# Applications of Littlewood-Paley Theory forB˙σ-Morrey Spaces to the Boundedness of Integral Operators

Authors: Yasuo Komori-Furuya; Katsuo Matsuoka; Eiichi Nakai; Yoshihiro Sawano
Journal: Journal of Function Spaces and Applications
 (2013)

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## Abstract

The boundedness of the various operators onB˙σ-Morrey spaces is considered in the framework of the Littlewood-Paley decompositions. First, the Littlewood-Paley characterization of B˙σ-Morrey-Campanato spaces is established. As an application, the boundedness of Riesz potential operators is revisted. Also, a characterization of B˙σ-Lipschitz spaces is obtained: and, as an application, the boundedness of Riesz potential operators on B˙σ-Lipschitz spaces is discussed.


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## Body

## 1. Introduction

One of the main aims of the present paper is to investigate the structure of theBσ-spaces B˙σ(Lp,λ)(ℝn) by proving Theorem 1 below. To state the main result, we need the following setup. We write
(1)Q(x,r)≔{y=(y1,y2,…,yn)∈ℝn:maxj=1,…,n|xi-yi|<r}
for x=(x1,x2,…,xn)∈ℝn and r>0. We abbreviate Q(0,r) to Qr. For a set E⊂ℝn, we denote by χE its indicator function. The space Ccomp∞(ℝn∖{0}) denotes the set of all compactly supported C∞(ℝn)-functions supported on ℝn∖{0}. Now we formulate our main result in the simplest form.Theorem 1.
Let1<p<∞, σ>0 and λ∈[-n/p,-σ). Let φ∈Ccomp∞(ℝn∖{0}) be a function such that
(2)∑j=-∞∞φj(ξ)=∑j=-∞∞φ(2-jξ)=χℝn∖{0}(ξ)(ξ∈ℝn).
If a measurable function f belongs to B˙σ(Lp,λ)(ℝn), that is, f satisfies
(3)∥f∥B˙σ(Lp,λ)∶=supr>01rσ{supx∈ℝn,s>0Q(x,s)⊂Qr1(2s)λ(1sn∫Q(x,s)|f(y)|pdy)1/p}<∞,
then f∈𝒮′(ℝn) and
(4)∥f∥B˙σ(Lp,λ)~supr>01rσ×[supz∈ℝn,k∈ℤQ(z,2-k)⊂Qr2kλ×{2kn∫Q(z,2-k)(∑j=k-1∞|ℱ-1φj*f(y)|2)p/2dy}1/p],
where ℱ-1 denotes the inverse Fourier transform and the implicit constant in ~ does not depend on f.Whenσ=0, then this is due to Mazzucato [1]. See [2] as well.The spaceB˙σ(Lp,λ) appearing in Theorem 1 is a special case of Bσ spaces. Theory of Bσ spaces stems from the Beurling work [3]. Beurling introduced the space Bp(ℝn) together with its predual Ap(ℝn) so-called the Beurling algebra [3]. Later, to extend Wiener’s ideas [4, 5] which describe the behavior of functions at infinity, Feichtinger [6] gave an equivalent norm on Bp(ℝn), which is a special case of norms to describe nonhomogeneous Herz spaces Kp,rα(ℝn) introduced in [7]. On the other hand, Lu and Yang [8, 9] introduced the central bounded mean oscillation space CBMOp(ℝn) with the norm
(5)∥f∥CBMOp=supr>0(1|Qr|∫Qr|f(x)-fQr|pdx)1/p.
Recently, in [10], Bσ-Morrey-Campanato spaces have been introduced to unify central Morrey spaces, central bounded mean oscillation spaces, and usual Morrey-Campanato spaces. Using Bσ-spaces, we can study both local and global regularities of functions simultaneously. For example, when we consider B˙σ(Lp,λ), the underlying norm given by
(6)supx∈ℝn,s>0Q(x,s)⊂Qr1(2s)λ(1sn∫Q(x,s)|f(y)|pdy)1/p
measures local regularities of functions, and the parameter σ plays the role of global regularity. Moreover, the proof of Theorem 13 reveals this aspect of our studying both local and global regularities of functions simultaneously.Theorem1 concerns the Littlewood-Paley theory. The Littlewood-Paley theory is one of the most powerful tools in harmonic analysis. Roughly speaking, this is a technique of transforming functions into good ones in order to measure the norms. Another side of Littlewood-Paley theory is that the functions are broken into good pieces of functions. We shall illustrate that the Littlewood-Paley theory is very useful by applying this to the fractional integral operator Iα of order α. The fractional integral operator Iα(0<α<n), which is given by
(7)Iαf(x):=∫ℝnf(y)|x-y|n-αdyf∈L1(ℝn),
plays an important role not only in harmonic analysis but also in partial differential equations. It is well known that Iα can be seen as the inverse operator of (-Δ)α/2 modulo a multiplicative constant, and hence Iα has the smoothing effect. However, due to this smoothing effect, Iα still seems to have a lot to be investigated.To describe the related function spaces and formulate Theorem1 in the full statement, we now fix some more notations. Since cubes play the central role, let us fix our notation of cubes. We denote by 𝒬 the set of all cubes of the form Q(x,r) with x∈ℝn and r>0. Also, we denote by 𝒟 the set of dyadic cubes:
(8)𝒟:={2-νm+2-ν[0,1)n:ν∈ℤ,m∈ℤn},
where to define the right-hand side we used the Minkowski sum. Given Q∈𝒟∪𝒬, we denote by c(Q) the center of a cube Q and by |Q| its volume. The notation ℓ(Q) stands for |Q|1/n. Given a set E⊂ℝn, we define
(9)𝒟(E):={Q∈𝒟:Q⊂E},𝒬(E):={Q∈𝒬:Q⊂E}.With these notations in mind, let us recall the definition of the function spaceB˙σ(Lp,λ)(ℝn) and related function spaces defined in [10]. Here we redefine B˙σ(Lp,λ)(ℝn) in terms of dyadic cubes. However, a geometric observation shows that the definition of B˙σ(Lp,λ)(ℝn) through dyadic cubes and that through cubes are equivalent.Definition 2.
Let1≤p≤∞, σ≥0, and -n/p≤λ≤0. Define B˙σ(Lp,λ)(ℝn) as the set of all f∈Llocp(ℝn) such that ∥f∥B˙σ(Lp,λ)<∞, where
(10)∥f∥B˙σ(Lp,λ):=supr>01rσ{supQ∈𝒟(Qr)1ℓ(Q)λ(1|Q|∫Q|f(y)|pdy)1/p}.
Define B˙σ(WLp,λ)(ℝn) in analogy with B˙σ(Lp,λ)(ℝn). The norm is given by
(11)∥f∥B˙σ(WLp,λ)∶=supr>01rσ{supQ∈𝒟(Qr)1ℓ(Q)λ+n/psupλ>0λ|{z∈Q:|f(x)|>λ}|1/p}.As we remarked above, (3) and (10) are equivalent definitions, and the same can be said for weak type spaces. Namely,
(12)∥f∥B˙σ(Lp,λ)~supr>01rσ{supQ∈𝒬(Qr)1ℓ(Q)λ(1|Q|∫Q|f(y)|pdy)1/p},∥f∥B˙σ(WLp,λ)~supr>01rσ{supQ∈𝒬(Qr)1ℓ(Q)λ+n/psupλ>0λ|{z∈Q:|f(x)|>λ}|1/p}
for any measurable function f on ℝn. In view of (12), we identify the right-hand side and the left-hand side in these formulas. Note that from the definition of norms, we have
(13)∥f∥B˙σ(Lp,λ)~supx∈ℝn,r>01(|x|+r)σ{1rλ+n/p(∫Q(x,r)|f(y)|pdy)1/p},(14)∥f∥B˙σ(WLp,λ)~supx∈ℝn,r>01(|x|+r)σ×{1rλ+n/psupλ>0λ|{z∈Q(x,r):|f(x)|>λ}|1/p}
for any measurable function f on ℝn.We need to pay attention to the word “local,” when we discussB˙σ(Lp,λ)(ℝn). Burenkov and Guliyev, together with their successors, investigated local Morrey-type spaces in [11]. By  “local” in [11], they meant that they defined the  “local Morrey-type” norm by
(15)supr>0r-λ(1|Qr|∫Qr|f(y)|pdy)1/p,
which indicates that sup appearing in the definition is restricted to all balls centered at the origin. However, by  “local” in the present paper we meant that we measure the regularity of functions by (6). Nowadays there are many different definitions related to classical Morrey spaces, so we need to carefully distinguish the different definitions and the names given to the definitions. See [12] for related usage of the word  “local” such as central mean oscillation, central BMO, λ-central bounded mean oscillation, and λ-central Morrey spaces.The goal of the present paper is to show that these function spaces fall under the scope of the Littlewood-Paley theory. As an application of this fact we show the boundedness property ofIα and singular integral operators. The Littlewood-Paley theory is a powerful tool to investigate the boundedness property of Iα. To consider the connection between B˙σ(Lp,λ)(ℝn) and the Littlewood-Paley theory, we present definitions. Here and below we use the definition of the Fourier transform below for definiteness:
(16)ℱf(ξ):=1(2π)n/2∫Rnf(x)e-ix·ξdx,ℱ-1f(x):=1(2π)n/2∫Rnf(ξ)eix·ξdξ.
Let φ∈Ccomp∞(ℝn∖{0}) be a function such that
(17)∑j=-∞∞φ(2-jξ)=χℝn∖{0}(ξ)(ξ∈ℝn).
Following [13], we define
(18)τj:=τ(2-j·)(j∈ℤ,τ∈𝒮(ℝn)),τ(D)f:=ℱ-1[τ·ℱf](τ∈𝒮(ℝn),f∈𝒮′(ℝn)).
It may be helpful to observe that
(19)ℱ-1τj(x)=2jnℱ-1τ(2jx)(x∈ℝn,j∈ℤ)
for each τ∈𝒮(ℝn).In the present paper, the following function space of Littlewood-Paley type will play a key role. Here and below we denoteJQ:=-log2ℓ(Q) for a cube Q∈𝒟. Observe that
(20)2JQ=ℓ(Q)-1(Q∈𝒟).Definition 3.
Let1<p<∞,σ>0,-n/p≤λ≤0. Given f∈𝒮′(ℝn), set
(21)∥f∥B˙σ(Lp,λD)≔supr>01rσ[sup,R:Q(x,R)∈𝒟(Qr)x1(2R)λsupQ∈𝒟(Qr)1ℓ(Q)λ×{1|Q|∫Q(∑j=JQ∞|φj(D)f(y)|2)p/2dy}1/psup,R:Q(x,R)∈𝒟(Qr)x1(2R)λ].
The function space B˙σ(Lp,λD)(ℝn) denotes the set of all tempered distributions f∈𝒮′(ℝn) for which the quantity ∥f∥B˙σ(Lp,λD) is finite.Note that Definition3 is closely related to the space F˙p,qs,τ(ℝn) defined by Yang and Yuan and the function space F˙p,qs,τ(ℝn) is investigated in [14–21].In Theorem1, we did not mention what happens if f∈𝒮′(ℝn), and the right-hand side of (4) is finite. Here including this problem, we reformulate and reinforce Theorem 1.Theorem 4.
Let1<p<∞,σ>0,-n/p≤λ<-σ. Then the spaces B˙σ(Lp,λD)(ℝn) and B˙σ(Lp,λ)(ℝn) are isomorphic. More precisely, we have the following.(a)
B˙σ(Lp,λ)(ℝn)↪𝒮′(ℝn) in the sense of continuous embedding.(b)
B˙σ(Lp,λD)(ℝn)↪Llocp(ℝn) in the sense of continuous embedding.(c)
Letf∈𝒮'(ℝn)∩Llocp(ℝn). If f∈B˙σ(Lp,λ)(ℝn), then f∈B˙σ(Lp,λD)(ℝn). Conversely, if f∈B˙σ(Lp,λD)(ℝn), then there exists a polynomial P such that f-P∈B˙σ(Lp,λ)(ℝn), and in this case the norms ∥f-P∥B˙σ(Lp,λ) and ∥f∥B˙σ(Lp,λD) are equivalent.(d)
Different choices ofφ satisfying (17) will yield the equivalent norms in the definition of ∥·∥B˙σ(Lp,λD).Instead of Theorem1 itself, we shall prove Theorem 4 in the present paper.Let1<p<∞ and λ≥-n/p. The Morrey space Lp,λ(ℝn), which can be realized as B˙0(Lp,λ)(ℝn), admits two different Littlewood-Paley characterizations. In [1], Mazzucato established
(22)∥f∥Lp,λ~supQ∈𝒟1ℓ(Q)λ{1|Q|∫Q(∑j=-∞∞|φj(D)f(x)|2)p/2dx}1/p.
Meanwhile, combining this equivalence with what we proved in [15], we can say that
(23)∥f∥Lp,λ~supQ∈𝒟1ℓ(Q)λ{1|Q|∫Q(∑j=JQ∞|φj(D)f(x)|2)p/2dx}1/p.
Thus, (23) is closer to Definition 3 than (22). Also, (23) seems to have stemmed from the famous technique due to Uchiyama [22]. We take advantage of equivalence of Theorem 4 in the proof of Theorem 6.To establish Theorem4, we will need an auxiliary vector-valued estimate of the Hardy-Littlewood maximal operator M. Define the Hardy-Littlewood maximal operator M by
(24)Mf(x):=supr>01|Q(x,r)|∫Q(x,r)|f(y)|dy.
The following proposition is proven in our earlier paper [12]. This is an extension of [23] to B˙(Lp,λ)(ℝn).Proposition 5.
Let1<p<∞,1<u≤∞,σ≥0, and λ<0. Assume in addition that σ+λ<0. Then we have
(25)∥(∑j=1∞Mfju)1/u∥B˙σ(Lp,λ)≤C∥(∑j=1∞|fj|u)1/u∥B˙σ(Lp,λ)
for some C>0 independent of {fj}j=1∞⊂B˙σ(Lp,λ)(ℝn), where we modify (25) obviously when u=∞.Chiarenza and Frasca obtained the boundedness of Hardy-Littlewood maximal operators on global Morrey spaces in [24]. In [11] Burenkov and Guliyev considered local Morrey-type spaces, where they showed that maximal operators are bounded [25–27].With Theorem4 and Proposition 5 in mind, we investigate the boundedness property of Iα again as we announced in the beginning. More precisely, we shall provide an alternative proof of the following theorems, which were proven earlier in [10, 12].Theorem 6 (see [10, 12]).
Suppose that the parametersα,σ,p,λ,q, and μ satisfy
(26)0<α<n,σ≥0,λ,μ<0,1<p,q<∞.
Assume in addition that
(27)μ=λ+α,q=λpμ=λpλ+α,λ+σ+α≤0,λ+np≥0.
Then Iα is a bounded operator from B˙σ(Lp,λ)(ℝn) to B˙σ(Lq,μ)(ℝn).Theorem 7 (see [10, 12]).
Suppose that the parametersα,σ,λ,q, and μ satisfy (26) and
(28)μ=λ+α,q=λμ=λλ+α,λ+σ+α<0,λ+n≥0
with p=1. Then Iα is a bounded linear operator from B˙σ(Lp,λ)(ℝn) to B˙σ(WLq,μ)(ℝn).We can also consider Campanato spaces and Lipschitz spaces in this framework. First, let𝒫d(ℝn) be the set of all polynomials having a degree at most d∈{0,1,2,…}. For a cube Q, a locally integrable function f over Q and a nonnegative integer d, there exists a unique polynomial P∈𝒫d(ℝn) such that, for all q∈𝒫d(ℝn),
(29)∫Q(f(x)-P(x))q(x)dx=0.
Denote this unique polynomial P by PQdf. It follows immediately from the definition that PQdg=g if g∈𝒫d(ℝn). We can characterize Λ spaces and Λ˙ spaces (cf. [28]).Definition 8.
Let1≤p≤∞ and -n/p≤λ<d. Then, for f∈Llocp(ℝn), define that
(30)∥f∥Λp,λ(d):=supr≥11rλ(1|Qr|∫Qr|f(y)-PQrdf(y)|pdy)1/p,∥f∥Λ˙p,λ(d):=supr>01rλ(1|Qr|∫Qr|f(y)-PQrdf(y)|pdy)1/p,
where we use the obvious modification when p=∞. The spaces Λp,λ(d)(ℝn) and Λ˙p,λ(d)(ℝn) are the sets (modulo 𝒫d(ℝn)) of all f∈Llocp(ℝn) for which the quantities ∥f∥Λp,λ(d) and ∥f∥Λ˙p,λ(d) are finite, respectively.Note that, in particular,(31)Λp,λ(0)(ℝn)=CMOp,λ(ℝn),Λ˙p,λ(0)(ℝn)=CBMOp,λ(ℝn)
with norm coincidence. For the definition of CMOp,λ(ℝn) and CBMOp,λ(ℝn), we refer to [12].Definition 9.
Forσ≥0, 1≤p≤∞, λ≥-n/p and d∈{0,1,2,…} with d≥λ, let B˙σ(ℒp,λ(d))(ℝn) be the set of all f∈Llocp(ℝn) such that ∥f∥B˙σ(ℒp,λ(d))<∞, where
(32)∥f∥B˙σ(ℒp,λ(d))∶=supr>01rσ×{sup,R:Q(x,R)∈𝒟(Qr)x1(2R)λsupQ∈𝒬(Qr)1ℓ(Q)λ×(1|Q|∫Q|f(y)-PQdf(y)|pdy)1/psup,R:Q(x,R)∈𝒟(Qr)x1(2R)λ},
where we use the obvious modification when p=∞.Observe that the quotient spaceB˙σ(ℒp,λ(d))(ℝn)/𝒫d(ℝn) is a Banach space equipped with the norm ∥f∥B˙σ(ℒp,λ(d)).Proposition 10.
Let1≤p≤∞, σ≥0, λ<0. Assume that -n/p≤λ and that σ+λ<0. Then
(33)B˙σ(ℒp,λ(0))(ℝn)/𝒫0(ℝn)≈B˙σ(Lp,λ)(ℝn),
with equivalent norms.Now we define a function space by way of difference. Forh∈ℝn, an integer L∈ℕ and a function f:ℝn→ℂ, we define
(34)Δh0f:=f,ΔhLf:=ΔhL-1f(·+h)-ΔhL-1f
inductively.Definition 11.
Forσ≥0, θ>0 and L∈{0,1,2,…} with L>θ, let B˙σ(Lipθ(L))(ℝn) be the set of all continuous functions f such that ∥f∥B˙σ(Lipθ(L))<∞, where
(35)∥f∥B˙σ(Lipθ(L))∶=supr>01rσ×{supQ∈𝒬(Qr)(supx∈Qsup|y|≤ℓ(Q)1ℓ(Q)θ|ΔyLf(x)|)}.
ThenB˙σ(Lipθ(L))(ℝn) is a Banach space equipped with the norm
(36)∥f∥B˙σ(Lipθ(L))+|f(0)|,
and also the quotient space B˙σ(Lipθ(L))(ℝn)/𝒫L-1(ℝn) is a Banach space equipped with the norm ∥f∥B˙σ(Lipθ(L)).Definition 12.
Forσ≥0 and θ>0 let B˙σ(LipθD)(ℝn) be the set of all f∈𝒮′(ℝn) such that ∥f∥B˙σ(LipθD)<∞, where
(37)∥f∥B˙σ(LipθD)∶=supr>01rσ{supQ∈𝒟(Qr)(supx∈Q2JQθ|φJQ(D)f(x)|)}.
In the present paper, we aim to show the following equivalence as well. Here, fort∈ℝ, we let [t] the largest integer m such that m≤t.Theorem 13.
Letσ≥0 and θ>0. Assume that the integer d satisfies d>[σ+θ]. (1)
Iff:ℝn→ℂ is a continuous function such that the quantity ∥f∥B˙σ(Lipθ(d+1)) is finite, then f∈𝒮′(ℝn) and the inequality ∥f∥B˙σ(LipθD)≲∥f∥B˙σ(Lipθ(d+1)) holds.(2)
Iff∈B˙σ(LipθD)(ℝn), then f can be represented by a continuous function and the inequality ∥f∥B˙σ(Lipθ(d+1))≲∥f∥B˙σ(LipθD) holds.Remark 14.
It is absolutely necessary to assume thatf is a continuous function in Theorem 13, when d≥1. We remark that there exists a discontinuous function f such that Δhd+1f(x)=0 for all x,h∈ℝn. See [29, Proposition A1] for such an example constructed algebraically.Now we explain notations and we describe its organization of the present paper. We use the following notations for the inequalities. First, we use standard notation for inequalities. For example, in the present paper a chain of inequalities of the form(38)A1≲A2≤A3~A4
appears in (110) below. The inequality (38) means that there exist C1,C2,C3>0 such that
(39)A1≤C1A2,A2≤A3,A3≤C2A4,A4≤C3A3.
If the implicit constants in ≲ or ~ do depend on some important parameters α,β,…, then we write ≲α,β,… or ~α,β,…. We shall prove Theorem 4 in Section 3. We prove Theorem 6 in Section 4. Theorem 7 is covered in Section 5. We prove Theorem 13 in Section 6. Finally in Section 7 we present another application of Theorem 4 by showing that the Fourier multiplier is bounded on B˙σ(Lp,λ)(ℝn).


