If 1<p≤∞,
by Hölder's inequality and Lemma 5.2, we have
(5.87)Z1≲{∑τ′∈I0∑ν′=1N(0,τ′)[mQτ′0,ν′(|Q0(f)|)]pμ(Qτ′0,ν′)infz∈Qτ0,νinfy∈Qτ′0,ν′1V1(z)+V1(y)+V(z,y)1(1+d(z,y))ϵ}1/p, which together with Lemma 5.2
again yields that (5.88){∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)[Z1]p}1/p≲{∑τ′∈I0∑ν′=1N(0,τ′)μ(Qτ′0,ν′)[mQτ′0,ν′(|Q0(f)|)]p}1/p.This is the desired estimate.
From (5.82), it also follows that (5.89)supz∈Qτ0,ν|(D0D˜k′)(z,yτ′k′,ν′)|≲2−k′ϵinfz∈Qτ0,ν1V1(z)+V1(yτ′k′,ν′)+V(z,yτ′k′,ν′)1(1+d(z,yτ′k′,ν′))ϵ, which proves
that (5.90)Z2≲∑k′=1∞2−k′ϵ∑τ′∈Ik′∑ν′=1N(k′,τ′)μ(Qτ′k′,ν′)|Qk′(f)(yτ′k′,ν′)|×infz∈Qτ0,ν1V1(z)+V1(yτ′k′,ν′)+V(z,yτ′k′,ν′)1(1+d(z,yτ′k′,ν′))ϵ. Notice also that by Lemma 2.19
and (1.3), for any yτ′k′,ν′∈Qτ′k′,ν′, V1(yτ′k′,ν′)≲2−k′nμ(Qτ′k′,ν′). If p(s,ϵ)<p≤1,
then this fact together with (5.5) and Lemma 5.2 proves that(5.91){∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)[Z2]p}1/p≲{∑k′=1∞2k′sq(∑τ′∈Ik′∑ν′=1N(k′,τ′)μ(Qτ′k′,ν′)|Qk′(f)(yτ′k′,ν′)|p)q/p}1/q, where in the last step, we used
(5.5) when q/p≤1 or Hölder's inequality when q/p>1.
If1<p≤∞,
then by Hölder's inequality and Lemma 5.2, we first have(5.92)Z2≲{∑k′=1∞2−k′ϵ′p∑τ′∈Ik′∑ν′=1N(k′,τ′)μ(Qτ′k′,ν′)|Qk′(f)(yτ′k′,ν′)|p×infz∈Qτ0,ν1V1(z)+V1(yτ′k′,ν′)+V(z,yτ′k′,ν′)1(1+d(z,yτ′k′,ν′))ϵ}1/p, where max{−s,0}<ϵ′<ϵ.
This together with Lemma 5.2 again shows that (5.93){∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)[Z2]p}1/p≲{∑k′=1∞2k′sq(∑τ′∈Ik′∑ν′=1N(k′,τ′)μ(Qτ′k′,ν′)|Qk′(f)(yτ′k′,ν′)|p)q/p}1/q, where we used (5.5) when q/p≤1 or Hölder's inequality when q/p>1.
All the above estimates together with the arbitrary
choice ofyτ′k′,ν′ yield that the first term in the left-hand
side of (5.76) is controlled by its right-hand side.
We now verify that the second term of the left-hand
side of (5.76) is also controlled by its right-hand side. To this end, by
(5.78), (5.94)|Dk(f)(z)|≲∑τ′∈I0∑ν′=1N(0,τ′)mQτ′0,ν′(|Q0(f)|)∫Qτ′0,ν′|(DkD˜0)(z,y)|dμ(y)+∑k′=1∞∑τ′∈Ik′∑ν′=1N(k′,τ′)μ(Qτ′k′,ν′)|(DkD˜k′)(z,yτ′k′,ν′)||Qk′(f)(yτ′k′,ν′)|≡Y1+Y2. The estimate for Y2 is as in (5.7), and we only need to estimate Y1.
In this case, by Lemma 3.19 together with Remark 3.3, for any ϵ∈(0,ϵ1⋀ϵ2),
we have (5.95)supz∈Qτk,νY1≲2−kϵ∑τ′∈I0∑ν′=1N(0,τ′)mQτ′0,ν′(|Q0(f)|)μ(Qτ′0,ν′)infz∈Qτk,νinfy∈Qτ′0,ν′1V1(z)+V1(y)+V(z,y)1(1+d(z,y))ϵ, and therefore, if n/(n+ϵ)<p≤1 and s<ϵ,
by (5.5) and Lemma 5.2 together with (5.86), we obtain(5.96){∑k=1∞2ksq(∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)[supz∈Qτk,νY1]p)q/p}1/q≲{∑τ′∈I0∑ν′=1N(0,τ′)μ(Qτ′0,ν′)[mQτ′0,ν′(|Q0(f)|)]p}1/p, while when 1<p≤∞,
by Hölder's inequality and Lemma 5.2,(5.97){∑k=1∞2ksq(∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)[supz∈Qτk,νY1]p)q/p}1/q≲{∑τ′∈I0∑ν′=1N(0,τ′)μ(Qτ′0,ν′)[mQτ′0,ν′(|Q0(f)|)]p}1/p.
All these estimates imply that the left-hand side of
(5.76) is controlled by its right-hand side, which together with the symmetry
verifies (5.76).
Similarly, to establish (5.77), we only need to verify
that its left-hand side is controlled by its right-hand side. To this end, the
estimates for Z1 are still valid for the current case. To
estimate Z2,
by Lemma 5.3 and (5.5) when q≤1 or Hölder's inequality when 1<q≤∞,
we have that for p(s,ϵ)<r<min{1,p,q}, (5.98){∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)[Z2]p}1/p≲∥{∑k′=1∞[M(∑τ′∈Ik′∑ν′=1N(k′,τ′)2k′sr|Qk′(f)(yτ′k′,ν′)|rχQτ′k′,ν′)]q/r}1/q∥Lp(𝒳)≲∥{∑k′=1∞∑τ′∈Ik′∑ν′=1N(k′,τ′)2k′sq|Qk′(f)(yτ′k′,ν′)|qχQτ′k′,ν′}1/q∥Lp(𝒳), where in the last step, we used
Lemma 3.14. This is the desired  estimate.
The estimation forY2 is as in the proof of (5.8). To finish the
proof of (5.77), we still need to estimate Y1.
In what follows, we set a=1 when p/q≤1 and a=q/p when p/q>1.
The estimates (5.95) and (5.5) when p/q≤1 or Hölder's inequality when p/q>1 show that (5.99)∥{∑k=1∞∑τ∈Ik∑ν=1N(k,τ)2ksq[supz∈Qτk,νY1]qχQτk,ν}1/q∥Lp(𝒳)p≲∑k=1∞2k(s−ϵ)pa∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)×[∑τ′∈I0∑ν′=1N(0,τ′)μ(Qτ′0,ν′)mQτ′0,ν′(|Q0(f)|)infz∈Qτk,νinfy∈Qτ′0,ν′1V1(z)+V1(y)+V(z,y)1(1+d(z,y))ϵ]p. If n/(n+ϵ)<p≤1,
by (5.5), s<ϵ, and Lemma 5.2 together with
(5.82),(5.100)∥{∑k=1∞∑τ∈Ik∑ν=1N(k,τ)2ksq[supz∈Qτk,νY1]qχQτk,ν}1/q∥Lp(𝒳)p≲∑τ′∈I0∑ν′=1N(0,τ′)μ(Qτ′0,ν′)[mQτ′0,ν′(|Q0(f)|)]p; and if 1<p≤∞,
by Hölder's inequality and Lemma 5.2,
(5.101)∥{∑k=1∞∑τ∈Ik∑ν=1N(k,τ)2ksq[supz∈Qτk,νY1]qχQτk,ν}1/q∥Lp(𝒳)p≲∑τ′∈I0∑ν′=1N(0,τ′)μ(Qτ′0,ν′)[mQτ′0,ν′(|Q0(f)|)]p, which completes the proof of
(5.77), and hence, that of Proposition 5.25.Remark 5.26.
We point out that Remark5.5 applies in a similar way to Proposition
5.25.Similarly to Proposition5.6, using Proposition 5.4,
we can verify that the definitions of ∥⋅∥Bp,qs(𝒳) and ∥⋅∥Fp,qs(𝒳) are independent of the choices of IATIs. We omit the details.Proposition 5.27.
Let all the notation be as in Proposition5.25.
(i)
For allf∈(𝒢0ϵ(β,γ))′ with 0<β,γ<ϵ, |s|<ϵ, p(s,ϵ)<p≤∞, and 0<q≤∞, (5.102){∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)[mQτ0,ν(|D0(f)|)]p}1/p+{∑k=1∞2ksq∥Dk(f)∥Lp(𝒳)q}1/q~{∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)[mQτ0,ν(|Q0(f)|)]p}1/p+{∑k=1∞2ksq∥Qk(f)∥Lp(𝒳)q}1/q with the usual modification made
when p=∞ or q=∞.(ii)
For allf∈(𝒢0ϵ(β,γ))′ with 0<β,γ<ϵ, |s|<ϵ, p(s,ϵ)<p<∞, and p(s,ϵ)<q≤∞, (5.103){∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)[mQτ0,ν(|D0(f)|)]p}1/p+∥{∑k=1∞2ksq|Dk(f)|q}1/q∥Lp(𝒳)~{∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)[mQτ0,ν(|Q0(f)|)]p}1/p+∥{∑k=1∞2ksq|Qk(f)|q}1/q∥Lp(𝒳) with the usual modification made
when q=∞.We now verify that the definitions of the norm∥⋅∥Bp,qs(𝒳) and the norm ∥⋅∥Fp,qs(𝒳) are independent of the choice of the
underlying space of distributions as follows.Proposition 5.28.
Let all the notation be as in Definition5.24.
(i)
Let|s|<ϵ, p(s,ϵ)<p≤∞, and 0<q≤∞.
If f∈(𝒢0ϵ(β1,γ1))′ with (5.104)max{0,−s+n(1p−1)+}<β1<ϵ,n(1p−1)+<γ1<ϵ, and if ∥f∥Bp,qs(𝒳)<∞,
then f∈(𝒢0ϵ(β2,γ2))′ for every β2,γ2  satisfying (5.104).(ii)
Let|s|<ϵ, p(s,ϵ)<p<∞, and p(s,ϵ)<q≤∞.
If f∈(𝒢0ϵ(β1,γ1))′ with β1,γ1 as in (5.104), and if ∥f∥Fp,qs(𝒳)<∞,
then f∈(𝒢0ϵ(β2,γ2))′ for every β2,γ2 satisfying (5.104).Proof.
Similarly to the proof of Proposition5.7, we only need to verify (i).
Let ψ∈𝒢(ϵ,ϵ) and all the notation as in Theorem 4.14 with N=0.
We first claim that for k∈ℤ+, (5.105)|〈D˜k(⋅,y),ψ〉|≲2−kβ2∥ψ∥𝒢(β2,γ2)1V1(x1)+V(x1,y)1(1+d(x1,y))γ2. In fact, if k∈ℕ,
(5.105) is just (5.24). If k=0,
then (5.106)|〈D˜0(⋅,y),ψ〉|=|∫𝒳D˜0(z,y)ψ(z)dμ(z)|≲∥ψ∥𝒢(β2,γ2)∫𝒳1V1(z)+V1(y)+V(z,y)1(1+d(z,y))ϵ1V1(x1)+V(x1,z)1(1+d(x1,z))γ2dμ(z). Notice that V1(x1)+V(x1,z)~V1(x1)+V1(z)+V(x1,z). By Lemma 4.4, we have (5.107)|〈D˜0(⋅,y),ψ〉|≲∥ψ∥𝒢(β2,γ2)1V1(x1)+V1(y)+V(x1,y){1(1+d(x1,y))ϵ+1(1+d(x1,y))γ2}≲∥ψ∥𝒢(β2,γ2)1V1(x1)+V(x1,y)1(1+d(x1,y))γ2, which shows that (5.105) also
holds when k=0.
From Theorem4.16 together with (5.80), (5.105), and
Lemma 2.1(iii), it follows that(5.108)|〈f,ψ〉|=|∑τ∈I0∑ν=1N(0,τ)∫Qτ0,ν〈D˜0(⋅,y),ψ〉dμ(y)Dτ,10,ν(f)+∑k=1∞∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)〈D˜k(⋅,yτk,ν),ψ〉Dk(f)(yτk,ν)|≲∥ψ∥𝒢(β2,γ2){∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)mQτ0,ν(|D0(f)|)1V1(x1)+V(x1,yτ0,ν)1(1+d(x1,yτ0,ν))γ2+∑k=1∞2−kβ2∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)|Dk(f)(yτk,ν)|1V1(x1)+V(x1,yτk,ν)1(1+d(x1,yτk,ν))γ2}.
Notice that ifp≤1 and γ2>n(1/p−1),
then (5.30) with β2=0 still holds. Thus, when p≤1,
by this fact and (5.5), we have(5.109)|〈f,ψ〉|≲∥ψ∥𝒢(β2,γ2){∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)[mQτ0,ν(|D0(f)|)]p+∑k=1∞2−k[β2p−n(1−p)+sp]2ksp(∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)|Dk(f)(yτk,ν)|p)}1/p≲∥ψ∥𝒢(β2,γ2)∥f∥Bp,qs(𝒳), where in the last inequality, we
used the assumption that β2>−s+n(1/p−1) when p≤1,
and (5.5) when q/p≤1 or Hölder's inequality when q/p>1.
If1<p≤∞,
by Hölder's inequality, we have (5.110)|〈f,ψ〉|≲∥ψ∥𝒢(β2,γ2){(∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)[mQτ0,ν(|D0(f)|)]p)1/p×[∫𝒳1V1(x1)+V(x1,y)1(1+d(x1,y))γ2dμ(y)]1/p′+∑k=1∞2−kβ2(∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)|Dk(f)(yτk,ν)|p)1/p×[∫𝒳1V1(x1)+V(x1,y)1(1+d(x1,y))γ2dμ(y)]1/p′}≲∥ψ∥𝒢(β2,γ2)∥f∥Bp,qs(𝒳), here, again, we used the
assumption β2>−s in this case, and (5.5) when q≤1 or Hölder's inequality when q>1.
Using (5.109) and (5.110) together with an argument
similar to that used in the proof of Proposition 5.7 then completes the proof
of Proposition 5.28.We can now introduce the Besov spacesBp,qs(𝒳) and Triebel-Lizorkin spaces Fp,qs(𝒳).Definition 5.29.
Letϵ1∈(0,1], ϵ2>0, ϵ3>0, ϵ∈(0,ϵ1⋀ϵ2), and let {Sk}k∈ℤ+ be an (ϵ1,ϵ2,ϵ3)-IATI.
Set D0=S0 and Dk=Sk−Sk−1 for k∈ℕ.
Let {Qτ0,ν:τ∈I0,ν=1,…,N(0,τ)} with a fixed large j∈ℕ be dyadic cubes as in Section
4.
(i)
Let|s|<ϵ,p(s,ϵ)<p≤∞, and 0<q≤∞.
The spaceBp,qs(𝒳) is defined to be the set of all f∈(𝒢0ϵ(β,γ))′,
for some β,γ satisfying (5.111)max{s,0,−s+n(1p−1)+}<β<ϵ,n(1p−1)+<γ<ϵ such that (5.112)∥f∥Bp,qs(𝒳)={∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)[mQτ0,ν(|D0(f)|)]p}1/p+{∑k=1∞2ksq∥Dk(f)∥Lp(𝒳)q}1/q<∞ with the usual modifications
made when p=∞ or q=∞.(ii)
Let|s|<ϵ,p(s,ϵ)<p<∞, and p(s,ϵ)<q≤∞. The spaceFp,qs(𝒳) is defined to be the set of all f∈(𝒢0ϵ(β,γ))′ for some β,γ satisfying (5.111) such that(5.113)∥f∥Fp,qs(𝒳)={∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)[mQτ0,ν(|D0(f)|)]p}1/p+∥{∑k=1∞2ksq|Dk(f)|q}1/q∥Lp(𝒳)<∞ with the usual modification made
when q=∞.Remark 5.30.
Propositions5.27 and 5.28 show that the definitions of the
spaces
Bp,qs(𝒳) and Fp,qs(𝒳) are independent of the choice of inhomogeneous
approximations of the identity as in Definition 2.2 and the distribution space (𝒢0ϵ(β,γ))′ with β and γ satisfying (5.111). We also remark that to guarantee
that the definitions of the spaces Bp,qs(𝒳) and Fp,qs(𝒳) are independent of the choice of the
distribution space (𝒢0ϵ(β,γ))′,
we only need the restriction that β and γ satisfy (5.104). Moreover, if we assume that max{0,s}<β<ϵ and n(1/p−1)+<γ<ϵ,
then we can verify that 𝒢(β,γ) is contained in Bp,qs(𝒳) and Fp,qs(𝒳);
see Proposition 5.31 below.


### 5.4. Properties ofBp,qs(𝒳) and Fp,qs(𝒳) and Boundedness of Singular Integrals

In this subsection, we first present some basic
properties ofBp,qs(𝒳) and Fp,qs(𝒳).
Then we establish a Lusin-area characterization of the spaces Fp,qs(𝒳) and the relations between the spaces B˙p,qs(𝒳) and Bp,qs(𝒳) and between the spaces F˙p,qs(𝒳) and
Fp,qs(𝒳).
Using the Lusin-area characterization of Fp,qs(𝒳),
we also obtain the relation between the spaces Fp,qs(𝒳) and the local Hardy spaces hp(𝒳) in the sense of Goldberg [73]. Finally, we obtain the
boundedness on Bp,qs(𝒳)
and Fp,qs(𝒳) of singular integrals with some natural extra
size and regularity conditions of Nagel-Stein type in [44].The following proposition is an inhomogeneous version
of Proposition5.10.Proposition 5.31.
Letϵ1∈(0,1], ϵ2>0, ϵ∈(0,ϵ1⋀ϵ2), and |s|<ϵ.
(i)
Forp(s,ϵ)<p≤∞ and
0<q0≤q1≤∞, Bp,q0s(𝒳)⊂Bp,q1s(𝒳); and for p(s,ϵ)<p<∞ and p(s,ϵ)<q0≤q1≤∞, Fp,q0s(𝒳)⊂Fp,q1s(𝒳).(ii)
Let−ϵ<s+θ<ϵ and θ>0.
Then for p(s,ϵ)<p≤∞ and 0<q0,q1≤∞, Bp,q0s+θ(𝒳)⊂Bp,q1s(𝒳); and for p(s,ϵ)<p<∞ and p(s,ϵ)<q0,q1≤∞, Fp,q0s+θ(𝒳)⊂Fp,q1s(𝒳).(iii)
Ifp(s,ϵ)<p<∞ and p(s,ϵ)<q≤∞,
then Bp,min(p,q)s(𝒳)⊂Fp,qs(𝒳)⊂Bp,max(p,q)s(𝒳).(iv)
Ifβ,γ as in (5.104), then Bp,qs(𝒳)⊂(𝒢0ϵ(β,γ))′ when p(s,ϵ)<p≤∞ and 0<q≤∞,
and Fp,qs(𝒳)⊂(𝒢0ϵ(β,γ))′ when p(s,ϵ)<p<∞ and p(s,ϵ)<q≤∞.(v)
Ifmax{s,0}<β<ϵ and n(1/p−1)+<γ<ϵ,
then 𝒢(β,γ)⊂Bp,qs(𝒳) when p(s,ϵ)<p≤∞ and 0<q≤∞,
and 𝒢(β,γ)⊂Fp,qs(𝒳) when p(s,ϵ)<p<∞ and p(s,ϵ)<q≤∞.(vi)
If1<p<∞,
then Fp,20(𝒳)=Lp(𝒳) with equivalent norms.(vii)
The spacesBp,qs(𝒳) with p(s,ϵ)<p≤∞ and 0<q≤∞ and the spaces Fp,qs(𝒳) with p(s,ϵ)<p<∞ and p(s,ϵ)<q≤∞ are complete.Proof.
Property (i) is a simple corollary of (5.5). To see (ii), we notice that(5.114){∑k=0∞2skq1|bk|q1}1/q1≤supk∈ℤ+2(s+θ)k|bk|{∑l=0∞2−θlq1}1/q1≲supk∈ℤ+2(s+θ)k|bk|,
which combined with (i) verifies (ii); see also [3, the proof of Proposition 2.3.2/2].
The proof of (iii) is similar to that of Property (ii)
in Proposition5.10; see also [3, 6].
Property (iv) is implied by the proof of Proposition5.28, and Property (vii) can be easily deduced from Property (iv) and Property
(vi) is just Proposition 3.30.
To see Property (v), similarly to the proof of (5.105),
for f∈𝒢(β,γ),
we have that for all k∈ℤ+ and x∈𝒳, (5.115)|Dk(f)(x)|≲2−kβ∥f∥𝒢(β,γ)1V1(x1)+V(x1,x)1(1+d(x1,x))γ. Notice that (5.39) with β=0 is still true when γ>n(1/p−1)+.
From this fact and β>s,
it follows that (5.116)∥f∥Bp,qs(𝒳)~{∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)[mQτ0,ν(|D0(f)|)]p}1/p+{∑k=1∞2ksq∥Dk(f)∥Lp(𝒳)q}1/q≲∥f∥𝒢(β,γ){[∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)1(V1(x1)+V(x1,yτ0,ν))p1(1+d(x1,yτ0,ν))γp]1/p+[∑k=1∞2ksq2−kβq]1/q}≲∥f∥𝒢(β,γ). Thus, 𝒢(β,γ)⊂Bp,qs(𝒳),
which together with (i) also proves that 𝒢(β,γ)⊂Fp,qs(𝒳).
This verifies (v) and hence, finishes the proof of Proposition 5.31.Whenp,q≥1,
the following theorem implies that the norms of ∥⋅∥Bp,qs(𝒳) and ∥⋅∥Fp,qs(𝒳) have the following equivalent and simple
version.Proposition 5.32.
Let all the notation be as in Proposition5.25.
(i)
For allf∈(𝒢0ϵ(β,γ))′ with 0<β,γ<ϵ, |s|<ϵ, 1≤p≤∞, and 0<q≤∞, (5.117){∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)[mQτ0,ν(|D0(f)|)]p}1/p+{∑k=1∞2ksq∥Dk(f)∥Lp(𝒳)q}1/q~{∑k=0∞2ksq∥Dk(f)∥Lp(𝒳)q}1/q with the usual modification made
when p=∞ or q=∞.(ii)
For allf∈(𝒢0ϵ(β,γ))′ with 0<β,γ<ϵ, |s|<ϵ, 1≤p<∞, and p(s,ϵ)<q≤∞, (5.118){∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)[mQτ0,ν(|D0(f)|)]p}1/p+∥{∑k=1∞2ksq|Dk(f)|q}1/q∥Lp(𝒳)~∥{∑k=0∞2ksq|Dk(f)|q}1/q∥Lp(𝒳) with the usual modification made
when q=∞.Proof.
We first verify (5.117). To see this, by p≥1,
Hölder's inequality, and Lemma 2.19, we have (5.119){∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)[mQτ0,ν(|D0(f)|)]p}1/p≤{∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)mQτ0,ν(|D0(f)|p)}1/p=∥D0(f)∥Lp(𝒳), which shows that the left-hand
side of (5.117) is controlled by its right-hand side.
To see the converse, by Lemma2.19 and Theorem 4.16
together with a proof similar to that of (5.76) in Proposition 5.25, we have
that for all f∈(𝒢0ϵ(β,γ))′ with 0<β,γ<ϵ,
and |s|<ϵ, (5.120)∥D0(f)∥Lp(𝒳)={∑τ∈I0∑ν=1N(0,τ)∫Qτ0,ν|D0(f)(z)|pdμ(z)}1/p≤{∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)[supz∈Qτ0,ν|D0(f)(z)|]p}1/p≲{∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)[mQτ0,ν(|D0(f)|)]p}1/p+{∑k=1∞2ksq(∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)[infz∈Qτk,ν|Dk(f)(z)|]p)q/p}1/q, which is, up to a bounded
multiplicative constant, controlled by the left-hand side of (5.117). This
completes the proof of (5.117).
The estimate (5.119) also proves that the left-hand
side of (5.118) is controlled by its right-hand side. The converse inequality
can be proved by a way similar to that of (5.77) in Proposition 5.25, which
completes the proof of Proposition 5.32.We next give an inhomogeneous Lusin-area
characterization for the Triebel-Lizorkin  spacesFp,qs(𝒳).Definition 5.33.
Lets∈ℝ, a>0, q∈(0,∞], and let ϵ1∈(0,1], ϵ2>0, ϵ3>0, ϵ∈(0,ϵ1⋀ϵ2]. Let {Sk}k∈ℤ+ be an (ϵ1,ϵ2,ϵ3)-IATI.
Set D0=S0 and Dk=Sk−Sk−1 for k∈ℕ.
The inhomogeneous Lusin-area function (also called the inhomogeneous
Littlewood-PaleyS-function) Sq,as(f)(x) for any f∈(𝒢0ϵ(β,γ))′ with 0<β,γ≤ϵ and x∈𝒳 is given by (5.121)Sq,as(f)(x)={∑k=0∞∫d(x,y)<a2−k2ksq|Dk(f)(y)|qdμ(y)Va2−k(x)}1/q, where the usual modification is
made when q=∞.Theorem 5.34.
Leta>0, ϵ1∈(0,1], ϵ2>0, ϵ3>0, ϵ∈(0,ϵ1⋀ϵ2),
and let {Sk}k∈ℤ+ be an (ϵ1,ϵ2,ϵ3)-IATI with C621−j≤a.
Set D0=S0 and Dk=Sk−Sk−1 for k∈ℤ.
Let |s|<ϵ, p(s,ϵ)<p<∞, 1≤q≤∞, and let Sq,as(f) be as in Definition 5.33 for any f∈(𝒢0ϵ(β,γ))′ with β,γ as in (5.111). Then f∈Fp,qs(𝒳) if and only if f∈(𝒢0ϵ(β,γ))′ for some
β,γ as in (5.111) and Sq,as(f)∈Lp(𝒳).
Moreover, in this case, (5.122)∥f∥Fp,qs(𝒳)~∥Sq,as(f)∥Lp(𝒳).Proof.
Similarly to the proof of Theorem5.13, there exists a constant C>0 such that (5.123)∥Sq,as(f)∥Lp(𝒳)=∥{∑k=0∞∑τ∈Ik∑ν=0N(k,τ)2ksq∫d(⋅,y)<a2−k|Dk(f)(y)|qdμ(y)Va2−k(⋅)χQτk,ν(⋅)}1/q∥Lp(𝒳)≲∥{∑k=0∞∑τ∈Ik∑ν=0N(k,τ)2ksq[supy∈B(zτk,ν,C2−k)|Dk(f)(y)|]qχQτk,ν}1/q∥Lp(𝒳)≲{∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)[supz∈B(zτ0,ν,C)|D0(f)(z)|]p}1/p+∥{∑k=1∞∑τ∈Ik∑ν=0N(k,τ)2ksq[supz∈B(zτk,ν,C2−k)|Dk(f)(z)|]qχQτk,ν}1/q∥Lp(𝒳)≲{∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)[mQτ0,ν(|D0(f)|)]p}1/p+∥{∑k=1∞∑τ∈Ik∑ν=0N(k,τ)2ksq[infz∈Qτk,ν|Dk(f)(z)|]qχQτk,ν}1/q∥Lp(𝒳)~∥f∥Fp,qs(𝒳), where the last inequality can be
proved by a way similar to the proof of (5.76) in Proposition 5.25; see also the
proof of Theorem 5.13.
On the other hand, since1≤q≤∞,
by Hölder's inequality and Lemma 2.19 together with C621−j≤a,
we have that for x∈Qτk,ν with k∈ℤ+, Qτk,ν⊂{y∈𝒳:d(y,x)≤a2−k} and Va(x)~μ(Qτ0,ν),
and (5.124){∫d(x,y)<a|D0(f)(y)|qdμ(y)Va(x)}1/q≥∫d(x,y)<a|D0(f)(y)|dμ(y)Va(x)=∑τ∈I0∑ν=1N(0,τ)∫d(x,y)<a|D0(f)(y)|dμ(y)Va(x)χQτ0,ν(x)≳∑τ∈I0∑ν=1N(0,τ)mQτ0,ν(|D0(f)|)χQτ0,ν(x). Therefore, similarly to the
proof of Theorem 5.13, by the estimate as above and Lemma 2.19
again, (5.125)Sq,as(f)(x)={∑k=0∞∫d(x,y)<a2−k2ksq|Dk(f)(y)|qdμ(y)Va2−k(x)}1/q≳∑τ∈I0∑ν=1N(0,τ)mQτ0,ν(|D0(f)|)χQτ0,ν(x)+{∑k=1∞∑τ∈Ik∑ν=1N(k,τ)2ksq[infy∈Qτk,ν|Dk(f)(y)|]qχQτk,ν(x)}1/q, which together with Proposition
5.25 and Lemma 2.19 proves that (5.126)∥Sq,as(f)∥Lp(𝒳)≳{∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)[mQτ0,ν(|D0(f)|)]p}1/p+∥{∑k=1∞2ksq|Dk(f)|q}1/q∥Lp(𝒳)~∥f∥Fp,qs(𝒳). This finishes the proof of
Theorem 5.34.Remark 5.35.
Comparing Theorem5.34 with Theorem 5.13, we here need to require that 1≤q≤∞ and C621−j≤a due to the inhomogeneity of Triebel-Lizorkin
spaces Fp,qs(𝒳).From Theorem2.29, we immediately obtain the following
technical lemma, which will be useful in applications.Lemma 5.36.
LetS0 be as in Definition 2.2 and ϵ∈(0,ϵ1⋀ϵ2].
Let 0<β,γ≤ϵ, and let S0t be the integral operator with the kernel S0t(x,y)=S0(y,x) for all x,y∈𝒳.
Then S0t is bounded on 𝒢(β,γ),
namely, there exists a constant C>0 such that for all g∈𝒢(β,γ), S0t(g)∈𝒢(β,γ), and (5.127)∥S0t(g)∥𝒢(β,γ)≤C∥g∥𝒢(β,γ).We now establish some relations between homogeneous
Besov and Triebel-Lizorkin spaces with the corresponding inhomogeneous
ones.Proposition 5.37.
Let all the notation be as in Definition5.29 and μ(𝒳)=∞.
Then there exists a constant C>0 such that if f∈Bp,qs(𝒳) or Fp,qs(𝒳),
then f−S0(f)∈B˙p,qs(𝒳) or F˙p,qs(𝒳),
and (5.128)∥f−S0(f)∥B˙p,qs(𝒳)≤C∥f∥Bp,qs(𝒳) or (5.129)∥f−S0(f)∥F˙p,qs(𝒳)≤C∥f∥Fp,qs(𝒳), respectively.Proof.
Letf∈Bp,qs(𝒳) or f∈Fp,qs(𝒳).
By Proposition 5.28, without loss of generality, we may assume that f∈(𝒢0ϵ(β,γ))′ with β,γ as in (5.35). From 𝒢°(β,γ)⊂𝒢(β,γ),
it follows that f∈(𝒢°0ϵ(β,γ))′.
On the other hand, for any g∈𝒢°0ϵ(β,γ),
we have 〈S0(f),g〉=〈f,S0t(g)〉, which together with Lemma 5.36 also shows that S0(f)∈(𝒢°0ϵ(β,γ))′.
Thus, f−S0(f)∈(𝒢°0ϵ(β,γ))′ with β,γ as in (5.35).
To verify the norm inequalities in the proposition, in
what follows, we letI be the identity operator on Bp,qs(𝒳) or Fp,qs(𝒳).
Let
β and γ be as in (5.35), let {Dk}k∈ℤ be as in Definition 5.8, and let {D˜k′}k′∈ℤ+ be as in Theorem 4.14. We first claim that for
all k∈ℤ and k′∈ℤ+, (5.130)|(Dk(I−S0)D˜k′)(x,y)|≲2−|k−k′|ϵ1V2−(k⋀k′)(x)+V2−(k⋀k′)(y)+V(x,y)2−(k⋀k′)ϵ(2−(k⋀k′)+d(x,y))ϵ. We verify (5.130) by considering
the following three cases.
Case 1  (k′≥0≥k).
In this case, we have(5.131)|(Dk(I−S0)D˜k′)(x,y)|=|∫𝒳[Dk(x,z)−Dk(x,y)]((I−S0)D˜k′)(z,y)dμ(z)|≤|∫𝒳[Dk(x,z)−Dk(x,y)]D˜k′(z,y)dμ(z)|+|∫𝒳[Dk(x,z)−Dk(x,y)](S0D˜k′)(z,y)dμ(z)|. On the first term, an argument
similar to the proof of (3.2) in Lemma 3.2 gives the desired estimate. For the
second term, by Lemma 3.19, we first have (5.132)|(S0D˜k′)(z,y)|≲2−k′ϵ1V1(z)+V1(y)+V(z,y)1(1+d(z,y))ϵ, which together with an argument
similar to the proof of (3.2) in Lemma 3.2 also gives the desired estimate for
the second term.
Case 2  (k′≥k>0).
In this case, we write(5.133)|(Dk(I−S0)D˜k′)(x,y)|≤|(DkD˜k′)(x,y)|+|(DkS0D˜k′)(x,y)|. The estimate (3.2) in Lemma 3.2
directly gives the desired estimate for the first term. Denote the second term
by J and write (5.134)J≡|∬𝒳×𝒳Dk(x,z)[S0(z,u)−S0(z,y)]D˜k′(u,y)dμ(u)dμ(z)|≤∫𝒳∫d(u,y)≤(1+d(z,y))/2|Dk(x,z)||[S0(z,u)−S0(z,y)]D˜k′(u,y)|dμ(u)dμ(z)+∫𝒳∫d(u,y)>(1+d(z,y))/2|Dk(x,z)||S0(z,u)D˜k′(u,y)|dμ(u)dμ(z)+∫𝒳∫d(u,y)>(1+d(z,y))/2|Dk(x,z)||S0(z,y)D˜k′(u,y)|dμ(u)dμ(z)≡J1+J2+J3. The regularity of S0 together with Lemma 2.1(ii) yields
that (5.135)J1≲∫𝒳∫d(u,y)≤(1+d(z,y))/2|Dk(x,z)|(d(u,y)1+d(z,y))ϵ1V1(z)+V1(y)+V(z,y)×1(1+d(z,y))ϵ|D˜k′(u,y)|dμ(u)dμ(z)≲2−k′ϵ∫𝒳|Dk(x,z)|1V1(z)+V1(y)+V(z,y)1(1+d(z,y))ϵdμ(z). If d(x,y)≤2−k,
then by Lemma 2.1(ii) again, J1≲2−k′ϵ(1/V2−k(x)), which is the desired estimate. When d(x,y)>2−k,
we further control J1 by (5.136)J1≲2−k′ϵ{∫d(x,z)≥d(x,y)/21V2−k(x)+V2−k(z)+V(x,z)2−kϵ(2−k+d(x,z))ϵ×1V1(z)+V1(y)+V(z,y)1(1+d(z,y))ϵdμ(z)+∫d(x,z)<d(x,y)/2⋯}. Notice that d(x,z)<d(x,y)/2 also implies that d(z,y)≥d(x,y)/2.
From this together with Lemma 2.1(ii), it follows that (5.137)J1≲2−k′ϵ{1V(x,y)2−kϵd(x,y)ϵ+1V(x,y)1d(x,y)ϵ}≲2−(k′−k)ϵ1V(x,y)2−kϵd(x,y)ϵ, which is also the desired
estimate.
The estimate forJ3 is similar to that for J1.
To estimateJ2,
choosing ϵ′>ϵ and using Lemma 2.1(ii), we first
have (5.138)∫d(u,y)>(1+d(z,y))/2|S0(z,u)D˜k′(u,y)|dμ(u)≲1V1(z)2−k′ϵ(1+d(z,y))ϵ; and by the fact that for d(u,y)>(1+d(z,y))/2, V(u,y)~V(y,u)≳V1(y)+V(y,z), we also obtain (5.139)∫d(u,y)>(1+d(z,y))/2|S0(z,u)D˜k′(u,y)|dμ(u)≲1V1(y)+V(y,z)2−k′ϵ(1+d(z,y))ϵ. Therefore, combining these
estimates gives (5.140)J2≲2−k′ϵ∫𝒳|Dk(x,z)|1V1(z)+V1(y)+V(z,y)1(1+d(z,y))ϵdμ(z), which together with some
computations the same as for J1 gives the desired estimate for J2.
Case 3  (k>k′≥0).
The proof for this case is similar to Case2 by symmetry. We omit the details
for simplicity, which completes the proof of (5.130).
Theorem4.16 together with (5.80), (5.130), and Lemma
5.3 yields that for n/(n+ϵ)<r≤1, k∈ℤ, and x∈𝒳, (5.141)|Dk(f−S0(f))(x)|≲2−|k|ϵ2(k⋀0)n(1−1/r){M(∑τ′∈I0∑ν′=1N(0,τ′)[mQτ′0,ν′(|D0(f)|)]rχQτ′0,ν′)(x)}1/r+∑k′=1∞2−|k−k′|ϵ2[(k⋀k′)−k′]n(1−1/r){M(∑τ′∈Ik′∑ν′=1N(k′,τ′)|Dk′(f)(yτ′k′,ν′)|rχQτ′k′,ν′)(x)}1/r≡Z1+Z2.
We now first consider the case of Besov spaces.
Choosingp(s,ϵ)<r<min{1,p} and using s<ϵ and the boundedness of M yield (5.142){∑k∈ℤ2ksq∥Z1∥Lp(𝒳)q}1/q≲{∑τ′∈I0∑ν′=1N(0,τ′)μ(Qτ′0,ν′)[mQτ′0,ν′(|D0(f)|)]p}1/p≲∥f∥Bp,qs(𝒳), which is the desired estimate.
Similarly, choosingr as above, when 1≤p≤∞,
by Minkowski's inequality and the boundedness of M together with the assumption s<ϵ,
we have (5.143){∑k∈ℤ2ksq∥Z2∥Lp(𝒳)q}1/q≲{∑k∈ℤ∥∑k′=1∞2(k−k′)s−|k−k′|ϵ2[(k⋀k′)−k′]n(1−1/r)2k′s×{M(∑τ′∈Ik′∑ν′=1N(k′,τ′)|Dk′(f)(yτ′k′,ν′)|rχQτ′k′,ν′)}1/r∥Lp(𝒳)q}1/q≲{∑k′=1∞2k′sq∥Dk′(f)∥Lp(𝒳)q}1/q≲∥f∥Bp,qs(𝒳), where in the second-to-last
inequality, we used (5.5) when q≤1 and Hölder's inequality when q>1,
while when r<p<1,
instead of Minkowski's inequality by (5.5), we have (5.144){∑k∈ℤ2ksq∥Z2∥Lp(𝒳)q}1/q≲{∑k′=1∞2k′sq∥Dk′(f)∥Lp(𝒳)q}1/q≲∥f∥Bp,qs(𝒳), where in the second-to-last
inequality, we used (5.5) when q/p≤1 and Hölder's inequality when q/p>1.
Combining the above estimates completes the proof of Besov spaces.
We now turn to the case of Triebel-Lizorkin spaces. In
this case, we also choosep(s,ϵ)<r<min{1,p} and use s<ϵ and the boundedness of M to obtain (5.145)∥{∑k∈ℤ2ksq|Z1|q}1/q∥Lp(𝒳)≲{∑τ′∈I0∑ν′=1N(0,τ′)μ(Qτ′0,ν′)[mQτ′0,ν′(|D0(f)|)]p}1/p≲∥f∥Fp,qs(𝒳), while some computations similar
to the proof of (5.8) in Proposition 5.4 also show that(5.146)∥{∑k∈ℤ2ksq|Z2|q}1/q∥Lp(𝒳)≲∥{∑k′=1∞2k′sq|Dk′(f)|q}1/q∥Lp(𝒳)≲∥f∥Fp,qs(𝒳), which completes the proof of
Proposition 5.37.Remark 5.38.
Obviously,S0 in Proposition 5.37 can be replaced by Sk0 with k0∈ℤ or any integral operator with a kernel having
similar properties.Proposition 5.39.
Letϵ be as in Definition 5.29, 0<s<ϵ, and μ(𝒳)=∞.
Then,
(i)
if1≤p≤∞ and 0<q≤∞,
then Bp,qs(𝒳)=B˙p,qs(𝒳)∩Lp(𝒳) and moreover, for any f∈Bp,qs(𝒳), (5.147)∥f∥Bp,qs(𝒳)~∥f∥B˙p,qs(𝒳)+∥f∥Lp(𝒳);(ii)
if1≤p<∞ and p(s,ϵ)<q≤∞,
then Fp,qs(𝒳)=F˙p,qs(𝒳)∩Lp(𝒳) and moreover, for any f∈Fp,qs(𝒳), (5.148)∥f∥Fp,qs(𝒳)~∥f∥F˙p,qs(𝒳)+∥f∥Lp(𝒳).Proof.
We use the same notation as in Definition5.8. Let f∈Bp,qs(𝒳) or f∈Fp,qs(𝒳).
Then, by Proposition 5.28, it is easy to see that f∈(𝒢°0ϵ(β,γ))′ with β,γ as in (5.35).
To verify (i), letf∈Bp,qs(𝒳).
If 1≤p<∞,
Proposition 3.18, Minkowski's inequality, and Hölder's inequality when 1≤q≤∞ or (5.5) when 0<q<1 show that (5.149)∥f∥Lp(𝒳)≤∥S0(f)∥Lp(𝒳)+∑k=1∞∥Dk(f)∥Lp(𝒳)≲{∥S0(f)∥Lp(𝒳)q+∑k=1∞2ksq∥Dk(f)∥Lp(𝒳)q}1/q≲∥f∥Bp,qs(𝒳). If p=∞,
by Theorem 3.29, we have that f=D˜0S0(f)+∑k=1∞D˜kDk(f) holds in (𝒢(β,γ))′ with s<β<ϵ and 0<γ<ϵ,
where D˜k with k∈ℤ+ is as in Theorem 3.26. From this and f∈B∞,qs(𝒳),
it follows that for almost all x∈𝒳, (5.150)|f(x)|≲∥f∥B∞,qs(𝒳){∫𝒳|D˜0(x,y)|dμ(y)+∑k=1∞2−ks∫𝒳|D˜k(x,y)|dμ(y)}≲∥f∥B∞,qs(𝒳).
Moreover, when1≤p≤∞,
Proposition 2.7(iii) further yields (5.151)∥f∥B˙p,qs(𝒳)~{∑k=−∞∞2ksq∥Dk(f)∥Lp(𝒳)q}1/q≲{∑k=−∞02ksq∥Dk(f)∥Lp(𝒳)q}1/q+{∑k=1∞2ksq∥Dk(f)∥Lp(𝒳)q}1/q≲∥f∥Bp,qs(𝒳). Thus, f∈B˙p,qs(𝒳)∩Lp(𝒳) and ∥f∥Lp(𝒳)+∥f∥B˙p,qs(𝒳)≲∥f∥Bp,qs(𝒳).
Conversely, iff∈B˙p,qs(𝒳)∩Lp(𝒳),
it is obvious that f∈(𝒢(β,γ))′ with β,γ as in (5.111); and moreover, Proposition 2.7(iii) also proves (5.152)∥f∥Bp,qs(𝒳)~{∥S0(f)∥Lp(𝒳)q+∑k=1∞2ksq∥Dk(f)∥Lp(𝒳)q}1/q≲∥f∥Lp(𝒳)+∥f∥B˙p,qs(𝒳), which completes the proof of (i).
To prove (ii), letf∈Fp,qs(𝒳).
From Proposition 3.18 and Hölder's inequality when 1≤q≤∞ or (5.5) when p(s,ϵ)<q<1,
it follows that (5.153)∥f∥Lp(𝒳)≤∥|S0(f)|+∑k=1∞|Dk(f)|∥Lp(𝒳)≲∥{|S0(f)|+∑k=1∞2ksq|Dk(f)|q}1/q∥Lp(𝒳)≲∥f∥Fp,qs(𝒳),∥f∥F˙p,qs(𝒳)~∥{∑k∈ℤ2ksq|Dk(f)|q}1/q∥Lp(𝒳)≲∥{∑k=−∞02ksq|Dk(f)|q}1/q∥Lp(𝒳)+∥{∑k=1∞2ksq|Dk(f)|q}1/q∥Lp(𝒳)≲Z+∥f∥Fp,qs(𝒳). If p/q≤1,
by (5.5), Proposition 2.7(iii), and s>0, (5.154)(Z)p≲∑k=−∞02ksp∥Dk(f)∥Lp(𝒳)p≲∥f∥Lp(𝒳)p∑k=−∞02ksp≲∥f∥Lp(𝒳)p, while when p/q>1,
by Minkowski's inequality, Proposition 2.7(iii), and s>0, (5.155)Z≲{∑k=−∞02ksq∥Dk(f)∥Lp(𝒳)q}1/q≲∥f∥Lp(𝒳){∑k=−∞02ksq}1/q≲∥f∥Lp(𝒳). Thus, f∈F˙p,qs(𝒳)∩Lp(𝒳) and ∥f∥Lp(𝒳)+∥f∥F˙p,qs(𝒳)≲∥f∥Fp,qs(𝒳).
Conversely, iff∈F˙p,qs(𝒳)∩Lp(𝒳),
it is again obvious that f∈(𝒢(β,γ))′ with β,γ as in (5.111); and moreover, Proposition 2.7(iii) yields that(5.156)∥f∥Fp,qs(𝒳)~∥{|S0(f)|q+∑k=1∞2ksq|Dk(f)|q}1/q∥Lp(𝒳)≲∥f∥Lp(𝒳)+∥f∥F˙p,qs(𝒳), which completes the proof of
(ii) and hence, the proof of Proposition 5.39.We now introduce the local Hardy spaces in the sense
of Goldberg [73].Definition 5.40.
Let all the notation be as in Definition5.29. The local Hardy spacehp(𝒳),
for n/(n+ϵ)<p≤1,
is defined to be the inhomogeneous Triebel-Lizorkin space Fp,20(𝒳),
with norm ∥f∥hp(𝒳)≡∥f∥Fp,20(𝒳).We now introduce the definitions ofhp(𝒳)-atoms and hp(𝒳)-blocks.Definition 5.41.
Letj∈ℕ be as in Theorem 5.34 and 0<p≤1.
A function a on 𝒳 is called an hp(𝒳)-atom if a satisfies (i) through (iii) of Definition 5.15
with r<C62−j,
and a is called an hp(𝒳)-block if a satisfies (i) and (ii) of Definition 5.15 with r≥C62−j.Applying Theorem5.34 and Proposition 5.37, we obtain
the following atomic and block decomposition characterization for the local
Hardy spaces hp(𝒳),
which is similar to Theorem 5.16.Theorem 5.42.
Letϵ and β,γ be as in Definition 5.29. If n/(n+ϵ)<p≤1,
then f∈hp(𝒳) if and only if there exist two sequences of
numbers {λk}k∈ℤ+ and {μk}k∈ℤ+ with ∑k=0∞|λk|p<∞ and ∑k=0∞|μk|p<∞,
a sequence of hp(𝒳)-atoms {ak}k∈ℤ+ and a sequence of hp(𝒳)-blocks {bk}k∈ℤ+ such that f=∑k=0∞λkak+∑k=0∞μkbk in (𝒢0ϵ(β,γ))′.
Moreover, in this case, (5.157)∥f∥hp(𝒳)~inf{(∑k=0∞|λk|p)1/p+(∑k=0∞|μk|p)1/p}, where the infimum is taken over
all the above decompositions of f.Proof.
Letf∈hp(𝒳).
In order to derive the decomposition of f into atoms and blocks, we will assume for
simplicity that μ(𝒳)=∞.
The case where μ(𝒳)<∞ can be proved by using Theorems 5.34,
2.6, and 3.29 together with an argument similar to that used for the
proof of the necessity of Theorem 2.21 in [48].
By Proposition5.37, we have f−S0(f)∈Hp(𝒳) and moreover, (5.158)∥f−S0(f)∥Hp(𝒳)≲∥f∥hp(𝒳), where S0 is as in Definition 5.29. By Theorem 5.16, there
exist a sequence of numbers {λk}k=0∞⊂ℂ with ∑k=0∞|λk|p<∞ and a sequence of Hp(𝒳)-atoms {ak}k=0∞ such that in (𝒢°0ϵ(β,γ))′, (5.159)f−S0(f)=∑k=0∞λkak,(∑k=0∞|λk|p)1/p≲∥f−S0(f)∥Hp(𝒳). Let S0t denote the adjoint operator to S0 with integral kernel S0t(x,y)=S0(y,x).
Notice that if g∈𝒢0ϵ(β,γ),
then g−S0t(g)∈𝒢°0ϵ(β,γ) by Lemma 5.36, which together with (5.159) shows
that (5.160)〈f,g〉=〈2S0(f)−S0S0(f),g〉+∑k=0∞λk〈ak−S0(ak),g〉. Using Lemma 2.19, we
have (5.161)2S0(f)(x)−S0S0(f)(x)=∑τ∈I0∑ν=1N(0,τ)[2S0(f)(x)−S0S0(f)(x)]χQτ0,ν(x)≡∑τ∈I0∑ν=1N(0,τ)λτ0,νbτ0,ν(x), where (5.162)λτ0,ν≡[μ(Qτ0,ν)]1/psupz∈Qτ0,ν|2S0(f)(z)−S0S0(f)(z)|, and bτ0,ν(x)≡0 when λτ0,ν=0, otherwise, (5.163)bτ0,ν(x)≡1λτ0,ν[2S0(f)(x)−S0S0(f)(x)]χQτ0,ν(x). It is easy to see that suppbτ0,ν⊂Qτ0,ν⊂B(zτ0,ν,C62−j) and(5.164)∥bτ0,ν∥L2(𝒳)≤[μ(Qτ0,ν)]1/21λτ0,νsupz∈Qτ0,ν|2S0(f)(z)−S0S0(f)(z)|≲[μ(B(zτ0,ν,C62−j))]1/2−1/p. Thus, bτ0,ν is an hp(𝒳)-block multiplied with a normalizing constant.
Moreover, noticing that the kernel of S0S0 has the properties similar to S0 by Lemma 3.19, applying Theorem 4.16, and using
an argument similar to the proof of (5.77) in Proposition 5.25, we obtain(5.165){∑τ∈I0∑ν=1N(0,τ)|λτ0,ν|p}1/p≲{∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)[supz∈Qτ0,ν|2S0(f)(z)−S0S0(f)(z)|]p}1/p≲{∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)[mQτ0,ν(|D0(f)|)]p}1/p+∥{∑k=1∞∑τ∈Ik∑ν=1N(k,τ)[infz∈Qτk,ν|Dk(f)(z)|]2χQτk,ν}1/2∥Lp(𝒳)~∥f∥hp(𝒳). We next show that (5.166)〈2S0(f)−S0S0(f),g〉=∑τ∈I0∑ν=0N(0,τ)λτ0,ν〈bτ0,ν,g〉, which can be deduced by |〈bτ0,ν,g〉|≲1. To see this, by Hölder's inequality, Lemma 2.1(iii), and (5.30) in the proof of Proposition 5.7 together with the assumption γ>n(1/p−1),
we have (5.167)|〈bτ0,ν,g〉|≤∥bτ0,ν∥L2(𝒳)∥gχQτ0,ν∥L2(𝒳)≤[μ(Qτ0,ν)]1/2−1/p{∫Qτ0,ν[1V1(x1)+V(x1,x)1(1+d(x1,x))γ]2dμ(x)}1/2≲1[V1(x1)]1/p≲1.
Notice that by (5.168) below, we have ∥ak∥hp(𝒳)≲1.
We can obtain a desired hp(𝒳)-block decomposition of S0(ak) in the same way as 2S0(f)−S0S0(f),
which completes the proof of the necessity.
We now use Theorem5.34 to verify the sufficiency of
the condition in the theorem. To this end, by Fatou's lemma, it suffices to
prove that for any hp(𝒳)-atom or any hp(𝒳)-block b
, (5.168)∥S2,a0(b)∥Lp(𝒳)≲1, where a≥C621−j is as in Theorem 5.34. If b is an hp(𝒳)-atom, the estimate (5.168) can be established
by an argument similar to the proof of the sufficiency of Theorem 2.21 in
[48]. We omit the
details. Now, we suppose b is an hp(𝒳)-block supported on B(x0,r) with r≥C62−j.
Choose N∈ℕ such that N≥max{2,2/a} and write (5.169)∥S2,a0(b)∥Lp(𝒳)p=∫B(x0,Nar)[S2,a0(b)(x)]pdμ(x)+∫𝒳∖B(x0,Nar)…≡J1+J2. Hölder's inequality together
with the size condition of b shows that (5.170)J1≤[μ(B(x0,Nar))]1−p/2∥S2,a0(b)∥L2(𝒳)p≲[μ(B(x0,r))]1−p/2∥b∥L2(𝒳)p≲1. Now if x∈𝒳∖B(x0,Nar), d(y,x)≤a2−k with k∈ℤ+ and u∈B(x0,r),
we then have d(u,x0)≤(1/2)(2−k+d(y,x0)),
which together with the size conditions of Dk and b,
Lemma 2.1(iii), and Hölder's inequality yields that (5.171)|Dk(b)(y)|≲[μ(B(x0,r))]1−1/p1V2−k(y)+V2−k(x0)+V(y,x0)2−kϵ(2−k+d(y,x0))ϵ. From this, it follows
that (5.172)S2,a0(b)(x)≲[μ(B(x0,r))]1−1/p{∑k=0∞[1V2−k(x)+V2−k(x0)+V(x,x0)2−kϵ(2−k+d(x,x0))ϵ]2}1/2. Therefore, if we choose ϵ′∈(0,ϵ) such that p>n/(n+ϵ′),
by (5.5), we then have (5.173)J2≲1r(ϵ−ϵ′)p[μ(B(x0,r))]p−1∑k=0∞2−k(ϵ−ϵ′)p×∫𝒳∖B(x0,Nar)(1V2−k(x)+V2−k(x0)+V(x,x0))p2−kϵ′p(2−k+d(x,x0))ϵ′pdμ(x)≲1r(ϵ−ϵ′)p[μ(B(x0,r))]p−1∑k=0∞2−k(ϵ−ϵ′)p×∑l=k∞∫2l2−kNar≤d(x,x0)<2l+12−kNar(1V2−k(x)+V2−k(x0)+V(x,x0))p2−kϵ′p(2−k+d(x,x0))ϵ′pdμ(x)≲1, which completes the proof
of
Theorem 5.42.We now recall the definition of the Lipschitz spaceLips(𝒳) with s>0;
see [74].Definition 5.43.
Lets>0.
The Lipschitz (or Hölder) spaceLips(𝒳) is defined to be the set of all functions f on X such that (5.174)∥f∥Lips(𝒳)=supx∈B,rB≥1|f(x)|μ(B)s+supx≠y|f(x)−f(y)|V(x,y)s<∞, where the first supremum is
taken over all balls B of 𝒳 with radius rB≥1.Observe that these classes are rather Lipschitz (or
Hölder) classes with respect to the measure distanceρ(x,y)≡inf{μ(B):x,y∈B,Ba  ball},
not the distance d.The dual spaces ofh1(𝒳) and hp(𝒳) when p<1 are proved, respectively, to be bmo(𝒳) and Lip1/p−1(𝒳) in [74] as follows.Theorem 5.44.
(i) The spacebmo(𝒳) is the dual space of h1(𝒳), in the following sense: if f=∑k=0∞λkak+∑k=0∞μkbk∈h1(𝒳),
with atom's ak and block's bk,
is as in Theorem 5.42, then for each g∈bmo(𝒳), (5.175)limN→∞{∑k=0Nλk∫𝒳ak(x)g(x)dμ(x)+∑k=0Nμk∫𝒳bk(x)g(x)dμ(x)} is a well-defined continuous
linear functional ℒg:f↦〈f,g〉 with norm ≲∥g∥bmo(𝒳).
Conversely, each continuous linear functionalℒ on h1(𝒳) has the form ℒ=ℒg for some g∈bmo(𝒳) with ∥g∥bmo(𝒳)≲∥ℒ∥.
(ii) Assume thatϵ is as in Definition 5.29, n/(n+ϵ)<p<1, and s=1/p−1.
Then Lips(𝒳) is the dual space of hp(𝒳) in the sense of (i).Remark 5.45.
We point out that Remark5.20 applies in a similar way to Theorem 5.44.In what follows, for anyβ,γ>0,
we let (5.176)𝒢b(β,γ)={f∈𝒢(β,γ):fhasboundedsupport}.Using Proposition5.31, by an argument similar to the
proof of Proposition 5.21, we establish the following density result for Bp,qs(𝒳) and Fp,qs(𝒳).
We omit the details.Proposition 5.46.
Letϵ1,ϵ2,ϵ, and let |s|<ϵ be as in Definition 5.29. Then 𝒢b(ϵ1,ϵ2) is dense in Bp,qs(𝒳) when p(s,ϵ)<p<∞ and 0<q<∞,
and in Fp,qs(𝒳) when p(s,ϵ)<p,q<∞.We now turn to boundedness results for singular
integral operators onBp,qs(𝒳) and Fp,qs(𝒳) spaces. In what follows, it will be convenient
to put ∥f∥C˙η(𝒳)≡∥f∥L∞(𝒳) when
η=0.Letϵ>0 and σ>0.
A linear operator T,
which is initially assumed to be continuous from Cbη(𝒳) to (Cbη(𝒳))′ for all η∈(0,ϵ),
is called an inhomogeneous singular
integral of order(ϵ,σ) if T has a distributional kernel K which satisfies the conditions (I-1) through
(I-4) of the kernel of a singular integral of order ϵ in Subsection 5.2, and the following additional
“vanishing" condition that(I-2)1
the property (I-2) also holds in the limiting
caseη=η˜=0,
for T and its adjoint operator Tt,
that is, there exists a constant C>0 such that for every normalized ϵ-bump function φ, ∥Tφ∥L∞(𝒳)+∥Ttφ∥L∞(𝒳)≲C, as well as the following additional size condition that(I-3)3
for allx,y∈𝒳 with d(x,y)≥1, |K(x,y)|≤C(1/V(x,y))(1/d(x,y)σ).Remark 5.47.
We point out that ifT is a singular integral operator of order (ϵ,σ),
then T extends to a continuous linear operator from Cη(𝒳) to (𝒢b(η,γ))′ for all η∈(0,ϵ] and all γ>0;
see Proposition 2.25.
We also claim that forf∈𝒢b(η,δ) with η∈(0,ϵ] and δ>0, Tf can be defined as a distribution in (𝒢0ϵ(β,γ))′ with 0<β,γ≤ϵ.
We first define Tf as a distribution in (𝒢(β,γ))′ with 0<β≤ϵ and γ>0.
In fact, for any given β∈(0,ϵ],
noticing that 𝒢b(η1,δ)⊂𝒢b(η2,δ) when η1≥η2,
without loss of generality, we may assume that η≤β.
Assume that suppf⊂B(x0,r) for some x0∈𝒳 and r>0.
Let ψ∈Cbη(𝒳) such that ψ(x)=1 when x∈B(x0,2r) and ψ(x)=0 when x∉B(x0,4r).
For any g∈𝒢(β,γ) with η≤β≤ϵ and γ>0,
it is easy to see that ψg∈Cbη(𝒳).
From this and 𝒢b(η,δ)⊂Cbη(𝒳),
it follows that 〈Tf,ψg〉 is well defined. On the other hand, we define 〈Tf,(1−ψ)g〉 by (5.177)〈Tf,(1−ψ)g〉=∬𝒳×𝒳K(x,y)f(y)(1−ψ(x))g(x)dμ(x)dμ(y).Clearly, if suppf∩supp{(1−ψ)g}=∅,
this definition coincides with (I-1). Moreover, by (I-3)3 and Lemma 2.1(i), we have (5.178)|〈Tf,(1−ψ)g〉|≲∥f∥L∞(𝒳)∥g∥L1(𝒳)≲∥f∥𝒢(η,δ)∥g∥𝒢(β,γ). It is also easy to verify that 〈Tf,ψg〉+〈Tf,(1−ψ)g〉 is independent of the choice of ψ.
Thus, we can define Tf by 〈Tf,g〉=〈Tf,ψg〉+〈Tf,(1−ψ)g〉, so that Tf∈(𝒢°(β,γ))′ with 0<β≤ϵ and γ>0.
Now for any g∈𝒢0ϵ(β,γ) with 0<β,γ≤ϵ,
let {gn}n∈ℕ⊂𝒢(ϵ,ϵ) such that ∥gn−g∥𝒢(β,γ)→0 as n→∞.
We then define Tf∈(𝒢0ϵ(β,γ))′ with 0<β,γ≤ϵ by 〈Tf,g〉=limn→∞〈Tf,gn〉. It is easy to check that 〈Tf,g〉 is independent of the choice of {gn}n∈ℕ⊂𝒢(ϵ,ϵ).
In this sense, we have Tf∈(𝒢0ϵ(β,γ))′ with 0<β,γ≤ϵ.In what follows, forϵ∈(0,1] and 0<β,γ≤ϵ,
put (5.179)𝒢0,bϵ(β,γ)={f∈𝒢0ϵ(β,γ):fhasbounded  support}.Theorem 5.48.
Letϵ1,ϵ2, ϵ, and |s|<ϵ be as in Definition 5.29. Let σ>0 and let T be a singular integral of order (ϵ,σ) with σ>n(1/p−1)+.
Then T is bounded on Bp,qs(𝒳) when p(s,ϵ)<p<∞ and 0<q<∞ and bounded from Bp,qs(𝒳)∩𝒢b(ϵ1,ϵ2) to Bp,qs(𝒳) when max{p,q}=∞,
and T is also bounded on Fp,qs(𝒳) when p(s,ϵ)<p,q<∞,
and bounded from Fp,qs(𝒳)∩𝒢b(ϵ1,ϵ2) to Fp,qs(𝒳) when p(s,ϵ)<p<∞ and q=∞.Proof.
By Propositions5.46 and 5.31(vii) together with a density
argument, it suffices to verify the conclusions of the theorem for all f∈𝒢0,bϵ(β,γ) with β,γ as in (5.111).
Let{Sk}k∈ℤ+ be an IATI with bounded support as constructed in Theorem
2.6. Put D0=S0 and Dk=Sk−Sk−1 for k∈ℕ.
Using Remark 5.47 and Theorem 4.15 together with an argument similar to the
proof of Theorem 5.23, we see that for all k∈ℤ+ and x∈𝒳, (5.180)DkTf(x)=∑τ′∈I0∑ν′=1N(0,τ′)D¯τ′,10,ν′(f)∫Qτ′0,ν′(DkTD0(⋅,y))(x)dμ(y)+∑k′=1∞∑τ′∈Ik′∑ν′=1N(k′,τ′)μ(Qτ′k′,ν′)(DkTDk′(⋅,yτ′k′,ν′))(x)D¯k′(f)(yτ′k′,ν′), where all the notation is as in
Theorem 4.15.
Fork,k′∈ℤ+, τ′∈Ik′, ν′=1,…,N(k′,τ′), y∈Qτ′k′,ν′, and x∈𝒳,
let Z(x,y)≡(DkTDk′(⋅,y))(x). We now claim that when k=k′=0, (5.181)|Z(x,y)|≲(11+d(x,y))σ1V1(x)+V1(y)+V(x,y); when k=0 and k′∈ℕ,
for any fixed η∈(0,ϵ), (5.182)|Z(x,y)|≲2−k′η(11+d(x,y))η1V1(x)+V1(y)+V(x,y); when k∈ℕ and k′=0,
for any fixed η∈(0,ϵ), (5.183)|Z(x,y)|≲2−kη(11+d(x,y))η1V1(x)+V1(y)+V(x,y); and when k,k′∈ℕ,
for any fixed η∈(0,ϵ), (5.184)|Z(x,y)|≲2−|k−k′|η(2−(k⋀k′)2−(k⋀k′)+d(x,y))η1V2−(k⋀k′)(x)+V2−(k⋀k′)(y)+V(x,y).
Obviously, (5.184) is just (5.65). The estimate (5.183)
is easily deduced from (5.182) by symmetry. Thus, we only need to verify (5.181)
and (5.182). Assume that k′∈ℤ+.
To prove (5.181) and (5.182), it suffices to prove that (5.185)|Z(x,y)|≲2−k′η11V1(x), and that when d(x,y)≥25,
then (5.186)|Z(x,y)|≲(2−k′d(x,y))η21V(x,y), where when k′=0, η1=0, and η2=σ,
and when k′∈ℕ, η1=η2=η.
To see (5.185), for z∈𝒳,
set Ψ(z)≡∫𝒳K(w,z)D0(w,x)dμ(w). From the properties of S0 in Theorem 2.6, Condition (I-2), Condition
(I-4), and Condition (I-2)1, it follows that for any fixed η∈[0,ϵ), ∥Ψ∥C˙η(𝒳)≲1/V1(x), which shows that for all z∈𝒳, |Ψ(z)|≲1/V1(x), and for η∈(0,ϵ) and all z,y∈𝒳, (5.187)|Ψ(z)−Ψ(y)|≲d(z,y)η1V1(x).Thus, when k′=0,
by size conditions of both Ψ and S0,
then (5.188)|Z(x,y)|=|∫𝒳Ψ(z)S0(z,y)dμ(z)|≲1V1(x), while when k′∈ℕ,
by the vanishing moment and the size condition of Dk′ together with the regularity of Ψ,
for any fixed η∈(0,ϵ), (5.189)|Z(x,y)|=|∫𝒳[Ψ(z)−Ψ(y)]Dk′(z,y)dμ(z)|≲2−k′η1V1(x),which verifies (5.185).
To see (5.186), we first notice that when d(z,y)<22−k′, d(w,x)<22, and d(x,y)≥25,
then d(w,z)≥d(x,y)−d(w,x)−d(z,y)>max{d(x,y)/2,8,8d(z,y)}, d(w,y)>d(x,y)/2, and d(w,z)≥d(w,y)−d(z,y)>d(w,y)/2.
Therefore, (5.190)V(w,z)≳V(w,y)≳V(y,x)~V(x,y).Thus, if k′=0,
by (I-3)3, we have (5.191)|Z(x,y)|≲∬𝒳×𝒳|D0(w,x)|1V(w,z)1d(z,y)σ|D0(z,y)|dμ(w)dμ(z)≲1V(x,y)1d(x,y)σ, while when k′∈ℕ,
by the regularity on K and the vanishing moment of Dk′,
for any fixed η∈(0,ϵ),
we have (5.192)|Z(x,y)|≲∬𝒳×𝒳|D0(w,x)|d(z,y)ϵV(w,z)d(w,z)ϵ|Dk′(z,y)|dμ(w)dμ(z)≲(2−k′d(x,y))ϵ1V(x,y),which implies (5.186). Thus, the
estimates (5.181) and (5.182) hold.
Using these estimates and Remark5.26, by a procedure
essentially similar to the proof of Proposition 5.25, we then obtain the
boundedness on Bp,qs(𝒳) and Fp,qs(𝒳) of T.
The details are left to the reader.


