# Commutators of Singular Integral Operators and Lipschitz Functions on Amalgam Spaces

Authors: Jiao Ma; Zongguang Liu
Journal: Journal of Function Spaces
 (2020)

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## Abstract

In this paper, we obtain the boundedness of commutators generated by singular integral operators and Lipschitz functions on amalgam spaces and weighted amalgam spaces.


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## Body

## 1. Introduction

Letb be a locally integrable function on ℝn and TΩ be a Calderón–Zygmund singular integral operator, then the commutator b,TΩ generated by b and TΩ is defined as(1)b,TΩfx=bxTΩfx−TΩbfx.Coifman et al. [1] proved that b,TΩ is bounded on Lpℝn for 1<p<∞ if and only if b∈BMO. Then, Janson [2] obtained that b,TΩ is bounded from Lp to Lq if and only if b∈Lipβ, where 1<p<q<∞, 0<β<1, and 1/q=1/p−β/n. Recently, Wang [3] showed that b,TΩ is bounded on weighted amalgam spaces when b∈BMO. On the contrary, in [4], the boundedness for the commutator generated by the singular integral operator and the weighted Lipschitz function on Lebesgue spaces is obtained. The aim of this paper is to establish the boundedness of the commutator b,TΩ generated by the Calderón–Zygmund singular integral operator and the weighted Lipschitz function on amalgam spaces.Amalgam spaces were first introduced by Wiener [5, 6]. He considered the special cases L1,l2, L2,l∞ and L∞,l1, L1,l∞. A systematic study on general amalgam spaces Lp,lq goes back to the work of Holland [7]. He defined the spaces Lp,lq1≤p,q≤∞ to be the sets of all locally p-th power summable functions f on ℝ such that(2)fLp,lq≔∑n=−∞∞∫nn+1fxpdxq/p1/q<∞,with the usual convention applied when p or q is infinity.Stewart [8] gave the definition of spaces Lp,lq over a locally compact Abelian group G. By the structure theorem, we may write G=ℝa×G1, where a is a nonnegative integer and G1 is a group which contains a compact open subgroup H (if G1 is compact, we take H=G1; if G1 is discrete and infinite, we take H=0; otherwise, H is arbitrary but fixed). Let I=0,1 and Iα=α+I, where α=n1,…,na,t, with nl∈ℤ l=1,…,a and t∈G1 (the collection of t’s being a transversal of G1 in H, i.e., G1=∪tt+H is a coset decomposition of G1). Then, in terms of the disjoint union G=∪α∈JIα, where J = {α:α=n1,…,na,t,nl∈ℤ for l=1,…,a and t∈G1}, the amalgam space of Lp and lq on G was defined by(3)Lp,lqG=f:fp,q<∞,where(4)fp,q=∑α∈JfLpIαq1/q,0<p≤∞,0<q<∞,fp,∞=supα∈JfLpIα,0<p≤∞.We refer the reader to the work of Fournier and Steward [9] for more information about these spaces.Fofana introduced in [10] the amalgam spaces Lp,Lqαℝn as follows:(5)Lp,Lqαℝn=f∈Llocpℝn:fLp,Lqαℝn<∞,1≤p,q,α≤∞,where(6)fLp,Lqαℝn=supr>0By,r1/α−1/p−1/qfχBy,rLpℝnLqℝn,here By,r=x∈ℝn:x−y<r and By,r is its Lebesgue measure. In [10], the author showed that the space Lp,Lqαℝn is nontrivial if and only if p≤α≤q; thus, we will always assume p≤α≤q throughout the present article.We recall the following interesting facts about the amalgam spacesLp,Lqαℝn:(1)
For1≤p≤α≤q≤∞, we can easily see that Lp,Lqαℝn⊆Lp,Lqℝn, where Lp,Lqℝn is Wiener’s amalgam space (see [7])(2)
If1≤p<α and q=∞, then Lp,Lqαℝn is just the classical Morrey space Lp,1−p/αℝn(3)
Ifp=α and q=∞, then Lp,Lqαℝn is just the usual Lebesgue space LpℝnThe first main result of this paper is as follows.Theorem 1.
Let1<p<q<∞, 0<γ<1, 1/q=1/p−γ/n, and 1/β=1/α−γ/n. Assume that p<α<s<∞ and q<β<s<∞. If b∈Lipγ, then b,TΩ is bounded from Lp,Lsαℝn to Lq,Lsβℝn.
We also need the definition of the weighted amalgam spaces.Definition 1 (see [3]).
Let1≤p≤α≤q≤∞ and w and μ be two weights on ℝn. Then, the weighted amalgam spaces Lp,Lqαw;μ are defined as(7)Lp,Lqαw;μ=f∈Llocpw:fLp,Lqαw;μ<∞,where(8)fLp,Lqαw;μ=supr>0wBy,r1/α−1/p−1/qfχBy,rLwpLμq,with the usual modification when q=∞. Moreover, the spaces defined above are Banach spaces.Definition 2 (see [3]).
Let1≤p≤α≤q≤∞ and u, v, and μ be three weights on ℝn. Then, the weighted amalgam spaces Lp,Lqαv,u;μ are defined as(9)Lp,Lqαv,u;μ=f∈Llocpv:fLp,Lqαv,u;μ<∞,where(10)fLp,Lqαv,u;μ=supr>0uBy,r1/α−1/p−1/qfχBy,rLvpLμq,with the usual modification when q=∞. Moreover, the spaces defined above are Banach spaces.
The second main result is the following:Theorem 2.
Let1<p<q<∞, 0<β<1, p<α1<s<∞, q<α2<s<∞, 1/q=1/p−β/n, and 1/α2=1/α1−β/n. Suppose that u∈A1, v is a weight, and b∈Lipβ,u, then b,TΩ is bounded from Lp,Lsα1u,u1−q;ν to Lq,Lsα2u1−q;ν.
Throughout the paper, we will assume that the Calderón–Zygmund singular integral operatorTΩ is defined by(11)TΩfx=p.v.∫ℝnΩx−yx−ynfydy,where Ω is a homogeneous function of degree zero, ∫Sn−1Ωx′dσx′=0 and Ω∈C∞Sn−1.


