# 𝒬

Authors: Guanlong Bao; Hasi Wulan
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

We introduce a new space,𝒬K(ℝn) space, of several real variables with nondecreasing functions K. By giving basic properties of the weighted function K, by establishing a Stegenga-type estimate, and by introducing the K-Carleson measure on ℝ+n+1, we obtain various characterizations of 𝒬K(ℝn) space.


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## Body

## 1. Introduction

Recall that a locally integrable functionf belongs to BMO(ℝn) if
(1)∥f∥BMO(ℝn)=sup⁡I⊂ℝn|I|-1∫I‍|f(x)-fI|dx<∞,
where I denotes a cube in ℝn with edges parallel to the coordinate axes and |I| denotes the Lebesgue measure of I and
(2)fI=|I|-1∫I‍f(x)dx.
Via the John-Nirenberg inequality [1], one can show an equivalent condition of BMO(ℝn) as follows:
(3)∥f∥BMO(ℝn)2≈sup⁡I|I|-2∬I|f(x)-f(y)|2dxdy.C. Fefferman’s famous equation,(H1)*=BMO, describes a deep relation between BMO and the Hardy space (cf. [2, 3]). This leads quite naturally to increased study of these functions from the point of real variable theory and complex function theory views in the recent fifty years. See [2–9] for more results about BMO(ℝn) space.As a generalization ofBMO(ℝn), the space 𝒬α(ℝn), α∈ℝ, introduced by Essén et al. in [10], is defined to be the class of all locally integrable functions f∈Lloc⁡2(ℝn) such that
(4)∥f∥𝒬α(ℝn)2=sup⁡I[ℓ(I)]2α-n∬I|f(x)-f(y)|2|x-y|n+2αdxdy<∞,
where ℓ(I)=|I|1/n denotes the edge length of the cube I.It is easy to see that𝒬α(ℝn) is always a subclass of BMO(ℝn) and 𝒬α(ℝn)=BMO(ℝn) by choosing α=-n/2. Moreover, we know by [10] that 𝒬α(ℝn)=BMO(ℝn) if and only if α<0. Also, we see that 𝒬α(ℝ) is trivial (containing a.e. constant functions only) if and only if α>1/2 and 𝒬α(ℝn), n≥2, is trivial if and only if α≥1.In this paper, we introduce and develop a more general space𝒬K(ℝn) of several real variables, which can be viewed as an extension and improvement of 𝒬α(ℝn) spaces as well as BMO(ℝn). A theory of 𝒬K(𝔻) spaces on unit disc 𝔻 has been extensively studied for recent years in the context of a wide class of function spaces; see, for example, [11–15]. Motivated by the theory of analytic 𝒬K(𝔻) spaces, we define the following.Definition 1.
LetK:[0,∞)→[0,∞) be a nondecreasing function. A function f∈Lloc⁡2(ℝn) is said to belong to the space 𝒬K(ℝn) if
(5)∥f∥𝒬K(ℝn)2=sup⁡I⊂ℝn∬I|f(x)-f(y)|2|x-y|2nK(|x-y|ℓ(I))dxdy<∞.
If we takeK(t)=tn-2α, for α∈ℝ, then 𝒬K(ℝn)=𝒬α(ℝn). Modulo constants, 𝒬K(ℝn) is a Banach space under the norm defined in (5).
Our paper is organized as follows.
In Section2, we investigate the relationship between 𝒬K(ℝn) and BMO(ℝn) and give a sufficient and necessary condition for space 𝒬K(ℝn) which is nontrivial.
In Section3, we give several results about the weight function K on which 𝒬K(ℝn) obviously depends. In the study of 𝒬K(ℝn), the auxiliary function φK defined by
(6)φK(s)=sup⁡0<t≤1K(st)K(t),0<s<∞,
still works well as the analytic 𝒬K(𝔻) spaces.
In Section4, we define the K-Carleson measure on ℝ+n+1. By establishing a Stegenga-type estimate, we obtain a characterization of 𝒬K(ℝn) spaces in terms of the K-Carleson measure.
Throughout this note,a≲b means that there is a positive constant C such that a≤Cb. Moreover, if both a≲b and b≲a hold, then one says that a≈b. For the convenience of calculation, in this paper, we always assume that K:[0,∞)→[0,∞) is nondecreasing and K(2t)≈K(t).


## 2. Basic Properties of𝒬K(ℝn)

Our first observation is that𝒬K(ℝn) is invariant under the conformal mappings and rotations; that is, for any conformal map ϕ(x)=λx+x0, λ≠0 and x0∈ℝn, or any rotation ψ(x)=xM for an orthogonal matrix M of order n,
(7)∥f∘ϕ∥𝒬K(ℝn)=∥f∥𝒬K(ℝn),∥f∘ψ∥𝒬K(ℝn)≈∥f∥𝒬K(ℝn)
hold for any f∈𝒬K(ℝn).We say that the space𝒬K(ℝn) is trivial if 𝒬K(ℝn) contains only a.e. constant functions. To discuss this problem, we recall the space CIS(ℝn), the class of all functions f∈C1(ℝn), with
(8)∥f∥CIS2=sup⁡I|I|((2-n)/n)∫I‍|∇f(x)|2dx<∞,
where ∇f(x)=(∂f(x)/∂x1,…,∂f(x)/∂xn). By [16], we know that
(9)(1+|x|2)-1∈CIS(ℝ2),ln⁡(1+|x|2)∈CIS(ℝn),n>2.
Thus CIS(ℝn) is not trivial for n≥2. However, CIS(ℝ) is trivial.For any cubeI, if x,y∈I, then |x-y|≤nℓ(I). For t>0, tI means that the cube has the same center as I and the edge length tℓ(I). If x∈I and |y|<ℓ(I), then x+y∈3I. By the change of variable, f∈𝒬K(ℝn) if and only if
(10)sup⁡I∫|y|<ℓ(I)‍∫I‍|f(x+y)-f(x)|2×K(|y|ℓ(I))|y|-2ndxdy<∞.Theorem 2.
The following statements are true.(a)
𝒬1/2(ℝ)⊆𝒬K(ℝ). Moreover, 𝒬K(ℝ) is never trivial.(b)
Forn≥2, 𝒬K(ℝn) is not trivial if and only if
(11)∫0n‍K(t)tn-1dt<+∞.Moreover, if (11) holds, then
(12)CIS(ℝn)⊂𝒬K(ℝn).Proof.