### 5.5.T(1)-Theorems

In this subsection, we will establish analogues of David
and Journé'sT(1)-theorem [68] and of Stein's variant of this theorem in [75] for RD-spaces.We begin with generalizing [71, Theorem 1, page 114] on ℝn to the setting of spaces of homogeneous type
which is of independent of interest; see also [69, Proposition 2].Theorem 5.49.
Letϵ∈(0,1], β∈(0,ϵ), and let T be as in Proposition 2.12 with a distributional
kernel K satisfying the size condition (2.59). Then T can be extended as a continuous linear
operator on C˙β(𝒳) if and only if T∈WBP(β) and T(1)=0 in (C°bβ(𝒳))′.Proof.
We first prove the sufficiency. To this end, for anyk1∈ℤ,
let (5.193)Bk1={x∈𝒳:2k1−1≤d(x,x1)<2k1}. Fix any x0∈Bk1.
Let θ be as in Lemma 2.15. For any k2∈ℤ,
put θk2(y)=θ(d(x0,y)/3⋅2k2), and define ωk2=1−θk2.
For any f∈Cβ(𝒳),
following an argument as in the proof of Lemma 2.20 (see also [69, the proof of Proposition 2]), for a.e.x∈B(x0,(3/2)2k2),
we have (5.194)Tf(x)=∫𝒳K(x,y)[f(y)−f(x)]θk2(y)dμ(y)+[∫𝒳K(x,y)f(y)ωk2(y)dμ(y)+f(x)T(θk2)(x)]=Γ1(x)+Γ2(x). By Lemma 2.15, for a.e.x∈B(x0,(3/2)2k2),
we also have (5.195)|T(θk2)(x)|≲CT+∥T∥WBP(β),(5.196)T(θk2)(x)=Ck2−∫𝒳[K(x,y)−K(x0,y)]ωk2(y)dμ(y), where Ck2 is a constant independent of x.
For anyx∈B(x0,2k2/2),
we then consider x′∈𝒳 satisfying 2k2−1≤d(x,x′)<2k2.
Notice that if x∈B(x0,2k2/2) and d(x,x′)<2k2,
then x′∈B(x0,(3/2)2k2),
and that θk2(y)≠0 implies d(x,y)<14d(x,x′).
Thus, if x,x′ satisfy (5.194), (5.195), and (5.196), by the
size condition (2.59) of the kernel K,
the definition of C˙β(𝒳) and Lemma 2.1(i), we then have (5.197)|Γ1(x)|≤∫d(x,y)<14d(x,x′)|K(x,y)||f(y)−f(x)|dμ(y)≲d(x,x′)βCT∥f∥C˙β(𝒳) and a similar estimate also
holds for Γ1(x′),
which clearly implies (5.198)|Γ1(x)−Γ1(x′)|≤∥f∥C˙β(𝒳)CTd(x,x′)β.
Moreover, by (5.196), we also have (5.199)Γ2(x)−Γ2(x′)=∫𝒳[K(x,y)−K(x′,y)][f(y)−f(x)]ωk2(y)dμ(y)+[f(x)−f(x′)]∫𝒳K(x′,y)θk2(y)dμ(y)=Γ2,1+Γ2,2. The estimate (5.195) yields (5.200)|Γ2,2|≲d(x,x′)β(CT+∥T∥BMO(β))∥f∥C˙β(𝒳). Notice that ωk2(y)≠0 implies that d(x,y)>2d(x,x′).
The regularity (2.49) on K and the definition of C˙β(𝒳) together with Lemma 2.1(i) and β<ϵ then yield (5.201)|Γ2,1|≲CT∥f∥C˙β(𝒳)∫d(x,y)>2d(x,x′)d(x,x′)ϵV(x,y)d(x,y)ϵd(x,y)βdμ(y)≲d(x,x′)βCT∥f∥C˙β(𝒳). Combining all the above
estimates shows that for a.e.x∈B(x0,2k2/4) and a.e.x′∈𝒳 satisfying 2k2−1≤d(x,x′)<2k2,
we have (5.202)|Tf(x)−Tf(x′)|≲d(x,x′)β(CT+∥T∥WBP(β))∥f∥C˙β(𝒳). Then an argument via the
Besicovitch covering lemma further shows that there exists an extension of Tf such that (5.203)∥Tf∥C˙β(𝒳)≲(CT+∥T∥WBP(β))∥f∥C˙β(𝒳), which completes the proof of the
sufficiency.
We now prove the necessity. SinceT is extended as a continuous linear operator on C˙β(𝒳) and 1=0 in C˙β(𝒳).
Thus, T(1)=0 in C˙β(𝒳).
Since C˙β(𝒳)=B˙∞,∞β(𝒳)=(B˙1,1−β(𝒳))′ by Theorems 6.11 and 8.11(i), and C°bβ(𝒳)⊂B˙1,1−β(𝒳) by Proposition 5.10(iv), we then have that for
all f∈C°bβ(𝒳), 〈T(1),f〉=0,
which just means that T(1)=0 in (C°bβ(𝒳))′.
Let nowϕ,ψ∈Cbβ(𝒳) as in Definition 2.13. Fix x0′∈𝒳 such that 2r≤d(x0′,x0)<3r.
Since x0′∉suppϕ,
by the size condition (2.59) of the kernel K and (2.55), we then have  (5.204)|T(ϕ)(x0′)|=|∫B(x0,r)K(x0′,y)ϕ(y)dμ(y)|≲1Vr(x0′)∥ϕ∥L∞(𝒳)μ(B(x0,r))≲1. Since T is bounded on C˙β(𝒳),
when d(x,x0′)≤5r,
we then have (5.205)|T(ϕ)(x)|≲|T(ϕ)(x0′)|+d(x,x0′)β∥ϕ∥C˙β(𝒳)≲1. When d(x,x0′)>5r,
by the size condition (2.59) of the kernel K and (2.55) again, we also have (5.206)|T(ϕ)(x)|=|∫B(x0,r)K(x,y)ϕ(y)dμ(y)|≲1V(x,x0′)∥ϕ∥L∞(𝒳)μ(B(x0,r))≲1. Thus, ∥T(ϕ)∥L∞(𝒳)≲1, which gives that |〈T(ϕ),ψ〉|≲∥ψ∥L1(𝒳)≲μ(B(x0,r)). That is, T∈WBP(β),
which completes the proof of Theorem 5.49.Remark 5.50.
(i) The proof of Theorem5.49 in combination with Corollary 2.23 shows
that if T is bounded on C˙β(𝒳),
then there exists a constant C>0 such that for all ϕ∈C˙bβ(𝒳) and all x∈𝒳, (5.207)|T(ϕ)(x)|≤C(CT+∥T∥C˙β(𝒳)→C˙β(𝒳))[diam(suppϕ)]β∥ϕ∥C˙β(𝒳).
(ii) LetT be as in Theorem 5.49 and T(1)=0 in (C°bβ(𝒳))′.
Then from (i) and Theorem 5.49, it is easy to see that T∈WBP(β) if and only if (5.207) holds.Now we recall the notion of Carleson measures and
establish their connection withBMO(𝒳) functions.Definition 5.51.
A positive measureν on 𝒳×(0,∞) is said to be a Carleson measure if there exists a
constant C>0 such that for every ball B(x,r) for some x∈𝒳 and r>0, (5.208)ν(B(x,r)×(0,r))≤Cμ(B(x,r)). The smallest bound C as above is defined to be the Carleson norm of ν and is denoted by ∥ν∥ℭ.For any given open setE of 𝒳,
let (5.209)E^={(x,t)∈𝒳×(0,∞):B(x,t)⊂E}.We first establish a basic property of Carleson
measures; see [75, pages 59-60] or [81, page 198] for the case of ℝn.Lemma 5.52.
Ifν is a Carleson measure in 𝒳×(0,∞) and E⊂𝒳 is open, then ν(E^)≤C∥ν∥ℭμ(E).Proof.
Without loss of generality, we may assume
thatE is a bounded open set of 𝒳.
By the Whitney-type covering lemma (see [28, Theorem (3.2)]), there exist a constant C>0 and a sequence of balls {B(yj,rj)}j satisfying E=∪jB(yj,rj), ∑jχB(yj,rj)≤C and B(yj,3rj)∩(𝒳∖E)≠∅ for each j.
If (x,t)∈E^,
then there exists j0 such that x∈B(yj0,rj0), B(x,t)⊂E, and B(yj0,3rj0)∩(𝒳∖E)≠∅.
From this, it follows that t<6rj0,
and therefore, (x,t)∈B(yj0,rj0)×(0,6rj0).
Thus, E^⊂∪j{B(yj,rj)×(0,6rj)}, which together with the definition of Carleson
measures implies that (5.210)ν(E^)≤∑jν(B(yj,rj)×(0,6rj))≲∥ν∥ℭ∑jμ(B(yj,rj)). On the other hand, from E=∪jB(yj,rj) and ∑jχB(yj,rj)≤C,
it follows that (5.211)μ(E)=μ(⋃jB(yj,rj))≳∫𝒳∑jχB(yj,rj)(y)dμ(y)~∑jμ(B(yj,rj)). Combining both estimates yields ν(E^)≲∥ν∥ℭμ(E), which completes the proof of Lemma 5.52.Let{Sk}k∈ℤ be an (ϵ1,ϵ2,ϵ3)-ATI as in Definition 2.2. For (x,t)∈𝒳×(0,∞),
we define (5.212)𝔖(f)(x,t)=∑j=−∞∞Sj(f)(x)χ(2−j−1,2−j](t).Proposition 5.53.
Let𝔖 be as in (5.212). For any p∈(1,∞),
there exists a constant Cp>0 such that for all f∈Lp(𝒳) and all Carleson measures ν, (5.213)∫𝒳×(0,∞)|𝔖(f)(x,t)|pdν(x,t)≤Cp∥ν∥ℭ∫𝒳|f(x)|pdμ(x).Proof.
Let𝔖 be as in above. For any x∈𝒳,
we define (5.214)M𝔖f(x)=sup{|𝔖(f)(y,t)|:d(x,y)<t}. We first claim that for all x∈𝒳, M𝔖f(x)≲Mf(x), where M is the Hardy-Littlewood maximal function. To
see this, for any (y,t)∈𝒳×(0,∞) satisfying d(x,y)<t,
assuming 2−j0−1<t≤2−j0 for some j0∈ℤ,
by Lemma 2.1(vi), we then have (5.215)|𝔖(f)(y,t)|≲∫d(y,z)<2−j01V2−j0(y)+V2−j0(z)+V(y,z)(2−j02−j0+d(y,z))ϵ2|f(z)|dμ(z)+∑l=1∞∫2l−12−j0≤d(y,z)<2l2−j0⋯≲Mf(x), which implies the claim.
Notice that(5.216)∫𝒳×(0,∞)|𝔖(f)(x,t)|pdν(x,t)=p∫0∞λp−1ν({(x,t)∈𝒳×(0,∞):|𝔖(f)(x,t)|>λ})dλ. Let Eλ={x∈𝒳:M𝔖f(x)>λ}. We then claim (5.217){(x,t)∈𝒳×(0,∞):|𝔖(f)(x,t)|>λ}⊂E^λ. In fact, for any (x,t)∈𝒳×(0,∞) such that |𝔖(f)(x,t)|>λ,
assume that 2−j0−1<t≤2−j0 for some j0∈ℤ.
Then |𝔖(f)(x,t)|>λ if and only if |Sj0(f)(x)|>λ.
If d(y,x)<t,
then M𝔖f(y)>λ and hence B(x,t)⊂Eλ,
which implies the claim.
From this claim and Lemma5.52, it follows that (5.218)ν({(x,t)∈𝒳×(0,∞):|𝔖(f)(x,t)|>λ})≤ν(E^λ)≲∥ν∥ℭμ(Eλ), which together with the Lp(𝒳)-boundedness for p∈(1,∞) of M yields (5.219)∫𝒳×(0,∞)|𝔖(f)(x,t)|pdν(x,t)≲∥ν∥ℭ∫0∞λp−1μ(Eλ)dλ~∥ν∥ℭ∫𝒳|M𝔖(f)(x)|pdμ(x)≲∥ν∥ℭ∫𝒳|M(f)(x)|pdμ(x)≲∥ν∥ℭ∫𝒳|f(x)|pdμ(x), which completes the proof of
Proposition 5.53.The relation between Carleson measures andBMO(𝒳) functions can be stated as below.Proposition 5.54.
Letb∈BMO(𝒳) and let {Sk}k∈ℤ be an (ϵ1,ϵ2,ϵ3)-ATI as in Definition 2.2. For k∈ℤ,
let Dk=Sk−Sk−1.
Then the measure ν defined by (5.220)dν(x,t)=∑j=−∞∞|Dj(b)(x)|2χ(2−j−1,2−j](t)dμ(x)dtt is a Carleson measure such that ∥ν∥ℭ is dominated by ∥b∥BMO(𝒳)2.Proof.
For any ballB=B(x0,r) with some x0∈𝒳 and r>0,
assume that 2−j0−1<r≤2−j0 for some j0∈ℤ.
We then have (5.221)ν(B(x0,r)×(0,r))=∫B(x0,r)∫0rdν(x,t)=∑j=−∞∞∫B(x0,r)∫0r|Dj(b)(x)|2χ(2−j−1,2−j](t)dμ(x)dtt≲∑j=j0−1∞∫B(x0,r)|Dj(b)(x)|2dμ(x). Let B˜=B(x0,3r), B˜0=B˜, and B˜k=B(x0,2k3r) for k∈ℕ.
Set also (5.222)bB˜=1μ(B˜)∫B˜b(y)dμ(y). Since ∫𝒳Dj(x,y)dμ(y)=0,
we then have (5.223)ν(B(x0,r)×(0,r))≲∑j=j0−1∞∫B(x0,r)|Dj((b−bB˜)χB˜)(x)|2dμ(x)+∑j=j0−1∞∫B(x0,r)|Dj((b−bB˜)χ𝒳∖B˜)(x)|2dμ(x)≡Y1+Y2. By Lemma 3.9, we
have (5.224)Y1≲∑j=j0−1∞∫𝒳|Dj((b−bB˜)χB˜)(x)|2dμ(x)≲∫𝒳|((b(x)−bB˜)χB˜(x))|2dμ(x)≲∥b∥BMO(𝒳)2μ(B(x0,r)). To estimate Y2,
we first notice that if d(x,x0)<r and y∉B˜k for k∈ℤ+,
then d(y,x)≥2k3r−r>2k−13r and hence (5.225)|Dj(x,y)|≲1V2−j(x)+V2−j(y)+V(x,y)(2−j2−j+d(x,y))ϵ2≲1V2k+23r(x)(2−j2kr)ϵ2. From this, it follows that (5.226)Y2≲∑j=j0−1∞∫B(x0,r)[∑k=0∞∫B˜k+1∖B˜k|b(y)−bB˜||Dj(x,y)|dμ(y)]2dμ(x)≲1r2ϵ2∑j=j0−1∞2−2jϵ2∫B(x0,r)[∑k=0∞12kϵ21V2k+23r(x)∫B(x,2k+23r)|b(y)−bB˜|dμ(y)]2dμ(x)≲∥b∥BMO(𝒳)2μ(B(x0,r)), where we used the well-known
fact that (5.227)∥b∥BMO(𝒳)~infC∈ℂsupx∈𝒳,r>01μ(B(x,r))∫B(x,r)|f(y)−C|dμ(y). This finishes the proof of
Proposition 5.54.Combining Proposition5.53 with Proposition 5.54
yields the following conclusion, which will be used in the proof of T(1)-theorem.Corollary 5.55.
Let
all the notation be as in Propositions5.53 and 5.54. For any p∈(1,∞),
there exists a constant Cp>0 such that for all f∈Lp(𝒳) and b∈BMO(𝒳), (5.228)∑j=−∞∞∫𝒳|Sj(f)(x)|p|Dj(b)(x)|2dμ(x)≤Cp∥b∥BMO(𝒳)2∫𝒳|f(x)|pdμ(x).Proof.
From Propositions5.53 and 5.54,
it follows that (5.229)∫𝒳×(0,∞)|𝔖(f)(x,t)|pdν(x,t)≲∥b∥BMO(𝒳)2∫𝒳|f(x)|pdμ(x), where 𝔖(f) is as in Proposition 5.53 and dν(x,t) is as in  Proposition 5.54. Moreover,  we
have (5.230)∫𝒳×(0,∞)|𝔖(f)(x,t)|pdν(x,t)=∑k=−∞∞∫𝒳∫2−k−12−k|𝔖(f)(x,t)|pdν(x,t)=∑k=−∞∞∫𝒳|Sk(f)(x)|p∫0∞χ(2−k−1,2−k](t)(∑j=−∞∞|Dj(b)(x)|2χ(2−j−1,2−j](t))dμ(x)dtt=log2{∑k=−∞∞∫𝒳|Sk(f)(x)|p|Dk(b)(x)|2dμ(x)}. Combining both estimates
completes the proof of Corollary 5.55.We now can state a variant ofDavid-JournéT(1)-theorem on spaces of homogeneous type. In
what follows, for any ϵ∈(0,1],
a continuous function on 𝒳×𝒳∖{(x,x):x∈𝒳} is said to be a standard kernel of orderϵ if it satisfies (I-3) in Subsection 5.2. Let β∈(0,ϵ).
A continuous linear operator T from Cbβ(𝒳) to (Cbβ(𝒳))′ is said to have a standard distributional kernelKof orderϵ if T and K satisfy (2.48). Also, the adjoint T∗ of T is given by that for all f,g∈Cbβ(𝒳), (Tf,g)=(f,T∗g). Then T∗:Cbβ(𝒳)→(Cbβ(𝒳))′ is a continuous mapping; moreover, T∗ is associated to the kernel K∗(x,y)=K¯(y,x) for all x,y∈𝒳.Theorem 5.56.
Letϵ∈(0,1], β∈(0,ϵ), and let T be a continuous linear operator from Cbβ(𝒳) to (Cbβ(𝒳))′.
Assume that T has a standard distributional kernel K of order ϵ as in (2.48). Then T extends to a bounded operator on L2(𝒳) if and only if the following conditions are
true:
(i)
T(1)∈BMO(𝒳),(ii)
T∗(1)∈BMO(𝒳),(iii)
T∈WBP(β).Proof.
We first verify the sufficiency. By Proposition2.12, both T and T∗ can be extended to a continuous linear
operator from Cβ(𝒳) to (C°bβ(𝒳))′.
If T(1)=T∗(1)=0 in (C°bβ(𝒳))′,
then by Theorem 5.49, T and T∗ can be extended as continuous linear operators
on C˙β(𝒳).
By this fact, Theorem 5.23 and Proposition 5.10(v), we know that T extends to a bounded operator on L2(𝒳).
We now consider the general case. Let{Sj}j∈ℤ be as in Definition 2.2 and Dj=Sj−Sj−1 for j∈ℤ.
Let {D˜j}j∈ℤ be as in Theorem 3.10. Let b∈BMO(𝒳).
For any f∈Cbβ(𝒳),
we define the paraproduct(5.231)Pb(f)(x)=∑j=−∞∞D˜j(Dj(b)Sj(f))(x). We will show that the kernel of Pb is a standard kernel of order ϵ,
that Pb is bounded on L2(𝒳),
and that Pb(1)=b and Pb∗(1)=0.
In order to be rigorous in the following calculations, we should consider ∑j=−NN instead of ∑j=−∞∞ and then let N→∞.
However, we will omit these details; see [75, pages 302–305].
(1) The size of the kernel. The kernel ofPb is (5.232)K(x,y)=∑j=−∞∞∫𝒳D˜j(x,z)Dj(b)(z)Sj(z,y)dμ(z). Let Bk=B(z,2k+12−j), for k∈ℤ+,
and bB0=(1/μ(B0))∫B0b(z)dμ(z).
Since (5.233)∫𝒳Dj(z,y)dμ(y)=0, by the size condition of Dj and (5.227), for any z∈𝒳,
we have (5.234)|Dj(b)(z)|≤∫B0|Dj(z,y)||b(y)−bB0|dμ(y)+∑k=1∞∫Bk∖Bk−1|Dj(z,y)||b(y)−bB0|dμ(y)≲∑k=0∞12kϵ21μ(Bk)∫Bk|b(y)−bB0|dμ(y)≲∥b∥BMO(𝒳). Let ϵ′∈(ϵ,ϵ1⋀ϵ2).
By the size conditions of D˜j and Sj,
(5.234), Lemmas 4.4, and 3.5, we have (5.235)|K(x,y)|≲∥b∥BMO(𝒳)∑j=−∞∞∫𝒳1V2−j(x)+V2−j(z)+V(x,z)(2−j2−j+d(x,z))ϵ′×1V2−j(z)+V2−j(y)+V(z,y)(2−j2−j+d(z,y))ϵ2dμ(z)≲∥b∥BMO(𝒳)∑j=−∞∞1V2−j(x)+V2−j(y)+V(x,y)(2−j2−j+d(x,y))ϵ′≲∥b∥BMO(𝒳)1V(x,y), which verifies the size
condition of the kernel K.
Ifd(x,x′)≤d(x,y)/2 with x≠y,
by (5.234),
(5.236)|K(x,y)−K(x′,y)|=|∑j=−∞∞∫𝒳[D˜j(x,z)−D˜j(x′,z)]Dj(b)(z)Sj(z,y)dμ(z)|≲∥b∥BMO(𝒳)∑j=−∞∞{∫d(x,x′)≤(2−j+d(x,z))/2|D˜j(x,z)−D˜j(x′,z)||Sj(z,y)|dμ(z)+∫d(x,x′)>(2−j+d(x,z))/2|D˜j(x,z)||Sj(z,y)|dμ(z)+∫d(x,x′)>(2−j+d(x,z))/2|D˜j(x′,z)||Sj(z,y)|dμ(z)}≡∥b∥BMO(𝒳){Y1+Y2+Y3}. The regularity of D˜j and the size condition of Sj together with Lemmas 4.4 and 3.5
give (5.237)Y1≲∑j=−∞∞∫𝒳(d(x,x′)2−j+d(x,z))ϵ′1V2−j(x)+V2−j(z)+V(x,z)(2−j2−j+d(x,z))ϵ′×1V2−j(z)+V2−j(y)+V(z,y)(2−j2−j+d(z,y))ϵ2dμ(z)≲∑j=−∞∞(d(x,x′)2−j+d(x,y))ϵ1V2−j(x)+V2−j(y)+V(x,y)(2−j2−j+d(x,y))ϵ′−ϵ≲d(x,x′)ϵV(x,y)d(x,y)ϵ. The size conditions of D˜j and Sj together with Lemmas 4.4 and 3.5 also
yield (5.238)Y2≲∑j=−∞∞∫d(x,x′)>(2−j+d(x,z))/21V2−j(x)+V2−j(z)+V(x,z)(2−j2−j+d(x,z))ϵ′×1V2−j(z)+V2−j(y)+V(z,y)(2−j2−j+d(z,y))ϵ2dμ(z)≲∑j=−∞∞1V2−j(x)+V2−j(y)+V(x,y)d(x,x′)ϵ2−j(ϵ′−ϵ)(2−j+d(x,y))ϵ′≲d(x,x′)ϵV(x,y)d(x,y)ϵ. Similarly, we have (5.239)Y3≲d(x,x′)ϵV(x,y)d(x,y)ϵ, which shows that K has the desired regularity on the first
variable.
An argument similar to above also proves that ifd(y,y′)≤d(x,y)/2 with x≠y, (5.240)|K(x,y)−K(x,y′)|≲∥b∥BMO(𝒳)d(y,y′)ϵV(x,y)d(x,y)ϵ. Thus, K is a standard kernel of order ϵ.
(2) Boundedness ofPb on L2(𝒳).
For any f,g∈L2(𝒳),
by Hölder's inequality, Corollary 5.55, and Lemma 3.9 together with Remark 3.16,
we have (5.241)|〈Pb(f),g〉|=|∑j=−∞∞∫𝒳Dj(b)(x)Sj(f)(x)D˜j∗(g)(x)dμ(x)|≲{∑j=−∞∞∫𝒳|Dj(b)(x)|2|Sj(f)(x)|2dμ(x)}1/2{∑j=−∞∞∫𝒳|D˜jt(g)(x)|2dμ(x)}1/2≲∥b∥BMO(𝒳)∥f∥L2(𝒳)∥g∥L2(𝒳), which together with a duality
argument yields that (5.242)∥Pb(f)∥L2(𝒳)≲∥b∥BMO(𝒳)∥f∥L2(𝒳).
(3)Pb(1)=b and Pb∗(1)=0 in (C°bβ(𝒳))′.
Since ∫𝒳D˜j(x,z)dμ(x)=0,
from this, it follows that Pb∗(1)=0 in (C°bβ(𝒳))′.
Also, since ∫𝒳Sj(y,z)dμ(z)=1,
from this and Theorem 3.29, it follows that Pb(1)=b.
We can now finish the proof of the sufficiency. For
any given operatorT which satisfies (i), (ii), and (iii), let b1=T(1) and b2=T∗(1).
Then, there exist paraproducts Pbi such that Pbi(1)=bi and Pbi∗(1)=0 for i=1,2.
Then the operator T˜=T−Pb1−Pb2∗ lies in WBP(β) and T˜(1)=T˜∗(1)=0.
Thus, by Theorem 5.23, we know that T˜ is bounded on L2(𝒳),
which together with the boundedness of Pb1 and Pb2∗ on L2(𝒳) also yields the boundedness of T on L2(𝒳).
This completes the proof of the sufficiency.
We now check the necessity. By Remark2.14(iii), we
know T∈WBP(β).
To verify T(1),T∗(1)∈BMO(𝒳),
we first claim that if T is as in Proposition 2.12 and T is bounded on L2(𝒳),
then T is also bounded from Lb∞(𝒳) to BMO(𝒳),
namely, for all f∈Lb∞(𝒳), (5.243)∥T(f)∥BMO(𝒳)≲∥f∥L∞(𝒳). The proof of this claim is
standard; see, for example, [75, pages 156-157] or [81, pages 118-119]. For any ball B=B(x0,r) with some x0∈𝒳 and r>0,
let CB=T(fχ𝒳∖B(x0,2r))(x0).
Since f∈Lb∞(𝒳) and K is locally integrable away from the diagonal
of 𝒳×𝒳,
it is easy to see that |CB|<∞.
By the boundedness of T on L2(𝒳) and (2.49) together with Lemma 2.1(i), we have(5.244)1μ(B)∫B|T(f)(x)−CB|dμ(x)≤1μ(B)∫B|T(fχB(x0,2r))(x)|dμ(x)+1μ(B)∫B|T(fχ𝒳∖B(x0,2r))(x)−CB|dμ(x)≲∥T∥L2(𝒳)→L2(𝒳)∥f∥L∞(𝒳)+1μ(B)∫B∫𝒳∖B(x0,2r)|K(x,y)−K(x0,y)||f(y)|dμ(y)dμ(x)≲(∥T∥L2(𝒳)→L2(𝒳)+CT)∥f∥L∞(𝒳), which proves (5.243).
Using (5.243), we then can verify that if T is bounded on L2(𝒳),
then T(1)∈BMO(𝒳).
To see this, for any g∈C°bβ(𝒳) with suppg⊂B(x0,r) for some x0∈𝒳 and r>0,
let ψ∈Cbβ(𝒳) be as in the proof of Proposition 2.12. By
(5.243), T(ψ)∈BMO(𝒳),
which together with Theorem 5.19(i) below yields that (5.245)|〈T(ψ),g〉|≤∥T(ψ)∥BMO(𝒳)∥g∥H1(𝒳)≲(∥T∥L2(𝒳)→L2(𝒳)+CT)∥g∥H1(𝒳). On the other hand, by (2.49) and
Lemma 2.1(i), we also have (5.246)|〈T(1−ψ),g〉|=|∫𝒳∫d(y,x0)≥2r[K(x,y)−K(x0,y)](1−ψ(y))g(x)dμ(y)dμ(x)|≲CT∫𝒳{∫d(y,x0)≥2rrϵV(x0,y)d(x0,y)ϵdμ(y)}|g(x)|dμ(x)≲CT∥g∥L1(𝒳)≲CT∥g∥H1(𝒳). Thus, |〈T(1),g〉|≲(∥T∥L2(𝒳)→L2(𝒳)+CT)∥g∥H1(𝒳), which together with Corollary 2.11(i), Proposition 5.21, Theorems 6.11,
and 5.19(i) implies that T(1)∈BMO(𝒳) and ∥T(1)∥BMO(𝒳)≲∥T∥L2(𝒳)→L2(𝒳)+CT. An argument similar to this also proves that T∗(1)∈BMO(𝒳) and ∥T∗(1)∥BMO(𝒳)≲∥T∥L2(𝒳)→L2(𝒳)+CT, which completes the proof of Theorem 5.56.We now state a variant of theT(1)-theorem in the sense of Stein [75, page 294]. Let ϵ∈(0,1], β∈(0,ϵ), and let T be a continuous linear operator from Cbβ(𝒳) to (Cbβ(𝒳))′.
We assume that associated to T,
there is a standard kernel of order ϵ,
in the sense that if f∈Cbβ(𝒳),
then, outside the support of f,
the distribution Tf agrees with the function (5.247)T(f)(x)=∫𝒳K(x,y)f(y)dμ(y). Similarly to [75, page 294], we assume that T and T∗ are restrictedly bounded. Whenever ϕR,x0 is a normalized bump function for the ball B(x0,R) with some x0∈𝒳 and R>0,
the distributions T(ϕR,x0) and T∗(ϕR,x0) belong to L2(𝒳),
and the estimates (5.248)∥T(ϕR,x0)∥L2(𝒳)≤A[μ(B(x0,R))]1/2,(5.249)∥T∗(ϕR,x0)∥L2(𝒳)≤A[μ(B(x0,R))]1/2 hold with an A>0 that is independent of R, x0 and ϕR,x0.Theorem 5.57.
Letϵ∈(0,1], β∈(0,ϵ), and let T be a continuous linear operator from Cbβ(𝒳) to (Cbβ(𝒳))′ associated with a standard kernel of order ϵ in the sense of (5.247). Then T extends to a bounded linear operator on L2(𝒳) if and only if both T and T∗ are restrictedly bounded in the sense of
(5.248) and (5.249).Proof.
The necessity is obvious. We only need to
prove the sufficiency. We first make the following claim that iff∈C°bβ(𝒳),
then (5.250)Tf∈L1(𝒳). In fact, assume that suppf⊂B(x0,r) for some x0∈𝒳 and r>0.
Since f is a multiple of a bump function, Tf∈L2(𝒳) by (5.248), and hence ∫B(x0,2r)|Tf(x)|dμ(x)<∞.
If x∉B(x0,2r),
then by (5.247), ∫𝒳f(x)dμ(x)=0, and the regularity on K,
we have (5.251)|Tf(x)|=|∫𝒳K(x,y)f(y)dμ(y)|=|∫𝒳[K(x,y)−K(x,x0)]f(y)dμ(y)|≲CT∫B(x0,r)d(y,x0)ϵV(x,x0)d(x,x0)ϵ|f(y)|dμ(y)≲CT∥f∥L∞(𝒳)μ(B(x0,r))rϵV(x,x0)d(x,x0)ϵ, which implies that ∫𝒳∖B(x0,2r)|Tf(x)|dμ(x)<∞.
Thus, the claim  (5.250) holds.
We now verify thatT(1)∈BMO(𝒳).
To this end, we first prove that there exists a constant A>0 such that whenever ϕR,x0 is a normalized bump function for the ball B(x0,R) with x0∈𝒳 and R>0,
then T(ϕR,x0)∈BMO(𝒳) with (5.252)∥T(ϕR,x0)∥BMO(𝒳)≤A. Let B˜1=B(x˜0,R˜) with some x˜0∈𝒳 and R˜>0 be any ball, and let B˜2=B(x˜0,2R˜) and B˜3=B(x˜0,3R˜).
Fix a function θ∈Cbβ(𝒳) with θ(x)=1 for d(x,x˜0)≤2R˜ and θ(x)=0 for d(x,x˜0)≥3R˜.
Write (5.253)ϕR,x0(x)=ϕR,x0(x)θ(x)+ϕR,x0(x)(1−θ(x))=f1(x)+f2(x). Observe that f1 is, up to a bounded multiplicative constant, a
normalized bump function for either the ball B(x0,R) or the ball B˜3,
whichever has the smaller radius. Thus, by (5.248), we have (5.254)∫B˜1|Tf1(x)|2dμ(x)≤∥Tf1∥L2(𝒳)2≤A2min{μ(B(x0,R)),μ(B(x˜0,3R˜))}≤A′μ(B(x˜0,R˜)). Since suppf2⊂(𝒳∖B˜2),
for x∈B˜1,
by (5.247), we have (5.255)Tf2(x)=∫𝒳K(x,y)f2(y)dμ(y). Let CB˜1=∫𝒳K(x˜0,y)f2(y)dμ(y).
Then |CB˜1|<∞,
and for x∈B˜1,
by the regularity on K on the first variable and Lemma 2.1(i), (5.256)|Tf2(x)−CB˜1|≲∫d(x˜0,y)>2R˜|K(x,y)−K(x˜0,y)|dμ(y)≲∫d(x˜0,y)>2R˜d(x,x˜0)ϵV(x˜0,y)d(x˜0,y)ϵdμ(y)≲1. Combining the estimates for f1 and f2 gives that for any ball B˜1, (5.257)∫B˜1|T(ϕR,x0)−CB˜1|2dμ(x)≤Aμ(B˜1), which shows (5.252).
Letθ∈Cb1(ℝ) and θ(0)=1.
For any ν>0 and x∈𝒳,
set θν(x)=θ(νd(x,x1)).
By (5.252), {T(θν)}ν>0 is uniformly bounded in BMO(𝒳).
Since BMO(𝒳)=(H1(𝒳))′ (see Theorem 5.19(i)), as is well known, BMO(𝒳) is weakly* compact in the dual topology. Thus, every
sequence {T(θνk)}k∈ℕ has a subsequence which weakly* converges. Let f∈C°bβ(𝒳).
By (5.250) and 〈T(θν),f〉=〈θν,T∗(f)〉,
we know that whatever limit (denote it by a), we extract from a subsequence of {T(θν)}ν>0,
then 〈a,f〉=〈1,T∗(f)〉. This shows that the limit a is independent of the subsequence, and we are
justified in setting a=T(1)=limν→0T(θν) and (5.258)〈T(1),f〉=∫𝒳T∗(f)(x)dμ(x), with a similar statement when
the roles of T and T∗ are reversed. Thus, T(1), T∗(1)∈BMO(𝒳).
Moreover, if T is restrictedly bounded, then T∈WBP(β).
Thus, by Theorem 5.56, we know T is bounded on L2(𝒳),
which completes the proof of Theorem 5.57.Remark 5.58.
From (5.258), we see that if T is bounded on L2(𝒳),
then T(1)∈(C°bβ(𝒳))′ for a certain β∈(0,1] is constant if and only if for any f∈C°bβ(𝒳),(5.259)∫𝒳T∗(f)(x)dμ(x)=0.


## 5.1. Plancherel-Pôlya Inequalities and Definition ofB˙p,qs(𝒳) and F˙p,qs(𝒳)

Throughout this and the next subsection, we will always
assume thatμ(𝒳)=∞.
We first introduce the norms in B˙p,qs(𝒳) and F˙p,qs(𝒳) via certain ATIs. We then prove that they are independent of
the choices of ATIs and spaces of distributions. To this end, we
need to establish homogeneous Plancherel-Pôlya inequalities; see also [79].Adapting Triebel's approach to homogeneous Besov and
Triebel-Lizorkin spaces onℝn in [3], we make the following.Definition 5.1.
Letϵ1∈(0,1],ϵ2>0,ϵ3>0,ϵ∈(0,ϵ1⋀ϵ2), and let {Sk}k∈ℤ be an (ϵ1,ϵ2,ϵ3)-ATI.
For k∈ℤ,
set Dk=Sk−Sk−1.
(i)
For allf∈(𝒢°0ϵ(β,γ))′ with 0<β,γ<ϵ,|s|<ϵ,p(s,ϵ)<p≤∞, and 0<q≤∞,
one sets ∥f∥B˙p,qs(𝒳)≡{∑k∈ℤ2ksq∥Dk(f)∥Lp(𝒳)q}1/q with the usual modification made when p=∞ or q=∞.(ii)
For allf∈(𝒢°0ϵ(β,γ))′ with 0<β,γ<ϵ,|s|<ϵ,p(s,ϵ)<p<∞, and p(s,ϵ)<q≤∞,
one defines
∥f∥F˙p,qs(𝒳)≡∥{∑k∈ℤ2ksq|Dk(f)|q}1/q∥Lp(𝒳)  with the usual modification made when q=∞.To verify that the definitions of∥⋅∥B˙p,qs(𝒳) and ∥⋅∥F˙p,qs(𝒳) are independent of the choice of ATIs, we need two technical lemmas, which have independent interest.Lemma 5.2.
Letϵ>0,k′,k∈ℤ,
and yτk,ν be any point in Qτk,ν for τ∈Ik and ν=1,…,N(k,τ).
If n/(n+ϵ)<p≤∞,
then for any x∈𝒳,(5.1)∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)[1V2−(k′⋀k)(x)+V(x,yτk,ν)](p⋀1)[2−(k⋀k′)2−(k⋀k′)+d(x,yτk,ν)]ϵ(p⋀1)≤C[V2−(k′⋀k)(x)]1−(p⋀1), where C>0 is independent of x∈𝒳,k,k′,τ,  and ν.Proof.
Notice that for anyz∈Qτk,ν,
by Lemma 2.1(iii), we have(5.2)V2−(k′⋀k)(yτk,ν)+V(x,yτk,ν)~V2−(k′⋀k)(z)+V(x,z),2−(k⋀k′)+d(x,yτk,ν)~2−(k⋀k′)+d(x,z). These estimates together with
Lemma 2.19 and the second inequality in (1.3) yield that the left-hand side of
(5.1) is, up to a bounded multiplicative constant, controlled
by (5.3)∫𝒳[1V2−(k′⋀k)(x)+V(x,z)](p⋀1)[2−(k⋀k′)2−(k⋀k′)+d(x,z)]ϵ(p⋀1)dμ(z)≲[V2−(k′⋀k)(x)]1−(p⋀1)∑l=0∞2l[n−n(p⋀1)−ϵ(p⋀1)]≲[V2−(k′⋀k)(x)]1−(p⋀1), which completes the proof of
Lemma 5.2.Lemma 5.3.
Letϵ>0,k′,k∈ℤ,
and yτk,ν be any point in Qτk,ν for τ∈Ik and ν=1,…,N(k,τ).
If n/(n+ϵ)<r≤1,
then there exists a constant C>0 depending on r such that for all aτk,ν∈ℂ and all x∈𝒳, (5.4)∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)1V2−(k′⋀k)(x)+V(x,yτk,ν)2−(k⋀k′)ϵ(2−(k⋀k′)+d(x,yτk,ν))ϵ|aτk,ν|≤C2[(k⋀k′)−k]n(1−1/r){M(∑τ∈Ik∑ν=1N(k,τ)|aτk,ν|rχQτk,ν)(x)}1/r, where C>0 is also independent of k,k′,τ, and ν.Proof.
We first recall the following well-known inequality that for allr∈(0,1] and all aj∈ℂ with j in some countable set of indices,(5.5)(∑j|aj|)r≤∑j|aj|r. From this inequality, the fact
that for all τ∈Ik and ν=1,…,N(k,τ),
and all z∈𝒳, μ(Qτk,ν)χQτk,ν(z)~V2−k(z)χQτk,ν(z), and the fact that for all z∈𝒳 and all k,k′∈ℤ,
by (1.3), V2−(k⋀k′)(z)≲2[k−(k⋀k′)]nV2−k(z), together with (5.2), Lemma 2.1(vi),
and the second inequality of (1.3), it follows that the left-hand  side of (5.4)
is, up to a bounded multiplicative constant, controlled by(5.6){∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)r(1V2−(k′⋀k)(x)+V(x,yτk,ν))r2−(k⋀k′)rϵ(2−(k⋀k′)+d(x,yτk,ν))rϵ|aτk,ν|r}1/r≲2[(k⋀k′)−k]n(1−1/r){1V2−(k′⋀k)(x)∫d(x,z)<2−(k′⋀k)(∑τ∈Ik∑ν=1N(k,τ)|aτk,ν|rχQτk,ν(z))dμ(z)+∑l=0∞12l[rϵ+n(r−1)]1V2l+12−(k′⋀k)(x)×∫d(x,z)<2l+12−(k′⋀k)(∑τ∈Ik∑ν=1N(k,τ)|aτk,ν|rχQτk,ν(z))dμ(z)}1/r, which implies the desired
conclusion.Using these technical lemmas and the discrete Calderón
reproducing formulae, we can now establish thePlancherel-Pôlya inequality.Proposition 5.4.
Letϵ1∈(0,1],ϵ2>0,ϵ3>0,ϵ∈(0,ϵ1⋀ϵ2), and let {Sk}k∈ℤ and {Pk}k∈ℤ be two (ϵ1,ϵ2,ϵ3)-ATIs. For k∈ℤ,
set Dk=Sk−Sk−1 and Qk=Pk−Pk−1.
(i)
For allf∈(𝒢°0ϵ(β,γ))′ with 0<β,γ<ϵ, |s|<ϵ, p(s,ϵ)<p≤∞, and 0<q≤∞, (5.7){∑k∈ℤ2ksq(∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)[supz∈Qτk,ν|Dk(f)(z)|]p)q/p}1/q~{∑k∈ℤ2ksq(∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)[infz∈Qτk,ν|Qk(f)(z)|]p)q/p}1/q.(ii)
For allf∈(𝒢°0ϵ(β,γ))′ with 0<β,γ<ϵ,|s|<ϵ,p(s,ϵ)<p<∞, and p(s,ϵ)<q≤∞,(5.8)∥{∑k∈ℤ∑τ∈Ik∑ν=1N(k,τ)2ksq[supz∈Qτk,ν|Dk(f)(z)|]qχQτk,ν}1/q∥Lp(𝒳)~∥{∑k∈ℤ∑τ∈Ik∑ν=1N(k,τ)2ksq[infz∈Qτk,ν|Qk(f)(z)|]qχQτk,ν}1/q∥Lp(𝒳).Proof.
We first verify (5.7). By Theorem 4.13, there exist functions {D˜k}k∈ℤ such that for all f∈(𝒢°0ϵ(β,γ))′ with 0<β,γ<ϵ and all z∈𝒳, (5.9)f(z)=∑k′∈ℤ∑τ′∈Ik′∑ν′=1N(k′,τ′)μ(Qτ′k′,ν′)D˜k′(z,yτ′k′,ν′)Qk′(f)(yτ′k′,ν′)holds in (𝒢°0ϵ(β,γ))′,
where D˜k′ satisfies the same conditions as D˜k in Theorem 4.13 and yτ′k′,ν′ is any point in Qτ′k′,ν′.
For any ϵ′∈(0,ϵ),
by Lemma 3.2 (see also Remark 3.3), for all z∈𝒳,(5.10)|(DkD˜k′)(z,yτ′k′,ν′)|≲2−|k−k′|ϵ′1V2−(k⋀k′)(z)+V2−(k⋀k′)(yτ′k′,ν′)+V(z,yτ′k′,ν′)2−(k⋀k′)ϵ(2−(k⋀k′)+d(z,yτ′k′,ν′))ϵ. If p(s,ϵ)<p≤1,
by applying Dk to (5.9), and making use of (5.10), (5.5),
and (5.2) together with Lemma 5.2, we obtain(5.11)∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)[supz∈Qτk,ν|Dk(f)(z)|]p≲∑k′∈ℤ∑τ′∈Ik′∑ν′=1N(k′,τ′)2−|k−k′|ϵ′pμ(Qτ′k′,ν′)p[V2−(k⋀k′)(yτ′k′,ν′)]1−p|Qk′(f)(yτ′k′,ν′)|p. From this and the fact that (5.12)V2−(k⋀k′)(yτ′k′,ν′)≲2[k′−(k⋀k′)]nV2−k′(yτ′k′,ν′)~2[k′−(k⋀k′)]nμ(Qτ′k′,ν′) together with Hölder's
inequality when q/p≥1 and (5.5) when q/p<1,
it follows that if we choose ϵ′∈(0,ϵ) such that max{n/(n+ϵ′),n/(n+ϵ′+s)}<p≤1 and |s|<ϵ′,
the left-hand side of (5.7) is, up to a bounded multiplicative constant,
dominated by (5.13){∑k∈ℤ[∑k′∈ℤ∑τ′∈Ik′∑ν′=1N(k′,τ′)2−|k−k′|ϵ′p2(k−k′)spμ(Qτ′k′,ν′)p[V2−(k⋀k′)(yτ′k′,ν′)]1−p2k′sp|Qk′(f)(yτ′k′,ν′)|p]q/p}1/q≲{∑k′∈ℤ2k′sq(∑τ′∈Ik′∑ν′=1N(k′,τ′)μ(Qτ′k′,ν′)|Qk′(f)(yτ′k′,ν′)|p)q/p}1/q, which together with the
arbitrary choice of yτ′k′,ν′ yields (5.14){∑k∈ℤ2ksq(∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)[supz∈Qτk,ν|Dk(f)(z)|]p)q/p}1/q≲{∑k∈ℤ2ksq(∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)[infz∈Qτk,ν|Qk(f)(z)|]p)q/p}1/q. By symmetry, we then obtain
(5.7) when p(s,ϵ)<p≤1.
Ifp∈(1,∞] and if we choose ϵ′∈(0,ϵ) such that |s|<ϵ′,
by Hölder's inequality and Lemma 5.2, we obtain (5.15)∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)[supz∈Qτk,ν|Dk(f)(z)|]p≲∑k′∈ℤ∑τ′∈Ik′∑ν′=1N(k′,τ′)2−|k−k′|ϵ′2(k′−k)s(p−1)μ(Qτ′k′,ν′)|Qk′(f)(yτ′k′,ν′)|p, which together with Hölder's
inequality when q/p≥1 and (5.5) when q/p<1 yields that the left-hand side of (5.7) is, up
to a bounded multiplicative constant, controlled by (5.16){∑k∈ℤ[∑k′∈ℤ∑τ′∈Ik′∑ν′=1N(k′,τ′)2−|k−k′|ϵ′2(k−k′)sμ(Qτ′k′,ν′)2k′sp|Qk′(f)(yτ′k′,ν′)|p]q/p}1/q≲{∑k′∈ℤ2k′sq(∑τ′∈Ik′∑ν′=1N(k′,τ′)μ(Qτ′k′,ν′)|Qk′(f)(yτ′k′,ν′)|p)q/p}1/q. Then, since yτ′k′,ν′ was an arbitrary point in Qτ′k′,ν′,
we see that the estimate (5.14) also holds when p∈(1,∞],
which by symmetry then completes the proof of (5.7).
We now verify (5.8). By applying Dk to (5.9), and making use of (5.10) together
with (5.2), and Lemma 5.3, we obtain that for ϵ′>|s| and r>max{p,p(s,ϵ′),q}, (5.17){∑k∈ℤ∑τ∈Ik∑ν=1N(k,τ)2ksq[supz∈Qτk,ν|Dk(f)(z)|]qχQτk,ν(x)}1/q≲{∑k∈ℤ[∑k′∈ℤ2(k−k′)s−|k−k′|ϵ′2[(k⋀k′)−k′]n(1−1/r)×{M(∑τ′∈Ik′∑ν′=1N(k′,τ′)2k′sr|Qk′(f)(yτ′k′,ν′)|rχQτ′k′,ν′)(x)}1/r]q}1/q, which together with Hölder's
inequality when q∈(1,∞] or (5.5) when q≤1,
and Lemma 3.14 further implies that the left-hand side of (5.8) is, up to a bounded
multiplicative constant, dominated by (5.18)∥{∑k′∈ℤ[M(∑τ′∈Ik′∑ν′=1N(k′,τ′)2k′sr|Qk′(f)(yτ′k′,ν′)|rχQτ′k′,ν′)]q/r}r/q∥Lp/r(𝒳)1/r≲∥{∑k′∈ℤ∑τ′∈Ik′∑ν′=1N(k′,τ′)2k′sq|Qk′(f)(yτ′k′,ν′)|qχQτ′k′,ν′}1/q∥Lp(𝒳). Then, by symmetry and the fact that yτ′k′,ν′ was an arbitrary point in Qτ′k′,ν′,
we obtain (5.8), which completes the proof of Proposition 5.4.The following remark is useful in
applications.Remark 5.5.
Let all the notation be as in Proposition5.4, except that Sk (and therefore Dk) for k∈ℤ has regularity only in the second variable.
Then, there exists a constant C>0 such that
(i)
for allf∈(𝒢°0ϵ(β,γ))′ with 0<β,γ<ϵ,|s|<ϵ,p(s,ϵ)<p≤∞, and 0<q≤∞, (5.19){∑k∈ℤ2ksq(∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)[supz∈Qτk,ν|Dk(f)(z)|]p)q/p}1/q≤C{∑k∈ℤ2ksq(∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)[infz∈Qτk,ν|Qk(f)(z)|]p)q/p}1/q;(ii)
for allf∈(𝒢°0ϵ(β,γ))′ with 0<β,γ<ϵ,|s|<ϵ,p(s,ϵ)<p<∞, and p(s,ϵ)<q≤∞, (5.20)∥{∑k∈ℤ∑τ∈Ik∑ν=1N(k,τ)2ksq[supz∈Qτk,ν|Dk(f)(z)|]qχQτk,ν}1/q∥Lp(𝒳)≤C∥{∑k∈ℤ∑τ∈Ik∑ν=1N(k,τ)2ksq[infz∈Qτk,ν|Qk(f)(z)|]qχQτk,ν}1/q∥Lp(𝒳).Using Proposition5.4, we can easily verify that the
definitions of ∥⋅∥B˙p,qs(𝒳) and ∥⋅∥F˙p,qs(𝒳) are independent of the choices of ATIs. We omit the details.Proposition 5.6.
Let all the notation be as in Proposition5.4.
(i)
For allf∈(𝒢°0ϵ(β,γ))′ with 0<β,γ<ϵ,|s|<ϵ,p(s,ϵ)<p≤∞, and 0<q≤∞, (5.21){∑k∈ℤ2ksq∥Dk(f)∥Lp(𝒳)q}1/q~{∑k∈ℤ2ksq∥Qk(f)∥Lp(𝒳)q}1/q.(ii)
For allf∈(𝒢°0ϵ(β,γ))′ with 0<β,γ<ϵ,|s|<ϵ,p(s,ϵ)<p<∞, and p(s,ϵ)<q≤∞, (5.22)∥{∑k∈ℤ2ksq|Dk(f)|q}1/q∥Lp(𝒳)~∥{∑k∈ℤ2ksq|Qk(f)|q}1/q∥Lp(𝒳).We now verify that the definition of the norm∥⋅∥B˙p,qs(𝒳) and the norm ∥⋅∥F˙p,qs(𝒳) is independent of the choice of the underlying
space of distributions. We recall that a+=max{a,0} for any a∈ℝ.Proposition 5.7.
Let all the notation be as in Definition5.1.
(i)
Let|s|<ϵ,p(s,ϵ)<p≤∞, and 0<q≤∞.
If f∈(𝒢°0ϵ(β1,γ1))′ with(5.23)max{0,−s+n(1p−1)+}<β1<ϵ,max{n(1p−1)+,s−κp}<γ1<ϵ, and if ∥f∥B˙p,qs(𝒳)<∞,
then f∈(𝒢°0ϵ(β2,γ2))′ for every β2,γ2 satisfying (5.23).(ii)
Let|s|<ϵ,p(s,ϵ)<p<∞, and p(s,ϵ)<q≤∞.
If f∈(𝒢°0ϵ(β1,γ1))′ with β1,γ1 as in (5.23), and if ∥f∥F˙p,qs(𝒳)<∞,
then f∈(𝒢°0ϵ(β2,γ2))′ for every β2,γ2 satisfying (5.23).Proof.
Letψ∈𝒢°(ϵ,ϵ).
Adopting the notation from Theorem 4.11, we first claim that for k∈ℤ+, (5.24)|〈D˜k(⋅,y),ψ〉|≲2−kβ2∥ψ∥𝒢(β2,γ2)1V1(x1)+V(x1,y)1(1+d(x1,y))γ2, and that for k=−1,−2,…,(5.25)|〈D˜k(⋅,y),ψ〉|≲2kγ2′∥ψ∥𝒢(β2,γ2)1V2−k(x1)+V(x1,y)2−kγ2(2−k+d(x1,y))γ2, where γ2′ can be any positive number in (0,γ2).
In fact, to verify (5.24), by the vanishing moment of D˜k,
Lemma 2.1(ii), we have  (5.26)|〈D˜k(⋅,y),ψ〉|=|∫𝒳D˜k(z,y)[ψ(z)−ψ(y)]dμ(z)|≲∥ψ∥𝒢(β2,γ2){∫d(z,y)≤(1+d(y,x1))/2|D˜k(z,y)|1V1(x1)+V(x1,y)d(z,y)β2(1+d(x1,y))β2+γ2dμ(z)+∫d(z,y)>(1+d(y,x1))/2|D˜k(z,y)|[1V1(x1)+V(x1,z)1(1+d(x1,z))γ2+1V1(x1)+V(x1,y)1(1+d(x1,y))γ2]dμ(z)}≲2−kβ2∥ψ∥𝒢(β2,γ2)1V1(x1)+V(x1,y)1(1+d(x1,y))γ2×{1+∫𝒳1V1(x1)+V(x1,z)1(1+d(x1,z))γ2dμ(z)+∫𝒳1V2−k(y)+V(z,y)2−k(ϵ−β2)(2−k+d(z,y))ϵ−β2dμ(z)}, where the last quantity is, up
to a bounded multiplicative constant, controlled by the right-hand side of
(5.24), and in the last inequality, we used the  fact that for d(z,y)>(1+d(y,x1))/2,V(z,y)≳V1(y)+V(x1,y)~V1(x1)+V(x1,y);
see  also Lemma 2.1(vii). Thus, (5.24)  holds.
To see  (5.25), by ∫𝒳ψ(z)dμ(z)=0,
(i) and (ii) of Lemma 2.1, and the fact that for d(z,x1)>(2−k+d(x1,y))/2, V(x1,z)≳V2−k(x1)+V(x1,y), we obtain that for k=−1,−2,…, (5.27)|〈D˜k(⋅,y),ψ〉|=|∫𝒳[D˜k(z,y)−D˜k(x1,y)]ψ(z)dμ(z)|≲∥ψ∥𝒢(β2,γ2){∫d(z,x1)≤(2−k+d(y,x1))/2d(z,x1)ϵ(2−k+d(x1,y))ϵ2−kϵ(2−k+d(x1,y))ϵ×1V2−k(x1)+V2−k(y)+V(x1,y)×1V1(x1)+V(x1,z)1(1+d(x1,z))γ2dμ(z)+∫d(z,x1)>(2−k+d(y,x1))/2[1V2−k(z)+V2−k(y)+V(z,y)2−kϵ(2−k+d(z,y))ϵ+1V2−k(x1)+V2−k(y)+V(x1,y)2−kϵ(2−k+d(x1,y))ϵ]×1V1(x1)+V(x1,z)1(1+d(x1,z))γ2dμ(z)}≲1V2−k(x1)+V(x1,y)×{1(2−k+d(x1,y))γ2+2−kϵ(2−k+d(x1,y))ϵ1(2−k+d(x1,y))γ2′}, where the last quantity is, up
to a bounded multiplicative constant, dominated by the right-hand side of
(5.25); namely, (5.25)  holds.
Thus, Theorem4.13 together with (5.24), (5.25),  and
Lemma 2.1(iii) yield that
(5.28)|〈f,ψ〉|=|∑k=−∞∞∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)Dk(f)(yτk,ν)〈D˜k(⋅,yτk,ν),ψ〉|≲∥ψ∥𝒢(β2,γ2){∑k=0∞∑τ∈Ik∑ν=1N(k,τ)2−kβ2μ(Qτk,ν)|Dk(f)(yτk,ν)|×1V1(x1)+V(x1,yτk,ν)1(1+d(x1,yτk,ν))γ2+∑k=−∞−1∑τ∈Ik∑ν=1N(k,τ)2kγ2′μ(Qτk,ν)|Dk(f)(yτk,ν)|×1V2−k(x1)+V(x1,yτk,ν)2−kγ2(2−k+d(x1,yτk,ν))γ2}.
If p≤1,
by (5.5), (5.29)|〈f,ψ〉|≲∥ψ∥𝒢(β2,γ2){∑k=0∞2−k(β2+s)[∑τ∈Ik∑ν=1N(k,τ)2kspμ(Qτk,ν)p|Dk(f)(yτk,ν)|p×1(V1(x1)+V(x1,yτk,ν))p1(1+d(x1,yτk,ν))γ2p]1/p+∑k=−∞−12k(γ2′−s)[∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)p2ksp|Dk(f)(yτk,ν)|p×1(V2−k(x1)+V(x1,yτk,ν))p2−kγ2p(2−k+d(x1,yτk,ν))γ2p]1/p}. Notice that when p≤1,
by Lemma 2.1(vii) and γ2>n(1/p−1),
for k∈ℤ+, τ∈Ik and ν=1,…,N(k,τ)
, (5.30)μ(Qτk,ν)p−1(V1(x1)+V(x1,yτk,ν))p1(1+d(x1,yτk,ν))γ2p≲1V1(x1)[V1(yτk,ν)+V(x1,yτk,ν)μ(Qτk,ν)]1−p1(1+d(x1,yτk,ν))γ2p≲2kn(1−p)1V1(x1); and similarly, by the first
inequality of (1.3),  for k=−1,−2,…,τ∈Ik and ν=1,…,N(k,τ), (5.31)μ(Qτk,ν)p−1(V2−k(x1)+V(x1,yτk,ν))p2−kγ2p(2−k+d(x1,yτk,ν))γ2p≲2kκ1V1(x1)[V2−k(yτk,ν)+V(x1,yτk,ν)μ(Qτk,ν)]1−p2−kγ2p(2−k+d(x1,yτk,ν))γ2p≲2kκ1V1(x1). Thus, if p≤1,
then by Hölder's inequality when q≥1 or (5.5) when q<1 together with Proposition 5.4, (5.32)|〈f,ψ〉|≲1V1(x1)1/p∥ψ∥𝒢(β2,γ2){∑k=0∞2k[n(1/p−1)−(β2+s)]+∑k=−∞−12k[κ/p+γ2′−s]}×[∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)2ksp|Dk(f)(yτk,ν)|p]1/p≲1V1(x1)1/p∥ψ∥𝒢(β2,γ2)∥f∥B˙p,qs(𝒳), where we used the assumption
that β2>−s+n(1/p−1) and we chose γ2′∈((s−κ/p)+,γ2).
Similarly, ifp>1,
by Lemma 2.1(ii), we have  (5.33)|〈f,ψ〉|≲1V1(x1)1/p∥ψ∥𝒢(β2,γ2)×{∑k=0∞2−k(β2+s)[∫𝒳1V1(x1)+V(x1,y)1(1+d(x1,y))γ2p′dμ(y)]1/p′+∑k=0∞2k(κ/p−γ2′−s)[∫𝒳1V2−k(x1)+V(x1,y)2−kγ2p2′(2−k+d(x1,y))γ2p′dμ(y)]1/p′}×[∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)2ksp|Dk(f)(yτk,ν)|p]1/p≲1V1(x1)1/p∥ψ∥𝒢(β2,γ2)∥f∥B˙p,qs(𝒳), where we used β2>−s and chose γ2′>max{0,s−κ/p}.
Let nowh∈𝒢°0ϵ(β2,γ2).
Then there exists {hn}n=1∞⊂𝒢°(ϵ,ϵ) such that as n→∞,∥h−hn∥𝒢(β2,γ2)→0.
By (5.32) and (5.33), we obtain(5.34)|〈f,hn−hm〉|≲∥f∥B˙p,qs(𝒳)∥hn−hm∥𝒢(β2,γ2), which shows that limn→∞〈f,hn〉 exists and the limit is independent of the
choice of {hn}.
Thus, if we define 〈f,h〉=limn→∞〈f,hn〉, by (5.32) and (5.33), we have |〈f,h〉|≲∥f∥B˙p,qs(𝒳)∥h∥𝒢°0ϵ(β2,γ2). That is, f∈(𝒢°0ϵ(β2,γ2))′,
which completes the  proof of (i).
The conclusion (ii) can be deduced from (i) and the
fact that∥f∥B˙p,max(p,q)s(𝒳)≲∥f∥F˙p,qs(𝒳); see [3, Proposition 2.3.2/2] or Proposition 5.10(ii)
below.Now we can introduce the Besov spaces,B˙p,qs(𝒳),
and the Triebel-Lizorkin spaces, F˙p,qs(𝒳).Definition 5.8.
Letϵ1∈(0,1],ϵ2>0,ϵ3>0,ϵ∈(0,ϵ1⋀ϵ2), and let {Sk}k∈ℤ be an (ϵ1,ϵ2,ϵ3)-ATI.
For k∈ℤ,
set Dk=Sk−Sk−1.
(i)
Let|s|<ϵ,p(s,ϵ)<p≤∞, and 0<q≤∞.
The spaceB˙p,qs(𝒳) is defined to be the set of all f∈(𝒢°0ϵ(β,γ))′,
for some β,γ satisfying(5.35)max{s,0,−s+n(1p−1)+}<β<ϵ,max{s−κp,n(1p−1)+,−s+n(1p−1)+−κ(1−1p)+}<γ<ϵsuch that ∥f∥B˙p,qs(𝒳)={∑k=−∞∞2ksq∥Dk(f)∥Lp(𝒳)q}1/q<∞ with the usual modifications made when p=∞ or q=∞.(ii)
Let|s|<ϵ,p(s,ϵ)<p<∞, and p(s,ϵ)<q≤∞. The spaceF˙p,qs(𝒳) is defined to be the set of all f∈(𝒢°0ϵ(β,γ))′ for some β,γ satisfying (5.35) such that ∥f∥F˙p,qs(𝒳)=∥{∑k=−∞∞2ksq|Dk(f)|q}1/q∥Lp(𝒳)<∞ with the usual modification made when q=∞.Propositions5.6 and 5.7 show that the
definitions of the spaces B˙p,qs(𝒳) and F˙p,qs(𝒳) are independent of the choice of the
approximations of the identity and the distribution space, (𝒢°0ϵ(β,γ))′,
with β,γ as in (5.35).Remark 5.9.
To guarantee that the definitions of the spacesB˙p,qs(𝒳) and F˙p,qs(𝒳) are independent of the choice of the
distribution space (𝒢°0ϵ(β,γ))′,
we only need the restriction that β,γ satisfy (5.23); see Proposition 5.7. Moreover,
if we assume that max{s,0}<β<ϵ and max{n(1/p−1)+,−s+n(1/p−1)+−κ(1−1/p)+}<γ<ϵ,
we can verify that the space of test functions 𝒢°(β,γ) is contained in B˙p,qs(𝒳) and F˙p,qs(𝒳);
see Proposition 5.10 below. Thus, the spaces B˙p,qs(𝒳) and F˙p,qs(𝒳) are nonempty if we restrict β,γ as in (5.35).


## 5.2. Properties ofB˙p,qs(𝒳) and F˙p,qs(𝒳) and Boundedness of Singular Integrals

In this subsection, we first present some basic properties
ofB˙p,qs(𝒳) and F˙p,qs(𝒳).
Then we establish a Lusin-area characterization of F˙p,qs(𝒳),
and as an application, we discuss the relation between the spaces F˙p,qs(𝒳) and the atomic Hardy spaces Hatp(𝒳) of Coifman and Weiss in [28] and the Hardy spaces Hp(𝒳) in [48]. Finally, we show that the singular integrals
considered by Nagel and Stein in [44] act boundedly on B˙p,qs(𝒳) and F˙p,qs(𝒳).Proposition 5.10.
Letϵ1∈(0,1],ϵ2>0,ϵ∈(0,ϵ1⋀ϵ2), and |s|<ϵ.
(i)
Forp(s,ϵ)<p≤∞ and 0<q0≤q1≤∞,B˙p,q0s(𝒳)⊂B˙p,q1s(𝒳); and for p(s,ϵ)<p<∞ and p(s,ϵ)<q0≤q1≤∞,F˙p,q0s(𝒳)⊂F˙p,q1s(𝒳).(ii)
Ifp(s,ϵ)<p<∞ and p(s,ϵ)<q≤∞,
then B˙p,min(p,q)s(𝒳)⊂F˙p,qs(𝒳)⊂B˙p,max(p,q)s(𝒳).(iii)
Ifβ,γ as in (5.23), then B˙p,qs(𝒳)⊂(𝒢°0ϵ(β,γ))′ when p(s,ϵ)<p≤∞ and 0<q≤∞,
and F˙p,qs(𝒳)⊂(𝒢°0ϵ(β,γ))′ when p(s,ϵ)<p<∞ and p(s,ϵ)<q≤∞.(iv)
Ifmax{s,0}<β<ϵ and max{n(1/p−1)+,−s+n(1/p−1)+−κ(1−1/p)+}<γ<ϵ,
then 𝒢°(β,γ)⊂B˙p,qs(𝒳) when p(s,ϵ)<p≤∞ and 0<q≤∞,
and 𝒢°(β,γ)⊂F˙p,qs(𝒳) when p(s,ϵ)<p<∞ and p(s,ϵ)<q≤∞.(v)
If1<p<∞,
then F˙p,20(𝒳)=Lp(𝒳) with equivalent norms.(vi)
The spacesB˙p,qs(𝒳)/𝒩 with p(s,ϵ)<p≤∞ and 0<q≤∞ and the spaces F˙p,qs(𝒳)/𝒩 with p(s,ϵ)<p<∞ and p(s,ϵ)<q≤∞ are complete.Proof.
Property (i) is a simple corollary of (5.5). Property (ii) can be
deduced from Minkowski's inequality, (5.5), and the following generalized
Minkowski inequality that for all 1≤q≤∞, (5.36){∑k=−∞∞[∫𝒳|ak(x)|dμ(x)]q}1/q≲∫𝒳{∑k=−∞∞|ak(x)|q}1/qdμ(x); see also [3, Proposition 2.3.2/2] and [6, Proposition 2.3].
Property (iii) is implied by the proof of Proposition5.7 and Property (vi) can be easily deduced from Property (iii).
To verify (iv), similarly to the proofs of (5.24) and
(5.25), for f∈𝒢°(β,γ),
we have that for k∈ℤ+, (5.37)|Dk(f)(x)|≲2−kβ∥f∥𝒢(β,γ)1V1(x1)+V(x1,x)1(1+d(x1,x))γ, and that for k=−1,−2,…,(5.38)|Dk(f)(x)|≲2kγ'∥f∥𝒢(β,γ)1V2−k(x1)+V(x1,x)2−kγ(2−k+d(x1,x))γ, where γ′ can be any positive number in (0,γ).
Moreover, since γ>n(1/p−1)+, (5.39){∫𝒳1(V1(x1)+V(x1,x))p1(1+d(x1,x))γpdμ(x)}1/p≲{1V1(x1)p−1+∑l=0∞1V2lr(x1)p−112lγp}1/p≲1V1(x1)1−1/p, and for k=−1,−2,…,(5.40){∫𝒳1(V2−k(x1)+V(x1,x))p2−kγp(2−k+d(x1,x))γpdμ(x)}1/p≲{1V2−k(x1)p−1+∑l=0∞1V2l2−k(x1)p−112lγp}1/p≲2kκ(1−1/p)+−kn(1/p−1)+V1(x1)1/p−1. Choose γ′∈(0,γ) such that γ′>n(1/p−1)+−s−κ(1−1/p)+.
From the above estimates, it follows that (5.41)∥f∥B˙p,qs(𝒳)≲∥f∥𝒢(β,γ)1V1(x1)1−1/p{∑k=0∞2k(s−β)q+∑k=−∞−12k[s+γ'−n(1/p−1)++κ(1−1/p)+]q}1/q≲∥f∥𝒢(β,γ). Thus,
𝒢°(β,γ)⊂B˙p,qs(𝒳),
which together with (ii) implies that 𝒢°(β,γ)⊂F˙p,qs(𝒳).
Property (v) is a simple conclusion of Proposition3.15, which completes the proof of Proposition 5.10.We next give a Lusin-area characterization for
Triebel-Lizorkin spacesF˙p,qs(𝒳),
which will allow to establish some relations between the Triebel-Lizorkin
spaces F˙p,qs(𝒳),
the atomic Hardy spaces of Coifman and Weiss in [28], and the Hardy spaces Hp(𝒳) in [48].Definition 5.11.
Lets∈ℝ,a>0,q∈(0,∞], and let ϵ1∈(0,1],ϵ2>0,ϵ3>0,ϵ∈(0,ϵ1⋀ϵ2]. Let {Sk}k∈ℤ be an (ϵ1,ϵ2,ϵ3)-ATI.
For k∈ℤ,
set Dk=Sk−Sk−1.
The Lusin-area function (also
called the Littlewood-PaleyS-function) S˙q,as(f) for any f∈(𝒢°0ϵ(β,γ))′ with 0<β,γ≤ϵ and x∈𝒳 is given by (5.42)S˙q,as(f)(x)={∑k∈ℤ∫d(x,y)<a2−k2ksq|Dk(f)(y)|qdμ(y)Va2−k(x)}1/q, where the usual modification is
made when q=∞.Remark 5.12.
(i) By Lemma2.1(vi), if we replace Va2−k(x),
respectively, by Va2−k(y),Va2−k(y)+Va2−k(x),V2−k(x),V2−k(y),
or V2−k(x)+V2−k(y) in the definition of S˙q,as(f),
then the corresponding Littlewood-Paley S-functions S˙q,as(f) are pointwise equivalent. This is often useful
in applications.
(ii) It is easy to see thatS˙2,a0(f) is a discrete version of the corresponding
Littlewood-Paley S-function in [48].Theorem 5.13.
Letϵ1∈(0,1],ϵ2>0,ϵ3>0,ϵ∈(0,ϵ1⋀ϵ2),
and let {Sk}k∈ℤ be an (ϵ1,ϵ2,ϵ3)-ATI.
For k∈ℤ,
set Dk=Sk−Sk−1.
Let a>0,|s|<ϵ,p(s,ϵ)<p<∞,p(s,ϵ)<q≤∞, and let S˙q,as(f) be as in Definition 5.11 for any f∈(𝒢°0ϵ(β,γ))′ with β,γ as in (5.35). Then f∈F˙p,qs(𝒳) if and only if f∈(𝒢°0ϵ(β,γ))′ for some β,γ as in (5.35), and S˙q,as(f)∈Lp(𝒳).
Moreover, in this case, ∥f∥F˙p,qs(𝒳)~∥S˙q,as(f)∥Lp(𝒳).Proof.
We use the notation as in the proof of
Proposition5.4. It is easy to check that for any fixed constant C>0, (5.43)supz∈B(zτk,ν,C2−k)|(DkD˜k′)(z,yτ′k′,ν′)|≲2−|k−k′|ϵ′1V2−(k⋀k′)(zτk,ν)+V2−(k⋀k′)(yτ′k′,ν′)+V(zτk,ν,yτ′k′,ν′)2−(k⋀k′)ϵ(2−(k⋀k′)+d(zτk,ν,yτ′k′,ν′))ϵ. Using this to replace (5.10) and
then repeating the proof of (5.8)
in Proposition 5.4
yield (5.44)∥{∑k∈ℤ∑τ∈Ik∑ν=1N(k,τ)2ksq[supz∈B(zτk,ν,C2−k)|Dk(f)(z)|]qχQτk,ν}1/q∥Lp(𝒳)~∥{∑k∈ℤ∑τ∈Ik∑ν=1N(k,τ)2ksq[infz∈Qτk,ν|Dk(f)(z)|]qχQτk,ν}1/q∥Lp(𝒳)~∥f∥F˙p,qs(𝒳),which shows that(5.45)∥S˙q,as(f)∥Lp(𝒳)≤∥{∑k∈ℤ∑τ∈Ik∑ν=1N(k,τ)2ksq[supz∈B(zτk,ν,C2−k)|Dk(f)(z)|]qχQτk,ν}1/q∥Lp(𝒳)~∥f∥F˙p,qs(𝒳).
On the other hand, noticing that ifx∈B(zτk,ν,min{a,1}C72−(k+j)),
then (5.46)B(zτk,ν,min{a,1}C72−(k+j))⊂{y∈𝒳:d(x,y)<a2−k}, therefore, by Lemma
2.19, (5.47)S˙q,as(f)(x)={∑k∈ℤ∑τ∈Ik∑ν=1N(k,τ)∫d(x,y)<a2−k2ksq|Dk(f)(y)|qχQτk,ν(x)dμ(y)Va2−k(x)}1/q≥{∑k∈ℤ∑τ∈Ik∑ν=1N(k,τ)2ksq[infy∈B(zτk,ν,min{a,1}C72−(k+j))|Dk(f)(y)|]qχB(zτk,ν,min{a,1}C72−(k+j))(x)}1/q, which together with the
following estimate (5.48)∥{∑k∈ℤ2ksq|Dk(f)|q}1/q∥Lp(𝒳)≲∥{∑k∈ℤ∑τ∈Ik∑ν=1N(k,τ)2ksq[infy∈B(zτk,ν,min{a,1}C72−(k+j))|Dk(f)(y)|]qχB(zτk,ν,min{a,1}C72−(k+j))}1/q∥Lp(𝒳) yields that
∥S˙q,as(f)∥Lp(𝒳)≳∥f∥F˙p,qs(𝒳). Using the notation as in Lemma 5.3, we can
verify the estimate (5.48) by repeating the proof of (5.8) in Proposition 5.4
and replacing  Lemma 5.3 by the following estimate that for any fixed constant C>0, (5.49)∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)1V2−(k′⋀k)(x)+V2−(k′⋀k)(yτk,ν)+V(x,yτk,ν)2−(k⋀k′)ϵ(2−(k⋀k′)+d(x,yτk,ν))ϵ|aτk,ν|≲2[(k⋀k′)−k]n(1−1/r){M(∑τ∈Ik∑ν=1N(k,τ)|aτk,ν|rχB(zτk,ν,C2−k))(x)}1/r, which can be proved by the same
way as the proof of Lemma 5.3 via the following facts μ(B(zτk,ν,C2−k))~μ(Qτk,ν) and (5.50)μ(B(zτk,ν,C2−k))χB(zτk,ν,C2−k)(z)~V2−k(z)χB(zτk,ν,C2−k)(z).This finishes the proof of
Theorem 5.13.Definition 5.14.
Let all the notation be as in
Definition5.8. The Hardy spaceHp(𝒳) when n/(n+ϵ)<p≤1 is defined to be the Triebel-Lizorkin spaces F˙p,20(𝒳),
with norm ∥f∥Hp(𝒳)≡∥f∥F˙p,20(𝒳).We now recall the definition of atoms on spaces of
homogeneous type in [28].Definition 5.15.
A functiona on 𝒳 is called an Hp(𝒳)-atom if
(i)
suppa⊂B(x0,r) for some x0∈𝒳 and some r>0;(ii)
∥a∥L2(𝒳)≤[μ(B(x0,r))]1/2−1/p;(iii)
∫𝒳a(x)dμ(x)=0.If (i) to (iii) apply, we also say thata is an Hp(𝒳)-atom supported on B(x0,r).Theorem 5.16.
Assume thatϵ and β,γ are as in Definition 5.8. If n/(n+ϵ)<p≤1,
then f∈Hp(𝒳) if and only if there exist a sequence of
numbers {λk}k=0∞⊂ℂ with ∑k=0∞|λk|p<∞ and a sequence of Hp(𝒳)-atoms {ak}k=0∞ such that f=∑k=0∞λkak in (𝒢°0ϵ(β,γ))′.
Moreover, in this case, ∥f∥Hp(𝒳)~inf{(∑k=0∞|λk|p)1/p}, where the infimum is taken over all the above
decompositions  of f.This theorem can be proved by a literal repetition of Theorem 2.21 in [48],
with Lemma 2.22 therein replaced by Theorem 2.6 in this paper. We omit the details.Remark 5.17.
Theorem5.16 shows that the Hardy spaces
defined here are the same as those in [48]. Moreover, by [48, Remark 2.27], we know that the Hardy space H1(𝒳) also coincides with the atomic Hardy space Hat1(𝒳) of Coifman and Weiss in [28]. Moreover, when n/(n+ϵ)<p<1 and 𝒳 is an Ahlfors n-regular metric measure space, Hp(𝒳) also coincides with the atomic Hardy space Hatp(𝒳) of Coifman and Weiss in [28]. However, if 𝒳 is a general space of homogeneous type, it is
still unclear so far to us if Hp(𝒳)=Hatp(𝒳) when p<1;
see [48, Remark 2.30].We now recall the definition of the Lipschitz spaceL˙ips(𝒳) with s>0;
see [28].Definition 5.18.
Lets>0.
The Lipschitz (or Hölder) spaceL˙ips(𝒳) is defined to be the set of all functions f on X such that (5.51)∥f∥L˙ips(𝒳)=supx≠y|f(x)−f(y)|V(x,y)s<∞.Observe that these classes are rather Lipschitz (or
Hölder) classes with respect to the measure distanceρ(x,y)≡inf{μ(B):x,y∈B,Baball},
not the distance d.Following Coifman and Weiss [28], we immediately obtain from
Theorem 5.16 in this paper and [48, Remark 2.27] the duality between H1(𝒳) and BMO(𝒳) and between Hatp(𝒳) and L˙ip1/p−1(𝒳) when p<1,
which is [48, Corollary 2.29].Theorem 5.19.
(i) The spaceBMO(𝒳)/𝒩 is the dual space of H1(𝒳), in the following sense: if f=∑k=0∞λkak∈H1(𝒳) is as in Theorem 5.16, then for each g∈BMO(𝒳), limN→∞∑k=0Nλk∫𝒳ak(x)g(x)dμ(x) is a well-defined continuous linear functional ℒg:f↦〈f,g〉 with norm ≲∥g∥BMO(𝒳).
Conversely, each continuous linear functionalℒ on H1(𝒳)
has the form
ℒ=ℒg for some g∈BMO(𝒳) with ∥g∥BMO(𝒳)≲∥ℒ∥.
Moreover,ℒg=0 if and only if g∈𝒩.
(ii) Assume thatϵ is as in Definition 5.8, n/(n+ϵ)<p<1, and s=1/p−1.
Then L˙ips(𝒳) (more precisely, L˙ips(𝒳)/𝒩) is the dual space of Hp(𝒳) in the sense of (i).Remark 5.20.
It is easy to see that if𝒳=ℝn and μ is the n-dimensional Lebesgue measure, then C˙ns(𝒳)=L˙ips(𝒳) with equivalent norms. Thus,(5.52)B˙∞,∞ns(𝒳)=F˙∞,∞ns(𝒳)=L˙ips(𝒳) with  equivalent norms.
Therefore, the dual space of Hp(𝒳) with n/(n+ϵ)<p<1 is the space B˙∞,∞n(1/p−1)(𝒳);
see also [3, Theorem
2.11.3(ii)]. However, for a general space 𝒳 of homogeneous type, it seems that C˙ns(𝒳)≠L˙ips(𝒳),
unless μ(B(x,r))~rn for all x∈𝒳 and r>0,
namely, 𝒳 is an Ahlfors n-regular metric measure space. Thus, one
cannot expect that [3, Theorem 2.11.3] still holds when 𝒳 is a general space of homogeneous type
considered in this paper, which demonstrates an essential difference between
function spaces on general spaces of homogeneous type considered in this paper
and those on Ahlfors regular metric measure space.Using Proposition5.10 and Theorem 2.6, we obtain the
following density property of B˙p,qs(𝒳) and F˙p,qs(𝒳).Proposition 5.21.
Letϵ1, ϵ2, ϵ, and let |s|<ϵ be as in Definition 5.8. Then 𝒢°b(ϵ1,ϵ2) is dense in B˙p,qs(𝒳) when p(s,ϵ)<p<∞ and 0<q<∞,
and in F˙p,qs(𝒳) when p(s,ϵ)<p,q<∞.Proof.
By similarity, we only verify the conclusion onB˙p,qs(𝒳).
Let f∈B˙p,qs(𝒳) with s,p,q as in the proposition. By Proposition 5.10(iii), we know f∈(𝒢°0ϵ(β,γ))′ with β,γ as in (5.23). Let {Dk}k∈ℤ be an ATI with bounded support as constructed in Theorem
2.6. By Theorem 4.13 and using the notation as in Theorem 4.12, we have (5.53)f(x)=∑k′=−∞∞∑τ′∈Ik′∑ν′=1N(k′,τ′)μ(Qτ′k′,ν′)Dk′(x,yτ′k′,ν′)D¯k′(f)(yτ′k′,ν′) in (𝒢°0ϵ(β,γ))′.
It is easy to check that for any fixed k′∈ℤ and τ′∈Ik′,N(k′,τ′) is a finite set of indices. For any fixed k′∈ℤ,
choose {Ik′N}N∈ℕ such that Ik′N⊂Ik′N+1,Ik′N is a finite set of indices, and Ik′N→Ik′ as N→∞.
If for any N∈ℕ,
setting (5.54)fN(x)=∑|k′|≤N∑τ′∈Ik′N∑ν′=1N(k′,τ′)μ(Qτ′k′,ν′)Dk′(x,yτ′k′,ν′)D¯k′(f)(yτ′k′,ν′), then it is easy to check fN∈𝒢°b(ϵ1,ϵ2).
From the proof of Proposition 5.4 together with Remark 5.5, it is easy to see
that ∥f−fN∥B˙p,qs(𝒳)→0, as N→∞ (here we need p,q<∞). Thus, 𝒢°b(ϵ1,ϵ2) is dense in B˙p,qs(𝒳) with s,p,q as in the proposition, which completes the
proof of Proposition 5.21.We now discuss the boundedness onB˙p,qs(𝒳) and F˙p,qs(𝒳) of the singular integral operators introduced
by Nagel and Stein in [44]. To state their definition, we need first to recall
the notion of a normalized bump function on 𝒳.
A function φ on 𝒳 is said to be an ϵ-bump function associated to a ball B(x0,δ) for some x0∈𝒳 and δ>0,
if it is supported in that ball, and if there exists a constant C>0 independent of φ such that ∥φ∥C˙η(𝒳)≤Cδ−η for all η∈(0,ϵ].
Notice that then ∥φ∥L∞(𝒳)≤C (see Remark 2.14(ii)). If C=1 in this definition, then such a bump function
is called a normalizedϵ-bump function for the ball B(x0,δ).Letϵ∈(0,1].
A linear operator T,
which is initially assumed to be continuous from Cbη(𝒳) to (Cbη(𝒳))′ for all η∈(0,ϵ),
is called a singular integral of
orderϵ if T has a distributional kernel K(x,y) which is locally integrable away from the
diagonal of 𝒳×𝒳,
and satisfies the following conditions.(I-1)
Ifφ,ψ∈Cbη(𝒳) have disjoint supports, then (5.55)〈Tφ,ψ〉=∬𝒳×𝒳K(x,y)φ(y)ψ(x)dμ(x)dμ(y).(I-2)
Ifφ is a normalized ϵ-bump function associated to a ball of radius r,
then ∥Tφ∥C˙η(𝒳)≤Cr−η for all η∈(0,ϵ),
where C>0 is independent of φ.
More precisely, for each η∈(0,ϵ),
there is another η˜∈(0,ϵ) and a constant Cη,η˜>0 so that whenever φ is a Cbη˜(𝒳) function supported in a ball B(x0,r),
then rη∥Tφ∥C˙η(𝒳)≤Cη,η˜supη0≤η˜rη0∥φ∥C˙η0(𝒳).(I-3)
There exists a constantC>0 such that
(I-3)1
for
allx,y∈𝒳 with x≠y, |K(x,y)|≤C(1/V(x,y));(I-3)2
for
allx,x′,y∈𝒳 with d(x,x′)≤d(x,y)/2 and x≠y, (5.56)|K(x,y)−K(x′,y)|+|K(y,x)−K(y,x′)|≤Cd(x,x′)ϵV(x,y)d(x,y)ϵ.(I-4)
Properties (I-1) through (I-3) also hold withx and y interchanged. That is, these properties also
hold for the adjoint operator Tt defined by 〈Ttφ,ψ〉=〈φ,Tψ〉.Remark 5.22.
We remark that ifT is a singular integral operator of order ϵ,
then T extends to a continuous linear operator from Cη(𝒳) to (𝒢°b(η,γ))′ for all η∈(0,ϵ] and all γ>0 by Proposition 2.12.
We also claim that forf∈𝒢°b(η,δ) with η∈(0,ϵ] and δ>0, Tf can be defined as a distribution in (𝒢°0ϵ(β,γ))′ with 0<β,γ≤ϵ.
We first define Tf as a distribution in (𝒢°(β,γ))′ with 0<β≤ϵ and γ>0.
In fact, for any given β∈(0,ϵ],
noticing that 𝒢°b(η1,δ)⊂𝒢°b(η2,δ) when η1≥η2,
without loss of generality, we may assume that η≤β.
Assume that suppf⊂B(x0,r) for some x0∈𝒳 and r>0.
Let ψ∈Cbη(𝒳) such that ψ(x)=1 when x∈B(x0,2r) and ψ(x)=0 when x∉B(x0,4r).
Choose any g∈𝒢°(β,γ) with 0<β≤ϵ and γ>0.
It is easy to check that ψg∈Cbη(𝒳).
Notice that 𝒢°b(η,δ)⊂Cbη(𝒳). Both facts show that 〈Tf,ψg〉 is well defined. On the other hand, we define 〈Tf,(1−ψ)g〉 by (5.57)〈Tf,(1−ψ)g〉=∬𝒳×𝒳[K(x,y)−K(x,x0)]f(y)(1−ψ(x))g(x)dμ(x)dμ(y). By (I-3)2 and Lemma 2.1(i), it is easy to check that(5.58)|〈Tf,(1−ψ)g〉|≲∥f∥L1(𝒳)∥g∥L∞(𝒳)≲∥f∥𝒢(η,δ)∥g∥𝒢(β,γ). Moreover, if g∈𝒢°b(β,γ),
since ∫𝒳f(y)dμ(y)=0,
we then obtain (5.59)〈Tf,(1−ψ)g〉=∬𝒳×𝒳K(x,y)f(y)(1−ψ(x))g(x)dμ(x)dμ(y), which coincides with (I-1).
Furthermore, it is easy to verify that 〈Tf,ψg〉+〈Tf,(1−ψ)g〉 is independent of the choice of ψ.
Thus, we can define Tf by (5.60)〈Tf,g〉=〈Tf,ψg〉+〈Tf,(1−ψ)g〉. In this sense, we have Tf∈(𝒢°(β,γ))′ with 0<β≤ϵ and γ>0.
Now for any g∈𝒢°0ϵ(β,γ) with 0<β,γ≤ϵ,
let {gn}n∈ℕ⊂𝒢°(ϵ,ϵ) such that ∥gn−g∥𝒢(β,γ)→0 as n→∞.
We then define Tf∈(𝒢°0ϵ(β,γ))′ with 0<β,γ≤ϵ by 〈Tf,g〉=limn→∞〈Tf,gn〉. It is easy to check that 〈Tf,g〉 is independent of the choice of {gn}n∈ℕ⊂𝒢°(ϵ,ϵ).
In this sense, we have Tf∈(𝒢°0ϵ(β,γ))′ with 0<β,γ≤ϵ.In what follows, for anyϵ∈(0,1] and 0<β,γ≤ϵ,
we set (5.61)𝒢°0,bϵ(β,γ)={f∈𝒢°0ϵ(β,γ):fhasboundedsupport}.Theorem 5.23.
Letϵ1,ϵ2,ϵ, and |s|<ϵ be as in Definition 5.8. Let T be a singular integral of order ϵ satisfying (I-1) through (I-4). Then T is bounded on B˙p,qs(𝒳) when p(s,ϵ)<p<∞ and 0<q<∞ and bounded from B˙p,qs(𝒳)∩𝒢°b(ϵ1,ϵ2) to B˙p,qs(𝒳) when max{p,q}=∞,
and T is also bounded on F˙p,qs(𝒳) when p(s,ϵ)<p,q<∞, and bounded from F˙p,qs(𝒳)∩𝒢°b(ϵ1,ϵ2) to F˙p,qs(𝒳) when p(s,ϵ)<p<∞ and q=∞.Proof.
By Propositions5.21 and 5.10(vi) together with a density
argument, to prove the theorem, it suffices to verify the conclusions of the
theorem only for all f∈𝒢°0,bϵ(β,γ) with 0<β,γ<ϵ as in (5.35). Let {Sk}k∈ℤ be an ATI with bounded support as constructed in Theorem
2.6. Put Dk=Sk−Sk−1 for k∈ℤ.
We then interpret DkT=Tt(Dkt),
where Dkt(x,y)=Dk(y,x) for all x,y∈𝒳 (cf. Lemma 3.12). By Remark 5.22, we have DkT∈(𝒢°0ϵ(β,γ))′.
On the other hand, let f∈𝒢°0,bϵ(β,γ).
By Theorem 4.12, we have (5.62)f(z)=∑k′∈ℤ∑τ′∈Ik′∑ν′=1N(k′,τ′)μ(Qτ′k′,ν′)Dk′(z,yτ′k′,ν′)D¯k′(f)(yτ′k′,ν′), where the series converges in 𝒢°0ϵ(β,γ) and all the notation is as in Theorem 4.12. Thus, for all x∈𝒳,
we have (5.63)DkTf(x)=∑k′∈ℤ∑τ′∈Ik′∑ν′=1N(k′,τ′)μ(Qτ′k′,ν′)(DkTDk′(⋅,yτ′k′,ν′))(x)D¯k′(f)(yτ′k′,ν′). For k,k′∈ℤ,τ′∈Ik′,ν′=1,…,N(k′,τ′),yτ′k′,ν′∈Qτ′k′,ν′,
and x∈𝒳,
let (5.64)J(x,yτ′k′,ν′)≡(DkTDk′(⋅,yτ′k′,ν′))(x). We now claim that for any fixed η∈(0,ϵ), (5.65)|J(x,yτ′k′,ν′)|≲2−|k−k′|η(2−(k⋀k′)2−(k⋀k′)+d(x,yτ′k′,ν′))η1V2−(k⋀k′)(x)+V2−(k⋀k′)(yτ′k′,ν′)+V(x,yτ′k′,ν′).
To verify (5.65), by symmetry, we only need to verify
that if k≤k′,
then (5.66)|J(x,yτ′k′,ν′)|≲2(k−k′)η1V2−k(x), and moreover, if d(x,yτ′k′,ν′)≥25−k,
then (5.67)|J(x,yτ′k′,ν′)|≲(2−k′d(x,yτ′k′,ν′))η1V(x,yτ′k′,ν′). To see this, for y∈𝒳,
set (5.68)Ψ(y)=∫𝒳K(z,y)Dk(x,z)dμ(z). Notice that for any fixed x∈𝒳, Dk(x,z) is an adjusted bump function in z associated with the ball B(x,22−k).
Conditions (I-2) and (I-4) show that for any fixed η∈(0,ϵ),(5.69)2−kη∥Ψ∥C˙η(𝒳)≲1V2−k(x), which further implies that for
all y∈𝒳, (5.70)|Ψ(y)−Ψ(yτ′k′,ν′)|≲d(y,yτ′k′,ν′)η2−kη1V2−k(x). Thus, (5.71)|J(x,yτ′k′,ν′)|=|∫𝒳Ψ(y)Dk′(y,yτ′k′,ν′)dμ(y)|=|∫𝒳[Ψ(y)−Ψ(yτ′k′,ν′)]Dk′(y,yτ′k′,ν′)dμ(y)|≲2(k−k′)η1V2−k(x). That is, (5.66) holds.
On the other hand, whenk≤k′, z∈B(x,22−k), y∈B(yτ′k′,ν′,22−k′), and d(x,yτ′k′,ν′)≥25−k,
then by Condition (I-3), we have (5.72)|K(z,y)−K(z,yτ′k′,ν′)|≲2−k′ϵd(x,yτ′k′,ν′)ϵV(x,yτ′k′,ν′), which yields
that (5.73)|J(x,yτ′k′,ν′)|=|∬𝒳×𝒳[K(z,y)−K(z,yτ′k′,ν′)]Dk(z,x)Dk′(y,yτ′k′,ν′)dμ(y)dμ(z)|≲(2−k′d(x,yτ′k′,ν′))η1V(x,yτ′k′,ν′), namely, (5.67) holds. Thus,
(5.65) holds.
Notice that the estimate (5.65) is analogous to the
estimate (5.10) in the proof of Proposition 5.4. We can further proceed from
here on exactly as in the proof of Proposition 5.4 in order to obtain the
boundedness of T on B˙p,qs(𝒳) and F˙p,qs(𝒳).
This finishes the proof of Theorem 5.16.