### 1.1. Some Preliminaries and Notations

We denote byΔhk the kth difference operator. That is,(12)Δh1fx=Δhfx=fx+h−fx,Δhk+1fx=Δhkfx+h−Δhkfx,k≥1.For a cubeQ and a locally integrable function f, we write(13)fQ=1Q∫Qfxdx.Forβ>0, the Lipschitz space Λβ consists of functions f such that(14)fΛβ=supx,h∈ℝn,h≠0Δhβ+1fxhβ<∞.Forβ>0 and f∈Lloc1, we say f∈Lipβ if there is a constant C>0 such that(15)fx−fy≤Cx−yβ,for any x, y∈ℝn. The minimal constant C mentioned above is called the Lipβ norm of f, denoted by fLipβ.Remark 1.
When0<β<1, Λβ=Lipβ; when β≥1, Lipβ⊊Λβ.Lemma 1 (see [11]).
If0<β<1 and 1≤p<∞, then(16)fLipβ≈supQ1Q1+β/n∫Qfx−fQdx≈supQ1Qβ/n1Q∫Qfx−fQpdx1/p.Lemma 2 (see [11]).
IfQ∗⊆Q, then fQ∗−fQ≤CfLipβQβ/n.
From Lemma1, we see that(17)fy−fQ≤CfLipβQβ/n,a.e.y∈Q.
The following result was established in [2].Theorem 3 (see [2]).
Let0<β<1 and 1≤p<∞. Then, the following statements are equivalent:(a)
b∈Lipβ(b)
b,TΩ is bounded from Lpℝn to Lqℝn, where 1/q=1/p−β/n
A weightw is said to belong to the Muckenhoupt class Ap for 1<p<∞, 1/p+1/p′=1, if there exists a constant C such that(18)1B∫Bwxdx1B∫Bwx1−p′dxp−1≤C,for any ball B⊂ℝn. The class A1 is defined by replacing the above inequality with(19)1B∫Bwxdx≤Cwx,a.e.x∈B,for any ball B⊂ℝn and A∞=∪p≥1Ap (see [12]).
Let0<β<1 and w be a weight function, then the weighted Lipschitz space Lipβ,w is defined by(20)Lipβ,w=f∈Lloc1ℝn:fLipβ,w<∞,where(21)fLipβ,w=supB1wB1+β/n∫Bfy−fBdy.Lemma 3 (see [4, 12]).
If0<β<1, u∈A1, b∈Lipβ,u, and 1≤p≤∞, then we have(22)bLipβ,u≈supQuQ−β/nuQ−1∫Qbx−bBpux1−pdx1/p.Lemma 4 (see [4, 12]).
If0<β<1, u∈A1, b∈Lipβ,u, and Q is a cube, then there exists a constant C>0, independent of b and Q, such that(23)supx∈Qbx−bQ≤CbLipβ,uuQ1+β/nQ−1.Lemma 5 (see [4, 12]).
If0<β<1, u∈A1, b∈Lipβ,u, and Q is a cube, then there exists a constant C>0, independent of b and Q, such that(24)b2kQ−bQ≤Ckuxu2kQβ/nbLipβ,u,for allx∈Q.Theorem 4 (see [4]).
Letu∈A1, 1/q=1/p−β/n for 0<β<1 and 1<p<q<∞. Then, b∈Lipβ,u if and only if b,TΩ is bounded from Lpu to Lqu1−q.


