# A Characterization of Central BMO Space via the Commutator of Fractional Hardy Operator

Authors: Lei Zhang; Shaoguang Shi
Journal: Journal of Function Spaces
 (2020)

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## Abstract

This paper is devoted in characterizing the central BMOℝn space via the commutator of the fractional Hardy operator with rough kernel. Precisely, by a more explicit decomposition of the operator and the kernel function, we will show that if the symbol function belongs to the central BMO ℝn space, then the commutator are bounded on Lebesgue space. Conversely, the boundedness of the commutator implies that the symbol function belongs to the central BMO ℝn space by exploiting the center symmetry of the Hardy operator deeply.


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## Body

## 1. Introduction

In this paper, we focus on the need for characterizing the central BMOℝn space via the boundedness of the commutators of the following fractional Hardy operators
(1)HΩ,αfx=1xn−α∫y<xΩx−yfydy,HΩ,α∗fx=∫y≥xΩx−yfyyn−αdy,0<α<n.HΩ,α∗ is the dual operator of HΩ,α. Here, Ω satisfies
(2)Ωtx=Ωx,∀t>0,x∈ℝn;(3)∫Sn−1Ωx′dσx′=0;(4)Ω∈LqSn−1∀q>1;and Sn−1=x∈ℝn:∣x∣=1 denotes the unit sphere in ℝn.For a functionb, the commutators of HΩ,α and HΩ,α∗ can be written as
(5)b,HΩ,αf≔bHΩ,αf−HΩ,αbf,b,HΩ,α∗f≔bHΩ,α∗f−HΩ,α∗bf.In [1], Fu et al. considered the boundedness of HΩ,α and b,HΩ,α on homogeneous Herz spaces and Lebesgue spaces under the assumption that Ω satisfies (2) and (4). We recall the results from ([1], Proposition 3.1 and Theorem 3.1) asSuppose that(6)1<p1,p2<∞with1p1=1p2+αn;1p1+1s+1q=1forsomes>p1′=p1p1−1;Ωsatisfies1.1and1.3;b∈CBMOmaxp2,sℝn.Then,(7)HΩ,αisboundedfromLp1ℝntoLp2ℝn;b;HΩ,αisboundedfromLp1ℝntoLp2ℝn.For the boundedness of the classical fractional Hardy operator, see [2]. For a ball Br≔B0,r⊂ℝn (i.e., a ball centered at 0 with radius r>0) and p≥1, CBMOpℝn is the central BMO ℝn function space, which was introduced by Lu and Yang [3] via the norm
(8)bCBMOpℝn=supr>01Br∫Brbx−bBrpdx1/pwithbBr=1Br∫Brbxdx.It is easy to see that CBMOℝn can be understood as a local version of the classical BMO ℝn space at the origin [4] and
(9)BMOℝn⊂CBMOℝn,CBMOp>1ℝn⊂CBMOℝn.Hence, the famous John-Nirenberg inequality for BMOℝn space is not true for CBMO ℝn space, which reveals that they have quite different properties.In the study of harmonic analysis, the characterization of function spaces via the boundedness of the commutators plays an important role in the field of PDEs, see, for example, [5–11] and the references therein. However, there are less attention paid for the commutators with rough kernels since the characterization depends heavily on the smoothness of the kernel function Ω. Under the premise that Ω is smooth enough, i.e., Ω∈C∞Sn−1 or Ω∈Lip1Sn−1, see, for example, [12–15]. It is difficult to weaken the smoothness of Ω, Chen and Ding [16] considered a characterization of BMOℝn space under the condition that Ω satisfies the following Hölder condition of log type
(10)Ωx′−Ωy′≤Alog2/x′−y′γwithA>0,γ>1,x′,y′∈Sn−1.It is easy to check that (10) is weaker than the Lipschitz condition and stronger than the condition (4). For q≥1 and Ω∈LqSn−1, we call Ω satisfies the Lq–Dini condition if ∫01wqδ/δ<∞, where wqδ is defined as
(11)wqδ=supτ≤δ∫Sn−1Ωτx′−Ωx′qdσx′1/qwithτ=supx′∈Sn−1τx′−x′.As a useful supplement of [1], we give a characterization of the CBMO ℝn space via the boundedness of b,HΩ,α and b,HΩ,α∗ as follows.Theorem 1.
Let1<p1,p2<∞with1/p1=1/p2+α/nand1/p1+1/s1/q+1forsomes>p1′.
(a)
If(12)Ωsatisfies1.1and1.3;b∈CBMOmaxp2,sℝn,then
(13)b,HΩ,αisboundedfromLp1ℝntoLp2ℝn;b,HΩ,αisboundedfromLp1ℝntoLp2ℝn.(b)
if(14)Ωsatisfies1.1,1.2,and1.4;b,HΩ,αisboundedfromLp1ℝntoLp2ℝn;b,HΩ,αisboundedfromLp1ℝntoLp2ℝn,thenb∈CBMOℝn.A part of Theorem1 has been proven in [1], we will show the rest of Theorem 1 in Section 2. In what follows, we will denote C by a positive constant which may vary from line to line. The symbol A≲B means A≤CB and ℤ for the set of all integers. Last, but not least, Br≔B0,r, Bk≔B2k, Ck≔Bk\Bk−1, and χk≔χCk with k∈ℤ.