## 5.3. Inhomogeneous Plancherel-Pôlya Inequalities and Definition ofBp,qs(𝒳) and Fp,qs(𝒳)

Throughout this and the next subsection,μ(𝒳) can be finite or infinite. We first introduce
the norms in Bp,qs(𝒳) and Fp,qs(𝒳)
via certain IATI and then verify that these norms are
independent of the choices of IATIs and spaces of distributions; see [37]. To this end, we need to
establish inhomogeneous Plancherel-Pôlya inequalities; see [80]. In what follows, for any
dyadic cube Q,
we set mQ(f)=(1/μ(Q))∫Qf(x)dμ(x).Definition 5.24.
Letϵ1∈(0,1],ϵ2>0,ϵ3>0,ϵ∈(0,ϵ1⋀ϵ2), and let {Sk}k∈ℤ+ be an (ϵ1,ϵ2,ϵ3)-IATI.
Set D0=S0 and Dk=Sk−Sk−1 for k∈ℕ.
Let {Qτ0,ν:τ∈I0,ν=1,…,N(0,τ)} with a fixed large j∈ℕ be dyadic cubes as in Section
4.
(i)
For allf∈(𝒢0ϵ(β,γ))′ with 0<β,γ<ϵ,|s|<ϵ,p(s,ϵ)<p≤∞, and 0<q≤∞,
we define(5.74)∥f∥Bp,qs(𝒳)={∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)[mQτ0,ν(|D0(f)|)]p}1/p+{∑k=1∞2ksq∥Dk(f)∥Lp(𝒳)q}1/qwith the usual
modification made when p=∞ or q=∞.(ii)
For allf∈(𝒢0ϵ(β,γ))′ with 0<β,γ<ϵ,|s|<ϵ,p(s,ϵ)<p<∞, and p(s,ϵ)<q≤∞,
we define (5.75)∥f∥Fp,qs(𝒳)={∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)[mQτ0,ν(|D0(f)|)]p}1/p+∥{∑k=1∞2ksq|Dk(f)|q}1/q∥Lp(𝒳) with the usual
modification made when q=∞.The following theorem is theinhomogeneous Plancherel-Pôlya inequalities.Proposition 5.25.
Letϵ1∈(0,1],ϵ2>0,ϵ3>0,ϵ∈(0,ϵ1⋀ϵ2), and let {Sk}k∈ℤ+ and {Pk}k∈ℤ+ be two (ϵ1,ϵ2,ϵ3)-IATIs. Set D0=S0 and Q0=P0,
and Dk=Sk−Sk−1 and Qk=Pk−Pk−1 for k∈ℕ.
Let {Qτ0,ν:τ∈I0,ν=1,…,N(0,τ)} with a fixed large j∈ℕ be dyadic cubes as in Section 4.
(i)
For allf∈(𝒢0ϵ(β,γ))′ with 0<β,γ<ϵ, |s|<ϵ, p(s,ϵ)<p≤∞, and 0<q≤∞,
{∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)[mQτ0,ν(|D0(f)|)]p}1/p+{∑k=1∞2ksq(∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)[supz∈Qτk,ν|Dk(f)(z)|]p)q/p}1/q(5.76)~{∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)[mQτ0,ν(|Q0(f)|)]p}1/p+{∑k=1∞2ksq(∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)[infz∈Qτk,ν|Qk(f)(z)|]p)q/p}1/q
with the usual modification made
when p=∞ or q=∞.(ii)
For allf∈(𝒢0ϵ(β,γ))′ with 0<β,γ<ϵ, |s|<ϵ, p(s,ϵ)<p<∞, and p(s,ϵ)<q≤∞,(5.77){∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)[mQτ0,ν(|D0(f)|)]p}1/p+∥{∑k=1∞∑τ∈Ik∑ν=1N(k,τ)2ksq[supz∈Qτk,ν|Dk(f)(z)|]qχQτk,ν}1/q∥Lp(𝒳)~{∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)[mQτ0,ν(|Q0(f)|)]p}1/p+∥{∑k=1∞∑τ∈Ik∑ν=1N(k,τ)2ksq[infz∈Qτk,ν|Qk(f)(z)|]qχQτk,ν}1/q∥Lp(𝒳)with the usual modification made
when q=∞.Proof.
We first verify (5.76). By Theorem 4.16 together with Remark 4.17, there
exists a family of functions {D˜k′(x,y)}k′∈ℤ+ such that for all f∈(𝒢0ϵ(β,γ))′ with 0<β,γ<ϵ and all z∈𝒳, (5.78)f(z)=∑τ′∈I0∑ν′=1N(0,τ′)∫Qτ′0,ν′D˜0(x,y)dμ(y)Qτ′,10,ν′(f)+∑k′=1∞∑τ′∈Ik′∑ν′=1N(k′,τ′)μ(Qτ′k′,ν′)D˜k′(z,yτ′k′,ν′)Qk′(f)(yτ′k′,ν′), where Qτ′,10,ν′ denotes the integral operator with kernel (5.79)Qτ′,10,ν′(z)=1μ(Qτ′0,ν′)∫Qτ′0,ν′Q0(u,z)dμ(u) and {D˜k′}k′∈ℤ+ satisfy the same conditions as {D˜k}k∈ℤ+ in
Theorem 4.14. From (5.78) together with (5.80)|Qτ′,10,ν′(f)|=|∫𝒳Qτ′,10,ν′(z)f(z)dμ(z)|=|1μ(Qτ′0,ν′)∫Qτ′0,ν′Q0(f)(u)dμ(u)|≤mQτ′0,ν′(|Q0(f)|), it follows that (5.81)mQτ0,ν(|D0(f)|)=1μ(Qτ0,ν)∫Qτ0,ν|D0(f)(z)|dμ(z)≤∑τ′∈I0∑ν′=1N(0,τ′)mQτ′0,ν′(|Q0(f)|)1μ(Qτ0,ν)∫Qτ0,ν∫Qτ′0,ν′|(D0D˜0)(z,y)|dμ(y)dμ(z)+∑k′=1∞∑τ′∈Ik′∑ν′=1N(k′,τ′)|Qk′(f)(yτ′k′,ν′)|μ(Qτ′k′,ν′)1μ(Qτ0,ν)∫Qτ0,ν|(D0D˜k′)(z,yτ′k′,ν′)|dμ(z)≡Z1+Z2. By Lemma 3.19 together with
Remark 3.3, we have that for any ϵ∈(0,ϵ1⋀ϵ2),
all k′∈ℤ+ and all z,y∈𝒳,(5.82)|(D0D˜k′)(z,y)|≲2−k′ϵ1V1(z)+V1(y)+V(z,y)1(1+d(z,y))ϵ. From this together with Lemma
2.1(iii), it follows that (5.83)supz∈Qτ0,νsupy∈Qτ′0,ν′|(D0D˜0)(z,y)|≲infz∈Qτ0,νinfy∈Qτ′0,ν′1V1(z)+V1(y)+V(z,y)1(1+d(z,y))ϵ. Thus, (5.84)Z1≲∑τ′∈I0∑ν′=1N(0,τ′)mQτ′0,ν′(|Q0(f)|)μ(Qτ′0,ν′)infz∈Qτ0,νinfy∈Qτ′0,ν′1V1(z)+V1(y)+V(z,y)1(1+d(z,y))ϵ. Therefore, if n/(n+ϵ)<p≤1,
by (5.5) and Lemma 5.2, we then have(5.85){∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)[Z1]p}1/p≲{∑τ′∈I0∑ν′=1N(0,τ′)[mQτ′0,ν′(|Q0(f)|)]p[μ(Qτ′0,ν′)]p[V1(zτ′0,ν′)]1−p}1/p≲{∑τ′∈I0∑ν′=1N(0,τ′)μ(Qτ′0,ν′)[mQτ′0,ν′(|Q0(f)|)]p}1/p, where in the last step, we used
the fact that for any yτ′0,ν′∈Qτ′0,ν′, (5.86)V1(yτ′0,ν′)~μ(Qτ′0,ν′). If 1<p≤∞,
by Hölder's inequality and Lemma 5.2, we have
(5.87)Z1≲{∑τ′∈I0∑ν′=1N(0,τ′)[mQτ′0,ν′(|Q0(f)|)]pμ(Qτ′0,ν′)infz∈Qτ0,νinfy∈Qτ′0,ν′1V1(z)+V1(y)+V(z,y)1(1+d(z,y))ϵ}1/p, which together with Lemma 5.2
again yields that (5.88){∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)[Z1]p}1/p≲{∑τ′∈I0∑ν′=1N(0,τ′)μ(Qτ′0,ν′)[mQτ′0,ν′(|Q0(f)|)]p}1/p.This is the desired estimate.
From (5.82), it also follows that (5.89)supz∈Qτ0,ν|(D0D˜k′)(z,yτ′k′,ν′)|≲2−k′ϵinfz∈Qτ0,ν1V1(z)+V1(yτ′k′,ν′)+V(z,yτ′k′,ν′)1(1+d(z,yτ′k′,ν′))ϵ, which proves
that (5.90)Z2≲∑k′=1∞2−k′ϵ∑τ′∈Ik′∑ν′=1N(k′,τ′)μ(Qτ′k′,ν′)|Qk′(f)(yτ′k′,ν′)|×infz∈Qτ0,ν1V1(z)+V1(yτ′k′,ν′)+V(z,yτ′k′,ν′)1(1+d(z,yτ′k′,ν′))ϵ. Notice also that by Lemma 2.19
and (1.3), for any yτ′k′,ν′∈Qτ′k′,ν′, V1(yτ′k′,ν′)≲2−k′nμ(Qτ′k′,ν′). If p(s,ϵ)<p≤1,
then this fact together with (5.5) and Lemma 5.2 proves that(5.91){∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)[Z2]p}1/p≲{∑k′=1∞2k′sq(∑τ′∈Ik′∑ν′=1N(k′,τ′)μ(Qτ′k′,ν′)|Qk′(f)(yτ′k′,ν′)|p)q/p}1/q, where in the last step, we used
(5.5) when q/p≤1 or Hölder's inequality when q/p>1.
If1<p≤∞,
then by Hölder's inequality and Lemma 5.2, we first have(5.92)Z2≲{∑k′=1∞2−k′ϵ′p∑τ′∈Ik′∑ν′=1N(k′,τ′)μ(Qτ′k′,ν′)|Qk′(f)(yτ′k′,ν′)|p×infz∈Qτ0,ν1V1(z)+V1(yτ′k′,ν′)+V(z,yτ′k′,ν′)1(1+d(z,yτ′k′,ν′))ϵ}1/p, where max{−s,0}<ϵ′<ϵ.
This together with Lemma 5.2 again shows that (5.93){∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)[Z2]p}1/p≲{∑k′=1∞2k′sq(∑τ′∈Ik′∑ν′=1N(k′,τ′)μ(Qτ′k′,ν′)|Qk′(f)(yτ′k′,ν′)|p)q/p}1/q, where we used (5.5) when q/p≤1 or Hölder's inequality when q/p>1.
All the above estimates together with the arbitrary
choice ofyτ′k′,ν′ yield that the first term in the left-hand
side of (5.76) is controlled by its right-hand side.
We now verify that the second term of the left-hand
side of (5.76) is also controlled by its right-hand side. To this end, by
(5.78), (5.94)|Dk(f)(z)|≲∑τ′∈I0∑ν′=1N(0,τ′)mQτ′0,ν′(|Q0(f)|)∫Qτ′0,ν′|(DkD˜0)(z,y)|dμ(y)+∑k′=1∞∑τ′∈Ik′∑ν′=1N(k′,τ′)μ(Qτ′k′,ν′)|(DkD˜k′)(z,yτ′k′,ν′)||Qk′(f)(yτ′k′,ν′)|≡Y1+Y2. The estimate for Y2 is as in (5.7), and we only need to estimate Y1.
In this case, by Lemma 3.19 together with Remark 3.3, for any ϵ∈(0,ϵ1⋀ϵ2),
we have (5.95)supz∈Qτk,νY1≲2−kϵ∑τ′∈I0∑ν′=1N(0,τ′)mQτ′0,ν′(|Q0(f)|)μ(Qτ′0,ν′)infz∈Qτk,νinfy∈Qτ′0,ν′1V1(z)+V1(y)+V(z,y)1(1+d(z,y))ϵ, and therefore, if n/(n+ϵ)<p≤1 and s<ϵ,
by (5.5) and Lemma 5.2 together with (5.86), we obtain(5.96){∑k=1∞2ksq(∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)[supz∈Qτk,νY1]p)q/p}1/q≲{∑τ′∈I0∑ν′=1N(0,τ′)μ(Qτ′0,ν′)[mQτ′0,ν′(|Q0(f)|)]p}1/p, while when 1<p≤∞,
by Hölder's inequality and Lemma 5.2,(5.97){∑k=1∞2ksq(∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)[supz∈Qτk,νY1]p)q/p}1/q≲{∑τ′∈I0∑ν′=1N(0,τ′)μ(Qτ′0,ν′)[mQτ′0,ν′(|Q0(f)|)]p}1/p.
All these estimates imply that the left-hand side of
(5.76) is controlled by its right-hand side, which together with the symmetry
verifies (5.76).
Similarly, to establish (5.77), we only need to verify
that its left-hand side is controlled by its right-hand side. To this end, the
estimates for Z1 are still valid for the current case. To
estimate Z2,
by Lemma 5.3 and (5.5) when q≤1 or Hölder's inequality when 1<q≤∞,
we have that for p(s,ϵ)<r<min{1,p,q}, (5.98){∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)[Z2]p}1/p≲∥{∑k′=1∞[M(∑τ′∈Ik′∑ν′=1N(k′,τ′)2k′sr|Qk′(f)(yτ′k′,ν′)|rχQτ′k′,ν′)]q/r}1/q∥Lp(𝒳)≲∥{∑k′=1∞∑τ′∈Ik′∑ν′=1N(k′,τ′)2k′sq|Qk′(f)(yτ′k′,ν′)|qχQτ′k′,ν′}1/q∥Lp(𝒳), where in the last step, we used
Lemma 3.14. This is the desired  estimate.
The estimation forY2 is as in the proof of (5.8). To finish the
proof of (5.77), we still need to estimate Y1.
In what follows, we set a=1 when p/q≤1 and a=q/p when p/q>1.
The estimates (5.95) and (5.5) when p/q≤1 or Hölder's inequality when p/q>1 show that (5.99)∥{∑k=1∞∑τ∈Ik∑ν=1N(k,τ)2ksq[supz∈Qτk,νY1]qχQτk,ν}1/q∥Lp(𝒳)p≲∑k=1∞2k(s−ϵ)pa∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)×[∑τ′∈I0∑ν′=1N(0,τ′)μ(Qτ′0,ν′)mQτ′0,ν′(|Q0(f)|)infz∈Qτk,νinfy∈Qτ′0,ν′1V1(z)+V1(y)+V(z,y)1(1+d(z,y))ϵ]p. If n/(n+ϵ)<p≤1,
by (5.5), s<ϵ, and Lemma 5.2 together with
(5.82),(5.100)∥{∑k=1∞∑τ∈Ik∑ν=1N(k,τ)2ksq[supz∈Qτk,νY1]qχQτk,ν}1/q∥Lp(𝒳)p≲∑τ′∈I0∑ν′=1N(0,τ′)μ(Qτ′0,ν′)[mQτ′0,ν′(|Q0(f)|)]p; and if 1<p≤∞,
by Hölder's inequality and Lemma 5.2,
(5.101)∥{∑k=1∞∑τ∈Ik∑ν=1N(k,τ)2ksq[supz∈Qτk,νY1]qχQτk,ν}1/q∥Lp(𝒳)p≲∑τ′∈I0∑ν′=1N(0,τ′)μ(Qτ′0,ν′)[mQτ′0,ν′(|Q0(f)|)]p, which completes the proof of
(5.77), and hence, that of Proposition 5.25.Remark 5.26.
We point out that Remark5.5 applies in a similar way to Proposition
5.25.Similarly to Proposition5.6, using Proposition 5.4,
we can verify that the definitions of ∥⋅∥Bp,qs(𝒳) and ∥⋅∥Fp,qs(𝒳) are independent of the choices of IATIs. We omit the details.Proposition 5.27.
Let all the notation be as in Proposition5.25.
(i)
For allf∈(𝒢0ϵ(β,γ))′ with 0<β,γ<ϵ, |s|<ϵ, p(s,ϵ)<p≤∞, and 0<q≤∞, (5.102){∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)[mQτ0,ν(|D0(f)|)]p}1/p+{∑k=1∞2ksq∥Dk(f)∥Lp(𝒳)q}1/q~{∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)[mQτ0,ν(|Q0(f)|)]p}1/p+{∑k=1∞2ksq∥Qk(f)∥Lp(𝒳)q}1/q with the usual modification made
when p=∞ or q=∞.(ii)
For allf∈(𝒢0ϵ(β,γ))′ with 0<β,γ<ϵ, |s|<ϵ, p(s,ϵ)<p<∞, and p(s,ϵ)<q≤∞, (5.103){∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)[mQτ0,ν(|D0(f)|)]p}1/p+∥{∑k=1∞2ksq|Dk(f)|q}1/q∥Lp(𝒳)~{∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)[mQτ0,ν(|Q0(f)|)]p}1/p+∥{∑k=1∞2ksq|Qk(f)|q}1/q∥Lp(𝒳) with the usual modification made
when q=∞.We now verify that the definitions of the norm∥⋅∥Bp,qs(𝒳) and the norm ∥⋅∥Fp,qs(𝒳) are independent of the choice of the
underlying space of distributions as follows.Proposition 5.28.
Let all the notation be as in Definition5.24.
(i)
Let|s|<ϵ, p(s,ϵ)<p≤∞, and 0<q≤∞.
If f∈(𝒢0ϵ(β1,γ1))′ with (5.104)max{0,−s+n(1p−1)+}<β1<ϵ,n(1p−1)+<γ1<ϵ, and if ∥f∥Bp,qs(𝒳)<∞,
then f∈(𝒢0ϵ(β2,γ2))′ for every β2,γ2  satisfying (5.104).(ii)
Let|s|<ϵ, p(s,ϵ)<p<∞, and p(s,ϵ)<q≤∞.
If f∈(𝒢0ϵ(β1,γ1))′ with β1,γ1 as in (5.104), and if ∥f∥Fp,qs(𝒳)<∞,
then f∈(𝒢0ϵ(β2,γ2))′ for every β2,γ2 satisfying (5.104).Proof.
Similarly to the proof of Proposition5.7, we only need to verify (i).
Let ψ∈𝒢(ϵ,ϵ) and all the notation as in Theorem 4.14 with N=0.
We first claim that for k∈ℤ+, (5.105)|〈D˜k(⋅,y),ψ〉|≲2−kβ2∥ψ∥𝒢(β2,γ2)1V1(x1)+V(x1,y)1(1+d(x1,y))γ2. In fact, if k∈ℕ,
(5.105) is just (5.24). If k=0,
then (5.106)|〈D˜0(⋅,y),ψ〉|=|∫𝒳D˜0(z,y)ψ(z)dμ(z)|≲∥ψ∥𝒢(β2,γ2)∫𝒳1V1(z)+V1(y)+V(z,y)1(1+d(z,y))ϵ1V1(x1)+V(x1,z)1(1+d(x1,z))γ2dμ(z). Notice that V1(x1)+V(x1,z)~V1(x1)+V1(z)+V(x1,z). By Lemma 4.4, we have (5.107)|〈D˜0(⋅,y),ψ〉|≲∥ψ∥𝒢(β2,γ2)1V1(x1)+V1(y)+V(x1,y){1(1+d(x1,y))ϵ+1(1+d(x1,y))γ2}≲∥ψ∥𝒢(β2,γ2)1V1(x1)+V(x1,y)1(1+d(x1,y))γ2, which shows that (5.105) also
holds when k=0.
From Theorem4.16 together with (5.80), (5.105), and
Lemma 2.1(iii), it follows that(5.108)|〈f,ψ〉|=|∑τ∈I0∑ν=1N(0,τ)∫Qτ0,ν〈D˜0(⋅,y),ψ〉dμ(y)Dτ,10,ν(f)+∑k=1∞∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)〈D˜k(⋅,yτk,ν),ψ〉Dk(f)(yτk,ν)|≲∥ψ∥𝒢(β2,γ2){∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)mQτ0,ν(|D0(f)|)1V1(x1)+V(x1,yτ0,ν)1(1+d(x1,yτ0,ν))γ2+∑k=1∞2−kβ2∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)|Dk(f)(yτk,ν)|1V1(x1)+V(x1,yτk,ν)1(1+d(x1,yτk,ν))γ2}.
Notice that ifp≤1 and γ2>n(1/p−1),
then (5.30) with β2=0 still holds. Thus, when p≤1,
by this fact and (5.5), we have(5.109)|〈f,ψ〉|≲∥ψ∥𝒢(β2,γ2){∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)[mQτ0,ν(|D0(f)|)]p+∑k=1∞2−k[β2p−n(1−p)+sp]2ksp(∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)|Dk(f)(yτk,ν)|p)}1/p≲∥ψ∥𝒢(β2,γ2)∥f∥Bp,qs(𝒳), where in the last inequality, we
used the assumption that β2>−s+n(1/p−1) when p≤1,
and (5.5) when q/p≤1 or Hölder's inequality when q/p>1.
If1<p≤∞,
by Hölder's inequality, we have (5.110)|〈f,ψ〉|≲∥ψ∥𝒢(β2,γ2){(∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)[mQτ0,ν(|D0(f)|)]p)1/p×[∫𝒳1V1(x1)+V(x1,y)1(1+d(x1,y))γ2dμ(y)]1/p′+∑k=1∞2−kβ2(∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)|Dk(f)(yτk,ν)|p)1/p×[∫𝒳1V1(x1)+V(x1,y)1(1+d(x1,y))γ2dμ(y)]1/p′}≲∥ψ∥𝒢(β2,γ2)∥f∥Bp,qs(𝒳), here, again, we used the
assumption β2>−s in this case, and (5.5) when q≤1 or Hölder's inequality when q>1.
Using (5.109) and (5.110) together with an argument
similar to that used in the proof of Proposition 5.7 then completes the proof
of Proposition 5.28.We can now introduce the Besov spacesBp,qs(𝒳) and Triebel-Lizorkin spaces Fp,qs(𝒳).Definition 5.29.
Letϵ1∈(0,1], ϵ2>0, ϵ3>0, ϵ∈(0,ϵ1⋀ϵ2), and let {Sk}k∈ℤ+ be an (ϵ1,ϵ2,ϵ3)-IATI.
Set D0=S0 and Dk=Sk−Sk−1 for k∈ℕ.
Let {Qτ0,ν:τ∈I0,ν=1,…,N(0,τ)} with a fixed large j∈ℕ be dyadic cubes as in Section
4.
(i)
Let|s|<ϵ,p(s,ϵ)<p≤∞, and 0<q≤∞.
The spaceBp,qs(𝒳) is defined to be the set of all f∈(𝒢0ϵ(β,γ))′,
for some β,γ satisfying (5.111)max{s,0,−s+n(1p−1)+}<β<ϵ,n(1p−1)+<γ<ϵ such that (5.112)∥f∥Bp,qs(𝒳)={∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)[mQτ0,ν(|D0(f)|)]p}1/p+{∑k=1∞2ksq∥Dk(f)∥Lp(𝒳)q}1/q<∞ with the usual modifications
made when p=∞ or q=∞.(ii)
Let|s|<ϵ,p(s,ϵ)<p<∞, and p(s,ϵ)<q≤∞. The spaceFp,qs(𝒳) is defined to be the set of all f∈(𝒢0ϵ(β,γ))′ for some β,γ satisfying (5.111) such that(5.113)∥f∥Fp,qs(𝒳)={∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)[mQτ0,ν(|D0(f)|)]p}1/p+∥{∑k=1∞2ksq|Dk(f)|q}1/q∥Lp(𝒳)<∞ with the usual modification made
when q=∞.Remark 5.30.
Propositions5.27 and 5.28 show that the definitions of the
spaces
Bp,qs(𝒳) and Fp,qs(𝒳) are independent of the choice of inhomogeneous
approximations of the identity as in Definition 2.2 and the distribution space (𝒢0ϵ(β,γ))′ with β and γ satisfying (5.111). We also remark that to guarantee
that the definitions of the spaces Bp,qs(𝒳) and Fp,qs(𝒳) are independent of the choice of the
distribution space (𝒢0ϵ(β,γ))′,
we only need the restriction that β and γ satisfy (5.104). Moreover, if we assume that max{0,s}<β<ϵ and n(1/p−1)+<γ<ϵ,
then we can verify that 𝒢(β,γ) is contained in Bp,qs(𝒳) and Fp,qs(𝒳);
see Proposition 5.31 below.


## 5.4. Properties ofBp,qs(𝒳) and Fp,qs(𝒳) and Boundedness of Singular Integrals

In this subsection, we first present some basic
properties ofBp,qs(𝒳) and Fp,qs(𝒳).
Then we establish a Lusin-area characterization of the spaces Fp,qs(𝒳) and the relations between the spaces B˙p,qs(𝒳) and Bp,qs(𝒳) and between the spaces F˙p,qs(𝒳) and
Fp,qs(𝒳).
Using the Lusin-area characterization of Fp,qs(𝒳),
we also obtain the relation between the spaces Fp,qs(𝒳) and the local Hardy spaces hp(𝒳) in the sense of Goldberg [73]. Finally, we obtain the
boundedness on Bp,qs(𝒳)
and Fp,qs(𝒳) of singular integrals with some natural extra
size and regularity conditions of Nagel-Stein type in [44].The following proposition is an inhomogeneous version
of Proposition5.10.Proposition 5.31.
Letϵ1∈(0,1], ϵ2>0, ϵ∈(0,ϵ1⋀ϵ2), and |s|<ϵ.
(i)
Forp(s,ϵ)<p≤∞ and
0<q0≤q1≤∞, Bp,q0s(𝒳)⊂Bp,q1s(𝒳); and for p(s,ϵ)<p<∞ and p(s,ϵ)<q0≤q1≤∞, Fp,q0s(𝒳)⊂Fp,q1s(𝒳).(ii)
Let−ϵ<s+θ<ϵ and θ>0.
Then for p(s,ϵ)<p≤∞ and 0<q0,q1≤∞, Bp,q0s+θ(𝒳)⊂Bp,q1s(𝒳); and for p(s,ϵ)<p<∞ and p(s,ϵ)<q0,q1≤∞, Fp,q0s+θ(𝒳)⊂Fp,q1s(𝒳).(iii)
Ifp(s,ϵ)<p<∞ and p(s,ϵ)<q≤∞,
then Bp,min(p,q)s(𝒳)⊂Fp,qs(𝒳)⊂Bp,max(p,q)s(𝒳).(iv)
Ifβ,γ as in (5.104), then Bp,qs(𝒳)⊂(𝒢0ϵ(β,γ))′ when p(s,ϵ)<p≤∞ and 0<q≤∞,
and Fp,qs(𝒳)⊂(𝒢0ϵ(β,γ))′ when p(s,ϵ)<p<∞ and p(s,ϵ)<q≤∞.(v)
Ifmax{s,0}<β<ϵ and n(1/p−1)+<γ<ϵ,
then 𝒢(β,γ)⊂Bp,qs(𝒳) when p(s,ϵ)<p≤∞ and 0<q≤∞,
and 𝒢(β,γ)⊂Fp,qs(𝒳) when p(s,ϵ)<p<∞ and p(s,ϵ)<q≤∞.(vi)
If1<p<∞,
then Fp,20(𝒳)=Lp(𝒳) with equivalent norms.(vii)
The spacesBp,qs(𝒳) with p(s,ϵ)<p≤∞ and 0<q≤∞ and the spaces Fp,qs(𝒳) with p(s,ϵ)<p<∞ and p(s,ϵ)<q≤∞ are complete.Proof.
Property (i) is a simple corollary of (5.5). To see (ii), we notice that(5.114){∑k=0∞2skq1|bk|q1}1/q1≤supk∈ℤ+2(s+θ)k|bk|{∑l=0∞2−θlq1}1/q1≲supk∈ℤ+2(s+θ)k|bk|,
which combined with (i) verifies (ii); see also [3, the proof of Proposition 2.3.2/2].
The proof of (iii) is similar to that of Property (ii)
in Proposition5.10; see also [3, 6].
Property (iv) is implied by the proof of Proposition5.28, and Property (vii) can be easily deduced from Property (iv) and Property
(vi) is just Proposition 3.30.
To see Property (v), similarly to the proof of (5.105),
for f∈𝒢(β,γ),
we have that for all k∈ℤ+ and x∈𝒳, (5.115)|Dk(f)(x)|≲2−kβ∥f∥𝒢(β,γ)1V1(x1)+V(x1,x)1(1+d(x1,x))γ. Notice that (5.39) with β=0 is still true when γ>n(1/p−1)+.
From this fact and β>s,
it follows that (5.116)∥f∥Bp,qs(𝒳)~{∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)[mQτ0,ν(|D0(f)|)]p}1/p+{∑k=1∞2ksq∥Dk(f)∥Lp(𝒳)q}1/q≲∥f∥𝒢(β,γ){[∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)1(V1(x1)+V(x1,yτ0,ν))p1(1+d(x1,yτ0,ν))γp]1/p+[∑k=1∞2ksq2−kβq]1/q}≲∥f∥𝒢(β,γ). Thus, 𝒢(β,γ)⊂Bp,qs(𝒳),
which together with (i) also proves that 𝒢(β,γ)⊂Fp,qs(𝒳).
This verifies (v) and hence, finishes the proof of Proposition 5.31.Whenp,q≥1,
the following theorem implies that the norms of ∥⋅∥Bp,qs(𝒳) and ∥⋅∥Fp,qs(𝒳) have the following equivalent and simple
version.Proposition 5.32.
Let all the notation be as in Proposition5.25.
(i)
For allf∈(𝒢0ϵ(β,γ))′ with 0<β,γ<ϵ, |s|<ϵ, 1≤p≤∞, and 0<q≤∞, (5.117){∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)[mQτ0,ν(|D0(f)|)]p}1/p+{∑k=1∞2ksq∥Dk(f)∥Lp(𝒳)q}1/q~{∑k=0∞2ksq∥Dk(f)∥Lp(𝒳)q}1/q with the usual modification made
when p=∞ or q=∞.(ii)
For allf∈(𝒢0ϵ(β,γ))′ with 0<β,γ<ϵ, |s|<ϵ, 1≤p<∞, and p(s,ϵ)<q≤∞, (5.118){∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)[mQτ0,ν(|D0(f)|)]p}1/p+∥{∑k=1∞2ksq|Dk(f)|q}1/q∥Lp(𝒳)~∥{∑k=0∞2ksq|Dk(f)|q}1/q∥Lp(𝒳) with the usual modification made
when q=∞.Proof.
We first verify (5.117). To see this, by p≥1,
Hölder's inequality, and Lemma 2.19, we have (5.119){∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)[mQτ0,ν(|D0(f)|)]p}1/p≤{∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)mQτ0,ν(|D0(f)|p)}1/p=∥D0(f)∥Lp(𝒳), which shows that the left-hand
side of (5.117) is controlled by its right-hand side.
To see the converse, by Lemma2.19 and Theorem 4.16
together with a proof similar to that of (5.76) in Proposition 5.25, we have
that for all f∈(𝒢0ϵ(β,γ))′ with 0<β,γ<ϵ,
and |s|<ϵ, (5.120)∥D0(f)∥Lp(𝒳)={∑τ∈I0∑ν=1N(0,τ)∫Qτ0,ν|D0(f)(z)|pdμ(z)}1/p≤{∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)[supz∈Qτ0,ν|D0(f)(z)|]p}1/p≲{∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)[mQτ0,ν(|D0(f)|)]p}1/p+{∑k=1∞2ksq(∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)[infz∈Qτk,ν|Dk(f)(z)|]p)q/p}1/q, which is, up to a bounded
multiplicative constant, controlled by the left-hand side of (5.117). This
completes the proof of (5.117).
The estimate (5.119) also proves that the left-hand
side of (5.118) is controlled by its right-hand side. The converse inequality
can be proved by a way similar to that of (5.77) in Proposition 5.25, which
completes the proof of Proposition 5.32.We next give an inhomogeneous Lusin-area
characterization for the Triebel-Lizorkin  spacesFp,qs(𝒳).Definition 5.33.
Lets∈ℝ, a>0, q∈(0,∞], and let ϵ1∈(0,1], ϵ2>0, ϵ3>0, ϵ∈(0,ϵ1⋀ϵ2]. Let {Sk}k∈ℤ+ be an (ϵ1,ϵ2,ϵ3)-IATI.
Set D0=S0 and Dk=Sk−Sk−1 for k∈ℕ.
The inhomogeneous Lusin-area function (also called the inhomogeneous
Littlewood-PaleyS-function) Sq,as(f)(x) for any f∈(𝒢0ϵ(β,γ))′ with 0<β,γ≤ϵ and x∈𝒳 is given by (5.121)Sq,as(f)(x)={∑k=0∞∫d(x,y)<a2−k2ksq|Dk(f)(y)|qdμ(y)Va2−k(x)}1/q, where the usual modification is
made when q=∞.Theorem 5.34.
Leta>0, ϵ1∈(0,1], ϵ2>0, ϵ3>0, ϵ∈(0,ϵ1⋀ϵ2),
and let {Sk}k∈ℤ+ be an (ϵ1,ϵ2,ϵ3)-IATI with C621−j≤a.
Set D0=S0 and Dk=Sk−Sk−1 for k∈ℤ.
Let |s|<ϵ, p(s,ϵ)<p<∞, 1≤q≤∞, and let Sq,as(f) be as in Definition 5.33 for any f∈(𝒢0ϵ(β,γ))′ with β,γ as in (5.111). Then f∈Fp,qs(𝒳) if and only if f∈(𝒢0ϵ(β,γ))′ for some
β,γ as in (5.111) and Sq,as(f)∈Lp(𝒳).
Moreover, in this case, (5.122)∥f∥Fp,qs(𝒳)~∥Sq,as(f)∥Lp(𝒳).Proof.
Similarly to the proof of Theorem5.13, there exists a constant C>0 such that (5.123)∥Sq,as(f)∥Lp(𝒳)=∥{∑k=0∞∑τ∈Ik∑ν=0N(k,τ)2ksq∫d(⋅,y)<a2−k|Dk(f)(y)|qdμ(y)Va2−k(⋅)χQτk,ν(⋅)}1/q∥Lp(𝒳)≲∥{∑k=0∞∑τ∈Ik∑ν=0N(k,τ)2ksq[supy∈B(zτk,ν,C2−k)|Dk(f)(y)|]qχQτk,ν}1/q∥Lp(𝒳)≲{∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)[supz∈B(zτ0,ν,C)|D0(f)(z)|]p}1/p+∥{∑k=1∞∑τ∈Ik∑ν=0N(k,τ)2ksq[supz∈B(zτk,ν,C2−k)|Dk(f)(z)|]qχQτk,ν}1/q∥Lp(𝒳)≲{∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)[mQτ0,ν(|D0(f)|)]p}1/p+∥{∑k=1∞∑τ∈Ik∑ν=0N(k,τ)2ksq[infz∈Qτk,ν|Dk(f)(z)|]qχQτk,ν}1/q∥Lp(𝒳)~∥f∥Fp,qs(𝒳), where the last inequality can be
proved by a way similar to the proof of (5.76) in Proposition 5.25; see also the
proof of Theorem 5.13.
On the other hand, since1≤q≤∞,
by Hölder's inequality and Lemma 2.19 together with C621−j≤a,
we have that for x∈Qτk,ν with k∈ℤ+, Qτk,ν⊂{y∈𝒳:d(y,x)≤a2−k} and Va(x)~μ(Qτ0,ν),
and (5.124){∫d(x,y)<a|D0(f)(y)|qdμ(y)Va(x)}1/q≥∫d(x,y)<a|D0(f)(y)|dμ(y)Va(x)=∑τ∈I0∑ν=1N(0,τ)∫d(x,y)<a|D0(f)(y)|dμ(y)Va(x)χQτ0,ν(x)≳∑τ∈I0∑ν=1N(0,τ)mQτ0,ν(|D0(f)|)χQτ0,ν(x). Therefore, similarly to the
proof of Theorem 5.13, by the estimate as above and Lemma 2.19
again, (5.125)Sq,as(f)(x)={∑k=0∞∫d(x,y)<a2−k2ksq|Dk(f)(y)|qdμ(y)Va2−k(x)}1/q≳∑τ∈I0∑ν=1N(0,τ)mQτ0,ν(|D0(f)|)χQτ0,ν(x)+{∑k=1∞∑τ∈Ik∑ν=1N(k,τ)2ksq[infy∈Qτk,ν|Dk(f)(y)|]qχQτk,ν(x)}1/q, which together with Proposition
5.25 and Lemma 2.19 proves that (5.126)∥Sq,as(f)∥Lp(𝒳)≳{∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)[mQτ0,ν(|D0(f)|)]p}1/p+∥{∑k=1∞2ksq|Dk(f)|q}1/q∥Lp(𝒳)~∥f∥Fp,qs(𝒳). This finishes the proof of
Theorem 5.34.Remark 5.35.
Comparing Theorem5.34 with Theorem 5.13, we here need to require that 1≤q≤∞ and C621−j≤a due to the inhomogeneity of Triebel-Lizorkin
spaces Fp,qs(𝒳).From Theorem2.29, we immediately obtain the following
technical lemma, which will be useful in applications.Lemma 5.36.
LetS0 be as in Definition 2.2 and ϵ∈(0,ϵ1⋀ϵ2].
Let 0<β,γ≤ϵ, and let S0t be the integral operator with the kernel S0t(x,y)=S0(y,x) for all x,y∈𝒳.
Then S0t is bounded on 𝒢(β,γ),
namely, there exists a constant C>0 such that for all g∈𝒢(β,γ), S0t(g)∈𝒢(β,γ), and (5.127)∥S0t(g)∥𝒢(β,γ)≤C∥g∥𝒢(β,γ).We now establish some relations between homogeneous
Besov and Triebel-Lizorkin spaces with the corresponding inhomogeneous
ones.Proposition 5.37.
Let all the notation be as in Definition5.29 and μ(𝒳)=∞.
Then there exists a constant C>0 such that if f∈Bp,qs(𝒳) or Fp,qs(𝒳),
then f−S0(f)∈B˙p,qs(𝒳) or F˙p,qs(𝒳),
and (5.128)∥f−S0(f)∥B˙p,qs(𝒳)≤C∥f∥Bp,qs(𝒳) or (5.129)∥f−S0(f)∥F˙p,qs(𝒳)≤C∥f∥Fp,qs(𝒳), respectively.Proof.
Letf∈Bp,qs(𝒳) or f∈Fp,qs(𝒳).
By Proposition 5.28, without loss of generality, we may assume that f∈(𝒢0ϵ(β,γ))′ with β,γ as in (5.35). From 𝒢°(β,γ)⊂𝒢(β,γ),
it follows that f∈(𝒢°0ϵ(β,γ))′.
On the other hand, for any g∈𝒢°0ϵ(β,γ),
we have 〈S0(f),g〉=〈f,S0t(g)〉, which together with Lemma 5.36 also shows that S0(f)∈(𝒢°0ϵ(β,γ))′.
Thus, f−S0(f)∈(𝒢°0ϵ(β,γ))′ with β,γ as in (5.35).
To verify the norm inequalities in the proposition, in
what follows, we letI be the identity operator on Bp,qs(𝒳) or Fp,qs(𝒳).
Let
β and γ be as in (5.35), let {Dk}k∈ℤ be as in Definition 5.8, and let {D˜k′}k′∈ℤ+ be as in Theorem 4.14. We first claim that for
all k∈ℤ and k′∈ℤ+, (5.130)|(Dk(I−S0)D˜k′)(x,y)|≲2−|k−k′|ϵ1V2−(k⋀k′)(x)+V2−(k⋀k′)(y)+V(x,y)2−(k⋀k′)ϵ(2−(k⋀k′)+d(x,y))ϵ. We verify (5.130) by considering
the following three cases.
Case 1  (k′≥0≥k).
In this case, we have(5.131)|(Dk(I−S0)D˜k′)(x,y)|=|∫𝒳[Dk(x,z)−Dk(x,y)]((I−S0)D˜k′)(z,y)dμ(z)|≤|∫𝒳[Dk(x,z)−Dk(x,y)]D˜k′(z,y)dμ(z)|+|∫𝒳[Dk(x,z)−Dk(x,y)](S0D˜k′)(z,y)dμ(z)|. On the first term, an argument
similar to the proof of (3.2) in Lemma 3.2 gives the desired estimate. For the
second term, by Lemma 3.19, we first have (5.132)|(S0D˜k′)(z,y)|≲2−k′ϵ1V1(z)+V1(y)+V(z,y)1(1+d(z,y))ϵ, which together with an argument
similar to the proof of (3.2) in Lemma 3.2 also gives the desired estimate for
the second term.
Case 2  (k′≥k>0).
In this case, we write(5.133)|(Dk(I−S0)D˜k′)(x,y)|≤|(DkD˜k′)(x,y)|+|(DkS0D˜k′)(x,y)|. The estimate (3.2) in Lemma 3.2
directly gives the desired estimate for the first term. Denote the second term
by J and write (5.134)J≡|∬𝒳×𝒳Dk(x,z)[S0(z,u)−S0(z,y)]D˜k′(u,y)dμ(u)dμ(z)|≤∫𝒳∫d(u,y)≤(1+d(z,y))/2|Dk(x,z)||[S0(z,u)−S0(z,y)]D˜k′(u,y)|dμ(u)dμ(z)+∫𝒳∫d(u,y)>(1+d(z,y))/2|Dk(x,z)||S0(z,u)D˜k′(u,y)|dμ(u)dμ(z)+∫𝒳∫d(u,y)>(1+d(z,y))/2|Dk(x,z)||S0(z,y)D˜k′(u,y)|dμ(u)dμ(z)≡J1+J2+J3. The regularity of S0 together with Lemma 2.1(ii) yields
that (5.135)J1≲∫𝒳∫d(u,y)≤(1+d(z,y))/2|Dk(x,z)|(d(u,y)1+d(z,y))ϵ1V1(z)+V1(y)+V(z,y)×1(1+d(z,y))ϵ|D˜k′(u,y)|dμ(u)dμ(z)≲2−k′ϵ∫𝒳|Dk(x,z)|1V1(z)+V1(y)+V(z,y)1(1+d(z,y))ϵdμ(z). If d(x,y)≤2−k,
then by Lemma 2.1(ii) again, J1≲2−k′ϵ(1/V2−k(x)), which is the desired estimate. When d(x,y)>2−k,
we further control J1 by (5.136)J1≲2−k′ϵ{∫d(x,z)≥d(x,y)/21V2−k(x)+V2−k(z)+V(x,z)2−kϵ(2−k+d(x,z))ϵ×1V1(z)+V1(y)+V(z,y)1(1+d(z,y))ϵdμ(z)+∫d(x,z)<d(x,y)/2⋯}. Notice that d(x,z)<d(x,y)/2 also implies that d(z,y)≥d(x,y)/2.
From this together with Lemma 2.1(ii), it follows that (5.137)J1≲2−k′ϵ{1V(x,y)2−kϵd(x,y)ϵ+1V(x,y)1d(x,y)ϵ}≲2−(k′−k)ϵ1V(x,y)2−kϵd(x,y)ϵ, which is also the desired
estimate.
The estimate forJ3 is similar to that for J1.
To estimateJ2,
choosing ϵ′>ϵ and using Lemma 2.1(ii), we first
have (5.138)∫d(u,y)>(1+d(z,y))/2|S0(z,u)D˜k′(u,y)|dμ(u)≲1V1(z)2−k′ϵ(1+d(z,y))ϵ; and by the fact that for d(u,y)>(1+d(z,y))/2, V(u,y)~V(y,u)≳V1(y)+V(y,z), we also obtain (5.139)∫d(u,y)>(1+d(z,y))/2|S0(z,u)D˜k′(u,y)|dμ(u)≲1V1(y)+V(y,z)2−k′ϵ(1+d(z,y))ϵ. Therefore, combining these
estimates gives (5.140)J2≲2−k′ϵ∫𝒳|Dk(x,z)|1V1(z)+V1(y)+V(z,y)1(1+d(z,y))ϵdμ(z), which together with some
computations the same as for J1 gives the desired estimate for J2.
Case 3  (k>k′≥0).
The proof for this case is similar to Case2 by symmetry. We omit the details
for simplicity, which completes the proof of (5.130).
Theorem4.16 together with (5.80), (5.130), and Lemma
5.3 yields that for n/(n+ϵ)<r≤1, k∈ℤ, and x∈𝒳, (5.141)|Dk(f−S0(f))(x)|≲2−|k|ϵ2(k⋀0)n(1−1/r){M(∑τ′∈I0∑ν′=1N(0,τ′)[mQτ′0,ν′(|D0(f)|)]rχQτ′0,ν′)(x)}1/r+∑k′=1∞2−|k−k′|ϵ2[(k⋀k′)−k′]n(1−1/r){M(∑τ′∈Ik′∑ν′=1N(k′,τ′)|Dk′(f)(yτ′k′,ν′)|rχQτ′k′,ν′)(x)}1/r≡Z1+Z2.
We now first consider the case of Besov spaces.
Choosingp(s,ϵ)<r<min{1,p} and using s<ϵ and the boundedness of M yield (5.142){∑k∈ℤ2ksq∥Z1∥Lp(𝒳)q}1/q≲{∑τ′∈I0∑ν′=1N(0,τ′)μ(Qτ′0,ν′)[mQτ′0,ν′(|D0(f)|)]p}1/p≲∥f∥Bp,qs(𝒳), which is the desired estimate.
Similarly, choosingr as above, when 1≤p≤∞,
by Minkowski's inequality and the boundedness of M together with the assumption s<ϵ,
we have (5.143){∑k∈ℤ2ksq∥Z2∥Lp(𝒳)q}1/q≲{∑k∈ℤ∥∑k′=1∞2(k−k′)s−|k−k′|ϵ2[(k⋀k′)−k′]n(1−1/r)2k′s×{M(∑τ′∈Ik′∑ν′=1N(k′,τ′)|Dk′(f)(yτ′k′,ν′)|rχQτ′k′,ν′)}1/r∥Lp(𝒳)q}1/q≲{∑k′=1∞2k′sq∥Dk′(f)∥Lp(𝒳)q}1/q≲∥f∥Bp,qs(𝒳), where in the second-to-last
inequality, we used (5.5) when q≤1 and Hölder's inequality when q>1,
while when r<p<1,
instead of Minkowski's inequality by (5.5), we have (5.144){∑k∈ℤ2ksq∥Z2∥Lp(𝒳)q}1/q≲{∑k′=1∞2k′sq∥Dk′(f)∥Lp(𝒳)q}1/q≲∥f∥Bp,qs(𝒳), where in the second-to-last
inequality, we used (5.5) when q/p≤1 and Hölder's inequality when q/p>1.
Combining the above estimates completes the proof of Besov spaces.
We now turn to the case of Triebel-Lizorkin spaces. In
this case, we also choosep(s,ϵ)<r<min{1,p} and use s<ϵ and the boundedness of M to obtain (5.145)∥{∑k∈ℤ2ksq|Z1|q}1/q∥Lp(𝒳)≲{∑τ′∈I0∑ν′=1N(0,τ′)μ(Qτ′0,ν′)[mQτ′0,ν′(|D0(f)|)]p}1/p≲∥f∥Fp,qs(𝒳), while some computations similar
to the proof of (5.8) in Proposition 5.4 also show that(5.146)∥{∑k∈ℤ2ksq|Z2|q}1/q∥Lp(𝒳)≲∥{∑k′=1∞2k′sq|Dk′(f)|q}1/q∥Lp(𝒳)≲∥f∥Fp,qs(𝒳), which completes the proof of
Proposition 5.37.Remark 5.38.
Obviously,S0 in Proposition 5.37 can be replaced by Sk0 with k0∈ℤ or any integral operator with a kernel having
similar properties.Proposition 5.39.
Letϵ be as in Definition 5.29, 0<s<ϵ, and μ(𝒳)=∞.
Then,
(i)
if1≤p≤∞ and 0<q≤∞,
then Bp,qs(𝒳)=B˙p,qs(𝒳)∩Lp(𝒳) and moreover, for any f∈Bp,qs(𝒳), (5.147)∥f∥Bp,qs(𝒳)~∥f∥B˙p,qs(𝒳)+∥f∥Lp(𝒳);(ii)
if1≤p<∞ and p(s,ϵ)<q≤∞,
then Fp,qs(𝒳)=F˙p,qs(𝒳)∩Lp(𝒳) and moreover, for any f∈Fp,qs(𝒳), (5.148)∥f∥Fp,qs(𝒳)~∥f∥F˙p,qs(𝒳)+∥f∥Lp(𝒳).Proof.
We use the same notation as in Definition5.8. Let f∈Bp,qs(𝒳) or f∈Fp,qs(𝒳).
Then, by Proposition 5.28, it is easy to see that f∈(𝒢°0ϵ(β,γ))′ with β,γ as in (5.35).
To verify (i), letf∈Bp,qs(𝒳).
If 1≤p<∞,
Proposition 3.18, Minkowski's inequality, and Hölder's inequality when 1≤q≤∞ or (5.5) when 0<q<1 show that (5.149)∥f∥Lp(𝒳)≤∥S0(f)∥Lp(𝒳)+∑k=1∞∥Dk(f)∥Lp(𝒳)≲{∥S0(f)∥Lp(𝒳)q+∑k=1∞2ksq∥Dk(f)∥Lp(𝒳)q}1/q≲∥f∥Bp,qs(𝒳). If p=∞,
by Theorem 3.29, we have that f=D˜0S0(f)+∑k=1∞D˜kDk(f) holds in (𝒢(β,γ))′ with s<β<ϵ and 0<γ<ϵ,
where D˜k with k∈ℤ+ is as in Theorem 3.26. From this and f∈B∞,qs(𝒳),
it follows that for almost all x∈𝒳, (5.150)|f(x)|≲∥f∥B∞,qs(𝒳){∫𝒳|D˜0(x,y)|dμ(y)+∑k=1∞2−ks∫𝒳|D˜k(x,y)|dμ(y)}≲∥f∥B∞,qs(𝒳).
Moreover, when1≤p≤∞,
Proposition 2.7(iii) further yields (5.151)∥f∥B˙p,qs(𝒳)~{∑k=−∞∞2ksq∥Dk(f)∥Lp(𝒳)q}1/q≲{∑k=−∞02ksq∥Dk(f)∥Lp(𝒳)q}1/q+{∑k=1∞2ksq∥Dk(f)∥Lp(𝒳)q}1/q≲∥f∥Bp,qs(𝒳). Thus, f∈B˙p,qs(𝒳)∩Lp(𝒳) and ∥f∥Lp(𝒳)+∥f∥B˙p,qs(𝒳)≲∥f∥Bp,qs(𝒳).
Conversely, iff∈B˙p,qs(𝒳)∩Lp(𝒳),
it is obvious that f∈(𝒢(β,γ))′ with β,γ as in (5.111); and moreover, Proposition 2.7(iii) also proves (5.152)∥f∥Bp,qs(𝒳)~{∥S0(f)∥Lp(𝒳)q+∑k=1∞2ksq∥Dk(f)∥Lp(𝒳)q}1/q≲∥f∥Lp(𝒳)+∥f∥B˙p,qs(𝒳), which completes the proof of (i).
To prove (ii), letf∈Fp,qs(𝒳).
From Proposition 3.18 and Hölder's inequality when 1≤q≤∞ or (5.5) when p(s,ϵ)<q<1,
it follows that (5.153)∥f∥Lp(𝒳)≤∥|S0(f)|+∑k=1∞|Dk(f)|∥Lp(𝒳)≲∥{|S0(f)|+∑k=1∞2ksq|Dk(f)|q}1/q∥Lp(𝒳)≲∥f∥Fp,qs(𝒳),∥f∥F˙p,qs(𝒳)~∥{∑k∈ℤ2ksq|Dk(f)|q}1/q∥Lp(𝒳)≲∥{∑k=−∞02ksq|Dk(f)|q}1/q∥Lp(𝒳)+∥{∑k=1∞2ksq|Dk(f)|q}1/q∥Lp(𝒳)≲Z+∥f∥Fp,qs(𝒳). If p/q≤1,
by (5.5), Proposition 2.7(iii), and s>0, (5.154)(Z)p≲∑k=−∞02ksp∥Dk(f)∥Lp(𝒳)p≲∥f∥Lp(𝒳)p∑k=−∞02ksp≲∥f∥Lp(𝒳)p, while when p/q>1,
by Minkowski's inequality, Proposition 2.7(iii), and s>0, (5.155)Z≲{∑k=−∞02ksq∥Dk(f)∥Lp(𝒳)q}1/q≲∥f∥Lp(𝒳){∑k=−∞02ksq}1/q≲∥f∥Lp(𝒳). Thus, f∈F˙p,qs(𝒳)∩Lp(𝒳) and ∥f∥Lp(𝒳)+∥f∥F˙p,qs(𝒳)≲∥f∥Fp,qs(𝒳).
Conversely, iff∈F˙p,qs(𝒳)∩Lp(𝒳),
it is again obvious that f∈(𝒢(β,γ))′ with β,γ as in (5.111); and moreover, Proposition 2.7(iii) yields that(5.156)∥f∥Fp,qs(𝒳)~∥{|S0(f)|q+∑k=1∞2ksq|Dk(f)|q}1/q∥Lp(𝒳)≲∥f∥Lp(𝒳)+∥f∥F˙p,qs(𝒳), which completes the proof of
(ii) and hence, the proof of Proposition 5.39.We now introduce the local Hardy spaces in the sense
of Goldberg [73].Definition 5.40.
Let all the notation be as in Definition5.29. The local Hardy spacehp(𝒳),
for n/(n+ϵ)<p≤1,
is defined to be the inhomogeneous Triebel-Lizorkin space Fp,20(𝒳),
with norm ∥f∥hp(𝒳)≡∥f∥Fp,20(𝒳).We now introduce the definitions ofhp(𝒳)-atoms and hp(𝒳)-blocks.Definition 5.41.
Letj∈ℕ be as in Theorem 5.34 and 0<p≤1.
A function a on 𝒳 is called an hp(𝒳)-atom if a satisfies (i) through (iii) of Definition 5.15
with r<C62−j,
and a is called an hp(𝒳)-block if a satisfies (i) and (ii) of Definition 5.15 with r≥C62−j.Applying Theorem5.34 and Proposition 5.37, we obtain
the following atomic and block decomposition characterization for the local
Hardy spaces hp(𝒳),
which is similar to Theorem 5.16.Theorem 5.42.
Letϵ and β,γ be as in Definition 5.29. If n/(n+ϵ)<p≤1,
then f∈hp(𝒳) if and only if there exist two sequences of
numbers {λk}k∈ℤ+ and {μk}k∈ℤ+ with ∑k=0∞|λk|p<∞ and ∑k=0∞|μk|p<∞,
a sequence of hp(𝒳)-atoms {ak}k∈ℤ+ and a sequence of hp(𝒳)-blocks {bk}k∈ℤ+ such that f=∑k=0∞λkak+∑k=0∞μkbk in (𝒢0ϵ(β,γ))′.
Moreover, in this case, (5.157)∥f∥hp(𝒳)~inf{(∑k=0∞|λk|p)1/p+(∑k=0∞|μk|p)1/p}, where the infimum is taken over
all the above decompositions of f.Proof.
Letf∈hp(𝒳).
In order to derive the decomposition of f into atoms and blocks, we will assume for
simplicity that μ(𝒳)=∞.
The case where μ(𝒳)<∞ can be proved by using Theorems 5.34,
2.6, and 3.29 together with an argument similar to that used for the
proof of the necessity of Theorem 2.21 in [48].
By Proposition5.37, we have f−S0(f)∈Hp(𝒳) and moreover, (5.158)∥f−S0(f)∥Hp(𝒳)≲∥f∥hp(𝒳), where S0 is as in Definition 5.29. By Theorem 5.16, there
exist a sequence of numbers {λk}k=0∞⊂ℂ with ∑k=0∞|λk|p<∞ and a sequence of Hp(𝒳)-atoms {ak}k=0∞ such that in (𝒢°0ϵ(β,γ))′, (5.159)f−S0(f)=∑k=0∞λkak,(∑k=0∞|λk|p)1/p≲∥f−S0(f)∥Hp(𝒳). Let S0t denote the adjoint operator to S0 with integral kernel S0t(x,y)=S0(y,x).
Notice that if g∈𝒢0ϵ(β,γ),
then g−S0t(g)∈𝒢°0ϵ(β,γ) by Lemma 5.36, which together with (5.159) shows
that (5.160)〈f,g〉=〈2S0(f)−S0S0(f),g〉+∑k=0∞λk〈ak−S0(ak),g〉. Using Lemma 2.19, we
have (5.161)2S0(f)(x)−S0S0(f)(x)=∑τ∈I0∑ν=1N(0,τ)[2S0(f)(x)−S0S0(f)(x)]χQτ0,ν(x)≡∑τ∈I0∑ν=1N(0,τ)λτ0,νbτ0,ν(x), where (5.162)λτ0,ν≡[μ(Qτ0,ν)]1/psupz∈Qτ0,ν|2S0(f)(z)−S0S0(f)(z)|, and bτ0,ν(x)≡0 when λτ0,ν=0, otherwise, (5.163)bτ0,ν(x)≡1λτ0,ν[2S0(f)(x)−S0S0(f)(x)]χQτ0,ν(x). It is easy to see that suppbτ0,ν⊂Qτ0,ν⊂B(zτ0,ν,C62−j) and(5.164)∥bτ0,ν∥L2(𝒳)≤[μ(Qτ0,ν)]1/21λτ0,νsupz∈Qτ0,ν|2S0(f)(z)−S0S0(f)(z)|≲[μ(B(zτ0,ν,C62−j))]1/2−1/p. Thus, bτ0,ν is an hp(𝒳)-block multiplied with a normalizing constant.
Moreover, noticing that the kernel of S0S0 has the properties similar to S0 by Lemma 3.19, applying Theorem 4.16, and using
an argument similar to the proof of (5.77) in Proposition 5.25, we obtain(5.165){∑τ∈I0∑ν=1N(0,τ)|λτ0,ν|p}1/p≲{∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)[supz∈Qτ0,ν|2S0(f)(z)−S0S0(f)(z)|]p}1/p≲{∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)[mQτ0,ν(|D0(f)|)]p}1/p+∥{∑k=1∞∑τ∈Ik∑ν=1N(k,τ)[infz∈Qτk,ν|Dk(f)(z)|]2χQτk,ν}1/2∥Lp(𝒳)~∥f∥hp(𝒳). We next show that (5.166)〈2S0(f)−S0S0(f),g〉=∑τ∈I0∑ν=0N(0,τ)λτ0,ν〈bτ0,ν,g〉, which can be deduced by |〈bτ0,ν,g〉|≲1. To see this, by Hölder's inequality, Lemma 2.1(iii), and (5.30) in the proof of Proposition 5.7 together with the assumption γ>n(1/p−1),
we have (5.167)|〈bτ0,ν,g〉|≤∥bτ0,ν∥L2(𝒳)∥gχQτ0,ν∥L2(𝒳)≤[μ(Qτ0,ν)]1/2−1/p{∫Qτ0,ν[1V1(x1)+V(x1,x)1(1+d(x1,x))γ]2dμ(x)}1/2≲1[V1(x1)]1/p≲1.
Notice that by (5.168) below, we have ∥ak∥hp(𝒳)≲1.
We can obtain a desired hp(𝒳)-block decomposition of S0(ak) in the same way as 2S0(f)−S0S0(f),
which completes the proof of the necessity.
We now use Theorem5.34 to verify the sufficiency of
the condition in the theorem. To this end, by Fatou's lemma, it suffices to
prove that for any hp(𝒳)-atom or any hp(𝒳)-block b
, (5.168)∥S2,a0(b)∥Lp(𝒳)≲1, where a≥C621−j is as in Theorem 5.34. If b is an hp(𝒳)-atom, the estimate (5.168) can be established
by an argument similar to the proof of the sufficiency of Theorem 2.21 in
[48]. We omit the
details. Now, we suppose b is an hp(𝒳)-block supported on B(x0,r) with r≥C62−j.
Choose N∈ℕ such that N≥max{2,2/a} and write (5.169)∥S2,a0(b)∥Lp(𝒳)p=∫B(x0,Nar)[S2,a0(b)(x)]pdμ(x)+∫𝒳∖B(x0,Nar)…≡J1+J2. Hölder's inequality together
with the size condition of b shows that (5.170)J1≤[μ(B(x0,Nar))]1−p/2∥S2,a0(b)∥L2(𝒳)p≲[μ(B(x0,r))]1−p/2∥b∥L2(𝒳)p≲1. Now if x∈𝒳∖B(x0,Nar), d(y,x)≤a2−k with k∈ℤ+ and u∈B(x0,r),
we then have d(u,x0)≤(1/2)(2−k+d(y,x0)),
which together with the size conditions of Dk and b,
Lemma 2.1(iii), and Hölder's inequality yields that (5.171)|Dk(b)(y)|≲[μ(B(x0,r))]1−1/p1V2−k(y)+V2−k(x0)+V(y,x0)2−kϵ(2−k+d(y,x0))ϵ. From this, it follows
that (5.172)S2,a0(b)(x)≲[μ(B(x0,r))]1−1/p{∑k=0∞[1V2−k(x)+V2−k(x0)+V(x,x0)2−kϵ(2−k+d(x,x0))ϵ]2}1/2. Therefore, if we choose ϵ′∈(0,ϵ) such that p>n/(n+ϵ′),
by (5.5), we then have (5.173)J2≲1r(ϵ−ϵ′)p[μ(B(x0,r))]p−1∑k=0∞2−k(ϵ−ϵ′)p×∫𝒳∖B(x0,Nar)(1V2−k(x)+V2−k(x0)+V(x,x0))p2−kϵ′p(2−k+d(x,x0))ϵ′pdμ(x)≲1r(ϵ−ϵ′)p[μ(B(x0,r))]p−1∑k=0∞2−k(ϵ−ϵ′)p×∑l=k∞∫2l2−kNar≤d(x,x0)<2l+12−kNar(1V2−k(x)+V2−k(x0)+V(x,x0))p2−kϵ′p(2−k+d(x,x0))ϵ′pdμ(x)≲1, which completes the proof
of
Theorem 5.42.We now recall the definition of the Lipschitz spaceLips(𝒳) with s>0;
see [74].Definition 5.43.
Lets>0.
The Lipschitz (or Hölder) spaceLips(𝒳) is defined to be the set of all functions f on X such that (5.174)∥f∥Lips(𝒳)=supx∈B,rB≥1|f(x)|μ(B)s+supx≠y|f(x)−f(y)|V(x,y)s<∞, where the first supremum is
taken over all balls B of 𝒳 with radius rB≥1.Observe that these classes are rather Lipschitz (or
Hölder) classes with respect to the measure distanceρ(x,y)≡inf{μ(B):x,y∈B,Ba  ball},
not the distance d.The dual spaces ofh1(𝒳) and hp(𝒳) when p<1 are proved, respectively, to be bmo(𝒳) and Lip1/p−1(𝒳) in [74] as follows.Theorem 5.44.
(i) The spacebmo(𝒳) is the dual space of h1(𝒳), in the following sense: if f=∑k=0∞λkak+∑k=0∞μkbk∈h1(𝒳),
with atom's ak and block's bk,
is as in Theorem 5.42, then for each g∈bmo(𝒳), (5.175)limN→∞{∑k=0Nλk∫𝒳ak(x)g(x)dμ(x)+∑k=0Nμk∫𝒳bk(x)g(x)dμ(x)} is a well-defined continuous
linear functional ℒg:f↦〈f,g〉 with norm ≲∥g∥bmo(𝒳).
Conversely, each continuous linear functionalℒ on h1(𝒳) has the form ℒ=ℒg for some g∈bmo(𝒳) with ∥g∥bmo(𝒳)≲∥ℒ∥.
(ii) Assume thatϵ is as in Definition 5.29, n/(n+ϵ)<p<1, and s=1/p−1.
Then Lips(𝒳) is the dual space of hp(𝒳) in the sense of (i).Remark 5.45.
We point out that Remark5.20 applies in a similar way to Theorem 5.44.In what follows, for anyβ,γ>0,
we let (5.176)𝒢b(β,γ)={f∈𝒢(β,γ):fhasboundedsupport}.Using Proposition5.31, by an argument similar to the
proof of Proposition 5.21, we establish the following density result for Bp,qs(𝒳) and Fp,qs(𝒳).
We omit the details.Proposition 5.46.
Letϵ1,ϵ2,ϵ, and let |s|<ϵ be as in Definition 5.29. Then 𝒢b(ϵ1,ϵ2) is dense in Bp,qs(𝒳) when p(s,ϵ)<p<∞ and 0<q<∞,
and in Fp,qs(𝒳) when p(s,ϵ)<p,q<∞.We now turn to boundedness results for singular
integral operators onBp,qs(𝒳) and Fp,qs(𝒳) spaces. In what follows, it will be convenient
to put ∥f∥C˙η(𝒳)≡∥f∥L∞(𝒳) when
η=0.Letϵ>0 and σ>0.
A linear operator T,
which is initially assumed to be continuous from Cbη(𝒳) to (Cbη(𝒳))′ for all η∈(0,ϵ),
is called an inhomogeneous singular
integral of order(ϵ,σ) if T has a distributional kernel K which satisfies the conditions (I-1) through
(I-4) of the kernel of a singular integral of order ϵ in Subsection 5.2, and the following additional
“vanishing" condition that(I-2)1
the property (I-2) also holds in the limiting
caseη=η˜=0,
for T and its adjoint operator Tt,
that is, there exists a constant C>0 such that for every normalized ϵ-bump function φ, ∥Tφ∥L∞(𝒳)+∥Ttφ∥L∞(𝒳)≲C, as well as the following additional size condition that(I-3)3
for allx,y∈𝒳 with d(x,y)≥1, |K(x,y)|≤C(1/V(x,y))(1/d(x,y)σ).Remark 5.47.
We point out that ifT is a singular integral operator of order (ϵ,σ),
then T extends to a continuous linear operator from Cη(𝒳) to (𝒢b(η,γ))′ for all η∈(0,ϵ] and all γ>0;
see Proposition 2.25.
We also claim that forf∈𝒢b(η,δ) with η∈(0,ϵ] and δ>0, Tf can be defined as a distribution in (𝒢0ϵ(β,γ))′ with 0<β,γ≤ϵ.
We first define Tf as a distribution in (𝒢(β,γ))′ with 0<β≤ϵ and γ>0.
In fact, for any given β∈(0,ϵ],
noticing that 𝒢b(η1,δ)⊂𝒢b(η2,δ) when η1≥η2,
without loss of generality, we may assume that η≤β.
Assume that suppf⊂B(x0,r) for some x0∈𝒳 and r>0.
Let ψ∈Cbη(𝒳) such that ψ(x)=1 when x∈B(x0,2r) and ψ(x)=0 when x∉B(x0,4r).
For any g∈𝒢(β,γ) with η≤β≤ϵ and γ>0,
it is easy to see that ψg∈Cbη(𝒳).
From this and 𝒢b(η,δ)⊂Cbη(𝒳),
it follows that 〈Tf,ψg〉 is well defined. On the other hand, we define 〈Tf,(1−ψ)g〉 by (5.177)〈Tf,(1−ψ)g〉=∬𝒳×𝒳K(x,y)f(y)(1−ψ(x))g(x)dμ(x)dμ(y).Clearly, if suppf∩supp{(1−ψ)g}=∅,
this definition coincides with (I-1). Moreover, by (I-3)3 and Lemma 2.1(i), we have (5.178)|〈Tf,(1−ψ)g〉|≲∥f∥L∞(𝒳)∥g∥L1(𝒳)≲∥f∥𝒢(η,δ)∥g∥𝒢(β,γ). It is also easy to verify that 〈Tf,ψg〉+〈Tf,(1−ψ)g〉 is independent of the choice of ψ.
Thus, we can define Tf by 〈Tf,g〉=〈Tf,ψg〉+〈Tf,(1−ψ)g〉, so that Tf∈(𝒢°(β,γ))′ with 0<β≤ϵ and γ>0.
Now for any g∈𝒢0ϵ(β,γ) with 0<β,γ≤ϵ,
let {gn}n∈ℕ⊂𝒢(ϵ,ϵ) such that ∥gn−g∥𝒢(β,γ)→0 as n→∞.
We then define Tf∈(𝒢0ϵ(β,γ))′ with 0<β,γ≤ϵ by 〈Tf,g〉=limn→∞〈Tf,gn〉. It is easy to check that 〈Tf,g〉 is independent of the choice of {gn}n∈ℕ⊂𝒢(ϵ,ϵ).
In this sense, we have Tf∈(𝒢0ϵ(β,γ))′ with 0<β,γ≤ϵ.In what follows, forϵ∈(0,1] and 0<β,γ≤ϵ,
put (5.179)𝒢0,bϵ(β,γ)={f∈𝒢0ϵ(β,γ):fhasbounded  support}.Theorem 5.48.
Letϵ1,ϵ2, ϵ, and |s|<ϵ be as in Definition 5.29. Let σ>0 and let T be a singular integral of order (ϵ,σ) with σ>n(1/p−1)+.
Then T is bounded on Bp,qs(𝒳) when p(s,ϵ)<p<∞ and 0<q<∞ and bounded from Bp,qs(𝒳)∩𝒢b(ϵ1,ϵ2) to Bp,qs(𝒳) when max{p,q}=∞,
and T is also bounded on Fp,qs(𝒳) when p(s,ϵ)<p,q<∞,
and bounded from Fp,qs(𝒳)∩𝒢b(ϵ1,ϵ2) to Fp,qs(𝒳) when p(s,ϵ)<p<∞ and q=∞.Proof.
By Propositions5.46 and 5.31(vii) together with a density
argument, it suffices to verify the conclusions of the theorem for all f∈𝒢0,bϵ(β,γ) with β,γ as in (5.111).
Let{Sk}k∈ℤ+ be an IATI with bounded support as constructed in Theorem
2.6. Put D0=S0 and Dk=Sk−Sk−1 for k∈ℕ.
Using Remark 5.47 and Theorem 4.15 together with an argument similar to the
proof of Theorem 5.23, we see that for all k∈ℤ+ and x∈𝒳, (5.180)DkTf(x)=∑τ′∈I0∑ν′=1N(0,τ′)D¯τ′,10,ν′(f)∫Qτ′0,ν′(DkTD0(⋅,y))(x)dμ(y)+∑k′=1∞∑τ′∈Ik′∑ν′=1N(k′,τ′)μ(Qτ′k′,ν′)(DkTDk′(⋅,yτ′k′,ν′))(x)D¯k′(f)(yτ′k′,ν′), where all the notation is as in
Theorem 4.15.
Fork,k′∈ℤ+, τ′∈Ik′, ν′=1,…,N(k′,τ′), y∈Qτ′k′,ν′, and x∈𝒳,
let Z(x,y)≡(DkTDk′(⋅,y))(x). We now claim that when k=k′=0, (5.181)|Z(x,y)|≲(11+d(x,y))σ1V1(x)+V1(y)+V(x,y); when k=0 and k′∈ℕ,
for any fixed η∈(0,ϵ), (5.182)|Z(x,y)|≲2−k′η(11+d(x,y))η1V1(x)+V1(y)+V(x,y); when k∈ℕ and k′=0,
for any fixed η∈(0,ϵ), (5.183)|Z(x,y)|≲2−kη(11+d(x,y))η1V1(x)+V1(y)+V(x,y); and when k,k′∈ℕ,
for any fixed η∈(0,ϵ), (5.184)|Z(x,y)|≲2−|k−k′|η(2−(k⋀k′)2−(k⋀k′)+d(x,y))η1V2−(k⋀k′)(x)+V2−(k⋀k′)(y)+V(x,y).
Obviously, (5.184) is just (5.65). The estimate (5.183)
is easily deduced from (5.182) by symmetry. Thus, we only need to verify (5.181)
and (5.182). Assume that k′∈ℤ+.
To prove (5.181) and (5.182), it suffices to prove that (5.185)|Z(x,y)|≲2−k′η11V1(x), and that when d(x,y)≥25,
then (5.186)|Z(x,y)|≲(2−k′d(x,y))η21V(x,y), where when k′=0, η1=0, and η2=σ,
and when k′∈ℕ, η1=η2=η.
To see (5.185), for z∈𝒳,
set Ψ(z)≡∫𝒳K(w,z)D0(w,x)dμ(w). From the properties of S0 in Theorem 2.6, Condition (I-2), Condition
(I-4), and Condition (I-2)1, it follows that for any fixed η∈[0,ϵ), ∥Ψ∥C˙η(𝒳)≲1/V1(x), which shows that for all z∈𝒳, |Ψ(z)|≲1/V1(x), and for η∈(0,ϵ) and all z,y∈𝒳, (5.187)|Ψ(z)−Ψ(y)|≲d(z,y)η1V1(x).Thus, when k′=0,
by size conditions of both Ψ and S0,
then (5.188)|Z(x,y)|=|∫𝒳Ψ(z)S0(z,y)dμ(z)|≲1V1(x), while when k′∈ℕ,
by the vanishing moment and the size condition of Dk′ together with the regularity of Ψ,
for any fixed η∈(0,ϵ), (5.189)|Z(x,y)|=|∫𝒳[Ψ(z)−Ψ(y)]Dk′(z,y)dμ(z)|≲2−k′η1V1(x),which verifies (5.185).
To see (5.186), we first notice that when d(z,y)<22−k′, d(w,x)<22, and d(x,y)≥25,
then d(w,z)≥d(x,y)−d(w,x)−d(z,y)>max{d(x,y)/2,8,8d(z,y)}, d(w,y)>d(x,y)/2, and d(w,z)≥d(w,y)−d(z,y)>d(w,y)/2.
Therefore, (5.190)V(w,z)≳V(w,y)≳V(y,x)~V(x,y).Thus, if k′=0,
by (I-3)3, we have (5.191)|Z(x,y)|≲∬𝒳×𝒳|D0(w,x)|1V(w,z)1d(z,y)σ|D0(z,y)|dμ(w)dμ(z)≲1V(x,y)1d(x,y)σ, while when k′∈ℕ,
by the regularity on K and the vanishing moment of Dk′,
for any fixed η∈(0,ϵ),
we have (5.192)|Z(x,y)|≲∬𝒳×𝒳|D0(w,x)|d(z,y)ϵV(w,z)d(w,z)ϵ|Dk′(z,y)|dμ(w)dμ(z)≲(2−k′d(x,y))ϵ1V(x,y),which implies (5.186). Thus, the
estimates (5.181) and (5.182) hold.
Using these estimates and Remark5.26, by a procedure
essentially similar to the proof of Proposition 5.25, we then obtain the
boundedness on Bp,qs(𝒳) and Fp,qs(𝒳) of T.
The details are left to the reader.