## 2. Preliminaries

In the present paper, we frequently use the following fundamental inequalities.Lemma 15 (see [30, page 466]).
Letν,μ∈ℤ,M,N>0, and L∈ℕ∪{0} satisfy
(40)ν≥μ,N>M+L+n.
Suppose that ϕ∈CL(ℝn) such that
(41)|∇Lϕ(x)|≤2μ(n+L)(1+2μ|x-xϕ|)M
for some xϕ∈ℝn. Assume in addition that ψ is a measurable function such that
(42)∫ℝnxβψ(x)dx=0,|β|≤L-1
and that
(43)|ψ(x)|≤2νn(1+2ν|x-xψ|)N
for some xψ∈ℝn. Then we have
(44)|∫ℝnϕ(x)ψ(x)dx|≲2μn-(ν-μ)L(1+2μ|xϕ-xψ|)M.To formulate the next lemma, we recall the definition ofAp with 1≤p<∞. A measurable function w, which takes values in (0,∞) almost everywhere, is said to be an Ap weight or belongs to the class Ap, if
(45)Ap(w)≔supQ∈𝒬(1|Q|∫Qw(x)dx)×(1|Q|∫Qw(x)-1/(p-1)dx)p-1<∞(1<p<∞),A1(w)∶=∥w-1Mw∥L∞<∞(p=1).
For all p>1, it is easy to see that A1⊂Ap and that Ap(w)≤A1(w).Lemma 16 (see [31]).
Let1<p<∞ and φ satisfy (17). Then, for w∈Ap, we have
(46)∥w1/pf∥Lp~p,Ap(w)∥w1/p(∑j=-∞∞|φj(D)f|2)1/2∥Lp
for all f∈Llocp(w)∩𝒮′(ℝn).We also need a piece of information on dilation of the spaceB˙σ(Lp,λ)(ℝn).Lemma 17.
Let1≤p<∞, λ≥-n/p and σ≥0. Let t>0 and f∈B˙σ(Lp,λ)(ℝn). Then
(47)∥f(t·)∥B˙σ(Lp,λ)=tσ+λ∥f∥B˙σ(Lp,λ).It is just a matter of handling the left-hand side carefully. But, for the sake of convenience, we supply the proof.Proof.
From the definition of∥f(t·)∥B˙σ(Lp,λ), we deduce
(48)∥f(t·)∥B˙σ(Lp,λ)=supr>01rσ{sup,R:Q(x,R)∈𝒟(Qr)x1(2R)λsupx,R:Q(x,R)∈𝒟(Qr)1(2R)λ×(1|Q(x,R)|∫Q(x,R)|f(ty)|pdy)1/psup,R:Q(x,R)∈𝒟(Qr)x1(2R)λ}=supr>01rσ{sup,R:Q(x,R)∈𝒟(Qr)x1(2R)λsupx,R:Q(x,R)∈𝒟(Qr)1(2R)λ×(1|Q(tx,tR)|∫Q(tx,tR)|f(y)|pdy)1/psup,R:Q(x,R)∈𝒟(Qr)x1(2R)λ}=supr>01rσ{sup,R:Q(x,R)∈𝒟(Qr)x1(2R)λsupx,R:Q(tx,tR)∈𝒟(Qtr)1(2R)λ×(1|Q(tx,tR)|∫Q(tx,tR)|f(y)|pdy)1/psup,R:Q(x,R)∈𝒟(Qr)x1(2R)λ}=tσ+λsupr>01(tr)σ{sup,R:Q(x,R)∈𝒟(Qr)x1(2R)λsupx,R:Q(tx,tR)∈𝒟(Qtr)1(2tR)λ×(1|Q(tx,tR)|∫Q(tx,tR)|f(y)|pdy)1/psup,R:Q(x,R)∈𝒟(Qr)x1(2R)λ}=tσ+λ∥f∥B˙σ(Lp,λ),
proving the lemma.In the course of the proof of Theorem6, we need another piece of information on the space Bσ(Lp,λ)(ℝn). Let 1≤p≤∞, σ≥0 and -n/p≤λ≤0. Define Bσ(Lp,λ)(ℝn) as the set of all f∈Llocp(ℝn) such that ∥f∥Bσ(Lp,λ)<∞, where
(49)∥f∥Bσ(Lp,λ)∶=supr≥11rσ{supQ∈𝒬(Qr)1ℓ(Q)λ(1|Q|∫Q|f(y)|pdy)1/p}.The next lemma concerns the norm of the translation operator.Lemma 18.
Let1≤p<∞, λ≥-n/p and σ≥0. Then
(50)∥f(·-x0)∥Bσ(Lp,λ)≲(1+|x0|)σ∥f∥Bσ(Lp,λ).Proof.
Letr≥1 and Q=Q(z,s)∈𝒬(Qr) be fixed and consider
(51)I:=1rσ{1(2s)λ(1|Q(z,s)|∫Q(z,s)|f(y-x0)|pdy)1/p}.
Then we have
(52)I=1rσ{1(2s)λ(1|Q(z-x0,s)|∫Q(z-x0,s)|f(y)|pdy)1/p}
and Q(z-x0,s)∈𝒬(Qr+|x0|). Note that
(53)rσ(1+|x0|)σ≥(r+|x0|)σ
since r≥1. This implies that
(54)I≤(1+|x0|)σ1(r+|x0|)σ×{1(2s)λ(1|Q(z-x0,s)|∫Q(z-x0,s)|f(y)|pdy)1/p}.
If we consider the supremum over r,s>0 and z∈ℝn such that Q(z,s)⊂Qr, then we obtain the desired result.Lemma 19.
Letφ∈𝒮(ℝn). Let σ≥0, 1≤p<∞ and λ∈(-n/p,-σ). Then
(55)∥φ(D)f∥B˙σ(Lp,λ)≲∥f∥B˙σ(Lp,λ)
for all f∈B˙σ(Lp,λ).Proof.
We have(56)∥f(·-y)∥Bσ(Lp,λ)≲(1+|y|)σ∥f∥Bσ(Lp,λ)(y∈ℝn)
from Lemma 18. Hence, by the triangle inequality, we obtain that
(57)∥φ(D)f∥Bσ(Lp,λ)=c∥∫ℝnℱ-1φ(y)f(·-y)dy∥Bσ(Lp,λ)≲∫ℝn|ℱ-1φ(y)|·∥f(·-y)∥Bσ(Lp,λ)dy≲∫ℝn|ℱ-1φ(y)|(1+|y|)σ·∥f∥Bσ(Lp,λ)dy~∥f∥Bσ(Lp,λ).
Meanwhile, since ∫Qr|f(x)|dx≤rσ+λ+n/p∥f∥B˙σ(Lp,λ), for r∈(0,∞), we have, on Q1,
(58)|φ(D)f(x)|≲∫ℝn|f(y)|(1+|x-y|)-σ-λ-(n/p)-1dy≲∫ℝn|f(y)|(1+|y|)-σ-λ-(n/p)-1dy≲∑k=0∞2-(σ+λ+(n/p)+1)k∫Q2k|f(y)|dy≲∑k=0∞2-(σ+λ+(n/p)+1)k2(σ+λ+(n/p))k∥f∥B˙σ(Lp,λ)~∥f∥B˙σ(Lp,λ).
Assuming that σ+λ<0, we have
(59)∥φ(D)f∥B˙σ(Lp,λ)≤∥φ(D)f∥Bσ(Lp,λ)+∥f∥L∞(Q1).
Putting (57)–(59) together, we obtain the desired result.Lemma 20.
Suppose that the parametersσ,p, and λ satisfy
(60)σ≥0,λ<0,1≤p<∞.
Let φ∈𝒮(ℝn). Then
(61)∥φk(D)f∥B˙σ(Lp,λ)=∥φ(2-kD)f∥B˙σ(Lp,λ)≲∥f∥B˙σ(Lp,λ).Proof.
Note thatφk(D)f(x)=φ(D)f†,k(2kx) for f†,k=f(2k·)∈𝒮′(ℝn). Recall that B˙σ(Lp,λ) has a scaling invariance, as we have verified in Lemma 17. So, to prove (61), we can assume that k=0. In this case, (61) is Lemma 19 itself.We need the following sequence of functions.Lemma 21.
There exists a sequence of functions{rj}j=-∞∞⊂L2([0,1]) such that, for any fixed constant p∈(0,∞),
(62)(∫01|∑j=-∞∞ajrj(t)|pdt)1/p~(∑j=-∞∞|aj|2)1/2
provided {aj}j=-∞∞∈ℓ2(ℤ).In the present paper the sequence{rj}j=-∞∞ above is called a Rademacher sequence.