### 1.2. Proofs of the Main Results

Proof of Theorem 1.
Let1<p<q<∞, 0<γ<1, p<α<s, q<β<s, 1/q=1/p−γ/n, 1/β=1/α−γ/n, and f∈Lp,Lsαℝn. Fix y∈ℝn and r>0 and set B=By,r and 2B=By,2r. Let f1=fχ2B and f2=f−f1. We write(25)By,r1/β−1/q−1/sb,TΩfχBy,rLq≤By,r1/β−1/q−1/sb,TΩf1χBy,rLq+By,r1/β−1/q−1/sb,TΩf2χBy,rLq≕J1y,r+J2y,r.
We will estimateJ1y,r and J2y,r separately. Using Theorem 3, we have(26)J1y,r≤By,r1/β−1/q−1/sb,TΩf1Lq≤CBy,r1/β−1/q−1/sfχ2BLp=CBy,2r1/α−1/p−1/sfχ2BLpBy,r1/β−1/q−1/sBy,2r1/α−1/p−1/s.
Since1/q=1/p−γ/n and 1/β=1/α−γ/n, we have 1/α−1/p−1/s=1/β−1/q−1/s<0. Consequently,(27)J1y,r≤CBy,2r1/α−1/p−1/sfχ2BLp.
Now we treatJ2y,r. It is obvious that x−z≈y−z, when x∈By,r and z∈By,2rc. We have(28)b,TΩf2x≤bx−bBTΩf2x+TΩb−bBf2x.
Decomposingℝn into a geometrically increasing sequence of concentric balls, we have(29)TΩf2x≤C∫ℝnf2zx−zndz≤C∫By,2rcfzy−zndz=C∑j=1∞∫By,2j+1r\By,2jrfzy−zndz≤C∑j=1∞1By,2j+1r∫By,2j+1rfzdzTΩb−bBf2x=∫ℝnbz−bBf2zΩx−zx−zndz≤C∫2Bcbz−bBfz1x−zndz≤C∑j=1∞1By,2j+1r∫By,2j+1rbz−bBfzdz.
From these estimates, it follows that(30)J2y,r=By,r1/β−1/q−1/s∫By,rb,TΩf2xqdx1/q≤CBy,r1/β−1/q−1/s∑j=1∞1By,2j+1r∫By,2j+1rfzdz×∫By,rbx−bBqdx1/q+CBy,r1/β−1/s∑j=1∞1By,2j+1r∫By,2j+1rbz−bBfzdz≕J3y,r+J4y,r.
Using Hölder’s inequality and Lemma1, we have(31)J3y,r≤CbLipγBy,r1/β−1/q−1/s∑j=1∞1By,2j+1r∫By,2j+1rfzdzBy,r1/p≤CbLipγ∑j=1∞By,r1/β−1/q−1/s+1/p1By,2j+1r∫By,2j+1rfzpdz1/p=CbLipγ∑j=1∞By,2j+1r1/α−1/p−1/sfχBy,2j+1rLpBy,r1/α−1/sBy,2j+1r1/α−1/s=CbLipγ∑j=1∞By,2j+1r1/α−1/p−1/sfχBy,2j+1rLp12j+1n1/α−1/s.
As forJ4y,r, we get(32)J4y,r≤CBy,r1/β−1/s∑j=1∞1By,2j+1r×∫By,2j+1rbz−b2j+1B+b2j+1B−bBfzdz=CBy,r1/β−1/s∑j=1∞1By,2j+1r∫By,2j+1rbz−b2j+1Bfzdz+CBy,r1/β−1/s∑j=1∞1By,2j+1r∫By,2j+1rb2j+1B−bBfzdz≕J5y,r+J6y,r.
It follows from Hölder’s inequality and Lemma1 that(33)J5y,r≤CBy,r1/β−1/s∑j=1∞1By,2j+1r∫By,2j+1rfzpdz1/p×∫By,2j+1rbz−b2j+1Bp′dz1/p′≤CbLipγBy,r1/β−1/s∑j=1∞fχBy,2j+1rLpBy,2j+1rγ/n−1/p=CbLipγ∑j=1∞By,2j+1r1/α−1/p−1/sfχBy,2j+1rLpBy,r1/β−1/sBy,2j+1r1/β−1/s=CbLipγ∑j=1∞By,2j+1r1/α−1/p−1/sfχBy,2j+1rLp12j+1n1/β−1/s.
For the last termJ6y,r, we use Hölder’s inequality and Lemma 2 to get(34)J6y,r≤CBy,r1/β−1/s∑j=1∞bLipγBy,2j+1rγ/n1By,2j+1r∫By,2j+1rfzdz≤CbLipγ∑j=1∞By,r1/β−1/s∫By,2j+1rfzpdz1/p⋅By,2j+1rγ/n−1/p=CbLipγ∑j=1∞By,2j+1r1/α−1/p−1/sfχBy,2j+1rLpBy,r1/β−1/sBy,2j+1r1/β−1/s=CbLipγ∑j=1∞By,2j+1r1/α−1/p−1/sfχBy,2j+1rLp12j+1n1/β−1/s.
Summarizing the above estimates, we conclude that(35)J2y,r≤C∑j=1∞By,2j+1r1/α−1/p−1/sfχBy,2j+1rLp12j+1n1/α−1/s+C∑j=1∞By,2j+1r1/α−1/p−1/sfχBy,2j+1rLp12j+1n1/β−1/s.
Taking theLs-norm and using Minkowski’s inequality, we get(36)By,r1/β−1/q−1/sb,TΩfχBy,rLqLs≤J1Ls+J2Ls≤CBy,2r1/α−1/p−1/sfχ2BLpLs+C∑j=1∞By,2j+1r1/α−1/p−1/sfχBy,2j+1rLpLs12j+1n1/α−1/s+C∑j=1∞By,2j+1r1/α−1/p−1/sfχBy,2j+1rLpLs12j+1n1/β−1/s≤CfLp,Lsαℝn.
Thus, by taking the supremum over allr>0, the proof is completed.Proof of Theorem 2.
Let1<p<q<∞, 0<β<1, p<α1<s, q<α2<s, 1/q=1/p−β/n, 1/α2=1/α1−β/n, and f∈Lp,Lsα1u,u1−q;ν. Fix y∈ℝn and r>0 and set B=By,r and 2B=By,2r. Let f1=fχ2B and f2=f−f1. We write(37)u1−qBy,r1/α2−1/q−1/sb,TΩfχBy,rLqu1−q≤u1−qBy,r1/α2−1/q−1/sb,TΩf1χBy,rLqu1−q+u1−qBy,r1/α2−1/q−1/sb,TΩf2χBy,rLqu1−q≕K1y,r+K2y,r.
Now we will estimateK1y,r and K2y,r separately. Using Theorem 4, we have(38)K1y,r≤u1−qBy,r1/α2−1/q−1/sb,TΩf1Lqu1−q≤Cu1−qBy,r1/α2−1/q−1/sfχ2BLpu=Cu1−qBy,2r1/α1−1/p−1/sfχ2BLpuu1−qBy,r1/α2−1/q−1/su1−qBy,2r1/α1−1/p−1/s.