## 5.5.T(1)-Theorems

In this subsection, we will establish analogues of David
and Journé'sT(1)-theorem [68] and of Stein's variant of this theorem in [75] for RD-spaces.We begin with generalizing [71, Theorem 1, page 114] on ℝn to the setting of spaces of homogeneous type
which is of independent of interest; see also [69, Proposition 2].Theorem 5.49.
Letϵ∈(0,1], β∈(0,ϵ), and let T be as in Proposition 2.12 with a distributional
kernel K satisfying the size condition (2.59). Then T can be extended as a continuous linear
operator on C˙β(𝒳) if and only if T∈WBP(β) and T(1)=0 in (C°bβ(𝒳))′.Proof.
We first prove the sufficiency. To this end, for anyk1∈ℤ,
let (5.193)Bk1={x∈𝒳:2k1−1≤d(x,x1)<2k1}. Fix any x0∈Bk1.
Let θ be as in Lemma 2.15. For any k2∈ℤ,
put θk2(y)=θ(d(x0,y)/3⋅2k2), and define ωk2=1−θk2.
For any f∈Cβ(𝒳),
following an argument as in the proof of Lemma 2.20 (see also [69, the proof of Proposition 2]), for a.e.x∈B(x0,(3/2)2k2),
we have (5.194)Tf(x)=∫𝒳K(x,y)[f(y)−f(x)]θk2(y)dμ(y)+[∫𝒳K(x,y)f(y)ωk2(y)dμ(y)+f(x)T(θk2)(x)]=Γ1(x)+Γ2(x). By Lemma 2.15, for a.e.x∈B(x0,(3/2)2k2),
we also have (5.195)|T(θk2)(x)|≲CT+∥T∥WBP(β),(5.196)T(θk2)(x)=Ck2−∫𝒳[K(x,y)−K(x0,y)]ωk2(y)dμ(y), where Ck2 is a constant independent of x.
For anyx∈B(x0,2k2/2),
we then consider x′∈𝒳 satisfying 2k2−1≤d(x,x′)<2k2.
Notice that if x∈B(x0,2k2/2) and d(x,x′)<2k2,
then x′∈B(x0,(3/2)2k2),
and that θk2(y)≠0 implies d(x,y)<14d(x,x′).
Thus, if x,x′ satisfy (5.194), (5.195), and (5.196), by the
size condition (2.59) of the kernel K,
the definition of C˙β(𝒳) and Lemma 2.1(i), we then have (5.197)|Γ1(x)|≤∫d(x,y)<14d(x,x′)|K(x,y)||f(y)−f(x)|dμ(y)≲d(x,x′)βCT∥f∥C˙β(𝒳) and a similar estimate also
holds for Γ1(x′),
which clearly implies (5.198)|Γ1(x)−Γ1(x′)|≤∥f∥C˙β(𝒳)CTd(x,x′)β.
Moreover, by (5.196), we also have (5.199)Γ2(x)−Γ2(x′)=∫𝒳[K(x,y)−K(x′,y)][f(y)−f(x)]ωk2(y)dμ(y)+[f(x)−f(x′)]∫𝒳K(x′,y)θk2(y)dμ(y)=Γ2,1+Γ2,2. The estimate (5.195) yields (5.200)|Γ2,2|≲d(x,x′)β(CT+∥T∥BMO(β))∥f∥C˙β(𝒳). Notice that ωk2(y)≠0 implies that d(x,y)>2d(x,x′).
The regularity (2.49) on K and the definition of C˙β(𝒳) together with Lemma 2.1(i) and β<ϵ then yield (5.201)|Γ2,1|≲CT∥f∥C˙β(𝒳)∫d(x,y)>2d(x,x′)d(x,x′)ϵV(x,y)d(x,y)ϵd(x,y)βdμ(y)≲d(x,x′)βCT∥f∥C˙β(𝒳). Combining all the above
estimates shows that for a.e.x∈B(x0,2k2/4) and a.e.x′∈𝒳 satisfying 2k2−1≤d(x,x′)<2k2,
we have (5.202)|Tf(x)−Tf(x′)|≲d(x,x′)β(CT+∥T∥WBP(β))∥f∥C˙β(𝒳). Then an argument via the
Besicovitch covering lemma further shows that there exists an extension of Tf such that (5.203)∥Tf∥C˙β(𝒳)≲(CT+∥T∥WBP(β))∥f∥C˙β(𝒳), which completes the proof of the
sufficiency.
We now prove the necessity. SinceT is extended as a continuous linear operator on C˙β(𝒳) and 1=0 in C˙β(𝒳).
Thus, T(1)=0 in C˙β(𝒳).
Since C˙β(𝒳)=B˙∞,∞β(𝒳)=(B˙1,1−β(𝒳))′ by Theorems 6.11 and 8.11(i), and C°bβ(𝒳)⊂B˙1,1−β(𝒳) by Proposition 5.10(iv), we then have that for
all f∈C°bβ(𝒳), 〈T(1),f〉=0,
which just means that T(1)=0 in (C°bβ(𝒳))′.
Let nowϕ,ψ∈Cbβ(𝒳) as in Definition 2.13. Fix x0′∈𝒳 such that 2r≤d(x0′,x0)<3r.
Since x0′∉suppϕ,
by the size condition (2.59) of the kernel K and (2.55), we then have  (5.204)|T(ϕ)(x0′)|=|∫B(x0,r)K(x0′,y)ϕ(y)dμ(y)|≲1Vr(x0′)∥ϕ∥L∞(𝒳)μ(B(x0,r))≲1. Since T is bounded on C˙β(𝒳),
when d(x,x0′)≤5r,
we then have (5.205)|T(ϕ)(x)|≲|T(ϕ)(x0′)|+d(x,x0′)β∥ϕ∥C˙β(𝒳)≲1. When d(x,x0′)>5r,
by the size condition (2.59) of the kernel K and (2.55) again, we also have (5.206)|T(ϕ)(x)|=|∫B(x0,r)K(x,y)ϕ(y)dμ(y)|≲1V(x,x0′)∥ϕ∥L∞(𝒳)μ(B(x0,r))≲1. Thus, ∥T(ϕ)∥L∞(𝒳)≲1, which gives that |〈T(ϕ),ψ〉|≲∥ψ∥L1(𝒳)≲μ(B(x0,r)). That is, T∈WBP(β),
which completes the proof of Theorem 5.49.Remark 5.50.
(i) The proof of Theorem5.49 in combination with Corollary 2.23 shows
that if T is bounded on C˙β(𝒳),
then there exists a constant C>0 such that for all ϕ∈C˙bβ(𝒳) and all x∈𝒳, (5.207)|T(ϕ)(x)|≤C(CT+∥T∥C˙β(𝒳)→C˙β(𝒳))[diam(suppϕ)]β∥ϕ∥C˙β(𝒳).
(ii) LetT be as in Theorem 5.49 and T(1)=0 in (C°bβ(𝒳))′.
Then from (i) and Theorem 5.49, it is easy to see that T∈WBP(β) if and only if (5.207) holds.Now we recall the notion of Carleson measures and
establish their connection withBMO(𝒳) functions.Definition 5.51.
A positive measureν on 𝒳×(0,∞) is said to be a Carleson measure if there exists a
constant C>0 such that for every ball B(x,r) for some x∈𝒳 and r>0, (5.208)ν(B(x,r)×(0,r))≤Cμ(B(x,r)). The smallest bound C as above is defined to be the Carleson norm of ν and is denoted by ∥ν∥ℭ.For any given open setE of 𝒳,
let (5.209)E^={(x,t)∈𝒳×(0,∞):B(x,t)⊂E}.We first establish a basic property of Carleson
measures; see [75, pages 59-60] or [81, page 198] for the case of ℝn.Lemma 5.52.
Ifν is a Carleson measure in 𝒳×(0,∞) and E⊂𝒳 is open, then ν(E^)≤C∥ν∥ℭμ(E).Proof.
Without loss of generality, we may assume
thatE is a bounded open set of 𝒳.
By the Whitney-type covering lemma (see [28, Theorem (3.2)]), there exist a constant C>0 and a sequence of balls {B(yj,rj)}j satisfying E=∪jB(yj,rj), ∑jχB(yj,rj)≤C and B(yj,3rj)∩(𝒳∖E)≠∅ for each j.
If (x,t)∈E^,
then there exists j0 such that x∈B(yj0,rj0), B(x,t)⊂E, and B(yj0,3rj0)∩(𝒳∖E)≠∅.
From this, it follows that t<6rj0,
and therefore, (x,t)∈B(yj0,rj0)×(0,6rj0).
Thus, E^⊂∪j{B(yj,rj)×(0,6rj)}, which together with the definition of Carleson
measures implies that (5.210)ν(E^)≤∑jν(B(yj,rj)×(0,6rj))≲∥ν∥ℭ∑jμ(B(yj,rj)). On the other hand, from E=∪jB(yj,rj) and ∑jχB(yj,rj)≤C,
it follows that (5.211)μ(E)=μ(⋃jB(yj,rj))≳∫𝒳∑jχB(yj,rj)(y)dμ(y)~∑jμ(B(yj,rj)). Combining both estimates yields ν(E^)≲∥ν∥ℭμ(E), which completes the proof of Lemma 5.52.Let{Sk}k∈ℤ be an (ϵ1,ϵ2,ϵ3)-ATI as in Definition 2.2. For (x,t)∈𝒳×(0,∞),
we define (5.212)𝔖(f)(x,t)=∑j=−∞∞Sj(f)(x)χ(2−j−1,2−j](t).Proposition 5.53.
Let𝔖 be as in (5.212). For any p∈(1,∞),
there exists a constant Cp>0 such that for all f∈Lp(𝒳) and all Carleson measures ν, (5.213)∫𝒳×(0,∞)|𝔖(f)(x,t)|pdν(x,t)≤Cp∥ν∥ℭ∫𝒳|f(x)|pdμ(x).Proof.
Let𝔖 be as in above. For any x∈𝒳,
we define (5.214)M𝔖f(x)=sup{|𝔖(f)(y,t)|:d(x,y)<t}. We first claim that for all x∈𝒳, M𝔖f(x)≲Mf(x), where M is the Hardy-Littlewood maximal function. To
see this, for any (y,t)∈𝒳×(0,∞) satisfying d(x,y)<t,
assuming 2−j0−1<t≤2−j0 for some j0∈ℤ,
by Lemma 2.1(vi), we then have (5.215)|𝔖(f)(y,t)|≲∫d(y,z)<2−j01V2−j0(y)+V2−j0(z)+V(y,z)(2−j02−j0+d(y,z))ϵ2|f(z)|dμ(z)+∑l=1∞∫2l−12−j0≤d(y,z)<2l2−j0⋯≲Mf(x), which implies the claim.
Notice that(5.216)∫𝒳×(0,∞)|𝔖(f)(x,t)|pdν(x,t)=p∫0∞λp−1ν({(x,t)∈𝒳×(0,∞):|𝔖(f)(x,t)|>λ})dλ. Let Eλ={x∈𝒳:M𝔖f(x)>λ}. We then claim (5.217){(x,t)∈𝒳×(0,∞):|𝔖(f)(x,t)|>λ}⊂E^λ. In fact, for any (x,t)∈𝒳×(0,∞) such that |𝔖(f)(x,t)|>λ,
assume that 2−j0−1<t≤2−j0 for some j0∈ℤ.
Then |𝔖(f)(x,t)|>λ if and only if |Sj0(f)(x)|>λ.
If d(y,x)<t,
then M𝔖f(y)>λ and hence B(x,t)⊂Eλ,
which implies the claim.
From this claim and Lemma5.52, it follows that (5.218)ν({(x,t)∈𝒳×(0,∞):|𝔖(f)(x,t)|>λ})≤ν(E^λ)≲∥ν∥ℭμ(Eλ), which together with the Lp(𝒳)-boundedness for p∈(1,∞) of M yields (5.219)∫𝒳×(0,∞)|𝔖(f)(x,t)|pdν(x,t)≲∥ν∥ℭ∫0∞λp−1μ(Eλ)dλ~∥ν∥ℭ∫𝒳|M𝔖(f)(x)|pdμ(x)≲∥ν∥ℭ∫𝒳|M(f)(x)|pdμ(x)≲∥ν∥ℭ∫𝒳|f(x)|pdμ(x), which completes the proof of
Proposition 5.53.The relation between Carleson measures andBMO(𝒳) functions can be stated as below.Proposition 5.54.
Letb∈BMO(𝒳) and let {Sk}k∈ℤ be an (ϵ1,ϵ2,ϵ3)-ATI as in Definition 2.2. For k∈ℤ,
let Dk=Sk−Sk−1.
Then the measure ν defined by (5.220)dν(x,t)=∑j=−∞∞|Dj(b)(x)|2χ(2−j−1,2−j](t)dμ(x)dtt is a Carleson measure such that ∥ν∥ℭ is dominated by ∥b∥BMO(𝒳)2.Proof.
For any ballB=B(x0,r) with some x0∈𝒳 and r>0,
assume that 2−j0−1<r≤2−j0 for some j0∈ℤ.
We then have (5.221)ν(B(x0,r)×(0,r))=∫B(x0,r)∫0rdν(x,t)=∑j=−∞∞∫B(x0,r)∫0r|Dj(b)(x)|2χ(2−j−1,2−j](t)dμ(x)dtt≲∑j=j0−1∞∫B(x0,r)|Dj(b)(x)|2dμ(x). Let B˜=B(x0,3r), B˜0=B˜, and B˜k=B(x0,2k3r) for k∈ℕ.
Set also (5.222)bB˜=1μ(B˜)∫B˜b(y)dμ(y). Since ∫𝒳Dj(x,y)dμ(y)=0,
we then have (5.223)ν(B(x0,r)×(0,r))≲∑j=j0−1∞∫B(x0,r)|Dj((b−bB˜)χB˜)(x)|2dμ(x)+∑j=j0−1∞∫B(x0,r)|Dj((b−bB˜)χ𝒳∖B˜)(x)|2dμ(x)≡Y1+Y2. By Lemma 3.9, we
have (5.224)Y1≲∑j=j0−1∞∫𝒳|Dj((b−bB˜)χB˜)(x)|2dμ(x)≲∫𝒳|((b(x)−bB˜)χB˜(x))|2dμ(x)≲∥b∥BMO(𝒳)2μ(B(x0,r)). To estimate Y2,
we first notice that if d(x,x0)<r and y∉B˜k for k∈ℤ+,
then d(y,x)≥2k3r−r>2k−13r and hence (5.225)|Dj(x,y)|≲1V2−j(x)+V2−j(y)+V(x,y)(2−j2−j+d(x,y))ϵ2≲1V2k+23r(x)(2−j2kr)ϵ2. From this, it follows that (5.226)Y2≲∑j=j0−1∞∫B(x0,r)[∑k=0∞∫B˜k+1∖B˜k|b(y)−bB˜||Dj(x,y)|dμ(y)]2dμ(x)≲1r2ϵ2∑j=j0−1∞2−2jϵ2∫B(x0,r)[∑k=0∞12kϵ21V2k+23r(x)∫B(x,2k+23r)|b(y)−bB˜|dμ(y)]2dμ(x)≲∥b∥BMO(𝒳)2μ(B(x0,r)), where we used the well-known
fact that (5.227)∥b∥BMO(𝒳)~infC∈ℂsupx∈𝒳,r>01μ(B(x,r))∫B(x,r)|f(y)−C|dμ(y). This finishes the proof of
Proposition 5.54.Combining Proposition5.53 with Proposition 5.54
yields the following conclusion, which will be used in the proof of T(1)-theorem.Corollary 5.55.
Let
all the notation be as in Propositions5.53 and 5.54. For any p∈(1,∞),
there exists a constant Cp>0 such that for all f∈Lp(𝒳) and b∈BMO(𝒳), (5.228)∑j=−∞∞∫𝒳|Sj(f)(x)|p|Dj(b)(x)|2dμ(x)≤Cp∥b∥BMO(𝒳)2∫𝒳|f(x)|pdμ(x).Proof.
From Propositions5.53 and 5.54,
it follows that (5.229)∫𝒳×(0,∞)|𝔖(f)(x,t)|pdν(x,t)≲∥b∥BMO(𝒳)2∫𝒳|f(x)|pdμ(x), where 𝔖(f) is as in Proposition 5.53 and dν(x,t) is as in  Proposition 5.54. Moreover,  we
have (5.230)∫𝒳×(0,∞)|𝔖(f)(x,t)|pdν(x,t)=∑k=−∞∞∫𝒳∫2−k−12−k|𝔖(f)(x,t)|pdν(x,t)=∑k=−∞∞∫𝒳|Sk(f)(x)|p∫0∞χ(2−k−1,2−k](t)(∑j=−∞∞|Dj(b)(x)|2χ(2−j−1,2−j](t))dμ(x)dtt=log2{∑k=−∞∞∫𝒳|Sk(f)(x)|p|Dk(b)(x)|2dμ(x)}. Combining both estimates
completes the proof of Corollary 5.55.We now can state a variant ofDavid-JournéT(1)-theorem on spaces of homogeneous type. In
what follows, for any ϵ∈(0,1],
a continuous function on 𝒳×𝒳∖{(x,x):x∈𝒳} is said to be a standard kernel of orderϵ if it satisfies (I-3) in Subsection 5.2. Let β∈(0,ϵ).
A continuous linear operator T from Cbβ(𝒳) to (Cbβ(𝒳))′ is said to have a standard distributional kernelKof orderϵ if T and K satisfy (2.48). Also, the adjoint T∗ of T is given by that for all f,g∈Cbβ(𝒳), (Tf,g)=(f,T∗g). Then T∗:Cbβ(𝒳)→(Cbβ(𝒳))′ is a continuous mapping; moreover, T∗ is associated to the kernel K∗(x,y)=K¯(y,x) for all x,y∈𝒳.Theorem 5.56.
Letϵ∈(0,1], β∈(0,ϵ), and let T be a continuous linear operator from Cbβ(𝒳) to (Cbβ(𝒳))′.
Assume that T has a standard distributional kernel K of order ϵ as in (2.48). Then T extends to a bounded operator on L2(𝒳) if and only if the following conditions are
true:
(i)
T(1)∈BMO(𝒳),(ii)
T∗(1)∈BMO(𝒳),(iii)
T∈WBP(β).Proof.
We first verify the sufficiency. By Proposition2.12, both T and T∗ can be extended to a continuous linear
operator from Cβ(𝒳) to (C°bβ(𝒳))′.
If T(1)=T∗(1)=0 in (C°bβ(𝒳))′,
then by Theorem 5.49, T and T∗ can be extended as continuous linear operators
on C˙β(𝒳).
By this fact, Theorem 5.23 and Proposition 5.10(v), we know that T extends to a bounded operator on L2(𝒳).
We now consider the general case. Let{Sj}j∈ℤ be as in Definition 2.2 and Dj=Sj−Sj−1 for j∈ℤ.
Let {D˜j}j∈ℤ be as in Theorem 3.10. Let b∈BMO(𝒳).
For any f∈Cbβ(𝒳),
we define the paraproduct(5.231)Pb(f)(x)=∑j=−∞∞D˜j(Dj(b)Sj(f))(x). We will show that the kernel of Pb is a standard kernel of order ϵ,
that Pb is bounded on L2(𝒳),
and that Pb(1)=b and Pb∗(1)=0.
In order to be rigorous in the following calculations, we should consider ∑j=−NN instead of ∑j=−∞∞ and then let N→∞.
However, we will omit these details; see [75, pages 302–305].
(1) The size of the kernel. The kernel ofPb is (5.232)K(x,y)=∑j=−∞∞∫𝒳D˜j(x,z)Dj(b)(z)Sj(z,y)dμ(z). Let Bk=B(z,2k+12−j), for k∈ℤ+,
and bB0=(1/μ(B0))∫B0b(z)dμ(z).
Since (5.233)∫𝒳Dj(z,y)dμ(y)=0, by the size condition of Dj and (5.227), for any z∈𝒳,
we have (5.234)|Dj(b)(z)|≤∫B0|Dj(z,y)||b(y)−bB0|dμ(y)+∑k=1∞∫Bk∖Bk−1|Dj(z,y)||b(y)−bB0|dμ(y)≲∑k=0∞12kϵ21μ(Bk)∫Bk|b(y)−bB0|dμ(y)≲∥b∥BMO(𝒳). Let ϵ′∈(ϵ,ϵ1⋀ϵ2).
By the size conditions of D˜j and Sj,
(5.234), Lemmas 4.4, and 3.5, we have (5.235)|K(x,y)|≲∥b∥BMO(𝒳)∑j=−∞∞∫𝒳1V2−j(x)+V2−j(z)+V(x,z)(2−j2−j+d(x,z))ϵ′×1V2−j(z)+V2−j(y)+V(z,y)(2−j2−j+d(z,y))ϵ2dμ(z)≲∥b∥BMO(𝒳)∑j=−∞∞1V2−j(x)+V2−j(y)+V(x,y)(2−j2−j+d(x,y))ϵ′≲∥b∥BMO(𝒳)1V(x,y), which verifies the size
condition of the kernel K.
Ifd(x,x′)≤d(x,y)/2 with x≠y,
by (5.234),
(5.236)|K(x,y)−K(x′,y)|=|∑j=−∞∞∫𝒳[D˜j(x,z)−D˜j(x′,z)]Dj(b)(z)Sj(z,y)dμ(z)|≲∥b∥BMO(𝒳)∑j=−∞∞{∫d(x,x′)≤(2−j+d(x,z))/2|D˜j(x,z)−D˜j(x′,z)||Sj(z,y)|dμ(z)+∫d(x,x′)>(2−j+d(x,z))/2|D˜j(x,z)||Sj(z,y)|dμ(z)+∫d(x,x′)>(2−j+d(x,z))/2|D˜j(x′,z)||Sj(z,y)|dμ(z)}≡∥b∥BMO(𝒳){Y1+Y2+Y3}. The regularity of D˜j and the size condition of Sj together with Lemmas 4.4 and 3.5
give (5.237)Y1≲∑j=−∞∞∫𝒳(d(x,x′)2−j+d(x,z))ϵ′1V2−j(x)+V2−j(z)+V(x,z)(2−j2−j+d(x,z))ϵ′×1V2−j(z)+V2−j(y)+V(z,y)(2−j2−j+d(z,y))ϵ2dμ(z)≲∑j=−∞∞(d(x,x′)2−j+d(x,y))ϵ1V2−j(x)+V2−j(y)+V(x,y)(2−j2−j+d(x,y))ϵ′−ϵ≲d(x,x′)ϵV(x,y)d(x,y)ϵ. The size conditions of D˜j and Sj together with Lemmas 4.4 and 3.5 also
yield (5.238)Y2≲∑j=−∞∞∫d(x,x′)>(2−j+d(x,z))/21V2−j(x)+V2−j(z)+V(x,z)(2−j2−j+d(x,z))ϵ′×1V2−j(z)+V2−j(y)+V(z,y)(2−j2−j+d(z,y))ϵ2dμ(z)≲∑j=−∞∞1V2−j(x)+V2−j(y)+V(x,y)d(x,x′)ϵ2−j(ϵ′−ϵ)(2−j+d(x,y))ϵ′≲d(x,x′)ϵV(x,y)d(x,y)ϵ. Similarly, we have (5.239)Y3≲d(x,x′)ϵV(x,y)d(x,y)ϵ, which shows that K has the desired regularity on the first
variable.
An argument similar to above also proves that ifd(y,y′)≤d(x,y)/2 with x≠y, (5.240)|K(x,y)−K(x,y′)|≲∥b∥BMO(𝒳)d(y,y′)ϵV(x,y)d(x,y)ϵ. Thus, K is a standard kernel of order ϵ.
(2) Boundedness ofPb on L2(𝒳).
For any f,g∈L2(𝒳),
by Hölder's inequality, Corollary 5.55, and Lemma 3.9 together with Remark 3.16,
we have (5.241)|〈Pb(f),g〉|=|∑j=−∞∞∫𝒳Dj(b)(x)Sj(f)(x)D˜j∗(g)(x)dμ(x)|≲{∑j=−∞∞∫𝒳|Dj(b)(x)|2|Sj(f)(x)|2dμ(x)}1/2{∑j=−∞∞∫𝒳|D˜jt(g)(x)|2dμ(x)}1/2≲∥b∥BMO(𝒳)∥f∥L2(𝒳)∥g∥L2(𝒳), which together with a duality
argument yields that (5.242)∥Pb(f)∥L2(𝒳)≲∥b∥BMO(𝒳)∥f∥L2(𝒳).
(3)Pb(1)=b and Pb∗(1)=0 in (C°bβ(𝒳))′.
Since ∫𝒳D˜j(x,z)dμ(x)=0,
from this, it follows that Pb∗(1)=0 in (C°bβ(𝒳))′.
Also, since ∫𝒳Sj(y,z)dμ(z)=1,
from this and Theorem 3.29, it follows that Pb(1)=b.
We can now finish the proof of the sufficiency. For
any given operatorT which satisfies (i), (ii), and (iii), let b1=T(1) and b2=T∗(1).
Then, there exist paraproducts Pbi such that Pbi(1)=bi and Pbi∗(1)=0 for i=1,2.
Then the operator T˜=T−Pb1−Pb2∗ lies in WBP(β) and T˜(1)=T˜∗(1)=0.
Thus, by Theorem 5.23, we know that T˜ is bounded on L2(𝒳),
which together with the boundedness of Pb1 and Pb2∗ on L2(𝒳) also yields the boundedness of T on L2(𝒳).
This completes the proof of the sufficiency.
We now check the necessity. By Remark2.14(iii), we
know T∈WBP(β).
To verify T(1),T∗(1)∈BMO(𝒳),
we first claim that if T is as in Proposition 2.12 and T is bounded on L2(𝒳),
then T is also bounded from Lb∞(𝒳) to BMO(𝒳),
namely, for all f∈Lb∞(𝒳), (5.243)∥T(f)∥BMO(𝒳)≲∥f∥L∞(𝒳). The proof of this claim is
standard; see, for example, [75, pages 156-157] or [81, pages 118-119]. For any ball B=B(x0,r) with some x0∈𝒳 and r>0,
let CB=T(fχ𝒳∖B(x0,2r))(x0).
Since f∈Lb∞(𝒳) and K is locally integrable away from the diagonal
of 𝒳×𝒳,
it is easy to see that |CB|<∞.
By the boundedness of T on L2(𝒳) and (2.49) together with Lemma 2.1(i), we have(5.244)1μ(B)∫B|T(f)(x)−CB|dμ(x)≤1μ(B)∫B|T(fχB(x0,2r))(x)|dμ(x)+1μ(B)∫B|T(fχ𝒳∖B(x0,2r))(x)−CB|dμ(x)≲∥T∥L2(𝒳)→L2(𝒳)∥f∥L∞(𝒳)+1μ(B)∫B∫𝒳∖B(x0,2r)|K(x,y)−K(x0,y)||f(y)|dμ(y)dμ(x)≲(∥T∥L2(𝒳)→L2(𝒳)+CT)∥f∥L∞(𝒳), which proves (5.243).
Using (5.243), we then can verify that if T is bounded on L2(𝒳),
then T(1)∈BMO(𝒳).
To see this, for any g∈C°bβ(𝒳) with suppg⊂B(x0,r) for some x0∈𝒳 and r>0,
let ψ∈Cbβ(𝒳) be as in the proof of Proposition 2.12. By
(5.243), T(ψ)∈BMO(𝒳),
which together with Theorem 5.19(i) below yields that (5.245)|〈T(ψ),g〉|≤∥T(ψ)∥BMO(𝒳)∥g∥H1(𝒳)≲(∥T∥L2(𝒳)→L2(𝒳)+CT)∥g∥H1(𝒳). On the other hand, by (2.49) and
Lemma 2.1(i), we also have (5.246)|〈T(1−ψ),g〉|=|∫𝒳∫d(y,x0)≥2r[K(x,y)−K(x0,y)](1−ψ(y))g(x)dμ(y)dμ(x)|≲CT∫𝒳{∫d(y,x0)≥2rrϵV(x0,y)d(x0,y)ϵdμ(y)}|g(x)|dμ(x)≲CT∥g∥L1(𝒳)≲CT∥g∥H1(𝒳). Thus, |〈T(1),g〉|≲(∥T∥L2(𝒳)→L2(𝒳)+CT)∥g∥H1(𝒳), which together with Corollary 2.11(i), Proposition 5.21, Theorems 6.11,
and 5.19(i) implies that T(1)∈BMO(𝒳) and ∥T(1)∥BMO(𝒳)≲∥T∥L2(𝒳)→L2(𝒳)+CT. An argument similar to this also proves that T∗(1)∈BMO(𝒳) and ∥T∗(1)∥BMO(𝒳)≲∥T∥L2(𝒳)→L2(𝒳)+CT, which completes the proof of Theorem 5.56.We now state a variant of theT(1)-theorem in the sense of Stein [75, page 294]. Let ϵ∈(0,1], β∈(0,ϵ), and let T be a continuous linear operator from Cbβ(𝒳) to (Cbβ(𝒳))′.
We assume that associated to T,
there is a standard kernel of order ϵ,
in the sense that if f∈Cbβ(𝒳),
then, outside the support of f,
the distribution Tf agrees with the function (5.247)T(f)(x)=∫𝒳K(x,y)f(y)dμ(y). Similarly to [75, page 294], we assume that T and T∗ are restrictedly bounded. Whenever ϕR,x0 is a normalized bump function for the ball B(x0,R) with some x0∈𝒳 and R>0,
the distributions T(ϕR,x0) and T∗(ϕR,x0) belong to L2(𝒳),
and the estimates (5.248)∥T(ϕR,x0)∥L2(𝒳)≤A[μ(B(x0,R))]1/2,(5.249)∥T∗(ϕR,x0)∥L2(𝒳)≤A[μ(B(x0,R))]1/2 hold with an A>0 that is independent of R, x0 and ϕR,x0.Theorem 5.57.
Letϵ∈(0,1], β∈(0,ϵ), and let T be a continuous linear operator from Cbβ(𝒳) to (Cbβ(𝒳))′ associated with a standard kernel of order ϵ in the sense of (5.247). Then T extends to a bounded linear operator on L2(𝒳) if and only if both T and T∗ are restrictedly bounded in the sense of
(5.248) and (5.249).Proof.
The necessity is obvious. We only need to
prove the sufficiency. We first make the following claim that iff∈C°bβ(𝒳),
then (5.250)Tf∈L1(𝒳). In fact, assume that suppf⊂B(x0,r) for some x0∈𝒳 and r>0.
Since f is a multiple of a bump function, Tf∈L2(𝒳) by (5.248), and hence ∫B(x0,2r)|Tf(x)|dμ(x)<∞.
If x∉B(x0,2r),
then by (5.247), ∫𝒳f(x)dμ(x)=0, and the regularity on K,
we have (5.251)|Tf(x)|=|∫𝒳K(x,y)f(y)dμ(y)|=|∫𝒳[K(x,y)−K(x,x0)]f(y)dμ(y)|≲CT∫B(x0,r)d(y,x0)ϵV(x,x0)d(x,x0)ϵ|f(y)|dμ(y)≲CT∥f∥L∞(𝒳)μ(B(x0,r))rϵV(x,x0)d(x,x0)ϵ, which implies that ∫𝒳∖B(x0,2r)|Tf(x)|dμ(x)<∞.
Thus, the claim  (5.250) holds.
We now verify thatT(1)∈BMO(𝒳).
To this end, we first prove that there exists a constant A>0 such that whenever ϕR,x0 is a normalized bump function for the ball B(x0,R) with x0∈𝒳 and R>0,
then T(ϕR,x0)∈BMO(𝒳) with (5.252)∥T(ϕR,x0)∥BMO(𝒳)≤A. Let B˜1=B(x˜0,R˜) with some x˜0∈𝒳 and R˜>0 be any ball, and let B˜2=B(x˜0,2R˜) and B˜3=B(x˜0,3R˜).
Fix a function θ∈Cbβ(𝒳) with θ(x)=1 for d(x,x˜0)≤2R˜ and θ(x)=0 for d(x,x˜0)≥3R˜.
Write (5.253)ϕR,x0(x)=ϕR,x0(x)θ(x)+ϕR,x0(x)(1−θ(x))=f1(x)+f2(x). Observe that f1 is, up to a bounded multiplicative constant, a
normalized bump function for either the ball B(x0,R) or the ball B˜3,
whichever has the smaller radius. Thus, by (5.248), we have (5.254)∫B˜1|Tf1(x)|2dμ(x)≤∥Tf1∥L2(𝒳)2≤A2min{μ(B(x0,R)),μ(B(x˜0,3R˜))}≤A′μ(B(x˜0,R˜)). Since suppf2⊂(𝒳∖B˜2),
for x∈B˜1,
by (5.247), we have (5.255)Tf2(x)=∫𝒳K(x,y)f2(y)dμ(y). Let CB˜1=∫𝒳K(x˜0,y)f2(y)dμ(y).
Then |CB˜1|<∞,
and for x∈B˜1,
by the regularity on K on the first variable and Lemma 2.1(i), (5.256)|Tf2(x)−CB˜1|≲∫d(x˜0,y)>2R˜|K(x,y)−K(x˜0,y)|dμ(y)≲∫d(x˜0,y)>2R˜d(x,x˜0)ϵV(x˜0,y)d(x˜0,y)ϵdμ(y)≲1. Combining the estimates for f1 and f2 gives that for any ball B˜1, (5.257)∫B˜1|T(ϕR,x0)−CB˜1|2dμ(x)≤Aμ(B˜1), which shows (5.252).
Letθ∈Cb1(ℝ) and θ(0)=1.
For any ν>0 and x∈𝒳,
set θν(x)=θ(νd(x,x1)).
By (5.252), {T(θν)}ν>0 is uniformly bounded in BMO(𝒳).
Since BMO(𝒳)=(H1(𝒳))′ (see Theorem 5.19(i)), as is well known, BMO(𝒳) is weakly* compact in the dual topology. Thus, every
sequence {T(θνk)}k∈ℕ has a subsequence which weakly* converges. Let f∈C°bβ(𝒳).
By (5.250) and 〈T(θν),f〉=〈θν,T∗(f)〉,
we know that whatever limit (denote it by a), we extract from a subsequence of {T(θν)}ν>0,
then 〈a,f〉=〈1,T∗(f)〉. This shows that the limit a is independent of the subsequence, and we are
justified in setting a=T(1)=limν→0T(θν) and (5.258)〈T(1),f〉=∫𝒳T∗(f)(x)dμ(x), with a similar statement when
the roles of T and T∗ are reversed. Thus, T(1), T∗(1)∈BMO(𝒳).
Moreover, if T is restrictedly bounded, then T∈WBP(β).
Thus, by Theorem 5.56, we know T is bounded on L2(𝒳),
which completes the proof of Theorem 5.57.Remark 5.58.
From (5.258), we see that if T is bounded on L2(𝒳),
then T(1)∈(C°bβ(𝒳))′ for a certain β∈(0,1] is constant if and only if for any f∈C°bβ(𝒳),(5.259)∫𝒳T∗(f)(x)dμ(x)=0.


## 6. Triebel-Lizorkin Spaces withp=∞

In this section, we will develop a theory for
Triebel-Lizorkin spaces withp=∞ by using the Carleson characterizations. We
again distinguish between the homogeneous and the inhomogeneous cases and
examine the relations between these cases as well as with BMO-type spaces.


### 6.1. Plancherel-Pôlya Inequality and Definition ofF˙∞,qs(𝒳)

Throughout this and the next subsection, we will assume
thatμ(𝒳)=∞.
In this subsection, we introduce the norm in
F˙∞,qs(𝒳) in a similar way as in [82] and, using the homogeneous
discrete Calderón reproducing formulae, Theorem 4.13, we will prove that the
norm
∥⋅∥F˙∞,qs(𝒳) is independent of the choices of ATIs and spaces of distributions via some
Plancherel-Pôlya inequality; see also [83, 84].Definition 6.1.
Letϵ1∈(0,1], ϵ2>0, ϵ3>0, ϵ∈(0,ϵ1⋀ϵ2), and {Sk}k∈ℤ be an (ϵ1,ϵ2,ϵ3)-ATI.
For k∈ℤ,
set Dk=Sk−Sk−1.
Let |s|<ϵ and p(s,ϵ)<q≤∞.
For any f∈(𝒢°0ϵ(β,γ))′ with 0<β,γ<ϵ,
define (6.1)∥f∥F˙∞,qs(𝒳)=supl∈ℤsupα∈Il{1μ(Qαl)∫Qαl∑k=l∞2ksq|Dk(f)(x)|qdμ(x)}1/q, where the supremum is
taken over all dyadic cubes as in Lemma 2.19 and the usual modification is made
when q=∞.Remark 6.2.
(i) From Lemma2.19 and the
doubling property (1.2), it is easy to see that an equivalent norm is obtained
if the supremum in Definition 6.1 is taken with respect to all balls with
positive radius instead of all dyadic cubes as in Lemma 2.19.
(ii) Letl0∈ℤ.
It is easy to see that there exists a constant Cl0>0 such that for all f∈(𝒢°0ϵ(β,γ))′, (6.2)supl∈ℤsupα∈Il{1μ(Qαl)∫Qαl∑k=l+l0∞2ksq|Dk(f)(x)|qdμ(x)}1/q≲Cl0∥f∥F˙∞,qs(𝒳). Thus, (6.3)∥f∥F˙∞,qs(𝒳)~supl∈ℤsupα∈Il{1μ(Qαl)∫Qαl∑k=l+l0∞2ksq|Dk(f)(x)|qdμ(x)}1/q.We now establish the following usefulPlancherel-Pôlya inequality, which
complements Proposition 5.4(ii) for the case p=∞.Proposition 6.3.
Letϵ1∈(0,1], ϵ2>0, ϵ3>0, ϵ∈(0,ϵ1⋀ϵ2), and let {Sk}k∈ℤ and {Pk}k∈ℤ be two (ϵ1,ϵ2,ϵ3)-ATIs. For k∈ℤ,
set Dk=Sk−Sk−1 and Qk=Pk−Pk−1.
Let |s|<ϵ and p(s,ϵ)<q≤∞.
Then for all f∈(𝒢°0ϵ(β,γ))′ with 0<β,γ<ϵ, (6.4)supl∈ℤsupα∈Il{1μ(Qαl)∑k=l∞∑τ∈Ik∑ν=1N(k,τ)2ksqμ(Qτk,ν)χ{(τ,ν):Qτk,ν⊂Qαl}(τ,ν)[supx∈Qτk,ν|Dk(f)(x)|]q}1/q~supl∈ℤsupα∈Il{1μ(Qαl)∑k=l∞∑τ∈Ik∑ν=1N(k,τ)2ksqμ(Qτk,ν)χ{(τ,ν):Qτk,ν⊂Qαl}(τ,ν)[infx∈Qτk,ν|Qk(f)(x)|]q}1/q.Proof.
To prove Proposition6.3, it suffices to show that for all f∈(𝒢°0ϵ(β,γ))′ with 0<β,γ<ϵ,
the left-hand side of (6.4) is controlled by its right-hand side.
Let all the notation be as in the proof of Proposition5.4. Then, by (5.9) and (5.10), we have1μ(Qαl)∑k=l∞∑τ∈Ik∑ν=1N(k,τ)2ksqμ(Qτk,ν)χ{(τ,ν):Qτk,ν⊂Qαl}(τ,ν)[supx∈Qτk,ν|Dk(f)(x)|]q≲1μ(Qαl)∑k=l∞∑τ∈Ik∑ν=1N(k,τ)2ksqμ(Qτk,ν)χ{(τ,ν):Qτk,ν⊂Qαl}(τ,ν)×[∑k′=l∞∑τ′∈Ik′∑ν′=1N(k′,τ′)2−|k−k′|ϵ′μ(Qτ′k′,ν′)|Qk′(f)(yτ′k′,ν′)|×1V2−(k⋀k′)(yτk,ν)+V2−(k⋀k′)(yτ′k′,ν′)+V(yτk,ν,yτ′k′,ν′)(2−(k⋀k′)2−(k⋀k′)+d(yτk,ν,yτ′k′,ν′))ϵ]q(6.5)+1μ(Qαl)∑k=l∞∑τ∈Ik∑ν=1N(k,τ)2ksqμ(Qτk,ν)χ{(τ,ν):Qτk,ν⊂Qαl}(τ,ν)×[∑k′=−∞l−1∑τ′∈Ik′∑ν′=1N(k′,τ′)2−|k−k′|ϵ′μ(Qτ′k′,ν′)|Qk′(f)(yτ′k′,ν′)|⋯]q≡Y1+Y2.
Then by Lemma 2.19, if x∉B(zαl,4C62−l) and y∈Qαl,
then d(x,y)≥3C62−l,
where and in what follows, zαl is the “center" of Qαl as in Lemma 2.19. By Lemma 2.19 again, we can
find m1∈ℕ such that B(zαl,4C62−l)⊂⋃i=1m1Qτil, τi∈Il, B(zαl,4C62−l)∩Qτil≠∅ and m1 is no more than a constant which is
independent of α and l;
see the details for a proof of the last fact in [85, pages 1385-1386].
Moreover, (6.6)μ(Qτil)~μ(Qαl). With these choices, we further
control Y1 by (6.7)Y1≲1μ(Qαl)∑k=l∞∑τ∈Ik∑ν=1N(k,τ)2ksqμ(Qτk,ν)χ{(τ,ν):Qτk,ν⊂Qαl}(τ,ν)×[∑k′=l∞∑τ′∈Ik′∑ν′=1N(k′,τ′)2−|k−k′|ϵ′μ(Qτ′k′,ν′)χ{(τ′,ν′):Qτ′k′,ν′⊂∪i=1m1Qτil}(τ′,ν′)×|Qk′(f)(yτ′k′,ν′)|1V2−(k⋀k′)(yτk,ν)+V2−(k⋀k′)(yτ′k′,ν′)+V(yτk,ν,yτ′k′,ν′)×(2−(k⋀k′)2−(k⋀k′)+d(yτk,ν,yτ′k′,ν′))ϵ]q+1μ(Qαl)∑k=l∞∑τ∈Ik∑ν=1N(k,τ)2ksqμ(Qτk,ν)χ{(τ,ν):Qτk,ν⊂Qαl}(τ,ν)×[∑k′=l∞∑τ′∈Ik′∑ν′=1N(k′,τ′)2−|k−k′|ϵ′μ(Qτ′k′,ν′)χ{(τ′,ν′):Qτ′k′,ν′∩(∪i=1m1Qτil)=∅}(τ′,ν′)×|Qk′(f)(yτ′k′,ν′)|1V2−(k⋀k′)(yτk,ν)+V2−(k⋀k′)(yτ′k′,ν′)+V(yτk,ν,yτ′k′,ν′)×(2−(k⋀k′)2−(k⋀k′)+d(yτk,ν,yτ′k′,ν′))ϵ]q≡Y1,1+Y1,2. We first estimate Y1,1.
If q≤1,
by (5.5), Lemma 5.2, (5.12), (6.6), and choosing ϵ′∈(0,ϵ) such that ϵ′>s and q>p(s,ϵ′), (6.8)Y1,1≲1μ(Qαl)∑i=1m1∑k′=l∞∑τ′∈Ik′∑ν′=1N(k′,τ′)∑k=l∞2ksqχ{(τ′,ν′):Qτ′k′,ν′⊂Qτil}(τ′,ν′)×[μ(Qτ′k′,ν′)|Qk′(f)(yτ′k′,ν′)|]q2−|k−k′|ϵ′q×∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)χ{(τ,ν):Qτk,ν⊂Qαl}(τ,ν)×[1V2−(k⋀k′)(yτk,ν)+V2−(k⋀k′)(yτ′k′,ν′)+V(yτk,ν,yτ′k′,ν′)(2−(k⋀k′)2−(k⋀k′)+d(yτk,ν,yτ′k′,ν′))ϵ]q≲∑i=1m11μ(Qτil)∑k′=l∞∑τ′∈Ik′∑ν′=1N(k′,τ′)2k′sqμ(Qτ′k′,ν′)|Qk′(f)(yτ′k′,ν′)|qχ{(τ′,ν′):Qτ′k′,ν′⊂Qτil}(τ′,ν′), which together with the
arbitrary choice of yτ′k′,ν′∈Qτ′k′,ν′ shows that (Y1,1)1/q is controlled by the right-hand side of (6.4)
in this case.
If1<q≤∞,
choosing ϵ′>|s| together with Hölder's inequality and Lemma
5.2 yields (6.9)∑k′=l∞∑τ′∈Ik′∑ν′=1N(k′,τ′)2k′s2−|k−k′|ϵ′2(k−k′)sμ(Qτ′k′,ν′)χ{(τ′,ν′):Qτ′k′,ν′⊂Qτil}(τ′,ν′)×|Qk′(f)(yτ′k′,ν′)|1V2−(k⋀k′)(yτk,ν)+V2−(k⋀k′)(yτ′k′,ν′)+V(yτk,ν,yτ′k′,ν′)×(2−(k⋀k′)2−(k⋀k′)+d(yτk,ν,yτ′k′,ν′))ϵ≲{∑k′=l∞∑τ′∈Ik′∑ν′=1N(k′,τ′)2k′sq2−|k−k′|ϵ′2(k−k′)sμ(Qτ′k′,ν′)χ{(τ′,ν′):Qτ′k′,ν′⊂Qτil}(τ′,ν′)×|Qk′(f)(yτ′k′,ν′)|q1V2−(k⋀k′)(yτk,ν)+V2−(k⋀k′)(yτ′k′,ν′)+V(yτk,ν,yτ′k′,ν′)×(2−(k⋀k′)2−(k⋀k′)+d(yτk,ν,yτ′k′,ν′))ϵ}1/q, which together with Minkowski's
inequality, (6.6), and Lemma 5.2 yields that (6.10)(Y1,1)1/q≲∑i=1m1{1μ(Qτil)∑k′=l∞∑τ′∈Ik′∑ν′=1N(k′,τ′)2k′sqμ(Qτ′k′,ν′)χ{(τ′,ν′):Qτ′k′,ν′⊂Qτil}(τ′,ν′)|Qk′(f)(yτ′k′,ν′)|q}1/q, which completes the estimate for Y1,1.
We now estimateY1,2.
To this end, for j∈ℤ+,
let (6.11)Ilj={α′∈Il:α′≠τ1,…,τm1,3C62j−l≤d(zα′l,zαl)<3C62j−l+1}. We first claim that there exists m2∈ℕ which is no more than a constant independent
of l and j such that (6.12)⋃α′∈IljQα′l⊂⋃i=1,…,m2αi∈Il−jQαil−j,Qαil−j∩(⋃α′∈IljQα′l)≠∅ for i=1,…,m2,
and moreover, if Qτ′k′,ν′⊂∪α′∈IljQα′l,
then (6.13)μ(⋃i=1,…,m2αi∈Il−jQαil−j)≲V2j−l(yτ′k′,ν′). In fact, by Lemma 2.19, there exists m2∈ℕ such that (6.12) holds. Notice that for any
fixed i0∈{1,…,m2}, ∪α′∈IljQα′l⊂B(zαi0j−l,14C62j−l). By an argument in [85, pages 1385-1386], we know
that the number of
α∈Il−j such that Qαl−j∩B(zαi0j−l,14C62j−l)≠∅ is no more than a constant which is
independent of j and l.
Thus, the claim (6.12) holds.
To see the claim (6.13), we only need to notice that (6.14)⋃i=1,…,m2αi∈Il−jQαil−j⊂B(yτ′k′,ν′,9C62j−l), which implies the claim (6.13).
We also notice that ifQτ′k′,ν′⊂Qα′l with α′∈Ili,
then for all y∈Qαl, (6.15)d(y,yτ′k′,ν′)≥C62j−l.
Using these properties, we now estimateY1,2 by first considering the case q≤1.
In fact, in this case, from (5.5), (6.15), (6.12), and (6.13), it follows
that (6.16)Y1,2≲∑j=0∞2−j[ϵq−n(1−q)]∑i=1m21μ(Qαil−j)∑k′=l∞∑τ′∈Ik′∑ν′=1N(k′,τ′)2k′sqμ(Qτ′k′,ν′)×χ{(τ′,ν′):Qτ′k′,ν′⊂Qαil−j}(τ′,ν′)|Qk′(f)(yτ′k′,ν′)|q×(∑k=l∞2−|k−k′|ϵ′q2(k−k′)sq2−(k⋀k′)ϵq2k′n(1−q)2−ln(1−q)+lϵq), which together with q>n/(n+ϵ) and choosing ϵ′>s further implies that (6.17)Y1,2≲∑j=0∞2−j[ϵq−n(1−q)]∑i=1m21μ(Qαil−j)∑k′=l∞∑τ′∈Ik′∑ν′=1N(k′,τ′)2k′sqμ(Qτ′k′,ν′)×χ{(τ′,ν′):Qτ′k′,ν′⊂Qαil−j}(τ′,ν′)|Qk′(f)(yτ′k′,ν′)|q. From this, it is easy to deduce
the desired estimate for Y1,2 in this case.
If1<q≤∞,
by Hölder's inequality and Lemma 5.2, we have (6.18)∑k′=l∞∑τ′∈Ik′∑ν′=1N(k′,τ′)2k′s2−|k−k′|ϵ′2(k−k′)sμ(Qτ′k′,ν′)χ{(τ′,ν′):Qτ′k′,ν′∩(∪i=1m1Qτil)=∅}(τ′,ν′)×|Qk′(f)(yτ′k′,ν′)|1V2−(k⋀k′)(yτk,ν)+V2−(k⋀k′)(yτ′k′,ν′)+V(yτk,ν,yτ′k′,ν′)×(2−(k⋀k′)2−(k⋀k′)+d(yτk,ν,yτ′k′,ν′))ϵ≲{∑k′=l∞∑τ′∈Ik′∑ν′=1N(k′,τ′)2k′sq2−|k−k′|ϵ′2(k−k′)sμ(Qτ′k′,ν′)|Qk′(f)(yτ′k′,ν′)|q×1V2−(k⋀k′)(yτk,ν)+V2−(k⋀k′)(yτ′k′,ν′)+V(yτk,ν,yτ′k′,ν′)×χ{(τ′,ν′):Qτ′k′,ν′∩(∪i=1m1Qτil)=∅}(τ′,ν′)(2−(k⋀k′)2−(k⋀k′)+d(yτk,ν,yτ′k′,ν′))ϵ}1/q, which together with (6.15),
(6.12), and (6.13) yields that (6.19)Y1,2≲∑j=0∞2−jϵ∑i=1m21μ(Qαil−j)∑k′=l∞∑τ′∈Ik′∑ν′=1N(k′,τ′)2k′sqμ(Qτ′k′,ν′)×χ{(τ′,ν′):Qτ′k′,ν′⊂Qαil−j}(τ′,ν′)|Qk′(f)(yτ′k′,ν′)|q, where we choose ϵ′>s.
From this, we deduce the desired estimate for Y1,2 when 1<q≤∞,
which completes the estimate for Y1.
We now estimateY2 by using the following trivial estimate that (6.20){∑τ′∈Ik′∑ν′=1N(k′,τ′)2k′sq|Qk′(f)(yτ′k′,ν′)|q}1/q≲RHSof(6.4), where and in the sequel, RHS
stands for “right-hand side." From this, it follows that if q≤1,
by choosing ϵ′>s,
we then have (6.21)Y2≲1μ(Qαl)∑k=l∞∑τ∈Ik∑ν=1N(k,τ)2ksqμ(Qτk,ν)χ{(τ,ν):Qτk,ν⊂Qαl}(τ,ν)×(∑k′=−∞l−1∑τ′∈Ik′∑ν′=1N(k′,τ′)2−(k−k′)ϵ′q|Qk′(f)(yτ′k′,ν′)|q)≲(RHSof(6.4))q; if 1<q≤∞,
by Hölder's inequality, Lemma 5.2, and (6.20), (6.22)Y2≲1μ(Qαl)∑k=l∞∑τ∈Ik∑ν=1N(k,τ)2ksqμ(Qτk,ν)χ{(τ,ν):Qτk,ν⊂Qαl}(τ,ν)×{∑k′=−∞l−1∑τ′∈Ik′∑ν′=1N(k′,τ′)2−(k−k′)ϵ′2(k−k′)sμ(Qτ′k′,ν′)|Qk′(f)(yτ′k′,ν′)|q×1V2−(k⋀k′)(yτk,ν)+V2−(k⋀k′)(yτ′k′,ν′)+V(yτk,ν,yτ′k′,ν′)×(2−(k⋀k′)2−(k⋀k′)+d(yτk,ν,yτ′k′,ν′))ϵ}≲(RHSof(6.4))q, which completes the proof of
Proposition 6.3.Remark 6.4.
We point out that Remark5.5 applies in a similar way to Proposition 6.3.From Proposition6.3, it is easy to deduce that the
definition of the norm ∥⋅∥F˙∞,qs(𝒳) is independent of the choice of ATIs. We omit the  details.Proposition 6.5.
Adopting the
notation from Proposition6.3, one has for all f∈(𝒢°0ϵ(β,γ))′ with 0<β,γ<ϵ, (6.23)supl∈ℤsupα∈Il{1μ(Qαl)∫Qαl∑k=l∞2ksq|Dk(f)(x)|qdμ(x)}1/q~supl∈ℤsupα∈Il{1μ(Qαl)∫Qαl∑k=l∞2ksq|Qk(f)(x)|qdμ(x)}1/q.The following theorem will show that the definition of
the norm∥⋅∥F˙∞,qs(𝒳) is independent of the choice of the space of
distributions.Proposition 6.6.
Let all the notation be as in Definition6.1. Let |s|<ϵ and p(s,ϵ)<q≤∞.
If f∈(𝒢°0ϵ(β1,γ1))′ with max{0,−s}<β1<ϵ and max{0,s}<γ1<ϵ,
and if ∥f∥F˙∞,qs(𝒳)<∞,
then f∈(𝒢°0ϵ(β2,γ2))′ for every max{0,−s}<β2<ϵ and
max{0,s}<γ2<ϵ.Proof.
We use the same notation as in the proof of Proposition5.7. Let ψ∈𝒢°(ϵ,ϵ).
For any f∈(𝒢°0ϵ(β1,γ1))′ with max{0,−s}<β1<ϵ and max{0,s}<γ1<ϵ,
when q≤1,
by Theorem 4.13 together with (5.24), (5.25), and (5.5),  we have  (6.24)|〈f,ψ〉|=|∑k=−∞∞∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)Dk(f)(yτk,ν)〈D˜k(⋅,yτk,ν),ψ〉|≲∥ψ∥𝒢(β2,γ2){∑k=0∞∑τ∈Ik∑ν=1N(k,τ)2−kβ2q[μ(Qτk,ν)|Dk(f)(yτk,ν)|]q×[1V1(x1)+V(x1,yτk,ν)1(1+d(x1,yτk,ν))γ2]q+∑k=−∞−1∑τ∈Ik∑ν=1N(k,τ)2kγ2′q[μ(Qτk,ν)|Dk(f)(yτk,ν)|]q×[1V2−k(x1)+V(x1,yτk,ν)2−kγ2(2−k+d(x1,yτk,ν))γ2]q}1/q.
Notice that fork∈ℤ+,
when d(x1,yτk,ν)<1,
then V1(x1)~V1(yτk,ν)≳2kκV2−k(yτk,ν)≳μ(Qτk,ν), while when 2l≤d(x1,yτk,ν)<2l+1 for some l∈ℤ+,
then V(x1,yτk,ν)≳V2l(yτk,ν)≳V2−k(yτk,ν)~μ(Qτk,ν). Therefore,  for k∈ℤ+, (6.25)μ(Qτk,ν)V1(x1)+V(x1,yτk,ν)≲1.
Whenk∈ℤ∖ℤ+,
noticing that if d(x1,yτk,ν)<2−k,
then (6.26)V2−k(x1)≳V2−k(yτk,ν)≳μ(Qτk,ν); if 2l2−k≤d(x1,yτk,ν)<2l+12−k for some l∈ℤ+,
then (6.27)V(x1,yτk,ν)≳V2l2−k(yτk,ν)≳2lκV2−k(yτk,ν)≳μ(Qτk,ν), we also have (6.28)μ(Qτk,ν)V2−k(x1)+V(x1,yτk,ν)≲1.
We also need the following trivial estimate that(6.29)∑τ∈Ik∑ν=1N(k,τ)2ksq|Dk(f)(yτk,ν)|q≲∥f∥F˙∞,qs(𝒳)q.
Using (6.25), (6.28), and (6.29) yields that (6.30)|〈f,ψ〉|≲∥ψ∥𝒢(β2,γ2)∥f∥F˙∞,qs(𝒳){∑k=0∞2−k(β2+s)q+∑k=−∞−12k(γ2′−s)q}1/q≲∥ψ∥𝒢(β2,γ2)∥f∥F˙∞,qs(𝒳), where we chose s<γ2′<γ2.
If1<q≤∞,
Hölder's inequality, (6.25), (6.28), (6.29), and Lemma 2.1(ii) prove that (6.31)|〈f,ψ〉|≲∥ψ∥𝒢(β2,γ2){∑k=0∞2−kβ22−ks[∑τ∈Ik∑ν=1N(k,τ)2ksqμ(Qτk,ν)|Dk(f)(yτk,ν)|q×1V1(x1)+V(x1,yτk,ν)1(1+d(x1,yτk,ν))γ2]1/q×[∫𝒳1V1(x1)+V(x1,y)1(1+d(x1,y))γ2dμ(y)]1/q′+∑k=−∞−12kγ2′2−ks[∑τ∈Ik∑ν=1N(k,τ)2ksqμ(Qτk,ν)|Dk(f)(yτk,ν)|q×1V2−k(x1)+V(x1,yτk,ν)2−kγ2(2−k+d(x1,yτk,ν))γ2]1/q×[1V2−k(x1)+V(x1,y)2−kγ2(2−k+d(x1,y))γ2dμ(y)]1/q′}≲∥ψ∥𝒢(β2,γ2)∥f∥F˙∞,qs(𝒳).
Combining (6.30) and (6.31) with an argument similar
to the proof of Proposition 5.7 then completes the proof of Proposition
6.6.Now we can introduce the homogeneous Triebel-Lizorkin
spacesF˙∞,qs(𝒳).Definition 6.7.
Letϵ1∈(0,1], ϵ2>0, ϵ3>0, ϵ∈(0,ϵ1⋀ϵ2) and let {Sk}k∈ℤ be an (ϵ1,ϵ2,ϵ3)-ATI.
For k∈ℤ,
set Dk=Sk−Sk−1.
Let |s|<ϵ and p(s,ϵ)<q≤∞.
The spaceF˙∞,qs(𝒳) is defined to be the set of all f∈(𝒢°0ϵ(β,γ))′,
for some β,γ satisfying (6.32)|s|<β<ϵ,max{s,0,−s−κ}<γ<ϵ such that (6.33)∥f∥F˙∞,qs(𝒳)=supl∈ℤsupα∈Il{1μ(Qαl)∫Qαl∑k=l∞2ksq|Dk(f)(x)|qdμ(x)}1/q<∞, where the supremum is taken over
all dyadic cubes as in Lemma 2.19 and the usual modification is made when q=∞.Propositions6.5 and 6.6 show that the definition of
the spaces F˙∞,qs(𝒳) is independent of the choice of the ATI and the distribution space (𝒢°(β,γ))′,
with β and γ satisfying (6.32).Remark 6.8.
To guarantee that the definition of the spaceF˙∞,qs(𝒳) is independent of the choice of the
distribution space (𝒢°(β,γ))′,
we only need the restriction that max{0,−s}<β<ϵ and max{0,s}<γ<ϵ.
Moreover, if we assume that max{0,s}<β<ϵ and max{0,s−κ}<γ<ϵ,
we can then verify that 𝒢°(β,γ)⊂F˙∞,qs(𝒳);
see Proposition 6.9 below.


### 6.2. Properties ofF˙∞,qs(𝒳) and Boundedness of Singular Integrals

In this subsection, we first present some basic
properties ofF˙∞,qs(𝒳).
By establishing a maximal function characterization of F˙∞,qs(𝒳),
we then establish some relations between the spaces F˙∞,qs(𝒳) and the spaces C˙s(𝒳) and between the spaces F˙∞,qs(𝒳) and the space BMO(𝒳).
Finally, we obtain the boundedness on F˙∞,qs(𝒳) of the singular integrals considered by Nagel
and Stein in [44].Proposition 6.9.
Letϵ be as in Definition 6.7,  |s|<ϵ, and p(s,ϵ)<q≤∞. Then,
(i)
F˙∞,q0s(𝒳)⊂F˙∞,q1s(𝒳) if p(s,ϵ)<q0≤q1≤∞;(ii)
B˙∞,qs(𝒳)⊂F˙∞,qs(𝒳)⊂B˙∞,∞s(𝒳);(iii)
ifmax{0,−s}<β<ϵ and max{0,s}<γ<ϵ,
then F˙∞,qs(𝒳)⊂(𝒢°0ϵ(β,γ))′;(iv)
ifmax{s,0}<β<ϵ and max{0,−s−κ}<γ<ϵ,
then 𝒢°(β,γ)⊂F˙∞,qs(𝒳);(v)
the spacesF˙∞,qs(𝒳)/𝒩 are  complete.Proof.
Property (i) is a simple corollary of (5.5).
To see (ii),B˙∞,qs(𝒳)⊂F˙∞,qs(𝒳) is obvious by the definitions of the both
spaces B˙∞,qs(𝒳) and F˙∞,qs(𝒳),
while F˙∞,qs(𝒳)⊂B˙∞,∞s(𝒳) can be obtained by their definitions together
with the Lebesgue differential theorem; we omit the details.
Property (iii) is a consequence of the second
inclusion in Property (ii) of this proposition and Proposition5.10(iii) on B˙∞,∞s(𝒳),
while Property (v) can be easily deduced from Property (iii), Property (iv) is
a conclusion of the first inclusion in Property (ii) of this proposition and
Proposition 5.10(iv) on B˙∞,qs(𝒳),
which completes the proof of this proposition.To obtain some relations between the spacesF˙∞,qs(𝒳) and the space BMO(𝒳),
we need the following technical result which in fact gives a new
characterization of Triebel-Lizorkin space F˙∞,qs(𝒳).
We first introduce a maximal function.For anyx∈𝒳 and l∈ℤ,
using Lemma 2.19, it is easy to prove that there exists a finite number of α∈Il such that (6.34)Qαl∩B(x,2−l+1)≠∅,which will denote by Qαlil with i=1,…,ml(x),
and moreover, ml(x) is no more than a positive integer m∈ℕ which is independent of l and x;
see [85, pages 1385-1386] for a detailed proof. In what follows, for convenience sake, we will
always assume that ml(x)=m by letting Qαlil=∅ for i=ml(x)+1,…,m when ml(x)<m.
Let {Dk}k∈ℤ be as in Definition 6.7. Now for any s∈ℝ, q∈(0,∞], f∈(𝒢°(β,γ))′ with 0<β,γ<ϵ and x∈𝒳,
we define the maximal functionℭ˙qs(f)(x) by (6.35)ℭ˙qs(f)(x)=supl∈ℤ[1μ(∪i=1mQαlil)∫∪i=1mQαlil∑k=l∞2ksq|Dk(f)(y)|qdμ(y)]1/q, where the usual modification is
made when q=∞.Proposition 6.10.
Letϵ, s, and q be as in Definition 6.7. Then f∈F˙∞,qs(𝒳) if and only if f∈(𝒢°(β,γ))′ with β,γ as in (6.32) and ℭ˙qs(f)∈L∞(𝒳).
Moreover, in this case, (6.36)∥f∥F˙∞,qs(𝒳)~∥ℭ˙qs(f)∥L∞(𝒳).Proof.
For anys∈ℝ, q∈(0,∞], f∈(𝒢°(β,γ))′ with β,γ as in (6.32) and x∈𝒳,
let (6.37)ℭ˙q,1s(f)(x)=supl∈ℤsupQαl∋xα∈Il[1μ(Qαl)∫Qαl∑k=l∞2ksq|Dk(f)(y)|qdμ(y)]1/q, where the usual modification is
made when q=∞.
Obviously, (6.38)∥f∥F˙∞,qs(𝒳)~∥ℭ˙q,1s(f)∥L∞(𝒳). To prove the conclusion of this
proposition, it suffices to verify that when (6.39)⋃i=1ml(x)Qαlil⊃B(x,2−l+1), for any i1,i2∈{1,…,ml(x)}, (6.40)μ(Qαli1l)~μ(Qαli2l). To verify (6.40), by symmetry,
it suffices to verify that (6.41)μ(Qαli1l)≲μ(Qαli2l). To this end, suppose zk∈B(x,2−l+1)∩Qαlikl with k=1,2.
Then for any w∈B(zαli1l,C62−l),
we have (6.42)d(w,zαli2l)≤d(w,zαli1l)+d(zαli1l,z1)+d(z1,x)+d(x,z2)+d(z2,zαli2l)≤3C62−l+2−l+2. Thus, Qαli1l⊂B(zαli1l,C62−l)⊂B(zαli2l,(3C6+4)2−l), which together with Lemma 2.19 and the double
property of μ gives (6.41), and hence, completes the proof
of Proposition 6.10.Using Theorem5.19(i) and Proposition 6.10, and an
argument similar to that in [86], we can establish the connections between F˙∞,qs(𝒳) with BMO(𝒳) and C˙s(𝒳) with s>0 as follows.Theorem 6.11.
Letϵ be as in Definition 6.7. Then,
(i)
if0<s<ϵ,
then C˙s(𝒳)=F˙∞,∞s(𝒳)=B˙∞,∞s(𝒳) with equivalent norms;(ii)
BMO(𝒳)=F˙∞,20(𝒳) with equivalent norm.Proof.
We first verify (i). Letf∈C˙s(𝒳) and let {Dk}k∈ℤ be as in Definition 6.7. We first claim that f∈(𝒢°0ϵ(β,γ))′ when 0<β<ϵ and s<γ<ϵ.
In fact, from f∈C˙s(𝒳),
it follows that for all x∈𝒳, (6.43)|f(x)−f(x1)|≲∥f∥C˙s(𝒳)d(x,x1)s, which implies that for any g∈𝒢°(β,γ) with 0<β<ϵ and s<γ<ϵ, (6.44)|〈f,g〉|=|∫𝒳[f(x)−f(x1)]g(x)dμ(x)|≲∥f∥C˙s(𝒳)∥g∥𝒢(β,γ)∫𝒳d(x,x1)s1V1(x1)+V(x1,x)(11+d(x1,x))γdμ(x)≲∥f∥C˙s(𝒳)∥g∥𝒢(β,γ), where in the last step, we used
Lemma 2.1(ii).
Moreover, for allk∈ℤ and x∈𝒳,
by Lemma 2.1(ii) and 0<s<ϵ<ϵ2,
we then have (6.45)|Dk(f)(x)|=|∫𝒳Dk(x,y)f(y)dμ(y)|=|∫𝒳Dk(x,y)[f(y)−f(x)]dμ(y)|≲∥f∥C˙s(𝒳)∫𝒳1V2−k(x)+V2−k(y)+V(x,y)(2−k2−k+d(x,y))ϵ2d(x,y)sdμ(y)≲2−ks∥f∥C˙s(𝒳), which proves that (6.46)∥f∥F˙∞,∞s(𝒳)=supk∈ℤsupx∈𝒳2ks|Dk(f)(x)|≲∥f∥C˙s(𝒳). Thus, C˙s(𝒳)⊂F˙∞,∞s(𝒳).
Conversely, letf∈F˙∞,∞s(𝒳).
By Proposition 6.6, without loss of generality, we may assume that f∈(𝒢°(β,γ))′ with β,γ as in (6.32). Let all the notation as in
Theorem 3.10. Then, by Theorem 3.13, for any g∈𝒢°(β,γ) with β,γ as in (6.32), since ∫𝒳g(x)dμ(x)=0,
we then have (6.47)〈f,g〉=∑k=−∞∞〈D˜kDk(f)(⋅),g(⋅)〉=∑k=−∞∞〈D˜kDk(f)(⋅)−D˜kDk(f)(x1),g(⋅)〉, which means that in (𝒢°(β,γ))′, f(x)=∑k=−∞∞[D˜kDk(f)(⋅)−D˜kDk(f)(x1)]. For x∈𝒳,
we now let h(x)≡∑k=−∞∞[D˜kDk(f)(x)−D˜kDk(f)(x1)] and we first verify that h is a function satisfying the following growth
condition that (6.48)|h(x)|≲∥f∥F˙∞,∞s(𝒳)d(x,x1)s. In fact, for any x∈𝒳,
assume that 2−l0−1<d(x,x1)≤2−l0 with some l0∈ℤ and write that (6.49)h(x)=∑k=−∞∞∫𝒳[D˜k(x,z)−D˜k(x1,z)]Dk(f)(z)dμ(z)=∑k=−∞l0−1∫𝒳[D˜k(x,z)−D˜k(x1,z)]Dk(f)(z)dμ(z)+∑k=l0∞⋯≡J1+J2. For J1,
by the regularity of D˜k,
Lemma 2.1(ii) and s<ϵ<ϵ1,
we have (6.50)|J1|≲∑k=−∞l0−1∫𝒳(d(x,x1)2−k+d(x,z))ϵ11V2−k(x)+V2−k(z)+V(x,z)(2−k2−k+d(x,z))ϵ2|Dk(f)(z)|dμ(z)≲∥f∥F˙∞,∞s(𝒳)d(x,x1)s. To estimate J2,
by s>0 and Proposition 2.7(i), we
obtain (6.51)J2≲∥f∥F˙∞,∞s(𝒳)∑k=l0∞2−ks∫𝒳[|D˜k(x,z)|+|D˜k(x1,z)|]dμ(z)≲∥f∥F˙∞,∞s(𝒳)d(x,x1)s. Thus, our claim is true.
Notice that for anyx,y∈𝒳, (6.52)h(x)−h(y)=∑k=−∞∞[D˜kDk(f)(x)−D˜kDk(f)(y)]. Repeating the above proof yields
that for all x,y∈𝒳, (6.53)|h(x)−h(y)|≲∥f∥F˙∞,∞s(𝒳)d(x,y)s. Thus, h∈C˙s(𝒳) and (6.54)∥h∥C˙s(𝒳)≲∥f∥F˙∞,∞s(𝒳). In this sense, we say that F˙∞,∞s(𝒳)⊂C˙s(𝒳), which finishes the proof of (i).
To see (ii), letf∈BMO(𝒳) and {Dk}k∈ℤ be as in Definition 6.7. Proposition 5.10(iv)
and Theorem 5.19(i) immediately imply that f∈(𝒢°0ϵ(β,γ))′ with
0<β,γ<ϵ.
Let now Qαl for l∈ℤ and α∈Il be a dyadic cube as in Lemma 2.19. Set Bαl=B(zαl,2C62−l) and write (6.55)f=(f−mBαl(f))χBαl+(f−mBαl(f))χ{𝒳∖Bαl}+mBαl(f)≡f1+f2+f3. Obviously, μ(Qαl)~μ(Bαl) and Dk(f3)=0;
in combination with Proposition 3.15, this shows that (6.56){1μ(Qαl)∫Qαl∑k=l∞|Dk(f1)(x)|2dμ(x)}1/2≲{1μ(Qαl)∫𝒳|f1(x)|2dμ(x)}1/2≲{1μ(Bαl)∫Bαl|f−mBαl(f)|2dμ(x)}1/2≲∥f∥BMO(𝒳). To estimate f2,
notice that for x∈Qαl and y∈𝒳∖Bαl,
then d(x,y)≳2−l+d(y,zαl),
which together with the size condition of Dk yields that for x∈Qαl and k≥l, (6.57)|Dk(f2)(x)|=|∫𝒳∖BαlDk(x,y)[f(y)−mBαl(f)]dμ(y)|≲2(l−k)ϵ2∑j=0∞12jϵ21V(zαl,2C62j−l+1)∫d(y,zαl)<2C62j−l+1|f(y)−mBαl(f)|dμ(y). The definition of BMO(𝒳) together with the double property of μ gives that (6.58)|mB(zαl,2C62j−l+1)(f)−mBαl(f)|≲(j+1)∥f∥BMO(𝒳), which further implies that for x∈Qαl and k≥l, (6.59)|Dk(f2)(x)|≲2(l−k)ϵ2∑j=0∞12jϵ2{1V(zαl,2C62j−l+1)∫d(y,zαl)<2C62j−l+1|f(y)−mB(zαl,2C62j−l+1)(f)|dμ(y)+|mB(zαl,2C62j−l+1)(f)−mBαl(f)|}≲2(l−k)ϵ2∥f∥BMO(𝒳). From this, it follows
that (6.60){1μ(Qαl)∫Qαl∑k=l∞|Dk(f2)(x)|2dμ(x)}1/2≲∥f∥BMO(𝒳){∑k=l∞22(l−k)ϵ2}1/2≲∥f∥BMO(𝒳). Combining the above estimates,
we know that f∈F˙∞,20(𝒳) and (6.61)∥f∥F˙∞,20(𝒳)≲∥f∥BMO(𝒳).
We now prove the converse. Letf∈F˙∞,20(𝒳).
By Proposition 6.6, without loss of generality, we may assume that f∈(𝒢°(β,γ))′ with β,γ as in (6.32). In the rest of the proof of this
theorem, we denote ℭ˙20 and S˙2,10 simply by ℭ˙ and S˙.
Moreover, for j∈ℤ, f as above and x∈𝒳,
we define (6.62)S˙j(f)(x)={∑k=j∞∫d(x,y)<2−k|Dk(f)(y)|2dμ(y)V2−k(x)}1/2. Obviously, S˙∞(f)(x)=S˙(f)(x). Let the notation as in Theorem 3.11. For any f as above and x∈𝒳,
we also set (6.63)S¯˙(f)(x)={∑k=−∞∞∫d(x,y)<2−k|D¯k(f)(y)|2dμ(y)V2−k(x)}1/2. Theorem 5.13 together with Remark
5.5 shows that for all f∈H1(𝒳), (6.64)∥S¯˙(f)∥L1(𝒳)≲∥f∥H1(𝒳).
For any fixedf as above and x∈𝒳,
we define the “stopping-time" j(x) by (6.65)j(x)=inf{j∈ℤ:S˙j(f)(x)≤Aℭ˙(f)(x)}, where A>0 is a large constant to be determined later. We
first claim that for any y∈𝒳 and l∈ℤ,
if we choose A to be large enough, then there exists a
constant C12>0 such that (6.66)μ({x∈𝒳:d(x,y)<2−l,l≥j(x)})≥C12μ(B(y,2−l)). In fact, let B0=B(y,2−l).
Then ⋃x∈B0B(x,2−l)⊂B(y,2−l+1). Let (6.67)B(y,2−l+1)⊂⋃i=1mQαlil≡P as in the definition of ℭ˙.
Let w∈B(y,2−l+1)∩Qαlil. Then for any x∈Qαlil, d(x,y)≤d(x,w)+d(w,y)≤C62−l+2−l+1, which shows that (6.68)P⊂B(y,(C6+2)2−l).Thus, μ(B0)~μ(P). From this, it follows that(6.69)1μ(B0)∫B0[S˙l(f)(x)]2dμ(x)=∑k=l∞1μ(B0)∫B0∫d(x,z)<2−k|Dk(f)(z)|2dμ(z)V2−k(x)dμ(x)≤C1μ(P)∑k=l∞∫P∫B0|Dk(f)(z)|2χB(z,2−k)(x)dμ(x)V2−k(x)dμ(z)≤C131μ(P)∫P∑k=l∞|Dk(f)(z)|2dμ(z)≤C13[infx∈B0ℭ˙(f)(x)]2, where C,C13>0 are constants independent of l and x.
Thus, if A2>C13,
then (6.70)μ({x∈B0:Sl(f)(x)>Aℭ˙(f)(x)})≤C13A2μ(B0), which in turn shows (6.66) with C12=1−C13/A2>0 if A2>C13.
Letg∈𝒢°b(ϵ1,ϵ2) and ∥g∥H1(𝒳)≤1.
By Theorem 3.11, (6.66), the Fubini theorem, Hölder's inequality, (6.64), and
Proposition 6.10, we then have (6.71)|〈f,g〉|=|〈f,∑k=−∞∞DkD¯k(g)〉|=|∑k=−∞∞〈Dkt(f),D¯k(g)〉|≤∑k=−∞∞∫𝒳|Dkt(f)(y)D¯k(g)(y)|dμ(y)≲∫𝒳[∑k=l(x)∞∫d(x,y)<2−k|Dkt(f)(y)D¯k(g)(y)|dμ(y)V2−k(x)]dμ(x)≲∫𝒳S˙l(x)(f)(x)S¯˙(g)(x)dμ(x)≲∫𝒳ℭ˙(f)(x)S¯˙(g)(x)dμ(x)≲∥ℭ˙(f)∥L∞(𝒳)∥g∥H1(𝒳)≲∥f∥F˙∞,20(𝒳), which together with Theorem 5.19(i) further implies that f∈BMO(𝒳) and (6.72)∥f∥BMO(𝒳)≲∥f∥F˙∞,20(𝒳). This finishes the proof of
Theorem 6.11.We end this subsection by the boundedness onF˙∞,qs(𝒳) of singular integral operators of order ϵ,
which satisfy (I-1) through (I-4) in Subsection 5.2.Theorem 6.12.
Letϵ and q be as in Definition 6.7. If T is a singular integral operator of order ϵ,
then T is bounded from 𝒢°b(β,γ) (with 0<β≤ϵ and γ>0) to F˙∞,qs(𝒳).
Moreover, there exists a constant C>0 such that for all f∈𝒢°b(β,γ) with 0<β≤ϵ and γ>0, (6.73)∥Tf∥F˙∞,qs(𝒳)≤C∥f∥F˙∞,qs(𝒳).Proof.
Combining some estimates and technics used in the proof of Proposition5.25 with those used in the proof of Proposition 6.3 gives the desired
conclusions. The details are left to the reader.Remark 6.13.
By Theorem8.15 below, if p(s,ϵ)<q≤∞,
then F˙∞,qs(𝒳) is the dual space of F˙1,q′−s(𝒳).
In this case, Theorem 6.12 can be deduced from Theorem 5.23 together with a
duality argument. This provides another proof of Theorem 6.12.