## 3. Littlewood-Paley Characterization ofB˙σ(Lp,λ)(ℝn)

### 3.1. Proof of Proposition10

It follows from the definition of the norm∥f∥B˙σ(Lp,λ) that
(63)∥f+C∥B˙σ(ℒp,λ(0))≲∥f∥B˙σ(Lp,λ)
for any constant function C. Note that this implies that B˙σ(Lp,λ) is a subset of B˙σ(ℒp,λ(0)).Letf∈B˙σ(ℒp,λ(0))(ℝn). Then, since p>1, we can use the Hölder inequality and we have
(64)|PQs0f(0)-PQr0f(0)|≤1|Qs|∫Qs|f(x)-PQr0f(0)|dx≲rσ+λ∥f∥B˙σ(ℒp,λ(0))(ℝn)
for r≤s≤2r. Assuming that σ+λ<0, the sequence {PQr0f(0)}r>0 is convergent by the Cauchy test. Thus, we can consider the mapping
(65)ι:f∈B˙σ(ℒp,λ(0))(ℝn)↦f-limR→∞PQR0f(0).
Let us check that the range of ι is in B˙σ(Lp,λ)(ℝn).Letr>0 and Q⊂Qr be fixed. Let N be the largest integer such that 2NQ⊂Q2r. Then observe that 2N+2Q⊃Qr. We decompose
(66)f-limR→∞PQR0f(0)=f-PQ0f(0)+P2NQ0f(0)-PQr0f(0)+∑i=1∞(P2i-1Qr0f(0)-P2iQr0f(0))-∑l=1N(P2lQ0f(0)-P2l-1Q0f(0)).
We consider the Lp-norm over Q and multiply 1/(rσℓ(Q)λ+n/p).As for the first term, directly from the definition of∥f∥ℒp,λ(0), we obtain that
(67)1rσℓ(Q)λ+n/p∫Q|f(x)-PQ0f(0)|dx≤∥f∥ℒp,λ(0).
Also, a geometric observation shows that the second term can be estimated similarly. Since λ<0, we obtain that
(68)1rσℓ(Q)λ|P2NQ0f(0)-PQr0f(0)|≤1rσ(2Nℓ(Q))λ|P2NQ0f(0)-PQr0f(0)|≲∥f∥ℒp,λ(0).
By using (64) and λ<0, we can handle the third term:
(69)1rσℓ(Q)λ|∑i=1∞(P2i-1Qr0f(0)-P2iQr0f(0))|≤1rσℓ(Q)λ∑i=1∞|P2i-1Qr0f(0)-P2iQr0f(0)|≤1rσ(2Nℓ(Q))λ∑i=1∞|P2i-1Qr0f(0)-P2iQr0f(0)|.
In view of the way in which we chose N, we obtain that
(70)1rσℓ(Q)λ|∑i=1∞(P2i-1Qr0f(0)-P2iQr0f(0))|≲1rσ+λ∑i=1∞|P2i-1Qr0f(0)-P2iQr0f(0)|≲∥f∥ℒp,λ(0).
In summary,
(71)1rσℓ(Q)λ|∑i=1∞(P2i-1Qr0f(0)-P2iQr0f(0))|≲∥f∥ℒp,λ(0).
It remains to estimate the fourth term. We employ the following estimate:
(72)|P2lQ0f(0)-P2l-1Q0f(0)|≲rσℓ(2lQ)λ∥f∥ℒp,λ(0).
Since λ<0, (72) is summable and we obtain that
(73)1rσℓ(Q)λ|∑l=1N(P2lQ0f(0)-P2l-1Q0f(0))|≲∥f∥ℒp,λ(0),
where the implicit constant in (73) is independent of N.It follows from (67), (68), (71), and (73) that ι sends B˙σ(ℒp,λ(0))(ℝn) to B˙σ(Lp,λ)(ℝn) boundedly;
(74)∥ι(f)∥B˙σ(Lp,λ)≲∥f∥B˙σ(ℒp,λ(0))
for all f∈B˙σ(Lp,λ)(ℝn).Meanwhile, it follows from the norm∥·∥B˙σ(Lp,λ) that
(75)limR→∞PQR0f(0)=0
for all f∈B˙σ(Lp,λ)(ℝn). In view of (63), ι is a surjection:
(76)ι(B˙σ(ℒp,λ(0))(ℝn))=B˙σ(Lp,λ)(ℝn).Finally observe that, forf∈B˙σ(ℒp,λ(0))(ℝn), ι(f)=0 if and only if f is a constant function, that is, f∈𝒫0(ℝn). Namely,
(77)Ker(ι)=𝒫0(ℝn).
Thus, from (75), (76) and (77), we conclude (33).


### 3.2. Proof of Theorem4 Part (a)

For1<p<∞ and σ∈ℝ, let us define the function space u-Llocp(〈·〉σ)(ℝn) of uniformly Llocp(〈·〉σ)(ℝn) functions by
(78)u-Llocp(〈·〉σ)(ℝn)={f∈Llocp(ℝn):∥f∥u-Llocp(〈·〉σ)<∞},
where, for m∈ℤn, we write 〈m〉:=1+|m|2, and the norm is given by
(79)∥f∥u-Llocp(〈·〉σ)=supm∈ℤn(〈m〉σp∫m+[0,1]n|f(y)|pdy)1/p.
Then from the definition of the norms (10) and (78), the following chain of continuous inclusion holds. For -n/p≤λ,
(80)B˙σ(Lp,λ)(ℝn)↪u-Llocp(〈·〉-σ)(ℝn)↪𝒮′(ℝn).
Thus, (a) follows.Remark 22.
The spaceu-Llocp(〈·〉-σ)(ℝn) is a special case of amalgams investigated in [32].


### 3.3. Proof of Theorem4 Part (b)

Letφ∈𝒮(ℝn) be a fixed function satisfying (17). It follows from (17) that there exist J- and J+ such that
(81)∑j=J-J+φ(ξ)φ(2-jξ)=φ(ξ).
Let f∈B˙σ(Lp,λD)(ℝn) and fix a dyadic cube Q* such that 0∈Q*¯. We are going to show that
(82)g1:=limL→∞∑j=-LJQ*φj(D)f∈L∞(Q*)∩𝒮′(ℝn)
converges in the topology of L∞(Q*)∩𝒮′(ℝn), and that
(83)g2:=limL→∞∑j=JQ*Lφj(D)f∈Lp(Q*)∩𝒮′(ℝn)
converges in the topology of Lp(Q*)∩𝒮′(ℝn).The presice meaning of (82) and (83) is that as follows: (82) means
(84)∥g1-∑j=-LJQ*φj(D)f∥L∞(Q*)→0,〈∑j=-LJQ*φj(D)f,τ〉→〈g1,τ〉(τ∈𝒮(ℝn))
as L→∞ and (83) means that
(85)∥g2-∑j=JQ*L-1φj(D)f∥Lp(Q*)→0〈∑j=JQ*L-1φj(D)f,τ〉→〈g2,τ〉(τ∈𝒮(ℝn))
as L→∞. Once (82) and (83) are proven, we will have
(86)g:=∑j=-∞∞φj(D)f∈Lp(Q*)∩𝒮′(ℝn)
converges in the topology of Lp(Q*)∩𝒮′(ℝn) and that f-g is a polynomial. Hence it follows that f∈Llocp(ℝn). Remark also that the convergence in 𝒮′(ℝn) of the sum defining g2 is a generality. So let us prove (82) and (83).Let us begin with proving (82). To this end we take a dyadic cube Q containing Q*. Since we are assuming (81), we deduce that
(87)|φJQ(D)f(x)|≲∑l=J-J+|ℱ-1φl+JQ*φJQ(D)f(x)|≤∑l=J-J+∫ℝn|ℱ-1φl+JQ(x-y)φJQ(D)f(y)|dy(x∈ℝn).
Since ℱ-1φ∈𝒮(ℝn), we have
(88)|φJQ(D)f(x)|≲1|Q|∫ℝn|φJQ(D)f(y)|(1+ℓ(Q)-1|y|)2n+2σ+2dy(x∈Q)
from (19) and (20). By the Hölder inequality and the fact that σ≥0, we have
(89)|φJQ(D)f(x)|≲1|Q|(∫ℝndy(1+ℓ(Q)-1|y|)p′(n+σ+1))1/p′×(∫ℝn|φJQ(D)f(y)|pdy(1+ℓ(Q)-1|y|)p(n+σ+1))1/p~(1|Q|∫ℝn|φJQ(D)f(y)|p(1+ℓ(Q)-1|y|)p(n+σ+1)dy)1/p.
for all x∈Q. By decomposing the last integral dyadically, we obtain that
(90)|φJQ(D)f(x)|≲∑l=0∞(2-lp(n+σ+1)1|Q|∫2lQ|φJQ(D)f(y)|pdy)1/p.
Observe from Definition 3 that
(91)∥φJQ(D)f∥Lp(2lQ)≲|2lQ|((σ+λ)/n)+(1/p)∥f∥B˙σ(Lp,λD)
for all l≥0. Thus, we obtain
(92)|φJQ(D)f(x)|=|Q|(λ+σ)/n∑l=0∞2-l(n+σ+1)+l(σ+λ+(n/p))∥f∥B˙σ(Lp,λD)≲|Q|(σ+λ)/n∥f∥B˙σ(Lp,λD)(x∈Q).
Hence, from (92) and the fact that λ+σ<0, we have
(93)∥∑j=-∞JQ*-1|φj(D)f|∥L∞(Q*)≤∑j=-∞JQ*-1supx∈Q*|φj(D)f(x)|≲|Q*|(σ+λ)/n∥f∥B˙σ(Lp,λD),
which proves the embedding into L∞(Q*). To prove that the sum defining g1 converges in the topology of 𝒮′(ℝn), we first fix a cube R containing Q* and observe
(94)∥∑j=JRJQ*|φj(D)f|∥Lp(R)≲|R|((σ+λ)/n)+(1/p)×log(2+ℓ(R)ℓ(Q*))∥f∥B˙σ(Lp,λD)
from (91) and the fact that σ+λ+n/p≥0. Meanwhile, again by (92), we have
(95)∥φJQ(D)f∥L∞(Q)≲|Q|(σ+λ)/n∥f∥B˙σ(Lp,λD)
for all cubes Q containing R. As a result, we obtain that
(96)∥∑j=-∞JR|φj(D)f|∥Lp(R)≲|R|((σ+λ)/n)+(1/p)∥f∥B˙σ(Lp,λD)
from the fact that σ+λ<0. It follows from (94) and (96) that the sum defining g1 converges in 𝒮′(ℝn).Now let us prove (83). First, choose θ∈(1+(λ+σ)p/n,1). By virtue of Lemma 16 and the fact that Ap((MχQ*)θ) has a bound independent of Q*∈𝒟 [33], we obtain that
(97)∥∑j=JQ*∞φj(D)f∥Lp(Q*)≤∥(MχQ*)θ/p∑j=JQ*∞φj(D)f∥Lp≲∥(MχQ*)θ/p(∑j=JQ*-j0∞|φj(D)f|2)1/2∥Lp.
Here for the last inequality, we employed a geometric observation of the support of φ. Therefore, j0 of the last inequality depends only on φ.SinceMχQ*(x)~(1+ℓ(Q*)-1|x-c(Q*)|)-n, we have
(98)∥∑j=JQ*∞φj(D)f∥Lp(Q*)≲∑l=-j0∞2-lnθ/p∥(∑j=JQ*-j0∞|φj(D)f|2)1/2∥Lp(2lQ*)≲∑l=-j0∞2-lnθ/pℓ(2lQ*)σ+λ+(n/p)∥f∥B˙σ(Lp,λD).
Recall that we are assuming that θ∈(1+(λ+σ)p/n,1). So (98) is summable and we have
(99)∥∑j=JQ*∞φj(D)f∥Lp(Q*)≲ℓ(Q*)σ+λ+(n/p)∥f∥B˙σ(Lp,λD),
showing (83).With (82) and (83), the proof of (b) is now complete.Remark 23.
We did not use the structure ofφj(D)f in the proof. Therefore, we can deduce the following variant.Proposition 24.
Suppose that we are given a collection{Fj}j=-∞∞⊂𝒮′(ℝn) such that supp(ℱFj)⊂supp(φj) and that
(100)supr>01rσ[supQ∈𝒟(Qr)1ℓ(Q)λ{1|Q|∫Q(∑j=JQ∞|Fj(y)|2)p/2dy}1/p]<∞.
Then G:=∑j=-∞∞Fj converges in the topology of 𝒮′(ℝn).