Noticing thatu1−q∈A∞ implies that u1−q is a doubling measure and that 1/q=1/p−β/n and 1/α2=1/α1−β/n, and it follows that(39)u1−qBy,r1/α2−1/q−1/su1−qBy,2r1/α1−1/q−1/s≤C.
So we have(40)K1y,r≤Cu1−qBy,2r1/α1−1/p−1/sfχ2BLpu.
Similar to the proof of Theorem1, we decompose K2y,r as(41)K2y,r=u1−qBy,r1/α2−1/q−1/sb,TΩf2χBy,rLqu1−q≤Cu1−qBy,r1/α2−1/q−1/s∑j=1∞1By,2j+1r∫By,2j+1rfzdz×∫By,rbx−bBqu1−qxdx1/q+Cu1−qBy,r1/α2−1/s∑j=1∞1By,2j+1r∫By,2j+1rbz−bBfzdz≕K3y,r+K4y,r.
Notice thatu∈A1 implies u∈Aq for all q>1. Using Hölder’s inequality and Lemma 4, we have(42)K3y,r≤Cu1−qBy,r1/α2−1/s∑j=1∞1By,2j+1r∫By,2j+1rfzdz×bLipβ,uuBy,rβ/nuBy,rBy,r≤CbLipβ,uu1−qBy,r1/α2−1/s∑j=1∞1By,2j+1r×∫By,2j+1rfzpuzdz1/p∫By,2j+1ruz−p′/pdz1/p′×uBy,rβ/nuBy,rBy,r≤CbLipβ,u∑j=1∞u1−qBy,r1/α2−1/sfχBy,2j+1rLpu×uBy,2j+1r−1/puBy,rβ/n⋅essinfx∈Bux.
From the fact that(43)uBy,rβ/n≤uBy,2j+1rβ/n,and 1/q=1/p−β/n, we get(44)K3y,r≤CbLipβ,u∑j=1∞u1−qBy,r1/α2−1/sfχBy,2j+1rLpu×uBy,2j+1r−1/qessinfx∈Bux=CbLipβ,u∑j=1∞u1−qBy,2j+1r1/α1−1/p−1/sfχBy,2j+1rLpu×u1−qBy,r1/α2−1/q−1/su1−qBy,2j+1r1/α1−1/p−1/suBy,2j+1r−1/q×u1−qBy,r1/qessinfx∈Bux.
Using the estimate(45)u1−qBy,r1/qessinfx∈Bux=∫By,ru1−qxdx1/qessinfx∈Bux≤essinfx∈Buqx∫By,ru1−qxdx1/q≤∫By,ruxdx1/q=uBy,r1/q,we conclude that(46)K3y,r≤CbLipβ,u∑j=1∞u1−qBy,2j+1r1/α1−1/p−1/sfχBy,2j+1rLpu×u1−qBy,r1/α2−1/q−1/su1−qBy,2j+1r1/α1−1/p−1/suBy,ruBy,2j+1r1/q.
Sinceu1−q,u∈A∞, there exists a constant δ>0 such that(47)u1−qBy,r1/α2−1/q−1/su1−qBy,2j+1r1/α1−1/p−1/s≤CBy,rδ1/α2−1/q−1/sBy,2j+1rδ1/α2−1/q−1/s≤C12j+1nδ1/α2−1/q−1/suBy,ruBy,2j+1r1/q≤CBy,rδ/qBy,2j+1rδ/q≤C12j+1nδ/q.
Hence, we have(48)K3y,r≤CbLipβ,u∑j=1∞u1−qBy,2j+1r1/α1−1/p−1/sfχBy,2j+1rLpu×12j+1nδ1/α2−1/s.
As forK4y,r, we have(49)K4y,r≤Cu1−qBy,r1/α2−1/s∑j=1∞1By,2j+1r×∫By,2j+1rbz−b2j+1Bfzdz+Cu1−qBy,r1/α2−1/s∑j=1∞1By,2j+1r×∫By,2j+1rfzdz⋅bB−b2j+1B≕K5y,r+K6y,r.
It follows from Hölder’s inequality, the fact thatu∈A1, and Lemma 4, then we have(50)K5y,r≤Cu1−qBy,r1/α2−1/s∑j=1∞1By,2j+1r∫By,2j+1rfzdz×bLipβ,uuBy,2j+1rβ/nuBy,2j+1rBy,2j+1r≤CbLipβ,u∑j=1∞u1−qBy,r1/α2−1/sfχBy,2j+1rLpu×uBy,2j+1r−1/quBy,2j+1rBy,2j+1r≤CbLipβ,u∑j=1∞u1−qBy,2j+1r1/α1−1/p−1/sfχBy,2j+1rLpu×u1−qBy,r1/α1−1/p−1/su1−qBy,2j+1r1/α1−1/p−1/s×u1−qBy,r1/q⋅ess infx∈2j+1BuxuBy,2j+1r−1/q≤CbLipβ,u∑j=1∞u1−qBy,2j+1r1/α1−1/p−1/sfχBy,2j+1rLpu×u1−qBy,r1/α2−1/q−1/su1−qBy,2j+1r1/α1−1/p−1/suBy,ruBy,2j+1r1/q≤CbLipβ,u∑j=1∞u1−qBy,2j+1r1/α1−1/p−1/sfχBy,2j+1rLpu×12j+1nδ1/α2−1/s.
For the last term, applying Lemma5 and the A1 property, we have(51)K6y,r≤CbLipβ,uu1−qBy,r1/α2−1/s∑j=1∞j+1uBy,2j+1rβ/n×essinfx∈BuxfχBy,2j+1rLpuuBy,2j+1r−1/p=CbLipβ,u∑j=1∞j+1u1−qBy,2j+1r1/α1−1/p−1/sfχBy,2j+1rLpu×u1−qBy,r1/α2−1/q−1/su1−qBy,2j+1r1/α1−1/p−1/su1−qBy,r1/q×essinfx∈Bux⋅uBy,2j+1r−1/q.
By an argument similar to that used in estimatingK5, we get(52)K6y,r≤CbLipβ,u∑j=1∞j+1u1−qBy,2j+1r1/α1−1/p−1/sfχBy,2j+1rLpu×u1−qBy,r1/α2−1/q−1/su1−qBy,2j+1r1/α1−1/p−1/suBy,ruBy,2j+1r1/q≤CbLipβ,u∑j=1∞j+1u1−qBy,2j+1r1/α1−1/p−1/sfχBy,2j+1rLpu×12j+1nδ1/α2−1/s.
Taking theLsν-norm and using Minkowski’s inequality, we get(53)u1−qBy,r1/α2−1/q−1/sb,TΩfχBy,rLqu1−qLsν≤K1Lsν+K2Lsν≤Cu1−qBy,2r1/α1−1/p−1/sfχ2BLpuLsν+C∑j=1∞u1−qBy,2j+1r1/α1−1/p−1/sfχBy,2j+1rLpuLsν×j+112j+1nδ1/α2−1/s≤CfLp,Lsα1u,u1−q;ν.
Finally, by taking the supremum over allr>0, the proof is completed.