### 6.3. Inhomogeneous Plancherel-Pôlya Inequality and Definition ofF∞,qs(𝒳)

In this and the next subsection,μ(𝒳) can be finite or infinite. We first introduce
the norm in ∥⋅∥F∞,qs(𝒳) via some IATI and then verify that this norm is independent
of the choices of IATIs and the distribution spaces; see also
[84]. Similarly to the
case of the space F˙∞,qs(𝒳),
we need also first to establish an inhomogeneous Plancherel-Pôlya inequality
related to the norm ∥⋅∥F∞,qs(𝒳).Definition 6.14.
Letϵ1∈(0,1], ϵ2>0, ϵ3>0, ϵ∈(0,ϵ1⋀ϵ2) and let {Sk}k∈ℤ+ be an (ϵ1,ϵ2,ϵ3)-IATI.
Set Dk=Sk−Sk−1 for k∈ℕ,
and D0=S0.
Let {Qτ0,ν:τ∈I0,ν=1,…,N(0,τ)} with a fixed large j∈ℕ be dyadic cubes as in Section 4. Let |s|<ϵ and p(s,ϵ)<q≤∞.
For any f∈(𝒢0ϵ(β,γ))′ with 0<β,γ<ϵ,
define (6.74)∥f∥F∞,qs(𝒳)=max{supτ∈I0ν=1,…,N(0,τ)mQτ0,ν(|D0(f)|),supl∈ℕsupα∈Il[1μ(Qαl)∫Qαl∑k=l∞2ksq|Dk(f)(x)|qdμ(x)]1/q}, where the supremum is
taken over all dyadic cubes as in Lemma 2.19 and Section 4, and the usual
modification is made when q=∞.To verify that the definition of∥⋅∥F∞,qs(𝒳) is independent of the choice of IATIs, we need the following inequality of Plancherel-Pôlya type.Proposition 6.15.
Letϵ1∈(0,1], ϵ2>0, ϵ3>0, ϵ∈(0,ϵ1⋀ϵ2) and let {Sk}k∈ℤ and {Pk}k∈ℤ be two (ϵ1,ϵ2,ϵ3)-ATIs. Set Dk=Sk−Sk−1 and Qk=Pk−Pk−1 for k∈ℕ, D0=S0,
and Q0=S0.
Let {Qτ0,ν:τ∈I0,ν=1,…,N(0,τ)} with a fixed large j∈ℕ be dyadic cubes as in Section 4. Let |s|<ϵ and p(s,ϵ)<q≤∞.
Then for all f∈(𝒢0ϵ(β,γ))′ with 0<β,γ<ϵ, (6.75)max{supτ∈I0ν=1,…,N(0,τ)mQτ0,ν(|D0(f)|),supl∈ℕsupα∈Il[1μ(Qαl)∑k=l∞∑τ∈Ik∑ν=1N(k,τ)2ksqμ(Qτk,ν)χ{(τ,ν):Qτk,ν⊂Qαl}(τ,ν)(supx∈Qτk,ν|Dk(f)(x)|)q]1/q}~max{supτ∈I0ν=1,…,N(0,τ)mQτ0,ν(|Q0(f)|),supl∈ℕsupα∈Il[1μ(Qαl)∑k=l∞∑τ∈Ik∑ν=1N(k,τ)2ksqμ(Qτk,ν)χ{(τ,ν):Qτk,ν⊂Qαl}(τ,ν)(infx∈Qτk,ν|Qk(f)(x)|)q]1/q}.Proof.
To prove Proposition6.15, it suffices to verify that for all f∈(𝒢0ϵ(β,γ))′ with 0<β,γ<ϵ,
the left-hand side of (6.75) is controlled by its right side.
Let all the notation be as in Proposition5.25. Then,
as in the proof of Proposition 5.25, by (5.78) and (5.80), we still control mQτ0,ν(|D0(f)|) by Z1+Z2.
Moreover, by (5.82) and Lemma 2.1(ii), we have (6.76)Z1≲supτ′∈I0ν′=1,…,N(0,τ′)mQτ′0,ν′(|Q0(f)|), which is the desired estimate.
To estimateZ2,
we need the following trivial estimate that for k′∈ℕ, (6.77)[∑τ′∈Ik′∑ν′=1N(k′,τ′)2k′sq|Qk′(f)(yτ′k′,ν′)|q]1/q≲supl∈ℕsupα∈Il[1μ(Qαl)∑k=l∞∑τ∈Ik∑ν=1N(k,τ)2ksqμ(Qτk,ν)χ{(τ,ν):Qτk,ν⊂Qαl}(τ,ν)(infx∈Qτk,ν|Qk(f)(x)|)q]1/q. By this and (5.82) together with μ(Qτ′k′,ν′)≲V1(yτ′k′,ν′) and (5.5), when q≤1,
we have (6.78)Z2≲∑k′=1∞2−k′(ϵ+s){∑τ′∈Ik′∑ν′=1N(k′,τ′)2k′sq|Qk′(f)(yτ′k′,ν′)|q}1/q≲RHSof(6.75), where we used the assumption
that |s|<ϵ.
Similarly, when1<q≤∞,
by Hölder's inequality and Lemma 2.1(ii), (6.79)Z2≲∑k′=1∞2−k′(ϵ+s){∑τ′∈Ik′∑ν′=1N(k′,τ′)2k′sq|Qk′(f)(yτ′k′,ν′)|q}1/q×{∫𝒳1V1(yτ0,ν)+V1(y)+V(yτ0,ν,y)1(1+d(yτ0,ν,y))ϵdμ(y)}1/q′≲RHSof(6.75). Thus, mQτ0,ν(|D0(f)|) for τ∈I0 and ν=1,…,N(0,τ) is controlled by the right-hand side of
(6.75).
We now verify that the second term of the left-hand
side of (6.75) is also controlled by its right-hand side. To this end, for any k∈ℕ and z∈Qτk,ν,
we also control |Dk(f)(z)| by Y1+Y2.
The estimate for Y2 is similar to the estimates for Y1+Y2 in the proof of Proposition 6.3 and we omit
the details. To estimate Y1,
by (5.95) and Lemma 2.1(ii), we have (6.80)supx∈Qτk,νY1≲2−kϵ[supτ′∈I0ν′=1,…,N(0,τ′)mQτ′0,ν′(|Q0(f)|)]×∑τ′∈I0∑ν′=1N(0,τ′)μ(Qτ′0,ν′)1V1(yτk,ν)+V1(yτ′0,ν′)+V(yτk,ν,yτ′0,ν′)1(1+d(yτk,ν,yτ′0,ν′))ϵ≲2−kϵ[supτ′∈I0ν′=1,…,N(0,τ′)mQτ′0,ν′(|Q0(f)|)], which together with Lemma 2.19
and |s|<ϵ gives that for l∈ℕ and α∈Il, (6.81)[1μ(Qαl)∑k=l∞∑τ∈Ik∑ν=1N(k,τ)2ksqμ(Qτk,ν)χ{(τ,ν):Qτk,ν⊂Qαl}(τ,ν)supx∈Qτk,ν|Y1|q]1/q≲supτ′∈I0ν′=1,…,N(0,τ′)mQτ′0,ν′(|Q0(f)|). This is the desired estimate for Y1 and hence, we complete the proof of Proposition
6.15.Remark 6.16.
We point out that Remark5.5 applies in a similar way to  Proposition
6.15.From Proposition6.15 and Lemma 2.19, it is easy to
deduce the following proposition. We omit the  details.Proposition 6.17.
Adopting the notation from Proposition6.15,
one has for all for f∈(𝒢0ϵ(β,γ))′ with
0<β,γ<ϵ, (6.82)max{supτ∈I0ν=1,…,N(0,τ)mQτ0,ν(|D0(f)|),supl∈ℕsupα∈Il[1μ(Qαl)∫Qαl∑k=l∞2ksq|Dk(f)(x)|qdμ(x)]1/q}~max{supτ∈I0ν=1,…,N(0,τ)mQτ0,ν(|Q0(f)|),supl∈ℕsupα∈Il[1μ(Qαl)∫Qαl∑k=l∞2ksq|Qk(f)(x)|qdμ(x)]1/q}.Proposition6.17 shows that the definition of the norm ∥⋅∥F∞,qs(𝒳) is independent of the choice of IATIs. We now verify that under some restrictions
on β and γ,
it is also independent of the choice of distribution  spaces.Proposition 6.18.
Let all the notation be as in Definition6.14. Let |s|<ϵ and p(s,ϵ)<q≤∞.
If f∈(𝒢0ϵ(β1,γ1))′ with max{0,−s}<β1<ϵ and 0<γ1<ϵ,
and if ∥f∥F∞,qs(𝒳)<∞,
then f∈(𝒢0ϵ(β2,γ2))′ for every max{0,−s}<β2<ϵ and
0<γ2<ϵ.Proof.
We use the notation from the proof of Proposition5.28. Let ψ∈𝒢(ϵ,ϵ),
and f∈(𝒢0ϵ(β1,γ1))′ with max{0,d(1−1/q)+−s−d}<β1<ϵ and 0<γ1<ϵ,
and ∥f∥F∞,qs(𝒳)<∞.
To verify that f∈(𝒢0ϵ(β2,γ2))′ with max{0,−s}<β2<ϵ and 0<γ2<ϵ,
we need the following trivial estimates that for τ∈I0 and ν=1,…,N(0,τ),(6.83)mQτ0,ν(|D0(f)|)≲∥f∥F∞,qs(𝒳) and that for k∈ℕ, (6.84)[∑τ∈Ik∑ν=1N(k,τ)2ksq|Dk(f)(yτk,ν)|q]1/q≲∥f∥F∞,qs(𝒳).
Whenq≤1,
by Theorem 4.16, (5.80), (5.105), (5.5), (6.25), (6.83), and (6.84),  we
have (6.85)|〈f,ψ〉|=|∑τ∈I0∑ν=1N(0,τ)∫Qτ0,ν〈D˜0(⋅,y),ψ〉dμ(y)Dτ,10,ν(f)+∑k=1∞∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)〈D˜k(⋅,yτk,ν),ψ〉Dk(f)(yτk,ν)|≲∥ψ∥𝒢(β2,γ2)∥f∥F∞,qs(𝒳), where we used the assumption
that γ2>0 and β2>−s.
If1<q≤∞,
by (6.83), Hölder's inequality, (6.25), and (6.84), we obtain (6.86)|〈f,ψ〉|≲∥ψ∥𝒢(β2,γ2){∥f∥F∞,qs(𝒳)+∑k=1∞2−kβ2[∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)|Dk(f)(yτk,ν)|q×1V1(x1)+V(x1,yτk,ν)1(1+d(x1,yτk,ν))γ2]1/q×[∫𝒳1V1(x1)+V(x1,y)1(1+d(x1,y))γ2dμ(y)]1/q′}≲∥ψ∥𝒢(β2,γ2)∥f∥F∞,qs(𝒳), where we used the assumption β2>−s and γ2>0 again. This finishes the proof of Proposition
6.18.Based on Propositions6.17 and 6.18, we can
now introduce the inhomogeneous Triebel-Lizorkin spaces F∞,qs(𝒳).Definition 6.19.
Letϵ1∈(0,1], ϵ2>0, ϵ3>0, ϵ∈(0,ϵ1⋀ϵ2) and let {Sk}k∈ℤ be an (ϵ1,ϵ2,ϵ3)-IATI.
Set Dk=Sk−Sk−1 for k∈ℕ and D0=S0.
Let {Qτ0,ν:τ∈I0,ν=1,…,N(0,τ)} with a fixed large j∈ℕ be dyadic cubes as in Section 4. Let |s|<ϵ and p(s,ϵ)<q≤∞.
The spaceF∞,qs(𝒳) is defined to be the set of all f∈(𝒢0ϵ(β,γ))′,
for some |s|<β<ϵ and 0<γ<ϵ,
such that (6.87)∥f∥F∞,qs(𝒳)=max{supτ∈I0ν=1,…,N(0,τ)mQτ0,ν(|D0(f)|),supl∈ℕsupα∈Il[1μ(Qαl)∫Qαl∑k=l∞2ksq|Dk(f)(x)|qdμ(x)]1/q}<∞, where the supremum is taken over
all dyadic cubes as in Lemma 2.19 and Section 4, and the usual modification is
made when q=∞.Propositions6.17 and 6.18 show that the
definition of the space F∞,qs(𝒳) is independent of the choices of IATIs and the distribution spaces, (𝒢(β,γ))′ with β and γ with |s|<β<ϵ and 0<γ<ϵ.Remark 6.20.
To guarantee that the definition of the spaceF∞,qs(𝒳) is independent of the choice of the
distribution space (𝒢(β,γ))′,
we only need to restrict max{0,−s}<β<ϵ and 0<γ<ϵ.
However, if max{0,s}<β<ϵ and 0<γ<ϵ,
we can then verify that 𝒢(β,γ)⊂F∞,qs(𝒳); see Proposition 6.21 below.


### 6.4. Properties ofF∞,qs(𝒳) and Boundedness of Singular Integrals

In this subsection, we first present some basic
properties ofF∞,qs(𝒳) and the relation between F∞,qs(𝒳) and F˙∞,qs(𝒳).
By establishing a maximal function characterization of F∞,qs(𝒳),
we then derive some relations between the spaces F∞,qs(𝒳) and Cs(𝒳) and between the spaces F∞,qs(𝒳) and bmo(𝒳).
Finally, we prove boundedness results on the spaces F∞,qs(𝒳) for the classes of singular integral operators
considered in Subsection 5.4.Proposition 6.21.
Letϵ1∈(0,1], ϵ2>0, ϵ∈(0,ϵ1⋀ϵ2), |s|<ϵ, and p(s,ϵ)<q≤∞.
Then, the following hold.
(i)
F∞,q0s(𝒳)⊂F∞,q1s(𝒳) if p(s,ϵ)<q0≤q1≤∞.(ii)
Let−ϵ<s+θ<ϵ and θ>0.
Then for p(s,ϵ)<q0,q1≤∞,(6.88)F∞,q0s+θ(𝒳)⊂F∞,q1s(𝒳).(iii)
B∞,qs(𝒳)⊂F∞,qs(𝒳)⊂B∞,∞s(𝒳).(iv)
Ifmax{0,−s}<β<ϵ and 0<γ<ϵ,
then F∞,qs(𝒳)⊂(𝒢0ϵ(β,γ))′.(v)
Ifmax{s,0}<β<ϵ and 0<γ<ϵ,
then 𝒢(β,γ)⊂F∞,qs(𝒳).(vi)
The spacesF∞,qs(𝒳) are complete.Proof.
Property (ii) can be established by an argument similar to that used for
property (ii) of Proposition5.31, while property (i) and property (iii) through
property (vi) can be proved by an argument similar to those used for
Proposition 6.9 via Proposition 5.31, which completes the proof of Proposition 6.21.The following proposition gives a new characterization
of the spacesF∞,qs(𝒳) when 1≤q≤∞.Proposition 6.22.
Letϵ and {Dk}k∈ℤ+ be as in Definition 6.19. If 1≤q≤∞,
then f∈F∞,qs(𝒳) if and only if f∈(𝒢0ϵ(β,γ))′,
for some |s|<β<ϵ and 0<γ<ϵ,
and (6.89)supl∈ℤ+supα∈Il[1μ(Qαl)∫Qαl∑k=l∞2ksq|Dk(f)(x)|qdμ(x)]1/q<∞. Moreover, in this case, (6.90)∥f∥F∞,qs(𝒳)~supl∈ℤ+supα∈Il[1μ(Qαl)∫Qαl∑k=l∞2ksq|Dk(f)(x)|qdμ(x)]1/q.Proof.
Fixτ∈I0 and
ν=1,…,N(0,τ).
Since Qτ0,ν⊂Qτ0,
then it is easy to verify that Qτ0⊂B(zτ0,ν,2C6),
and hence μ(Qτ0,ν)≲μ(Qτ0)≲μ(B(zτ0,ν,2C6))≲μ(Qτ0,ν). From this and
1≤q≤∞ together with Hölder's inequality, it follows
that (6.91)mQτ0,ν(|D0(f)|)≤[1μ(Qτ0,ν)∫Qτ0,ν|D0(f)(x)|qdμ(x)]1/q≲[1μ(Qτ0)∫Qτ0|D0(f)(x)|qdμ(x)]1/q. Thus, (6.92)∥f∥F∞,qs(𝒳)≲supl∈ℤ+supα∈Il[1μ(Qαl)∫Qαl∑k=l∞2ksq|Dk(f)(x)|qdμ(x)]1/q.
To see the converse, it suffices to verify that for
allf∈F∞,qs(𝒳) and τ∈I0, (6.93)[1μ(Qτ0)∫Qτ0|D0(f)(x)|qdμ(x)]1/q≲∥f∥F∞,qs(𝒳). To see this, by the construction
of {Qτ0,ν:τ∈I0,ν=1,…,N(0,τ)},
we have(6.94)[1μ(Qτ0)∫Qτ0|D0(f)(x)|qdμ(x)]1/q≲{1μ(Qτ0)∑ν=1N(0,τ)μ(Qτ0,ν)[supx∈Qτ0,ν|D0(f)(x)|]q}1/q≲supν=1,…,N(0,τ)supx∈Qτ0,ν|D0(f)(x)|. Using Theorem 4.16 together with
some estimates similar to those for Z1+Z2 in the proof of Proposition 6.15, we can then
verify that (6.95)supν=1,…,N(0,τ)supx∈Qτ0,ν|D0(f)(x)|≲∥f∥F∞,qs(𝒳), which gives the desired estimate
and hence, we complete the proof of Proposition 6.22.Using Proposition6.22, we now complement Proposition
5.39(ii) for the case p=∞ as follows.Proposition 6.23.
Letϵ>0 be as in Definition 6.19, 0<s<ϵ, 1≤q≤∞ and μ(𝒳)=∞.
Then F∞,qs(𝒳)=F˙∞,qs(𝒳)∩L∞(𝒳),
and moreover, for any f∈F∞,qs(𝒳), (6.96)∥f∥F∞,qs(𝒳)~∥f∥F˙∞,qs(𝒳)+∥f∥L∞(𝒳).Proof.
Takef∈F∞,qs(𝒳).
By Proposition 6.18, without loss of generality, we may assume that f∈(𝒢(β,γ))′ with s<β,γ<ϵ as in (6.32). Thus, we also have f∈(𝒢°(β,γ))′ with s<β,γ<ϵ as in (6.32); see the proof of Proposition
5.37. By Theorem 3.29, we have that (6.97)f=D˜0S0(f)+∑k=1∞D˜kDk(f) holds in ∈(𝒢(β,γ))′ with s<β<ϵ and 0<γ<ϵ,
where D˜k with k∈ℤ+ is as in Theorem 3.26. From this and Lemma 2.19,
it follows that for any x∈𝒳, (6.98)|f(x)|≲∑τ∈I0∑ν=1N(0,τ)∫Qτ0,ν|D˜0(x,y)||S0(f)(y)|dμ(y)+∑k=1∞∑τ∈Ik∑ν=1N(k,τ)∫Qτk,ν|D˜k(x,y)||Dk(f)(y)|dμ(y)≲∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)1V1(x)+V1(yτ0,ν)+V(x,yτ0,ν)(11+d(x,yτ0,ν))ϵmQτ0,ν(|S0(f)|)+∑k=1∞∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)1V2−k(x)+V2−k(yτk,ν)+V(x,yτk,ν)×(2−k2−k+d(x,yτk,ν))ϵ[supy∈Qτk,ν|Dk(f)(y)|]. Since f∈F∞,qs(𝒳),
by its definition, we have that for τ∈I0 and ν=1,…,N(0,τ), (6.99)mQτ0,ν(|S0(f)|)≲∥f∥F∞,qs(𝒳), and the definition of
F∞,qs(𝒳) together with Proposition 6.15 also implies
that for any k∈ℕ, τ∈Ik,
and ν=1,…,N(k,τ), (6.100){∑τ∈Ik∑ν=1N(k,τ)2ksq[supy∈Qτk,ν|Dk(f)(y)|]q}1/q≲∥f∥F∞,qs(𝒳). Notice that 1≤q≤∞.
Both estimates via Hölder's inequality, the fact that μ(Qτk,ν)~V2−k(yτk,ν),
Lemma 5.2, and the assumption that s>0 further yield that (6.101)|f(x)|≲∥f∥F∞,qs(𝒳){1+∑k=1∞2−ks(∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)(2−k2−k+d(x,yτk,ν))ϵq′×1V2−k(x)+V2−k(yτk,ν)+V(x,yτk,ν))1/q′}≲∥f∥F∞,qs(𝒳), namely, f∈L∞(𝒳) and (6.102)∥f∥L∞(𝒳)≲∥f∥F∞,qs(𝒳).
Moreover,(6.103)∥f∥F˙∞,qs(𝒳)≲supl∈ℕsupα∈Il[1μ(Qαl)∫Qαl∑k=l∞2ksq|Dk(f)(x)|qdμ(x)]1/q+supl∈ℤ∖ℕsupα∈Il[1μ(Qαl)∫Qαl∑k=l∞2ksq|Dk(f)(x)|qdμ(x)]1/q≲∥f∥F∞,qs(𝒳)+supl∈ℤ∖ℕsupα∈Il[1μ(Qαl)∫Qαl∑k=l02ksq|Dk(f)(x)|qdμ(x)]1/q+supl∈ℤ∖ℕsupα∈Il[1μ(Qαl)∫Qαl∑k=1∞2ksq|Dk(f)(x)|qdμ(x)]1/q. To estimate the second term, by
Proposition 2.7(i), we have (6.104)|Dk(f)(x)|≲∥f∥L∞(𝒳)≲∥f∥F∞,qs(𝒳), which together with s>0 shows (6.105)supl∈ℤ∖ℕsupα∈Il[1μ(Qαl)∫Qαl∑k=l02ksq|Dk(f)(x)|qdμ(x)]1/q≲∥f∥F∞,qs(𝒳)supl∈ℤ∖ℕ[∑k=l02ksq]1/q≲∥f∥F∞,qs(𝒳). To estimate the third term, for
any l∈ℤ∖ℕ and α∈Il,
set (6.106)Ilα={β∈I1:Qβ1⊂Qαl}. Lemma 2.19 proves that (6.107)∑β∈Ilαμ(Qβ1)=μ(Qαl), which together with Lemma 2.19
further yields that (6.108)supl∈ℤ∖ℕsupα∈Il[1μ(Qαl)∫Qαl∑k=1∞2ksq|Dk(f)(x)|qdμ(x)]1/q≲supl∈ℤ∖ℕsupα∈Il[1μ(Qαl)∑β∈Ilαμ(Qβ1)1μ(Qβ1)∫Qβ1∑k=1∞2ksq|Dk(f)(x)|qdμ(x)]1/q≲∥f∥F∞,qs(𝒳). Thus, f∈F˙∞,qs(𝒳)∩L∞(𝒳) and(6.109)∥f∥L∞(𝒳)+∥f∥F˙∞,qs(𝒳)≲∥f∥F∞,qs(𝒳).
Conversely, letf∈F˙∞,qs(𝒳)∩L∞(𝒳).
Obviously, f∈(𝒢(β,γ))′ with s<β<ϵ and 0<γ<ϵ.
On the other hand, Proposition 2.7(i) shows that for any x∈𝒳, (6.110)|S0(f)(x)|≲∥f∥L∞(𝒳). Since 1≤q≤∞,
by Proposition 6.22, (5.5) together with (6.110), and Lemma 2.19 together with
(6.107), we obtain (6.111)∥f∥F∞,qs(𝒳)~supl∈ℕsupα∈Il[1μ(Qαl)∫Qαl∑k=l∞2ksq|Dk(f)(x)|qdμ(x)]1/q+supα∈I0[1μ(Qα0)∫Qα0(|S0(f)(x)|q+∑k=1∞2ksq|Dk(f)(x)|q)dμ(x)]1/q≲∥f∥F˙∞,qs(𝒳)+supα∈I0{1μ(Qα0)∑β∈I0αμ(Qβ0)[1μ(Qβ0)∫Qβ0∑k=1∞2ksq|Dk(f)(x)|qdμ(x)]}1/q≲∥f∥F˙∞,qs(𝒳), which means that f∈F∞,qs(𝒳).
Thus, (6.112)F˙∞,qs(𝒳)∩L∞(𝒳)⊂F∞,qs(𝒳),which completes the proof of
Proposition 6.23.From Proposition6.23 and Theorem 6.11(i), we easily
deduce the following results.Corollary 6.24.
Letϵ>0 be as in Definition 6.19. Then,
(i)
for anys∈(0,1], Cs(𝒳)=C˙s(𝒳)∩L∞(𝒳),
and moreover, for all f∈Cs(𝒳),(6.113)∥f∥Cs(𝒳)=∥f∥L∞(𝒳)+∥f∥C˙s(𝒳);(ii)
for anys∈(0,ϵ), Cs(𝒳)=B∞,∞s(𝒳)=F∞,∞s(𝒳) with equivalent norms.Proof.
Property (i) is a simple consequence of the
definitions of bothCs(𝒳) and C˙s(𝒳).
Whenμ(𝒳)=∞,
then Property (ii) can be deduced from Property (i), Theorem 6.11(i), and
Proposition 6.23, which completes the proof of Corollary 6.24 in this case.
An alternative way to prove Property (ii), which works
whenμ(𝒳)=∞ and also when μ(𝒳)<∞,
follows the line of reasoning in the proof of Theorem 6.11(i). In fact, let f∈Cs(𝒳) and {Dk}k∈ℤ+ be as in Definition 6.19. Then from f∈L∞(𝒳),
it follows that f∈(𝒢(β,γ))′ with s<β<ϵ and 0<γ<ϵ.
Moreover, by Proposition 2.7(i), we have that for all x∈𝒳, (6.114)|D0(f)(x)|=|∫𝒳D0(x,y)f(y)dμ(y)|≲∥f∥L∞(𝒳), which together with (6.45) shows
that f∈F∞,∞s(𝒳) and (6.115)∥f∥F∞,∞s(𝒳)≲∥f∥L∞(𝒳).
Conversely, supposef∈F∞,∞s(𝒳).
By Proposition 6.18, we can assume that f∈(𝒢(β,γ))′ with s<β<ϵ and 0<γ<ϵ.
Using the same notation as in Theorem 3.26, by Theorem 3.29 and the definition of ∥⋅∥F∞,∞s(𝒳) together with s>0,
we further obtain that for all x∈𝒳, (6.116)|f(x)|=|∑k=0∞D˜kDk(f)(x)|≲∥f∥F∞,∞s(𝒳)∑k=0∞2−ks≲∥f∥F∞,∞s(𝒳). Thus, f∈L∞(𝒳) and (6.117)∥f∥L∞(𝒳)≲∥f∥F∞,∞s(𝒳). Moreover, if d(x,y)≥1/2,
then (6.118)|f(x)−f(y)|≲∥f∥L∞(𝒳)≲∥f∥F∞,∞s(𝒳)d(x,y)s. Assume that 2−l0−1≤d(x,y)<2−l0 with l0∈ℕ.
By the regularity of D˜k,
Lemma 2.1(ii) and Proposition 2.7(i) together with 0<s<ϵ<ϵ1,
we then obtain (6.119)|f(x)−f(y)|≲|∑k=0l0−1∫𝒳[D˜k(x,z)−D˜k(y,z)]Dk(f)(z)dμ(z)|+∑k=l0∞{∫𝒳|D˜k(x,z)||Dk(f)(z)|dμ(z)+∫𝒳|D˜k(y,z)||Dk(f)(z)|dμ(z)}≲∑k=0l0−1∫𝒳(d(x,y)2−k+d(x,z))ϵ11V2−k(x)+V2−k(y)+V(x,y)×(2−k2−k+d(x,z))ϵ2|Dk(f)(z)|dμ(z)+∥f∥F∞,∞s(𝒳)∑k=l0∞2−ks≲∥f∥F∞,∞s(𝒳)d(x,y)s. Thus, f∈Cs(𝒳) and (6.120)∥f∥Cs(𝒳)≲∥f∥F∞,∞s(𝒳),which completes the proof of
Corollary 6.24(ii).We now establish the connection betweenF˙∞,qs(𝒳) and F∞,qs(𝒳) for all admissible s and q.Proposition 6.25.
Letϵ, s, q, and S0 be as in Definition 6.19 and let μ(𝒳)=∞.
Then there exists a constant C>0 such that for all f∈F˙∞,qs(𝒳), f−S0(f)∈F∞,qs(𝒳) and (6.121)∥f−S0(f)∥F∞,qs(𝒳)≤C∥f∥F˙∞,qs(𝒳).Proof.
Letf∈F˙∞,qs(𝒳).
By Proposition 6.6, we may assume that f∈(𝒢°0ϵ(β,γ))′ with β,γ as in (6.32). On the other hand, for any g∈𝒢0ϵ(β,γ) with β,γ as in (6.32), from Lemma 5.36 and ∫𝒳S0(x,y)dμ(x)=1,
it is easy to deduce that g−S0t(g)∈𝒢°(β,γ) with β,γ as in (6.32). Thus, from (6.122)〈f−S0(f),g〉=〈f,g−S0t(g)〉, we deduce that f−S0(f)∈(𝒢0ϵ(β,γ))′ with β,γ as in (6.32).
We now verify (6.121). In what follows, let I be the identity operator, let {Dk′}k′∈ℤ be as in Definition 6.7, and let {D˜k′}k′∈ℤ be as in Theorem 4.11. We first claim that for
all k′∈ℤ and x,y∈𝒳, (6.123)|(S0(I−S0)D˜k′)(x,y)|≲2−|k′|ϵ1V2−(0⋀k′)(x)+V2−(0⋀k′)(y)+V(x,y)(2−(0⋀k′)2−(0⋀k′)+d(x,y))ϵ and that for all k∈ℕ, k′∈ℤ,
and x,y∈ℤ, (6.124)|(Dk(I−S0)D˜k′)(x,y)|≲2−|k−k′|ϵ1V2−(k⋀k′)(x)+V2−(k⋀k′)(y)+V(x,y)(2−(k⋀k′)2−(k⋀k′)+d(x,y))ϵ. The estimate (6.124) is
essentially the same as the estimate (5.130) by symmetry, while the estimate
(6.123) can essentially be obtained by an argument similar to Cases 1 and 2
of the proof of the estimate (5.130).
Using the estimate (6.123) and Theorem 4.13, we have
that for any x∈𝒳, (6.125)|S0(I−S0)(f)(x)|=|∑k′∈ℤ∑τ′∈Ik′∑ν′=1N(k′,τ′)μ(Qτ′k′,ν′)(S0(I−S0)D˜k′)(x,yτ′k′,ν′)Dk′(f)(yτ′k′,ν′)|≲∑k′∈ℤ∑τ′∈Ik′∑ν′=1N(k′,τ′)2−|k′|ϵμ(Qτ′k′,ν′)1V2−(0⋀k′)(x)+V2−(0⋀k′)(yτ′k′,ν′)+V(x,yτ′k′,ν′)×(2−(0⋀k′)2−(0⋀k′)+d(x,yτ′k′,ν′))ϵ|Dk′(f)(yτ′k′,ν′)|. From Proposition 6.3, we deduce
the following trivial estimate that for all k′∈ℤ, (6.126){∑τ′∈Ik′∑ν′=1N(k′,τ′)2k′sq|Dk′(f)(yτ′k′,ν′)|q}1/q≲∥f∥F˙∞,qs(𝒳), and from Lemma 2.19, it follows
that (6.127)μ(Qτ′k′,ν′)V2−(0⋀k′)(x)+V2−(0⋀k′)(yτ′k′,ν′)+V(x,yτ′k′,ν′)≲1. Both estimates together with
(5.5) show that when q≤1,
for any x∈𝒳, (6.128)|S0(I−S0)(f)(x)|≲∑k′∈ℤ2−|k′|ϵ2−k′s{∑τ′∈Ik′∑ν′=1N(k′,τ′)2k′sq|Dk′(f)(yτ′k′,ν′)|q}1/q≲∥f∥F˙∞,qs(𝒳), while when 1<q≤∞,
both estimates together with Hölder's inequality, Lemma 5.2, and Lemma 2.1(ii)
still yield that for all x∈𝒳, (6.129)|S0(I−S0)(f)(x)|≲∑k′∈ℤ2−|k′|ϵ2−k′s[∑τ′∈Ik′∑ν′=1N(k′,τ′)μ(Qτ′k′,ν′)1V2−(0⋀k′)(x)+V2−(0⋀k′)(yτ′k′,ν′)+V(x,yτ′k′,ν′)×(2−(0⋀k′)2−(0⋀k′)+d(x,yτ′k′,ν′))ϵ]1/q′[∑τ′∈Ik′∑ν′=1N(k′,τ′)2k′sq|Dk′(f)(yτ′k′,ν′)|q]1/q≲∥f∥F˙∞,qs(𝒳). Thus, for τ∈I0 and ν=1,…,N(0,τ), (6.130)mQτ0,ν(|S0(I−S0)(f)|)=1μ(Qτ0,ν)∫Qτ0,ν|S0(I−S0)(f)(x)|dμ(x)≲∥f∥F˙∞,qs(𝒳),which is the desired estimate.
Forl∈ℕ and α∈Il,
from Theorem 5.16 together with (6.124), it follows that  (6.131)1μ(Qαl)∫Qαl∑k=l∞2ksq|Dk(I−S0)(f)(x)|qdμ(x)≲1μ(Qαl)∫Qαl∑k=l∞2ksq[∑k′∈ℤ∑τ′∈Ik′∑ν′=1N(k′,τ′)2−|k−k′|ϵμ(Qτ′k′,ν′)1V2−(k⋀k′)(x)+V2−(k⋀k′)(yτ′k′,ν′)+V(x,yτ′k′,ν′)×(2−(k⋀k′)2−(k⋀k′)+d(x,yτ′k′,ν′))ϵ|Dk′(f)(yτ′k′,ν′)|]qdμ(x). Then, an argument similar to
that used to estimate Y1+Y2 in the proof of Proposition 6.3 together with
Proposition 6.3 yields that (6.132)1μ(Qαl)∫Qαl∑k=l∞2ksq|Dk(I−S0)(f)(x)|qdμ(x)≲∥f∥F˙∞,qs(𝒳)q. Thus, (I−S0)(f)∈F∞,qs(𝒳) and (6.133)∥(I−S0)(f)∥F∞,qs(𝒳)≲∥f∥F˙∞,qs(𝒳), which completes the proof of
Proposition 6.25.Remark 6.26.
We point out that Remark5.38 applies in a similar way to Proposition
6.25.To establish the equivalence betweenbmo(𝒳) and
F∞,20(𝒳),
we need the following technical result which in fact gives a new
characterization of Triebel-Lizorkin space F∞,qs(𝒳) when 1≤q≤∞.
We first introduce an inhomogeneous maximal function.For anyx∈𝒳 and l∈ℤ+,
we choose m∈ℕ by a way similar to that in the definition of ℭ˙qs.
Let {Dk}k∈ℤ+
be as in Definition 6.19. Then for any s∈ℝ, q∈(0,∞], f∈(𝒢(β,γ))′ with 0<β,γ<ϵ and x∈𝒳,
we define the inhomogeneous maximal
functionℭqs(f)(x) by (6.134)ℭqs(f)(x)=supl∈ℤ+[1μ(∪i=1mQαlil)∫∪i=1mQαlil∑k=l∞2ksq|Dk(f)(y)|qdμ(y)]1/q, where the usual modification is
made when q=∞.Using Proposition6.22 and an argument similar to the
proof of Proposition 6.10 yields the following characterization of F∞,qs(𝒳) with 1≤q≤∞;
we omit the details.Proposition 6.27.
Letϵ and s be as in Definition 6.7 and let 1≤q≤∞.
Then f∈F∞,qs(𝒳) if and only if f∈(𝒢(β,γ))′ with |s|<β<ϵ and 0<γ<ϵ,
and ℭqs(f)∈L∞(𝒳).
Moreover, in this case, (6.135)∥f∥F∞,qs(𝒳)~∥ℭqs(f)∥L∞(𝒳).Now, from Theorem5.44(i), Proposition 6.22, and
Proposition 6.27, we can deduce the following relation between bmo(𝒳) and F∞,20(𝒳).Theorem 6.28.
bmo(𝒳)=F∞,20(𝒳),
with equivalent norms.Proof.
Letf∈bmo(𝒳) and {Dk}k∈ℤ+ be as in Definition 6.19. Proposition 6.21(v)
and Theorem 5.44(i) immediately imply that f∈(𝒢0ϵ(β,γ))′ with 0<β,γ<ϵ.
Let now Qαl for l∈ℤ+ and α∈Il be a dyadic cube as in Lemma 2.19. Let Bαl be as in the proof of Theorem 6.11(ii). We
then decompose f=f1+f2+f3 in the same way as in the proof of Theorem 6.11(ii). The estimations for f1 and f2 are as in the proof of Theorem 6.11(ii) by
replacing Proposition 3.15 by Proposition 3.30. If l∈ℕ,
then (6.136){1μ(Qαl)∫Qαl∑k=l∞|Dk(f3)(x)|dμ(x)}1/2=0, while when l=0,
we then have (6.137){1μ(Qαl)∫Qαl∑k=l∞|Dk(f3)(x)|dμ(x)}1/2=|f3|≤∥f∥bmo(𝒳). Thus, f∈F∞,20(𝒳) and
∥f∥F∞,20(𝒳)≲∥f∥bmo(𝒳).
Conversely, using  Theorem5.44(i), Proposition 6.22,
and Proposition 6.27 together with an argument similar to the proof of Theorem
6.11(ii), we can prove that if f∈F∞,20(𝒳),
then f∈bmo(𝒳) and ∥f∥bmo(𝒳)≲∥f∥F∞,20(𝒳), which completes the proof of Theorem 6.28.We end this subsection by considering the boundedness onF∞,qs(𝒳) of singular integrals of order (ϵ,σ).Theorem 6.29.
Letϵ, s, and q be as in Definition 6.19. Let σ>0 and let T be a singular integral of order (ϵ,σ).
Then T is bounded from 𝒢b(β,γ) with 0<β≤ϵ and γ>0 to F∞,qs(𝒳).
Moreover, there exists a constant C>0 such that for all f∈𝒢b(β,γ) with 0<β≤ϵ and γ>0, (6.138)∥Tf∥F∞,qs(𝒳)≤C∥f∥F∞,qs(𝒳).Proof.
Combining some estimates and technics used in the proof of Proposition5.54 with those used in the proof of Proposition 6.15 gives the desired
conclusions. The details are left to the reader.Remark 6.30.
By Theorem8.18 below, if p(s,ϵ)<q≤∞,
then F∞,qs(𝒳) is the dual space of F1,q′−s(𝒳).
In this case, Theorem 6.29 can be deduced from Theorem 5.48 together with a
duality argument, which provides another proof of Theorem 6.29.


## 6.1. Plancherel-Pôlya Inequality and Definition ofF˙∞,qs(𝒳)

Throughout this and the next subsection, we will assume
thatμ(𝒳)=∞.
In this subsection, we introduce the norm in
F˙∞,qs(𝒳) in a similar way as in [82] and, using the homogeneous
discrete Calderón reproducing formulae, Theorem 4.13, we will prove that the
norm
∥⋅∥F˙∞,qs(𝒳) is independent of the choices of ATIs and spaces of distributions via some
Plancherel-Pôlya inequality; see also [83, 84].Definition 6.1.
Letϵ1∈(0,1], ϵ2>0, ϵ3>0, ϵ∈(0,ϵ1⋀ϵ2), and {Sk}k∈ℤ be an (ϵ1,ϵ2,ϵ3)-ATI.
For k∈ℤ,
set Dk=Sk−Sk−1.
Let |s|<ϵ and p(s,ϵ)<q≤∞.
For any f∈(𝒢°0ϵ(β,γ))′ with 0<β,γ<ϵ,
define (6.1)∥f∥F˙∞,qs(𝒳)=supl∈ℤsupα∈Il{1μ(Qαl)∫Qαl∑k=l∞2ksq|Dk(f)(x)|qdμ(x)}1/q, where the supremum is
taken over all dyadic cubes as in Lemma 2.19 and the usual modification is made
when q=∞.Remark 6.2.
(i) From Lemma2.19 and the
doubling property (1.2), it is easy to see that an equivalent norm is obtained
if the supremum in Definition 6.1 is taken with respect to all balls with
positive radius instead of all dyadic cubes as in Lemma 2.19.
(ii) Letl0∈ℤ.
It is easy to see that there exists a constant Cl0>0 such that for all f∈(𝒢°0ϵ(β,γ))′, (6.2)supl∈ℤsupα∈Il{1μ(Qαl)∫Qαl∑k=l+l0∞2ksq|Dk(f)(x)|qdμ(x)}1/q≲Cl0∥f∥F˙∞,qs(𝒳). Thus, (6.3)∥f∥F˙∞,qs(𝒳)~supl∈ℤsupα∈Il{1μ(Qαl)∫Qαl∑k=l+l0∞2ksq|Dk(f)(x)|qdμ(x)}1/q.We now establish the following usefulPlancherel-Pôlya inequality, which
complements Proposition 5.4(ii) for the case p=∞.Proposition 6.3.
Letϵ1∈(0,1], ϵ2>0, ϵ3>0, ϵ∈(0,ϵ1⋀ϵ2), and let {Sk}k∈ℤ and {Pk}k∈ℤ be two (ϵ1,ϵ2,ϵ3)-ATIs. For k∈ℤ,
set Dk=Sk−Sk−1 and Qk=Pk−Pk−1.
Let |s|<ϵ and p(s,ϵ)<q≤∞.
Then for all f∈(𝒢°0ϵ(β,γ))′ with 0<β,γ<ϵ, (6.4)supl∈ℤsupα∈Il{1μ(Qαl)∑k=l∞∑τ∈Ik∑ν=1N(k,τ)2ksqμ(Qτk,ν)χ{(τ,ν):Qτk,ν⊂Qαl}(τ,ν)[supx∈Qτk,ν|Dk(f)(x)|]q}1/q~supl∈ℤsupα∈Il{1μ(Qαl)∑k=l∞∑τ∈Ik∑ν=1N(k,τ)2ksqμ(Qτk,ν)χ{(τ,ν):Qτk,ν⊂Qαl}(τ,ν)[infx∈Qτk,ν|Qk(f)(x)|]q}1/q.Proof.
To prove Proposition6.3, it suffices to show that for all f∈(𝒢°0ϵ(β,γ))′ with 0<β,γ<ϵ,
the left-hand side of (6.4) is controlled by its right-hand side.
Let all the notation be as in the proof of Proposition5.4. Then, by (5.9) and (5.10), we have1μ(Qαl)∑k=l∞∑τ∈Ik∑ν=1N(k,τ)2ksqμ(Qτk,ν)χ{(τ,ν):Qτk,ν⊂Qαl}(τ,ν)[supx∈Qτk,ν|Dk(f)(x)|]q≲1μ(Qαl)∑k=l∞∑τ∈Ik∑ν=1N(k,τ)2ksqμ(Qτk,ν)χ{(τ,ν):Qτk,ν⊂Qαl}(τ,ν)×[∑k′=l∞∑τ′∈Ik′∑ν′=1N(k′,τ′)2−|k−k′|ϵ′μ(Qτ′k′,ν′)|Qk′(f)(yτ′k′,ν′)|×1V2−(k⋀k′)(yτk,ν)+V2−(k⋀k′)(yτ′k′,ν′)+V(yτk,ν,yτ′k′,ν′)(2−(k⋀k′)2−(k⋀k′)+d(yτk,ν,yτ′k′,ν′))ϵ]q(6.5)+1μ(Qαl)∑k=l∞∑τ∈Ik∑ν=1N(k,τ)2ksqμ(Qτk,ν)χ{(τ,ν):Qτk,ν⊂Qαl}(τ,ν)×[∑k′=−∞l−1∑τ′∈Ik′∑ν′=1N(k′,τ′)2−|k−k′|ϵ′μ(Qτ′k′,ν′)|Qk′(f)(yτ′k′,ν′)|⋯]q≡Y1+Y2.
Then by Lemma 2.19, if x∉B(zαl,4C62−l) and y∈Qαl,
then d(x,y)≥3C62−l,
where and in what follows, zαl is the “center" of Qαl as in Lemma 2.19. By Lemma 2.19 again, we can
find m1∈ℕ such that B(zαl,4C62−l)⊂⋃i=1m1Qτil, τi∈Il, B(zαl,4C62−l)∩Qτil≠∅ and m1 is no more than a constant which is
independent of α and l;
see the details for a proof of the last fact in [85, pages 1385-1386].
Moreover, (6.6)μ(Qτil)~μ(Qαl). With these choices, we further
control Y1 by (6.7)Y1≲1μ(Qαl)∑k=l∞∑τ∈Ik∑ν=1N(k,τ)2ksqμ(Qτk,ν)χ{(τ,ν):Qτk,ν⊂Qαl}(τ,ν)×[∑k′=l∞∑τ′∈Ik′∑ν′=1N(k′,τ′)2−|k−k′|ϵ′μ(Qτ′k′,ν′)χ{(τ′,ν′):Qτ′k′,ν′⊂∪i=1m1Qτil}(τ′,ν′)×|Qk′(f)(yτ′k′,ν′)|1V2−(k⋀k′)(yτk,ν)+V2−(k⋀k′)(yτ′k′,ν′)+V(yτk,ν,yτ′k′,ν′)×(2−(k⋀k′)2−(k⋀k′)+d(yτk,ν,yτ′k′,ν′))ϵ]q+1μ(Qαl)∑k=l∞∑τ∈Ik∑ν=1N(k,τ)2ksqμ(Qτk,ν)χ{(τ,ν):Qτk,ν⊂Qαl}(τ,ν)×[∑k′=l∞∑τ′∈Ik′∑ν′=1N(k′,τ′)2−|k−k′|ϵ′μ(Qτ′k′,ν′)χ{(τ′,ν′):Qτ′k′,ν′∩(∪i=1m1Qτil)=∅}(τ′,ν′)×|Qk′(f)(yτ′k′,ν′)|1V2−(k⋀k′)(yτk,ν)+V2−(k⋀k′)(yτ′k′,ν′)+V(yτk,ν,yτ′k′,ν′)×(2−(k⋀k′)2−(k⋀k′)+d(yτk,ν,yτ′k′,ν′))ϵ]q≡Y1,1+Y1,2. We first estimate Y1,1.
If q≤1,
by (5.5), Lemma 5.2, (5.12), (6.6), and choosing ϵ′∈(0,ϵ) such that ϵ′>s and q>p(s,ϵ′), (6.8)Y1,1≲1μ(Qαl)∑i=1m1∑k′=l∞∑τ′∈Ik′∑ν′=1N(k′,τ′)∑k=l∞2ksqχ{(τ′,ν′):Qτ′k′,ν′⊂Qτil}(τ′,ν′)×[μ(Qτ′k′,ν′)|Qk′(f)(yτ′k′,ν′)|]q2−|k−k′|ϵ′q×∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)χ{(τ,ν):Qτk,ν⊂Qαl}(τ,ν)×[1V2−(k⋀k′)(yτk,ν)+V2−(k⋀k′)(yτ′k′,ν′)+V(yτk,ν,yτ′k′,ν′)(2−(k⋀k′)2−(k⋀k′)+d(yτk,ν,yτ′k′,ν′))ϵ]q≲∑i=1m11μ(Qτil)∑k′=l∞∑τ′∈Ik′∑ν′=1N(k′,τ′)2k′sqμ(Qτ′k′,ν′)|Qk′(f)(yτ′k′,ν′)|qχ{(τ′,ν′):Qτ′k′,ν′⊂Qτil}(τ′,ν′), which together with the
arbitrary choice of yτ′k′,ν′∈Qτ′k′,ν′ shows that (Y1,1)1/q is controlled by the right-hand side of (6.4)
in this case.
If1<q≤∞,
choosing ϵ′>|s| together with Hölder's inequality and Lemma
5.2 yields (6.9)∑k′=l∞∑τ′∈Ik′∑ν′=1N(k′,τ′)2k′s2−|k−k′|ϵ′2(k−k′)sμ(Qτ′k′,ν′)χ{(τ′,ν′):Qτ′k′,ν′⊂Qτil}(τ′,ν′)×|Qk′(f)(yτ′k′,ν′)|1V2−(k⋀k′)(yτk,ν)+V2−(k⋀k′)(yτ′k′,ν′)+V(yτk,ν,yτ′k′,ν′)×(2−(k⋀k′)2−(k⋀k′)+d(yτk,ν,yτ′k′,ν′))ϵ≲{∑k′=l∞∑τ′∈Ik′∑ν′=1N(k′,τ′)2k′sq2−|k−k′|ϵ′2(k−k′)sμ(Qτ′k′,ν′)χ{(τ′,ν′):Qτ′k′,ν′⊂Qτil}(τ′,ν′)×|Qk′(f)(yτ′k′,ν′)|q1V2−(k⋀k′)(yτk,ν)+V2−(k⋀k′)(yτ′k′,ν′)+V(yτk,ν,yτ′k′,ν′)×(2−(k⋀k′)2−(k⋀k′)+d(yτk,ν,yτ′k′,ν′))ϵ}1/q, which together with Minkowski's
inequality, (6.6), and Lemma 5.2 yields that (6.10)(Y1,1)1/q≲∑i=1m1{1μ(Qτil)∑k′=l∞∑τ′∈Ik′∑ν′=1N(k′,τ′)2k′sqμ(Qτ′k′,ν′)χ{(τ′,ν′):Qτ′k′,ν′⊂Qτil}(τ′,ν′)|Qk′(f)(yτ′k′,ν′)|q}1/q, which completes the estimate for Y1,1.
We now estimateY1,2.
To this end, for j∈ℤ+,
let (6.11)Ilj={α′∈Il:α′≠τ1,…,τm1,3C62j−l≤d(zα′l,zαl)<3C62j−l+1}. We first claim that there exists m2∈ℕ which is no more than a constant independent
of l and j such that (6.12)⋃α′∈IljQα′l⊂⋃i=1,…,m2αi∈Il−jQαil−j,Qαil−j∩(⋃α′∈IljQα′l)≠∅ for i=1,…,m2,
and moreover, if Qτ′k′,ν′⊂∪α′∈IljQα′l,
then (6.13)μ(⋃i=1,…,m2αi∈Il−jQαil−j)≲V2j−l(yτ′k′,ν′). In fact, by Lemma 2.19, there exists m2∈ℕ such that (6.12) holds. Notice that for any
fixed i0∈{1,…,m2}, ∪α′∈IljQα′l⊂B(zαi0j−l,14C62j−l). By an argument in [85, pages 1385-1386], we know
that the number of
α∈Il−j such that Qαl−j∩B(zαi0j−l,14C62j−l)≠∅ is no more than a constant which is
independent of j and l.
Thus, the claim (6.12) holds.
To see the claim (6.13), we only need to notice that (6.14)⋃i=1,…,m2αi∈Il−jQαil−j⊂B(yτ′k′,ν′,9C62j−l), which implies the claim (6.13).
We also notice that ifQτ′k′,ν′⊂Qα′l with α′∈Ili,
then for all y∈Qαl, (6.15)d(y,yτ′k′,ν′)≥C62j−l.
Using these properties, we now estimateY1,2 by first considering the case q≤1.
In fact, in this case, from (5.5), (6.15), (6.12), and (6.13), it follows
that (6.16)Y1,2≲∑j=0∞2−j[ϵq−n(1−q)]∑i=1m21μ(Qαil−j)∑k′=l∞∑τ′∈Ik′∑ν′=1N(k′,τ′)2k′sqμ(Qτ′k′,ν′)×χ{(τ′,ν′):Qτ′k′,ν′⊂Qαil−j}(τ′,ν′)|Qk′(f)(yτ′k′,ν′)|q×(∑k=l∞2−|k−k′|ϵ′q2(k−k′)sq2−(k⋀k′)ϵq2k′n(1−q)2−ln(1−q)+lϵq), which together with q>n/(n+ϵ) and choosing ϵ′>s further implies that (6.17)Y1,2≲∑j=0∞2−j[ϵq−n(1−q)]∑i=1m21μ(Qαil−j)∑k′=l∞∑τ′∈Ik′∑ν′=1N(k′,τ′)2k′sqμ(Qτ′k′,ν′)×χ{(τ′,ν′):Qτ′k′,ν′⊂Qαil−j}(τ′,ν′)|Qk′(f)(yτ′k′,ν′)|q. From this, it is easy to deduce
the desired estimate for Y1,2 in this case.
If1<q≤∞,
by Hölder's inequality and Lemma 5.2, we have (6.18)∑k′=l∞∑τ′∈Ik′∑ν′=1N(k′,τ′)2k′s2−|k−k′|ϵ′2(k−k′)sμ(Qτ′k′,ν′)χ{(τ′,ν′):Qτ′k′,ν′∩(∪i=1m1Qτil)=∅}(τ′,ν′)×|Qk′(f)(yτ′k′,ν′)|1V2−(k⋀k′)(yτk,ν)+V2−(k⋀k′)(yτ′k′,ν′)+V(yτk,ν,yτ′k′,ν′)×(2−(k⋀k′)2−(k⋀k′)+d(yτk,ν,yτ′k′,ν′))ϵ≲{∑k′=l∞∑τ′∈Ik′∑ν′=1N(k′,τ′)2k′sq2−|k−k′|ϵ′2(k−k′)sμ(Qτ′k′,ν′)|Qk′(f)(yτ′k′,ν′)|q×1V2−(k⋀k′)(yτk,ν)+V2−(k⋀k′)(yτ′k′,ν′)+V(yτk,ν,yτ′k′,ν′)×χ{(τ′,ν′):Qτ′k′,ν′∩(∪i=1m1Qτil)=∅}(τ′,ν′)(2−(k⋀k′)2−(k⋀k′)+d(yτk,ν,yτ′k′,ν′))ϵ}1/q, which together with (6.15),
(6.12), and (6.13) yields that (6.19)Y1,2≲∑j=0∞2−jϵ∑i=1m21μ(Qαil−j)∑k′=l∞∑τ′∈Ik′∑ν′=1N(k′,τ′)2k′sqμ(Qτ′k′,ν′)×χ{(τ′,ν′):Qτ′k′,ν′⊂Qαil−j}(τ′,ν′)|Qk′(f)(yτ′k′,ν′)|q, where we choose ϵ′>s.
From this, we deduce the desired estimate for Y1,2 when 1<q≤∞,
which completes the estimate for Y1.
We now estimateY2 by using the following trivial estimate that (6.20){∑τ′∈Ik′∑ν′=1N(k′,τ′)2k′sq|Qk′(f)(yτ′k′,ν′)|q}1/q≲RHSof(6.4), where and in the sequel, RHS
stands for “right-hand side." From this, it follows that if q≤1,
by choosing ϵ′>s,
we then have (6.21)Y2≲1μ(Qαl)∑k=l∞∑τ∈Ik∑ν=1N(k,τ)2ksqμ(Qτk,ν)χ{(τ,ν):Qτk,ν⊂Qαl}(τ,ν)×(∑k′=−∞l−1∑τ′∈Ik′∑ν′=1N(k′,τ′)2−(k−k′)ϵ′q|Qk′(f)(yτ′k′,ν′)|q)≲(RHSof(6.4))q; if 1<q≤∞,
by Hölder's inequality, Lemma 5.2, and (6.20), (6.22)Y2≲1μ(Qαl)∑k=l∞∑τ∈Ik∑ν=1N(k,τ)2ksqμ(Qτk,ν)χ{(τ,ν):Qτk,ν⊂Qαl}(τ,ν)×{∑k′=−∞l−1∑τ′∈Ik′∑ν′=1N(k′,τ′)2−(k−k′)ϵ′2(k−k′)sμ(Qτ′k′,ν′)|Qk′(f)(yτ′k′,ν′)|q×1V2−(k⋀k′)(yτk,ν)+V2−(k⋀k′)(yτ′k′,ν′)+V(yτk,ν,yτ′k′,ν′)×(2−(k⋀k′)2−(k⋀k′)+d(yτk,ν,yτ′k′,ν′))ϵ}≲(RHSof(6.4))q, which completes the proof of
Proposition 6.3.Remark 6.4.
We point out that Remark5.5 applies in a similar way to Proposition 6.3.From Proposition6.3, it is easy to deduce that the
definition of the norm ∥⋅∥F˙∞,qs(𝒳) is independent of the choice of ATIs. We omit the  details.Proposition 6.5.
Adopting the
notation from Proposition6.3, one has for all f∈(𝒢°0ϵ(β,γ))′ with 0<β,γ<ϵ, (6.23)supl∈ℤsupα∈Il{1μ(Qαl)∫Qαl∑k=l∞2ksq|Dk(f)(x)|qdμ(x)}1/q~supl∈ℤsupα∈Il{1μ(Qαl)∫Qαl∑k=l∞2ksq|Qk(f)(x)|qdμ(x)}1/q.The following theorem will show that the definition of
the norm∥⋅∥F˙∞,qs(𝒳) is independent of the choice of the space of
distributions.Proposition 6.6.
Let all the notation be as in Definition6.1. Let |s|<ϵ and p(s,ϵ)<q≤∞.
If f∈(𝒢°0ϵ(β1,γ1))′ with max{0,−s}<β1<ϵ and max{0,s}<γ1<ϵ,
and if ∥f∥F˙∞,qs(𝒳)<∞,
then f∈(𝒢°0ϵ(β2,γ2))′ for every max{0,−s}<β2<ϵ and
max{0,s}<γ2<ϵ.Proof.
We use the same notation as in the proof of Proposition5.7. Let ψ∈𝒢°(ϵ,ϵ).
For any f∈(𝒢°0ϵ(β1,γ1))′ with max{0,−s}<β1<ϵ and max{0,s}<γ1<ϵ,
when q≤1,
by Theorem 4.13 together with (5.24), (5.25), and (5.5),  we have  (6.24)|〈f,ψ〉|=|∑k=−∞∞∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)Dk(f)(yτk,ν)〈D˜k(⋅,yτk,ν),ψ〉|≲∥ψ∥𝒢(β2,γ2){∑k=0∞∑τ∈Ik∑ν=1N(k,τ)2−kβ2q[μ(Qτk,ν)|Dk(f)(yτk,ν)|]q×[1V1(x1)+V(x1,yτk,ν)1(1+d(x1,yτk,ν))γ2]q+∑k=−∞−1∑τ∈Ik∑ν=1N(k,τ)2kγ2′q[μ(Qτk,ν)|Dk(f)(yτk,ν)|]q×[1V2−k(x1)+V(x1,yτk,ν)2−kγ2(2−k+d(x1,yτk,ν))γ2]q}1/q.
Notice that fork∈ℤ+,
when d(x1,yτk,ν)<1,
then V1(x1)~V1(yτk,ν)≳2kκV2−k(yτk,ν)≳μ(Qτk,ν), while when 2l≤d(x1,yτk,ν)<2l+1 for some l∈ℤ+,
then V(x1,yτk,ν)≳V2l(yτk,ν)≳V2−k(yτk,ν)~μ(Qτk,ν). Therefore,  for k∈ℤ+, (6.25)μ(Qτk,ν)V1(x1)+V(x1,yτk,ν)≲1.
Whenk∈ℤ∖ℤ+,
noticing that if d(x1,yτk,ν)<2−k,
then (6.26)V2−k(x1)≳V2−k(yτk,ν)≳μ(Qτk,ν); if 2l2−k≤d(x1,yτk,ν)<2l+12−k for some l∈ℤ+,
then (6.27)V(x1,yτk,ν)≳V2l2−k(yτk,ν)≳2lκV2−k(yτk,ν)≳μ(Qτk,ν), we also have (6.28)μ(Qτk,ν)V2−k(x1)+V(x1,yτk,ν)≲1.
We also need the following trivial estimate that(6.29)∑τ∈Ik∑ν=1N(k,τ)2ksq|Dk(f)(yτk,ν)|q≲∥f∥F˙∞,qs(𝒳)q.
Using (6.25), (6.28), and (6.29) yields that (6.30)|〈f,ψ〉|≲∥ψ∥𝒢(β2,γ2)∥f∥F˙∞,qs(𝒳){∑k=0∞2−k(β2+s)q+∑k=−∞−12k(γ2′−s)q}1/q≲∥ψ∥𝒢(β2,γ2)∥f∥F˙∞,qs(𝒳), where we chose s<γ2′<γ2.
If1<q≤∞,
Hölder's inequality, (6.25), (6.28), (6.29), and Lemma 2.1(ii) prove that (6.31)|〈f,ψ〉|≲∥ψ∥𝒢(β2,γ2){∑k=0∞2−kβ22−ks[∑τ∈Ik∑ν=1N(k,τ)2ksqμ(Qτk,ν)|Dk(f)(yτk,ν)|q×1V1(x1)+V(x1,yτk,ν)1(1+d(x1,yτk,ν))γ2]1/q×[∫𝒳1V1(x1)+V(x1,y)1(1+d(x1,y))γ2dμ(y)]1/q′+∑k=−∞−12kγ2′2−ks[∑τ∈Ik∑ν=1N(k,τ)2ksqμ(Qτk,ν)|Dk(f)(yτk,ν)|q×1V2−k(x1)+V(x1,yτk,ν)2−kγ2(2−k+d(x1,yτk,ν))γ2]1/q×[1V2−k(x1)+V(x1,y)2−kγ2(2−k+d(x1,y))γ2dμ(y)]1/q′}≲∥ψ∥𝒢(β2,γ2)∥f∥F˙∞,qs(𝒳).
Combining (6.30) and (6.31) with an argument similar
to the proof of Proposition 5.7 then completes the proof of Proposition
6.6.Now we can introduce the homogeneous Triebel-Lizorkin
spacesF˙∞,qs(𝒳).Definition 6.7.
Letϵ1∈(0,1], ϵ2>0, ϵ3>0, ϵ∈(0,ϵ1⋀ϵ2) and let {Sk}k∈ℤ be an (ϵ1,ϵ2,ϵ3)-ATI.
For k∈ℤ,
set Dk=Sk−Sk−1.
Let |s|<ϵ and p(s,ϵ)<q≤∞.
The spaceF˙∞,qs(𝒳) is defined to be the set of all f∈(𝒢°0ϵ(β,γ))′,
for some β,γ satisfying (6.32)|s|<β<ϵ,max{s,0,−s−κ}<γ<ϵ such that (6.33)∥f∥F˙∞,qs(𝒳)=supl∈ℤsupα∈Il{1μ(Qαl)∫Qαl∑k=l∞2ksq|Dk(f)(x)|qdμ(x)}1/q<∞, where the supremum is taken over
all dyadic cubes as in Lemma 2.19 and the usual modification is made when q=∞.Propositions6.5 and 6.6 show that the definition of
the spaces F˙∞,qs(𝒳) is independent of the choice of the ATI and the distribution space (𝒢°(β,γ))′,
with β and γ satisfying (6.32).Remark 6.8.
To guarantee that the definition of the spaceF˙∞,qs(𝒳) is independent of the choice of the
distribution space (𝒢°(β,γ))′,
we only need the restriction that max{0,−s}<β<ϵ and max{0,s}<γ<ϵ.
Moreover, if we assume that max{0,s}<β<ϵ and max{0,s−κ}<γ<ϵ,
we can then verify that 𝒢°(β,γ)⊂F˙∞,qs(𝒳);
see Proposition 6.9 below.


## 6.2. Properties ofF˙∞,qs(𝒳) and Boundedness of Singular Integrals

In this subsection, we first present some basic
properties ofF˙∞,qs(𝒳).
By establishing a maximal function characterization of F˙∞,qs(𝒳),
we then establish some relations between the spaces F˙∞,qs(𝒳) and the spaces C˙s(𝒳) and between the spaces F˙∞,qs(𝒳) and the space BMO(𝒳).
Finally, we obtain the boundedness on F˙∞,qs(𝒳) of the singular integrals considered by Nagel
and Stein in [44].Proposition 6.9.
Letϵ be as in Definition 6.7,  |s|<ϵ, and p(s,ϵ)<q≤∞. Then,
(i)
F˙∞,q0s(𝒳)⊂F˙∞,q1s(𝒳) if p(s,ϵ)<q0≤q1≤∞;(ii)
B˙∞,qs(𝒳)⊂F˙∞,qs(𝒳)⊂B˙∞,∞s(𝒳);(iii)
ifmax{0,−s}<β<ϵ and max{0,s}<γ<ϵ,
then F˙∞,qs(𝒳)⊂(𝒢°0ϵ(β,γ))′;(iv)
ifmax{s,0}<β<ϵ and max{0,−s−κ}<γ<ϵ,
then 𝒢°(β,γ)⊂F˙∞,qs(𝒳);(v)
the spacesF˙∞,qs(𝒳)/𝒩 are  complete.Proof.
Property (i) is a simple corollary of (5.5).
To see (ii),B˙∞,qs(𝒳)⊂F˙∞,qs(𝒳) is obvious by the definitions of the both
spaces B˙∞,qs(𝒳) and F˙∞,qs(𝒳),
while F˙∞,qs(𝒳)⊂B˙∞,∞s(𝒳) can be obtained by their definitions together
with the Lebesgue differential theorem; we omit the details.
Property (iii) is a consequence of the second
inclusion in Property (ii) of this proposition and Proposition5.10(iii) on B˙∞,∞s(𝒳),
while Property (v) can be easily deduced from Property (iii), Property (iv) is
a conclusion of the first inclusion in Property (ii) of this proposition and
Proposition 5.10(iv) on B˙∞,qs(𝒳),
which completes the proof of this proposition.To obtain some relations between the spacesF˙∞,qs(𝒳) and the space BMO(𝒳),
we need the following technical result which in fact gives a new
characterization of Triebel-Lizorkin space F˙∞,qs(𝒳).
We first introduce a maximal function.For anyx∈𝒳 and l∈ℤ,
using Lemma 2.19, it is easy to prove that there exists a finite number of α∈Il such that (6.34)Qαl∩B(x,2−l+1)≠∅,which will denote by Qαlil with i=1,…,ml(x),
and moreover, ml(x) is no more than a positive integer m∈ℕ which is independent of l and x;
see [85, pages 1385-1386] for a detailed proof. In what follows, for convenience sake, we will
always assume that ml(x)=m by letting Qαlil=∅ for i=ml(x)+1,…,m when ml(x)<m.
Let {Dk}k∈ℤ be as in Definition 6.7. Now for any s∈ℝ, q∈(0,∞], f∈(𝒢°(β,γ))′ with 0<β,γ<ϵ and x∈𝒳,
we define the maximal functionℭ˙qs(f)(x) by (6.35)ℭ˙qs(f)(x)=supl∈ℤ[1μ(∪i=1mQαlil)∫∪i=1mQαlil∑k=l∞2ksq|Dk(f)(y)|qdμ(y)]1/q, where the usual modification is
made when q=∞.Proposition 6.10.
Letϵ, s, and q be as in Definition 6.7. Then f∈F˙∞,qs(𝒳) if and only if f∈(𝒢°(β,γ))′ with β,γ as in (6.32) and ℭ˙qs(f)∈L∞(𝒳).
Moreover, in this case, (6.36)∥f∥F˙∞,qs(𝒳)~∥ℭ˙qs(f)∥L∞(𝒳).Proof.
For anys∈ℝ, q∈(0,∞], f∈(𝒢°(β,γ))′ with β,γ as in (6.32) and x∈𝒳,
let (6.37)ℭ˙q,1s(f)(x)=supl∈ℤsupQαl∋xα∈Il[1μ(Qαl)∫Qαl∑k=l∞2ksq|Dk(f)(y)|qdμ(y)]1/q, where the usual modification is
made when q=∞.
Obviously, (6.38)∥f∥F˙∞,qs(𝒳)~∥ℭ˙q,1s(f)∥L∞(𝒳). To prove the conclusion of this
proposition, it suffices to verify that when (6.39)⋃i=1ml(x)Qαlil⊃B(x,2−l+1), for any i1,i2∈{1,…,ml(x)}, (6.40)μ(Qαli1l)~μ(Qαli2l). To verify (6.40), by symmetry,
it suffices to verify that (6.41)μ(Qαli1l)≲μ(Qαli2l). To this end, suppose zk∈B(x,2−l+1)∩Qαlikl with k=1,2.
Then for any w∈B(zαli1l,C62−l),
we have (6.42)d(w,zαli2l)≤d(w,zαli1l)+d(zαli1l,z1)+d(z1,x)+d(x,z2)+d(z2,zαli2l)≤3C62−l+2−l+2. Thus, Qαli1l⊂B(zαli1l,C62−l)⊂B(zαli2l,(3C6+4)2−l), which together with Lemma 2.19 and the double
property of μ gives (6.41), and hence, completes the proof
of Proposition 6.10.Using Theorem5.19(i) and Proposition 6.10, and an
argument similar to that in [86], we can establish the connections between F˙∞,qs(𝒳) with BMO(𝒳) and C˙s(𝒳) with s>0 as follows.Theorem 6.11.
Letϵ be as in Definition 6.7. Then,
(i)
if0<s<ϵ,
then C˙s(𝒳)=F˙∞,∞s(𝒳)=B˙∞,∞s(𝒳) with equivalent norms;(ii)
BMO(𝒳)=F˙∞,20(𝒳) with equivalent norm.Proof.
We first verify (i). Letf∈C˙s(𝒳) and let {Dk}k∈ℤ be as in Definition 6.7. We first claim that f∈(𝒢°0ϵ(β,γ))′ when 0<β<ϵ and s<γ<ϵ.
In fact, from f∈C˙s(𝒳),
it follows that for all x∈𝒳, (6.43)|f(x)−f(x1)|≲∥f∥C˙s(𝒳)d(x,x1)s, which implies that for any g∈𝒢°(β,γ) with 0<β<ϵ and s<γ<ϵ, (6.44)|〈f,g〉|=|∫𝒳[f(x)−f(x1)]g(x)dμ(x)|≲∥f∥C˙s(𝒳)∥g∥𝒢(β,γ)∫𝒳d(x,x1)s1V1(x1)+V(x1,x)(11+d(x1,x))γdμ(x)≲∥f∥C˙s(𝒳)∥g∥𝒢(β,γ), where in the last step, we used
Lemma 2.1(ii).
Moreover, for allk∈ℤ and x∈𝒳,
by Lemma 2.1(ii) and 0<s<ϵ<ϵ2,
we then have (6.45)|Dk(f)(x)|=|∫𝒳Dk(x,y)f(y)dμ(y)|=|∫𝒳Dk(x,y)[f(y)−f(x)]dμ(y)|≲∥f∥C˙s(𝒳)∫𝒳1V2−k(x)+V2−k(y)+V(x,y)(2−k2−k+d(x,y))ϵ2d(x,y)sdμ(y)≲2−ks∥f∥C˙s(𝒳), which proves that (6.46)∥f∥F˙∞,∞s(𝒳)=supk∈ℤsupx∈𝒳2ks|Dk(f)(x)|≲∥f∥C˙s(𝒳). Thus, C˙s(𝒳)⊂F˙∞,∞s(𝒳).
Conversely, letf∈F˙∞,∞s(𝒳).
By Proposition 6.6, without loss of generality, we may assume that f∈(𝒢°(β,γ))′ with β,γ as in (6.32). Let all the notation as in
Theorem 3.10. Then, by Theorem 3.13, for any g∈𝒢°(β,γ) with β,γ as in (6.32), since ∫𝒳g(x)dμ(x)=0,
we then have (6.47)〈f,g〉=∑k=−∞∞〈D˜kDk(f)(⋅),g(⋅)〉=∑k=−∞∞〈D˜kDk(f)(⋅)−D˜kDk(f)(x1),g(⋅)〉, which means that in (𝒢°(β,γ))′, f(x)=∑k=−∞∞[D˜kDk(f)(⋅)−D˜kDk(f)(x1)]. For x∈𝒳,
we now let h(x)≡∑k=−∞∞[D˜kDk(f)(x)−D˜kDk(f)(x1)] and we first verify that h is a function satisfying the following growth
condition that (6.48)|h(x)|≲∥f∥F˙∞,∞s(𝒳)d(x,x1)s. In fact, for any x∈𝒳,
assume that 2−l0−1<d(x,x1)≤2−l0 with some l0∈ℤ and write that (6.49)h(x)=∑k=−∞∞∫𝒳[D˜k(x,z)−D˜k(x1,z)]Dk(f)(z)dμ(z)=∑k=−∞l0−1∫𝒳[D˜k(x,z)−D˜k(x1,z)]Dk(f)(z)dμ(z)+∑k=l0∞⋯≡J1+J2. For J1,
by the regularity of D˜k,
Lemma 2.1(ii) and s<ϵ<ϵ1,
we have (6.50)|J1|≲∑k=−∞l0−1∫𝒳(d(x,x1)2−k+d(x,z))ϵ11V2−k(x)+V2−k(z)+V(x,z)(2−k2−k+d(x,z))ϵ2|Dk(f)(z)|dμ(z)≲∥f∥F˙∞,∞s(𝒳)d(x,x1)s. To estimate J2,
by s>0 and Proposition 2.7(i), we
obtain (6.51)J2≲∥f∥F˙∞,∞s(𝒳)∑k=l0∞2−ks∫𝒳[|D˜k(x,z)|+|D˜k(x1,z)|]dμ(z)≲∥f∥F˙∞,∞s(𝒳)d(x,x1)s. Thus, our claim is true.
Notice that for anyx,y∈𝒳, (6.52)h(x)−h(y)=∑k=−∞∞[D˜kDk(f)(x)−D˜kDk(f)(y)]. Repeating the above proof yields
that for all x,y∈𝒳, (6.53)|h(x)−h(y)|≲∥f∥F˙∞,∞s(𝒳)d(x,y)s. Thus, h∈C˙s(𝒳) and (6.54)∥h∥C˙s(𝒳)≲∥f∥F˙∞,∞s(𝒳). In this sense, we say that F˙∞,∞s(𝒳)⊂C˙s(𝒳), which finishes the proof of (i).
To see (ii), letf∈BMO(𝒳) and {Dk}k∈ℤ be as in Definition 6.7. Proposition 5.10(iv)
and Theorem 5.19(i) immediately imply that f∈(𝒢°0ϵ(β,γ))′ with
0<β,γ<ϵ.
Let now Qαl for l∈ℤ and α∈Il be a dyadic cube as in Lemma 2.19. Set Bαl=B(zαl,2C62−l) and write (6.55)f=(f−mBαl(f))χBαl+(f−mBαl(f))χ{𝒳∖Bαl}+mBαl(f)≡f1+f2+f3. Obviously, μ(Qαl)~μ(Bαl) and Dk(f3)=0;
in combination with Proposition 3.15, this shows that (6.56){1μ(Qαl)∫Qαl∑k=l∞|Dk(f1)(x)|2dμ(x)}1/2≲{1μ(Qαl)∫𝒳|f1(x)|2dμ(x)}1/2≲{1μ(Bαl)∫Bαl|f−mBαl(f)|2dμ(x)}1/2≲∥f∥BMO(𝒳). To estimate f2,
notice that for x∈Qαl and y∈𝒳∖Bαl,
then d(x,y)≳2−l+d(y,zαl),
which together with the size condition of Dk yields that for x∈Qαl and k≥l, (6.57)|Dk(f2)(x)|=|∫𝒳∖BαlDk(x,y)[f(y)−mBαl(f)]dμ(y)|≲2(l−k)ϵ2∑j=0∞12jϵ21V(zαl,2C62j−l+1)∫d(y,zαl)<2C62j−l+1|f(y)−mBαl(f)|dμ(y). The definition of BMO(𝒳) together with the double property of μ gives that (6.58)|mB(zαl,2C62j−l+1)(f)−mBαl(f)|≲(j+1)∥f∥BMO(𝒳), which further implies that for x∈Qαl and k≥l, (6.59)|Dk(f2)(x)|≲2(l−k)ϵ2∑j=0∞12jϵ2{1V(zαl,2C62j−l+1)∫d(y,zαl)<2C62j−l+1|f(y)−mB(zαl,2C62j−l+1)(f)|dμ(y)+|mB(zαl,2C62j−l+1)(f)−mBαl(f)|}≲2(l−k)ϵ2∥f∥BMO(𝒳). From this, it follows
that (6.60){1μ(Qαl)∫Qαl∑k=l∞|Dk(f2)(x)|2dμ(x)}1/2≲∥f∥BMO(𝒳){∑k=l∞22(l−k)ϵ2}1/2≲∥f∥BMO(𝒳). Combining the above estimates,
we know that f∈F˙∞,20(𝒳) and (6.61)∥f∥F˙∞,20(𝒳)≲∥f∥BMO(𝒳).
We now prove the converse. Letf∈F˙∞,20(𝒳).
By Proposition 6.6, without loss of generality, we may assume that f∈(𝒢°(β,γ))′ with β,γ as in (6.32). In the rest of the proof of this
theorem, we denote ℭ˙20 and S˙2,10 simply by ℭ˙ and S˙.
Moreover, for j∈ℤ, f as above and x∈𝒳,
we define (6.62)S˙j(f)(x)={∑k=j∞∫d(x,y)<2−k|Dk(f)(y)|2dμ(y)V2−k(x)}1/2. Obviously, S˙∞(f)(x)=S˙(f)(x). Let the notation as in Theorem 3.11. For any f as above and x∈𝒳,
we also set (6.63)S¯˙(f)(x)={∑k=−∞∞∫d(x,y)<2−k|D¯k(f)(y)|2dμ(y)V2−k(x)}1/2. Theorem 5.13 together with Remark
5.5 shows that for all f∈H1(𝒳), (6.64)∥S¯˙(f)∥L1(𝒳)≲∥f∥H1(𝒳).
For any fixedf as above and x∈𝒳,
we define the “stopping-time" j(x) by (6.65)j(x)=inf{j∈ℤ:S˙j(f)(x)≤Aℭ˙(f)(x)}, where A>0 is a large constant to be determined later. We
first claim that for any y∈𝒳 and l∈ℤ,
if we choose A to be large enough, then there exists a
constant C12>0 such that (6.66)μ({x∈𝒳:d(x,y)<2−l,l≥j(x)})≥C12μ(B(y,2−l)). In fact, let B0=B(y,2−l).
Then ⋃x∈B0B(x,2−l)⊂B(y,2−l+1). Let (6.67)B(y,2−l+1)⊂⋃i=1mQαlil≡P as in the definition of ℭ˙.
Let w∈B(y,2−l+1)∩Qαlil. Then for any x∈Qαlil, d(x,y)≤d(x,w)+d(w,y)≤C62−l+2−l+1, which shows that (6.68)P⊂B(y,(C6+2)2−l).Thus, μ(B0)~μ(P). From this, it follows that(6.69)1μ(B0)∫B0[S˙l(f)(x)]2dμ(x)=∑k=l∞1μ(B0)∫B0∫d(x,z)<2−k|Dk(f)(z)|2dμ(z)V2−k(x)dμ(x)≤C1μ(P)∑k=l∞∫P∫B0|Dk(f)(z)|2χB(z,2−k)(x)dμ(x)V2−k(x)dμ(z)≤C131μ(P)∫P∑k=l∞|Dk(f)(z)|2dμ(z)≤C13[infx∈B0ℭ˙(f)(x)]2, where C,C13>0 are constants independent of l and x.
Thus, if A2>C13,
then (6.70)μ({x∈B0:Sl(f)(x)>Aℭ˙(f)(x)})≤C13A2μ(B0), which in turn shows (6.66) with C12=1−C13/A2>0 if A2>C13.
Letg∈𝒢°b(ϵ1,ϵ2) and ∥g∥H1(𝒳)≤1.
By Theorem 3.11, (6.66), the Fubini theorem, Hölder's inequality, (6.64), and
Proposition 6.10, we then have (6.71)|〈f,g〉|=|〈f,∑k=−∞∞DkD¯k(g)〉|=|∑k=−∞∞〈Dkt(f),D¯k(g)〉|≤∑k=−∞∞∫𝒳|Dkt(f)(y)D¯k(g)(y)|dμ(y)≲∫𝒳[∑k=l(x)∞∫d(x,y)<2−k|Dkt(f)(y)D¯k(g)(y)|dμ(y)V2−k(x)]dμ(x)≲∫𝒳S˙l(x)(f)(x)S¯˙(g)(x)dμ(x)≲∫𝒳ℭ˙(f)(x)S¯˙(g)(x)dμ(x)≲∥ℭ˙(f)∥L∞(𝒳)∥g∥H1(𝒳)≲∥f∥F˙∞,20(𝒳), which together with Theorem 5.19(i) further implies that f∈BMO(𝒳) and (6.72)∥f∥BMO(𝒳)≲∥f∥F˙∞,20(𝒳). This finishes the proof of
Theorem 6.11.We end this subsection by the boundedness onF˙∞,qs(𝒳) of singular integral operators of order ϵ,
which satisfy (I-1) through (I-4) in Subsection 5.2.Theorem 6.12.
Letϵ and q be as in Definition 6.7. If T is a singular integral operator of order ϵ,
then T is bounded from 𝒢°b(β,γ) (with 0<β≤ϵ and γ>0) to F˙∞,qs(𝒳).
Moreover, there exists a constant C>0 such that for all f∈𝒢°b(β,γ) with 0<β≤ϵ and γ>0, (6.73)∥Tf∥F˙∞,qs(𝒳)≤C∥f∥F˙∞,qs(𝒳).Proof.
Combining some estimates and technics used in the proof of Proposition5.25 with those used in the proof of Proposition 6.3 gives the desired
conclusions. The details are left to the reader.Remark 6.13.
By Theorem8.15 below, if p(s,ϵ)<q≤∞,
then F˙∞,qs(𝒳) is the dual space of F˙1,q′−s(𝒳).
In this case, Theorem 6.12 can be deduced from Theorem 5.23 together with a
duality argument. This provides another proof of Theorem 6.12.