### 3.4. Proof of Theorem4 Part (c)

Keeping what we proved in Parts (a) and (b) in mind, we shall now describe the isomorphism betweenB˙σ(Lp,λ)(ℝn) and B˙σ(Lp,λD)(ℝn). If f∈B˙σ(Lp,λ)(ℝn), then we claim that f∈B˙σ(Lp,λD)(ℝn) and that
(101)∥f∥B˙σ(Lp,λD)≲∥f∥B˙σ(Lp,λ).
The proof of (101) is comparably easy.Let us start to prove (101). Let r>0 and Q∈𝒟(Qr) be fixed. Also, fix a Rademacher sequence {rj}j=-∞∞⊂L2([0,1]). Define
(102)KQ(x;t):=∑j=JQ∞2jnrj(t)ℱ-1φ(2jx)(x∈ℝn,t>0).
Then we have
(103)∑j=JQ∞rj(t)φj(D)f=KQ(·;t)*f,1rσ(∫Q(∑j=JQ∞|φj(D)f(y)|2)p/2dy|Q|(λp/n)+1)1/p~1rσ(∬Q×[0,1]|KQ(·;t)*f(y)|pdydt|Q|(λp/n)+1)1/p
by the property of the Rademacher sequence. Let us set
(104)f1:=fχ2Q,f2:=f-f1.
Then we have
(105)|KQ(x;t)|≲|x|-n,|∂kKQ(x;t)|≲|x|-n-1,|ℱ[KQ(·;t)](ξ)|≲1,
since rj(t)∈{0,1,-1} with constant independent of Q.By the Calderón-Zygmund theory forLp(ℝn), we have
(106)1rσ(∫01∫Q|KQ(·;t)*f1(y)|pdydt|Q|(λp/n)+1)1/p≲1rσ(∫2Q|f(y)|pdy|Q|(λp/n)+1)1/p.
As for f2, we invoke the following pointwise estimate obtained from (105):
(107)|KQ(·;t)*f2(y)|≲∫ℝn∖2Q|f(z)||z-c(Q)|ndz(y∈Q).
We write
(108)Ik≔∫2k+1Q∖2kQ|f(z)||z-c(Q)|ndz~1|2kQ|∫2k+1Q∖2kQ|f(z)|dz,k=1,2,….
Notice that
(109)Ik≲(2k+1r)σ(2k+1ℓ(Q))λ∥f∥B˙σ(Lp,λ),
since 2k+1Q⊂Q2k+1r. Assuming that σ+λ<0, we have
(110)∑k=1∞Ikrσ|Q|λ/n≲∑k=1∞(2k+1r)σ(2k+1ℓ(Q))λrσ|Q|λ/n∥f∥B˙σ(Lp,λ)≲∑k=1∞2(σ+λ)k∥f∥B˙σ(Lp,λ)~∥f∥B˙σ(Lp,λ).
Consequently, from (106) and (110), we obtain (101).The heart of the matter is how to construct the inverse mapping fromB˙σ(Lp,λD)(ℝn) to B˙σ(Lp,λ)(ℝn). Let f~∈B˙σ(Lp,λD)(ℝn). Then we have established in (b) that
(111)f:=∑j=-∞∞φj(D)f~
exists in the topology of Llocp(ℝn)∩𝒮′(ℝn) in view of (92) and (99). We claim that
(112)∥f∥B˙σ(Lp,λ)≲∥f~∥B˙σ(Lp,λD).
Here and below in proving (112), we assume that f~=f by replacing f with f~.Let us prove (112). Suppose that we are given a dyadic cube Q and r>0 such that Q∈𝒟(Qr). We estimate
(113)1rσ|Q|(λ/n)+(1/p)(∫Q|f(x)|pdx)1/p.
First, we expand f by using (111). If we denote
(114)φ~j(ξ):=∑l=J-J+φ(2-j-lξ),ξ∈ℝn
for j∈ℤ, then we have
(115)f(x)=∑j=-∞∞φ~j(D)φj(D)f(x),
where the convergence, as we have established in Part (b), takes places in Llocp(ℝn)∩𝒮′(ℝn).We set(116)J1:=1rσ|Q|(λ/n)+(1/p)(∫Q|∑j=JQ∞φ~j(D)φj(D)f(x)|pdx)1/p,J2,k:=1rσ|Q|(λ/n)+(1/p)(∫Q|φ~k(D)φk(D)f(x)|pdx)1/p
for k∈ℤ.We decompose (83) by using J1 and J2,k. 								(117)1rσ|Q|(λ/n)+(1/p)(∫Q|f(x)|pdx)1/p≤J1+∑k=-∞JQ-1J2,k.
by using (115).We estimateJ1. Let θ be an auxiliary parameter again, which is taken so that
(118)θ∈(1+(λ+σ)pn,1).
By using the maximal operator M, we have
(119)J1≤1rσ|Q|(λ/n)+(1/p)×(∫ℝn|∑j=JQ∞φ~j(D)φj(D)f(x)|pMχQ(x)θdx)1/p.
By Lemma 16, we have
(120)J1≲1rσ|Q|(λ/n)+(1/p)×(∫ℝn(∑j=JQ∞|φj(D)f(x)|2)p/2MχQ(x)θdx)1/p.
We write
(121)ak∶=1rσ|Q|(λ/n)+(1/p)×(∫2kQ(∑j=JQ∞|φj(D)f(x)|2)p/2dx)1/p.
We decompose again the right-hand side of (120) dyadically.
(122)J1≲∑k=1∞2-knθ/pak.
For each k, we have 2kQ⊂Q2kr, since Q⊂Qr. Thus,
(123)ak=2kσ+k(λ+(n/p))(2kr)σ|2kQ|(λ/n)+(1/p)×(∫2kQ(∑j=JQ∞|φj(D)f(x)|2)p/2dx)1/p≲2k(σ+λ+(n/p))∥f∥B˙σ(Lp,λD).
Consequently, from (118), (122), and (123), we obtain that
(124)J1≲∑k=K∞2-(knθ/p)ak≲∑k=1∞2k(σ+λ+(n(1-θ)/p))∥f∥B˙σ(Lp,λD)≲∥f∥B˙σ(Lp,λD).
Thus, the estimate for J1 is now valid.As forJ2,k we choose L*≫1 first. Since φ∈𝒮, we have
(125)J2,k=(2π)-n/2·1rσ|Q|(λ/n)+(1/p)×(∫Q|∫ℝnℱ-1φ~k(x-y)φk(D)f(y)dy|pdx)1/p≲1rσ|Q|(λ/n)+(1/p)×(∫Q(∫ℝn2kn|φk(D)f(y)|〈2k(x-y)〉2L*dy)pdx)1/p.
By the Hölder inequality, we have
(126)J2,k≲1rσ|Q|(λ/n)+(1/p)×(sup,R:Q(x,R)∈𝒟(Qr)x1(2R)λ∫Q(∫ℝn2kn〈2k(x-y)〉L*p′dy)p/p′×(∫ℝn2kn|φk(D)f(y)|p〈2k(x-y)〉L*pdy)dxsup,R:Q(x,R)∈𝒟(Qr)x1(2R)λ)1/p~1rσ|Q|λ/n×(sup,R:Q(x,R)∈𝒟(Qr)x1(2R)λ∫ℝn(1|Q|∫Q2kn〈2k(x-y)〉L*pdx)×|φk(D)f(y)|pdysup,R:Q(x,R)∈𝒟(Qr)x1(2R)λ)1/p.
We have, for k≤JQ and y∈ℝn,
(127)1|Q|∫Q2kn〈2k(x-y)〉L*pdx≲2kn〈2k(y-c(Q))〉L*p.
Consequently, from (126) and (127), we have
(128)J2,k≲1rσ|Q|λ/n(∫ℝn2kn|φk(D)f(y)|p〈2k(y-c(Q))〉L*pdy)1/p.
We decompose the integral (128) dyadically:
(129)J2,k≲1rσ|Q|λ/n×∑l=1∞2(kn/p)-lL*(∫Q(c(Q),2-k+l)|φk(D)f(y)|pdy)1/p.
Notice that Q(c(Q),2-k+l)⊂Qr+2-k+l, since c(Q)∈Q⊂Qr. Consequently, assuming that L*≫1, we have
(130)J2,k≲1rσ|Q|λ/n∑l=1∞2(ln/p)-lL*(r+2-k+l)σ×|Q(c(Q),2-k+l)|λ/n∥f∥B˙σ(Lp,λD)=1rσ|Q|λ/n(r+2-k)σ2-kλ∥f∥B˙σ(Lp,λD).
We are going to add (130) over k≤JQ. To this end, observe that
(131)∑k=-∞JQ-1(r+2-k)σ2-kλ=1log2∑k=-∞JQ-1∫2-k2-k+1(r+2-k)σ2-kλdtt~∑k=-∞JQ-1∫2-k2-k+1(r+t)σtλdtt=∫2ℓ(Q)∞(r+t)σtλdtt.
If we change variables, then we have
(132)∑k=-∞JQ-1(r+2-k)σ2-kλ~∫2ℓ(Q)∞(r+t)σtλdtt=rσ+λ∫2ℓ(Q)/r∞(1+v)σvλdvv.
Now that ℓ(Q)≤r and σ+λ<0, we obtain
(133)∑k=-∞JQ-1(r+2-k)σ2-kλ~rσ+λ(∫2ℓ(Q)/r2vλdvv+∫2∞(1+v)σvλdvv)~rσℓ(Q)λ.
As a result, from (133) and the fact that ℓ(Q)≤r, the estimate (130) is summable over k≤JQ-1, and we obtain that
(134)∑k=-∞JQ-1J2,k≲∑k=-∞JQ-1(r+2-k)σ2-kλrσ|Q|λ/n~(r+ℓ(Q))σℓ(Q)λrσ|Q|λ/n∥f∥Lp,λD~∥f∥Lp,λD.
Consequently, (83) is proved, and we conclude the proof of Theorem 4(c).