(a) For any cubeI of ℝ and x,y∈I, we have
(13)|x-y|≤ℓ(I).
By assumption on K we have
(14)∬I|f(x)-f(y)|2|x-y|2K(|x-y|ℓ(I))dxdy≤K(1)∬I|f(x)-f(y)|2|x-y|2dxdy.
Hence, 𝒬1/2(ℝ)⊆𝒬K(ℝ). 𝒬1/2(ℝ) is not trivial and so is 𝒬K(ℝ).
(b)Necessity. It is enough to show that if
(15)∫0n‍K(t)tn-1dt=+∞,
then 𝒬K(ℝn) is trivial. We will prove the necessity by two steps.
Step
1. If f∈𝒬K(ℝn)∩C1(ℝn) and f is nonconstant, we may assume that f is real. Then there exists a point x0=(x10,x20,…,xn0) such that ∇f(x0)≠0. By the Householder reflector [17, p. 71], there exists an orthogonal matrix M=(aij),  i,j=1,2,…,n, such that
(16)∇f(x0)M=(|∇f(x0)|,0,…,0),
which gives
(17)∑i=1n‍∂f∂xi(x0)ai1=|∇f(x0)|,∑i=1n‍∂f∂xi(x0)aij=0,j=2,3,…,n.
Denote by MT the transpose of the matrix M. Set g(x)=f(xMT). Since det⁡(MT)≠0, there exists a point y0=(y10,y20,…,yn0) such that y0MT=x0. Write y=xMT for convenience as follows:
(18)yj=∑i=1n‍xiaji,j=1,2,…,n.
Consequently,
(19)∂g∂x1(y0)=∑j=1n‍∂f∂yj(y0MT)aj1=|∇f(x0)|.
Similarly,
(20)∂g∂xi(y0)=0,i=2,3,…,n.
Thus
(21)∇g(y0)=(|∇f(x0)|,0,…,0).
Note that g∈C1(ℝn). Then there exist a positive constant δ and a small cube I centered at y0 on which ∂g(x)/∂x1>2δ and ∂g(x)/∂xj<δ,  j≥2. Define
(22)D={ℓ(I)4x=(x1,…,xn)∈ℝn:|x2|+⋯+|xn|<x1<ℓ(I)4}.
If x,y∈I and x-y∈D, using the mean value theorem, we get
(23)g(x)-g(y)>δ(x1-y1).
Thus
(24)∬I|g(x)-g(y)|2|x-y|2nK(|x-y|ℓ(I))dxdy≥∫I/2‍dx∫I-x‍|g(x+z)-g(x)|2|z|2nK(|z|ℓ(I))dz≥∫I/2‍dx∫D‍δ2|z1|2|z|2nK(|z|ℓ(I))dz.
If z∈D, then |z|≈z1. Hence
(25)∬I|g(x)-g(y)|2|x-y|2nK(|x-y|ℓ(I))dxdy≳∫I/2‍dx∫D‍|z1|2|z1|2nK(|z1|ℓ(I))dz≈∫0ℓ(I)/4‍1z12n-2K(z1ℓ(I))dz1×∫{(z2,…,zn):|z2|+⋯+|zn|<z1}‍dz2⋯dzn≈∫0n‍K(t)tn-1dt=+∞.
It means that g is not an element of 𝒬K(ℝn). On the other hand, g∈𝒬K(ℝn) since 𝒬K(ℝn) is invariant under rotations. This is a contraction.
Step
2. Note that 𝒬K(ℝn) is conformal invariant. By Minkowski’s inequality, if f∈𝒬K(ℝn) and g∈L1(ℝn), then f*g∈𝒬K(ℝn) and
(26)∥f*g∥𝒬K(ℝn)≤∥f∥𝒬K(ℝn)∫ℝn‍|g(y)|dy,
where
(27)(f*g)(x)=∫ℝn‍f(x-y)g(y)dy.
In particular, if g∈C0∞, the class of smooth functions with compact support, then f*g∈𝒬K(ℝn)∩C1(ℝn). Thus f*g is constant by Step  1. By [10], there exits a sequence {gn}⊂C0∞ with gn≥0, ∫‍gn=1, and supp⁡gn shrinking to 0 such that f*gn→f a.e. It follows that f is constant a.e. Thus, we complete the proof of necessity.
Sufficiency. Give a cube I and suppose f∈CIS(ℝn). By
(28)|f(x+y)-f(y)|≤∫01‍|∇f(y+tx)||x|dt,
we have
(29)∬I|f(x)-f(y)|2|x-y|2nK(|x-y|ℓ(I))dxdy≤∫I‍∫|z|<nℓ(I)‍(∫01‍|∇f(x+tz)|dt)2×1|z|2n-2K(|z|ℓ(I))dzdx.
It follows by Minkowski’s inequality that
(30)(∬I|f(x)-f(y)|2|x-y|2nK(|x-y|ℓ(I))dxdy)1/2≤∫01‍(∫I‍∫|z|<nℓ(I)‍|∇f(x+tz)|21|z|2n-2×K(|z|ℓ(I))dzdx)1/2dt≤(∫|z|<nℓ(I)‍1|z|2n-2K(|z|ℓ(I))dz×∫3nI‍|∇f(w)|2dw)1/2≲∥f∥CIS×(ℓ(I)n-2∫0nℓ(I)‍1tn-1K(tℓ(I))dt)1/2≈∥f∥CIS(∫0n‍K(t)tn-1dt)1/2.
Thus, CIS(ℝn)⊆𝒬K(ℝn) and 𝒬K(ℝn) is not trivial.Theorem 3.
The space𝒬K(ℝn) is a subset of BMO(ℝn). Furthermore, (a)
if∫0n‍(K(t)/tn+1)dt<∞, then 𝒬K(ℝn)=BMO(ℝn);(b)
if𝒬K(ℝn)=BMO(ℝn), then, for all β<1, ∫0n‍(K(t)/tn+β)dt<∞.Proof.