## 6.3. Inhomogeneous Plancherel-Pôlya Inequality and Definition ofF∞,qs(𝒳)

In this and the next subsection,μ(𝒳) can be finite or infinite. We first introduce
the norm in ∥⋅∥F∞,qs(𝒳) via some IATI and then verify that this norm is independent
of the choices of IATIs and the distribution spaces; see also
[84]. Similarly to the
case of the space F˙∞,qs(𝒳),
we need also first to establish an inhomogeneous Plancherel-Pôlya inequality
related to the norm ∥⋅∥F∞,qs(𝒳).Definition 6.14.
Letϵ1∈(0,1], ϵ2>0, ϵ3>0, ϵ∈(0,ϵ1⋀ϵ2) and let {Sk}k∈ℤ+ be an (ϵ1,ϵ2,ϵ3)-IATI.
Set Dk=Sk−Sk−1 for k∈ℕ,
and D0=S0.
Let {Qτ0,ν:τ∈I0,ν=1,…,N(0,τ)} with a fixed large j∈ℕ be dyadic cubes as in Section 4. Let |s|<ϵ and p(s,ϵ)<q≤∞.
For any f∈(𝒢0ϵ(β,γ))′ with 0<β,γ<ϵ,
define (6.74)∥f∥F∞,qs(𝒳)=max{supτ∈I0ν=1,…,N(0,τ)mQτ0,ν(|D0(f)|),supl∈ℕsupα∈Il[1μ(Qαl)∫Qαl∑k=l∞2ksq|Dk(f)(x)|qdμ(x)]1/q}, where the supremum is
taken over all dyadic cubes as in Lemma 2.19 and Section 4, and the usual
modification is made when q=∞.To verify that the definition of∥⋅∥F∞,qs(𝒳) is independent of the choice of IATIs, we need the following inequality of Plancherel-Pôlya type.Proposition 6.15.
Letϵ1∈(0,1], ϵ2>0, ϵ3>0, ϵ∈(0,ϵ1⋀ϵ2) and let {Sk}k∈ℤ and {Pk}k∈ℤ be two (ϵ1,ϵ2,ϵ3)-ATIs. Set Dk=Sk−Sk−1 and Qk=Pk−Pk−1 for k∈ℕ, D0=S0,
and Q0=S0.
Let {Qτ0,ν:τ∈I0,ν=1,…,N(0,τ)} with a fixed large j∈ℕ be dyadic cubes as in Section 4. Let |s|<ϵ and p(s,ϵ)<q≤∞.
Then for all f∈(𝒢0ϵ(β,γ))′ with 0<β,γ<ϵ, (6.75)max{supτ∈I0ν=1,…,N(0,τ)mQτ0,ν(|D0(f)|),supl∈ℕsupα∈Il[1μ(Qαl)∑k=l∞∑τ∈Ik∑ν=1N(k,τ)2ksqμ(Qτk,ν)χ{(τ,ν):Qτk,ν⊂Qαl}(τ,ν)(supx∈Qτk,ν|Dk(f)(x)|)q]1/q}~max{supτ∈I0ν=1,…,N(0,τ)mQτ0,ν(|Q0(f)|),supl∈ℕsupα∈Il[1μ(Qαl)∑k=l∞∑τ∈Ik∑ν=1N(k,τ)2ksqμ(Qτk,ν)χ{(τ,ν):Qτk,ν⊂Qαl}(τ,ν)(infx∈Qτk,ν|Qk(f)(x)|)q]1/q}.Proof.
To prove Proposition6.15, it suffices to verify that for all f∈(𝒢0ϵ(β,γ))′ with 0<β,γ<ϵ,
the left-hand side of (6.75) is controlled by its right side.
Let all the notation be as in Proposition5.25. Then,
as in the proof of Proposition 5.25, by (5.78) and (5.80), we still control mQτ0,ν(|D0(f)|) by Z1+Z2.
Moreover, by (5.82) and Lemma 2.1(ii), we have (6.76)Z1≲supτ′∈I0ν′=1,…,N(0,τ′)mQτ′0,ν′(|Q0(f)|), which is the desired estimate.
To estimateZ2,
we need the following trivial estimate that for k′∈ℕ, (6.77)[∑τ′∈Ik′∑ν′=1N(k′,τ′)2k′sq|Qk′(f)(yτ′k′,ν′)|q]1/q≲supl∈ℕsupα∈Il[1μ(Qαl)∑k=l∞∑τ∈Ik∑ν=1N(k,τ)2ksqμ(Qτk,ν)χ{(τ,ν):Qτk,ν⊂Qαl}(τ,ν)(infx∈Qτk,ν|Qk(f)(x)|)q]1/q. By this and (5.82) together with μ(Qτ′k′,ν′)≲V1(yτ′k′,ν′) and (5.5), when q≤1,
we have (6.78)Z2≲∑k′=1∞2−k′(ϵ+s){∑τ′∈Ik′∑ν′=1N(k′,τ′)2k′sq|Qk′(f)(yτ′k′,ν′)|q}1/q≲RHSof(6.75), where we used the assumption
that |s|<ϵ.
Similarly, when1<q≤∞,
by Hölder's inequality and Lemma 2.1(ii), (6.79)Z2≲∑k′=1∞2−k′(ϵ+s){∑τ′∈Ik′∑ν′=1N(k′,τ′)2k′sq|Qk′(f)(yτ′k′,ν′)|q}1/q×{∫𝒳1V1(yτ0,ν)+V1(y)+V(yτ0,ν,y)1(1+d(yτ0,ν,y))ϵdμ(y)}1/q′≲RHSof(6.75). Thus, mQτ0,ν(|D0(f)|) for τ∈I0 and ν=1,…,N(0,τ) is controlled by the right-hand side of
(6.75).
We now verify that the second term of the left-hand
side of (6.75) is also controlled by its right-hand side. To this end, for any k∈ℕ and z∈Qτk,ν,
we also control |Dk(f)(z)| by Y1+Y2.
The estimate for Y2 is similar to the estimates for Y1+Y2 in the proof of Proposition 6.3 and we omit
the details. To estimate Y1,
by (5.95) and Lemma 2.1(ii), we have (6.80)supx∈Qτk,νY1≲2−kϵ[supτ′∈I0ν′=1,…,N(0,τ′)mQτ′0,ν′(|Q0(f)|)]×∑τ′∈I0∑ν′=1N(0,τ′)μ(Qτ′0,ν′)1V1(yτk,ν)+V1(yτ′0,ν′)+V(yτk,ν,yτ′0,ν′)1(1+d(yτk,ν,yτ′0,ν′))ϵ≲2−kϵ[supτ′∈I0ν′=1,…,N(0,τ′)mQτ′0,ν′(|Q0(f)|)], which together with Lemma 2.19
and |s|<ϵ gives that for l∈ℕ and α∈Il, (6.81)[1μ(Qαl)∑k=l∞∑τ∈Ik∑ν=1N(k,τ)2ksqμ(Qτk,ν)χ{(τ,ν):Qτk,ν⊂Qαl}(τ,ν)supx∈Qτk,ν|Y1|q]1/q≲supτ′∈I0ν′=1,…,N(0,τ′)mQτ′0,ν′(|Q0(f)|). This is the desired estimate for Y1 and hence, we complete the proof of Proposition
6.15.Remark 6.16.
We point out that Remark5.5 applies in a similar way to  Proposition
6.15.From Proposition6.15 and Lemma 2.19, it is easy to
deduce the following proposition. We omit the  details.Proposition 6.17.
Adopting the notation from Proposition6.15,
one has for all for f∈(𝒢0ϵ(β,γ))′ with
0<β,γ<ϵ, (6.82)max{supτ∈I0ν=1,…,N(0,τ)mQτ0,ν(|D0(f)|),supl∈ℕsupα∈Il[1μ(Qαl)∫Qαl∑k=l∞2ksq|Dk(f)(x)|qdμ(x)]1/q}~max{supτ∈I0ν=1,…,N(0,τ)mQτ0,ν(|Q0(f)|),supl∈ℕsupα∈Il[1μ(Qαl)∫Qαl∑k=l∞2ksq|Qk(f)(x)|qdμ(x)]1/q}.Proposition6.17 shows that the definition of the norm ∥⋅∥F∞,qs(𝒳) is independent of the choice of IATIs. We now verify that under some restrictions
on β and γ,
it is also independent of the choice of distribution  spaces.Proposition 6.18.
Let all the notation be as in Definition6.14. Let |s|<ϵ and p(s,ϵ)<q≤∞.
If f∈(𝒢0ϵ(β1,γ1))′ with max{0,−s}<β1<ϵ and 0<γ1<ϵ,
and if ∥f∥F∞,qs(𝒳)<∞,
then f∈(𝒢0ϵ(β2,γ2))′ for every max{0,−s}<β2<ϵ and
0<γ2<ϵ.Proof.
We use the notation from the proof of Proposition5.28. Let ψ∈𝒢(ϵ,ϵ),
and f∈(𝒢0ϵ(β1,γ1))′ with max{0,d(1−1/q)+−s−d}<β1<ϵ and 0<γ1<ϵ,
and ∥f∥F∞,qs(𝒳)<∞.
To verify that f∈(𝒢0ϵ(β2,γ2))′ with max{0,−s}<β2<ϵ and 0<γ2<ϵ,
we need the following trivial estimates that for τ∈I0 and ν=1,…,N(0,τ),(6.83)mQτ0,ν(|D0(f)|)≲∥f∥F∞,qs(𝒳) and that for k∈ℕ, (6.84)[∑τ∈Ik∑ν=1N(k,τ)2ksq|Dk(f)(yτk,ν)|q]1/q≲∥f∥F∞,qs(𝒳).
Whenq≤1,
by Theorem 4.16, (5.80), (5.105), (5.5), (6.25), (6.83), and (6.84),  we
have (6.85)|〈f,ψ〉|=|∑τ∈I0∑ν=1N(0,τ)∫Qτ0,ν〈D˜0(⋅,y),ψ〉dμ(y)Dτ,10,ν(f)+∑k=1∞∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)〈D˜k(⋅,yτk,ν),ψ〉Dk(f)(yτk,ν)|≲∥ψ∥𝒢(β2,γ2)∥f∥F∞,qs(𝒳), where we used the assumption
that γ2>0 and β2>−s.
If1<q≤∞,
by (6.83), Hölder's inequality, (6.25), and (6.84), we obtain (6.86)|〈f,ψ〉|≲∥ψ∥𝒢(β2,γ2){∥f∥F∞,qs(𝒳)+∑k=1∞2−kβ2[∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)|Dk(f)(yτk,ν)|q×1V1(x1)+V(x1,yτk,ν)1(1+d(x1,yτk,ν))γ2]1/q×[∫𝒳1V1(x1)+V(x1,y)1(1+d(x1,y))γ2dμ(y)]1/q′}≲∥ψ∥𝒢(β2,γ2)∥f∥F∞,qs(𝒳), where we used the assumption β2>−s and γ2>0 again. This finishes the proof of Proposition
6.18.Based on Propositions6.17 and 6.18, we can
now introduce the inhomogeneous Triebel-Lizorkin spaces F∞,qs(𝒳).Definition 6.19.
Letϵ1∈(0,1], ϵ2>0, ϵ3>0, ϵ∈(0,ϵ1⋀ϵ2) and let {Sk}k∈ℤ be an (ϵ1,ϵ2,ϵ3)-IATI.
Set Dk=Sk−Sk−1 for k∈ℕ and D0=S0.
Let {Qτ0,ν:τ∈I0,ν=1,…,N(0,τ)} with a fixed large j∈ℕ be dyadic cubes as in Section 4. Let |s|<ϵ and p(s,ϵ)<q≤∞.
The spaceF∞,qs(𝒳) is defined to be the set of all f∈(𝒢0ϵ(β,γ))′,
for some |s|<β<ϵ and 0<γ<ϵ,
such that (6.87)∥f∥F∞,qs(𝒳)=max{supτ∈I0ν=1,…,N(0,τ)mQτ0,ν(|D0(f)|),supl∈ℕsupα∈Il[1μ(Qαl)∫Qαl∑k=l∞2ksq|Dk(f)(x)|qdμ(x)]1/q}<∞, where the supremum is taken over
all dyadic cubes as in Lemma 2.19 and Section 4, and the usual modification is
made when q=∞.Propositions6.17 and 6.18 show that the
definition of the space F∞,qs(𝒳) is independent of the choices of IATIs and the distribution spaces, (𝒢(β,γ))′ with β and γ with |s|<β<ϵ and 0<γ<ϵ.Remark 6.20.
To guarantee that the definition of the spaceF∞,qs(𝒳) is independent of the choice of the
distribution space (𝒢(β,γ))′,
we only need to restrict max{0,−s}<β<ϵ and 0<γ<ϵ.
However, if max{0,s}<β<ϵ and 0<γ<ϵ,
we can then verify that 𝒢(β,γ)⊂F∞,qs(𝒳); see Proposition 6.21 below.


## 6.4. Properties ofF∞,qs(𝒳) and Boundedness of Singular Integrals

In this subsection, we first present some basic
properties ofF∞,qs(𝒳) and the relation between F∞,qs(𝒳) and F˙∞,qs(𝒳).
By establishing a maximal function characterization of F∞,qs(𝒳),
we then derive some relations between the spaces F∞,qs(𝒳) and Cs(𝒳) and between the spaces F∞,qs(𝒳) and bmo(𝒳).
Finally, we prove boundedness results on the spaces F∞,qs(𝒳) for the classes of singular integral operators
considered in Subsection 5.4.Proposition 6.21.
Letϵ1∈(0,1], ϵ2>0, ϵ∈(0,ϵ1⋀ϵ2), |s|<ϵ, and p(s,ϵ)<q≤∞.
Then, the following hold.
(i)
F∞,q0s(𝒳)⊂F∞,q1s(𝒳) if p(s,ϵ)<q0≤q1≤∞.(ii)
Let−ϵ<s+θ<ϵ and θ>0.
Then for p(s,ϵ)<q0,q1≤∞,(6.88)F∞,q0s+θ(𝒳)⊂F∞,q1s(𝒳).(iii)
B∞,qs(𝒳)⊂F∞,qs(𝒳)⊂B∞,∞s(𝒳).(iv)
Ifmax{0,−s}<β<ϵ and 0<γ<ϵ,
then F∞,qs(𝒳)⊂(𝒢0ϵ(β,γ))′.(v)
Ifmax{s,0}<β<ϵ and 0<γ<ϵ,
then 𝒢(β,γ)⊂F∞,qs(𝒳).(vi)
The spacesF∞,qs(𝒳) are complete.Proof.
Property (ii) can be established by an argument similar to that used for
property (ii) of Proposition5.31, while property (i) and property (iii) through
property (vi) can be proved by an argument similar to those used for
Proposition 6.9 via Proposition 5.31, which completes the proof of Proposition 6.21.The following proposition gives a new characterization
of the spacesF∞,qs(𝒳) when 1≤q≤∞.Proposition 6.22.
Letϵ and {Dk}k∈ℤ+ be as in Definition 6.19. If 1≤q≤∞,
then f∈F∞,qs(𝒳) if and only if f∈(𝒢0ϵ(β,γ))′,
for some |s|<β<ϵ and 0<γ<ϵ,
and (6.89)supl∈ℤ+supα∈Il[1μ(Qαl)∫Qαl∑k=l∞2ksq|Dk(f)(x)|qdμ(x)]1/q<∞. Moreover, in this case, (6.90)∥f∥F∞,qs(𝒳)~supl∈ℤ+supα∈Il[1μ(Qαl)∫Qαl∑k=l∞2ksq|Dk(f)(x)|qdμ(x)]1/q.Proof.
Fixτ∈I0 and
ν=1,…,N(0,τ).
Since Qτ0,ν⊂Qτ0,
then it is easy to verify that Qτ0⊂B(zτ0,ν,2C6),
and hence μ(Qτ0,ν)≲μ(Qτ0)≲μ(B(zτ0,ν,2C6))≲μ(Qτ0,ν). From this and
1≤q≤∞ together with Hölder's inequality, it follows
that (6.91)mQτ0,ν(|D0(f)|)≤[1μ(Qτ0,ν)∫Qτ0,ν|D0(f)(x)|qdμ(x)]1/q≲[1μ(Qτ0)∫Qτ0|D0(f)(x)|qdμ(x)]1/q. Thus, (6.92)∥f∥F∞,qs(𝒳)≲supl∈ℤ+supα∈Il[1μ(Qαl)∫Qαl∑k=l∞2ksq|Dk(f)(x)|qdμ(x)]1/q.
To see the converse, it suffices to verify that for
allf∈F∞,qs(𝒳) and τ∈I0, (6.93)[1μ(Qτ0)∫Qτ0|D0(f)(x)|qdμ(x)]1/q≲∥f∥F∞,qs(𝒳). To see this, by the construction
of {Qτ0,ν:τ∈I0,ν=1,…,N(0,τ)},
we have(6.94)[1μ(Qτ0)∫Qτ0|D0(f)(x)|qdμ(x)]1/q≲{1μ(Qτ0)∑ν=1N(0,τ)μ(Qτ0,ν)[supx∈Qτ0,ν|D0(f)(x)|]q}1/q≲supν=1,…,N(0,τ)supx∈Qτ0,ν|D0(f)(x)|. Using Theorem 4.16 together with
some estimates similar to those for Z1+Z2 in the proof of Proposition 6.15, we can then
verify that (6.95)supν=1,…,N(0,τ)supx∈Qτ0,ν|D0(f)(x)|≲∥f∥F∞,qs(𝒳), which gives the desired estimate
and hence, we complete the proof of Proposition 6.22.Using Proposition6.22, we now complement Proposition
5.39(ii) for the case p=∞ as follows.Proposition 6.23.
Letϵ>0 be as in Definition 6.19, 0<s<ϵ, 1≤q≤∞ and μ(𝒳)=∞.
Then F∞,qs(𝒳)=F˙∞,qs(𝒳)∩L∞(𝒳),
and moreover, for any f∈F∞,qs(𝒳), (6.96)∥f∥F∞,qs(𝒳)~∥f∥F˙∞,qs(𝒳)+∥f∥L∞(𝒳).Proof.
Takef∈F∞,qs(𝒳).
By Proposition 6.18, without loss of generality, we may assume that f∈(𝒢(β,γ))′ with s<β,γ<ϵ as in (6.32). Thus, we also have f∈(𝒢°(β,γ))′ with s<β,γ<ϵ as in (6.32); see the proof of Proposition
5.37. By Theorem 3.29, we have that (6.97)f=D˜0S0(f)+∑k=1∞D˜kDk(f) holds in ∈(𝒢(β,γ))′ with s<β<ϵ and 0<γ<ϵ,
where D˜k with k∈ℤ+ is as in Theorem 3.26. From this and Lemma 2.19,
it follows that for any x∈𝒳, (6.98)|f(x)|≲∑τ∈I0∑ν=1N(0,τ)∫Qτ0,ν|D˜0(x,y)||S0(f)(y)|dμ(y)+∑k=1∞∑τ∈Ik∑ν=1N(k,τ)∫Qτk,ν|D˜k(x,y)||Dk(f)(y)|dμ(y)≲∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)1V1(x)+V1(yτ0,ν)+V(x,yτ0,ν)(11+d(x,yτ0,ν))ϵmQτ0,ν(|S0(f)|)+∑k=1∞∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)1V2−k(x)+V2−k(yτk,ν)+V(x,yτk,ν)×(2−k2−k+d(x,yτk,ν))ϵ[supy∈Qτk,ν|Dk(f)(y)|]. Since f∈F∞,qs(𝒳),
by its definition, we have that for τ∈I0 and ν=1,…,N(0,τ), (6.99)mQτ0,ν(|S0(f)|)≲∥f∥F∞,qs(𝒳), and the definition of
F∞,qs(𝒳) together with Proposition 6.15 also implies
that for any k∈ℕ, τ∈Ik,
and ν=1,…,N(k,τ), (6.100){∑τ∈Ik∑ν=1N(k,τ)2ksq[supy∈Qτk,ν|Dk(f)(y)|]q}1/q≲∥f∥F∞,qs(𝒳). Notice that 1≤q≤∞.
Both estimates via Hölder's inequality, the fact that μ(Qτk,ν)~V2−k(yτk,ν),
Lemma 5.2, and the assumption that s>0 further yield that (6.101)|f(x)|≲∥f∥F∞,qs(𝒳){1+∑k=1∞2−ks(∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)(2−k2−k+d(x,yτk,ν))ϵq′×1V2−k(x)+V2−k(yτk,ν)+V(x,yτk,ν))1/q′}≲∥f∥F∞,qs(𝒳), namely, f∈L∞(𝒳) and (6.102)∥f∥L∞(𝒳)≲∥f∥F∞,qs(𝒳).
Moreover,(6.103)∥f∥F˙∞,qs(𝒳)≲supl∈ℕsupα∈Il[1μ(Qαl)∫Qαl∑k=l∞2ksq|Dk(f)(x)|qdμ(x)]1/q+supl∈ℤ∖ℕsupα∈Il[1μ(Qαl)∫Qαl∑k=l∞2ksq|Dk(f)(x)|qdμ(x)]1/q≲∥f∥F∞,qs(𝒳)+supl∈ℤ∖ℕsupα∈Il[1μ(Qαl)∫Qαl∑k=l02ksq|Dk(f)(x)|qdμ(x)]1/q+supl∈ℤ∖ℕsupα∈Il[1μ(Qαl)∫Qαl∑k=1∞2ksq|Dk(f)(x)|qdμ(x)]1/q. To estimate the second term, by
Proposition 2.7(i), we have (6.104)|Dk(f)(x)|≲∥f∥L∞(𝒳)≲∥f∥F∞,qs(𝒳), which together with s>0 shows (6.105)supl∈ℤ∖ℕsupα∈Il[1μ(Qαl)∫Qαl∑k=l02ksq|Dk(f)(x)|qdμ(x)]1/q≲∥f∥F∞,qs(𝒳)supl∈ℤ∖ℕ[∑k=l02ksq]1/q≲∥f∥F∞,qs(𝒳). To estimate the third term, for
any l∈ℤ∖ℕ and α∈Il,
set (6.106)Ilα={β∈I1:Qβ1⊂Qαl}. Lemma 2.19 proves that (6.107)∑β∈Ilαμ(Qβ1)=μ(Qαl), which together with Lemma 2.19
further yields that (6.108)supl∈ℤ∖ℕsupα∈Il[1μ(Qαl)∫Qαl∑k=1∞2ksq|Dk(f)(x)|qdμ(x)]1/q≲supl∈ℤ∖ℕsupα∈Il[1μ(Qαl)∑β∈Ilαμ(Qβ1)1μ(Qβ1)∫Qβ1∑k=1∞2ksq|Dk(f)(x)|qdμ(x)]1/q≲∥f∥F∞,qs(𝒳). Thus, f∈F˙∞,qs(𝒳)∩L∞(𝒳) and(6.109)∥f∥L∞(𝒳)+∥f∥F˙∞,qs(𝒳)≲∥f∥F∞,qs(𝒳).
Conversely, letf∈F˙∞,qs(𝒳)∩L∞(𝒳).
Obviously, f∈(𝒢(β,γ))′ with s<β<ϵ and 0<γ<ϵ.
On the other hand, Proposition 2.7(i) shows that for any x∈𝒳, (6.110)|S0(f)(x)|≲∥f∥L∞(𝒳). Since 1≤q≤∞,
by Proposition 6.22, (5.5) together with (6.110), and Lemma 2.19 together with
(6.107), we obtain (6.111)∥f∥F∞,qs(𝒳)~supl∈ℕsupα∈Il[1μ(Qαl)∫Qαl∑k=l∞2ksq|Dk(f)(x)|qdμ(x)]1/q+supα∈I0[1μ(Qα0)∫Qα0(|S0(f)(x)|q+∑k=1∞2ksq|Dk(f)(x)|q)dμ(x)]1/q≲∥f∥F˙∞,qs(𝒳)+supα∈I0{1μ(Qα0)∑β∈I0αμ(Qβ0)[1μ(Qβ0)∫Qβ0∑k=1∞2ksq|Dk(f)(x)|qdμ(x)]}1/q≲∥f∥F˙∞,qs(𝒳), which means that f∈F∞,qs(𝒳).
Thus, (6.112)F˙∞,qs(𝒳)∩L∞(𝒳)⊂F∞,qs(𝒳),which completes the proof of
Proposition 6.23.From Proposition6.23 and Theorem 6.11(i), we easily
deduce the following results.Corollary 6.24.
Letϵ>0 be as in Definition 6.19. Then,
(i)
for anys∈(0,1], Cs(𝒳)=C˙s(𝒳)∩L∞(𝒳),
and moreover, for all f∈Cs(𝒳),(6.113)∥f∥Cs(𝒳)=∥f∥L∞(𝒳)+∥f∥C˙s(𝒳);(ii)
for anys∈(0,ϵ), Cs(𝒳)=B∞,∞s(𝒳)=F∞,∞s(𝒳) with equivalent norms.Proof.
Property (i) is a simple consequence of the
definitions of bothCs(𝒳) and C˙s(𝒳).
Whenμ(𝒳)=∞,
then Property (ii) can be deduced from Property (i), Theorem 6.11(i), and
Proposition 6.23, which completes the proof of Corollary 6.24 in this case.
An alternative way to prove Property (ii), which works
whenμ(𝒳)=∞ and also when μ(𝒳)<∞,
follows the line of reasoning in the proof of Theorem 6.11(i). In fact, let f∈Cs(𝒳) and {Dk}k∈ℤ+ be as in Definition 6.19. Then from f∈L∞(𝒳),
it follows that f∈(𝒢(β,γ))′ with s<β<ϵ and 0<γ<ϵ.
Moreover, by Proposition 2.7(i), we have that for all x∈𝒳, (6.114)|D0(f)(x)|=|∫𝒳D0(x,y)f(y)dμ(y)|≲∥f∥L∞(𝒳), which together with (6.45) shows
that f∈F∞,∞s(𝒳) and (6.115)∥f∥F∞,∞s(𝒳)≲∥f∥L∞(𝒳).
Conversely, supposef∈F∞,∞s(𝒳).
By Proposition 6.18, we can assume that f∈(𝒢(β,γ))′ with s<β<ϵ and 0<γ<ϵ.
Using the same notation as in Theorem 3.26, by Theorem 3.29 and the definition of ∥⋅∥F∞,∞s(𝒳) together with s>0,
we further obtain that for all x∈𝒳, (6.116)|f(x)|=|∑k=0∞D˜kDk(f)(x)|≲∥f∥F∞,∞s(𝒳)∑k=0∞2−ks≲∥f∥F∞,∞s(𝒳). Thus, f∈L∞(𝒳) and (6.117)∥f∥L∞(𝒳)≲∥f∥F∞,∞s(𝒳). Moreover, if d(x,y)≥1/2,
then (6.118)|f(x)−f(y)|≲∥f∥L∞(𝒳)≲∥f∥F∞,∞s(𝒳)d(x,y)s. Assume that 2−l0−1≤d(x,y)<2−l0 with l0∈ℕ.
By the regularity of D˜k,
Lemma 2.1(ii) and Proposition 2.7(i) together with 0<s<ϵ<ϵ1,
we then obtain (6.119)|f(x)−f(y)|≲|∑k=0l0−1∫𝒳[D˜k(x,z)−D˜k(y,z)]Dk(f)(z)dμ(z)|+∑k=l0∞{∫𝒳|D˜k(x,z)||Dk(f)(z)|dμ(z)+∫𝒳|D˜k(y,z)||Dk(f)(z)|dμ(z)}≲∑k=0l0−1∫𝒳(d(x,y)2−k+d(x,z))ϵ11V2−k(x)+V2−k(y)+V(x,y)×(2−k2−k+d(x,z))ϵ2|Dk(f)(z)|dμ(z)+∥f∥F∞,∞s(𝒳)∑k=l0∞2−ks≲∥f∥F∞,∞s(𝒳)d(x,y)s. Thus, f∈Cs(𝒳) and (6.120)∥f∥Cs(𝒳)≲∥f∥F∞,∞s(𝒳),which completes the proof of
Corollary 6.24(ii).We now establish the connection betweenF˙∞,qs(𝒳) and F∞,qs(𝒳) for all admissible s and q.Proposition 6.25.
Letϵ, s, q, and S0 be as in Definition 6.19 and let μ(𝒳)=∞.
Then there exists a constant C>0 such that for all f∈F˙∞,qs(𝒳), f−S0(f)∈F∞,qs(𝒳) and (6.121)∥f−S0(f)∥F∞,qs(𝒳)≤C∥f∥F˙∞,qs(𝒳).Proof.
Letf∈F˙∞,qs(𝒳).
By Proposition 6.6, we may assume that f∈(𝒢°0ϵ(β,γ))′ with β,γ as in (6.32). On the other hand, for any g∈𝒢0ϵ(β,γ) with β,γ as in (6.32), from Lemma 5.36 and ∫𝒳S0(x,y)dμ(x)=1,
it is easy to deduce that g−S0t(g)∈𝒢°(β,γ) with β,γ as in (6.32). Thus, from (6.122)〈f−S0(f),g〉=〈f,g−S0t(g)〉, we deduce that f−S0(f)∈(𝒢0ϵ(β,γ))′ with β,γ as in (6.32).
We now verify (6.121). In what follows, let I be the identity operator, let {Dk′}k′∈ℤ be as in Definition 6.7, and let {D˜k′}k′∈ℤ be as in Theorem 4.11. We first claim that for
all k′∈ℤ and x,y∈𝒳, (6.123)|(S0(I−S0)D˜k′)(x,y)|≲2−|k′|ϵ1V2−(0⋀k′)(x)+V2−(0⋀k′)(y)+V(x,y)(2−(0⋀k′)2−(0⋀k′)+d(x,y))ϵ and that for all k∈ℕ, k′∈ℤ,
and x,y∈ℤ, (6.124)|(Dk(I−S0)D˜k′)(x,y)|≲2−|k−k′|ϵ1V2−(k⋀k′)(x)+V2−(k⋀k′)(y)+V(x,y)(2−(k⋀k′)2−(k⋀k′)+d(x,y))ϵ. The estimate (6.124) is
essentially the same as the estimate (5.130) by symmetry, while the estimate
(6.123) can essentially be obtained by an argument similar to Cases 1 and 2
of the proof of the estimate (5.130).
Using the estimate (6.123) and Theorem 4.13, we have
that for any x∈𝒳, (6.125)|S0(I−S0)(f)(x)|=|∑k′∈ℤ∑τ′∈Ik′∑ν′=1N(k′,τ′)μ(Qτ′k′,ν′)(S0(I−S0)D˜k′)(x,yτ′k′,ν′)Dk′(f)(yτ′k′,ν′)|≲∑k′∈ℤ∑τ′∈Ik′∑ν′=1N(k′,τ′)2−|k′|ϵμ(Qτ′k′,ν′)1V2−(0⋀k′)(x)+V2−(0⋀k′)(yτ′k′,ν′)+V(x,yτ′k′,ν′)×(2−(0⋀k′)2−(0⋀k′)+d(x,yτ′k′,ν′))ϵ|Dk′(f)(yτ′k′,ν′)|. From Proposition 6.3, we deduce
the following trivial estimate that for all k′∈ℤ, (6.126){∑τ′∈Ik′∑ν′=1N(k′,τ′)2k′sq|Dk′(f)(yτ′k′,ν′)|q}1/q≲∥f∥F˙∞,qs(𝒳), and from Lemma 2.19, it follows
that (6.127)μ(Qτ′k′,ν′)V2−(0⋀k′)(x)+V2−(0⋀k′)(yτ′k′,ν′)+V(x,yτ′k′,ν′)≲1. Both estimates together with
(5.5) show that when q≤1,
for any x∈𝒳, (6.128)|S0(I−S0)(f)(x)|≲∑k′∈ℤ2−|k′|ϵ2−k′s{∑τ′∈Ik′∑ν′=1N(k′,τ′)2k′sq|Dk′(f)(yτ′k′,ν′)|q}1/q≲∥f∥F˙∞,qs(𝒳), while when 1<q≤∞,
both estimates together with Hölder's inequality, Lemma 5.2, and Lemma 2.1(ii)
still yield that for all x∈𝒳, (6.129)|S0(I−S0)(f)(x)|≲∑k′∈ℤ2−|k′|ϵ2−k′s[∑τ′∈Ik′∑ν′=1N(k′,τ′)μ(Qτ′k′,ν′)1V2−(0⋀k′)(x)+V2−(0⋀k′)(yτ′k′,ν′)+V(x,yτ′k′,ν′)×(2−(0⋀k′)2−(0⋀k′)+d(x,yτ′k′,ν′))ϵ]1/q′[∑τ′∈Ik′∑ν′=1N(k′,τ′)2k′sq|Dk′(f)(yτ′k′,ν′)|q]1/q≲∥f∥F˙∞,qs(𝒳). Thus, for τ∈I0 and ν=1,…,N(0,τ), (6.130)mQτ0,ν(|S0(I−S0)(f)|)=1μ(Qτ0,ν)∫Qτ0,ν|S0(I−S0)(f)(x)|dμ(x)≲∥f∥F˙∞,qs(𝒳),which is the desired estimate.
Forl∈ℕ and α∈Il,
from Theorem 5.16 together with (6.124), it follows that  (6.131)1μ(Qαl)∫Qαl∑k=l∞2ksq|Dk(I−S0)(f)(x)|qdμ(x)≲1μ(Qαl)∫Qαl∑k=l∞2ksq[∑k′∈ℤ∑τ′∈Ik′∑ν′=1N(k′,τ′)2−|k−k′|ϵμ(Qτ′k′,ν′)1V2−(k⋀k′)(x)+V2−(k⋀k′)(yτ′k′,ν′)+V(x,yτ′k′,ν′)×(2−(k⋀k′)2−(k⋀k′)+d(x,yτ′k′,ν′))ϵ|Dk′(f)(yτ′k′,ν′)|]qdμ(x). Then, an argument similar to
that used to estimate Y1+Y2 in the proof of Proposition 6.3 together with
Proposition 6.3 yields that (6.132)1μ(Qαl)∫Qαl∑k=l∞2ksq|Dk(I−S0)(f)(x)|qdμ(x)≲∥f∥F˙∞,qs(𝒳)q. Thus, (I−S0)(f)∈F∞,qs(𝒳) and (6.133)∥(I−S0)(f)∥F∞,qs(𝒳)≲∥f∥F˙∞,qs(𝒳), which completes the proof of
Proposition 6.25.Remark 6.26.
We point out that Remark5.38 applies in a similar way to Proposition
6.25.To establish the equivalence betweenbmo(𝒳) and
F∞,20(𝒳),
we need the following technical result which in fact gives a new
characterization of Triebel-Lizorkin space F∞,qs(𝒳) when 1≤q≤∞.
We first introduce an inhomogeneous maximal function.For anyx∈𝒳 and l∈ℤ+,
we choose m∈ℕ by a way similar to that in the definition of ℭ˙qs.
Let {Dk}k∈ℤ+
be as in Definition 6.19. Then for any s∈ℝ, q∈(0,∞], f∈(𝒢(β,γ))′ with 0<β,γ<ϵ and x∈𝒳,
we define the inhomogeneous maximal
functionℭqs(f)(x) by (6.134)ℭqs(f)(x)=supl∈ℤ+[1μ(∪i=1mQαlil)∫∪i=1mQαlil∑k=l∞2ksq|Dk(f)(y)|qdμ(y)]1/q, where the usual modification is
made when q=∞.Using Proposition6.22 and an argument similar to the
proof of Proposition 6.10 yields the following characterization of F∞,qs(𝒳) with 1≤q≤∞;
we omit the details.Proposition 6.27.
Letϵ and s be as in Definition 6.7 and let 1≤q≤∞.
Then f∈F∞,qs(𝒳) if and only if f∈(𝒢(β,γ))′ with |s|<β<ϵ and 0<γ<ϵ,
and ℭqs(f)∈L∞(𝒳).
Moreover, in this case, (6.135)∥f∥F∞,qs(𝒳)~∥ℭqs(f)∥L∞(𝒳).Now, from Theorem5.44(i), Proposition 6.22, and
Proposition 6.27, we can deduce the following relation between bmo(𝒳) and F∞,20(𝒳).Theorem 6.28.
bmo(𝒳)=F∞,20(𝒳),
with equivalent norms.Proof.
Letf∈bmo(𝒳) and {Dk}k∈ℤ+ be as in Definition 6.19. Proposition 6.21(v)
and Theorem 5.44(i) immediately imply that f∈(𝒢0ϵ(β,γ))′ with 0<β,γ<ϵ.
Let now Qαl for l∈ℤ+ and α∈Il be a dyadic cube as in Lemma 2.19. Let Bαl be as in the proof of Theorem 6.11(ii). We
then decompose f=f1+f2+f3 in the same way as in the proof of Theorem 6.11(ii). The estimations for f1 and f2 are as in the proof of Theorem 6.11(ii) by
replacing Proposition 3.15 by Proposition 3.30. If l∈ℕ,
then (6.136){1μ(Qαl)∫Qαl∑k=l∞|Dk(f3)(x)|dμ(x)}1/2=0, while when l=0,
we then have (6.137){1μ(Qαl)∫Qαl∑k=l∞|Dk(f3)(x)|dμ(x)}1/2=|f3|≤∥f∥bmo(𝒳). Thus, f∈F∞,20(𝒳) and
∥f∥F∞,20(𝒳)≲∥f∥bmo(𝒳).
Conversely, using  Theorem5.44(i), Proposition 6.22,
and Proposition 6.27 together with an argument similar to the proof of Theorem
6.11(ii), we can prove that if f∈F∞,20(𝒳),
then f∈bmo(𝒳) and ∥f∥bmo(𝒳)≲∥f∥F∞,20(𝒳), which completes the proof of Theorem 6.28.We end this subsection by considering the boundedness onF∞,qs(𝒳) of singular integrals of order (ϵ,σ).Theorem 6.29.
Letϵ, s, and q be as in Definition 6.19. Let σ>0 and let T be a singular integral of order (ϵ,σ).
Then T is bounded from 𝒢b(β,γ) with 0<β≤ϵ and γ>0 to F∞,qs(𝒳).
Moreover, there exists a constant C>0 such that for all f∈𝒢b(β,γ) with 0<β≤ϵ and γ>0, (6.138)∥Tf∥F∞,qs(𝒳)≤C∥f∥F∞,qs(𝒳).Proof.
Combining some estimates and technics used in the proof of Proposition5.54 with those used in the proof of Proposition 6.15 gives the desired
conclusions. The details are left to the reader.Remark 6.30.
By Theorem8.18 below, if p(s,ϵ)<q≤∞,
then F∞,qs(𝒳) is the dual space of F1,q′−s(𝒳).
In this case, Theorem 6.29 can be deduced from Theorem 5.48 together with a
duality argument, which provides another proof of Theorem 6.29.


## 7. Frame Characterizations

In this section, using the discrete Calderón
reproducing formulae, we establish a frame characterization of Besov spaces and
Triebel-Lizorkin spaces.


### 7.1. Frame Characterization ofB˙p,qs(𝒳) and F˙p,qs(𝒳)

In this subsection, we assume thatμ(𝒳)=∞.
We first introduce some spaces of sequences, b˙p,qs(𝒳) and f˙p,qs(𝒳).Let(7.1)λ={λτk,ν:k∈ℤ,τ∈Ik,ν=1,…,N(k,τ)} be a sequence of complex
numbers. The spaceb˙p,qs(𝒳) with s∈ℝ and 0<p,q≤∞ is the set of all λ as in (7.1) such that  (7.2)∥λ∥b˙p,qs(𝒳)={∑k=−∞∞2ksq[∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)|λτk,ν|p]q/p}1/q<∞, and the spacef˙p,qs(𝒳) with s∈ℝ, 0<p<∞, and 0<q≤∞ is the set of all λ as in (7.1) such that  (7.3)∥λ∥f˙p,qs(𝒳)=∥{∑k=−∞∞∑τ∈Ik∑ν=1N(k,τ)2ksq|λτk,ν|qχQτk,ν}1/q∥Lp(𝒳)<∞. Moreover, the spacef˙∞,qs(𝒳) with s∈ℝ and 0<q≤∞ is the set of all λ as in (7.1) such that (7.4)∥λ∥f˙∞,qs(𝒳)=supl∈ℤsupα∈Il{1μ(Qαl)[∑k=l∞∑τ∈Ik∑ν=1N(k,τ)2ksqμ(Qτk,ν)|λτk,ν|qχ{(τ,ν):Qτk,ν⊂Qαl}(τ,ν)]}1/q<∞, where {Qαl}l∈ℤ,α∈Il are dyadic cubes as in Lemma 2.19.Proposition 7.1.
Letϵ be as in Definition 5.8, let |s|<ϵ,
and let p(s,ϵ)<p≤∞. Let λ be a sequence of numbers as in (7.1) and all
the other notation as in Theorem 4.11. Then, the following
hold.
(i)
If0<q≤∞ and ∥λ∥b˙p,qs(𝒳)<∞, then the series (7.5)∑k=−∞∞∑τ∈Ik∑ν=1N(k,τ)λτk,νμ(Qτk,ν)D˜k(x,yτk,ν) converges to some f∈B˙p,qs(𝒳) both in the norm of B˙p,qs(𝒳) and in (𝒢°0ϵ(β,γ))′ with (7.6)max{0,−s+n(1p−1)+}<β<ϵ,max{n(1p−1)+,s−κp}<γ<ϵ when p,q<∞ and only in (𝒢°0ϵ(β,γ))′ with β and γ as in (7.6) when max(p,q)=∞.
Moreover, in all cases, (7.7)∥f∥B˙p,qs(𝒳)≤C∥λ∥b˙p,qs(𝒳).(ii)
Ifp(s,ϵ)<q≤∞ and ∥λ∥f˙p,qs(𝒳)<∞, then the series in (7.5) converges to some f∈F˙p,qs(𝒳) both in the norm of F˙p,qs(𝒳) and in (𝒢°0ϵ(β,γ))′ with β,γ as in (7.6) when p,q<∞ and only in (𝒢°0ϵ(β,γ))′ with β and γ as in (7.6) when max(p,q)=∞.
Moreover, in all cases, (7.8)∥f∥F˙p,qs(𝒳)≤C∥λ∥f˙p,qs(𝒳).Proof.
We first verify that the series in (7.5) converges in (𝒢°0ϵ(β,γ))′ with β and γ as in (7.6) if λ∈b˙p,qs(𝒳) with s,p,q as in (i). By Lemma 2.19 and the definition of N(k,τ) together with (2.59) in [85, page 1385], we know that
for all k∈ℤ and τ∈Ik, N(k,τ) is a finite set. Without loss of generality,
since μ(𝒳)=∞,
we may assume that Ik=ℕ for all k∈ℤ.
With this assumption, for L∈ℕ,
we define (7.9)fL(x)=∑k=−LL∑τ=1L∑ν=1N(k,τ)μ(Qτk,ν)λτk,νD˜k(x,yτk,ν). Then fL∈𝒢°(ϵ′,ϵ′) with ϵ′∈(0,ϵ1⋀ϵ2),
and fL∈(𝒢°0ϵ(β,γ))′ with any β,γ∈(0,ϵ). For any ψ∈𝒢°(β,γ) with β,γ as in (7.6), L1,L2∈ℕ and L1<L2,
we have (7.10)|〈fL2−fL1,ψ〉|≤∑k=−L2−L1−1∑τ=1L1∑ν=1N(k,τ)μ(Qτk,ν)|λτk,ν||〈D˜k(⋅,yτk,ν),ψ〉|+∑k=L1+1L2∑τ=1L1∑ν=1N(k,τ)μ(Qτk,ν)|λτk,ν||〈D˜k(⋅,yτk,ν),ψ〉|+∑k=−L2L2∑τ=L1+1L2∑ν=1N(k,τ)μ(Qτk,ν)|λτk,ν||〈D˜k(⋅,yτk,ν),ψ〉|≡Z1+Z2+Z3.
Let us now consider two cases, respectively. We first
consider the casep≤1.
In this case, letting γ′∈(max{0,s−κ/p},γ),
by (5.25) and γ>n(1/p−1) together with (5.31) and Hölder's inequality,
we have (7.11)Z1≲∥ψ∥𝒢(β,γ)∑k=−L2−L1−1∑τ=1L1∑ν=1N(k,τ)2kγ′μ(Qτk,ν)|λτk,ν|1V2−k(x1)+V(x1,yτk,ν)2−kγ(2−k+d(x1,yτk,ν))γ≲∥ψ∥𝒢(β,γ)∑k=−L2−L1−12k(γ′−s+κ/p)[∑τ=1L1∑ν=1N(k,τ)2kspμ(Qτk,ν)|λτk,ν|p]1/p. If q≤1,
by (5.5), we further obtain (7.12)Z1≲∥ψ∥𝒢(β,γ){∑k=−L2−L1−12k(γ′−s+κ/p)q[∑τ=1L1∑ν=1N(k,τ)2kspμ(Qτk,ν)|λτk,ν|p]q/p}1/q≲∥ψ∥𝒢(β,γ){∑k=−L2−L1−1[∑τ=1L1∑ν=1N(k,τ)2kspμ(Qτk,ν)|λτk,ν|p]q/p}1/q, while if 1<q≤∞,
by Hölder's inequality, we have (7.13)Z1≲∥ψ∥𝒢(β,γ){∑k=−L2−L1−1[∑τ=1L1∑ν=1N(k,τ)2kspμ(Qτk,ν)|λτk,ν|p]q/p}1/q{∑k=−L2−L1−12k(γ′−s+κ/p)q′}1/q′≲∥ψ∥𝒢(β,γ)2−L1(γ′−s+κ/p){∑k=−L2−L1−1[∑τ=1L1∑ν=1N(k,τ)2kspμ(Qτk,ν)|λτk,ν|p]q/p}1/q. Thus, (7.14)limL1,L2→∞Z1=0.
Replacing the estimates (5.25) and (5.31) respectively
by the estimates (5.24) and (5.30), and using some similar computations to the
estimate for Z1,
we obtain (7.15)Z2≲∥ψ∥𝒢(β,γ)∑k=L1+1L2∑τ=1L1∑ν=1N(k,τ)2−kβμ(Qτk,ν)|λτk,ν|1V1(x1)+V(x1,yτk,ν)1(1+d(x1,yτk,ν))γ≲∥ψ∥𝒢(β,γ)∑k=L1+1L22−k[s+β−d(1/p−1)][∑τ=1L1∑ν=1N(k,τ)2kspμ(Qτk,ν)|λτk,ν|p]1/p≲∥ψ∥𝒢(β,γ){{∑k=L1+1L2[∑τ=1L1∑ν=1N(k,τ)2kspμ(Qτk,ν)|λτk,ν|p]q/p}1/q,q≤12−L1[s+β−d(1/p−1)]{∑k=L1+1L2[∑τ=1L1∑ν=1N(k,τ)2kspμ(Qτk,ν)|λτk,ν|p]q/p}1/q,1<q≤∞. where we used the assumption
that β>−s+d(1/p−1).
Thus, (7.16)limL1,L2→∞Z2=0.
The estimates (5.24) and (5.25) together with (5.5),
(5.30), and (5.31) further yield that (7.17)Z3≲∥ψ∥𝒢(β,γ){∑k=0L2∑τ=L1+1L2∑ν=1N(k,τ)2−kβμ(Qτk,ν)|λτk,ν|1V1(x1)+V(x1,yτk,ν)1(1+d(x1,yτk,ν))γ+∑k=−L2−1∑τ=L1+1L2∑ν=1N(k,τ)2kγ′μ(Qτk,ν)|λτk,ν|1V2−k(x1)+V(x1,yτk,ν)2−kγ(2−k+d(x1,yτk,ν))γ}≲∥ψ∥𝒢(β,γ){∑k=0L22−k[β+s−d(1/p−1)]+∑k=−L2−12k(γ′−s+κ/p)}[∑τ=L1+1L2∑ν=1N(k,τ)2kspμ(Qτk,ν)|λτk,ν|p]1/p,
where we chose γ′>0 as in the estimate for Z1.
For any given δ>0,
since γ′>s−κ/p and β>d(1/p−1)−s,
we can fix L20∈ℕ such that (7.18)∑k=L20+1∞2−k[β+s−d(1/p−1)]+∑k=−∞−L20−12k(γ′−s+κ/p)<δ. Since for all k∈ℤ, (7.19)[∑τ=L1+1L2∑ν=1N(k,τ)2kspμ(Qτk,ν)|λτk,ν|p]1/p≤[∑τ∈Ik∑ν=1N(k,τ)2kspμ(Qτk,ν)|λτk,ν|p]1/p<∥λ∥b˙p,qs(𝒳), we can choose N∈ℕ such that if L1>N,
then (7.20)[∑τ=L1+1L2∑ν=1N(k,τ)2kspμ(Qτk,ν)|λτk,ν|p]1/p<δ for all k=−L20,−L20+1,…,L20.
From (7.18), (7.19), and (7.20), it follows that if L2>L1>N,
then (7.21)Z3≲∥ψ∥𝒢(β,γ){∑k=0L202−k[β+s−d(1/p−1)]+∑k=−L20−12k(γ′−s+κ/p)}δ+C∥ψ∥𝒢(β,γ)∥λ∥b˙p,qs(𝒳){∑k=L20+1∞2−k[β+s−d(1/p−1)]+∑k=−∞−L20−12k(γ′−s+κ/p)}≲∥ψ∥𝒢(β,γ)δ, which just means that (7.22)limL1,L2→∞Z3=0.
We now consider the case1<p≤∞.
Replacing (5.5) by Hölder's inequality and using Lemma 5.2, similarly to the
estimate for the case p≤1,
we obtain (7.23)Z1≲∥ψ∥𝒢(β,γ)∑k=−L2−L1−12k(γ′−s)[∑τ=1L1∑ν=1N(k,τ)2kspμ(Qτk,ν)|λτk,ν|p×1V2−k(x1)+V(x1,yτk,ν)2−kγ(2−k+d(x1,yτk,ν))γ]1/p×[∫X1V2−k(x1)+V(x1,y)2−kγ(2−k+d(x1,y))γdμ(y)]1/p′≲∥ψ∥𝒢(β,γ)∑k=−L2−L1−12k(γ′−s+κ/p)[∑τ=1L1∑ν=1N(k,τ)2kspμ(Qτk,ν)|λτk,ν|p]1/p. Then repeating the proof of the
case p≤1 yields that (7.24)limL1,L2→∞Z1=0.
Similarly, forZ2,
Hölder's inequality and Lemma 5.2 imply that (7.25)Z2≲∥ψ∥𝒢(β,γ)2−k(s+β)[∑k=L1+1L2∑τ=1L1∑ν=1N(k,τ)2kspμ(Qτk,ν)|λτk,ν|p]1/p×[∫X1V1(x1)+V(x1,y)1(1+d(x1,y))γdμ(y)]1/p′≲∥ψ∥𝒢(β,γ)∑k=L1+1L22−k(s+β)[∑τ=1L1∑ν=1N(k,τ)2kspμ(Qτk,ν)|λτk,ν|p]1/p. Using the fact that β>−s and repeating the proof of the case p≤1 show that (7.26)limL1,L2→∞Z2=0.
To estimateZ3,
let (7.27)𝒳L1L2,k=⋃τ=L1+1L2⋃ν=1N(k,τ)Qτk,ν. The Hölder inequality shows
that (7.28)Z3≲∥ψ∥𝒢(β,γ){∑k=0L22−k(β+s)[∫𝒳L1L2,k1V1(x1)+V(x1,y)1(1+d(x1,y))γdμ(y)]1/p′+∑k=−L2−12k(γ′−s+κ/p)[∫𝒳L1L2,k1V2−k(x1)+V(x1,y)2−kγ(2−kγ+d(x1,y))γdμ(y)]1/p′}×[∑τ=L1+1L2∑ν=1N(k,τ)2kspμ(Qτk,ν)|λτk,ν|p]1/p, where we chose γ′∈(max{0,s−κ/p},γ).
If p∈(1,∞),
using the facts that γ′>s−κ/p and β>−s,
(7.19), (7.20), (7.29)∫𝒳1V1(x1)+V(x1,y)1(1+d(x1,y))γdμ(y)≲1,∫𝒳1V2−k(x1)+V(x1,y)2−kγ(2−kγ+d(x1,y))γdμ(y)≲1, and repeating the argument for
the case p≤1,
we can verify that (7.30)limL1,L2→∞Z3=0. If p=∞,
replacing (7.20) by (7.31)limL1,L2→∞∫𝒳L1L2,k1V1(x1)+V(x1,y)1(1+d(x1,y))γdμ(y)=0,limL1,L2→∞∫𝒳L1L2,k1V2−k(x1)+V(x1,y)2−kγ(2−kγ+d(x1,y))γdμ(y)=0 for any given k∈ℤ,
by an argument similar to the case p≤1,
we still obtain that (7.32)limL1,L2→∞Z3=0.
Thus, for any giveψ∈𝒢(β,γ), {〈fL,ψ〉}L∈ℕ
is a Cauchy sequence, which means that the
series in (7.5) converges to some f∈(𝒢°0ϵ(β,γ))′ with β,γ as in (7.6) if λ∈b˙p,qs(𝒳) with s,p,q as in the theorem.
Ifλ∈f˙p,qs(𝒳), by the proved fact on b˙p,qs(𝒳) and (7.33)b˙p,min(p,q)s(𝒳)⊂f˙p,qs(𝒳)⊂b˙p,max(p,q)s(𝒳) (see [87, Proposition 2.3]), we also
obtain that the series in (7.5) converges in (𝒢°0ϵ(β,γ))′ with β and γ as in (7.6).
Let us now verify that the series in (7.5) converges
in the norm of B˙p,qs(𝒳) or F˙p,qs(𝒳) when p,q<∞ if λ∈b˙p,qs(𝒳) or λ∈f˙p,qs(𝒳), respectively. To this end, let f be the series in (7.5). For L∈ℕ,
in (𝒢°0ϵ(β,γ))′ with β and γ as in (7.6), we then have (7.34)f−fL=∑k=L+1∞∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)λτk,νD˜k(x,yτk,ν)+∑k=−∞−L−1∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)λτk,νD˜k(x,yτk,ν)+∑k=−LL∑τ=L+1∞∑ν=1N(k,τ)μ(Qτk,ν)λτk,νD˜k(x,yτk,ν). Replacing Qk′(f)(yτ′k′,ν′) in the proof of Proposition 5.4 by λτk,ν here and repeating the proof of Proposition
5.4, we can verify that  (7.35)∥f−fL∥B˙p,qs(𝒳)≲{∑k=L+1∞2ksq[∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)|λτk,ν|p]q/p}1/q+{∑k=−∞−L−12ksq[∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)|λτk,ν|p]q/p}1/q+{∑k=−LL2ksq[∑τ=L+1∞∑ν=1N(k,τ)μ(Qτk,ν)|λτk,ν|p]q/p}1/q,∥f−fL∥F˙p,qs(𝒳)≲∥{∑k=L+1∞∑τ∈Ik∑ν=1N(k,τ)2ksq|λτk,ν|qχQτk,ν}1/q∥Lp(𝒳)+∥{∑k=−∞−L−1∑τ∈Ik∑ν=1N(k,τ)2ksq|λτk,ν|qχQτk,ν}1/q∥Lp(𝒳)+∥{∑k=−LL∑τ=L+1∞∑ν=1N(k,τ)2ksq|λτk,ν|qχQτk,ν}1/q∥Lp(𝒳). Thus, from Lebesgue's dominated
convergence theorems on the integral and the series, it is easy to deduce that (7.36)∥f−fL∥B˙p,qs(𝒳)→0,∥f−fL∥F˙p,qs(𝒳)→0 as L→∞.
Moreover, by Proposition 5.10(iv), we know that fL∈B˙p,qs(𝒳) and fL∈F˙p,qs(𝒳) if we choose ϵ′ large enough. Thus, f∈B˙p,qs(𝒳) if λ∈b˙p,qs(𝒳) and f∈F˙p,qs(𝒳) if λ∈f˙p,qs(𝒳) when p,q<∞.
The same arguments as in the proof of Propositions5.4
and 6.3 for the space B˙p,qs(𝒳) and the space F˙p,qs(𝒳) with all p,q as in the assumption of the theorem yield
(7.7) and (7.8), respectively, which completes the proof of Proposition
7.1.From Theorem4.13, Proposition 7.1, and the
Plancherel-Pôlya inequalities, Propositions 5.4 and 6.3, we obtain
the following frame characterizations of the spaces B˙p,qs(𝒳) and F˙p,qs(𝒳).Theorem 7.2.
Letϵ be as in Definition 5.8, let |s|<ϵ, and let p(s,ϵ)<p≤∞. Let all the other notation be as in Theorem
4.11 and λτk,ν=Dk(f)(yτk,ν) for k∈ℤ, τ∈Ik and ν=1,…,N(k,τ),
where yτk,ν is any fixed element in Qτk,ν.
Then, the following hold.
(i)
If0<q≤∞,
then f∈B˙p,qs(𝒳), if and only if f∈(𝒢°0ϵ(β,γ))′ for some β,γ as in (5.35), (7.37)f(x)=∑k=−∞∞∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)Dk(f)(yτk,ν)D˜k(x,yτk,ν), holds in (𝒢°0ϵ(β,γ))′ and λ∈b˙p,qs(𝒳). Moreover, in this case, (7.38)∥f∥B˙p,qs(𝒳)~∥λ∥b˙p,qs(𝒳).(ii)
Ifp(s,ϵ)<q≤∞,
then f∈F˙p,qs(𝒳) if and only if f∈(𝒢°0ϵ(β,γ))′ for some β,γ as in (5.35), (7.37), holds in (𝒢°0ϵ(β,γ))′ and λ∈f˙p,qs(𝒳). Moreover, in this case, (7.39)∥f∥F˙p,qs(𝒳)~∥λ∥f˙p,qs(𝒳).


### 7.2. Frame Characterization ofBp,qs(𝒳) and Fp,qs(𝒳)

Again, in this subsection,μ(𝒳) can be finite or infinite. We also first
introduce some spaces of sequences, bp,qs(𝒳) and fp,qs(𝒳).Let(7.40)λ={λτk,ν:k∈ℤ+,τ∈Ik,ν=1,…,N(k,τ)} be a sequence of complex
numbers. The spacebp,qs(𝒳) with s∈ℝ and 0<p,q≤∞ is the set of all λ as in (7.40) such that (7.41)∥λ∥bp,qs(𝒳)={∑k=0∞2ksq[∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)|λτk,ν|p]q/p}1/q<∞, and the spacefp,qs(𝒳) with s∈ℝ, 0<p<∞, and 0<q≤∞ is the set of all λ as in (7.40) such that  (7.42)∥λ∥fp,qs(𝒳)=∥{∑k=0∞∑τ∈Ik∑ν=1N(k,τ)2ksq|λτk,ν|qχQτk,ν}1/q∥Lp(𝒳)<∞. Moreover, the spacef∞,qs(𝒳) with s∈ℝ and 0<q≤∞ is the set of all λ as in (7.40) such that (7.43)∥λ∥f∞,qs(𝒳)=max[supτ∈I0ν=1,…,N(0,τ)|λτ0,ν|,supl∈ℕsupα∈Il{1μ(Qαl)[∑k=l∞∑τ∈Ik∑ν=1N(k,τ)2ksqμ(Qτk,ν)|λτk,ν|qχ{(τ,ν):Qτk,ν⊂Qαl}(τ,ν)]}1/q]<∞, where {Qαl}l∈ℕ,α∈Il are dyadic cubes as in Lemma 2.19.Proposition 7.3.
Letϵ be as in Definition 5.29, let |s|<ϵ,
and let p(s,ϵ)<p≤∞. Let λ be a sequence of numbers as in (7.40) and all
the other notation as in Theorem 4.14. Then, the following hold.
(i)
If0<q≤∞ and ∥λ∥bp,qs(𝒳)<∞, then the series (7.44)∑τ∈I0∑ν=1N(0,τ)λτ0,ν∫Qτ0,νD˜0(x,y)dμ(y)+∑k=1∞∑τ∈Ik∑ν=1N(k,τ)λτk,νμ(Qτk,ν)D˜k(x,yτk,ν) converges to some f∈Bp,qs(𝒳) both in the norm of Bp,qs(𝒳) and in (𝒢0ϵ(β,γ))′ with (7.45)max{0,−s+n(1p−1)+}<β<ϵ,n(1p−1)+<γ<ϵ, when p,q<∞ and only in (𝒢0ϵ(β,γ))′ with β and γ as in (7.45) when max(p,q)=∞.
Moreover, in all cases, (7.46)∥f∥Bp,qs(𝒳)≤C∥λ∥bp,qs(𝒳).(ii)
Ifp(s,ϵ)<q≤∞ and ∥λ∥fp,qs(𝒳)<∞, then the series in (7.44) converges to some f∈Fp,qs(𝒳) both in the norm of Fp,qs(𝒳) and in (𝒢0ϵ(β,γ))′ with β,γ as in (7.45) when p,q<∞ and only in (𝒢0ϵ(β,γ))′ with β and γ as in (7.45) when max(p,q)=∞.
Moreover, in all cases, (7.47)∥f∥Fp,qs(𝒳)≤C∥λ∥fp,qs(𝒳).Proof.
In what follows, for simplicity,  we set(7.48)D˜Qτ0,ν(x)=1μ(Qτ0,ν)∫Qτ0,νD˜0(x,y)dμ(y). Let us first show that the
series in (7.44) converges in (𝒢(β,γ))′ with β and γ as in (7.45). As in the proof of Proposition
7.1, we know that for all k∈ℤ+ and τ∈Ik, N(k,τ) is a finite set. Let us suppose Ik=ℕ for all k∈ℤ+;
the other cases are easier. With this assumption, for L∈ℕ,
we define (7.49)fL(x)=∑τ=1L∑ν=1N(0,τ)μ(Qτ0,ν)λτ0,νD˜Qτ0,ν(x)+∑k=1L∑τ=1L∑ν=1N(k,τ)μ(Qτk,ν)λτk,νD˜k(x,yτk,ν). Then fL∈𝒢(ϵ,ϵ) and fL∈(𝒢0ϵ(β,γ))′ with any β,γ∈(0,ϵ),
where ϵ can be any positive number in (0,ϵ1⋀ϵ2).
In what follows, we choose ϵ>max(β,γ) such that p>p(s,ϵ) for the spaces bp,qs(𝒳), and p,q>p(s,ϵ) for the spaces fp,qs(𝒳).
For anyψ∈𝒢(β,γ) with (β,γ) as in (7.45), L1,L2∈ℕ and L1<L2,
we have (7.50)|〈fL2−fL1,ψ〉|≤∑τ=L1+1L2∑ν=1N(0,τ)μ(Qτ0,ν)|λτ0,ν||〈D˜Qτ0,ν,ψ〉|+∑k=L1+1L2∑τ=1L2∑ν=1N(k,τ)μ(Qτk,ν)|λτk,ν||〈D˜k(⋅,yτk,ν),ψ〉|+∑k=1L1∑τ=L1+1L2∑ν=1N(k,τ)μ(Qτk,ν)|λτk,ν||〈D˜k(⋅,yτk,ν),ψ〉|≡Y1+Y2+Y3.
From (5.105), (5.5) when p≤1 or Hölder's inequality when 1<p≤∞,
and γ>n(1/p−1)+ together with (5.30), it follows that when
p≤1, (7.51)|Y1|≲∥ψ∥𝒢(β,γ){∑τ=L1+1L2∑ν=1N(0,τ)μ(Qτ0,ν)|λτ0,ν|p}1/p, while when
1<p≤∞,(7.52)|Y1|≲∥ψ∥𝒢(β,γ){∑τ=L1+1L2∑ν=1N(0,τ)μ(Qτ0,ν)|λτ0,ν|p}1/p×{∫𝒳L1L2,01V1(x1)+V(x1,y)1(1+d(x1,y))γdμ(y)}1/p′, where 𝒳L1L2,0 is as in the proof of Proposition 7.1.
ForY2,
by (5.105), (5.5) when p≤1 or Hölder's inequality when 1<p≤∞, γ>n(1/p−1)+ together with (5.30),  and Lemma 2.1(ii), we
obtain (7.53)|Y2|≲∥ψ∥𝒢(β,γ){∑k=L1+1L22−k[β+s−n(1/p−1)]2ks[∑τ=1L2∑ν=1N(k,τ)μ(Qτk,ν)|λτk,ν|p]1/p,p≤1∑k=L1+1L22−k(β+s)2ks[∑τ=1L2∑ν=1N(k,τ)μ(Qτk,ν)|λτk,ν|p]1/p,1<p≤∞. From this and (5.5) when q≤1 or Hölder's  inequality when 1<q≤∞ again, it further follows that  when p≤1 and q≤1, (7.54)|Y2|≲∥ψ∥𝒢(β,γ){∑k=L1+1L22ksq[∑τ=1L2∑ν=1N(k,τ)μ(Qτk,ν)|λτk,ν|p]q/p}1/q, while  when 1<q≤∞, (7.55)|Y2|≲∥ψ∥𝒢(β,γ){∑k=L1+1L22ksq[∑τ=1L2∑ν=1N(k,τ)μ(Qτk,ν)|λτk,ν|p]q/p}1/q{∑k=L1+1L22−k[β+s−n(1/p−1)]q′}1/q′, and that  when 1<p≤∞, (7.56)|Y2|≲∥ψ∥𝒢(β,γ){{∑k=L1+1L22ksq[∑τ=1L2∑ν=1N(k,τ)μ(Qτk,ν)|λτk,ν|p]q/p}1/q,q≤1{∑k=L1+1L22ksq[∑τ=1L2∑ν=1N(k,τ)μ(Qτk,ν)|λτk,ν|p]q/p}1/q{∑k=L1+1L22−k(β+s)q′}1/q′,1<q≤∞. where we used the fact that β>max(0,n(1/p−1)+−s).
Similarly, by (5.105), (5.5) when p≤1 or Hölder's inequality when 1<p≤∞,
and γ>n(1/p−1)+ together  with (5.30), we have (7.57)|Y3|≲∥ψ∥𝒢(β,γ){∑k=1L12−k[β−n(1/p−1)][∑τ=L1+1L2∑ν=1N(k,τ)μ(Qτk,ν)|λτk,ν|p]1/p,p≤1∑k=1L12−kβ[∑τ=L1+1L2∑ν=1N(k,τ)μ(Qτk,ν)|λτk,ν|p]1/p×{∫𝒳L1L2,k1V1(x1)+V(x1,y)1(1+d(x1,y))γdμ(y)}1/p′,1<p≤∞, where 𝒳L1L2,k is as in the proof  of Proposition 7.1.
From this and (5.5) when q≤1 or Hölder's inequality when 1<q≤∞,
we further deduce that when p≤1, (7.58)|Y3|≲∥ψ∥𝒢(β,γ){∑k=1L12ksq[∑τ=L1+1L2∑ν=1N(k,τ)μ(Qτk,ν)|λτk,ν|p]q/p}1/q, and that when 1<p≤∞ and q≤1, (7.59)|Y3|≲∥ψ∥𝒢(β,γ){∑k=1L12ksq[∑τ=L1+1L2∑ν=1N(k,τ)μ(Qτk,ν)|λτk,ν|p]q/p}1/q, or that when 1<p≤∞ and 1<q≤∞, (7.60)|Y3|≲∥ψ∥𝒢(β,γ){∑k=1L12ksq[∑τ=L1+1L2∑ν=1N(k,τ)μ(Qτk,ν)|λτk,ν|p]q/p}1/q×{∑k=1L22−k(β+s)q′[∫𝒳L1L2,k1V1(x1)+V(x1,y)1(1+d(x1,y))γdμ(y)]q′/p′}1/q′, where we used the fact that β>max(0,n(1/p−1)+−s).
Combining (7.51) through (7.60), by λ∈bp,qs(𝒳), (7.61)∫𝒳1V1(x1)+V(x1,y)1(1+d(x1,y))γdμ(y)<∞ when p=∞,
and (7.62)∑k=1∞2−k[β+s−n(1/p−1)]<∞ when p≤1 and q=∞,
or (7.63)∑k=1∞2−k(β+s)<∞ when 1<p≤∞ and q=∞,
it is easy to see that
{〈fL,ψ〉}L∈ℕ is a Cauchy sequence. This just means that the
series in (7.44) converges to some f∈(𝒢0ϵ(β,γ))′ with β,γ satisfying (7.45) if λ∈bp,qs(𝒳) with s,p,q as in the theorem. If λ∈fp,qs(𝒳) with s,p,q as in the theorem, by this fact and (7.64)bp,min(p,q)s(𝒳)⊂fp,qs(𝒳)⊂bp,max(p,q)s(𝒳)(see [87, Proposition 2.3]), we also
obtain that the series in (7.44) converges in (𝒢0ϵ(β,γ))′ with β and γ as in (7.45).
Let us now verify that the series in (7.44) converges
in the norm of
Bp,qs(𝒳) when p,q<∞,
if λ∈bp,qs(𝒳). Let f be the series in (7.44). We estimate the norm
in Bp,qs(𝒳) of f−fL by writing (7.65)f−fL=∑τ=L+1∞∑ν=1N(0,τ)μ(Qτ0,ν)λτ0,νD˜Qτ0,ν(x)+∑k=1∞∑τ=L+1∞∑ν=1N(k,τ)μ(Qτk,ν)λτk,νD˜k(x,yτk,ν)+∑k=L+1∞∑τ=1L∑ν=1N(k,τ)μ(Qτk,ν)λτk,νD˜k(x,yτk,ν). Replacing Qτ′,10,ν′(f) and Qk′(f)(yτ′k′,ν′) in the proof of Proposition 5.25 respectively
by λτ0,ν and λτk,ν here, and repeating the proof of Proposition
5.25, we then obtain  (7.66)∥f−fL∥Bp,qs(𝒳)≲{∑τ=L+1∞∑ν=1N(0,τ)μ(Qτ0,ν)|λτ0,ν|p}1/p+{∑k=1∞2ksq[∑τ=L+1∞∑ν=1N(k,τ)μ(Qτk,ν)|λτk,ν|p]q/p}1/q+{∑k=L+1∞2ksq[∑τ=1L∑ν=1N(k,τ)μ(Qτk,ν)|λτk,ν|p]q/p}1/q,∥f−fL∥Fp,qs(𝒳)≲{∑τ=L+1∞∑ν=1N(0,τ)μ(Qτ0,ν)|λτ0,ν|p}1/p+∥{∑k=1∞∑τ=L+1∞∑ν=1N(k,τ)2ksq|λτk,ν|qχQτk,ν}1/q∥Lp(𝒳)+∥{∑k=L+1∞∑τ=1L∑ν=1N(k,τ)2ksq|λτk,ν|qχQτk,ν}1/q∥Lp(𝒳). Then Lebesgue's dominated
convergence theorems on the integral and the series show that (7.67)∥f−fL∥Bp,qs(𝒳)→0,∥f−fL∥Fp,qs(𝒳)→0 as L→∞.
Moreover, by Proposition 5.31(v), we know that
fL∈Bp,qs(𝒳) and fL∈Fp,qs(𝒳) if we choose ϵ large enough. Thus, f∈Bp,qs(𝒳) if λ∈bp,qs(𝒳) and f∈Fp,qs(𝒳) if λ∈fp,qs(𝒳) when p,q<∞.
The same arguments as in the proof of Propositions5.25
and 6.15 for the space Bp,qs(𝒳) and the space Fp,qs(𝒳) with all p,q as in the assumption of the theorem yield (7.46)
and (7.47), respectively, which completes the proof of Proposition 7.3.From Theorem4.16, Proposition 7.3, and the
Plancherel-Pôlya inequalities, Propositions  5.25 and 6.15, we obtain
the following frame characterizations of the spaces Bp,qs(𝒳) and Fp,qs(𝒳).Theorem 7.4.
Letϵ be as in Definition 5.29, let |s|<ϵ,
and let p(s,ϵ)<p≤∞. Let all the other notation be as in Theorem
4.14, λτ0,ν=mQτ0,ν(D0(f)) for τ∈I0 and ν=1,…,N(0,τ),
and λτk,ν=Dk(f)(yτk,ν) for k∈ℕ, τ∈Ik and ν=1,…,N(k,τ),
where yτk,ν is any fixed element in Qτk,ν.
Then, the following hold.
(i)
If0<q≤∞,
then f∈Bp,qs(𝒳) if and only if f∈(𝒢0ϵ(β,γ))′ for some β,γ as in (5.111),  (7.68)f(x)=∑τ∈I0∑ν=1N(0,τ)λτ0,ν∫Qτk,νD˜0(x,y)dμ(y)+∑k=1∞∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)Dk(f)(yτk,ν)D˜k(x,yτk,ν) holds in (𝒢0ϵ(β,γ))′,
and λ∈bp,qs(𝒳). Moreover, in this case, (7.69)∥f∥Bp,qs(𝒳)~∥λ∥bp,qs(𝒳).(ii)
Ifp(s,ϵ)<q≤∞,
then
f∈Fp,qs(𝒳) if and only if f∈(𝒢0ϵ(β,γ))′ for some β,γ as in (5.111), (7.68) holds in (𝒢0ϵ(β,γ))′,
and λ∈fp,qs(𝒳). Moreover, in this case, (7.70)∥f∥Fp,qs(𝒳)~∥λ∥fp,qs(𝒳).