### 3.5. Proof of Theorem4 Part (d)

An important corollary of Proposition5 is that we can relax the condition (17) on φ, which is stronger than (d) in Theorem 4.Corollary 25.
Letφ*∈Ccomp∞(ℝn∖{0}) be such that
(135)∑j=-∞∞φ*(2-jξ)≳1(ξ∈ℝn∖{0}).
Define
(136)∥f∥B˙σ(Lp,λD)*≔supr>01rσ[sup,R:Q(x,R)∈𝒟(Qr)x1(2R)λsupQ∈𝒟(Qr)1ℓ(Q)λ×{1|Q|∫Q(∑j=JQ∞|φj*(D)f(y)|2)p/2dy}1/psup,R:Q(x,R)∈𝒟(Qr)x1(2R)λ].
Then the norm equivalence
(137)∥f∥B˙σ(Lp,λD)~∥f∥B˙σ(Lp,λD)*
holds.Proof.
Choose integersJ- and J+ so that
(138)Θ(ξ):=∑k=J-J+φ*(2-kξ)>0,ξ∈supp(φ).
Define
(139)φ**(ξ)≔{φ(ξ)Θ(ξ)(onsupp(φ)),0(outsidesupp(φ)).
Then φ**∈Ccomp∞(ℝn∖{0}) by virtue of (138). Observe also that
(140)φ**Θ=φ
in view of the size of supports. Hence, we have
(141)|φj(D)f(x)|≲∑k=J-J+|φj+k*(D)φj**(D)f(x)|≲∑k=J-J+M[φj+k*(D)f](x).
Since the numbers J± appearing in (138) are finite, we have
(142)∥f∥B˙σ(Lp,λD)*≲∥f∥B˙σ(Lp,λD)
from (141) and Proposition 5. The proof of
(143)∥f∥B˙σ(Lp,λD)*≳∥f∥B˙σ(Lp,λD).
is analogous; just swap the role of φ and φ*. With (142) and (143), we see that the norms in (137) are equivalent.


## 3.1. Proof of Proposition10

It follows from the definition of the norm∥f∥B˙σ(Lp,λ) that
(63)∥f+C∥B˙σ(ℒp,λ(0))≲∥f∥B˙σ(Lp,λ)
for any constant function C. Note that this implies that B˙σ(Lp,λ) is a subset of B˙σ(ℒp,λ(0)).Letf∈B˙σ(ℒp,λ(0))(ℝn). Then, since p>1, we can use the Hölder inequality and we have
(64)|PQs0f(0)-PQr0f(0)|≤1|Qs|∫Qs|f(x)-PQr0f(0)|dx≲rσ+λ∥f∥B˙σ(ℒp,λ(0))(ℝn)
for r≤s≤2r. Assuming that σ+λ<0, the sequence {PQr0f(0)}r>0 is convergent by the Cauchy test. Thus, we can consider the mapping
(65)ι:f∈B˙σ(ℒp,λ(0))(ℝn)↦f-limR→∞PQR0f(0).
Let us check that the range of ι is in B˙σ(Lp,λ)(ℝn).Letr>0 and Q⊂Qr be fixed. Let N be the largest integer such that 2NQ⊂Q2r. Then observe that 2N+2Q⊃Qr. We decompose
(66)f-limR→∞PQR0f(0)=f-PQ0f(0)+P2NQ0f(0)-PQr0f(0)+∑i=1∞(P2i-1Qr0f(0)-P2iQr0f(0))-∑l=1N(P2lQ0f(0)-P2l-1Q0f(0)).
We consider the Lp-norm over Q and multiply 1/(rσℓ(Q)λ+n/p).As for the first term, directly from the definition of∥f∥ℒp,λ(0), we obtain that
(67)1rσℓ(Q)λ+n/p∫Q|f(x)-PQ0f(0)|dx≤∥f∥ℒp,λ(0).
Also, a geometric observation shows that the second term can be estimated similarly. Since λ<0, we obtain that
(68)1rσℓ(Q)λ|P2NQ0f(0)-PQr0f(0)|≤1rσ(2Nℓ(Q))λ|P2NQ0f(0)-PQr0f(0)|≲∥f∥ℒp,λ(0).
By using (64) and λ<0, we can handle the third term:
(69)1rσℓ(Q)λ|∑i=1∞(P2i-1Qr0f(0)-P2iQr0f(0))|≤1rσℓ(Q)λ∑i=1∞|P2i-1Qr0f(0)-P2iQr0f(0)|≤1rσ(2Nℓ(Q))λ∑i=1∞|P2i-1Qr0f(0)-P2iQr0f(0)|.
In view of the way in which we chose N, we obtain that
(70)1rσℓ(Q)λ|∑i=1∞(P2i-1Qr0f(0)-P2iQr0f(0))|≲1rσ+λ∑i=1∞|P2i-1Qr0f(0)-P2iQr0f(0)|≲∥f∥ℒp,λ(0).
In summary,
(71)1rσℓ(Q)λ|∑i=1∞(P2i-1Qr0f(0)-P2iQr0f(0))|≲∥f∥ℒp,λ(0).
It remains to estimate the fourth term. We employ the following estimate:
(72)|P2lQ0f(0)-P2l-1Q0f(0)|≲rσℓ(2lQ)λ∥f∥ℒp,λ(0).
Since λ<0, (72) is summable and we obtain that
(73)1rσℓ(Q)λ|∑l=1N(P2lQ0f(0)-P2l-1Q0f(0))|≲∥f∥ℒp,λ(0),
where the implicit constant in (73) is independent of N.It follows from (67), (68), (71), and (73) that ι sends B˙σ(ℒp,λ(0))(ℝn) to B˙σ(Lp,λ)(ℝn) boundedly;
(74)∥ι(f)∥B˙σ(Lp,λ)≲∥f∥B˙σ(ℒp,λ(0))
for all f∈B˙σ(Lp,λ)(ℝn).Meanwhile, it follows from the norm∥·∥B˙σ(Lp,λ) that
(75)limR→∞PQR0f(0)=0
for all f∈B˙σ(Lp,λ)(ℝn). In view of (63), ι is a surjection:
(76)ι(B˙σ(ℒp,λ(0))(ℝn))=B˙σ(Lp,λ)(ℝn).Finally observe that, forf∈B˙σ(ℒp,λ(0))(ℝn), ι(f)=0 if and only if f is a constant function, that is, f∈𝒫0(ℝn). Namely,
(77)Ker(ι)=𝒫0(ℝn).
Thus, from (75), (76) and (77), we conclude (33).


## 3.2. Proof of Theorem4 Part (a)

For1<p<∞ and σ∈ℝ, let us define the function space u-Llocp(〈·〉σ)(ℝn) of uniformly Llocp(〈·〉σ)(ℝn) functions by
(78)u-Llocp(〈·〉σ)(ℝn)={f∈Llocp(ℝn):∥f∥u-Llocp(〈·〉σ)<∞},
where, for m∈ℤn, we write 〈m〉:=1+|m|2, and the norm is given by
(79)∥f∥u-Llocp(〈·〉σ)=supm∈ℤn(〈m〉σp∫m+[0,1]n|f(y)|pdy)1/p.
Then from the definition of the norms (10) and (78), the following chain of continuous inclusion holds. For -n/p≤λ,
(80)B˙σ(Lp,λ)(ℝn)↪u-Llocp(〈·〉-σ)(ℝn)↪𝒮′(ℝn).
Thus, (a) follows.Remark 22.
The spaceu-Llocp(〈·〉-σ)(ℝn) is a special case of amalgams investigated in [32].


## 3.3. Proof of Theorem4 Part (b)

Letφ∈𝒮(ℝn) be a fixed function satisfying (17). It follows from (17) that there exist J- and J+ such that
(81)∑j=J-J+φ(ξ)φ(2-jξ)=φ(ξ).
Let f∈B˙σ(Lp,λD)(ℝn) and fix a dyadic cube Q* such that 0∈Q*¯. We are going to show that
(82)g1:=limL→∞∑j=-LJQ*φj(D)f∈L∞(Q*)∩𝒮′(ℝn)
converges in the topology of L∞(Q*)∩𝒮′(ℝn), and that
(83)g2:=limL→∞∑j=JQ*Lφj(D)f∈Lp(Q*)∩𝒮′(ℝn)
converges in the topology of Lp(Q*)∩𝒮′(ℝn).The presice meaning of (82) and (83) is that as follows: (82) means
(84)∥g1-∑j=-LJQ*φj(D)f∥L∞(Q*)→0,〈∑j=-LJQ*φj(D)f,τ〉→〈g1,τ〉(τ∈𝒮(ℝn))
as L→∞ and (83) means that
(85)∥g2-∑j=JQ*L-1φj(D)f∥Lp(Q*)→0〈∑j=JQ*L-1φj(D)f,τ〉→〈g2,τ〉(τ∈𝒮(ℝn))
as L→∞. Once (82) and (83) are proven, we will have
(86)g:=∑j=-∞∞φj(D)f∈Lp(Q*)∩𝒮′(ℝn)
converges in the topology of Lp(Q*)∩𝒮′(ℝn) and that f-g is a polynomial. Hence it follows that f∈Llocp(ℝn). Remark also that the convergence in 𝒮′(ℝn) of the sum defining g2 is a generality. So let us prove (82) and (83).Let us begin with proving (82). To this end we take a dyadic cube Q containing Q*. Since we are assuming (81), we deduce that
(87)|φJQ(D)f(x)|≲∑l=J-J+|ℱ-1φl+JQ*φJQ(D)f(x)|≤∑l=J-J+∫ℝn|ℱ-1φl+JQ(x-y)φJQ(D)f(y)|dy(x∈ℝn).
Since ℱ-1φ∈𝒮(ℝn), we have
(88)|φJQ(D)f(x)|≲1|Q|∫ℝn|φJQ(D)f(y)|(1+ℓ(Q)-1|y|)2n+2σ+2dy(x∈Q)
from (19) and (20). By the Hölder inequality and the fact that σ≥0, we have
(89)|φJQ(D)f(x)|≲1|Q|(∫ℝndy(1+ℓ(Q)-1|y|)p′(n+σ+1))1/p′×(∫ℝn|φJQ(D)f(y)|pdy(1+ℓ(Q)-1|y|)p(n+σ+1))1/p~(1|Q|∫ℝn|φJQ(D)f(y)|p(1+ℓ(Q)-1|y|)p(n+σ+1)dy)1/p.
for all x∈Q. By decomposing the last integral dyadically, we obtain that
(90)|φJQ(D)f(x)|≲∑l=0∞(2-lp(n+σ+1)1|Q|∫2lQ|φJQ(D)f(y)|pdy)1/p.
Observe from Definition 3 that
(91)∥φJQ(D)f∥Lp(2lQ)≲|2lQ|((σ+λ)/n)+(1/p)∥f∥B˙σ(Lp,λD)
for all l≥0. Thus, we obtain
(92)|φJQ(D)f(x)|=|Q|(λ+σ)/n∑l=0∞2-l(n+σ+1)+l(σ+λ+(n/p))∥f∥B˙σ(Lp,λD)≲|Q|(σ+λ)/n∥f∥B˙σ(Lp,λD)(x∈Q).
Hence, from (92) and the fact that λ+σ<0, we have
(93)∥∑j=-∞JQ*-1|φj(D)f|∥L∞(Q*)≤∑j=-∞JQ*-1supx∈Q*|φj(D)f(x)|≲|Q*|(σ+λ)/n∥f∥B˙σ(Lp,λD),
which proves the embedding into L∞(Q*). To prove that the sum defining g1 converges in the topology of 𝒮′(ℝn), we first fix a cube R containing Q* and observe
(94)∥∑j=JRJQ*|φj(D)f|∥Lp(R)≲|R|((σ+λ)/n)+(1/p)×log(2+ℓ(R)ℓ(Q*))∥f∥B˙σ(Lp,λD)
from (91) and the fact that σ+λ+n/p≥0. Meanwhile, again by (92), we have
(95)∥φJQ(D)f∥L∞(Q)≲|Q|(σ+λ)/n∥f∥B˙σ(Lp,λD)
for all cubes Q containing R. As a result, we obtain that
(96)∥∑j=-∞JR|φj(D)f|∥Lp(R)≲|R|((σ+λ)/n)+(1/p)∥f∥B˙σ(Lp,λD)
from the fact that σ+λ<0. It follows from (94) and (96) that the sum defining g1 converges in 𝒮′(ℝn).Now let us prove (83). First, choose θ∈(1+(λ+σ)p/n,1). By virtue of Lemma 16 and the fact that Ap((MχQ*)θ) has a bound independent of Q*∈𝒟 [33], we obtain that
(97)∥∑j=JQ*∞φj(D)f∥Lp(Q*)≤∥(MχQ*)θ/p∑j=JQ*∞φj(D)f∥Lp≲∥(MχQ*)θ/p(∑j=JQ*-j0∞|φj(D)f|2)1/2∥Lp.
Here for the last inequality, we employed a geometric observation of the support of φ. Therefore, j0 of the last inequality depends only on φ.SinceMχQ*(x)~(1+ℓ(Q*)-1|x-c(Q*)|)-n, we have
(98)∥∑j=JQ*∞φj(D)f∥Lp(Q*)≲∑l=-j0∞2-lnθ/p∥(∑j=JQ*-j0∞|φj(D)f|2)1/2∥Lp(2lQ*)≲∑l=-j0∞2-lnθ/pℓ(2lQ*)σ+λ+(n/p)∥f∥B˙σ(Lp,λD).
Recall that we are assuming that θ∈(1+(λ+σ)p/n,1). So (98) is summable and we have
(99)∥∑j=JQ*∞φj(D)f∥Lp(Q*)≲ℓ(Q*)σ+λ+(n/p)∥f∥B˙σ(Lp,λD),
showing (83).With (82) and (83), the proof of (b) is now complete.Remark 23.
We did not use the structure ofφj(D)f in the proof. Therefore, we can deduce the following variant.Proposition 24.
Suppose that we are given a collection{Fj}j=-∞∞⊂𝒮′(ℝn) such that supp(ℱFj)⊂supp(φj) and that
(100)supr>01rσ[supQ∈𝒟(Qr)1ℓ(Q)λ{1|Q|∫Q(∑j=JQ∞|Fj(y)|2)p/2dy}1/p]<∞.
Then G:=∑j=-∞∞Fj converges in the topology of 𝒮′(ℝn).