## 1.1. Some Preliminaries and Notations

We denote byΔhk the kth difference operator. That is,(12)Δh1fx=Δhfx=fx+h−fx,Δhk+1fx=Δhkfx+h−Δhkfx,k≥1.For a cubeQ and a locally integrable function f, we write(13)fQ=1Q∫Qfxdx.Forβ>0, the Lipschitz space Λβ consists of functions f such that(14)fΛβ=supx,h∈ℝn,h≠0Δhβ+1fxhβ<∞.Forβ>0 and f∈Lloc1, we say f∈Lipβ if there is a constant C>0 such that(15)fx−fy≤Cx−yβ,for any x, y∈ℝn. The minimal constant C mentioned above is called the Lipβ norm of f, denoted by fLipβ.Remark 1.
When0<β<1, Λβ=Lipβ; when β≥1, Lipβ⊊Λβ.Lemma 1 (see [11]).
If0<β<1 and 1≤p<∞, then(16)fLipβ≈supQ1Q1+β/n∫Qfx−fQdx≈supQ1Qβ/n1Q∫Qfx−fQpdx1/p.Lemma 2 (see [11]).
IfQ∗⊆Q, then fQ∗−fQ≤CfLipβQβ/n.
From Lemma1, we see that(17)fy−fQ≤CfLipβQβ/n,a.e.y∈Q.
The following result was established in [2].Theorem 3 (see [2]).
Let0<β<1 and 1≤p<∞. Then, the following statements are equivalent:(a)
b∈Lipβ(b)
b,TΩ is bounded from Lpℝn to Lqℝn, where 1/q=1/p−β/n
A weightw is said to belong to the Muckenhoupt class Ap for 1<p<∞, 1/p+1/p′=1, if there exists a constant C such that(18)1B∫Bwxdx1B∫Bwx1−p′dxp−1≤C,for any ball B⊂ℝn. The class A1 is defined by replacing the above inequality with(19)1B∫Bwxdx≤Cwx,a.e.x∈B,for any ball B⊂ℝn and A∞=∪p≥1Ap (see [12]).
Let0<β<1 and w be a weight function, then the weighted Lipschitz space Lipβ,w is defined by(20)Lipβ,w=f∈Lloc1ℝn:fLipβ,w<∞,where(21)fLipβ,w=supB1wB1+β/n∫Bfy−fBdy.Lemma 3 (see [4, 12]).
If0<β<1, u∈A1, b∈Lipβ,u, and 1≤p≤∞, then we have(22)bLipβ,u≈supQuQ−β/nuQ−1∫Qbx−bBpux1−pdx1/p.Lemma 4 (see [4, 12]).
If0<β<1, u∈A1, b∈Lipβ,u, and Q is a cube, then there exists a constant C>0, independent of b and Q, such that(23)supx∈Qbx−bQ≤CbLipβ,uuQ1+β/nQ−1.Lemma 5 (see [4, 12]).
If0<β<1, u∈A1, b∈Lipβ,u, and Q is a cube, then there exists a constant C>0, independent of b and Q, such that(24)b2kQ−bQ≤Ckuxu2kQβ/nbLipβ,u,for allx∈Q.Theorem 4 (see [4]).
Letu∈A1, 1/q=1/p−β/n for 0<β<1 and 1<p<q<∞. Then, b∈Lipβ,u if and only if b,TΩ is bounded from Lpu to Lqu1−q.