Letf∈𝒬K(ℝn). For any cube I and x,y∈I, if r>0 is small enough, we know that the Lebesgue measure of the set
(31){z∈I:min⁡(|x-z|,|y-z|)>rℓ(I)}
is bigger than |I|(1-(4rnπn/2/nΓ(n/2))). Since K is nondecreasing,
(32)∫I‍min⁡(K(|x-z|ℓ(I)),K(|y-z|ℓ(I)))dz≥∫{z∈I:min⁡(|x-z|,|y-z|)>rℓ(I)}min⁡(K(|x-z|ℓ(I)),‍K(|y-z|ℓ(I)))dz≥K(r)|I|(1-4rnπn/2nΓ(n/2)).
Consequently,
(33)K(r)(1-4rnπn/2nΓ(n/2))|I|-2∬I|f(x)-f(y)|2dxdy≤|I|-3∭I|f(x)-f(y)|2×min⁡(K(|x-z|ℓ(I)),K(|y-z|ℓ(I)))dxdydz≤2|I|-3∭I|f(x)-f(z)|2K(|x-z|ℓ(I))dxdydz+2|I|-3∭I|f(y)-f(z)|2K(|y-z|ℓ(I))dxdydz≤4nn∬I|f(y)-f(z)|2|y-z|2nK(|y-z|ℓ(I))dydz.
For a small enough r>0, we obtain
(34)∥f∥BMO(ℝn)2≲sup⁡I∬I|f(y)-f(z)|2|y-z|2nK(|y-z|ℓ(I))dydz.
Thus 𝒬K(ℝn)⊆BMO(ℝn).
(a) Note that(35)∥f∥BMO(ℝn)2≈sup⁡I|I|-2∬I|f(x)-f(y)|2dxdy≈sup⁡I|I|-1∫I‍|f(x)-fI|2dx.
For a cube I and for every y∈ℝn with |y|<nl(I),
(36)∫I‍|f(x+y)-f(x)|2dx≲∫I‍|f(x+y)-f3nI|2+|f(x)-f3nI|2dx≲|I|∥f∥BMO(ℝn)2.
Therefore,
(37)∬I|f(x)-f(y)|2|x-y|2nK(|x-y|ℓ(I))dxdy≤∫|y|<nℓ(I)‍∫I‍|f(x+y)-f(x)|2×K(|y|/ℓ(I))|y|2ndxdy≲|I|∥f∥BMO(ℝn)2∫|y|<nℓ(I)‍K(|y|/ℓ(I))|y|2ndy≈∥f∥BMO(ℝn)2∫0n‍K(t)tn+1dt.
Thus BMO(ℝn)⊆𝒬K(ℝn), and this deduces that 𝒬K(ℝn)=BMO(ℝn).
(b) Consider the Schwartz space𝒮(ℝn) consisting of all those C∞ functions φ on ℝn such that
(38)sup⁡x∈ℝn|xα(Dβφ)(x)|<∞
for all multi-indices α=(α1,…,αn) and β=(β1,…,βn) of nonnegative integers, where xα=x1α1⋯xnαn and Dβ=(∂/∂x1)β1⋯(∂/∂xn)βn.
Letφ∈𝒮(ℝn) be a fixed function such that the Fourier transform φ^ of φ has support in the unit ball and φ≠0 on the cube [-3π,3π]n. Let {ak} be a sequence of real numbers and define
(39)g(x)=∑k=1∞‍akexp⁡(2kx1i),
where x1 is the first coordinate of x. By [10], f=φg∈BMO(ℝn) if and only if ∑k=1∞‍ak2<∞.
Suppose that𝒬K(ℝn)=BMO(ℝn). Set ak2=(1/2k(1-β)) for β<1. We know that f∈𝒬K(ℝn). Choosing I=[-π,π]n, we have
(40)∫|y|<2π‍∫I‍|f(x+y)-f(x)|2K(|y|)1|y|2ndxdy<∞.
Since |φ(x+y)|≥c>0 for x∈I and |y|<2π,
(41)|g(x+y)-g(x)|2≲|f(x+y)-f(x)|2+|g(x)|2|φ(x+y)-φ(x)|2.
Since φ∈𝒮(ℝn), we have |φ(x+y)-φ(x)|≲|y|. Hence,
(42)∫|y|<2π‍∫I‍|g(x)|2|φ(x+y)-φ(x)|2K(|y|)1|y|2ndxdy≲∫|y|<2π‍K(|y|)1|y|2n-2dy×∫[-π,π]n‍|∑k=1∞‍akexp⁡(2kx1i)|2dx≈∑k=1∞‍ak2∫0n‍K(t)tn-1dt<∞.
Writing y=(y1,y′), y′∈ℝn-1, we obtain
(43)∞>∫|y|<2π‍∫I‍|g(x+y)-g(x)|2K(|y|)1|y|2ndxdy≈∫|y|<2π‍∑k=1∞‍|ak(e2ky1i-1)|2K(|y|)1|y|2ndy≳∑k=1∞‍ak2∫|y′|<y1<2-k‍|2ky1|2K(|y|)1|y|2ndy≈∑k=1∞‍22kak2∫02-k‍K(t)tn-1dt≳∫01/2‍K(t)tn+βdt.For any cubeI of ℝn, when x,y∈I, we have that |x-y|≤nℓ(I). By the definition, the 𝒬K(ℝn) space depends on K(t) when 0≤t≤n. In fact, 𝒬K(ℝn) depends only on K(t) when t is near origin, which can be found by the following theorem. Here the proof of the theorem is left to the reader.Theorem 4.
The following statements are true.(a)
Suppose thatK(r)>0 for some r>0. One defines Kr(t)=min⁡(K(r),K(t)). Then 𝒬Kr(ℝn)=𝒬K(ℝn).(b)
Let(44)K1(t)={K(t),0<t≤1,K(1)tn-1,t≥1.Then 𝒬K1(ℝn)=𝒬K(ℝn).Remark 5.
IfK(0)>0, we have 𝒬K(ℝn)=𝒬n/2(ℝn). Since 𝒬n/2(ℝn) is trivial for n≥2, we only pay attention to the case K(0)=0.


## 3. Weighted Functions

The characterization of𝒬K(ℝn) depends on the properties of the weight function K obviously. In this section we give several results about the weight functions that are needed for the next section.In the analytic𝒬K(𝔻) spaces, the auxiliary function
(45)φK(s)=sup⁡0<t≤1K(st)K(t),0<s<∞,
plays a key role; see [12, 14, 15], for example.To study𝒬K(ℝn) spaces, we need some more constraints on K as follows:
(46)∫01‍φK(s)dsmin⁡(s,sn-1)<∞,(47)∫1∞‍φK(s)dssn+1<∞.