## 3.4. Proof of Theorem4 Part (c)

Keeping what we proved in Parts (a) and (b) in mind, we shall now describe the isomorphism betweenB˙σ(Lp,λ)(ℝn) and B˙σ(Lp,λD)(ℝn). If f∈B˙σ(Lp,λ)(ℝn), then we claim that f∈B˙σ(Lp,λD)(ℝn) and that
(101)∥f∥B˙σ(Lp,λD)≲∥f∥B˙σ(Lp,λ).
The proof of (101) is comparably easy.Let us start to prove (101). Let r>0 and Q∈𝒟(Qr) be fixed. Also, fix a Rademacher sequence {rj}j=-∞∞⊂L2([0,1]). Define
(102)KQ(x;t):=∑j=JQ∞2jnrj(t)ℱ-1φ(2jx)(x∈ℝn,t>0).
Then we have
(103)∑j=JQ∞rj(t)φj(D)f=KQ(·;t)*f,1rσ(∫Q(∑j=JQ∞|φj(D)f(y)|2)p/2dy|Q|(λp/n)+1)1/p~1rσ(∬Q×[0,1]|KQ(·;t)*f(y)|pdydt|Q|(λp/n)+1)1/p
by the property of the Rademacher sequence. Let us set
(104)f1:=fχ2Q,f2:=f-f1.
Then we have
(105)|KQ(x;t)|≲|x|-n,|∂kKQ(x;t)|≲|x|-n-1,|ℱ[KQ(·;t)](ξ)|≲1,
since rj(t)∈{0,1,-1} with constant independent of Q.By the Calderón-Zygmund theory forLp(ℝn), we have
(106)1rσ(∫01∫Q|KQ(·;t)*f1(y)|pdydt|Q|(λp/n)+1)1/p≲1rσ(∫2Q|f(y)|pdy|Q|(λp/n)+1)1/p.
As for f2, we invoke the following pointwise estimate obtained from (105):
(107)|KQ(·;t)*f2(y)|≲∫ℝn∖2Q|f(z)||z-c(Q)|ndz(y∈Q).
We write
(108)Ik≔∫2k+1Q∖2kQ|f(z)||z-c(Q)|ndz~1|2kQ|∫2k+1Q∖2kQ|f(z)|dz,k=1,2,….
Notice that
(109)Ik≲(2k+1r)σ(2k+1ℓ(Q))λ∥f∥B˙σ(Lp,λ),
since 2k+1Q⊂Q2k+1r. Assuming that σ+λ<0, we have
(110)∑k=1∞Ikrσ|Q|λ/n≲∑k=1∞(2k+1r)σ(2k+1ℓ(Q))λrσ|Q|λ/n∥f∥B˙σ(Lp,λ)≲∑k=1∞2(σ+λ)k∥f∥B˙σ(Lp,λ)~∥f∥B˙σ(Lp,λ).
Consequently, from (106) and (110), we obtain (101).The heart of the matter is how to construct the inverse mapping fromB˙σ(Lp,λD)(ℝn) to B˙σ(Lp,λ)(ℝn). Let f~∈B˙σ(Lp,λD)(ℝn). Then we have established in (b) that
(111)f:=∑j=-∞∞φj(D)f~
exists in the topology of Llocp(ℝn)∩𝒮′(ℝn) in view of (92) and (99). We claim that
(112)∥f∥B˙σ(Lp,λ)≲∥f~∥B˙σ(Lp,λD).
Here and below in proving (112), we assume that f~=f by replacing f with f~.Let us prove (112). Suppose that we are given a dyadic cube Q and r>0 such that Q∈𝒟(Qr). We estimate
(113)1rσ|Q|(λ/n)+(1/p)(∫Q|f(x)|pdx)1/p.
First, we expand f by using (111). If we denote
(114)φ~j(ξ):=∑l=J-J+φ(2-j-lξ),ξ∈ℝn
for j∈ℤ, then we have
(115)f(x)=∑j=-∞∞φ~j(D)φj(D)f(x),
where the convergence, as we have established in Part (b), takes places in Llocp(ℝn)∩𝒮′(ℝn).We set(116)J1:=1rσ|Q|(λ/n)+(1/p)(∫Q|∑j=JQ∞φ~j(D)φj(D)f(x)|pdx)1/p,J2,k:=1rσ|Q|(λ/n)+(1/p)(∫Q|φ~k(D)φk(D)f(x)|pdx)1/p
for k∈ℤ.We decompose (83) by using J1 and J2,k. 								(117)1rσ|Q|(λ/n)+(1/p)(∫Q|f(x)|pdx)1/p≤J1+∑k=-∞JQ-1J2,k.
by using (115).We estimateJ1. Let θ be an auxiliary parameter again, which is taken so that
(118)θ∈(1+(λ+σ)pn,1).
By using the maximal operator M, we have
(119)J1≤1rσ|Q|(λ/n)+(1/p)×(∫ℝn|∑j=JQ∞φ~j(D)φj(D)f(x)|pMχQ(x)θdx)1/p.
By Lemma 16, we have
(120)J1≲1rσ|Q|(λ/n)+(1/p)×(∫ℝn(∑j=JQ∞|φj(D)f(x)|2)p/2MχQ(x)θdx)1/p.
We write
(121)ak∶=1rσ|Q|(λ/n)+(1/p)×(∫2kQ(∑j=JQ∞|φj(D)f(x)|2)p/2dx)1/p.
We decompose again the right-hand side of (120) dyadically.
(122)J1≲∑k=1∞2-knθ/pak.
For each k, we have 2kQ⊂Q2kr, since Q⊂Qr. Thus,
(123)ak=2kσ+k(λ+(n/p))(2kr)σ|2kQ|(λ/n)+(1/p)×(∫2kQ(∑j=JQ∞|φj(D)f(x)|2)p/2dx)1/p≲2k(σ+λ+(n/p))∥f∥B˙σ(Lp,λD).
Consequently, from (118), (122), and (123), we obtain that
(124)J1≲∑k=K∞2-(knθ/p)ak≲∑k=1∞2k(σ+λ+(n(1-θ)/p))∥f∥B˙σ(Lp,λD)≲∥f∥B˙σ(Lp,λD).
Thus, the estimate for J1 is now valid.As forJ2,k we choose L*≫1 first. Since φ∈𝒮, we have
(125)J2,k=(2π)-n/2·1rσ|Q|(λ/n)+(1/p)×(∫Q|∫ℝnℱ-1φ~k(x-y)φk(D)f(y)dy|pdx)1/p≲1rσ|Q|(λ/n)+(1/p)×(∫Q(∫ℝn2kn|φk(D)f(y)|〈2k(x-y)〉2L*dy)pdx)1/p.
By the Hölder inequality, we have
(126)J2,k≲1rσ|Q|(λ/n)+(1/p)×(sup,R:Q(x,R)∈𝒟(Qr)x1(2R)λ∫Q(∫ℝn2kn〈2k(x-y)〉L*p′dy)p/p′×(∫ℝn2kn|φk(D)f(y)|p〈2k(x-y)〉L*pdy)dxsup,R:Q(x,R)∈𝒟(Qr)x1(2R)λ)1/p~1rσ|Q|λ/n×(sup,R:Q(x,R)∈𝒟(Qr)x1(2R)λ∫ℝn(1|Q|∫Q2kn〈2k(x-y)〉L*pdx)×|φk(D)f(y)|pdysup,R:Q(x,R)∈𝒟(Qr)x1(2R)λ)1/p.
We have, for k≤JQ and y∈ℝn,
(127)1|Q|∫Q2kn〈2k(x-y)〉L*pdx≲2kn〈2k(y-c(Q))〉L*p.
Consequently, from (126) and (127), we have
(128)J2,k≲1rσ|Q|λ/n(∫ℝn2kn|φk(D)f(y)|p〈2k(y-c(Q))〉L*pdy)1/p.
We decompose the integral (128) dyadically:
(129)J2,k≲1rσ|Q|λ/n×∑l=1∞2(kn/p)-lL*(∫Q(c(Q),2-k+l)|φk(D)f(y)|pdy)1/p.
Notice that Q(c(Q),2-k+l)⊂Qr+2-k+l, since c(Q)∈Q⊂Qr. Consequently, assuming that L*≫1, we have
(130)J2,k≲1rσ|Q|λ/n∑l=1∞2(ln/p)-lL*(r+2-k+l)σ×|Q(c(Q),2-k+l)|λ/n∥f∥B˙σ(Lp,λD)=1rσ|Q|λ/n(r+2-k)σ2-kλ∥f∥B˙σ(Lp,λD).
We are going to add (130) over k≤JQ. To this end, observe that
(131)∑k=-∞JQ-1(r+2-k)σ2-kλ=1log2∑k=-∞JQ-1∫2-k2-k+1(r+2-k)σ2-kλdtt~∑k=-∞JQ-1∫2-k2-k+1(r+t)σtλdtt=∫2ℓ(Q)∞(r+t)σtλdtt.
If we change variables, then we have
(132)∑k=-∞JQ-1(r+2-k)σ2-kλ~∫2ℓ(Q)∞(r+t)σtλdtt=rσ+λ∫2ℓ(Q)/r∞(1+v)σvλdvv.
Now that ℓ(Q)≤r and σ+λ<0, we obtain
(133)∑k=-∞JQ-1(r+2-k)σ2-kλ~rσ+λ(∫2ℓ(Q)/r2vλdvv+∫2∞(1+v)σvλdvv)~rσℓ(Q)λ.
As a result, from (133) and the fact that ℓ(Q)≤r, the estimate (130) is summable over k≤JQ-1, and we obtain that
(134)∑k=-∞JQ-1J2,k≲∑k=-∞JQ-1(r+2-k)σ2-kλrσ|Q|λ/n~(r+ℓ(Q))σℓ(Q)λrσ|Q|λ/n∥f∥Lp,λD~∥f∥Lp,λD.
Consequently, (83) is proved, and we conclude the proof of Theorem 4(c).


## 3.5. Proof of Theorem4 Part (d)

An important corollary of Proposition5 is that we can relax the condition (17) on φ, which is stronger than (d) in Theorem 4.Corollary 25.
Letφ*∈Ccomp∞(ℝn∖{0}) be such that
(135)∑j=-∞∞φ*(2-jξ)≳1(ξ∈ℝn∖{0}).
Define
(136)∥f∥B˙σ(Lp,λD)*≔supr>01rσ[sup,R:Q(x,R)∈𝒟(Qr)x1(2R)λsupQ∈𝒟(Qr)1ℓ(Q)λ×{1|Q|∫Q(∑j=JQ∞|φj*(D)f(y)|2)p/2dy}1/psup,R:Q(x,R)∈𝒟(Qr)x1(2R)λ].
Then the norm equivalence
(137)∥f∥B˙σ(Lp,λD)~∥f∥B˙σ(Lp,λD)*
holds.Proof.
Choose integersJ- and J+ so that
(138)Θ(ξ):=∑k=J-J+φ*(2-kξ)>0,ξ∈supp(φ).
Define
(139)φ**(ξ)≔{φ(ξ)Θ(ξ)(onsupp(φ)),0(outsidesupp(φ)).
Then φ**∈Ccomp∞(ℝn∖{0}) by virtue of (138). Observe also that
(140)φ**Θ=φ
in view of the size of supports. Hence, we have
(141)|φj(D)f(x)|≲∑k=J-J+|φj+k*(D)φj**(D)f(x)|≲∑k=J-J+M[φj+k*(D)f](x).
Since the numbers J± appearing in (138) are finite, we have
(142)∥f∥B˙σ(Lp,λD)*≲∥f∥B˙σ(Lp,λD)
from (141) and Proposition 5. The proof of
(143)∥f∥B˙σ(Lp,λD)*≳∥f∥B˙σ(Lp,λD).
is analogous; just swap the role of φ and φ*. With (142) and (143), we see that the norms in (137) are equivalent.