## 1.2. Proofs of the Main Results

Proof of Theorem 1.
Let1<p<q<∞, 0<γ<1, p<α<s, q<β<s, 1/q=1/p−γ/n, 1/β=1/α−γ/n, and f∈Lp,Lsαℝn. Fix y∈ℝn and r>0 and set B=By,r and 2B=By,2r. Let f1=fχ2B and f2=f−f1. We write(25)By,r1/β−1/q−1/sb,TΩfχBy,rLq≤By,r1/β−1/q−1/sb,TΩf1χBy,rLq+By,r1/β−1/q−1/sb,TΩf2χBy,rLq≕J1y,r+J2y,r.
We will estimateJ1y,r and J2y,r separately. Using Theorem 3, we have(26)J1y,r≤By,r1/β−1/q−1/sb,TΩf1Lq≤CBy,r1/β−1/q−1/sfχ2BLp=CBy,2r1/α−1/p−1/sfχ2BLpBy,r1/β−1/q−1/sBy,2r1/α−1/p−1/s.
Since1/q=1/p−γ/n and 1/β=1/α−γ/n, we have 1/α−1/p−1/s=1/β−1/q−1/s<0. Consequently,(27)J1y,r≤CBy,2r1/α−1/p−1/sfχ2BLp.
Now we treatJ2y,r. It is obvious that x−z≈y−z, when x∈By,r and z∈By,2rc. We have(28)b,TΩf2x≤bx−bBTΩf2x+TΩb−bBf2x.
Decomposingℝn into a geometrically increasing sequence of concentric balls, we have(29)TΩf2x≤C∫ℝnf2zx−zndz≤C∫By,2rcfzy−zndz=C∑j=1∞∫By,2j+1r\By,2jrfzy−zndz≤C∑j=1∞1By,2j+1r∫By,2j+1rfzdzTΩb−bBf2x=∫ℝnbz−bBf2zΩx−zx−zndz≤C∫2Bcbz−bBfz1x−zndz≤C∑j=1∞1By,2j+1r∫By,2j+1rbz−bBfzdz.
From these estimates, it follows that(30)J2y,r=By,r1/β−1/q−1/s∫By,rb,TΩf2xqdx1/q≤CBy,r1/β−1/q−1/s∑j=1∞1By,2j+1r∫By,2j+1rfzdz×∫By,rbx−bBqdx1/q+CBy,r1/β−1/s∑j=1∞1By,2j+1r∫By,2j+1rbz−bBfzdz≕J3y,r+J4y,r.
Using Hölder’s inequality and Lemma1, we have(31)J3y,r≤CbLipγBy,r1/β−1/q−1/s∑j=1∞1By,2j+1r∫By,2j+1rfzdzBy,r1/p≤CbLipγ∑j=1∞By,r1/β−1/q−1/s+1/p1By,2j+1r∫By,2j+1rfzpdz1/p=CbLipγ∑j=1∞By,2j+1r1/α−1/p−1/sfχBy,2j+1rLpBy,r1/α−1/sBy,2j+1r1/α−1/s=CbLipγ∑j=1∞By,2j+1r1/α−1/p−1/sfχBy,2j+1rLp12j+1n1/α−1/s.
As forJ4y,r, we get(32)J4y,r≤CBy,r1/β−1/s∑j=1∞1By,2j+1r×∫By,2j+1rbz−b2j+1B+b2j+1B−bBfzdz=CBy,r1/β−1/s∑j=1∞1By,2j+1r∫By,2j+1rbz−b2j+1Bfzdz+CBy,r1/β−1/s∑j=1∞1By,2j+1r∫By,2j+1rb2j+1B−bBfzdz≕J5y,r+J6y,r.
It follows from Hölder’s inequality and Lemma1 that(33)J5y,r≤CBy,r1/β−1/s∑j=1∞1By,2j+1r∫By,2j+1rfzpdz1/p×∫By,2j+1rbz−b2j+1Bp′dz1/p′≤CbLipγBy,r1/β−1/s∑j=1∞fχBy,2j+1rLpBy,2j+1rγ/n−1/p=CbLipγ∑j=1∞By,2j+1r1/α−1/p−1/sfχBy,2j+1rLpBy,r1/β−1/sBy,2j+1r1/β−1/s=CbLipγ∑j=1∞By,2j+1r1/α−1/p−1/sfχBy,2j+1rLp12j+1n1/β−1/s.
For the last termJ6y,r, we use Hölder’s inequality and Lemma 2 to get(34)J6y,r≤CBy,r1/β−1/s∑j=1∞bLipγBy,2j+1rγ/n1By,2j+1r∫By,2j+1rfzdz≤CbLipγ∑j=1∞By,r1/β−1/s∫By,2j+1rfzpdz1/p⋅By,2j+1rγ/n−1/p=CbLipγ∑j=1∞By,2j+1r1/α−1/p−1/sfχBy,2j+1rLpBy,r1/β−1/sBy,2j+1r1/β−1/s=CbLipγ∑j=1∞By,2j+1r1/α−1/p−1/sfχBy,2j+1rLp12j+1n1/β−1/s.