Note that (46) implies the following two conditions:
(48)∫01‍φK(s)dss<∞,(49)∫01‍φK(s)dssn-1<∞,n≥2.In particular, if we chooseK(t)=tn-2α, then condition (47) holds if and only if α>0; condition (48) holds if and only if n>2α; and condition (49) holds if and only if α<1.Lemma 6.
LetK satisfy
(50)sup⁡0<s<b∫sb‍K(t)tn+1dt∫0s‍tn-1K(t)dt<∞
for some 0<b≤∞. Then one can find another nonnegative weight function K* with K*(0)=0 such that 𝒬K*(ℝn)=𝒬K(ℝn), and the new weight function K* has the following properties.(a)
K* is nondecreasing on [0,∞).(b)
K*≈K on [0,∞), and thus K* satisfies condition (50).(c)
K*(2t)≈K*(t) on [0,∞).(d)
K* is differentiable (up to any given order) on (0,∞).(e)
For some small enoughc>0, K*(t)/tn-c is nonincreasing on (0,∞). Consequently, K*(t)/tn is also nonincreasing on (0,∞).Proof.
By Theorem4, we may assume that K(t)=K(1)tn-1 for t≥1. Since K satisfies condition (50), we claim that
(51)sup⁡0<s<∞∫s∞‍K(t)tn+1dt∫0s‍tn-1K(t)dt<∞.
If b=∞, the claim is true. If 0<b<∞, the claim will be confirmed by showing
(52)sup⁡0<s<b∫s∞‍K(t)tn+1dt∫0s‍tn-1K(t)dt<∞,sup⁡b≤s<∞∫s∞‍K(t)tn+1dt∫0s‍tn-1K(t)dt<∞.
For the case of 0<s<b, by (50),
(53)∫s∞‍K(t)tn+1dt∫0s‍tn-1K(t)dt=(∫sb‍K(t)tn+1dt+∫b∞‍K(t)tn+1dt)∫0s‍tn-1K(t)dt≤C+∫b∞‍K(t)tn+1dt∫0b‍tn-1K(t)dt.
Taking s=(b/2) in (50), we have
(54)∫0b‍tn-1K(t)dt<∞.
Since K(t)=K(1)tn-1, for t≥1, we have
(55)∫b∞‍K(t)tn+1dt∫0b‍tn-1K(t)dt<∞.
Thus
(56)sup⁡0<s<b∫s∞‍K(t)tn+1dt∫0s‍tn-1K(t)dt<∞.
Now, we prove
(57)sup⁡b≤s<∞∫s∞‍K(t)tn+1dt∫0s‍tn-1K(t)dt<∞.
In fact, if b≤s<∞, b≥1,
(58)∫s∞‍K(t)tn+1dt∫0s‍tn-1K(t)dt≤K(1)∫0b‍tn-1K(t)dt+K(1)s-1∫bs‍1K(1)dt≤K(1)∫0b‍tn-1K(t)dt+1.
If b≤s<∞, 0<b<1 and s<1,
(59)∫s∞‍K(t)tn+1dt∫0s‍tn-1K(t)dt≤∫b∞‍K(t)tn+1dt(∫0b‍tn-1K(t)dt+∫b1‍tn-1K(t)dt)<∞.
If b≤s<∞, 0<b<1 and s≥1,
(60)∫s∞‍K(t)tn+1dt∫0s‍tn-1K(t)dt≤∫b∞‍K(t)tn+1dt∫01‍tn-1K(t)dt+∫s∞‍K(t)tn+1dt∫1s‍tn-1K(t)dt≤∫b∞‍K(t)tn+1dt∫01‍tn-1K(t)dt+1.
Hence,
(61)sup⁡b≤s<∞∫s∞‍K(t)tn+1dt∫0s‍tn-1K(t)dt<∞.
Therefore, our claim above has been confirmed. Let
(62)K*(t)={tn∫t∞‍K(s)sn+1ds,0<t<∞,0,t=0.
Then K*(t)=K(1)tn-1 for t≥1.
(a) Fix0<t1<t2<∞ and consider the difference
(63)K*(t2)-K*(t1)=(t2n-t1n)×∫t2∞‍K(s)sn+1ds-t1n∫t1t2‍K(s)sn+1ds.
Since K is nondecreasing and nonnegative, we have
(64)K*(t2)-K*(t1)≥(t2n-t1n)K(t2)×∫t2∞‍dssn+1-t1nK(t2)∫t1t2‍dssn+1=0.
(b) Using the assumption thatK is nondecreasing again, we obtain
(65)K*(t)≥tnK(t)∫t∞‍dssn+1=K(t)n
for 0<t<∞. On the other hand,
(66)K*(t)≲tn(∫0t‍sn-1K(s)ds)-1≲K(t),0<t<∞.
Thus, K*≈K on (0,∞) and 𝒬K*(ℝn)=𝒬K(ℝn).
(c) For anyt>0, we have
(67)K*(2t)K*(t)=2n∫2t∞(K(s)/sn+1)‍ds∫t∞‍(K(s)/sn+1)ds≤2n.
Since K* is nondecreasing, K*(t)≤K*(2t).
(d) If we repeat the constructionK↦K*, then we can make the new weight function differentiable up to any desired order.
(e) Note that ifc>0 is sufficiently small, then we have
(68)(tc-nK*(t))′=tc-n-1(cK*(t)-K(t))<0,0<t<∞.
This means that K*(t)/tn-c is nonincreasing. The proof is complete.The following result shows that there is no essential difference between (47) and (50).Lemma 7.
The following are equivalent.(a)
Equation (50) holds for K.(b)
There exists a weightK1, comparable with K, such that, for some small enough c>0, K1(t)/tn-c is nonincreasing on (0,∞).(c)
Equation (47) holds for K.Proof.
We assume thatK(t)=K(1)tn-1 for t≥1.
(
a
)
⇒
(
b
) is obvious by Lemma 6.
Suppose that (b) holds. We have(69)∫sb‍K1(t)tn+1dt∫0s‍tn-1K1(t)dt≤∫s∞‍t-1-cdt∫0s‍tc-1dt=c-2.