## 4.B˙σ(Lp,λ)-B˙σ(Lq,μ) Boundedness of Iα, Proof of Theorem 6

To prove Theorem6, we need the lemmas. Note that Lemma 26 can be seen as the Plancherel-Polya-Nikolskij inequality for B˙σ(Lp,λD).Lemma 26.
Suppose that the parametersσ,p, and λ satisfy
(144)σ≥0,λ<0,1≤p<∞.
Let φ be chosen so that it satisfies (17). Then we have
(145)1rσ|Q|λ/n∥φJQ(D)f∥L∞(Q)≲∥φJQ(D)f∥B˙σ(Lp,λD)≲∥f∥B˙σ(Lp,λD)
for all r>0 and Q∈𝒟(Qr).Proof.
The right-hand inequality is a consequence of Lemma20.
Letx∈Q be fixed. We choose L∈ℕ sufficiently large. Also, take a compactly supported function φ¯ so that φ¯ equals 1 on supp(φ) as we did in (138). Then we have
(146)1rσ|Q|λ/n|φ¯JQ(D)φJQ(D)f(x)|=(2π)-n/2·1rσ|Q|(λ/n)+1×|∫ℝnℱ-1φ¯(x-yℓ(Q))φJQ(D)f(y)dy|≲1rσ|Q|(λ/n)+1×∫ℝn(1+|c(Q)-yℓ(Q)|)-L|φJQ(D)f(y)|dy.
We decompose ℝn dyadically. Then we have
(147)1rσ|Q|λ/n|φ¯JQ(D)φJQ(D)f(x)|≲∑l=1∞2-lLrσ|Q|(λ/n)+1∫2lQ|φJQ(D)f(y)|dy≲∑l=1∞2-l(L-σ-n-λ)(2lr)σ|2lQ|(λ/n)+1∫2lQ|φJQ(D)f(y)|dy≲∥φJQ(D)f∥B˙σ(Lp,λD).
Thus, the proof is now complete.In the course of the proof, we obtained the following chain of inequalities.Corollary 27.
Suppose that the parametersσ,p, and λ satisfy
(148)σ≥0,λ<0,1≤p<∞.
Let φ be chosen so that it satisfies (17). Then one has
(149)∥φj(D)f∥L∞(Qr)≲2-jλrσ∥φj(D)f∥B˙σ(Lp,λD)~2-jλrσ∥φj(D)f∥B˙σ(Lp,λ)≲2-jλrσ∥f∥B˙σ(Lp,λD)
for f∈B˙σ(Lp,λD)(ℝn), j∈ℤ and r>0 such that 2-j<r.The following estimate is somehow well known. Here we remark that the following form was recorded in [34]. Here we change variables j↦-j to transform the result into the one needed in the present paper.Lemma 28.
LetA,B,κ1,κ2>0. Then
(150)∑j=-∞∞min(2jκ1A,2-jκ2B)≲Aκ2/(κ1+κ2)Bκ1/(κ1+κ2).Now we prove Theorem6. Given f∈B˙σ(Lp,λ), we shall define
(151)Iαf(x)=∑j=-∞∞Iα[φj(D)f](x).
Note that, by using the Fourier transform, Iα[φj(D)f] can be defined. Once we show that {Iα[φj(D)f]}j=1∞ satisfies the condition in Proposition 24, then by Proposition 24  Iα can be defined. Since this essentially amounts to showing
(152)∥Iαf∥B˙σ(Lq,μD)<∞
by assuming that we can define Iαf, we omit the detail.Since1<q<∞, we have
(153)∥Iαf∥B˙σ(Lq,μ)~∥Iαf∥B˙σ(Lq,μD)
by virtue of Theorem 4. Likewise since 1<p<∞, we have
(154)∥f∥B˙σ(Lp,λ)~∥f∥B˙σ(Lp,λD).
Let Q be a fixed dyadic cube. Define φ*(ξ):=|ξ|αφ(ξ). Then we have (135) and
(155)φj*(ξ):=|2-jξ|αφj(ξ)
for ξ∈ℝn and j∈ℤ. Thus, we have
(156)1rσ|Q|μ/n(∑j=JQ∞|φj*(D)Iαf(x)|2)1/2=c1rσ|Q|μ/n(∑j=JQ∞2-2jα|φj(D)f(x)|2)1/2
since ℱIαf(ξ)=c|ξ|-αℱf(ξ). If we use the embedding ℓ1↪ℓ2, then we have
(157)1rσ|Q|μ/n(∑j=JQ∞2-2jα|φj(D)f(x)|2)1/2≲1rσ|Q|μ/n∑j=JQ∞2-jα|φj(D)f(x)|.
By Corollary 27 and μ=α+λ, we have
(158)1rσ|Q|μ/n∑j=JQ∞2-jα|φj(D)f(x)|≲1|Q|μ/n×∑j=JQ∞min(∑∥φj(D)f∥B˙σ(Lp,λD)2jμ,12jαrσ(∑ν=JQ∞|φν(D)f(x)|2)1/2∑)≲1|Q|μ/n×∑j=-∞∞min(∥f∥B˙σ(Lp,λD)2jμ,12jαrσ(∑ν=JQ∞|φν(D)f(x)|2)1/2).
Now we invoke Lemma 28 with
(159)A=∥f∥B˙σ(Lp,λD),B=1rσ(∑ν=JQ∞|φν(D)f(x)|2)1/2,κ1=-μ,κ2=α.
Since μ=α+λ, we have κ2/(κ1+κ2)=-α/λ,κ1/(κ1+κ2)=1+(α/λ). Thus, from Lemma 28, we obtain that
(160)1rσ|Q|μ/n(∑j=JQ∞2-jα|φj(D)f(x)|)≲Aκ2/(κ1+κ2)Bκ2/(κ1+κ2)=1|Q|μ/n(∥f∥B˙σ(Lp,λD))-α/λ×{1rσ(∑ν=JQ∞|φν(D)f(x)|2)1/2}1+(α/λ).
We consider the Lq(Q)-norm of the both sides. Then by the relation 1+α/λ=μ/λ=p/q, we deduce
(161)1rσ|Q|(μ/n)+(1/q)(∫Q(∑j=JQ∞2-jα|φj(D)f(y)|)qdy)1/q≲(∥f∥B˙σ(Lp,λD))-α/λ×{1rσp|Q|(λp/n)+1∫Q(∑ν=JQ∞|φν(D)f(y)|2)p/2dy}1/q=(∥f∥B˙σ(Lp,λD))-α/λ×{1rσp|Q|(λp/n)+1∫Q(∑ν=JQ∞|φν(D)f(y)|2)p/2dy}(λ+α)/(λp).
In view of the definition of ∥f∥B˙σ(Lp,λD), we have
(162)1rσ|Q|(μ/n)+(1/q)(∫Q(∑j=JQ∞2-jα|φj(D)f(y)|)qdy)1/q≲∥f∥B˙σ(Lp,λD).
Putting (156), (157) and (162) together, we obtain
(163)1rσ|Q|μ/n(1|Q|∫Q(∑j=JQ∞|φj*(D)Iαf(x)|2)q/2dx)1/q≲∥f∥B˙σ(Lp,λD).
Consequently, from (153), Corollary 25 and the definition of the norm B˙σ(Lq,μD)(ℝn), we obtain
(164)∥Iαf∥B˙σ(Lq,μ)≲∥Iαf∥B˙σ(Lq,μD)≲∥f∥B˙σ(Lp,λD).
By combining (164) with (154), we conclude that Iα is a bounded operator from B˙σ(Lp,λ)(ℝn) to B˙σ(Lq,μ)(ℝn).Remark 29.
Suppose that the parametersα,σ,p,q,μ, and λ satisfy
(165)0<α<n,σ≥0,λ,μ<0,1≤p<∞,1<q<∞
and (27). If we reexamine (157), we see that
(166)∥Iαf∥B˙σ(Lq,μ)~supr>0,Q∈𝒟(Qr)1rσ|Q|(μ/n)+(1/q)×(∫Q(∑j=JQ∞2-jα|φj(D)f(y)|)qdy)1/q≲∥(∑ν=-∞∞|φν(D)f|2)1/2∥B˙σ(Lp,λ)≲∥∑ν=-∞∞|φν(D)f|∥B˙σ(Lp,λ).


## 5.B˙σ(Lp,λ)-B˙σ(WLq,μ) Boundedness of Iα, Proof of Theorem 7

The definition ofIαf can be justified in the same was as Theorem 6. So we assume that Iαf is already defined.Letr>0 and a cube Q be fixed so that Q⊂Qr. Set q0:=λp/(λ+α-ε) and q1:=λp/(λ+α+ε) for some small ε>0. More precisely, we choose ε>0 so that
(167)μ+ε<0,q0,q1∈(1,∞),λ+σ+α+ε<0.
By interpolation described in [35] we have
(168)∥Iαf∥WLq(Q)~∥Iαf∥[Lq0(Q),Lq1(Q)]1/2,∞.
We calculate the K-functional. By the definition of the K-functional [35, Section 3.1] and by Remark 29, we have
(169)K(22jε|Q|-2ε/(λp)ℓ(Q)-2ε,Iαf;Lq0(Q),Lq1(Q))=inf{∫∥g0∥Lq0(Q)+22jε|Q|-2ε/(λp)r-2ε∥g1∥Lq1(Q):g0∈Lq0(Q),g1∈Lq1(Q),Iαf=g0+g1∫}.
Note that  (1/q)-(1/q0)=ε/(λp),(1/q)-(1/q1)=-ε/(λp). Hence we have
(170)1rσℓ(Q)(n/q)+μ|Q|ε/(λp)ℓ(Q)ε×K(22jε|Q|-2ε/(λp)ℓ(Q)-2ε,Iαf;Lq0(Q),Lq1(Q))≤inf{∥g0∥B˙σ(Lq0,μ-ε)+22jε∥g1∥B˙σ(Lq1,μ+ε):g0∈Lq0(Q),g1∈Lq1(Q),Iαf=g0+g1}.Let us define(171)g0:=Iα-ε(∑k=-∞j2kεφk(D)f),g1:=Iα+ε(∑k=j+1∞2-kεφk(D)f).
By Lemma 20 and Remark 29, we have
(172)∥g0∥B˙σ(Lq0,μ-ε)≲∥∑k=-∞j2kε|φk(D)f|∥B˙σ(Lp,λ),(173)∥g1∥B˙σ(Lq1,μ+ε)≲∥∑k=j+1∞2-kε|φk(D)f|∥B˙σ(Lp,λ),(174)∥φk(D)f∥B˙σ(Lp,λ)≲∥f∥B˙σ(Lp,λ).
From (172) and (173), we deduce that
(175)1rσℓ(Q)(n/q)+μ|Q|ε/(λp)ℓ(Q)ε×K(22jε|Q|-2ε/(λp)ℓ(Q)-2ε,Iαf;Lq0(Q),Lq1(Q))≲∑k=-∞∞2min(k,2j-k)ε∥φk(D)f∥B˙σ(Lp,λ)(j∈ℤ).
If we insert this estimate and (174), then we have
(176)1rσℓ(Q)(n/q)+μ∥Iαf∥[Lq0(Q),Lq1(Q)]1/2,∞~1rσℓ(Q)(n/q)+μ|Q|ε/(λp)ℓ(Q)ε×K(22jε|Q|-2ε/(λp)ℓ(Q)-2ε,Iαf;Lq0(Q),Lq1(Q))≲supj∈ℤ(2-jε∑k=-∞∞2min(k,2j-k)ε∥φk(D)f∥B˙σ(Lp,λ))≲∥f∥B˙σ(Lp,λ).
The proof is therefore complete.