Summarizing the above estimates, we conclude that(35)J2y,r≤C∑j=1∞By,2j+1r1/α−1/p−1/sfχBy,2j+1rLp12j+1n1/α−1/s+C∑j=1∞By,2j+1r1/α−1/p−1/sfχBy,2j+1rLp12j+1n1/β−1/s.
Taking theLs-norm and using Minkowski’s inequality, we get(36)By,r1/β−1/q−1/sb,TΩfχBy,rLqLs≤J1Ls+J2Ls≤CBy,2r1/α−1/p−1/sfχ2BLpLs+C∑j=1∞By,2j+1r1/α−1/p−1/sfχBy,2j+1rLpLs12j+1n1/α−1/s+C∑j=1∞By,2j+1r1/α−1/p−1/sfχBy,2j+1rLpLs12j+1n1/β−1/s≤CfLp,Lsαℝn.
Thus, by taking the supremum over allr>0, the proof is completed.Proof of Theorem 2.
Let1<p<q<∞, 0<β<1, p<α1<s, q<α2<s, 1/q=1/p−β/n, 1/α2=1/α1−β/n, and f∈Lp,Lsα1u,u1−q;ν. Fix y∈ℝn and r>0 and set B=By,r and 2B=By,2r. Let f1=fχ2B and f2=f−f1. We write(37)u1−qBy,r1/α2−1/q−1/sb,TΩfχBy,rLqu1−q≤u1−qBy,r1/α2−1/q−1/sb,TΩf1χBy,rLqu1−q+u1−qBy,r1/α2−1/q−1/sb,TΩf2χBy,rLqu1−q≕K1y,r+K2y,r.
Now we will estimateK1y,r and K2y,r separately. Using Theorem 4, we have(38)K1y,r≤u1−qBy,r1/α2−1/q−1/sb,TΩf1Lqu1−q≤Cu1−qBy,r1/α2−1/q−1/sfχ2BLpu=Cu1−qBy,2r1/α1−1/p−1/sfχ2BLpuu1−qBy,r1/α2−1/q−1/su1−qBy,2r1/α1−1/p−1/s.
Noticing thatu1−q∈A∞ implies that u1−q is a doubling measure and that 1/q=1/p−β/n and 1/α2=1/α1−β/n, and it follows that(39)u1−qBy,r1/α2−1/q−1/su1−qBy,2r1/α1−1/q−1/s≤C.
So we have(40)K1y,r≤Cu1−qBy,2r1/α1−1/p−1/sfχ2BLpu.
Similar to the proof of Theorem1, we decompose K2y,r as(41)K2y,r=u1−qBy,r1/α2−1/q−1/sb,TΩf2χBy,rLqu1−q≤Cu1−qBy,r1/α2−1/q−1/s∑j=1∞1By,2j+1r∫By,2j+1rfzdz×∫By,rbx−bBqu1−qxdx1/q+Cu1−qBy,r1/α2−1/s∑j=1∞1By,2j+1r∫By,2j+1rbz−bBfzdz≕K3y,r+K4y,r.
Notice thatu∈A1 implies u∈Aq for all q>1. Using Hölder’s inequality and Lemma 4, we have(42)K3y,r≤Cu1−qBy,r1/α2−1/s∑j=1∞1By,2j+1r∫By,2j+1rfzdz×bLipβ,uuBy,rβ/nuBy,rBy,r≤CbLipβ,uu1−qBy,r1/α2−1/s∑j=1∞1By,2j+1r×∫By,2j+1rfzpuzdz1/p∫By,2j+1ruz−p′/pdz1/p′×uBy,rβ/nuBy,rBy,r≤CbLipβ,u∑j=1∞u1−qBy,r1/α2−1/sfχBy,2j+1rLpu×uBy,2j+1r−1/puBy,rβ/n⋅essinfx∈Bux.
From the fact that(43)uBy,rβ/n≤uBy,2j+1rβ/n,and 1/q=1/p−β/n, we get(44)K3y,r≤CbLipβ,u∑j=1∞u1−qBy,r1/α2−1/sfχBy,2j+1rLpu×uBy,2j+1r−1/qessinfx∈Bux=CbLipβ,u∑j=1∞u1−qBy,2j+1r1/α1−1/p−1/sfχBy,2j+1rLpu×u1−qBy,r1/α2−1/q−1/su1−qBy,2j+1r1/α1−1/p−1/suBy,2j+1r−1/q×u1−qBy,r1/qessinfx∈Bux.