We obtain that (50) holds for K since K is comparable with K1. It means that (b)⇒(a) holds.
For(c)⇒(b), assume that (47) holds for K. We claim that
(70)lim⁡inf⁡t→0K(t)tn>0.
If s>1, it is clear that
(71)K(1)K(1/s)≤φK(s)
and by (47)
(72)∫01‍K(s)-1dss1-n=∫1∞‍K(1s)-1dssn+1<∞.
Thus, we have
(73)tnK(t)≲K(t)-1∫0t‍dss1-n≲∫01‍K(s)-1dss1-n<∞,
which gives the claim. We define
(74)K1(t)=tn∫t∞‍K(s)sn+1ds,0<t<∞.
It is easy to check that K1(t) is nondecreasing. Since K is nondecreasing, it follows that K(t)≲K1(t),  0<t<∞. We note that, for 0<t<1,
(75)∫t1‍K(s)sn+1ds≤K(t)∫t1‍φK(s/t)sn+1ds≤K(t)tn∫1∞‍φK(r)rn+1dr,∫1∞‍K(s)sn+1ds≈K(1)≲K(t)tn.
Thus, we obtain that
(76)K1(t)≲K(t)(∫1∞‍φK(s)sn+1ds+1),0<t<1.
For t∈[1,∞), we have
(77)K1(t)=tn∫t∞‍K(s)sn+1ds=K(1)tn-1=K(t).
Therefore, we get that K1≈K on (0,∞). Note that if c is sufficiently small, then we have
(78)(tc-nK1(t))′=tc-1-n(cK1(t)-K(t))<0,0<t<∞.
This means that K1(t)/tn-c is nonincreasing.
Suppose thatK1(t)/tn-c is nonincreasing on (0,∞). For s≥1,
(79)φK1(s)=sup⁡0<t≤1(st)n-cK1(st)(st)c-nK1(t)≤sup⁡0<t≤1(st)n-cK1(t)tc-nK1(t)=sn-c,
which gives
(80)∫1∞‍φK1(s)dssn+1<∞.
Thus (b)⇒(c) holds.Lemma 8.
LetK satisfy (49). Then one can find another nonnegative weight function K1 such that 𝒬K1(ℝn)=𝒬K(ℝn), and the new weight function K1 has the following properties. (a)
K1 is nondecreasing on (0,∞).(b)
K1≈K on (0,∞), and thus K1 satisfies condition (49).(c)
For some small enoughc>0, K1(t)/tn-2+c is nondecreasing on (0,∞). Consequently, K1(t)/tn-2 is also nondecreasing on (0,∞). Conversely, if K1(t)/tn-2+c is nondecreasing, for some c>0, then (49) holds for K1.(d)
(81)sup⁡0<s<∞∫0s‍K1(t)tn-1dt∫s∞‍tn-3K1(t)dt<∞.Proof.
Assume thatK(t)=K(1)tn-1 for t≥1. Define
(82)K1(t)={tn-2∫0t‍K(s)dssn-1,0<t≤1,tn-1∫01‍K(s)dssn-1,t≥1.
Note that (49) is a condition for n>1. Then, consider the following.
(a) is obvious.
(b) For0<t≤1,
(83)K1(t)=∫01‍K(st)dssn-1≤K(t)∫01‍φK(s)dssn-1.
On the other hand, since we always assume that K(2t)≈K(t), we obtain that
(84)K1(t)≥tn-2K(t2)∫t/2t‍dssn-1≳K(t).
Thus, K1≈K on (0,1). For t≥1, clearly, K1(t)≈K(t). Therefore, K1≈K on (0,∞) and we get that 𝒬K1(ℝn)=𝒬K(ℝn).
(c) If0<t≤1, for some small enough c>0,
(85)(K1(t)tn-2+c)′=t-c-n+1(K(t)-cK1(t))>0.
If t≥1, K1(t)/tn-2+c=K1(1)t1-c is nondecreasing. Thus K1(t)/tn-2+c is nondecreasing on (0,∞).
Conversely, ifK1(t)/tn-2+c is nondecreasing, for some c>0, then, for 0<s≤1,
(86)φK1(s)=sup⁡0<t≤1(st)n+c-2K1(st)(st)2-n-cK1(t)≤sup⁡0<t≤1(st)n+c-2K1(t)t2-n-cK1(t)=sn+c-2,
which gives
(87)∫01‍φK1(s)dssn-1<∞.
(d) Note thatK1(t)/tn-2+c is nondecreasing. For 0<s<∞, we have
(88)∫0s‍K1(t)tn-1dt∫s∞‍tn-3K1(t)dt≤∫0s‍1t1-cdt∫s∞‍1t1+cdt=c-2.
The proof is complete.We end this section by giving an example. Fix0<β<1, and set
(89)Kβ(t)={tn+β-1|log⁡t|,0<t≤1e,e-βtn-1,t>1e.
Since
(90)∫01/e‍Kβ(t)dttn-1<∞,∫01/e‍Kβ(t)dttn+β=∞,
we obtain that 𝒬Kβ(ℝn) is not trivial and
(91)𝒬Kβ(ℝn)⫋BMO(ℝn).
Moreover, a direct calculation shows that (46) and (47) hold for Kβ.


## 4. Carleson-Type Measures

LetI be a cube of ℝn and let ℝ+n+1 denote the upper half space based on ℝn. Define the Carleson box as follows:
(92)S(I)={(x,t)∈ℝ+n+1:x∈I,0<t<ℓ(I)}.
For p>0 and a positive Borel measure μ on ℝ+n+1, μ is said to be a p-Carleson measure if
(93)μ(S(I))≤Mℓ(I)pn
for some M<∞ and all cubes I⊆ℝn.Denote byδ(x) the distance of the point x∈ℝ+n+1 to the boundary ∂ℝ+n+1. Also y~ stands for the symmetric point of y∈ℝ+n+1 with respect to ℝn; that is, if y=(y1,…,yn,yn+1), then y~=(y1,…,yn,-yn+1).A positive Borel measureμ is said to be a K-Carleson measure on ℝ+n+1, as a modification of p-Carleson measure, provided
(94)sup⁡I⊆ℝn∫S(I)‍K(δ(x)ℓ(I))(δ(x))1-ndμ(x)<∞.Clearly, ifK(t)=tnp, then μ is a K-Carleson measure on ℝ+n+1 if and only if (δ(x))np+1-ndμ(x) is a p-Carleson measure on ℝ+n+1. Now, we give a characterization of K-Carleson measure as follows.Theorem 9.