## 7.1. Frame Characterization ofB˙p,qs(𝒳) and F˙p,qs(𝒳)

In this subsection, we assume thatμ(𝒳)=∞.
We first introduce some spaces of sequences, b˙p,qs(𝒳) and f˙p,qs(𝒳).Let(7.1)λ={λτk,ν:k∈ℤ,τ∈Ik,ν=1,…,N(k,τ)} be a sequence of complex
numbers. The spaceb˙p,qs(𝒳) with s∈ℝ and 0<p,q≤∞ is the set of all λ as in (7.1) such that  (7.2)∥λ∥b˙p,qs(𝒳)={∑k=−∞∞2ksq[∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)|λτk,ν|p]q/p}1/q<∞, and the spacef˙p,qs(𝒳) with s∈ℝ, 0<p<∞, and 0<q≤∞ is the set of all λ as in (7.1) such that  (7.3)∥λ∥f˙p,qs(𝒳)=∥{∑k=−∞∞∑τ∈Ik∑ν=1N(k,τ)2ksq|λτk,ν|qχQτk,ν}1/q∥Lp(𝒳)<∞. Moreover, the spacef˙∞,qs(𝒳) with s∈ℝ and 0<q≤∞ is the set of all λ as in (7.1) such that (7.4)∥λ∥f˙∞,qs(𝒳)=supl∈ℤsupα∈Il{1μ(Qαl)[∑k=l∞∑τ∈Ik∑ν=1N(k,τ)2ksqμ(Qτk,ν)|λτk,ν|qχ{(τ,ν):Qτk,ν⊂Qαl}(τ,ν)]}1/q<∞, where {Qαl}l∈ℤ,α∈Il are dyadic cubes as in Lemma 2.19.Proposition 7.1.
Letϵ be as in Definition 5.8, let |s|<ϵ,
and let p(s,ϵ)<p≤∞. Let λ be a sequence of numbers as in (7.1) and all
the other notation as in Theorem 4.11. Then, the following
hold.
(i)
If0<q≤∞ and ∥λ∥b˙p,qs(𝒳)<∞, then the series (7.5)∑k=−∞∞∑τ∈Ik∑ν=1N(k,τ)λτk,νμ(Qτk,ν)D˜k(x,yτk,ν) converges to some f∈B˙p,qs(𝒳) both in the norm of B˙p,qs(𝒳) and in (𝒢°0ϵ(β,γ))′ with (7.6)max{0,−s+n(1p−1)+}<β<ϵ,max{n(1p−1)+,s−κp}<γ<ϵ when p,q<∞ and only in (𝒢°0ϵ(β,γ))′ with β and γ as in (7.6) when max(p,q)=∞.
Moreover, in all cases, (7.7)∥f∥B˙p,qs(𝒳)≤C∥λ∥b˙p,qs(𝒳).(ii)
Ifp(s,ϵ)<q≤∞ and ∥λ∥f˙p,qs(𝒳)<∞, then the series in (7.5) converges to some f∈F˙p,qs(𝒳) both in the norm of F˙p,qs(𝒳) and in (𝒢°0ϵ(β,γ))′ with β,γ as in (7.6) when p,q<∞ and only in (𝒢°0ϵ(β,γ))′ with β and γ as in (7.6) when max(p,q)=∞.
Moreover, in all cases, (7.8)∥f∥F˙p,qs(𝒳)≤C∥λ∥f˙p,qs(𝒳).Proof.
We first verify that the series in (7.5) converges in (𝒢°0ϵ(β,γ))′ with β and γ as in (7.6) if λ∈b˙p,qs(𝒳) with s,p,q as in (i). By Lemma 2.19 and the definition of N(k,τ) together with (2.59) in [85, page 1385], we know that
for all k∈ℤ and τ∈Ik, N(k,τ) is a finite set. Without loss of generality,
since μ(𝒳)=∞,
we may assume that Ik=ℕ for all k∈ℤ.
With this assumption, for L∈ℕ,
we define (7.9)fL(x)=∑k=−LL∑τ=1L∑ν=1N(k,τ)μ(Qτk,ν)λτk,νD˜k(x,yτk,ν). Then fL∈𝒢°(ϵ′,ϵ′) with ϵ′∈(0,ϵ1⋀ϵ2),
and fL∈(𝒢°0ϵ(β,γ))′ with any β,γ∈(0,ϵ). For any ψ∈𝒢°(β,γ) with β,γ as in (7.6), L1,L2∈ℕ and L1<L2,
we have (7.10)|〈fL2−fL1,ψ〉|≤∑k=−L2−L1−1∑τ=1L1∑ν=1N(k,τ)μ(Qτk,ν)|λτk,ν||〈D˜k(⋅,yτk,ν),ψ〉|+∑k=L1+1L2∑τ=1L1∑ν=1N(k,τ)μ(Qτk,ν)|λτk,ν||〈D˜k(⋅,yτk,ν),ψ〉|+∑k=−L2L2∑τ=L1+1L2∑ν=1N(k,τ)μ(Qτk,ν)|λτk,ν||〈D˜k(⋅,yτk,ν),ψ〉|≡Z1+Z2+Z3.
Let us now consider two cases, respectively. We first
consider the casep≤1.
In this case, letting γ′∈(max{0,s−κ/p},γ),
by (5.25) and γ>n(1/p−1) together with (5.31) and Hölder's inequality,
we have (7.11)Z1≲∥ψ∥𝒢(β,γ)∑k=−L2−L1−1∑τ=1L1∑ν=1N(k,τ)2kγ′μ(Qτk,ν)|λτk,ν|1V2−k(x1)+V(x1,yτk,ν)2−kγ(2−k+d(x1,yτk,ν))γ≲∥ψ∥𝒢(β,γ)∑k=−L2−L1−12k(γ′−s+κ/p)[∑τ=1L1∑ν=1N(k,τ)2kspμ(Qτk,ν)|λτk,ν|p]1/p. If q≤1,
by (5.5), we further obtain (7.12)Z1≲∥ψ∥𝒢(β,γ){∑k=−L2−L1−12k(γ′−s+κ/p)q[∑τ=1L1∑ν=1N(k,τ)2kspμ(Qτk,ν)|λτk,ν|p]q/p}1/q≲∥ψ∥𝒢(β,γ){∑k=−L2−L1−1[∑τ=1L1∑ν=1N(k,τ)2kspμ(Qτk,ν)|λτk,ν|p]q/p}1/q, while if 1<q≤∞,
by Hölder's inequality, we have (7.13)Z1≲∥ψ∥𝒢(β,γ){∑k=−L2−L1−1[∑τ=1L1∑ν=1N(k,τ)2kspμ(Qτk,ν)|λτk,ν|p]q/p}1/q{∑k=−L2−L1−12k(γ′−s+κ/p)q′}1/q′≲∥ψ∥𝒢(β,γ)2−L1(γ′−s+κ/p){∑k=−L2−L1−1[∑τ=1L1∑ν=1N(k,τ)2kspμ(Qτk,ν)|λτk,ν|p]q/p}1/q. Thus, (7.14)limL1,L2→∞Z1=0.
Replacing the estimates (5.25) and (5.31) respectively
by the estimates (5.24) and (5.30), and using some similar computations to the
estimate for Z1,
we obtain (7.15)Z2≲∥ψ∥𝒢(β,γ)∑k=L1+1L2∑τ=1L1∑ν=1N(k,τ)2−kβμ(Qτk,ν)|λτk,ν|1V1(x1)+V(x1,yτk,ν)1(1+d(x1,yτk,ν))γ≲∥ψ∥𝒢(β,γ)∑k=L1+1L22−k[s+β−d(1/p−1)][∑τ=1L1∑ν=1N(k,τ)2kspμ(Qτk,ν)|λτk,ν|p]1/p≲∥ψ∥𝒢(β,γ){{∑k=L1+1L2[∑τ=1L1∑ν=1N(k,τ)2kspμ(Qτk,ν)|λτk,ν|p]q/p}1/q,q≤12−L1[s+β−d(1/p−1)]{∑k=L1+1L2[∑τ=1L1∑ν=1N(k,τ)2kspμ(Qτk,ν)|λτk,ν|p]q/p}1/q,1<q≤∞. where we used the assumption
that β>−s+d(1/p−1).
Thus, (7.16)limL1,L2→∞Z2=0.
The estimates (5.24) and (5.25) together with (5.5),
(5.30), and (5.31) further yield that (7.17)Z3≲∥ψ∥𝒢(β,γ){∑k=0L2∑τ=L1+1L2∑ν=1N(k,τ)2−kβμ(Qτk,ν)|λτk,ν|1V1(x1)+V(x1,yτk,ν)1(1+d(x1,yτk,ν))γ+∑k=−L2−1∑τ=L1+1L2∑ν=1N(k,τ)2kγ′μ(Qτk,ν)|λτk,ν|1V2−k(x1)+V(x1,yτk,ν)2−kγ(2−k+d(x1,yτk,ν))γ}≲∥ψ∥𝒢(β,γ){∑k=0L22−k[β+s−d(1/p−1)]+∑k=−L2−12k(γ′−s+κ/p)}[∑τ=L1+1L2∑ν=1N(k,τ)2kspμ(Qτk,ν)|λτk,ν|p]1/p,
where we chose γ′>0 as in the estimate for Z1.
For any given δ>0,
since γ′>s−κ/p and β>d(1/p−1)−s,
we can fix L20∈ℕ such that (7.18)∑k=L20+1∞2−k[β+s−d(1/p−1)]+∑k=−∞−L20−12k(γ′−s+κ/p)<δ. Since for all k∈ℤ, (7.19)[∑τ=L1+1L2∑ν=1N(k,τ)2kspμ(Qτk,ν)|λτk,ν|p]1/p≤[∑τ∈Ik∑ν=1N(k,τ)2kspμ(Qτk,ν)|λτk,ν|p]1/p<∥λ∥b˙p,qs(𝒳), we can choose N∈ℕ such that if L1>N,
then (7.20)[∑τ=L1+1L2∑ν=1N(k,τ)2kspμ(Qτk,ν)|λτk,ν|p]1/p<δ for all k=−L20,−L20+1,…,L20.
From (7.18), (7.19), and (7.20), it follows that if L2>L1>N,
then (7.21)Z3≲∥ψ∥𝒢(β,γ){∑k=0L202−k[β+s−d(1/p−1)]+∑k=−L20−12k(γ′−s+κ/p)}δ+C∥ψ∥𝒢(β,γ)∥λ∥b˙p,qs(𝒳){∑k=L20+1∞2−k[β+s−d(1/p−1)]+∑k=−∞−L20−12k(γ′−s+κ/p)}≲∥ψ∥𝒢(β,γ)δ, which just means that (7.22)limL1,L2→∞Z3=0.
We now consider the case1<p≤∞.
Replacing (5.5) by Hölder's inequality and using Lemma 5.2, similarly to the
estimate for the case p≤1,
we obtain (7.23)Z1≲∥ψ∥𝒢(β,γ)∑k=−L2−L1−12k(γ′−s)[∑τ=1L1∑ν=1N(k,τ)2kspμ(Qτk,ν)|λτk,ν|p×1V2−k(x1)+V(x1,yτk,ν)2−kγ(2−k+d(x1,yτk,ν))γ]1/p×[∫X1V2−k(x1)+V(x1,y)2−kγ(2−k+d(x1,y))γdμ(y)]1/p′≲∥ψ∥𝒢(β,γ)∑k=−L2−L1−12k(γ′−s+κ/p)[∑τ=1L1∑ν=1N(k,τ)2kspμ(Qτk,ν)|λτk,ν|p]1/p. Then repeating the proof of the
case p≤1 yields that (7.24)limL1,L2→∞Z1=0.
Similarly, forZ2,
Hölder's inequality and Lemma 5.2 imply that (7.25)Z2≲∥ψ∥𝒢(β,γ)2−k(s+β)[∑k=L1+1L2∑τ=1L1∑ν=1N(k,τ)2kspμ(Qτk,ν)|λτk,ν|p]1/p×[∫X1V1(x1)+V(x1,y)1(1+d(x1,y))γdμ(y)]1/p′≲∥ψ∥𝒢(β,γ)∑k=L1+1L22−k(s+β)[∑τ=1L1∑ν=1N(k,τ)2kspμ(Qτk,ν)|λτk,ν|p]1/p. Using the fact that β>−s and repeating the proof of the case p≤1 show that (7.26)limL1,L2→∞Z2=0.
To estimateZ3,
let (7.27)𝒳L1L2,k=⋃τ=L1+1L2⋃ν=1N(k,τ)Qτk,ν. The Hölder inequality shows
that (7.28)Z3≲∥ψ∥𝒢(β,γ){∑k=0L22−k(β+s)[∫𝒳L1L2,k1V1(x1)+V(x1,y)1(1+d(x1,y))γdμ(y)]1/p′+∑k=−L2−12k(γ′−s+κ/p)[∫𝒳L1L2,k1V2−k(x1)+V(x1,y)2−kγ(2−kγ+d(x1,y))γdμ(y)]1/p′}×[∑τ=L1+1L2∑ν=1N(k,τ)2kspμ(Qτk,ν)|λτk,ν|p]1/p, where we chose γ′∈(max{0,s−κ/p},γ).
If p∈(1,∞),
using the facts that γ′>s−κ/p and β>−s,
(7.19), (7.20), (7.29)∫𝒳1V1(x1)+V(x1,y)1(1+d(x1,y))γdμ(y)≲1,∫𝒳1V2−k(x1)+V(x1,y)2−kγ(2−kγ+d(x1,y))γdμ(y)≲1, and repeating the argument for
the case p≤1,
we can verify that (7.30)limL1,L2→∞Z3=0. If p=∞,
replacing (7.20) by (7.31)limL1,L2→∞∫𝒳L1L2,k1V1(x1)+V(x1,y)1(1+d(x1,y))γdμ(y)=0,limL1,L2→∞∫𝒳L1L2,k1V2−k(x1)+V(x1,y)2−kγ(2−kγ+d(x1,y))γdμ(y)=0 for any given k∈ℤ,
by an argument similar to the case p≤1,
we still obtain that (7.32)limL1,L2→∞Z3=0.
Thus, for any giveψ∈𝒢(β,γ), {〈fL,ψ〉}L∈ℕ
is a Cauchy sequence, which means that the
series in (7.5) converges to some f∈(𝒢°0ϵ(β,γ))′ with β,γ as in (7.6) if λ∈b˙p,qs(𝒳) with s,p,q as in the theorem.
Ifλ∈f˙p,qs(𝒳), by the proved fact on b˙p,qs(𝒳) and (7.33)b˙p,min(p,q)s(𝒳)⊂f˙p,qs(𝒳)⊂b˙p,max(p,q)s(𝒳) (see [87, Proposition 2.3]), we also
obtain that the series in (7.5) converges in (𝒢°0ϵ(β,γ))′ with β and γ as in (7.6).
Let us now verify that the series in (7.5) converges
in the norm of B˙p,qs(𝒳) or F˙p,qs(𝒳) when p,q<∞ if λ∈b˙p,qs(𝒳) or λ∈f˙p,qs(𝒳), respectively. To this end, let f be the series in (7.5). For L∈ℕ,
in (𝒢°0ϵ(β,γ))′ with β and γ as in (7.6), we then have (7.34)f−fL=∑k=L+1∞∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)λτk,νD˜k(x,yτk,ν)+∑k=−∞−L−1∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)λτk,νD˜k(x,yτk,ν)+∑k=−LL∑τ=L+1∞∑ν=1N(k,τ)μ(Qτk,ν)λτk,νD˜k(x,yτk,ν). Replacing Qk′(f)(yτ′k′,ν′) in the proof of Proposition 5.4 by λτk,ν here and repeating the proof of Proposition
5.4, we can verify that  (7.35)∥f−fL∥B˙p,qs(𝒳)≲{∑k=L+1∞2ksq[∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)|λτk,ν|p]q/p}1/q+{∑k=−∞−L−12ksq[∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)|λτk,ν|p]q/p}1/q+{∑k=−LL2ksq[∑τ=L+1∞∑ν=1N(k,τ)μ(Qτk,ν)|λτk,ν|p]q/p}1/q,∥f−fL∥F˙p,qs(𝒳)≲∥{∑k=L+1∞∑τ∈Ik∑ν=1N(k,τ)2ksq|λτk,ν|qχQτk,ν}1/q∥Lp(𝒳)+∥{∑k=−∞−L−1∑τ∈Ik∑ν=1N(k,τ)2ksq|λτk,ν|qχQτk,ν}1/q∥Lp(𝒳)+∥{∑k=−LL∑τ=L+1∞∑ν=1N(k,τ)2ksq|λτk,ν|qχQτk,ν}1/q∥Lp(𝒳). Thus, from Lebesgue's dominated
convergence theorems on the integral and the series, it is easy to deduce that (7.36)∥f−fL∥B˙p,qs(𝒳)→0,∥f−fL∥F˙p,qs(𝒳)→0 as L→∞.
Moreover, by Proposition 5.10(iv), we know that fL∈B˙p,qs(𝒳) and fL∈F˙p,qs(𝒳) if we choose ϵ′ large enough. Thus, f∈B˙p,qs(𝒳) if λ∈b˙p,qs(𝒳) and f∈F˙p,qs(𝒳) if λ∈f˙p,qs(𝒳) when p,q<∞.
The same arguments as in the proof of Propositions5.4
and 6.3 for the space B˙p,qs(𝒳) and the space F˙p,qs(𝒳) with all p,q as in the assumption of the theorem yield
(7.7) and (7.8), respectively, which completes the proof of Proposition
7.1.From Theorem4.13, Proposition 7.1, and the
Plancherel-Pôlya inequalities, Propositions 5.4 and 6.3, we obtain
the following frame characterizations of the spaces B˙p,qs(𝒳) and F˙p,qs(𝒳).Theorem 7.2.
Letϵ be as in Definition 5.8, let |s|<ϵ, and let p(s,ϵ)<p≤∞. Let all the other notation be as in Theorem
4.11 and λτk,ν=Dk(f)(yτk,ν) for k∈ℤ, τ∈Ik and ν=1,…,N(k,τ),
where yτk,ν is any fixed element in Qτk,ν.
Then, the following hold.
(i)
If0<q≤∞,
then f∈B˙p,qs(𝒳), if and only if f∈(𝒢°0ϵ(β,γ))′ for some β,γ as in (5.35), (7.37)f(x)=∑k=−∞∞∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)Dk(f)(yτk,ν)D˜k(x,yτk,ν), holds in (𝒢°0ϵ(β,γ))′ and λ∈b˙p,qs(𝒳). Moreover, in this case, (7.38)∥f∥B˙p,qs(𝒳)~∥λ∥b˙p,qs(𝒳).(ii)
Ifp(s,ϵ)<q≤∞,
then f∈F˙p,qs(𝒳) if and only if f∈(𝒢°0ϵ(β,γ))′ for some β,γ as in (5.35), (7.37), holds in (𝒢°0ϵ(β,γ))′ and λ∈f˙p,qs(𝒳). Moreover, in this case, (7.39)∥f∥F˙p,qs(𝒳)~∥λ∥f˙p,qs(𝒳).


## 7.2. Frame Characterization ofBp,qs(𝒳) and Fp,qs(𝒳)

Again, in this subsection,μ(𝒳) can be finite or infinite. We also first
introduce some spaces of sequences, bp,qs(𝒳) and fp,qs(𝒳).Let(7.40)λ={λτk,ν:k∈ℤ+,τ∈Ik,ν=1,…,N(k,τ)} be a sequence of complex
numbers. The spacebp,qs(𝒳) with s∈ℝ and 0<p,q≤∞ is the set of all λ as in (7.40) such that (7.41)∥λ∥bp,qs(𝒳)={∑k=0∞2ksq[∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)|λτk,ν|p]q/p}1/q<∞, and the spacefp,qs(𝒳) with s∈ℝ, 0<p<∞, and 0<q≤∞ is the set of all λ as in (7.40) such that  (7.42)∥λ∥fp,qs(𝒳)=∥{∑k=0∞∑τ∈Ik∑ν=1N(k,τ)2ksq|λτk,ν|qχQτk,ν}1/q∥Lp(𝒳)<∞. Moreover, the spacef∞,qs(𝒳) with s∈ℝ and 0<q≤∞ is the set of all λ as in (7.40) such that (7.43)∥λ∥f∞,qs(𝒳)=max[supτ∈I0ν=1,…,N(0,τ)|λτ0,ν|,supl∈ℕsupα∈Il{1μ(Qαl)[∑k=l∞∑τ∈Ik∑ν=1N(k,τ)2ksqμ(Qτk,ν)|λτk,ν|qχ{(τ,ν):Qτk,ν⊂Qαl}(τ,ν)]}1/q]<∞, where {Qαl}l∈ℕ,α∈Il are dyadic cubes as in Lemma 2.19.Proposition 7.3.
Letϵ be as in Definition 5.29, let |s|<ϵ,
and let p(s,ϵ)<p≤∞. Let λ be a sequence of numbers as in (7.40) and all
the other notation as in Theorem 4.14. Then, the following hold.
(i)
If0<q≤∞ and ∥λ∥bp,qs(𝒳)<∞, then the series (7.44)∑τ∈I0∑ν=1N(0,τ)λτ0,ν∫Qτ0,νD˜0(x,y)dμ(y)+∑k=1∞∑τ∈Ik∑ν=1N(k,τ)λτk,νμ(Qτk,ν)D˜k(x,yτk,ν) converges to some f∈Bp,qs(𝒳) both in the norm of Bp,qs(𝒳) and in (𝒢0ϵ(β,γ))′ with (7.45)max{0,−s+n(1p−1)+}<β<ϵ,n(1p−1)+<γ<ϵ, when p,q<∞ and only in (𝒢0ϵ(β,γ))′ with β and γ as in (7.45) when max(p,q)=∞.
Moreover, in all cases, (7.46)∥f∥Bp,qs(𝒳)≤C∥λ∥bp,qs(𝒳).(ii)
Ifp(s,ϵ)<q≤∞ and ∥λ∥fp,qs(𝒳)<∞, then the series in (7.44) converges to some f∈Fp,qs(𝒳) both in the norm of Fp,qs(𝒳) and in (𝒢0ϵ(β,γ))′ with β,γ as in (7.45) when p,q<∞ and only in (𝒢0ϵ(β,γ))′ with β and γ as in (7.45) when max(p,q)=∞.
Moreover, in all cases, (7.47)∥f∥Fp,qs(𝒳)≤C∥λ∥fp,qs(𝒳).Proof.
In what follows, for simplicity,  we set(7.48)D˜Qτ0,ν(x)=1μ(Qτ0,ν)∫Qτ0,νD˜0(x,y)dμ(y). Let us first show that the
series in (7.44) converges in (𝒢(β,γ))′ with β and γ as in (7.45). As in the proof of Proposition
7.1, we know that for all k∈ℤ+ and τ∈Ik, N(k,τ) is a finite set. Let us suppose Ik=ℕ for all k∈ℤ+;
the other cases are easier. With this assumption, for L∈ℕ,
we define (7.49)fL(x)=∑τ=1L∑ν=1N(0,τ)μ(Qτ0,ν)λτ0,νD˜Qτ0,ν(x)+∑k=1L∑τ=1L∑ν=1N(k,τ)μ(Qτk,ν)λτk,νD˜k(x,yτk,ν). Then fL∈𝒢(ϵ,ϵ) and fL∈(𝒢0ϵ(β,γ))′ with any β,γ∈(0,ϵ),
where ϵ can be any positive number in (0,ϵ1⋀ϵ2).
In what follows, we choose ϵ>max(β,γ) such that p>p(s,ϵ) for the spaces bp,qs(𝒳), and p,q>p(s,ϵ) for the spaces fp,qs(𝒳).
For anyψ∈𝒢(β,γ) with (β,γ) as in (7.45), L1,L2∈ℕ and L1<L2,
we have (7.50)|〈fL2−fL1,ψ〉|≤∑τ=L1+1L2∑ν=1N(0,τ)μ(Qτ0,ν)|λτ0,ν||〈D˜Qτ0,ν,ψ〉|+∑k=L1+1L2∑τ=1L2∑ν=1N(k,τ)μ(Qτk,ν)|λτk,ν||〈D˜k(⋅,yτk,ν),ψ〉|+∑k=1L1∑τ=L1+1L2∑ν=1N(k,τ)μ(Qτk,ν)|λτk,ν||〈D˜k(⋅,yτk,ν),ψ〉|≡Y1+Y2+Y3.
From (5.105), (5.5) when p≤1 or Hölder's inequality when 1<p≤∞,
and γ>n(1/p−1)+ together with (5.30), it follows that when
p≤1, (7.51)|Y1|≲∥ψ∥𝒢(β,γ){∑τ=L1+1L2∑ν=1N(0,τ)μ(Qτ0,ν)|λτ0,ν|p}1/p, while when
1<p≤∞,(7.52)|Y1|≲∥ψ∥𝒢(β,γ){∑τ=L1+1L2∑ν=1N(0,τ)μ(Qτ0,ν)|λτ0,ν|p}1/p×{∫𝒳L1L2,01V1(x1)+V(x1,y)1(1+d(x1,y))γdμ(y)}1/p′, where 𝒳L1L2,0 is as in the proof of Proposition 7.1.
ForY2,
by (5.105), (5.5) when p≤1 or Hölder's inequality when 1<p≤∞, γ>n(1/p−1)+ together with (5.30),  and Lemma 2.1(ii), we
obtain (7.53)|Y2|≲∥ψ∥𝒢(β,γ){∑k=L1+1L22−k[β+s−n(1/p−1)]2ks[∑τ=1L2∑ν=1N(k,τ)μ(Qτk,ν)|λτk,ν|p]1/p,p≤1∑k=L1+1L22−k(β+s)2ks[∑τ=1L2∑ν=1N(k,τ)μ(Qτk,ν)|λτk,ν|p]1/p,1<p≤∞. From this and (5.5) when q≤1 or Hölder's  inequality when 1<q≤∞ again, it further follows that  when p≤1 and q≤1, (7.54)|Y2|≲∥ψ∥𝒢(β,γ){∑k=L1+1L22ksq[∑τ=1L2∑ν=1N(k,τ)μ(Qτk,ν)|λτk,ν|p]q/p}1/q, while  when 1<q≤∞, (7.55)|Y2|≲∥ψ∥𝒢(β,γ){∑k=L1+1L22ksq[∑τ=1L2∑ν=1N(k,τ)μ(Qτk,ν)|λτk,ν|p]q/p}1/q{∑k=L1+1L22−k[β+s−n(1/p−1)]q′}1/q′, and that  when 1<p≤∞, (7.56)|Y2|≲∥ψ∥𝒢(β,γ){{∑k=L1+1L22ksq[∑τ=1L2∑ν=1N(k,τ)μ(Qτk,ν)|λτk,ν|p]q/p}1/q,q≤1{∑k=L1+1L22ksq[∑τ=1L2∑ν=1N(k,τ)μ(Qτk,ν)|λτk,ν|p]q/p}1/q{∑k=L1+1L22−k(β+s)q′}1/q′,1<q≤∞. where we used the fact that β>max(0,n(1/p−1)+−s).
Similarly, by (5.105), (5.5) when p≤1 or Hölder's inequality when 1<p≤∞,
and γ>n(1/p−1)+ together  with (5.30), we have (7.57)|Y3|≲∥ψ∥𝒢(β,γ){∑k=1L12−k[β−n(1/p−1)][∑τ=L1+1L2∑ν=1N(k,τ)μ(Qτk,ν)|λτk,ν|p]1/p,p≤1∑k=1L12−kβ[∑τ=L1+1L2∑ν=1N(k,τ)μ(Qτk,ν)|λτk,ν|p]1/p×{∫𝒳L1L2,k1V1(x1)+V(x1,y)1(1+d(x1,y))γdμ(y)}1/p′,1<p≤∞, where 𝒳L1L2,k is as in the proof  of Proposition 7.1.
From this and (5.5) when q≤1 or Hölder's inequality when 1<q≤∞,
we further deduce that when p≤1, (7.58)|Y3|≲∥ψ∥𝒢(β,γ){∑k=1L12ksq[∑τ=L1+1L2∑ν=1N(k,τ)μ(Qτk,ν)|λτk,ν|p]q/p}1/q, and that when 1<p≤∞ and q≤1, (7.59)|Y3|≲∥ψ∥𝒢(β,γ){∑k=1L12ksq[∑τ=L1+1L2∑ν=1N(k,τ)μ(Qτk,ν)|λτk,ν|p]q/p}1/q, or that when 1<p≤∞ and 1<q≤∞, (7.60)|Y3|≲∥ψ∥𝒢(β,γ){∑k=1L12ksq[∑τ=L1+1L2∑ν=1N(k,τ)μ(Qτk,ν)|λτk,ν|p]q/p}1/q×{∑k=1L22−k(β+s)q′[∫𝒳L1L2,k1V1(x1)+V(x1,y)1(1+d(x1,y))γdμ(y)]q′/p′}1/q′, where we used the fact that β>max(0,n(1/p−1)+−s).
Combining (7.51) through (7.60), by λ∈bp,qs(𝒳), (7.61)∫𝒳1V1(x1)+V(x1,y)1(1+d(x1,y))γdμ(y)<∞ when p=∞,
and (7.62)∑k=1∞2−k[β+s−n(1/p−1)]<∞ when p≤1 and q=∞,
or (7.63)∑k=1∞2−k(β+s)<∞ when 1<p≤∞ and q=∞,
it is easy to see that
{〈fL,ψ〉}L∈ℕ is a Cauchy sequence. This just means that the
series in (7.44) converges to some f∈(𝒢0ϵ(β,γ))′ with β,γ satisfying (7.45) if λ∈bp,qs(𝒳) with s,p,q as in the theorem. If λ∈fp,qs(𝒳) with s,p,q as in the theorem, by this fact and (7.64)bp,min(p,q)s(𝒳)⊂fp,qs(𝒳)⊂bp,max(p,q)s(𝒳)(see [87, Proposition 2.3]), we also
obtain that the series in (7.44) converges in (𝒢0ϵ(β,γ))′ with β and γ as in (7.45).
Let us now verify that the series in (7.44) converges
in the norm of
Bp,qs(𝒳) when p,q<∞,
if λ∈bp,qs(𝒳). Let f be the series in (7.44). We estimate the norm
in Bp,qs(𝒳) of f−fL by writing (7.65)f−fL=∑τ=L+1∞∑ν=1N(0,τ)μ(Qτ0,ν)λτ0,νD˜Qτ0,ν(x)+∑k=1∞∑τ=L+1∞∑ν=1N(k,τ)μ(Qτk,ν)λτk,νD˜k(x,yτk,ν)+∑k=L+1∞∑τ=1L∑ν=1N(k,τ)μ(Qτk,ν)λτk,νD˜k(x,yτk,ν). Replacing Qτ′,10,ν′(f) and Qk′(f)(yτ′k′,ν′) in the proof of Proposition 5.25 respectively
by λτ0,ν and λτk,ν here, and repeating the proof of Proposition
5.25, we then obtain  (7.66)∥f−fL∥Bp,qs(𝒳)≲{∑τ=L+1∞∑ν=1N(0,τ)μ(Qτ0,ν)|λτ0,ν|p}1/p+{∑k=1∞2ksq[∑τ=L+1∞∑ν=1N(k,τ)μ(Qτk,ν)|λτk,ν|p]q/p}1/q+{∑k=L+1∞2ksq[∑τ=1L∑ν=1N(k,τ)μ(Qτk,ν)|λτk,ν|p]q/p}1/q,∥f−fL∥Fp,qs(𝒳)≲{∑τ=L+1∞∑ν=1N(0,τ)μ(Qτ0,ν)|λτ0,ν|p}1/p+∥{∑k=1∞∑τ=L+1∞∑ν=1N(k,τ)2ksq|λτk,ν|qχQτk,ν}1/q∥Lp(𝒳)+∥{∑k=L+1∞∑τ=1L∑ν=1N(k,τ)2ksq|λτk,ν|qχQτk,ν}1/q∥Lp(𝒳). Then Lebesgue's dominated
convergence theorems on the integral and the series show that (7.67)∥f−fL∥Bp,qs(𝒳)→0,∥f−fL∥Fp,qs(𝒳)→0 as L→∞.
Moreover, by Proposition 5.31(v), we know that
fL∈Bp,qs(𝒳) and fL∈Fp,qs(𝒳) if we choose ϵ large enough. Thus, f∈Bp,qs(𝒳) if λ∈bp,qs(𝒳) and f∈Fp,qs(𝒳) if λ∈fp,qs(𝒳) when p,q<∞.
The same arguments as in the proof of Propositions5.25
and 6.15 for the space Bp,qs(𝒳) and the space Fp,qs(𝒳) with all p,q as in the assumption of the theorem yield (7.46)
and (7.47), respectively, which completes the proof of Proposition 7.3.From Theorem4.16, Proposition 7.3, and the
Plancherel-Pôlya inequalities, Propositions  5.25 and 6.15, we obtain
the following frame characterizations of the spaces Bp,qs(𝒳) and Fp,qs(𝒳).Theorem 7.4.
Letϵ be as in Definition 5.29, let |s|<ϵ,
and let p(s,ϵ)<p≤∞. Let all the other notation be as in Theorem
4.14, λτ0,ν=mQτ0,ν(D0(f)) for τ∈I0 and ν=1,…,N(0,τ),
and λτk,ν=Dk(f)(yτk,ν) for k∈ℕ, τ∈Ik and ν=1,…,N(k,τ),
where yτk,ν is any fixed element in Qτk,ν.
Then, the following hold.
(i)
If0<q≤∞,
then f∈Bp,qs(𝒳) if and only if f∈(𝒢0ϵ(β,γ))′ for some β,γ as in (5.111),  (7.68)f(x)=∑τ∈I0∑ν=1N(0,τ)λτ0,ν∫Qτk,νD˜0(x,y)dμ(y)+∑k=1∞∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)Dk(f)(yτk,ν)D˜k(x,yτk,ν) holds in (𝒢0ϵ(β,γ))′,
and λ∈bp,qs(𝒳). Moreover, in this case, (7.69)∥f∥Bp,qs(𝒳)~∥λ∥bp,qs(𝒳).(ii)
Ifp(s,ϵ)<q≤∞,
then
f∈Fp,qs(𝒳) if and only if f∈(𝒢0ϵ(β,γ))′ for some β,γ as in (5.111), (7.68) holds in (𝒢0ϵ(β,γ))′,
and λ∈fp,qs(𝒳). Moreover, in this case, (7.70)∥f∥Fp,qs(𝒳)~∥λ∥fp,qs(𝒳).


## 8. Real Interpolation and Dual Spaces

In this section, using the frame characterization of
Besov spaces and Triebel-Lizorkin spaces in the last section, we characterize
real interpolation spaces for our scales of Besov and Triebel-Lizorkin spaces
and determine their dual spaces (whenp≥1).


### 8.1. Real Interpolation Spaces

Let us first recall some general background on the
real interpolation method; see [3, pages 62–64] or [88, 89].Letℋ be a linear complex Hausdorff space, and let 𝒜0 and 𝒜1 be two complex quasi-Banach spaces such that 𝒜0⊂ℋ and 𝒜1⊂ℋ. Let 𝒜0+𝒜1 be the set of all elements a∈ℋ which can be represented as a=a0+a1 with a0∈𝒜0 and a1∈𝒜1. If 0<t<∞ and a∈𝒜0+𝒜1,
then Peetre's K-functional is given by(8.1)K(t,a)=K(t,a;𝒜0,𝒜1)=inf(∥a0∥𝒜0+t∥a1∥𝒜1),where the infimum is taken over
all representations of a of the form a=a0+a1 with a0∈𝒜0 and a1∈𝒜1.Definition 8.1.
Let0<σ<1. If 0<q<∞,
then one defines the interpolation
space(8.2)(𝒜0,𝒜1)σ,q≡{a:a∈𝒜0+𝒜1,∥a∥(𝒜0,𝒜1)σ,q≡{∫0∞[t−σK(t,a)]qdtt}1/q<∞}. If q=∞, then one defines
(8.3)(𝒜0,𝒜1)σ,∞≡{a:a∈𝒜0+𝒜1,∥a∥(𝒜0,𝒜1)σ,∞≡sup0<t<∞t−σK(t,a)<∞}.The following basic properties of(𝒜0,𝒜1)σ,q are proved in [3, pages 63-64] and [88, page 64].Proposition 8.2.
Let𝒜0 and 𝒜1 be two complex quasi-Banach spaces. Let 0<σ<1 and 0<q≤∞. Then,
(i)
(𝒜0,𝒜1)σ,q is a quasi-Banach space;(ii)
(𝒜0,𝒜1)σ,q=(𝒜1,𝒜0)1−σ,q;(iii)
letℋ be a linear complex Hausdorff space, and let ℬ0 and ℬ1 be two complex quasi-Banach spaces such that 𝒜0⊂ℬ0⊂ℋ and 𝒜1⊂ℬ1⊂ℋ. Then (𝒜0,𝒜1)σ,q⊂(ℬ0,ℬ1)σ,q.Using Theorems7.2 and 7.4 together with the
method of retraction and coretraction as in the proofs of Theorems 2.4.1 and 2.4.2 in [89],
we can easily deduce the following interpolation theorems; see also [90].Theorem 8.3.
Letϵ be as in Definition 5.8 and σ∈(0,1).
(i)
Let−ϵ<s0,s1<ϵ, s0≠s1, 1≤p≤∞,
and 1≤q0,q1,q≤∞. Then(8.4)(B˙p,q0s0(𝒳),B˙p,q1s1(𝒳))σ,q=B˙p,qs(𝒳),where s=(1−σ)s0+σs1.(ii)
Let−ϵ<s<ϵ, 1≤p≤∞, 1≤q0,q1≤∞, and q0≠q1. Then(8.5)(B˙p,q0s(𝒳),B˙p,q1s(𝒳))σ,q=B˙p,qs(𝒳),where 1/q=(1−σ)/q0+σ/q1.(iii)
Let−ϵ<s0,s1<ϵ, 1≤p0,p1≤∞, and p0≠p1. Then(8.6)(B˙p0,p0s0(𝒳),B˙p1,p1s1(𝒳))σ,p=B˙p,ps(𝒳),where 1/p=(1−σ)/p0+σ/p1 and s=(1−σ)s0+σs1.Theorem 8.4.
Letϵ be as in Definition 5.8, −ϵ<s0,s1<ϵ, 1≤p0,p1<∞, 1≤q0,q1≤∞, σ∈(0,1), s=(1−σ)s0+σs1, 1/p=(1−σ)/p0+σ/p1,
and 1/q=(1−σ)/q0+σ/q1.
(i)
Ifs0≠s1,
then(8.7)(F˙p0,q0s0(𝒳),F˙p1,q1s1(𝒳))σ,p=F˙p,ps(𝒳).(ii)
Ifs0=s1=s, p0=q0, p1=q1, and q0≠q1,
then(8.8)(F˙p0,p0s(𝒳),F˙p1,p1s(𝒳))σ,p=F˙p,ps(𝒳).(iii)
Ifs0=s1=s, q0=q1=q, and p0≠p1,
then(8.9)(F˙p0,qs(𝒳),F˙p1,qs(𝒳))σ,p=F˙p,qs(𝒳).Theorem 8.5.
Letϵ be as in Definition 5.29 and σ∈(0,1).
(i)
Let−ϵ<s0,s1<ϵ, s0≠s1, 1≤p≤∞,
and 1≤q0,q1,q≤∞. Then(8.10)(Bp,q0s0(𝒳),Bp,q1s1(𝒳))σ,q=Bp,qs(𝒳),where s=(1−σ)s0+σs1.(ii)
Let−ϵ<s<ϵ, 1≤p≤∞, 1≤q0,q1≤∞, and q0≠q1. Then(8.11)(Bp,q0s(𝒳),Bp,q1s(𝒳))σ,q=Bp,qs(𝒳),where 1/q=(1−σ)/q0+σ/q1.(iii)
Let−ϵ<s0,s1<ϵ, 1≤p0,p1≤∞, and p0≠p1. Then(8.12)(Bp0,p0s0(𝒳),Bp1,p1s1(𝒳))σ,p=Bp,ps(𝒳),where 1/p=(1−σ)/p0+σ/p1 and s=(1−σ)s0+σs1.Theorem 8.6.
Letϵ be as in Definition 5.29, −ϵ<s0,s1<ϵ, 1≤p0,p1<∞, 1≤q0,q1≤∞, σ∈(0,1), s=(1−σ)s0+σs1, 1/p=(1−σ)/p0+σ/p1, and 1/q=(1−σ)/q0+σ/q1.
(i)
Ifs0≠s1,
then(8.13)(Fp0,q0s0(𝒳),Fp1,q1s1(𝒳))σ,p=Fp,ps(𝒳).(ii)
Ifs0=s1=s, p0=q0, p1=q1, and q0≠q1,
then(8.14)(Bp0,p0s(𝒳),Bp1,p1s(𝒳))σ,p=Bp,ps(𝒳).(iii)
Ifs0=s1=s, q0=q1=q, and p0≠p1,
then(8.15)(Fp0,qs(𝒳),Fp1,qs(𝒳))σ,p=Fp,qs(𝒳).Proofs of Theorem8.3 through Theorem 8.6.
The proofs of Theorem8.3 through Theorem 8.6 are similar by using [89, Theorem 1.2.4]. We only give an outline here; see also [90, the proofs of Proposition
3.3 and Theorem 3.1].
To prove Theorems8.3 and 8.4, by Proposition 5.10(iii), we know that(8.16)B˙pi,qisi(𝒳),F˙pi,qisi(𝒳)⊂(𝒢°0ϵ(βi,γi))′,where max{0,−si+n(1/pi−1)+}<βi<ϵ and max{n(1/pi−1)+,si−κ/pi}<γi<ϵ with i=0,1. We then let β=max(β0,β1) and γ=(γ0,γ1). Then(8.17)B˙pi,qisi(𝒳),F˙pi,qisi(𝒳)⊂(𝒢°0ϵ(β,γ))′.In this sense, {B˙p0,q0s0(𝒳),B˙p1,q1s1(𝒳)} and {F˙p0,q0s0(𝒳),F˙p1,q1s1(𝒳)} are interpolation couples in the sense of
[89, Section 1.2.1]. Now, for f∈(𝒢°0ϵ(β,γ))′,
with the notation of Theorem 4.11, we can define the coretraction operator S˙ by(8.18)S˙(f)(x)={S˙(f)k(x)}k=−∞∞,where for k∈ℤ,(8.19)S˙(f)k(x)=∑τ∈Ik∑ν=1N(k,τ)Dk(f)(yτk,ν)χQτk,ν(x);and the corresponding retraction
operator R˙ by(8.20)R˙({fk})(x)=∑k=−∞∞∑τ∈Ik∑ν=1N(k,τ)[∫Qτk,νfk(y)dμ(y)]D˜k(x,yτk,ν).
By Theorem 4.13, for any f∈(𝒢°0ϵ(β,γ))′,
we have R˙S˙(f)(x)=f(x). In what follows, for s∈ℝ, 0<q≤∞, and 0<p≤∞,
we say {fk}k=−∞∞∈ℓ˙s,q(Lp)(𝒳),
if(8.21)∥{fk}k=−∞∞∥ℓ˙s,q(Lp)(𝒳)={∑k=−∞∞2ksq∥fk∥Lp(𝒳)q}1/q<∞;and we say {fk}k=−∞∞∈Lp(ℓ˙s,q)(𝒳),
if(8.22)∥{fk}k=−∞∞∥Lp(ℓ˙s,q)(𝒳)=∥{∑k=−∞∞2ksq|fk(x)|q}1/q∥Lp(𝒳)<∞,where the usual modifications
are made when p=∞ or q=∞. If F is an interpolation functor, then one obtains
by [89, Theorem 1.2.4] that(8.23)∥f∥F({B˙p0,q0s0(𝒳),B˙p1,q1s1(𝒳)})~∥S˙(f)∥F({ℓ˙s0,q0(Lp0)(𝒳),ℓ˙s1,q1(Lp1)(𝒳)}),∥f∥F({F˙p0,q0s0(𝒳),F˙p1,q1s1(𝒳)})~∥S˙(f)∥F({Lp0(ℓ˙s0,q0)(𝒳),Lp1(ℓ˙s1,q1)(𝒳)}). Using Proposition 7.1 and
Theorem 7.2, we can then finish the proofs of Theorems 8.3 and 8.4 by
the same procedures as those in [89, pages 182-183] and [89, pages 185-186].
To prove Theorems8.5 and 8.6, by Proposition 5.31(iv), we know that(8.24)Bpi,qisi(𝒳),Fpi,qisi(𝒳)⊂(𝒢0ϵ(βi,γi))′,where max{0,−si+n(1/pi−1)+}<βi<ϵ and n(1/pi−1)+<γi<ϵ with i=0,1. We then let β=max(β0,β1) and γ=(γ0,γ1). Then(8.25)Bpi,qisi(𝒳),Fpi,qisi(𝒳)⊂(𝒢0ϵ(β,γ))′.In this sense, {Bp0,q0s0(𝒳),Bp1,q1s1(𝒳)} and {Fp0,q0s0(𝒳),Fp1,q1s1(𝒳)} are interpolation couples in the sense of
[89, Section 1.2.1]. Now, for f∈(𝒢0ϵ(β,γ))′,
with the notation of Theorem 4.14, we can define the coretraction operator S by(8.26)S(f)(x)={S(f)k(x)}k=0∞,where(8.27)S(f)0(x)=∑τ∈I0∑ν=1N(0,τ)mQτ0,ν(D0(f))χQτ0,ν(x)and for k∈ℕ,(8.28)S(f)k(x)=∑τ∈Ik∑ν=1N(k,τ)Dk(f)(yτk,ν)χQτk,ν(x);and the corresponding retraction
operator R by(8.29)R({fk})(x)=∑τ∈I0∑ν=1N(0,τ)[∫Qτ0,νf0(y)dμ(y)]D˜Qτ0,ν(x)+∑k=1∞∑τ∈Ik∑ν=1N(k,τ)[∫Qτk,νfk(y)dμ(y)]D˜k(x,yτk,ν),where D˜Qτ0,ν(x) is as in the proof of Proposition 7.3. By
Theorem 4.16, for any f∈(𝒢0ϵ(β,γ))′,
we have RS(f)(x)=f(x). In what follows, for s∈ℝ, 0<q≤∞, and 0<p≤∞,
we say {fk}k=0∞∈ℓs,q(Lp)(𝒳),
if(8.30)∥{fk}k=0∞∥ℓs,q(Lp)(𝒳)={∑k=0∞2ksq∥fk∥Lp(𝒳)q}1/q<∞;and we say {fk}k=0∞∈Lp(ℓs,q)(𝒳),
if(8.31)∥{fk}k=0∞∥Lp(ℓs,q)(𝒳)=∥{∑k=0∞2ksq|fk(x)|q}1/q∥Lp(𝒳)<∞,where the usual modifications
are made when p=∞ or q=∞. If F is an interpolation functor, then one obtains
by [89, Theorem 1.2.4] that(8.32)∥f∥F({Bp0,q0s0(𝒳),Bp1,q1s1(𝒳)})~∥S(f)∥F({ℓs0,q0(Lp0)(𝒳),ℓs1,q1(Lp1)(𝒳)}),∥f∥F({Fp0,q0s0(𝒳),Fp1,q1s1(𝒳)})~∥S(f)∥F({Lp0(ℓs0,q0),Lp1(ℓs1,q1)}). Using Proposition 7.3 and
Theorem 7.4, we then finish the proofs of Theorems 8.5 and 8.6 by the
same procedures as those in [89, pages 182-183] and [89, pages 185-186], which
completes the proofs of Theorem 8.3 through Theorem 8.6.We remark that Theorem8.3 through Theorem 8.6 only deal with Besov spaces and Triebel-Lizorkin spaces which are Banach spaces,
since p,q≥1. Using Theorems 7.2 and 7.4 together with the following fact that for 0<p0≠p1≤∞ and σ∈(0,1),
if 1/p=(1−σ)/p0+σ/p1,(8.33)(ℓ˙p0,ℓ˙p1)σ,p=ℓ˙p,(ℓp0,ℓp1)σ,p=ℓp (see [88, Theorem 5.2.1] and also
[89, Remark 1.18.6/5]), by a method similar to the proofs of Theorem 8.3 through Theorem 8.6 (see
also [82, Corollary 6.6]), we can easily establish the following interpolation theorem which covers also
cases when p<1. We omit the details.Theorem 8.7.
Letϵ be as in Definition 5.8, s0,s1∈(−ϵ,ϵ), p(s0,ϵ)<p0≤∞, p(s1,ϵ)<p1≤∞, and p0≠p1. Let σ∈(0,1), 1/p=(1−σ)/p0+σ/p1, and s=(1−σ)s0+σs1. Then,
(i)
(B˙p0,p0s0(𝒳),B˙p1,p1s1(𝒳))σ,p=B˙p,ps(𝒳);(ii)
(Bp0,p0s0(𝒳),Bp1,p1s1(𝒳))σ,p=Bp,ps(𝒳).Making use of Calderón reproducing formulae in place of frame characterizations, we can also directly establish the followingreal interpolation theorems for Besov and Triebel-Lizorkin spaces which are only quasi-Banach spaces; see also [3, Theorem 2.4.2] and [91].Theorem 8.8.
Letϵ be as in Definition 5.8, σ∈(0,1), s0,s1∈(−ϵ,ϵ), s0≠s1, and s=(1−σ)s0+σs1. Then
(i)
ifmax{p(s0,ϵ),p(s1,ϵ)}<p≤∞ and 0<q0,q1,q≤∞,
then(8.34)(B˙p,q0s0(𝒳),B˙p,q1s1(𝒳))σ,q=B˙p,qs(𝒳);(ii)
ifmax{p(s0,ϵ),p(s1,ϵ)}<p≤∞,p(si,ϵ)<qi≤∞ for i=0,1 and 0<q≤∞,
then(8.35)(F˙p,q0s0(𝒳),F˙p,q1s1(𝒳))σ,q=B˙p,qs(𝒳).Proof.
We first verify (i). By Proposition5.10(iii), we know that(8.36)B˙p,q0s0(𝒳),B˙p,q1s1(𝒳)⊂(𝒢°0ϵ(β,γ))′with max{0,−s0+n(1/p−1)+,−s1+n(1/p−1)+}<β<ϵ and max{n(1/p−1)+,s0−κ/p,s1−κ/p}<γ<ϵ. Thus, we can take ℋ=(𝒢°0ϵ(β,γ))′ with β and γ as above.
We now verify that(8.37)(B˙p,∞s0(𝒳),B˙p,∞s1(𝒳))σ,q⊂B˙p,qs(𝒳).By Proposition 8.2(ii), without
loss of generality, we may assume that s0>s1.
Assume thatf∈(B˙p,∞s0(𝒳),B˙p,∞s1(𝒳))σ,q and f=f0+f1 with f0∈B˙p,∞s0(𝒳) and f1∈B˙p,∞s1(𝒳). Let {Dk}k∈ℤ be as in Definition 5.8. Then,(8.38)2ks0∥Dk(f)∥Lp(𝒳)≲2ks0∥Dk(f0)∥Lp(𝒳)+2k(s0−s1)2ks1∥Dk(f1)∥Lp(𝒳)≲∥f0∥B˙p,∞s0(𝒳)+2k(s0−s1)∥f1∥B˙p,∞s1(𝒳).Taking the infimum on all
representations f=f0+f1 yields that(8.39)2ks0∥Dk(f)∥Lp(𝒳)≲K(2k(s0−s1),f;B˙p,∞s0(𝒳),B˙p,∞s1(𝒳)).If 0<q<∞,
from (8.39), it follows that(8.40)∫0∞t−σq[K(t,f;B˙p,∞s0(𝒳),B˙p,∞s1(𝒳))]qdtt=∑k=−∞∞∫2(k−1)(s0−s1)2k(s0−s1)t−σq[K(t,f;B˙p,∞s0(𝒳),B˙p,∞s1(𝒳))]qdtt≳∑k=−∞∞2−σqk(s0−s1)[K(2k(s0−s1),f;B˙p,∞s0(𝒳),B˙p,∞s1(𝒳))]q≳∑k=−∞∞2ksq∥Dk(f)∥Lp(𝒳)q≳∥f∥B˙p,qs(𝒳)q;and if q=∞,
by (8.39), we then have(8.41)∥f∥B˙p,∞s(𝒳)=supk∈ℤ2ks∥Dk(f)∥Lp(𝒳)≲supk∈ℤ2k(s−s0)K(2k(s0−s1),f;B˙p,∞s0(𝒳),B˙p,∞s1(𝒳))≲sup0<t<∞t−σK(t,f;B˙p,∞s0(𝒳),B˙p,∞s1(𝒳)).Thus, (8.37) holds.
We now prove that if0<r<q,
then(8.42)B˙p,qs(𝒳)⊂(B˙p,rs0(𝒳),B˙p,rs1(𝒳))σ,q.Without loss of generality, we
may assume that s0>s1 again. Then we also have s0>s>s1. In what follows, we only consider the case q<∞ and we omit the details for the case q=∞ by similarity and simplicity. Let now f∈B˙p,qs(𝒳). We then write(8.43)∥f∥(B˙p,rs0(𝒳),B˙p,rs1(𝒳))σ,qq=∫0∞t−σq[K(t,f;B˙p,rs0(𝒳),B˙p,rs1(𝒳))]qdtt≲∑j=−∞∞2−jσq(s0−s1)[K(2j(s0−s1),f;B˙p,rs0(𝒳),B˙p,rs1(𝒳))]q.Let all the notation be as in
Theorem 4.11. For any j∈ℤ,
we write(8.44)f(x)=∑k=−∞j∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)Dk(f)(yτk,ν)D˜k(x,yτk,ν)+∑k=j+1∞∑τ∈Ik∑ν=1N(k,τ)⋯≡f0j+f1j.From this and the definition of K-functional, it follows that(8.45)∥f∥(B˙p,rs0(𝒳),B˙p,rs1(𝒳))σ,qq≲∑j=−∞∞2−jσq(s0−s1)(∥f0j∥B˙p,rs0(𝒳)q+2j(s0−s1)∥f1j∥B˙p,rs1(𝒳)q)≲∑j=−∞∞2jq(s−s0){∑k′=−∞∞2k′s0r[∑τ′∈Ik′∑ν′=1N(k′,τ′)μ(Qτ′k′,ν′)|Dk′(f0j)(yτ′k′,ν′)|p]r/p}q/r+∑j=−∞∞2jq(s−s1){∑k′=−∞∞2k′s1r[∑τ′∈Ik′∑ν′=1N(k′,τ′)μ(Qτ′k′,ν′)|Dk′(f1j)(yτ′k′,ν′)|p]r/p}q/r≡J1+J2.
We first estimateJ1 in the case p≤1. In this case, by (5.10), (5.5), Lemma 5.2, and (5.12), we have(8.46)J1≲∑j=−∞∞2jq(s−s0){∑k′=−∞∞2k′s0r(∑k=−∞j2−|k′−k|ϵp2n[k−(k⋀k′)](1−p)×∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)|Dk(f)(yτk,ν)|p)r/p}q/r.Now, we choose ϵ′∈(|s0|,ϵ) such that p>n/(n+s0+ϵ′). Using (5.5) when r/p≤1 or Hölder's inequality when 1<r/p<∞ then further shows that(8.47)J1≲∑j=−∞∞2jq(s−s0){∑k′=−∞∞2k′s0r[∑k=−∞j2−|k′−k|ϵ′r2n[k−(k⋀k′)](1/p−1)r×(∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)|Dk(f)(yτk,ν)|p)r/p]}q/r≲∑j=−∞∞2jq(s−s0){∑k=−∞j2ks0r(∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)|Dk(f)(yτk,ν)|p)r/p}q/r.Since q/r>1,
by Hölder's inequality and s0>s,
we have(8.48)J1≲∑j=−∞∞2jq(s−s0)/2{∑k=−∞j2k(s0−s)q/22ksq(∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)|Dk(f)(yτk,ν)|p)q/p}≲∥f∥B˙p,qs(𝒳)q,which is the desired estimate in
the  case p≤1.
While when1<p≤∞,
letting ϵ′∈(|s0|,ϵ),
by Hölder's inequality and Lemma 5.2, we have(8.49)J1≲∑j=−∞∞2jq(s−s0){∑k′=−∞∞2k′s0r(∑τ′∈Ik′∑ν′=1N(k′,τ′)μ(Qτ′k′,ν′)×[∑k=−∞j2−|k′−k|ϵ′p∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)|Dk(f)(yτk,ν)|p×1V2−(k⋀k′)(yτ′k′,ν′)+V2−(k⋀k′)(yτk,ν)+V(yτ′k′,ν′,yτk,ν)×(2−(k⋀k′)2−(k⋀k′)+d(yτ′k′,ν′,yτk,ν))ϵ])r/p}q/r≲∑j=−∞∞2jq(s−s0){∑k′=−∞∞2k′s0r[∑k=−∞j2−|k′−k|ϵ′p(∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)|Dk(f)(yτk,ν)|p)]r/p}q/r.Let ϵ′′=ϵ′ when r/p≤1 and ϵ′′∈(|s0|,ϵ′) when r/p>1. By (5.5) when r/p≤1 or Hölder's inequality when r/p>1,
we further obtain(8.50)J1≲∑j=−∞∞2jq(s−s0){∑k′=−∞∞2k′s0r[∑k=−∞j2−|k′−k|ϵ′′r(∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)|Dk(f)(yτk,ν)|p)r/p]}q/r≲∥f∥B˙p,qs(𝒳)q,where in the last step, we omit
some estimates similar to the case p≤1. This completes the estimate for J1.
We now turn to the estimate forJ2. We also need to consider two cases. We first assume that p≤1. In this case, the estimate (5.10) and Lemma 5.2 prove that(8.51)J2≲∑j=−∞∞2jq(s−s1){∑k′=−∞∞2k′s1r(∑k=j+1∞2−|k′−k|ϵp2n[k−(k⋀k′)](1−p)×∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)|Dk(f)(yτk,ν)|p)r/p}q/r.Now let ϵ′=0 when r/p≤1 or ϵ′∈(0,s−s1) when r/p>1. Using (5.5) when r/p≤1 or Hölder's inequality when r/p>1 gives that(8.52)J2≲∑j=−∞∞2jq(s−s1−ϵ′){∑k′=−∞∞2k′s1r[∑k=j+1∞2−|k′−k|ϵr2n[k−(k⋀k′)](1/p−1)r2kϵ′r×(∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)|Dk(f)(yτk,ν)|p)r/p]}q/r≲∑j=−∞∞2jq(s−s1−ϵ′){∑k=j+1∞2k(s1+ϵ′)r(∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)|Dk(f)(yτk,ν)|p)r/p}q/r.Using q/r>1 together with Hölder's inequality then further
yields that(8.53)J2≲∑j=−∞∞2jq(s−s1−ϵ′)/2{∑k=j+1∞2k(s1+ϵ′−s)q/22ksq(∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)|Dk(f)(yτk,ν)|p)q/p}≲∥f∥B˙p,qs(𝒳)q.If 1<p≤∞,
letting ϵ′′∈(0,s−s1),
by Hölder's inequality and Lemma 5.2, we obtain(8.54)J2≲∑j=−∞∞2jq(s−s1−ϵ′′){∑k′=−∞∞2k′s1r(∑k=j+1∞2−|k′−k|ϵp2kϵ′′p∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)|Dk(f)(yτk,ν)|p)r/p}q/r.Now, if r/p≤1,
we then take ϵ′=0,
and if r/p>1,
we take ϵ′∈(0,s−ϵ′′−s1);
by (5.5) or Hölder's inequality, we obtain(8.55)J2≲∑j=−∞∞2jq(s−s1−ϵ′′−ϵ′){∑k′=−∞∞2k′s1r[∑k=j+1∞2−|k′−k|ϵr2k(ϵ′′+ϵ′)r×(∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)|Dk(f)(yτk,ν)|p)r/p]}q/r≲∥f∥B˙p,qs(𝒳)q.Thus, (8.42) holds.
From (8.37), (8.42), and Proposition 8.2(iii) together
with Proposition 5.10(i), by taking 0<r<q0 and 0<r<q1,
we deduce that(8.56)B˙p,qs(𝒳)⊂(B˙p,rs0(𝒳),B˙p,rs1(𝒳))σ,q⊂(B˙p,q0s0(𝒳),B˙p,q1s1(𝒳))σ,q⊂(B˙p,∞s0(𝒳),B˙p,∞s1(𝒳))σ,q⊂B˙p,qs(𝒳).Thus, (i) holds.
To see (ii), by (i) and Proposition8.2(iii) together
with Propositions 5.10(ii) and 6.9(ii), we have(8.57)B˙p,qs(𝒳)=(B˙p,min{p,q0}s0(𝒳),B˙p,min{p,q1}s1(𝒳))σ,q⊂(F˙p,q0s0(𝒳),F˙p,q1s1(𝒳))σ,q⊂(B˙p,max{p,q0}s0(𝒳),B˙p,max{p,q1}s1(𝒳))σ,q=B˙p,qs(𝒳).Thus, (ii) holds, which
completes the proof of Theorem 8.8.Theorem 8.9.
Letϵ be as in Definition 5.29, σ∈(0,1), s0,s1∈(−ϵ,ϵ), s0≠s1, and s=(1−σ)s0+σs1. Then
(i)
ifmax{p(s0,ϵ),p(s1,ϵ)}<p≤∞ and 0<q0,q1,q≤∞,
then(8.58)(Bp,q0s0(𝒳),Bp,q1s1(𝒳))σ,q=Bp,qs(𝒳);(ii)
ifmax{p(s0,ϵ),p(s1,ϵ)}<p≤∞,p(si,ϵ)<qi≤∞ for i=0,1 and 0<q≤∞,
then(8.59)(Fp,q0s0(𝒳),Fp,q1s1(𝒳))σ,q=Bp,qs(𝒳).Proof.
Similarly to the proof of Theorem8.8, we
only need to verify (i), while (ii) can be deduced from (i), Propositions 8.10, 5.31, and 6.21.
To prove (i), by Proposition5.31(iv), we know that(8.60)Bp,q0s0(𝒳),Bp,q1s1(𝒳)⊂(𝒢0ϵ(β,γ))′with max{0,−s0+n(1/p−1)+,−s1+n(1/p−1)+}<β<ϵ and n(1/p−1)+<γ<ϵ. Thus, in this case, we can take ℋ=(𝒢0ϵ(β,γ))′ with β and γ as  above.
We now verify that(8.61)(Bp,∞s0(𝒳),Bp,∞s1(𝒳))σ,q⊂Bp,qs(𝒳).By Proposition 8.2(ii), without
loss of generality, we may assume that s0>s1.
Assume thatf∈(Bp,∞s0(𝒳),Bp,∞s1(𝒳))σ,q and f=f0+f1 with f0∈Bp,∞s0(𝒳) and f1∈Bp,∞s1(𝒳). Let {Dk}k∈ℤ+ be as in Definition 5.29. Then,(8.62)2ks0∥Dk(f)∥Lp(𝒳)≲2ks0∥Dk(f0)∥Lp(𝒳)+2k(s0−s1)2ks1∥Dk(f1)∥Lp(𝒳)≲∥f0∥Bp,∞s0(𝒳)+2k(s0−s1)∥f1∥Bp,∞s1(𝒳).Taking the infimum on all
representations f=f0+f1 yields that for all k∈ℤ+,(8.63)2ks0∥Dk(f)∥Lp(𝒳)≲K(2k(s0−s1),f;Bp,∞s0(𝒳),Bp,∞s1(𝒳)).If 0<q<∞,
from (8.63), it follows that(8.64)∫0∞t−σq[K(t,f;Bp,∞s0(𝒳),Bp,∞s1(𝒳))]qdtt=∑k=−∞∞∫2(k−1)(s0−s1)2k(s0−s1)t−σq[K(t,f;Bp,∞s0(𝒳),Bp,∞s1(𝒳))]qdtt≳∑k=0∞2−σqk(s0−s1)[K(2k(s0−s1),f;Bp,∞s0(𝒳),Bp,∞s1(𝒳))]q≳∑k=0∞2ksq∥Dk(f)∥Lp(𝒳)q≳∥f∥Bp,qs(𝒳)q;and if q=∞,
by (8.63), we then have(8.65)∥f∥Bp,∞s(𝒳)=supk∈ℤ+2ks∥Dk(f)∥Lp(𝒳)≲supk∈ℤ+2k(s−s0)K(2k(s0−s1),f;Bp,∞s0(𝒳),Bp,∞s1(𝒳))≲sup0<t<∞t−σK(t,f;Bp,∞s0(𝒳),Bp,∞s1(𝒳)).Thus, (8.61) holds.
We now prove that if0<r<q,
then(8.66)Bp,qs(𝒳)⊂(Bp,rs0(𝒳),Bp,rs1(𝒳))σ,q.Without loss of generality, we
may assume that s0>s1 again. Then we also have s0>s>s1. In what follows, we only consider the case q<∞ and we omit the details for the case q=∞ by similarity and simplicity. Let now f∈Bp,qs(𝒳). We then write(8.67)∥f∥(Bp,rs0(𝒳),Bp,rs1(𝒳))σ,qq=∫0∞t−σq[K(t,f;Bp,rs0(𝒳),Bp,rs1(𝒳))]qdtt≲∫01t−σq[K(t,f;Bp,rs0(𝒳),Bp,rs1(𝒳))]qdtt+∑j=0∞2−jσq(s0−s1)[K(2j(s0−s1),f;Bp,rs0(𝒳),Bp,rs1(𝒳))]q≡Y1+Y2.
To estimate the first term, by Proposition5.31(ii),
we have(8.68)K(t,f;Bp,rs0(𝒳),Bp,rs1(𝒳))≤t∥f∥Bp,rs1(𝒳)≲t∥f∥Bp,rs(𝒳),which shows that(8.69)Y1=∫01t−σq[K(t,f;Bp,rs0(𝒳),Bp,rs1(𝒳))]qdtt≲∥f∥Bp,rs(𝒳)q.
To estimate the second term, let all the notation be
as in Theorem4.14. For any j∈ℕ,
we write(8.70)f(x)={∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)mQτ0,ν(D0(f))D˜Qτ0,ν(x)+∑k=1j∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)Dk(f)(yτk,ν)D˜k(x,yτk,ν)}+∑k=j+1∞∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)Dk(f)(yτk,ν)D˜k(x,yτk,ν)≡f0j+f1j,where D˜Qτ0,ν(x) is as in the proof of Proposition 7.3. From
this and the definition of K-functional, it follows thatY2≲∑j=0∞2−jσq(s0−s1)(∥f0j∥Bp,rs0(𝒳)q+2j(s0−s1)∥f1j∥Bp,rs1(𝒳)q)≲∑j=0∞2jq(s−s0){∑τ′∈I0∑ν′=1N(0,τ′)μ(Qτ′0,ν′)[mQτ′0,ν′(|D0(f0j)|)]p}q/p+∑j=0∞2jq(s−s0){∑k′=1∞2k′s0r[∑τ′∈Ik′∑ν′=1N(k′,τ′)μ(Qτ′k′,ν′)|Dk′(f0j)(yτ′k′,ν′)|p]r/p}q/r(8.71)+∑j=0∞2jq(s−s1){∑τ′∈I0∑ν′=1N(0,τ′)μ(Qτ′0,ν′)[mQτ′0,ν′(|D0(f1j)|)]p}q/p+∑j=0∞2jq(s−s1){∑k′=1∞2k′s1r[∑τ′∈Ik′∑ν′=1N(k′,τ′)μ(Qτ′k′,ν′)|Dk′(f1j)(yτ′k′,ν′)|p]r/p}q/r≡J1+J2+J3+J4.
The estimate (5.82), (5.5), and Lemma 5.2 show that
when p≤1,(8.72)J1≲∑j=0∞2j(s−s0)q{∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)[mQτ0,ν(|D0(f)|)]p}q/p+∑j=0∞2j(s−s0)q{∑k=1j2−kϵp∑τ∈Ik∑ν=1N(k,τ)[μ(Qτk,ν)|Dk(f)(yτk,ν)|]p[V1(yτk,ν)]1−p}q/p≲∥f∥Bp,qs(𝒳)q,where in the last inequality, we
used (5.5) when q/p≤1 or Hölder's inequality when 1<q/p<∞.
When1<p≤∞,
from (5.82), Hölder's inequality, and Lemma 5.2, it follows
that(8.73)J1≲∑j=0∞2j(s−s0)q{∑τ′∈I0∑ν′=1N(0,τ′)μ(Qτ′0,ν′)×(∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)[mQτ0,ν(|D0(f)|)]p×1V1(yτ′0,ν′)+V1(yτ0,ν)+V(yτ′0,ν′,yτ0,ν)1(1+d(yτ′0,ν′,yτ0,ν))ϵ)}q/p+∑j=0∞2j(s−s0)q{∑τ′∈I0∑ν′=1N(0,τ′)μ(Qτ′0,ν′)×[∑k=1j2−kϵ′p∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)|Dk(f)(yτk,ν)|p×1V1(yτ′0,ν′)+V1(yτk,ν)+V(yτ′0,ν′,yτk,ν)1(1+d(yτ′0,ν′,yτk,ν))ϵ]}q/p≲∥f∥Bp,qs(𝒳)q,where we chose ϵ′∈(|s|,ϵ).
By (5.82), we write(8.74)J2≲∑j=0∞2j(s−s0)q{∑k′=1∞2k′s0r(∑τ′∈Ik′∑ν′=1N(k′,τ′)μ(Qτ′k′,ν′)×[∑τ∈I0∑ν=1N(0,τ)2−k′ϵμ(Qτ0,ν)mQτ0,ν(|D0(f)|)×1V1(yτ′k′,ν′)+V1(yτ0,ν)+V(yτ′k′,ν′,yτ0,ν)×1(1+d(yτ′k′,ν′,yτ0,ν))ϵ]p)r/p}q/r+∑j=0∞2j(s−s0)q{∑k′=1∞2k′s0r(∑τ′∈Ik′∑ν′=1N(k′,τ′)μ(Qτ′k′,ν′)[∑k=1j2−|k′−k|ϵ∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)|Dk(f)(yτk,ν)|×|(D˜k′Dk)(yτ′k′,ν′,yτk,ν)|]p)r/p}q/r≡J2,1+J2,2.The estimate J2,2 is completely similar to that for J1. Thus, we only need to estimate J2,1.
Whenp≤1,
by (5.5) and Lemma 5.2, we have(8.75)J2,1≲∑j=0∞2j(s−s0)q{∑k′=1∞2k′(s0−ϵ)r}q/r(∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)[mQτ0,ν(|D0(f)|)]p)q/p≲∥f∥Bp,qs(𝒳)q,while when 1<p≤∞,
Hölder's inequality and Lemma 5.2 give that(8.76)J2,1≲∑j=0∞2j(s−s0)q{∑k′=1∞2k′s0r(∑τ′∈Ik′∑ν′=1N(k′,τ′)μ(Qτ′k′,ν′)×[∑τ∈I0∑ν=1N(0,τ)2−k′ϵpμ(Qτ0,ν)×[mQτ0,ν(|D0(f)|)]p1V1(yτ′k′,ν′)+V1(yτ0,ν)+V(yτ′k′,ν′,yτ0,ν)×1(1+d(yτ′k′,ν′,yτ0,ν))ϵ])r/p}q/r≲∥f∥Bp,qs(𝒳)q,which is also the desired
estimate.
Similarly, from (5.82), (5.5), Hölder's inequality,
and Lemma 5.2, it follows that if p≤1,(8.77)J3≲∑j=0∞2j(s−s1)q{∑τ′∈I0∑ν′=1N(0,τ′)μ(Qτ′0,ν′)×[∑k=j+1∞∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)|Dk(f)(yτk,ν)|×2−kϵ1V1(yτ′0,ν′)+V1(yτk,ν)+V(yτ′0,ν′,yτk,ν)1(1+d(yτ′0,ν′,yτk,ν))d+ϵ]p}q/p≲∑j=0∞2j(s−s1)q[∑k=j+1∞2−k(ϵ+n−n/p)p∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)|Dk(f)(yτk,ν)|p]q/p≲∥f∥Bp,qs(𝒳)q,where we choose a1=1 if q/p≤1 and a1∈(0,1) if q/p>1 such that a1s>(1−a1)(ϵ+n−n/p)+s1,
while when 1<p≤∞,
by Hölder's inequality and Lemma 5.2, we  have(8.78)J3≲∑j=0∞2j(s−s1)q{∑τ′∈I0∑ν′=1N(0,τ′)μ(Qτ′0,ν′)×[∑k=j+1∞∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)|Dk(f)(yτk,ν)|p×2−kϵ′p1V1(yτ′0,ν′)+V1(yτk,ν)+V(yτ′0,ν′,yτk,ν)1(1+d(yτ′0,ν′,yτk,ν))ϵ]}q/p≲∥f∥Bp,qs(𝒳)q,where we chose ϵ′∈(|s1|,ϵ) and a2∈(0,1) such that a2ϵ′>(1−a2)s−s1.
The estimateJ4 is similar to that for J2 and we omit the details, which completes the
proof of Theorem 8.9.