Using the estimate(45)u1−qBy,r1/qessinfx∈Bux=∫By,ru1−qxdx1/qessinfx∈Bux≤essinfx∈Buqx∫By,ru1−qxdx1/q≤∫By,ruxdx1/q=uBy,r1/q,we conclude that(46)K3y,r≤CbLipβ,u∑j=1∞u1−qBy,2j+1r1/α1−1/p−1/sfχBy,2j+1rLpu×u1−qBy,r1/α2−1/q−1/su1−qBy,2j+1r1/α1−1/p−1/suBy,ruBy,2j+1r1/q.
Sinceu1−q,u∈A∞, there exists a constant δ>0 such that(47)u1−qBy,r1/α2−1/q−1/su1−qBy,2j+1r1/α1−1/p−1/s≤CBy,rδ1/α2−1/q−1/sBy,2j+1rδ1/α2−1/q−1/s≤C12j+1nδ1/α2−1/q−1/suBy,ruBy,2j+1r1/q≤CBy,rδ/qBy,2j+1rδ/q≤C12j+1nδ/q.
Hence, we have(48)K3y,r≤CbLipβ,u∑j=1∞u1−qBy,2j+1r1/α1−1/p−1/sfχBy,2j+1rLpu×12j+1nδ1/α2−1/s.
As forK4y,r, we have(49)K4y,r≤Cu1−qBy,r1/α2−1/s∑j=1∞1By,2j+1r×∫By,2j+1rbz−b2j+1Bfzdz+Cu1−qBy,r1/α2−1/s∑j=1∞1By,2j+1r×∫By,2j+1rfzdz⋅bB−b2j+1B≕K5y,r+K6y,r.
It follows from Hölder’s inequality, the fact thatu∈A1, and Lemma 4, then we have(50)K5y,r≤Cu1−qBy,r1/α2−1/s∑j=1∞1By,2j+1r∫By,2j+1rfzdz×bLipβ,uuBy,2j+1rβ/nuBy,2j+1rBy,2j+1r≤CbLipβ,u∑j=1∞u1−qBy,r1/α2−1/sfχBy,2j+1rLpu×uBy,2j+1r−1/quBy,2j+1rBy,2j+1r≤CbLipβ,u∑j=1∞u1−qBy,2j+1r1/α1−1/p−1/sfχBy,2j+1rLpu×u1−qBy,r1/α1−1/p−1/su1−qBy,2j+1r1/α1−1/p−1/s×u1−qBy,r1/q⋅ess infx∈2j+1BuxuBy,2j+1r−1/q≤CbLipβ,u∑j=1∞u1−qBy,2j+1r1/α1−1/p−1/sfχBy,2j+1rLpu×u1−qBy,r1/α2−1/q−1/su1−qBy,2j+1r1/α1−1/p−1/suBy,ruBy,2j+1r1/q≤CbLipβ,u∑j=1∞u1−qBy,2j+1r1/α1−1/p−1/sfχBy,2j+1rLpu×12j+1nδ1/α2−1/s.
For the last term, applying Lemma5 and the A1 property, we have(51)K6y,r≤CbLipβ,uu1−qBy,r1/α2−1/s∑j=1∞j+1uBy,2j+1rβ/n×essinfx∈BuxfχBy,2j+1rLpuuBy,2j+1r−1/p=CbLipβ,u∑j=1∞j+1u1−qBy,2j+1r1/α1−1/p−1/sfχBy,2j+1rLpu×u1−qBy,r1/α2−1/q−1/su1−qBy,2j+1r1/α1−1/p−1/su1−qBy,r1/q×essinfx∈Bux⋅uBy,2j+1r−1/q.
By an argument similar to that used in estimatingK5, we get(52)K6y,r≤CbLipβ,u∑j=1∞j+1u1−qBy,2j+1r1/α1−1/p−1/sfχBy,2j+1rLpu×u1−qBy,r1/α2−1/q−1/su1−qBy,2j+1r1/α1−1/p−1/suBy,ruBy,2j+1r1/q≤CbLipβ,u∑j=1∞j+1u1−qBy,2j+1r1/α1−1/p−1/sfχBy,2j+1rLpu×12j+1nδ1/α2−1/s.
Taking theLsν-norm and using Minkowski’s inequality, we get(53)u1−qBy,r1/α2−1/q−1/sb,TΩfχBy,rLqu1−qLsν≤K1Lsν+K2Lsν≤Cu1−qBy,2r1/α1−1/p−1/sfχ2BLpuLsν+C∑j=1∞u1−qBy,2j+1r1/α1−1/p−1/sfχBy,2j+1rLpuLsν×j+112j+1nδ1/α2−1/s≤CfLp,Lsα1u,u1−q;ν.
Finally, by taking the supremum over allr>0, the proof is completed.



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*Source: 4926328-2020-01-21.xml*