LetK satisfy (48). Let μ be a positive Borel measure on ℝ+n+1. Then μ is a K-Carleson measure if and only if
(95)sup⁡y∈ℝ+n+1∫ℝ+n+1‍K(δ(x)(δ(y))1/n|x-y~|(n+1)/n)(δ(x))1-ndμ(x)<∞.Proof (sufficiency).
LetI be a cube and take y to be the center of the Carleson box S(I). Then δ(y)=ℓ(I)/2. If x∈S(I), then |x-y~|≲ℓ(I) and hence
(96)∫ℝ+n+1‍K(δ(x)(δ(y))1/n|x-y~|(n+1)/n)(δ(x))1-ndμ(x)≥∫S(I)‍K(δ(x)(δ(y))1/n|x-y~|(n+1)/n)(δ(x))1-ndμ(x)≳∫S(I)‍K(δ(x)ℓ(I))(δ(x))1-ndμ(x).
Thus, if (95) holds, then μ is a K-Carleson measure.
Necessity. For y=(y′,yn+1)∈ℝ+n+1, let I⊆ℝn be the cube with center y′ and edge length δ(y). Set Em to be the Carleson box S(2mI) for each positive integer m. It is clear that
(97)|x-y~|≥δ(y),x∈E1,|x-y~|≈2mδ(y),x∈Em+1∖Em.
Then
(98)∫ℝ+n+1‍K(δ(x)(δ(y))1/n|x-y~|(n+1)/n)(δ(x))1-ndμ(x)=∫E1‍K(δ(x)(δ(y))1/n|x-y~|(n+1)/n)(δ(x))1-ndμ(x)+∑m=1∞‍∫Em+1∖Em‍K(δ(x)(δ(y))1/n|x-y~|(n+1)/n)(δ(x))1-ndμ(x)≲∫E1‍K(δ(x)ℓ(I))(δ(x))1-ndμ(x)+∑m=1∞‍∫Em+1‍K(δ(x)2m/n2m+1ℓ(I))(δ(x))1-ndμ(x).
Since μ is a K-Carleson measure,
(99)∫Em‍K(δ(x)2mℓ(I))(δ(x))1-ndμ(x)≲1.
This together with (48) yields
(100)∫ℝ+n+1‍K(δ(x)(δ(y))1/n|x-y~|(n+1)/n)(δ(x))1-ndμ(x)≲∑m=1∞‍φK(12m/n)≈∫01‍φK(s)dss<∞.Letf be a measurable function on ℝn satisfying
(101)∫ℝn‍|f(x)|1+|x|n+1dx<∞.
Its Poisson integral is defined by
(102)f(x,t)=∫ℝn‍Pt(x-y)f(y)dy,
where
(103)Pt(x)=cnt(t2+|x|2)(n+1)/2,cn=Γ((n+1)/2)π(n+1)/2.
The gradient of f(x,t) is
(104)∇f(x,t)=(∂f(x,t)∂x1,…,∂f(x,t)∂xn,∂f(x,t)∂t).
It is known that (101) holds for f∈BMO(ℝn) (see [9]).The following main theorem generalizes the result of𝒬α(ℝn) in [10].Theorem 10.
Suppose that (47) and (81) hold for K. Let f∈Lloc⁡2(ℝn) with (101). Then f∈𝒬K(ℝn) if and only if |∇f(x,t)|2dxdt is a K-Carleson measure.In order to prove Theorem10, we need the following Hardy-type inequalities.Lemma 11 (see [18]).
Let0<b≤∞, 1<p<∞, and p′=(p/(p-1)). Assume that functions μ and ν are measurable and nonnegative in the interval (0,b). Then
(105)∫0b‍(∫0s‍f(t)dt)pμ(s)ds≤C∫0b‍fp(s)ν(s)ds
holds for all measurable functions f≥0 if and only if
(106)A:=sup⁡0<s<b(∫sb‍μ(t)dt)1/p(∫0s‍(ν(t))1-p′dt)1/p′<∞,∫0b‍(∫sb‍f(t)dt)pμ(s)ds≤C∫0b‍fp(s)ν(s)ds
holds for all measurable functions f≥0 if and only if
(107)B:=sup⁡0<s<b(∫0s‍μ(t)dt)1/p(∫sb‍(ν(t))1-p′dt)1/p′<∞.
Here C depends only on p, A, or B.The following Stegenga-type estimate will be used in the proof of Theorem10.Lemma 12.
Suppose that (81) holds for K and
(108)sup⁡0<s<1∫s1‍K(t)t3n+1dt∫0s‍t3n-1K(t)dt<∞.
Let I and J be cubes in ℝn centered at x0 with ℓ(J)=3ℓ(I) and let f∈Lloc⁡1(ℝn) satisfy (101). Then, there is a constant C independent of f, I, and J, such that
(109)∫S(I)‍|∇f(x,t)|2K(tℓ(I))t1-ndxdt≤C∫|y|≤nℓ(J)‍∫J‍|f(x+y)-f(x)|2×K(|y|ℓ(J))1|y|2ndxdy+C(∫0n‍K(s)sn-1ds+∫1/8∞‍K(s)sn+1ds)×|J|-1∫J‍|f(x)-fJ|2dx+C∫0n‍K(s)sn-1ds(ℓ(J)∫ℝn∖(2/3)J‍|f(x)-fJ||x-x0|n+1dx)2.Proof.
Without loss of generality, we may assume thatx0=0. Let φ be a function with 0≤φ≤1 such that φ=1 on (2/3)J, supp⁡φ⊆(3/4)J, and
(110)|φ(x)-φ(y)|≲ℓ(J)-1|x-y|,x,y∈ℝn.
Following Stegenga [19], we write
(111)f=fJ+(f-fJ)φ+(f-fJ)(1-φ)=f1+f2+f3.
Then we have
(112)f(x,t)=f1(x,t)+f2(x,t)+f3(x,t)
for the corresponding Poisson integrals. Since f1 is constant, it contributes nothing to the integral with the gradient square.
Note that(113)∂Ps(y)∂yj=-(n+1)cnsyj(s2+|y|2)-(n+3)/2,j=1,…,n.