## 6. Structure ofB˙σ(LipθD), Proof of Theorem 13

We prove Theorem13. Let us recall (13), which we use now. To prove Theorem 13, we need the following lemma.Lemma 30.
For all multi-indicesβ, r>0, f∈B˙σ(LipθD)(ℝn), we have
(177)supr>01rσ{supx∈Q∈𝒬(Qr)1ℓ(Q)θ(2-JQ|β||∂βφJQ(D)f(x)|)}≲∥f∥B˙σ(LipθD).Proof.
To prove this, we fixx∈ℝn,r>0 and Q∈𝒬 so that
(178)x∈Q⊂Qr.
Also, we take τ∈Ccomp∞(ℝn∖{0}) so that τ≡1 on supp(φ). We set τJQ(ξ):=τ(2-JQξ) for ξ∈ℝn as usual. Then we have
(179)2-JQ|β|rσ|Q|θ/n|∂βφJQ(D)f(x)|~2-JQ|β|rσ|Q|θ/n|∫ℝnℱ-1[ξβτJQ](y)φJQ(D)f(x-y)dy|≲1rσ|Q|θ/n∫ℝn2JQn|φJQ(D)f(x-y)|(1+2JQ|y|)n+σ+1dy≲∥f∥B˙σ(LipθD)1rσ∫ℝn2JQn(r+|y|)σ(1+2JQ|y|)n+σ+1dy=∥f∥B˙σ(LipθD)∫ℝn2JQn(1+r-1|y|)σ(1+2JQ|y|)n+σ+1dy.
Now that 2r>2-JQ, we have
(180)2-JQ|β|rσ|Q|θ/n|∂βφJQ(D)f(x)|≲∥f∥B˙σ(LipθD)∫ℝn2JQn(1+2JQ|y|)σ(1+2JQ|y|)n+σ+1dy~∥f∥B˙σ(LipθD).
This proves the lemma.Proposition 31 (see [36]).
Letd∈ℕ∪{0}. Suppose that we are given a sequence {fj}j=1∞ of 𝒮′(ℝn) such that {∂βfj}j=1∞ is convergent for each β with |β|=d+1. Then there exists a sequence of polynomials {Pj}j=1∞⊂𝒫d such that {fj+Pj}j=1∞ is convergent in 𝒮′(ℝn).The next proposition reveals the fundamental structure ofB˙σ(LipθD)(ℝn).Proposition 32.
Assume that the parametersσ,θ∈ℝ and d∈ℤ satisfy
(181)σ≥0,θ>0,d>[σ+θ].
Let f∈B˙σ(LipθD)(ℝn). (1)There exists a sequence of polynomials{Pj}j=1∞ of degree at most d such that
(182)limj→∞(Pj+∑k=-j∞φk(D)f)converges in𝒮′(ℝn).(2) If we denote byg the limit (182), then
(183)∥g∥B˙σ(LipθD)=∥f∥B˙σ(LipθD)<∞.(3) The limitg is determined modulo a difference of polynomials of order d no matter how we choose {Pj}j=1∞.(4) The distributionf-g is a polynomial.(5) The distributionf is a continuous function.Proof.
(1) Letj≤0 and x∈ℝn. Choose m∈ℤ so that 2m≤|x|+2-j≤2m+1. Then we have x∈Q⊂Q4|x|+4r for some cube Q with ℓ(Q)=2m+1. By Lemma 30, we have
(184)2-j|β|(|x|+2-j)σ2-jθ|∂βφj(D)f(x)|≲∥f∥B˙σ(LipθD)(x∈ℝn).Consequently, then we have(185)|∂βφj(D)f(x)|≲2j(|β|-θ-σ)(1+|x|)σ∥f∥B˙σ(LipθD).The estimate (185) shows that ∑j=-∞1∂βφj(D)f(x) converges in the topology of 𝒮′(ℝn), provided that |β|=d+1. So we are in the position of resorting to Proposition 31.(2) This is clear sinceφj(D)f=φj(D)g.(3) Suppose that{P~j}j=1∞ is a sequence of polynomials of degree d such that
(186)h=limj→∞(P~j+∑k=-j∞φk(D)f)converges in the topology of𝒮′(ℝn). Then we claim ∂βh=∂βg for all β with length d+1. Indeed,
(187)∂βh=limj→∞∂β(P~j+∑k=-j∞φk(D)f)=limj→∞∂β(∑k=-j∞φk(D)f)=limj→∞∂β(Pj+∑k=-j∞φk(D)f)=∂βg.Consequently,g and h coincide modulo a polynomial of order d.(4) Just observe thatf-g has frequency support in the origin.(5) In view of (4), we have only to show thatg itself is a continuous function. As we can see from (185), if |β|=d+1, then the limit
(188)gβ(x)=∑j=-∞0∂βφj(D)f(x)defines aC∞(ℝn) function. If β,β′,γ,γ′ are multi-indices such that |β|=|γ|=d+1 and that β+β′=γ+γ′, then
(189)gβ+β′(x)=gγ+γ′(x).Consequently, there exists aC∞(ℝn) function G of polynomial growth such that ∂βG=gβ.Observe also that whenj≥0,
(190)|φj(D)f(x)|≲(|x|+2-j)σ2-jθ∥f∥B˙σ(LipθD)≤(|x|+1)σ2-jθ∥f∥B˙σ(LipθD).Consequently, sinceθ>0, by virtue of the Weierstrass test, we see the series
(191)1(1+|x|)σ∑j=0∞φj(D)f(x)converges uniformly to a continuous functionJ(x). Thus, g(x):=G(x)+(1+|x|)σJ(x) is a desired function.Here and below we assume thatf itself satisfies
(192)f=limj→∞(Pj+∑k=-j∞φk(D)f)
and that f is a continuous function.Lemma 33.
Letf∈B˙σ(LipθD)(ℝn) satisfy (192). Then we have
(193)∑m′=0d+1(-1)m′(d+1m′)f(x-m′y)=∑j=-∞∞∑m′=0d+1(-1)m′(d+1m′)φj(D)f(x-m′y)
in the topology of 𝒮′(ℝn).Remark that the left-hand side equals(-1)d+1Δ-yd+1f(x).Proof.
By (192), we have
(194)∑m′=0d+1(-1)m′(d+1m′)f(x-m′y)=limJ→∞∑j=-J∞∑m′=0d+1(-1)m′(d+1m′)×(Pj(x-m′y)+φj(D)f(x-m′y))=limJ→∞∑j=-J∞∑m′=0d+1(-1)m′(d+1m′)φj(D)f(x-m′y)=∑j=-∞∞∑m′=0d+1(-1)m′(d+1m′)φj(D)f(x-m′y).
The proof is therefore completed.Proof of Theorem13.
Chooseτ∈𝒮(ℝn) such that χQ1/2≤τ≤χQ1. Let us define
(195)Tj(x)∶=∑m′=1d+1∑m=1d+1(-1)d+m+m′(d+1)!(d+1m′)(d+1m)×md+1(2-jmm′)n[ℱ-1τ](x2-jmm′).
Then ℱTj equals a nonzero constant near a neighborhood of the origin and has compact support. Let us set φ≡ℱT1-ℱT0. From the above observation, φ can be used for the definition of B˙σ(LipθD)(ℝn).
Suppose thatf is a continuous function such that the quantity ∥f∥B˙σ(Lipθ) is finite. Let Q be a cube such that Q⊂Qr and fix x∈Q. Then we have
(196)Tj*f(x)=∑m′=1d+1∑m=1d+1(-1)d+m+m′(d+1)!(d+1m′)(d+1m)×md+1(2-jmm′)n[ℱ-1τ](·2-jmm′)*f(x)
and hence
(197)φj(D)f(x)=∑m=0d+1(d+1m)md+1(-1)d+m(d+1)!×∫ℝnℱ-1τ(y)∑m′=0d+1(-1)m′(d+1m′)×(f(x-2-j-1mm′y)-f(x-2-jmm′y))dy.
Observe that
(198)x∈Q(x,2-j|y|)⊂Q|x|+2-j|y|,2-jy∈Q2-j|y|,
and hence, for m=0,1,2,…,d+1,
(199)|∑m′=0d+1(-1)m′(d+1m′)f(x-2-jmm′y)|=|Δ2-jmyd+1f(x)|≤(|x|+2-j|y|)σ(2-j|y|)θ∥f∥B˙σ(Lipθ(d+1))≤(|x|+|x|·|y|+2-j|y|+2-j)σ×(2-j|y|)θ∥f∥B˙σ(Lipθ(d+1))=(|x|+2-j)σ(1+|y|)σ(2-j|y|)θ∥f∥B˙σ(Lipθ(d+1)).
From this estimate and the fact that θ>0 and σ≥0, we obtain that
(200)|φj(D)f(x)|≲(|x|+2-j)σ(2-j)θ∥f∥B˙σ(Lipθ(d+1)).
The estimate (200) shows that f∈B˙σ(LipθD)(ℝn).
Conversely, suppose thatf∈B˙σ(LipθD)(ℝn). We need to prove that f∈B˙σ(Lipθ(d+1))(ℝn), that is, we need to find a constant C>0 such that
(201)1rσℓ(Q)θ|Δ-yd+1f(x)|≤C,
for any r>0, any cube Q and any pair of points x,y satisfying x∈Q∈𝒬(Qr) and |y|≤ℓ(Q). Then, from f∈B˙σ(LipθD)(ℝn), we have
(202)∑m′=0d+1(-1)m′(d+1m′)f(x-m′y)=∑j=-∞∞∑m′=0d+1(-1)m′(d+1m′)φj(D)f(x-m′y)
by Lemma 33. Then by the mean value theorem, we have
(203)|Δ-yd+1f(x)|=|∑m′=0d+1(-1)m′(d+1m′)φj(D)f(x-m′y)|≲|y|d+1sup|z|≤(d+1)|y||∇d+1φj(D)f(x+z)|
and, by using the triangle inequality applied directly to the summation defining Δ-yd+1f(x), we have
(204)|Δ-yd+1f(x)|≲sup|z|≤(d+1)|y||φj(D)f(x+z)|.
Consequently, we have
(205)|Δ-yd+1f(x)|≲∑j=-∞∞min(|y|d+1sup|z|≤(d+1)|y||∇d+1φj(D)f(x+z)|,sup|z|≤(d+1)|y||φj(D)f(x+z)|).
It follows directly from the definition of the norm that
(206)sup|z|≤(d+1)|y||φj(D)f(x+z)|≲2-jθ(2-j+|x|+|y|)σ∥f∥B˙σ(LipθD)~(2-jθ(|x|+|y|)σ+2-j(θ+σ))∥f∥B˙σ(LipθD)
and from Lemma 30 and (181) that
(207)|y|d+1sup|z|≤(d+1)|y||∇d+1φj(D)f(x+z)|≲(2j(d+1-θ)(|x|+|y|)σ|y|d+1+2j(d+1-θ-σ)|y|d+1)×∥f∥B˙σ(LipθD).
Consequently, by Lemma 28, we have
(208)|Δ-yd+1f(x)|≲∑j=-∞∞(2-jθ(|x|+|y|)σ+2-j(θ+σ))×min(1,2j(d+1)|y|d+1)∥f∥B˙σ(LipθD)≲|y|θ(|x|+|y|)σ∥f∥B˙σ(LipθD).
Now that |y|≤ℓ(Q) and x∈Q⊂Qr, we deduce that
(209)|Δ-yd+1f(x)|≤ℓ(Q)θrσ∥f∥B˙σ(LipθD).
This concludes the proof of Theorem 13.Going through an argument similar to Theorem6, we obtain the following.Theorem 34.
Letα>0, σ≥0 and θ>0. Then the fractional integral operator Iα is bounded from B˙σ(LipθD)(ℝn) to B˙σ(Lipα+θD)(ℝn).As a corollary of Theorems13 and 34, we have the following conclusion.Corollary 35.
Letα>0, σ≥0 and θ>0. Assume in addition that the integer d satisfies d>[σ+θ+α]. Then the fractional integral operator Iα is bounded from B˙σ(Lipθ(d+1))(ℝn) to B˙σ(Lipα+θ(d+1))(ℝn).


## 7. Application of Theorem4  to the Fourier Multipliers

As a corollary of Theorem4, the boundedness of singular integral operators follows.Corollary 36.
Letm∈C∞(ℝn∖{0}) be such that
(210)pβ(m):=supξ∈ℝn∖{0}|ξ||β||∂βm(ξ)|<∞
for all multi-indices β. Then the mapping f∈L2(ℝn)↦ℱ-1[m·ℱf]∈L2(ℝn) extends naturally to a B˙σ(Lp,λ)(ℝn)-bounded operator.Proof.
To check this, by virtue of Theorem4, we need only to verify
(211)1rσ|Q|(λ/n)+(1/p)∥(∑j=JQ∞|φj(D)[ℱ-1[m·ℱf]]|2)1/2∥Lp(Q)≲∥f∥B˙σ(Lp,λD)
for all r>0 and Q∈𝒟(Qr), where φ∈Ccomp∞(ℝn∖{0}) satisfies (17). As usual we take J- and J+ so that it satisfies (138). Write
(212)mj≡∑l=J-J+φj+lm(j∈ℤ).
Then it is easy to verify that
(213)pβ(mj)=supξ∈ℝn∖{0}|ξ||β||∂βmj(ξ)|≲1
and that φ(2-j·)mj=φ(2-j·)m. Consequently, we have
(214)|ℱ-1mj(x)|≲2jn(1+|2jx|)-n-1(x∈ℝn).
Recall that we defined φj(D)[ℱ-1[m·ℱf]]≡ℱ-1[φj·m·ℱf]. Hence, a direct calculation yields
(215)φj(D)[ℱ-1[m·ℱf]]=(2π)n/2ℱ-1mj*ℱ-1[φj·ℱf]=(2π)n/2ℱ-1mj*φj(D)f.
If we insert (214) to (215), then we obtain that
(216)|φj(D)[ℱ-1[m·ℱf]](x)|≲M[φj(D)f](x).
If we combine Proposition 5 with (216), then we have
(217)∥(∑j=JQ∞|φj(D)[ℱ-1[m·ℱf]]|2)1/2∥Lp(Q)≲∥(∑j=JQ∞M[φj(D)f]2)1/2∥Lp(Q)≲rσ|Q|(λ/n)+(1/p)∥f∥B˙σ(Lp,λD).
Thus, (211) is established.



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*Source: 859402-2013-03-13.xml*