### 8.2. Dual Spaces

In this subsection, we are going to identify the dual
spaces of some classes of Besov spaces and Triebel-Lizorkin spaces. To this
end, we first recall the definitions of some auxiliary function spaces. Let0<p,q≤∞. The spacesLp(ℓ˙q)(𝒳)andℓ˙q(Lp)(𝒳) are  respectively defined to be the set of all
sequences f={fk}k∈ℤ of μ-measurable functions on 𝒳 such that(8.79)∥f∥Lp(ℓ˙q)(𝒳)=∥{∑k=−∞∞|fk|q}1/q∥Lp(𝒳)<∞,∥f∥ℓ˙q(Lp)(𝒳)={∑k=−∞∞∥fk∥Lp(𝒳)q}1/q<∞; and the spacesLp(ℓq)(𝒳)andℓq(Lp)(𝒳) are respectively defined to be the set of all
sequences f={fk}k∈ℤ+ of μ-measurable functions on 𝒳 such that(8.80)∥f∥Lp(ℓq)(𝒳)=∥{∑k=0∞|fk|q}1/q∥Lp(𝒳)<∞,∥f∥ℓq(Lp)(𝒳)={∑k=0∞∥fk∥Lp(𝒳)q}1/q<∞.The following result is well known (cf., e.g.,
[3, Proposition 2.11.1, pages 177-178]).Proposition 8.10.
Let1≤p<∞ and 0<q<∞. Then the following hold.
(i)
g∈(Lp(ℓ˙q)(𝒳))′ if and only if there exists a sequence {gj}j∈ℤ∈Lp′(ℓ˙q′)(𝒳) such that(8.81)g(f)=∑j=−∞∞∫𝒳gj(x)fj(x)dμ(x)for every f={fj}j∈ℤ∈Lp(ℓ˙q)(𝒳),
and ∥g∥~∥{gj}j∈ℤ∥Lp′(ℓ˙q′)(𝒳).(ii)
g∈(ℓ˙q(Lp)(𝒳))′ if and only if there is a sequence {gj}j∈ℤ∈ℓ˙q′(Lp′)(𝒳) such that(8.82)g(f)=∑j=−∞∞∫𝒳gj(x)fj(x)dμ(x)for every f={fj}j∈ℤ∈ℓ˙q(Lp)(𝒳),
and ∥g∥~∥{gj}j∈ℤ∥ℓ˙q′(Lp′)(𝒳).Using Proposition8.10, by a procedure similar to the
proof of Theorem 2.11.2 in [3], we can establish the following duality theorem for homogeneous Besov
spaces and Triebel-Lizorkin spaces. In what follows, when 0<q<1,
we also  let q′≡∞.Theorem 8.11.
Letϵ be as in Definition 5.8 and |s|<ϵ. Then, the following hold.
(i) If1≤p<∞ and 0<q<∞,
then(8.83)(B˙p,qs(𝒳))′=B˙p′,q′−s(𝒳).More precisely, given g∈B˙p′,q′−s(𝒳),
then ℒg(f)≡〈f,g〉 defines a linear functional on 𝒢°(ϵ,ϵ)∩B˙p,qs(𝒳) such that(8.84)|ℒg(f)|≤C∥f∥B˙p,qs(𝒳)∥g∥B˙p′,q′−s(𝒳),where C>0 is independent of f,
and this linear functional naturally extends to B˙p,qs(𝒳) by continuity with norm at most C∥g∥B˙p′,q′−s(𝒳).
Conversely, ifℒ is a linear functional on B˙p,qs(𝒳),
then there exists a unique g∈B˙p′,q′−s(𝒳) such that ℒ is the natural extension of ℒg,
with ∥g∥B˙p′,q′−s(𝒳)≤C∥ℒ∥.
(ii) If1<p,q<∞,
then(8.85)(F˙p,qs(𝒳))′=F˙p′,q′−s(𝒳).More precisely, given g∈F˙p′,q′−s(𝒳),
then ℒg(f)≡〈f,g〉 defines a linear functional on 𝒢°(ϵ,ϵ)∩F˙p,qs(𝒳) such that(8.86)|ℒg(f)|≤C∥f∥F˙p,qs(𝒳)∥g∥F˙p′,q′−s(𝒳),where C>0 is independent of f,
and this linear functional naturally extends to F˙p,qs(𝒳) by continuity with norm at most C∥g∥F˙p′,q′−s(𝒳).
Conversely, ifℒ is a linear functional on F˙p,qs(𝒳),
then there exists a unique g∈F˙p′,q′−s(𝒳) such that ℒ is the natural extension of ℒg,
with ∥g∥F˙p′,q′−s(𝒳)≤C∥ℒ∥.Proof.
Lets∈(−ϵ,ϵ). We first claim that if 1<p<∞ and p(s,ϵ)<q<∞,
then(8.87)F˙p′,q′−s(𝒳)⊂(F˙p,qs(𝒳))′,and that if 1≤p<∞ and 0<q<∞,(8.88)B˙p′,q′−s(𝒳)⊂(B˙p,qs(𝒳))′.
We first verify (8.87). Observe first that 1<p′<∞ and 1<q′≤∞. Let f∈F˙p′,q′−s(𝒳) and all the notation be as in Theorem 3.10. For
any φ∈𝒢°(ϵ1,ϵ2),
by the definition of F˙p′,q′−s(𝒳) and Theorem 3.13 together with Hölder's
inequality and Remark 5.5, we have(8.89)|f(φ)|=|∑k=−∞∞〈Dk(f),D˜kt(φ)〉|≤∫𝒳∑k=−∞∞|Dk(f)(x)||D˜kt(φ)(x)|dμ(x)≲∥f∥F˙p′,q′−s(𝒳)∥φ∥F˙p,qs(𝒳),which together with the
Hahn-Banach theorem and Proposition 5.21 shows (8.87).
Similarly, to see (8.88), for any f∈B˙p′,q′−s(𝒳) with 1<p′,q′≤∞ and φ∈𝒢°(ϵ1,ϵ2),
we have(8.90)|f(φ)|=|∑k=−∞∞〈Dk(f),D˜kt(φ)〉|≤∑k=−∞∞∥Dk(f)∥Lp′(𝒳)∥D˜kt(φ)∥Lp(𝒳)≤∥f∥B˙p′,q′−s(𝒳)∥φ∥B˙p,qs(𝒳),which together with the
Hahn-Banach theorem and Proposition 5.21 again proves (8.88).
We now complete the proof of (ii). Lets∈(−ϵ,ϵ) and let 1<p,q<∞. Let {Dk}k∈ℤ be as in Definition 5.8. Then(8.91)f∈F˙p,qs(𝒳)→{2ksDk(f)}k=−∞∞is a one-to-one mapping from F˙p,qs(𝒳) onto a subspace of Lp(ℓ˙q)(𝒳),
and every functional g∈(F˙p,qs(𝒳))′ can be interpreted as a functional on that
subspace. By the Hahn-Banach theorem, g can be extended to a continuous linear
functional on Lp(ℓ˙q)(𝒳),
where the norm of g is preserved. If φ∈𝒢°0ϵ(β,γ) with some fixed |s|<β<ϵ and max{s−κ/p,0,−s−κ(1−1/p)}<γ<ϵ,
then Proposition 8.10(ii) yields(8.92)g(φ)=∑k=−∞∞∫𝒳gk(x)Dk(φ)(x)dμ(x),(8.93)∥{2−skgk}k=−∞∞∥Lp′(ℓ˙q′)(𝒳)~∥g∥.The formula (8.92) can be written
as(8.94)g(φ)=∑k=−∞∞∫𝒳Dkt(gk)(x)φ(x)dμ(x),which means that in (𝒢°0ϵ(β,γ))′ with |s|<β<ϵ and max{s−κ/p,0,−s−κ(1−1/p)}<γ<ϵ,
we have(8.95)g(x)=∑k=−∞∞Dkt(gk)(x).Repeating the proof of
Proposition 5.4(ii), we find that(8.96)∥g∥F˙p′,q′−s(𝒳)≲∥{2−skgk}k=−∞∞∥Lp′(ℓ˙q′)(𝒳).Thus, g∈F˙p′,q′−s(𝒳),
which shows (ii).
To finish the proof of the theorem, we still need to
establish the converse of (8.88). To this end, we first let s∈(−ϵ,ϵ), 1≤p,q<∞,
and g∈(B˙p,qs(𝒳))′. Repeating the above proof with ℓ˙q′(Lp′)(𝒳) instead of Lp′(ℓ˙q′)(𝒳),
for any φ∈𝒢°0ϵ(β,γ) with some fixed |s|<β<ϵ and max{s−κ/p,0,−s−κ(1−1/p)}<γ<ϵ,
we see that (8.92) holds, and that(8.97)∥{2−skgk}k=−∞∞∥ℓ˙q′(Lp′)(𝒳)~∥g∥.Similarly, (8.92) means that
(8.95) holds in (𝒢°0ϵ(β,γ))′ with |s|<β<ϵ and max{s−κ/p,0,−s−κ(1−1/p)}<γ<ϵ. Repeating the proof of Proposition 5.6(ii), we obtain(8.98)∥g∥B˙p′,q′−s(𝒳)≲∥{2−skgk}k=−∞∞∥ℓ˙q′(Lp′)(𝒳).Thus, g∈B˙p′,q′−s(𝒳),
which proves (i) when s∈(−ϵ,ϵ) and 1≤p,q<∞.
Finally, lets∈(−ϵ,ϵ), 1≤p<∞, and 0<q<1. Then B˙p,qs(𝒳)⊂B˙p,1s(𝒳) by Proposition 5.10(i), which gives(8.99)B˙p′,∞−s(𝒳)=(B˙p,1s(𝒳))′⊂(B˙p,qs(𝒳))′.On the other hand, if g∈(B˙p,qs(𝒳))′ with s∈(−ϵ,ϵ), 1≤p<∞, and 0<q<1 and letting {Dk}k∈ℤ be as in Definition 5.8, we then have that for
all l∈ℤ,(8.100)|(Dlg)(φ)|=|g(Dlt(φ))|≲∥g∥∥Dlt(φ)∥B˙p,qs(𝒳).We now estimate ∥Dlt(φ)∥B˙p,qs(𝒳). For any k,l∈ℤ,
by Theorem 3.10, we have(8.101)DkDlt(φ)=∑k′=−∞∞DkDltD˜k′Dk′(φ)in 𝒢°0ϵ(β,γ) with β and γ as above. Using (3.2) of Lemma 3.2 and (iii) of Proposition 2.7 together with Hölder's inequality and the Fubini theorem gives(8.102)∥DkDltD˜k′Dk′(φ)∥Lp(𝒳)≲2−|k−l|ϵ1′∥D˜k′Dk′(φ)∥Lp(𝒳)≲2−|k−l|ϵ1′∥Dk′(φ)∥Lp(𝒳),∥DkDltD˜k′Dk′(φ)∥Lp(𝒳)≲∥DltD˜k′Dk′(φ)∥Lp(𝒳)≲2−|l−k′|ϵ1′′∥Dk′(φ)∥Lp(𝒳), where ϵ1′ and ϵ1′′ can be any positive number in (0,ϵ1). These estimates together with the geometric means and Proposition 2.7(iii)
again yield that for any σ∈(0,1),(8.103)∥DkDltD˜k′Dk′(φ)∥Lp(𝒳)≲2−|k−l|ϵ1′σ2−|l−k′|ϵ1′′(1−σ)∥Dk′(φ)∥Lp(𝒳)≲2−|k−l|ϵ1′σ2−|l−k′|ϵ1′′(1−σ)∥φ∥Lp(𝒳).If we choose σ∈(0,1) and ϵ1′∈(0,ϵ1) such that ϵ1′σ>|s|,
then from the above estimate and (8.101) together with Definition 5.1 and (5.5),
it follows that(8.104)∥Dlt(φ)∥B˙p,qs(𝒳)≲{∑k=−∞∞2ksq∥DkDlt(φ)∥Lp(𝒳)q}1/q≲{∑k=−∞∞2ksq[∑k′=−∞∞2−|k−l|ϵ1′σ2−|l−k′|ϵ1′′(1−σ)∥φ∥Lp(𝒳)]q}1/q≲2ls∥φ∥Lp(𝒳).Thus,(8.105)|(Dlg)(φ)|≲2−ls∥φ∥Lp(𝒳),which implies that(8.106)supl∈ℤ{2−ls∥Dl(g)∥Lp′(𝒳)}≲∥g∥.That is, g∈B˙p′∞−s(𝒳),
which completes the proof of (i), and hence, Theorem 8.11.To determine the dual space ofF˙p,qs(𝒳) when p=1,
following [82], we
first consider the corresponding spaces of sequences. Let λ be a sequence as in (7.1). Then for s∈ℝ, 0<q≤∞, and x∈𝒳,
we define(8.107)𝔊˙s,q(λ)(x)={∑k=−∞∞∑τ∈Ik∑ν=1N(k,τ)2ksq|λτk,ν|qχQτk,ν(x)}1/q,𝔊˙τ,k,νs,q(λ)(x)={∑k′=k+1∞∑Qτ′k′,ν′⊂Qτk,ν2k′sq|λτ′k′,ν′|qχQτ′k′,ν′(x)}1/q. Let m˙τ,k,νs,q(λ) denote the “1/4-median” of 𝔊˙τ,k,νs,q(λ) on Qτk,ν,
namely,(8.108)m˙τ,k,νs,q=inf{δ>0:μ({x∈Qτk,ν:𝔊˙τ,k,νs,q(λ)(x)>δ})<μ(Qτk,ν)4}.We also set(8.109)m˙s,q(λ)(x)=supk,τ,νm˙τ,k,νs,q(λ)χQτk,ν(x).The following conclusion is
trivial by the definition of f˙∞,qs(𝒳).Proposition 8.12.
Letδ∈(0,1], s∈ℝ, and 0<q≤∞. Assume that for each dyadic cube Q,
there exists a measurable set EQ⊂Q such that μ(EQ)/μ(Q)≥δ. Then(8.110)∥λ∥f˙∞,qs(𝒳)~supl∈ℤsupα∈Il{1μ(Qαl)[∑k=l∞∑τ∈Ik∑ν=1N(k,τ)2ksqμ(EQτk,ν)|λτk,ν|qχ{(τ,ν):Qτk,ν⊂Qαl}(τ,ν)]}1/q.We now establish a characterization off˙∞,qs(𝒳) by means of m˙s,q.Proposition 8.13.
Lets∈ℝ and 0<q≤∞. Then(8.111)∥λ∥f˙∞,qs(𝒳)~∥m˙s,q(λ)∥L∞(𝒳).Proof.
By Chebyshev's inequality, we see that(8.112)μ({x∈Qτk,ν:𝔊˙τ,k,νs,q(λ)(x)>δ})≤1δq∫Qτk,ν[𝔊˙τ,k,νs,q(λ)(x)]qdμ(x)≤μ(Qτk,ν)δq∥λ∥f˙∞,qs(𝒳)q<14μ(Qτk,ν),if δ>41/δ∥λ∥f˙∞,qs(𝒳). Hence,(8.113)∥m˙s,q(λ)∥L∞(𝒳)≲∥λ∥f˙∞,qs(𝒳).
To establish the converse, we introduce the extended
integer-valued stopping timek(x) for x∈𝒳 by(8.114)k(x)=inf{k∈ℤ:(∑k′=k∞∑τ′∈Ik′∑ν′=1N(k′,τ′)2k′sq|λτ′k′,ν′|qχQτ′k′,ν′)1/q≤m˙s,q(λ)(x)}.Also, for any Qτk,ν,
set(8.115)EQτk,ν={x∈Qτk,ν:k(x)≤k}.Then, obviously,(8.116)EQτk,ν={x∈Qτk,ν:𝔊˙τ,k,νs,q(λ)(x)≤m˙s,q(λ)(x)}.From (8.108), it follows
that(8.117)μ({x∈Qτk,ν:𝔊˙τ,k,νs,q(λ)(x)>m˙s,q(λ)(x)})≤μ({x∈Qτk,ν:𝔊˙τ,k,νs,q(λ)(x)>m˙τ,k,νs,q(λ)})≤14μ(Qτk,ν),and hence, μ(EQτk,ν)≥(3/4)μ(Qτk,ν),
and(8.118)(∑k=−∞∞∑τ∈Ik∑ν=1N(k,τ)2ksq|λτk,ν|qχEQτk,ν(x))1/q≲m˙s,q(λ)(x),which together with Proposition 8.12 yields(8.119)∥λ∥f˙∞,qs(𝒳)≲∥m˙s,q(λ)∥L∞(𝒳).This finishes the proof of
Proposition 8.13.We next prove the followingduality forf˙1,qs(𝒳).Proposition 8.14.
Assume thats∈ℝ and 0<q<∞. Then (f˙1,qs(𝒳))′=f˙∞,q′−s(𝒳). In particular, if λ={λτk,ν:k∈ℤ,τ∈Ik,ν=1,…,N(k,τ)}∈f˙∞,q′−s(𝒳),
then the map(8.120)t={tQτk,ν:k∈ℤ,τ∈Ik,ν=1,…,N(k,τ)}↦〈t,λ〉,where(8.121)〈t,λ〉=∑k=−∞∞∑τ∈Ik∑ν=1N(k,τ)tQτk,νλτk,ν¯μ(Qτk,ν),defines a continuous linear
functional on f˙1,qs(𝒳) with operator norm ∥λ∥(f˙1,qs(𝒳))′ equivalent to ∥λ∥f˙1,qs(𝒳),
and every ℓ∈(f˙1,qs(𝒳))′ is of this form for some  λ∈f˙1,qs(𝒳).Proof.
We first assume that1≤q<∞. Similarly to the proof of Proposition 8.13, let(8.122)EQτk,ν={x∈Qτk,ν:𝔊˙τ,k,ν−s,q′(λ)(x)≤m˙−s,q′(λ)(x)}.Then μ(EQτk,ν)≥(3/4)μ(Qτk,ν) and(8.123)(∑k=−∞∞∑τ∈Ik∑ν=1N(k,τ)2−ksq′|λτk,ν|q′χEQτk,ν(x))1/q′≲m˙−s,q′(λ)(x).From this, Hölder's inequality,
and Proposition 8.13, it follows that(8.124)|∑k=−∞∞∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)tQτk,νλτk,ν¯|≲∫𝒳∑k=−∞∞∑τ∈Ik∑ν=1N(k,τ)(2ks|tQτk,ν|χQτk,ν(x))(2−ks|λτk,ν|χEQτk,ν(x))dμ(x)≲∫𝒳{∑k=−∞∞∑τ∈Ik∑ν=1N(k,τ)2ksq|tQτk,ν|qχQτk,ν(x)}1/q{∑k=−∞∞∑τ∈Ik∑ν=1N(k,τ)2−ksq′|λτk,ν|q′χEQτk,ν(x)}1/q′dμ(x)≲∥t∥f˙1,qs(𝒳)∥m˙−s,q′(λ)∥L∞(𝒳)≲∥t∥f˙1,qs(𝒳)∥λ∥f˙∞,q′−s(𝒳),which yields(8.125)∥λ∥(f˙1,qs(𝒳))′≲∥λ∥f˙∞,q′−s(𝒳),if 1≤q<∞. The case 0<q<1 then follows from the trivial imbedding f˙1,qs(𝒳)⊂f˙1,1−s(𝒳).
Conversely, by Proposition8.10(i), it is easy to see
that every ℓ∈(f˙1,qs(𝒳))′ is of the form:(8.126)t↦∑k=−∞∞∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)tQτk,νλτk,νfor some λ as in (7.1). Assume first again that 1≤q<∞. Fix a dyadic cube Qαl. Let 𝒴 be the sequence space of all dyadic cubes Qτk,ν such that Qτk,ν⊂Qαl,
and let σ be a measure on  𝒴 such that the σ-measure of the point Qτk,ν is μ(Qτk,ν)/μ(Qαl). Then(8.127){1μ(Qαl)∑k=l∞∑τ∈Ik∑ν=1N(k,τ)2−ksqμ(Qτk,ν)|λτk,ν|q′χ{(τ,ν):Qτk,ν⊂Qαl}(τ,ν)}1/q′=∥{2−ksλτk,ν}Qτk,ν⊂Qαl∥ℓq′(𝒴,dσ)=sup∥t∥ℓq(𝒴,dσ)≤1|1μ(Qαl)∑k=l∞∑τ∈Ik∑ν=1N(k,τ)tQτk,ν2−ksλτk,ν¯μ(Qτk,ν)|≤∥λ∥(f˙1,qs(𝒳))′sup∥t∥ℓq(𝒴,dσ)≤1∥{2−kstQτk,ν1μ(Qαl)}Qτk,ν⊂Qαl∥f˙1,qs(𝒳).However, by Hölder's inequality,
we have(8.128)∥{2−kstQτk,ν1μ(Qαl)}Qτk,ν⊂Qαl∥f˙1,qs(𝒳)=1μ(Qαl)∫Qαl{∑k=l∞∑τ∈Ik∑ν=1N(k,τ)|tQτk,ν|qχ{(τ,ν):Qτk,ν⊂Qαl}(τ,ν)χQτk,ν(x)}1/qdμ(x)≲{1μ(Qαl)∑k=l∞∑τ∈Ik∑ν=1N(k,τ)|tQτk,ν|qχ{(τ,ν):Qτk,ν⊂Qαl}(τ,ν)μ(Qτk,ν)}1/q~∥t∥ℓq(𝒴,dσ).Thus, if 1≤q<∞,
then(8.129)∥λ∥f˙∞,q′−s(𝒳)≲∥λ∥(f˙1,qs(𝒳))′.For 0<q<1,
we have q′=∞. In this case, for any Qτk,ν,
we set (tQτk,ν)Qτ′k′,ν′=2−ks/μ(Qτk,ν) for Qτ′k′,ν′=Qτk,ν and 0 otherwise. Clearly, ∥tQτk,ν∥f˙1,qs(𝒳)=1 and hence,(8.130)∥λ∥f˙∞,∞−s(𝒳)≤supk,τ,ν|〈tQτk,ν,λ〉|≲∥λ∥(f˙1,qs(𝒳))′,which completes the proof of
Proposition 8.14.From Proposition7.1, Theorem 7.2, and Proposition 8.14, by a standard method as in [82, pages 79-80], we can obtain the dual space of F˙p,qs(𝒳) when p=1 as follows. We omit the  details.Theorem 8.15.
Letϵ be as in Definition 6.1, |s|<ϵ, and p(s,ϵ)<q<∞. Then (F˙1,qs(𝒳))′=F˙∞,q′−s(𝒳) in the sense of that in Theorem 8.11(ii).A slight modification of the proof of Proposition 2.11.1 in [3, pages
177-178] again gives the following inhomogeneous version of Propositions 8.10 and 8.14. We omit the details.Proposition 8.16.
Let1≤p<∞ and 0<q<∞. Then the following hold.
(i)
g∈(Lp(ℓq)(𝒳))′ if and only if there is a sequence {gj}j∈ℤ+∈Lp′(ℓq′)(𝒳) such that(8.131)g(f)=∑j=0∞∫𝒳gj(x)fj(x)dμ(x)for every f={fj}j∈ℤ+∈Lp(ℓq)(𝒳),
and ∥g∥~∥{gj}j∈ℤ+∥Lp′(ℓq′)(𝒳).(ii)
g∈(ℓq(Lp)(𝒳))′ if and only if there is a sequence {gj}j∈ℤ+∈ℓq′(Lp′)(𝒳)(8.132)g(f)=∑j=0∞∫𝒳gj(x)fj(x)dμ(x)for every f={fj}j∈ℤ+∈ℓq(Lp)(𝒳),
and ∥g∥~∥{gj}j∈ℤ+∥ℓq′(Lp′)(𝒳).Theorem 8.17.
Assume thats∈ℝ and 0<q<∞. Then (f1,qs(𝒳))′=f∞,q′−s(𝒳). In particular, if λ={λτk,ν:k∈ℤ+,τ∈Ik,ν=1,…,N(k,τ)}∈f∞,q′−s(𝒳),
then the map(8.133)t={tQτk,ν:k∈ℤ+,τ∈Ik,ν=1,…,N(k,τ)}→〈t,λ〉,where(8.134)〈t,λ〉=∑k=0∞∑τ∈Ik∑ν=1N(k,τ)tQτk,νλτk,ν¯μ(Qτk,ν),defines a continuous linear
functional on f1,qs(𝒳) with operator norm ∥λ∥(f1,qs(𝒳))′ equivalent to ∥λ∥f1,qs(𝒳),
and every ℓ∈(f1,qs(𝒳))′ is of this form for some λ∈f1,qs(𝒳).With Proposition8.16 and Theorem 8.17, respectively,
in place of Propositions 8.10 and 8.14, by a procedure similar to the
proof of Theorem 8.11 (see also [3, the proof of Theorem 2.11.2]) and Theorem 5.6 in
[82, pages 79-80], we
can establish the following inhomogeneous version of Theorems 8.11 and 8.15, which describes the dual spaces of some inhomogeneous Besov and
Triebel-Lizorkin spaces. We omit the  details.Theorem 8.18.
Letϵ be as in Definition 5.8 and |s|<ϵ. Then, the following hold.
(i) If1≤p<∞ and 0<q<∞,
then(8.135)(Bp,qs(𝒳))′=Bp′,q′−s(𝒳).More precisely, given g∈Bp′,q′−s(𝒳),
then ℒg(f)≡〈f,g〉 defines a linear functional on 𝒢(ϵ,ϵ)∩Bp,qs(𝒳) such that(8.136)|ℒg(f)|≤C∥f∥Bp,qs(𝒳)∥g∥Bp′,q′−s(𝒳),where C>0 is independent of f,
and this linear functional naturally extends to Bp,qs(𝒳) by continuity with norm at most C∥g∥Bp′,q′−s(𝒳).
Conversely, ifℒ is a linear functional on Bp,qs(𝒳),
then there exists a unique g∈Bp′,q′−s(𝒳) such that ℒ is the natural extension of ℒg,
with ∥g∥Bp′,q′−s(𝒳)≤C∥ℒ∥.
(ii) If1<p,q<∞,
or p=1 and p(s,ϵ)<q<∞,
then(8.137)(Fp,qs(𝒳))′=Fp′,q′−s(𝒳).More precisely, given g∈Fp′,q′−s(𝒳),
then ℒg(f)≡〈f,g〉 defines a linear functional on 𝒢(ϵ,ϵ)∩Fp,qs(𝒳) such that(8.138)|ℒg(f)|≤C∥f∥Fp,qs(𝒳)∥g∥Fp′,q′−s(𝒳),where C>0 is independent of f,
and this linear functional naturally extends to Fp,qs(𝒳) by continuity with norm at most C∥g∥Fp′,q′−s(𝒳).
Conversely, ifℒ is a linear functional on Fp,qs(𝒳),
then there exists a unique g∈Fp′,q′−s(𝒳) such that ℒ is the natural extension of ℒg,
with ∥g∥Fp′,q′−s(𝒳)≤C∥ℒ∥.


## 8.1. Real Interpolation Spaces

Let us first recall some general background on the
real interpolation method; see [3, pages 62–64] or [88, 89].Letℋ be a linear complex Hausdorff space, and let 𝒜0 and 𝒜1 be two complex quasi-Banach spaces such that 𝒜0⊂ℋ and 𝒜1⊂ℋ. Let 𝒜0+𝒜1 be the set of all elements a∈ℋ which can be represented as a=a0+a1 with a0∈𝒜0 and a1∈𝒜1. If 0<t<∞ and a∈𝒜0+𝒜1,
then Peetre's K-functional is given by(8.1)K(t,a)=K(t,a;𝒜0,𝒜1)=inf(∥a0∥𝒜0+t∥a1∥𝒜1),where the infimum is taken over
all representations of a of the form a=a0+a1 with a0∈𝒜0 and a1∈𝒜1.Definition 8.1.
Let0<σ<1. If 0<q<∞,
then one defines the interpolation
space(8.2)(𝒜0,𝒜1)σ,q≡{a:a∈𝒜0+𝒜1,∥a∥(𝒜0,𝒜1)σ,q≡{∫0∞[t−σK(t,a)]qdtt}1/q<∞}. If q=∞, then one defines
(8.3)(𝒜0,𝒜1)σ,∞≡{a:a∈𝒜0+𝒜1,∥a∥(𝒜0,𝒜1)σ,∞≡sup0<t<∞t−σK(t,a)<∞}.The following basic properties of(𝒜0,𝒜1)σ,q are proved in [3, pages 63-64] and [88, page 64].Proposition 8.2.
Let𝒜0 and 𝒜1 be two complex quasi-Banach spaces. Let 0<σ<1 and 0<q≤∞. Then,
(i)
(𝒜0,𝒜1)σ,q is a quasi-Banach space;(ii)
(𝒜0,𝒜1)σ,q=(𝒜1,𝒜0)1−σ,q;(iii)
letℋ be a linear complex Hausdorff space, and let ℬ0 and ℬ1 be two complex quasi-Banach spaces such that 𝒜0⊂ℬ0⊂ℋ and 𝒜1⊂ℬ1⊂ℋ. Then (𝒜0,𝒜1)σ,q⊂(ℬ0,ℬ1)σ,q.Using Theorems7.2 and 7.4 together with the
method of retraction and coretraction as in the proofs of Theorems 2.4.1 and 2.4.2 in [89],
we can easily deduce the following interpolation theorems; see also [90].Theorem 8.3.
Letϵ be as in Definition 5.8 and σ∈(0,1).
(i)
Let−ϵ<s0,s1<ϵ, s0≠s1, 1≤p≤∞,
and 1≤q0,q1,q≤∞. Then(8.4)(B˙p,q0s0(𝒳),B˙p,q1s1(𝒳))σ,q=B˙p,qs(𝒳),where s=(1−σ)s0+σs1.(ii)
Let−ϵ<s<ϵ, 1≤p≤∞, 1≤q0,q1≤∞, and q0≠q1. Then(8.5)(B˙p,q0s(𝒳),B˙p,q1s(𝒳))σ,q=B˙p,qs(𝒳),where 1/q=(1−σ)/q0+σ/q1.(iii)
Let−ϵ<s0,s1<ϵ, 1≤p0,p1≤∞, and p0≠p1. Then(8.6)(B˙p0,p0s0(𝒳),B˙p1,p1s1(𝒳))σ,p=B˙p,ps(𝒳),where 1/p=(1−σ)/p0+σ/p1 and s=(1−σ)s0+σs1.Theorem 8.4.
Letϵ be as in Definition 5.8, −ϵ<s0,s1<ϵ, 1≤p0,p1<∞, 1≤q0,q1≤∞, σ∈(0,1), s=(1−σ)s0+σs1, 1/p=(1−σ)/p0+σ/p1,
and 1/q=(1−σ)/q0+σ/q1.
(i)
Ifs0≠s1,
then(8.7)(F˙p0,q0s0(𝒳),F˙p1,q1s1(𝒳))σ,p=F˙p,ps(𝒳).(ii)
Ifs0=s1=s, p0=q0, p1=q1, and q0≠q1,
then(8.8)(F˙p0,p0s(𝒳),F˙p1,p1s(𝒳))σ,p=F˙p,ps(𝒳).(iii)
Ifs0=s1=s, q0=q1=q, and p0≠p1,
then(8.9)(F˙p0,qs(𝒳),F˙p1,qs(𝒳))σ,p=F˙p,qs(𝒳).Theorem 8.5.
Letϵ be as in Definition 5.29 and σ∈(0,1).
(i)
Let−ϵ<s0,s1<ϵ, s0≠s1, 1≤p≤∞,
and 1≤q0,q1,q≤∞. Then(8.10)(Bp,q0s0(𝒳),Bp,q1s1(𝒳))σ,q=Bp,qs(𝒳),where s=(1−σ)s0+σs1.(ii)
Let−ϵ<s<ϵ, 1≤p≤∞, 1≤q0,q1≤∞, and q0≠q1. Then(8.11)(Bp,q0s(𝒳),Bp,q1s(𝒳))σ,q=Bp,qs(𝒳),where 1/q=(1−σ)/q0+σ/q1.(iii)
Let−ϵ<s0,s1<ϵ, 1≤p0,p1≤∞, and p0≠p1. Then(8.12)(Bp0,p0s0(𝒳),Bp1,p1s1(𝒳))σ,p=Bp,ps(𝒳),where 1/p=(1−σ)/p0+σ/p1 and s=(1−σ)s0+σs1.Theorem 8.6.
Letϵ be as in Definition 5.29, −ϵ<s0,s1<ϵ, 1≤p0,p1<∞, 1≤q0,q1≤∞, σ∈(0,1), s=(1−σ)s0+σs1, 1/p=(1−σ)/p0+σ/p1, and 1/q=(1−σ)/q0+σ/q1.
(i)
Ifs0≠s1,
then(8.13)(Fp0,q0s0(𝒳),Fp1,q1s1(𝒳))σ,p=Fp,ps(𝒳).(ii)
Ifs0=s1=s, p0=q0, p1=q1, and q0≠q1,
then(8.14)(Bp0,p0s(𝒳),Bp1,p1s(𝒳))σ,p=Bp,ps(𝒳).(iii)
Ifs0=s1=s, q0=q1=q, and p0≠p1,
then(8.15)(Fp0,qs(𝒳),Fp1,qs(𝒳))σ,p=Fp,qs(𝒳).Proofs of Theorem8.3 through Theorem 8.6.
The proofs of Theorem8.3 through Theorem 8.6 are similar by using [89, Theorem 1.2.4]. We only give an outline here; see also [90, the proofs of Proposition
3.3 and Theorem 3.1].
To prove Theorems8.3 and 8.4, by Proposition 5.10(iii), we know that(8.16)B˙pi,qisi(𝒳),F˙pi,qisi(𝒳)⊂(𝒢°0ϵ(βi,γi))′,where max{0,−si+n(1/pi−1)+}<βi<ϵ and max{n(1/pi−1)+,si−κ/pi}<γi<ϵ with i=0,1. We then let β=max(β0,β1) and γ=(γ0,γ1). Then(8.17)B˙pi,qisi(𝒳),F˙pi,qisi(𝒳)⊂(𝒢°0ϵ(β,γ))′.In this sense, {B˙p0,q0s0(𝒳),B˙p1,q1s1(𝒳)} and {F˙p0,q0s0(𝒳),F˙p1,q1s1(𝒳)} are interpolation couples in the sense of
[89, Section 1.2.1]. Now, for f∈(𝒢°0ϵ(β,γ))′,
with the notation of Theorem 4.11, we can define the coretraction operator S˙ by(8.18)S˙(f)(x)={S˙(f)k(x)}k=−∞∞,where for k∈ℤ,(8.19)S˙(f)k(x)=∑τ∈Ik∑ν=1N(k,τ)Dk(f)(yτk,ν)χQτk,ν(x);and the corresponding retraction
operator R˙ by(8.20)R˙({fk})(x)=∑k=−∞∞∑τ∈Ik∑ν=1N(k,τ)[∫Qτk,νfk(y)dμ(y)]D˜k(x,yτk,ν).
By Theorem 4.13, for any f∈(𝒢°0ϵ(β,γ))′,
we have R˙S˙(f)(x)=f(x). In what follows, for s∈ℝ, 0<q≤∞, and 0<p≤∞,
we say {fk}k=−∞∞∈ℓ˙s,q(Lp)(𝒳),
if(8.21)∥{fk}k=−∞∞∥ℓ˙s,q(Lp)(𝒳)={∑k=−∞∞2ksq∥fk∥Lp(𝒳)q}1/q<∞;and we say {fk}k=−∞∞∈Lp(ℓ˙s,q)(𝒳),
if(8.22)∥{fk}k=−∞∞∥Lp(ℓ˙s,q)(𝒳)=∥{∑k=−∞∞2ksq|fk(x)|q}1/q∥Lp(𝒳)<∞,where the usual modifications
are made when p=∞ or q=∞. If F is an interpolation functor, then one obtains
by [89, Theorem 1.2.4] that(8.23)∥f∥F({B˙p0,q0s0(𝒳),B˙p1,q1s1(𝒳)})~∥S˙(f)∥F({ℓ˙s0,q0(Lp0)(𝒳),ℓ˙s1,q1(Lp1)(𝒳)}),∥f∥F({F˙p0,q0s0(𝒳),F˙p1,q1s1(𝒳)})~∥S˙(f)∥F({Lp0(ℓ˙s0,q0)(𝒳),Lp1(ℓ˙s1,q1)(𝒳)}). Using Proposition 7.1 and
Theorem 7.2, we can then finish the proofs of Theorems 8.3 and 8.4 by
the same procedures as those in [89, pages 182-183] and [89, pages 185-186].
To prove Theorems8.5 and 8.6, by Proposition 5.31(iv), we know that(8.24)Bpi,qisi(𝒳),Fpi,qisi(𝒳)⊂(𝒢0ϵ(βi,γi))′,where max{0,−si+n(1/pi−1)+}<βi<ϵ and n(1/pi−1)+<γi<ϵ with i=0,1. We then let β=max(β0,β1) and γ=(γ0,γ1). Then(8.25)Bpi,qisi(𝒳),Fpi,qisi(𝒳)⊂(𝒢0ϵ(β,γ))′.In this sense, {Bp0,q0s0(𝒳),Bp1,q1s1(𝒳)} and {Fp0,q0s0(𝒳),Fp1,q1s1(𝒳)} are interpolation couples in the sense of
[89, Section 1.2.1]. Now, for f∈(𝒢0ϵ(β,γ))′,
with the notation of Theorem 4.14, we can define the coretraction operator S by(8.26)S(f)(x)={S(f)k(x)}k=0∞,where(8.27)S(f)0(x)=∑τ∈I0∑ν=1N(0,τ)mQτ0,ν(D0(f))χQτ0,ν(x)and for k∈ℕ,(8.28)S(f)k(x)=∑τ∈Ik∑ν=1N(k,τ)Dk(f)(yτk,ν)χQτk,ν(x);and the corresponding retraction
operator R by(8.29)R({fk})(x)=∑τ∈I0∑ν=1N(0,τ)[∫Qτ0,νf0(y)dμ(y)]D˜Qτ0,ν(x)+∑k=1∞∑τ∈Ik∑ν=1N(k,τ)[∫Qτk,νfk(y)dμ(y)]D˜k(x,yτk,ν),where D˜Qτ0,ν(x) is as in the proof of Proposition 7.3. By
Theorem 4.16, for any f∈(𝒢0ϵ(β,γ))′,
we have RS(f)(x)=f(x). In what follows, for s∈ℝ, 0<q≤∞, and 0<p≤∞,
we say {fk}k=0∞∈ℓs,q(Lp)(𝒳),
if(8.30)∥{fk}k=0∞∥ℓs,q(Lp)(𝒳)={∑k=0∞2ksq∥fk∥Lp(𝒳)q}1/q<∞;and we say {fk}k=0∞∈Lp(ℓs,q)(𝒳),
if(8.31)∥{fk}k=0∞∥Lp(ℓs,q)(𝒳)=∥{∑k=0∞2ksq|fk(x)|q}1/q∥Lp(𝒳)<∞,where the usual modifications
are made when p=∞ or q=∞. If F is an interpolation functor, then one obtains
by [89, Theorem 1.2.4] that(8.32)∥f∥F({Bp0,q0s0(𝒳),Bp1,q1s1(𝒳)})~∥S(f)∥F({ℓs0,q0(Lp0)(𝒳),ℓs1,q1(Lp1)(𝒳)}),∥f∥F({Fp0,q0s0(𝒳),Fp1,q1s1(𝒳)})~∥S(f)∥F({Lp0(ℓs0,q0),Lp1(ℓs1,q1)}). Using Proposition 7.3 and
Theorem 7.4, we then finish the proofs of Theorems 8.5 and 8.6 by the
same procedures as those in [89, pages 182-183] and [89, pages 185-186], which
completes the proofs of Theorem 8.3 through Theorem 8.6.We remark that Theorem8.3 through Theorem 8.6 only deal with Besov spaces and Triebel-Lizorkin spaces which are Banach spaces,
since p,q≥1. Using Theorems 7.2 and 7.4 together with the following fact that for 0<p0≠p1≤∞ and σ∈(0,1),
if 1/p=(1−σ)/p0+σ/p1,(8.33)(ℓ˙p0,ℓ˙p1)σ,p=ℓ˙p,(ℓp0,ℓp1)σ,p=ℓp (see [88, Theorem 5.2.1] and also
[89, Remark 1.18.6/5]), by a method similar to the proofs of Theorem 8.3 through Theorem 8.6 (see
also [82, Corollary 6.6]), we can easily establish the following interpolation theorem which covers also
cases when p<1. We omit the details.Theorem 8.7.
Letϵ be as in Definition 5.8, s0,s1∈(−ϵ,ϵ), p(s0,ϵ)<p0≤∞, p(s1,ϵ)<p1≤∞, and p0≠p1. Let σ∈(0,1), 1/p=(1−σ)/p0+σ/p1, and s=(1−σ)s0+σs1. Then,
(i)
(B˙p0,p0s0(𝒳),B˙p1,p1s1(𝒳))σ,p=B˙p,ps(𝒳);(ii)
(Bp0,p0s0(𝒳),Bp1,p1s1(𝒳))σ,p=Bp,ps(𝒳).Making use of Calderón reproducing formulae in place of frame characterizations, we can also directly establish the followingreal interpolation theorems for Besov and Triebel-Lizorkin spaces which are only quasi-Banach spaces; see also [3, Theorem 2.4.2] and [91].Theorem 8.8.
Letϵ be as in Definition 5.8, σ∈(0,1), s0,s1∈(−ϵ,ϵ), s0≠s1, and s=(1−σ)s0+σs1. Then
(i)
ifmax{p(s0,ϵ),p(s1,ϵ)}<p≤∞ and 0<q0,q1,q≤∞,
then(8.34)(B˙p,q0s0(𝒳),B˙p,q1s1(𝒳))σ,q=B˙p,qs(𝒳);(ii)
ifmax{p(s0,ϵ),p(s1,ϵ)}<p≤∞,p(si,ϵ)<qi≤∞ for i=0,1 and 0<q≤∞,
then(8.35)(F˙p,q0s0(𝒳),F˙p,q1s1(𝒳))σ,q=B˙p,qs(𝒳).Proof.
We first verify (i). By Proposition5.10(iii), we know that(8.36)B˙p,q0s0(𝒳),B˙p,q1s1(𝒳)⊂(𝒢°0ϵ(β,γ))′with max{0,−s0+n(1/p−1)+,−s1+n(1/p−1)+}<β<ϵ and max{n(1/p−1)+,s0−κ/p,s1−κ/p}<γ<ϵ. Thus, we can take ℋ=(𝒢°0ϵ(β,γ))′ with β and γ as above.
We now verify that(8.37)(B˙p,∞s0(𝒳),B˙p,∞s1(𝒳))σ,q⊂B˙p,qs(𝒳).By Proposition 8.2(ii), without
loss of generality, we may assume that s0>s1.
Assume thatf∈(B˙p,∞s0(𝒳),B˙p,∞s1(𝒳))σ,q and f=f0+f1 with f0∈B˙p,∞s0(𝒳) and f1∈B˙p,∞s1(𝒳). Let {Dk}k∈ℤ be as in Definition 5.8. Then,(8.38)2ks0∥Dk(f)∥Lp(𝒳)≲2ks0∥Dk(f0)∥Lp(𝒳)+2k(s0−s1)2ks1∥Dk(f1)∥Lp(𝒳)≲∥f0∥B˙p,∞s0(𝒳)+2k(s0−s1)∥f1∥B˙p,∞s1(𝒳).Taking the infimum on all
representations f=f0+f1 yields that(8.39)2ks0∥Dk(f)∥Lp(𝒳)≲K(2k(s0−s1),f;B˙p,∞s0(𝒳),B˙p,∞s1(𝒳)).If 0<q<∞,
from (8.39), it follows that(8.40)∫0∞t−σq[K(t,f;B˙p,∞s0(𝒳),B˙p,∞s1(𝒳))]qdtt=∑k=−∞∞∫2(k−1)(s0−s1)2k(s0−s1)t−σq[K(t,f;B˙p,∞s0(𝒳),B˙p,∞s1(𝒳))]qdtt≳∑k=−∞∞2−σqk(s0−s1)[K(2k(s0−s1),f;B˙p,∞s0(𝒳),B˙p,∞s1(𝒳))]q≳∑k=−∞∞2ksq∥Dk(f)∥Lp(𝒳)q≳∥f∥B˙p,qs(𝒳)q;and if q=∞,
by (8.39), we then have(8.41)∥f∥B˙p,∞s(𝒳)=supk∈ℤ2ks∥Dk(f)∥Lp(𝒳)≲supk∈ℤ2k(s−s0)K(2k(s0−s1),f;B˙p,∞s0(𝒳),B˙p,∞s1(𝒳))≲sup0<t<∞t−σK(t,f;B˙p,∞s0(𝒳),B˙p,∞s1(𝒳)).Thus, (8.37) holds.
We now prove that if0<r<q,
then(8.42)B˙p,qs(𝒳)⊂(B˙p,rs0(𝒳),B˙p,rs1(𝒳))σ,q.Without loss of generality, we
may assume that s0>s1 again. Then we also have s0>s>s1. In what follows, we only consider the case q<∞ and we omit the details for the case q=∞ by similarity and simplicity. Let now f∈B˙p,qs(𝒳). We then write(8.43)∥f∥(B˙p,rs0(𝒳),B˙p,rs1(𝒳))σ,qq=∫0∞t−σq[K(t,f;B˙p,rs0(𝒳),B˙p,rs1(𝒳))]qdtt≲∑j=−∞∞2−jσq(s0−s1)[K(2j(s0−s1),f;B˙p,rs0(𝒳),B˙p,rs1(𝒳))]q.Let all the notation be as in
Theorem 4.11. For any j∈ℤ,
we write(8.44)f(x)=∑k=−∞j∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)Dk(f)(yτk,ν)D˜k(x,yτk,ν)+∑k=j+1∞∑τ∈Ik∑ν=1N(k,τ)⋯≡f0j+f1j.From this and the definition of K-functional, it follows that(8.45)∥f∥(B˙p,rs0(𝒳),B˙p,rs1(𝒳))σ,qq≲∑j=−∞∞2−jσq(s0−s1)(∥f0j∥B˙p,rs0(𝒳)q+2j(s0−s1)∥f1j∥B˙p,rs1(𝒳)q)≲∑j=−∞∞2jq(s−s0){∑k′=−∞∞2k′s0r[∑τ′∈Ik′∑ν′=1N(k′,τ′)μ(Qτ′k′,ν′)|Dk′(f0j)(yτ′k′,ν′)|p]r/p}q/r+∑j=−∞∞2jq(s−s1){∑k′=−∞∞2k′s1r[∑τ′∈Ik′∑ν′=1N(k′,τ′)μ(Qτ′k′,ν′)|Dk′(f1j)(yτ′k′,ν′)|p]r/p}q/r≡J1+J2.
We first estimateJ1 in the case p≤1. In this case, by (5.10), (5.5), Lemma 5.2, and (5.12), we have(8.46)J1≲∑j=−∞∞2jq(s−s0){∑k′=−∞∞2k′s0r(∑k=−∞j2−|k′−k|ϵp2n[k−(k⋀k′)](1−p)×∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)|Dk(f)(yτk,ν)|p)r/p}q/r.Now, we choose ϵ′∈(|s0|,ϵ) such that p>n/(n+s0+ϵ′). Using (5.5) when r/p≤1 or Hölder's inequality when 1<r/p<∞ then further shows that(8.47)J1≲∑j=−∞∞2jq(s−s0){∑k′=−∞∞2k′s0r[∑k=−∞j2−|k′−k|ϵ′r2n[k−(k⋀k′)](1/p−1)r×(∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)|Dk(f)(yτk,ν)|p)r/p]}q/r≲∑j=−∞∞2jq(s−s0){∑k=−∞j2ks0r(∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)|Dk(f)(yτk,ν)|p)r/p}q/r.Since q/r>1,
by Hölder's inequality and s0>s,
we have(8.48)J1≲∑j=−∞∞2jq(s−s0)/2{∑k=−∞j2k(s0−s)q/22ksq(∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)|Dk(f)(yτk,ν)|p)q/p}≲∥f∥B˙p,qs(𝒳)q,which is the desired estimate in
the  case p≤1.
While when1<p≤∞,
letting ϵ′∈(|s0|,ϵ),
by Hölder's inequality and Lemma 5.2, we have(8.49)J1≲∑j=−∞∞2jq(s−s0){∑k′=−∞∞2k′s0r(∑τ′∈Ik′∑ν′=1N(k′,τ′)μ(Qτ′k′,ν′)×[∑k=−∞j2−|k′−k|ϵ′p∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)|Dk(f)(yτk,ν)|p×1V2−(k⋀k′)(yτ′k′,ν′)+V2−(k⋀k′)(yτk,ν)+V(yτ′k′,ν′,yτk,ν)×(2−(k⋀k′)2−(k⋀k′)+d(yτ′k′,ν′,yτk,ν))ϵ])r/p}q/r≲∑j=−∞∞2jq(s−s0){∑k′=−∞∞2k′s0r[∑k=−∞j2−|k′−k|ϵ′p(∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)|Dk(f)(yτk,ν)|p)]r/p}q/r.Let ϵ′′=ϵ′ when r/p≤1 and ϵ′′∈(|s0|,ϵ′) when r/p>1. By (5.5) when r/p≤1 or Hölder's inequality when r/p>1,
we further obtain(8.50)J1≲∑j=−∞∞2jq(s−s0){∑k′=−∞∞2k′s0r[∑k=−∞j2−|k′−k|ϵ′′r(∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)|Dk(f)(yτk,ν)|p)r/p]}q/r≲∥f∥B˙p,qs(𝒳)q,where in the last step, we omit
some estimates similar to the case p≤1. This completes the estimate for J1.
We now turn to the estimate forJ2. We also need to consider two cases. We first assume that p≤1. In this case, the estimate (5.10) and Lemma 5.2 prove that(8.51)J2≲∑j=−∞∞2jq(s−s1){∑k′=−∞∞2k′s1r(∑k=j+1∞2−|k′−k|ϵp2n[k−(k⋀k′)](1−p)×∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)|Dk(f)(yτk,ν)|p)r/p}q/r.Now let ϵ′=0 when r/p≤1 or ϵ′∈(0,s−s1) when r/p>1. Using (5.5) when r/p≤1 or Hölder's inequality when r/p>1 gives that(8.52)J2≲∑j=−∞∞2jq(s−s1−ϵ′){∑k′=−∞∞2k′s1r[∑k=j+1∞2−|k′−k|ϵr2n[k−(k⋀k′)](1/p−1)r2kϵ′r×(∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)|Dk(f)(yτk,ν)|p)r/p]}q/r≲∑j=−∞∞2jq(s−s1−ϵ′){∑k=j+1∞2k(s1+ϵ′)r(∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)|Dk(f)(yτk,ν)|p)r/p}q/r.Using q/r>1 together with Hölder's inequality then further
yields that(8.53)J2≲∑j=−∞∞2jq(s−s1−ϵ′)/2{∑k=j+1∞2k(s1+ϵ′−s)q/22ksq(∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)|Dk(f)(yτk,ν)|p)q/p}≲∥f∥B˙p,qs(𝒳)q.If 1<p≤∞,
letting ϵ′′∈(0,s−s1),
by Hölder's inequality and Lemma 5.2, we obtain(8.54)J2≲∑j=−∞∞2jq(s−s1−ϵ′′){∑k′=−∞∞2k′s1r(∑k=j+1∞2−|k′−k|ϵp2kϵ′′p∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)|Dk(f)(yτk,ν)|p)r/p}q/r.Now, if r/p≤1,
we then take ϵ′=0,
and if r/p>1,
we take ϵ′∈(0,s−ϵ′′−s1);
by (5.5) or Hölder's inequality, we obtain(8.55)J2≲∑j=−∞∞2jq(s−s1−ϵ′′−ϵ′){∑k′=−∞∞2k′s1r[∑k=j+1∞2−|k′−k|ϵr2k(ϵ′′+ϵ′)r×(∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)|Dk(f)(yτk,ν)|p)r/p]}q/r≲∥f∥B˙p,qs(𝒳)q.Thus, (8.42) holds.
From (8.37), (8.42), and Proposition 8.2(iii) together
with Proposition 5.10(i), by taking 0<r<q0 and 0<r<q1,
we deduce that(8.56)B˙p,qs(𝒳)⊂(B˙p,rs0(𝒳),B˙p,rs1(𝒳))σ,q⊂(B˙p,q0s0(𝒳),B˙p,q1s1(𝒳))σ,q⊂(B˙p,∞s0(𝒳),B˙p,∞s1(𝒳))σ,q⊂B˙p,qs(𝒳).Thus, (i) holds.
To see (ii), by (i) and Proposition8.2(iii) together
with Propositions 5.10(ii) and 6.9(ii), we have(8.57)B˙p,qs(𝒳)=(B˙p,min{p,q0}s0(𝒳),B˙p,min{p,q1}s1(𝒳))σ,q⊂(F˙p,q0s0(𝒳),F˙p,q1s1(𝒳))σ,q⊂(B˙p,max{p,q0}s0(𝒳),B˙p,max{p,q1}s1(𝒳))σ,q=B˙p,qs(𝒳).Thus, (ii) holds, which
completes the proof of Theorem 8.8.Theorem 8.9.
Letϵ be as in Definition 5.29, σ∈(0,1), s0,s1∈(−ϵ,ϵ), s0≠s1, and s=(1−σ)s0+σs1. Then
(i)
ifmax{p(s0,ϵ),p(s1,ϵ)}<p≤∞ and 0<q0,q1,q≤∞,
then(8.58)(Bp,q0s0(𝒳),Bp,q1s1(𝒳))σ,q=Bp,qs(𝒳);(ii)
ifmax{p(s0,ϵ),p(s1,ϵ)}<p≤∞,p(si,ϵ)<qi≤∞ for i=0,1 and 0<q≤∞,
then(8.59)(Fp,q0s0(𝒳),Fp,q1s1(𝒳))σ,q=Bp,qs(𝒳).Proof.
Similarly to the proof of Theorem8.8, we
only need to verify (i), while (ii) can be deduced from (i), Propositions 8.10, 5.31, and 6.21.
To prove (i), by Proposition5.31(iv), we know that(8.60)Bp,q0s0(𝒳),Bp,q1s1(𝒳)⊂(𝒢0ϵ(β,γ))′with max{0,−s0+n(1/p−1)+,−s1+n(1/p−1)+}<β<ϵ and n(1/p−1)+<γ<ϵ. Thus, in this case, we can take ℋ=(𝒢0ϵ(β,γ))′ with β and γ as  above.
We now verify that(8.61)(Bp,∞s0(𝒳),Bp,∞s1(𝒳))σ,q⊂Bp,qs(𝒳).By Proposition 8.2(ii), without
loss of generality, we may assume that s0>s1.
Assume thatf∈(Bp,∞s0(𝒳),Bp,∞s1(𝒳))σ,q and f=f0+f1 with f0∈Bp,∞s0(𝒳) and f1∈Bp,∞s1(𝒳). Let {Dk}k∈ℤ+ be as in Definition 5.29. Then,(8.62)2ks0∥Dk(f)∥Lp(𝒳)≲2ks0∥Dk(f0)∥Lp(𝒳)+2k(s0−s1)2ks1∥Dk(f1)∥Lp(𝒳)≲∥f0∥Bp,∞s0(𝒳)+2k(s0−s1)∥f1∥Bp,∞s1(𝒳).Taking the infimum on all
representations f=f0+f1 yields that for all k∈ℤ+,(8.63)2ks0∥Dk(f)∥Lp(𝒳)≲K(2k(s0−s1),f;Bp,∞s0(𝒳),Bp,∞s1(𝒳)).If 0<q<∞,
from (8.63), it follows that(8.64)∫0∞t−σq[K(t,f;Bp,∞s0(𝒳),Bp,∞s1(𝒳))]qdtt=∑k=−∞∞∫2(k−1)(s0−s1)2k(s0−s1)t−σq[K(t,f;Bp,∞s0(𝒳),Bp,∞s1(𝒳))]qdtt≳∑k=0∞2−σqk(s0−s1)[K(2k(s0−s1),f;Bp,∞s0(𝒳),Bp,∞s1(𝒳))]q≳∑k=0∞2ksq∥Dk(f)∥Lp(𝒳)q≳∥f∥Bp,qs(𝒳)q;and if q=∞,
by (8.63), we then have(8.65)∥f∥Bp,∞s(𝒳)=supk∈ℤ+2ks∥Dk(f)∥Lp(𝒳)≲supk∈ℤ+2k(s−s0)K(2k(s0−s1),f;Bp,∞s0(𝒳),Bp,∞s1(𝒳))≲sup0<t<∞t−σK(t,f;Bp,∞s0(𝒳),Bp,∞s1(𝒳)).Thus, (8.61) holds.
We now prove that if0<r<q,
then(8.66)Bp,qs(𝒳)⊂(Bp,rs0(𝒳),Bp,rs1(𝒳))σ,q.Without loss of generality, we
may assume that s0>s1 again. Then we also have s0>s>s1. In what follows, we only consider the case q<∞ and we omit the details for the case q=∞ by similarity and simplicity. Let now f∈Bp,qs(𝒳). We then write(8.67)∥f∥(Bp,rs0(𝒳),Bp,rs1(𝒳))σ,qq=∫0∞t−σq[K(t,f;Bp,rs0(𝒳),Bp,rs1(𝒳))]qdtt≲∫01t−σq[K(t,f;Bp,rs0(𝒳),Bp,rs1(𝒳))]qdtt+∑j=0∞2−jσq(s0−s1)[K(2j(s0−s1),f;Bp,rs0(𝒳),Bp,rs1(𝒳))]q≡Y1+Y2.
To estimate the first term, by Proposition5.31(ii),
we have(8.68)K(t,f;Bp,rs0(𝒳),Bp,rs1(𝒳))≤t∥f∥Bp,rs1(𝒳)≲t∥f∥Bp,rs(𝒳),which shows that(8.69)Y1=∫01t−σq[K(t,f;Bp,rs0(𝒳),Bp,rs1(𝒳))]qdtt≲∥f∥Bp,rs(𝒳)q.
To estimate the second term, let all the notation be
as in Theorem4.14. For any j∈ℕ,
we write(8.70)f(x)={∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)mQτ0,ν(D0(f))D˜Qτ0,ν(x)+∑k=1j∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)Dk(f)(yτk,ν)D˜k(x,yτk,ν)}+∑k=j+1∞∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)Dk(f)(yτk,ν)D˜k(x,yτk,ν)≡f0j+f1j,where D˜Qτ0,ν(x) is as in the proof of Proposition 7.3. From
this and the definition of K-functional, it follows thatY2≲∑j=0∞2−jσq(s0−s1)(∥f0j∥Bp,rs0(𝒳)q+2j(s0−s1)∥f1j∥Bp,rs1(𝒳)q)≲∑j=0∞2jq(s−s0){∑τ′∈I0∑ν′=1N(0,τ′)μ(Qτ′0,ν′)[mQτ′0,ν′(|D0(f0j)|)]p}q/p+∑j=0∞2jq(s−s0){∑k′=1∞2k′s0r[∑τ′∈Ik′∑ν′=1N(k′,τ′)μ(Qτ′k′,ν′)|Dk′(f0j)(yτ′k′,ν′)|p]r/p}q/r(8.71)+∑j=0∞2jq(s−s1){∑τ′∈I0∑ν′=1N(0,τ′)μ(Qτ′0,ν′)[mQτ′0,ν′(|D0(f1j)|)]p}q/p+∑j=0∞2jq(s−s1){∑k′=1∞2k′s1r[∑τ′∈Ik′∑ν′=1N(k′,τ′)μ(Qτ′k′,ν′)|Dk′(f1j)(yτ′k′,ν′)|p]r/p}q/r≡J1+J2+J3+J4.
The estimate (5.82), (5.5), and Lemma 5.2 show that
when p≤1,(8.72)J1≲∑j=0∞2j(s−s0)q{∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)[mQτ0,ν(|D0(f)|)]p}q/p+∑j=0∞2j(s−s0)q{∑k=1j2−kϵp∑τ∈Ik∑ν=1N(k,τ)[μ(Qτk,ν)|Dk(f)(yτk,ν)|]p[V1(yτk,ν)]1−p}q/p≲∥f∥Bp,qs(𝒳)q,where in the last inequality, we
used (5.5) when q/p≤1 or Hölder's inequality when 1<q/p<∞.
When1<p≤∞,
from (5.82), Hölder's inequality, and Lemma 5.2, it follows
that(8.73)J1≲∑j=0∞2j(s−s0)q{∑τ′∈I0∑ν′=1N(0,τ′)μ(Qτ′0,ν′)×(∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)[mQτ0,ν(|D0(f)|)]p×1V1(yτ′0,ν′)+V1(yτ0,ν)+V(yτ′0,ν′,yτ0,ν)1(1+d(yτ′0,ν′,yτ0,ν))ϵ)}q/p+∑j=0∞2j(s−s0)q{∑τ′∈I0∑ν′=1N(0,τ′)μ(Qτ′0,ν′)×[∑k=1j2−kϵ′p∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)|Dk(f)(yτk,ν)|p×1V1(yτ′0,ν′)+V1(yτk,ν)+V(yτ′0,ν′,yτk,ν)1(1+d(yτ′0,ν′,yτk,ν))ϵ]}q/p≲∥f∥Bp,qs(𝒳)q,where we chose ϵ′∈(|s|,ϵ).
By (5.82), we write(8.74)J2≲∑j=0∞2j(s−s0)q{∑k′=1∞2k′s0r(∑τ′∈Ik′∑ν′=1N(k′,τ′)μ(Qτ′k′,ν′)×[∑τ∈I0∑ν=1N(0,τ)2−k′ϵμ(Qτ0,ν)mQτ0,ν(|D0(f)|)×1V1(yτ′k′,ν′)+V1(yτ0,ν)+V(yτ′k′,ν′,yτ0,ν)×1(1+d(yτ′k′,ν′,yτ0,ν))ϵ]p)r/p}q/r+∑j=0∞2j(s−s0)q{∑k′=1∞2k′s0r(∑τ′∈Ik′∑ν′=1N(k′,τ′)μ(Qτ′k′,ν′)[∑k=1j2−|k′−k|ϵ∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)|Dk(f)(yτk,ν)|×|(D˜k′Dk)(yτ′k′,ν′,yτk,ν)|]p)r/p}q/r≡J2,1+J2,2.The estimate J2,2 is completely similar to that for J1. Thus, we only need to estimate J2,1.
Whenp≤1,
by (5.5) and Lemma 5.2, we have(8.75)J2,1≲∑j=0∞2j(s−s0)q{∑k′=1∞2k′(s0−ϵ)r}q/r(∑τ∈I0∑ν=1N(0,τ)μ(Qτ0,ν)[mQτ0,ν(|D0(f)|)]p)q/p≲∥f∥Bp,qs(𝒳)q,while when 1<p≤∞,
Hölder's inequality and Lemma 5.2 give that(8.76)J2,1≲∑j=0∞2j(s−s0)q{∑k′=1∞2k′s0r(∑τ′∈Ik′∑ν′=1N(k′,τ′)μ(Qτ′k′,ν′)×[∑τ∈I0∑ν=1N(0,τ)2−k′ϵpμ(Qτ0,ν)×[mQτ0,ν(|D0(f)|)]p1V1(yτ′k′,ν′)+V1(yτ0,ν)+V(yτ′k′,ν′,yτ0,ν)×1(1+d(yτ′k′,ν′,yτ0,ν))ϵ])r/p}q/r≲∥f∥Bp,qs(𝒳)q,which is also the desired
estimate.
Similarly, from (5.82), (5.5), Hölder's inequality,
and Lemma 5.2, it follows that if p≤1,(8.77)J3≲∑j=0∞2j(s−s1)q{∑τ′∈I0∑ν′=1N(0,τ′)μ(Qτ′0,ν′)×[∑k=j+1∞∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)|Dk(f)(yτk,ν)|×2−kϵ1V1(yτ′0,ν′)+V1(yτk,ν)+V(yτ′0,ν′,yτk,ν)1(1+d(yτ′0,ν′,yτk,ν))d+ϵ]p}q/p≲∑j=0∞2j(s−s1)q[∑k=j+1∞2−k(ϵ+n−n/p)p∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)|Dk(f)(yτk,ν)|p]q/p≲∥f∥Bp,qs(𝒳)q,where we choose a1=1 if q/p≤1 and a1∈(0,1) if q/p>1 such that a1s>(1−a1)(ϵ+n−n/p)+s1,
while when 1<p≤∞,
by Hölder's inequality and Lemma 5.2, we  have(8.78)J3≲∑j=0∞2j(s−s1)q{∑τ′∈I0∑ν′=1N(0,τ′)μ(Qτ′0,ν′)×[∑k=j+1∞∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)|Dk(f)(yτk,ν)|p×2−kϵ′p1V1(yτ′0,ν′)+V1(yτk,ν)+V(yτ′0,ν′,yτk,ν)1(1+d(yτ′0,ν′,yτk,ν))ϵ]}q/p≲∥f∥Bp,qs(𝒳)q,where we chose ϵ′∈(|s1|,ϵ) and a2∈(0,1) such that a2ϵ′>(1−a2)s−s1.
The estimateJ4 is similar to that for J2 and we omit the details, which completes the
proof of Theorem 8.9.


## 8.2. Dual Spaces

In this subsection, we are going to identify the dual
spaces of some classes of Besov spaces and Triebel-Lizorkin spaces. To this
end, we first recall the definitions of some auxiliary function spaces. Let0<p,q≤∞. The spacesLp(ℓ˙q)(𝒳)andℓ˙q(Lp)(𝒳) are  respectively defined to be the set of all
sequences f={fk}k∈ℤ of μ-measurable functions on 𝒳 such that(8.79)∥f∥Lp(ℓ˙q)(𝒳)=∥{∑k=−∞∞|fk|q}1/q∥Lp(𝒳)<∞,∥f∥ℓ˙q(Lp)(𝒳)={∑k=−∞∞∥fk∥Lp(𝒳)q}1/q<∞; and the spacesLp(ℓq)(𝒳)andℓq(Lp)(𝒳) are respectively defined to be the set of all
sequences f={fk}k∈ℤ+ of μ-measurable functions on 𝒳 such that(8.80)∥f∥Lp(ℓq)(𝒳)=∥{∑k=0∞|fk|q}1/q∥Lp(𝒳)<∞,∥f∥ℓq(Lp)(𝒳)={∑k=0∞∥fk∥Lp(𝒳)q}1/q<∞.The following result is well known (cf., e.g.,
[3, Proposition 2.11.1, pages 177-178]).Proposition 8.10.
Let1≤p<∞ and 0<q<∞. Then the following hold.
(i)
g∈(Lp(ℓ˙q)(𝒳))′ if and only if there exists a sequence {gj}j∈ℤ∈Lp′(ℓ˙q′)(𝒳) such that(8.81)g(f)=∑j=−∞∞∫𝒳gj(x)fj(x)dμ(x)for every f={fj}j∈ℤ∈Lp(ℓ˙q)(𝒳),
and ∥g∥~∥{gj}j∈ℤ∥Lp′(ℓ˙q′)(𝒳).(ii)
g∈(ℓ˙q(Lp)(𝒳))′ if and only if there is a sequence {gj}j∈ℤ∈ℓ˙q′(Lp′)(𝒳) such that(8.82)g(f)=∑j=−∞∞∫𝒳gj(x)fj(x)dμ(x)for every f={fj}j∈ℤ∈ℓ˙q(Lp)(𝒳),
and ∥g∥~∥{gj}j∈ℤ∥ℓ˙q′(Lp′)(𝒳).Using Proposition8.10, by a procedure similar to the
proof of Theorem 2.11.2 in [3], we can establish the following duality theorem for homogeneous Besov
spaces and Triebel-Lizorkin spaces. In what follows, when 0<q<1,
we also  let q′≡∞.Theorem 8.11.
Letϵ be as in Definition 5.8 and |s|<ϵ. Then, the following hold.
(i) If1≤p<∞ and 0<q<∞,
then(8.83)(B˙p,qs(𝒳))′=B˙p′,q′−s(𝒳).More precisely, given g∈B˙p′,q′−s(𝒳),
then ℒg(f)≡〈f,g〉 defines a linear functional on 𝒢°(ϵ,ϵ)∩B˙p,qs(𝒳) such that(8.84)|ℒg(f)|≤C∥f∥B˙p,qs(𝒳)∥g∥B˙p′,q′−s(𝒳),where C>0 is independent of f,
and this linear functional naturally extends to B˙p,qs(𝒳) by continuity with norm at most C∥g∥B˙p′,q′−s(𝒳).
Conversely, ifℒ is a linear functional on B˙p,qs(𝒳),
then there exists a unique g∈B˙p′,q′−s(𝒳) such that ℒ is the natural extension of ℒg,
with ∥g∥B˙p′,q′−s(𝒳)≤C∥ℒ∥.
(ii) If1<p,q<∞,
then(8.85)(F˙p,qs(𝒳))′=F˙p′,q′−s(𝒳).More precisely, given g∈F˙p′,q′−s(𝒳),
then ℒg(f)≡〈f,g〉 defines a linear functional on 𝒢°(ϵ,ϵ)∩F˙p,qs(𝒳) such that(8.86)|ℒg(f)|≤C∥f∥F˙p,qs(𝒳)∥g∥F˙p′,q′−s(𝒳),where C>0 is independent of f,
and this linear functional naturally extends to F˙p,qs(𝒳) by continuity with norm at most C∥g∥F˙p′,q′−s(𝒳).
Conversely, ifℒ is a linear functional on F˙p,qs(𝒳),
then there exists a unique g∈F˙p′,q′−s(𝒳) such that ℒ is the natural extension of ℒg,
with ∥g∥F˙p′,q′−s(𝒳)≤C∥ℒ∥.Proof.
Lets∈(−ϵ,ϵ). We first claim that if 1<p<∞ and p(s,ϵ)<q<∞,
then(8.87)F˙p′,q′−s(𝒳)⊂(F˙p,qs(𝒳))′,and that if 1≤p<∞ and 0<q<∞,(8.88)B˙p′,q′−s(𝒳)⊂(B˙p,qs(𝒳))′.
We first verify (8.87). Observe first that 1<p′<∞ and 1<q′≤∞. Let f∈F˙p′,q′−s(𝒳) and all the notation be as in Theorem 3.10. For
any φ∈𝒢°(ϵ1,ϵ2),
by the definition of F˙p′,q′−s(𝒳) and Theorem 3.13 together with Hölder's
inequality and Remark 5.5, we have(8.89)|f(φ)|=|∑k=−∞∞〈Dk(f),D˜kt(φ)〉|≤∫𝒳∑k=−∞∞|Dk(f)(x)||D˜kt(φ)(x)|dμ(x)≲∥f∥F˙p′,q′−s(𝒳)∥φ∥F˙p,qs(𝒳),which together with the
Hahn-Banach theorem and Proposition 5.21 shows (8.87).
Similarly, to see (8.88), for any f∈B˙p′,q′−s(𝒳) with 1<p′,q′≤∞ and φ∈𝒢°(ϵ1,ϵ2),
we have(8.90)|f(φ)|=|∑k=−∞∞〈Dk(f),D˜kt(φ)〉|≤∑k=−∞∞∥Dk(f)∥Lp′(𝒳)∥D˜kt(φ)∥Lp(𝒳)≤∥f∥B˙p′,q′−s(𝒳)∥φ∥B˙p,qs(𝒳),which together with the
Hahn-Banach theorem and Proposition 5.21 again proves (8.88).
We now complete the proof of (ii). Lets∈(−ϵ,ϵ) and let 1<p,q<∞. Let {Dk}k∈ℤ be as in Definition 5.8. Then(8.91)f∈F˙p,qs(𝒳)→{2ksDk(f)}k=−∞∞is a one-to-one mapping from F˙p,qs(𝒳) onto a subspace of Lp(ℓ˙q)(𝒳),
and every functional g∈(F˙p,qs(𝒳))′ can be interpreted as a functional on that
subspace. By the Hahn-Banach theorem, g can be extended to a continuous linear
functional on Lp(ℓ˙q)(𝒳),
where the norm of g is preserved. If φ∈𝒢°0ϵ(β,γ) with some fixed |s|<β<ϵ and max{s−κ/p,0,−s−κ(1−1/p)}<γ<ϵ,
then Proposition 8.10(ii) yields(8.92)g(φ)=∑k=−∞∞∫𝒳gk(x)Dk(φ)(x)dμ(x),(8.93)∥{2−skgk}k=−∞∞∥Lp′(ℓ˙q′)(𝒳)~∥g∥.The formula (8.92) can be written
as(8.94)g(φ)=∑k=−∞∞∫𝒳Dkt(gk)(x)φ(x)dμ(x),which means that in (𝒢°0ϵ(β,γ))′ with |s|<β<ϵ and max{s−κ/p,0,−s−κ(1−1/p)}<γ<ϵ,
we have(8.95)g(x)=∑k=−∞∞Dkt(gk)(x).Repeating the proof of
Proposition 5.4(ii), we find that(8.96)∥g∥F˙p′,q′−s(𝒳)≲∥{2−skgk}k=−∞∞∥Lp′(ℓ˙q′)(𝒳).Thus, g∈F˙p′,q′−s(𝒳),
which shows (ii).
To finish the proof of the theorem, we still need to
establish the converse of (8.88). To this end, we first let s∈(−ϵ,ϵ), 1≤p,q<∞,
and g∈(B˙p,qs(𝒳))′. Repeating the above proof with ℓ˙q′(Lp′)(𝒳) instead of Lp′(ℓ˙q′)(𝒳),
for any φ∈𝒢°0ϵ(β,γ) with some fixed |s|<β<ϵ and max{s−κ/p,0,−s−κ(1−1/p)}<γ<ϵ,
we see that (8.92) holds, and that(8.97)∥{2−skgk}k=−∞∞∥ℓ˙q′(Lp′)(𝒳)~∥g∥.Similarly, (8.92) means that
(8.95) holds in (𝒢°0ϵ(β,γ))′ with |s|<β<ϵ and max{s−κ/p,0,−s−κ(1−1/p)}<γ<ϵ. Repeating the proof of Proposition 5.6(ii), we obtain(8.98)∥g∥B˙p′,q′−s(𝒳)≲∥{2−skgk}k=−∞∞∥ℓ˙q′(Lp′)(𝒳).Thus, g∈B˙p′,q′−s(𝒳),
which proves (i) when s∈(−ϵ,ϵ) and 1≤p,q<∞.
Finally, lets∈(−ϵ,ϵ), 1≤p<∞, and 0<q<1. Then B˙p,qs(𝒳)⊂B˙p,1s(𝒳) by Proposition 5.10(i), which gives(8.99)B˙p′,∞−s(𝒳)=(B˙p,1s(𝒳))′⊂(B˙p,qs(𝒳))′.On the other hand, if g∈(B˙p,qs(𝒳))′ with s∈(−ϵ,ϵ), 1≤p<∞, and 0<q<1 and letting {Dk}k∈ℤ be as in Definition 5.8, we then have that for
all l∈ℤ,(8.100)|(Dlg)(φ)|=|g(Dlt(φ))|≲∥g∥∥Dlt(φ)∥B˙p,qs(𝒳).We now estimate ∥Dlt(φ)∥B˙p,qs(𝒳). For any k,l∈ℤ,
by Theorem 3.10, we have(8.101)DkDlt(φ)=∑k′=−∞∞DkDltD˜k′Dk′(φ)in 𝒢°0ϵ(β,γ) with β and γ as above. Using (3.2) of Lemma 3.2 and (iii) of Proposition 2.7 together with Hölder's inequality and the Fubini theorem gives(8.102)∥DkDltD˜k′Dk′(φ)∥Lp(𝒳)≲2−|k−l|ϵ1′∥D˜k′Dk′(φ)∥Lp(𝒳)≲2−|k−l|ϵ1′∥Dk′(φ)∥Lp(𝒳),∥DkDltD˜k′Dk′(φ)∥Lp(𝒳)≲∥DltD˜k′Dk′(φ)∥Lp(𝒳)≲2−|l−k′|ϵ1′′∥Dk′(φ)∥Lp(𝒳), where ϵ1′ and ϵ1′′ can be any positive number in (0,ϵ1). These estimates together with the geometric means and Proposition 2.7(iii)
again yield that for any σ∈(0,1),(8.103)∥DkDltD˜k′Dk′(φ)∥Lp(𝒳)≲2−|k−l|ϵ1′σ2−|l−k′|ϵ1′′(1−σ)∥Dk′(φ)∥Lp(𝒳)≲2−|k−l|ϵ1′σ2−|l−k′|ϵ1′′(1−σ)∥φ∥Lp(𝒳).If we choose σ∈(0,1) and ϵ1′∈(0,ϵ1) such that ϵ1′σ>|s|,
then from the above estimate and (8.101) together with Definition 5.1 and (5.5),
it follows that(8.104)∥Dlt(φ)∥B˙p,qs(𝒳)≲{∑k=−∞∞2ksq∥DkDlt(φ)∥Lp(𝒳)q}1/q≲{∑k=−∞∞2ksq[∑k′=−∞∞2−|k−l|ϵ1′σ2−|l−k′|ϵ1′′(1−σ)∥φ∥Lp(𝒳)]q}1/q≲2ls∥φ∥Lp(𝒳).Thus,(8.105)|(Dlg)(φ)|≲2−ls∥φ∥Lp(𝒳),which implies that(8.106)supl∈ℤ{2−ls∥Dl(g)∥Lp′(𝒳)}≲∥g∥.That is, g∈B˙p′∞−s(𝒳),
which completes the proof of (i), and hence, Theorem 8.11.To determine the dual space ofF˙p,qs(𝒳) when p=1,
following [82], we
first consider the corresponding spaces of sequences. Let λ be a sequence as in (7.1). Then for s∈ℝ, 0<q≤∞, and x∈𝒳,
we define(8.107)𝔊˙s,q(λ)(x)={∑k=−∞∞∑τ∈Ik∑ν=1N(k,τ)2ksq|λτk,ν|qχQτk,ν(x)}1/q,𝔊˙τ,k,νs,q(λ)(x)={∑k′=k+1∞∑Qτ′k′,ν′⊂Qτk,ν2k′sq|λτ′k′,ν′|qχQτ′k′,ν′(x)}1/q. Let m˙τ,k,νs,q(λ) denote the “1/4-median” of 𝔊˙τ,k,νs,q(λ) on Qτk,ν,
namely,(8.108)m˙τ,k,νs,q=inf{δ>0:μ({x∈Qτk,ν:𝔊˙τ,k,νs,q(λ)(x)>δ})<μ(Qτk,ν)4}.We also set(8.109)m˙s,q(λ)(x)=supk,τ,νm˙τ,k,νs,q(λ)χQτk,ν(x).The following conclusion is
trivial by the definition of f˙∞,qs(𝒳).Proposition 8.12.
Letδ∈(0,1], s∈ℝ, and 0<q≤∞. Assume that for each dyadic cube Q,
there exists a measurable set EQ⊂Q such that μ(EQ)/μ(Q)≥δ. Then(8.110)∥λ∥f˙∞,qs(𝒳)~supl∈ℤsupα∈Il{1μ(Qαl)[∑k=l∞∑τ∈Ik∑ν=1N(k,τ)2ksqμ(EQτk,ν)|λτk,ν|qχ{(τ,ν):Qτk,ν⊂Qαl}(τ,ν)]}1/q.We now establish a characterization off˙∞,qs(𝒳) by means of m˙s,q.Proposition 8.13.
Lets∈ℝ and 0<q≤∞. Then(8.111)∥λ∥f˙∞,qs(𝒳)~∥m˙s,q(λ)∥L∞(𝒳).Proof.
By Chebyshev's inequality, we see that(8.112)μ({x∈Qτk,ν:𝔊˙τ,k,νs,q(λ)(x)>δ})≤1δq∫Qτk,ν[𝔊˙τ,k,νs,q(λ)(x)]qdμ(x)≤μ(Qτk,ν)δq∥λ∥f˙∞,qs(𝒳)q<14μ(Qτk,ν),if δ>41/δ∥λ∥f˙∞,qs(𝒳). Hence,(8.113)∥m˙s,q(λ)∥L∞(𝒳)≲∥λ∥f˙∞,qs(𝒳).
To establish the converse, we introduce the extended
integer-valued stopping timek(x) for x∈𝒳 by(8.114)k(x)=inf{k∈ℤ:(∑k′=k∞∑τ′∈Ik′∑ν′=1N(k′,τ′)2k′sq|λτ′k′,ν′|qχQτ′k′,ν′)1/q≤m˙s,q(λ)(x)}.Also, for any Qτk,ν,
set(8.115)EQτk,ν={x∈Qτk,ν:k(x)≤k}.Then, obviously,(8.116)EQτk,ν={x∈Qτk,ν:𝔊˙τ,k,νs,q(λ)(x)≤m˙s,q(λ)(x)}.From (8.108), it follows
that(8.117)μ({x∈Qτk,ν:𝔊˙τ,k,νs,q(λ)(x)>m˙s,q(λ)(x)})≤μ({x∈Qτk,ν:𝔊˙τ,k,νs,q(λ)(x)>m˙τ,k,νs,q(λ)})≤14μ(Qτk,ν),and hence, μ(EQτk,ν)≥(3/4)μ(Qτk,ν),
and(8.118)(∑k=−∞∞∑τ∈Ik∑ν=1N(k,τ)2ksq|λτk,ν|qχEQτk,ν(x))1/q≲m˙s,q(λ)(x),which together with Proposition 8.12 yields(8.119)∥λ∥f˙∞,qs(𝒳)≲∥m˙s,q(λ)∥L∞(𝒳).This finishes the proof of
Proposition 8.13.We next prove the followingduality forf˙1,qs(𝒳).Proposition 8.14.
Assume thats∈ℝ and 0<q<∞. Then (f˙1,qs(𝒳))′=f˙∞,q′−s(𝒳). In particular, if λ={λτk,ν:k∈ℤ,τ∈Ik,ν=1,…,N(k,τ)}∈f˙∞,q′−s(𝒳),
then the map(8.120)t={tQτk,ν:k∈ℤ,τ∈Ik,ν=1,…,N(k,τ)}↦〈t,λ〉,where(8.121)〈t,λ〉=∑k=−∞∞∑τ∈Ik∑ν=1N(k,τ)tQτk,νλτk,ν¯μ(Qτk,ν),defines a continuous linear
functional on f˙1,qs(𝒳) with operator norm ∥λ∥(f˙1,qs(𝒳))′ equivalent to ∥λ∥f˙1,qs(𝒳),
and every ℓ∈(f˙1,qs(𝒳))′ is of this form for some  λ∈f˙1,qs(𝒳).Proof.
We first assume that1≤q<∞. Similarly to the proof of Proposition 8.13, let(8.122)EQτk,ν={x∈Qτk,ν:𝔊˙τ,k,ν−s,q′(λ)(x)≤m˙−s,q′(λ)(x)}.Then μ(EQτk,ν)≥(3/4)μ(Qτk,ν) and(8.123)(∑k=−∞∞∑τ∈Ik∑ν=1N(k,τ)2−ksq′|λτk,ν|q′χEQτk,ν(x))1/q′≲m˙−s,q′(λ)(x).From this, Hölder's inequality,
and Proposition 8.13, it follows that(8.124)|∑k=−∞∞∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)tQτk,νλτk,ν¯|≲∫𝒳∑k=−∞∞∑τ∈Ik∑ν=1N(k,τ)(2ks|tQτk,ν|χQτk,ν(x))(2−ks|λτk,ν|χEQτk,ν(x))dμ(x)≲∫𝒳{∑k=−∞∞∑τ∈Ik∑ν=1N(k,τ)2ksq|tQτk,ν|qχQτk,ν(x)}1/q{∑k=−∞∞∑τ∈Ik∑ν=1N(k,τ)2−ksq′|λτk,ν|q′χEQτk,ν(x)}1/q′dμ(x)≲∥t∥f˙1,qs(𝒳)∥m˙−s,q′(λ)∥L∞(𝒳)≲∥t∥f˙1,qs(𝒳)∥λ∥f˙∞,q′−s(𝒳),which yields(8.125)∥λ∥(f˙1,qs(𝒳))′≲∥λ∥f˙∞,q′−s(𝒳),if 1≤q<∞. The case 0<q<1 then follows from the trivial imbedding f˙1,qs(𝒳)⊂f˙1,1−s(𝒳).
Conversely, by Proposition8.10(i), it is easy to see
that every ℓ∈(f˙1,qs(𝒳))′ is of the form:(8.126)t↦∑k=−∞∞∑τ∈Ik∑ν=1N(k,τ)μ(Qτk,ν)tQτk,νλτk,νfor some λ as in (7.1). Assume first again that 1≤q<∞. Fix a dyadic cube Qαl. Let 𝒴 be the sequence space of all dyadic cubes Qτk,ν such that Qτk,ν⊂Qαl,
and let σ be a measure on  𝒴 such that the σ-measure of the point Qτk,ν is μ(Qτk,ν)/μ(Qαl). Then(8.127){1μ(Qαl)∑k=l∞∑τ∈Ik∑ν=1N(k,τ)2−ksqμ(Qτk,ν)|λτk,ν|q′χ{(τ,ν):Qτk,ν⊂Qαl}(τ,ν)}1/q′=∥{2−ksλτk,ν}Qτk,ν⊂Qαl∥ℓq′(𝒴,dσ)=sup∥t∥ℓq(𝒴,dσ)≤1|1μ(Qαl)∑k=l∞∑τ∈Ik∑ν=1N(k,τ)tQτk,ν2−ksλτk,ν¯μ(Qτk,ν)|≤∥λ∥(f˙1,qs(𝒳))′sup∥t∥ℓq(𝒴,dσ)≤1∥{2−kstQτk,ν1μ(Qαl)}Qτk,ν⊂Qαl∥f˙1,qs(𝒳).However, by Hölder's inequality,
we have(8.128)∥{2−kstQτk,ν1μ(Qαl)}Qτk,ν⊂Qαl∥f˙1,qs(𝒳)=1μ(Qαl)∫Qαl{∑k=l∞∑τ∈Ik∑ν=1N(k,τ)|tQτk,ν|qχ{(τ,ν):Qτk,ν⊂Qαl}(τ,ν)χQτk,ν(x)}1/qdμ(x)≲{1μ(Qαl)∑k=l∞∑τ∈Ik∑ν=1N(k,τ)|tQτk,ν|qχ{(τ,ν):Qτk,ν⊂Qαl}(τ,ν)μ(Qτk,ν)}1/q~∥t∥ℓq(𝒴,dσ).Thus, if 1≤q<∞,
then(8.129)∥λ∥f˙∞,q′−s(𝒳)≲∥λ∥(f˙1,qs(𝒳))′.For 0<q<1,
we have q′=∞. In this case, for any Qτk,ν,
we set (tQτk,ν)Qτ′k′,ν′=2−ks/μ(Qτk,ν) for Qτ′k′,ν′=Qτk,ν and 0 otherwise. Clearly, ∥tQτk,ν∥f˙1,qs(𝒳)=1 and hence,(8.130)∥λ∥f˙∞,∞−s(𝒳)≤supk,τ,ν|〈tQτk,ν,λ〉|≲∥λ∥(f˙1,qs(𝒳))′,which completes the proof of
Proposition 8.14.From Proposition7.1, Theorem 7.2, and Proposition 8.14, by a standard method as in [82, pages 79-80], we can obtain the dual space of F˙p,qs(𝒳) when p=1 as follows. We omit the  details.Theorem 8.15.
Letϵ be as in Definition 6.1, |s|<ϵ, and p(s,ϵ)<q<∞. Then (F˙1,qs(𝒳))′=F˙∞,q′−s(𝒳) in the sense of that in Theorem 8.11(ii).A slight modification of the proof of Proposition 2.11.1 in [3, pages
177-178] again gives the following inhomogeneous version of Propositions 8.10 and 8.14. We omit the details.Proposition 8.16.
Let1≤p<∞ and 0<q<∞. Then the following hold.
(i)
g∈(Lp(ℓq)(𝒳))′ if and only if there is a sequence {gj}j∈ℤ+∈Lp′(ℓq′)(𝒳) such that(8.131)g(f)=∑j=0∞∫𝒳gj(x)fj(x)dμ(x)for every f={fj}j∈ℤ+∈Lp(ℓq)(𝒳),
and ∥g∥~∥{gj}j∈ℤ+∥Lp′(ℓq′)(𝒳).(ii)
g∈(ℓq(Lp)(𝒳))′ if and only if there is a sequence {gj}j∈ℤ+∈ℓq′(Lp′)(𝒳)(8.132)g(f)=∑j=0∞∫𝒳gj(x)fj(x)dμ(x)for every f={fj}j∈ℤ+∈ℓq(Lp)(𝒳),
and ∥g∥~∥{gj}j∈ℤ+∥ℓq′(Lp′)(𝒳).Theorem 8.17.
Assume thats∈ℝ and 0<q<∞. Then (f1,qs(𝒳))′=f∞,q′−s(𝒳). In particular, if λ={λτk,ν:k∈ℤ+,τ∈Ik,ν=1,…,N(k,τ)}∈f∞,q′−s(𝒳),
then the map(8.133)t={tQτk,ν:k∈ℤ+,τ∈Ik,ν=1,…,N(k,τ)}→〈t,λ〉,where(8.134)〈t,λ〉=∑k=0∞∑τ∈Ik∑ν=1N(k,τ)tQτk,νλτk,ν¯μ(Qτk,ν),defines a continuous linear
functional on f1,qs(𝒳) with operator norm ∥λ∥(f1,qs(𝒳))′ equivalent to ∥λ∥f1,qs(𝒳),
and every ℓ∈(f1,qs(𝒳))′ is of this form for some λ∈f1,qs(𝒳).With Proposition8.16 and Theorem 8.17, respectively,
in place of Propositions 8.10 and 8.14, by a procedure similar to the
proof of Theorem 8.11 (see also [3, the proof of Theorem 2.11.2]) and Theorem 5.6 in
[82, pages 79-80], we
can establish the following inhomogeneous version of Theorems 8.11 and 8.15, which describes the dual spaces of some inhomogeneous Besov and
Triebel-Lizorkin spaces. We omit the  details.Theorem 8.18.
Letϵ be as in Definition 5.8 and |s|<ϵ. Then, the following hold.
(i) If1≤p<∞ and 0<q<∞,
then(8.135)(Bp,qs(𝒳))′=Bp′,q′−s(𝒳).More precisely, given g∈Bp′,q′−s(𝒳),
then ℒg(f)≡〈f,g〉 defines a linear functional on 𝒢(ϵ,ϵ)∩Bp,qs(𝒳) such that(8.136)|ℒg(f)|≤C∥f∥Bp,qs(𝒳)∥g∥Bp′,q′−s(𝒳),where C>0 is independent of f,
and this linear functional naturally extends to Bp,qs(𝒳) by continuity with norm at most C∥g∥Bp′,q′−s(𝒳).
Conversely, ifℒ is a linear functional on Bp,qs(𝒳),
then there exists a unique g∈Bp′,q′−s(𝒳) such that ℒ is the natural extension of ℒg,
with ∥g∥Bp′,q′−s(𝒳)≤C∥ℒ∥.
(ii) If1<p,q<∞,
or p=1 and p(s,ϵ)<q<∞,
then(8.137)(Fp,qs(𝒳))′=Fp′,q′−s(𝒳).More precisely, given g∈Fp′,q′−s(𝒳),
then ℒg(f)≡〈f,g〉 defines a linear functional on 𝒢(ϵ,ϵ)∩Fp,qs(𝒳) such that(8.138)|ℒg(f)|≤C∥f∥Fp,qs(𝒳)∥g∥Fp′,q′−s(𝒳),where C>0 is independent of f,
and this linear functional naturally extends to Fp,qs(𝒳) by continuity with norm at most C∥g∥Fp′,q′−s(𝒳).
Conversely, ifℒ is a linear functional on Fp,qs(𝒳),
then there exists a unique g∈Fp′,q′−s(𝒳) such that ℒ is the natural extension of ℒg,
with ∥g∥Fp′,q′−s(𝒳)≤C∥ℒ∥.



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