We obtain
(114)∫ℝn‍∂Ps(y)∂yjdy=0,j=1,…,n.
Hence,
(115)∂f(x,s)∂xj=∫ℝn‍∂Ps(x-y)∂xjf(y)dy=∫ℝn‍(n+1)cns(yj-xj)×(s2+|y-x|2)-(n+3)/2f(y)dy=-∫ℝn‍∂Ps(y)∂yjf(x+y)dy=∫ℝn‍∂Ps(y)∂yj(f(x)-f(x+y))dy.
A direct calculation shows that
(116)|∂Ps(y)∂yj|≈s|yj|(s2+|y|2)-(n+3)/2≲(s2+|y|2)-(n+1)/2,
which gives
(117)|∂Ps(y)∂yj|≲s-n-1,|∂Ps(y)∂yj|≲|y|-n-1.
Therefore,
(118)∥∂f(x,s)∂xj∥L2(ℝn)≲s-n-1∫|y|≤s‍∥f(x+y)-f(x)∥L2(ℝn)dy+∫|y|>s‍∥f(x+y)-f(x)∥L2(ℝn)|y|-n-1dy.
We write y=rξ∈ℝn with r=|y| and |ξ|=1. Let
(119)A(r)=∫|ξ|=1‍∥f(x+rξ)-f(x)∥L2(ℝn)dξ.
Then
(120)∥∂f(x,s)∂xj∥L2(ℝn)≲s-n-1∫0s‍A(r)rn-1dr+∫s∞‍A(r)r-2dr.
Thus,
(121)∫S(I)‍|∂f(x,s)∂xj|2K(sℓ(I))s1-ndxds≤∫0ℓ(I)‍∥∂f(x,s)∂xj∥L2(ℝn)2K(sℓ(I))s1-nds≲∫0ℓ(I)‍K(sℓ(I))s-3n-1(∫0s‍A(r)rn-1dr)2ds+∫0ℓ(I)‍K(sℓ(I))s1-n(∫s∞‍A(r)r-2dr)2ds≈∫01‍K(s)1s3n+1ℓ(I)n(∫0s‍A(ℓ(I)r)rn-1dr)2ds+∫01‍K(s)1sn-1ℓ(I)n(∫s∞‍A(ℓ(I)r)r-2dr)2ds.
Note that
(122)sup⁡0<s<1∫s1‍K(t)t3n+1dt∫0s‍t3n-1K(t)dt<∞.
By Lemma 11,
(123)∫01‍K(s)1s3n+1(∫0s‍A(ℓ(I)r)rn-1dr)2ds≲∫01‍K(s)1sn+1A2(ℓ(I)s)ds.
Since (81) holds for K, by Lemma 11 again,
(124)∫01‍K(s)1sn-1(∫s∞‍A(ℓ(I)r)r-2dr)2ds≲∫0∞‍K(s)1sn+1A2(ℓ(I)s)ds.
Therefore,
(125)∫S(I)‍|∂f(x,s)∂xj|2K(sℓ(I))s1-ndxds≲∫0∞‍K(sℓ(I))s-1-nA2(s)ds.
By Hölder’s inequality,
(126)∫S(I)‍|∂f(x,s)∂xj|2K(sℓ(I))s1-ndxds≲∫0∞‍∫|ξ|=1‍∥f(x+sξ)-f(x)∥L2(ℝn)2×K(sℓ(I))s-1-ndξds≈∫ℝn‍[∫ℝn‍|f(x+y)-f(x)|2dx]×K(|y|ℓ(I))|y|-2ndy.
Note that
(127)∂Ps(y)∂s=cn[s2+|y|2-s2(n+1)](s2+|y|2)-(n+3)/2.
Hence,
(128)|∂Ps(y)∂s|≲(s2+|y|2)(s2+|y|2)-(n+3)/2.
It follows that
(129)|∂Ps(y)∂s|≲s-n-1,|∂Ps(y)∂s|≲|y|-n-1.
Since
(130)∫ℝn‍Ps(y)dy=1,
this deduces that
(131)∫ℝn‍∂Ps(y)∂sdy=0.
We have
(132)∂f(x,s)∂s=∫ℝn‍∂Ps(y)∂s(f(x+y)-f(x))dy.
Repeating the procedure above, we also can obtain
(133)∫S(I)‍|∂f(x,s)∂s|2K(sℓ(I))s1-ndxds≲∫ℝn‍[∫ℝn‍|f(x+y)-f(x)|2dx]K(|y|ℓ(I))|y|-2ndy.
Therefore,
(134)∫S(I)‍|∇f2(x,s)|2K(sℓ(I))s1-ndxds≲∬ℝn|f2(x+y)-f2(x)|2K(|y|ℓ(I))|y|-2ndxdy≈∬ℝn|f2(x)-f2(y)|2|x-y|2nK(|x-y|ℓ(I))dxdy≈∫‍∫x,y∈J‍⋯+∫‍∫x∉J,y∈3/4J‍⋯+∫‍∫y∉J,x∈3/4J≈B1+B2+B3.
To estimate B1, we note that
(135)|φ(x)-φ(y)|≲ℓ(J)-1|x-y|,x,y∈ℝn.
Thus
(136)|f2(x)-f2(y)|≲|f(x)-f(y)|+(ℓ(J))-1|x-y||f(y)-fJ|.
We have
(137)∬J|f(x)-f(y)|2|x-y|2nK(|x-y|ℓ(I))dxdy≲∫|y|≤nℓ(J)‍K(|y|ℓ(J))|y|-2n×∫J‍|f(x+y)-f(x)|2dxdy,(ℓ(J))-2∬J|f(y)-fJ|2|x-y|2n-2K(|x-y|ℓ(I))dxdy≲(ℓ(J))-2∫|y|≤nℓ(J)‍K(|y|ℓ(J))|y|2-2n×∫J‍|f(x+y)-fJ|2dxdy≲∫|y|≤nℓ(J)‍K(|y|ℓ(J))|y|-2n×∫J‍|f(x+y)-f(x)|2dxdy+(ℓ(J))-2∫|y|≤nℓ(J)‍K(|y|ℓ(J))|y|2-2n×∫J‍|f(x)-fJ|2dxdy≲∫|y|≤nℓ(J)‍K(|y|ℓ(J))|y|-2n×∫J‍|f(x+y)-f(x)|2dxdy+∫0n‍K(s)sn-1ds|J|-1∫J‍|f(x)-fJ|2dx.
Hence,
(138)B1≲∫|y|≤nℓ(J)‍K(|y|ℓ(J))|y|-2n∫J‍|f(x+y)-f(x)|2dxdy+∫0n‍K(s)sn-1ds|J|-1∫J‍|f(x)-fJ|2dx.
To handle B2, note that
(139)|x-y|>18ℓ(J),|f2(x)-f2(y)|≲|f(y)-fJ|,x∉J,y∈3/4J.
We obtain
(140)B2=∫x∉J‍∫y∈3/4J‍|f2(x)-f2(y)|2|x-y|2nK(|x-y|ℓ(I))dxdy≲∫y∈3/4J‍|f(y)-fJ|2dy∫|z|>(1/8)ℓ(J)‍K(|z|ℓ(J))|z|-2ndz≲∫1/8∞‍K(s)sn+1ds|J|-1∫J‍|f(y)-fJ|2dy.
Similarly,
(141)B3≲∫1/8∞‍K(s)sn+1ds|J|-1∫J‍|f(y)-fJ|2dy.
Moreover, if (x,s)∈S(I) and  y∈ℝn∖(2/3)J, then
(142)1(s+|x-y|)n+1≲|y|-n-1,
and, by |∇Ps(y)|≲(s+|y|)-n-1, we have
(143)∫S(I)‍|∇f3(x,s)|2K(sℓ(I))s1-ndxds≲∫S(I)‍(∫ℝn‍|∇Ps(x-y)||f3(y)|dy)2×K(sℓ(I))s1-ndxds≲∫S(I)‍(∫ℝn∖(2/3)J‍|f(y)-fJ|(s+|x-y|)n+1dy)2×K(sℓ(I))s1-ndxds≲∫01‍K(s)sn-1ds(ℓ(I)∫ℝn∖(2/3)J‍|f(y)-fJ||y|n+1dy)2.
Combining the inequalities above, Lemma 12 is proved.Proof of Theorem10.
We assume thatK(t)=K(1)tn-1 for t≥1. By Lemma 7, K satisfies (47) if and only if
(144)sup⁡0<s<1∫s1‍K(t)tn+1dt∫0s‍tn-1K(t)dt<∞.Sufficiency. Let I be a cube and f(x,t) is the Poisson integral of f. Note that
(145)f(x,|y|)-f(x)=∫0|y|‍∂f(x,t)∂tdt.
By Minkowski’s inequality, we have
(146)(∫I‍|f(x,|y|)-f(x)|2dx)1/2≤∫0|y|‍(∫I‍|∂f(x,t)∂t|2dx)1/2dt≤∫0|y|‍(∫I‍|∇f(x,t)|2dx)1/2dt.
Hence,
(147)∫|y|<ℓ(I)‍K(|y|ℓ(I))1|y|2n(∫I‍|f(x,|y|)-f(x)|2dx)dy≲(ℓ(I))2-n∫01‍K(r)rn+1(∫0r‍(∫I‍|∇f(x,ℓ(I)s)|2dx)1/2ds)2dr≲∫S(I)‍|∇f(x,t)|2K(tℓ(I))t1-ndxdt.
The last inequality above holds by Lemma 11 since K satisfies
(148)sup⁡0<s<1∫s1‍K(t)tn+1dt∫0s‍tn-1K(t)dt<∞.
Thus,
(149)sup⁡I∫|y|<ℓ(I)‍K(|y|ℓ(I))1|y|2n×(∫I‍|f(x,|y|)-f(x)|2dx)dy<∞.
For |y|<ℓ(I),
(150)∫I‍|f(x+y,|y|)-f(x+y)|2dx≤∫3I‍|f(x,|y|)-f(x)|2dx.
Similarly, we get
(151)sup⁡I∫|y|<ℓ(I)‍K(|y|ℓ(I))1|y|2n×(∫I‍|f(x+y,|y|)-f(x+y)|2dx)dy<∞.
Note that
(152)|f(x+y,|y|)-f(x,|y|)|≤∫0|y|‍|∇f(x+tey,|y|)|dt,ey=y|y|.
When |y|<ℓ(I), we employ Minkowski’s inequality to get
(153)(∫I‍|f(x+y,|y|)-f(x,|y|)|2dx)1/2≤∫0|y|‍(∫I‍|∇f(x+tey,|y|)|2dx)1/2dt≲|y|(∫3I‍|∇f(x,|y|)|2dx)1/2.
Hence,
(154)sup⁡I∫|y|<ℓ(I)‍K(|y|ℓ(I))1|y|2n×(∫I‍|f(x+y,|y|)-f(x,|y|)|2dx)dy≲sup⁡I∫S(I)‍|∇f(x,t)|2K(tℓ(I))t1-ndxdt<∞.
By the triangle inequality, we get
(155)sup⁡I∫|y|<ℓ(I)‍K(|y|ℓ(I))1|y|2n×(∫I‍|f(x+y)-f(x)|2dx)dy<∞.
Thus f∈𝒬K(ℝn).
Necessity. Note that
(156)sup⁡0<s<1∫s1‍K(t)t3n+1dt∫0s‍t3n-1K(t)dt≤sup⁡0<s<1∫s1‍K(t)tn+1dt∫0s‍tn-1K(t)dt.
Thus K satisfies the conditions of Lemma 12. Let I and J be cubes in ℝn centered at x0 with ℓ(J)=3ℓ(I). Since f∈𝒬K(ℝn)⊆BMO(ℝn), by [10, p. 590], we have
(157)ℓ(J)∫ℝn∖(2/3)J‍|f(x)-fJ||x-x0|n+1dx≲∥f∥BMO(ℝn).
Note that
(158)|J|-1∫J‍|f(x)-fJ|2dx≲∫|y|≤nℓ(J)‍∫J‍|f(x+y)-f(x)|2K(|y|ℓ(J))1|y|2ndxdy.
Lemma 12 gives
(159)∫S(I)‍|∇f(x,t)|2K(tℓ(I))t1-ndxdt≲∥f∥𝒬K(ℝn)2+∥f∥BMO(ℝn)2.
The proof of Theorem 10 is complete.



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*Source: 931937-2014-05-14.xml*