## 2. Proof of Theorem 1.1

We prove Theorem1 in this section. To do so, we need one lemma about the estimates of the kernel function Ω, which plays a key role in the proof.Lemma 2.
LetΩ satisfy (2) and (10). Then,
(a)
Ωx−y−Ωx≤C/logx/yγ with γ be given in (10) and x≥4y.(b)
IfΩ furthermore satisfies the Lq≥1–Dini condition, then, there is a C>0 such that for 0<C<1/2, r>0, x∈ℝn with x<Cr; we have(15)∫r<∣y∣<2rΩy−x−Ωyqdy1/q≤Crn/q∫x/2rx/rwqδδdδ;∫r<∣y∣<2rΩy−x−Ωyqyn−αqdy1/q≤Cr−n/q′+α∫x/2rx/rwqδδdδwith0<α<np1.Proof.
We give the proof by a slight modification from [17]. For x≥4y, we first show that
(16)x−y′−x′≤3yy,(17)1/x−1/x−y≤Cy/x2.
Indeed, the first inequality can be obtained immediately from ([18], Lemma 2). Since
(18)1/x−1/x−y≤4y/3x2≤Cy/x2,(a) can be shown by (2) and (10) as
(19)Ωx−y−Ωx≤Clog2/x−y′−x′γ≤Clogx/yγ.
We are left to show (b). The first estimate can be obtained directly by theLq–Dini condition as
(20)∫r<y<2rΩy−x−Ωyqdy1/q=∫r2r∫Sn−1Ωy′−t−1x′−Ωy′qdδy′tn−1dt1/q≤Crn/q∫x/2rx/rwqδδdδ,and the estimate
(21)∫Sn−1Ωy′−t−1x′−Ωy′qdδy′≤Cwqqxt.follows from ([19], p.65-77).
Accordingly, the second can be deduced similarly. In fact,(22)∫r<y<2rΩy−x−Ωyqyn−αqdy1/q=Cr−n/q′+α∫r2r∫Sn−1Ωy′−t−1x′−Ωy′qdδy′dtt1/q≤Cr−n/q′+α∫x/2rx/rwqδδdδ,which is the desired one.The following is the boundedness of the fractional Hardy operators.Theorem 3.
Let1<p1,p2<∞ with 1/p1=1/p2+α/n and Ω satisfy (2) and (4). Then, both HΩ,α and HΩ,α∗ are bounded from Lp1ℝn to Lp2ℝn.Proof.
Since theLp1ℝn,Lp2ℝn boundedness for HΩ,α is contained in ([1], Proposition 3.1), it is enough to check the boundedness for HΩ,α∗. Namely, the task is now left to show that there exist constants C>0 such that for any f∈Lp1ℝn, one has
(23)HΩ∗fLp2ℝn≤CfLp1ℝn.
To do so, we first recall a useful estimates from [1] as
(24)x−y≤x+2i≤2k+1&∫CiΩx−yqdy≤∫02k+1∫Sn−1Ωx′qdδx′tn−1dt≤C2knfor x∈Ck, y∈Ci, and i≤k. Since
(25)∫ℝnHΩ∗fxp2dx≤∑k=−∞+∞∫Ck∫x≤y≤2knΩx−yfyyn−αdyp2dx+∑k=−∞+∞∫Ck∫y>2knΩx−yfyyn−αdyp2dx≔I1+I2.
Applying Hölder’s inequality toq,p1,s for s>1 and (24), we have
(26)I1≤∑k=−∞+∞2−kn−αp2∫Ck∑i=−∞k∫CiΩx−yqdy1/q∫Cifyp1dy1/p1∫Cidy1/sp2dx=C∑k=−∞+∞∑i=−∞k2i−kn/s∫Cifyp1dy1/p1p2≤C∑k=−∞+∞∑i=−∞k2i−knp2/2sfLp1Bip2∑i=−∞k2i−knp2′/2sp2/p2′≤C∑k=−∞+∞∑i=−∞k2i−knp2/2sfLp1Bip2≤C∑i=−∞+∞fLp1Bip2∑k=i+∞2i−knp2/2s≤CfLp1ℝn,p2I2≤∑k=−∞+∞∫Bk∑i=k∞∫CiΩx−yqdy1/q∫Cifyp1dy1/p1∫Ciy−n−αsdy1/sp2dx≤C∑k=−∞+∞∑i=k+∞2k−in/p2fLp1Bip2≤C∑k=−∞+∞∑i=k+∞2k−in/2p2fLp1Bip2≤C∑i=−∞+∞fLp1Bip2∑k=−∞i2k−in/2p2≤CfLp1ℝnp2,which is our desired result.Now, we can prove Theorem1. Without loss of generality, we can assume that bCBMOmaxp2,sℝn=1 in the proof of (a) since b∈CBMOmaxp2,sℝn. We see at once that the Lp1ℝn,Lp2ℝn boundedness of b,HΩ,α is just ([1], Theorem 3.1). To complete the proof of (a), what is left is to show is that
(27)b,HΩ,α∗fLp2ℝn≤CfLp1ℝn.It is easy to check that(28)∫ℝnb,HΩ,α∗fxp2dx=∑k=−∞+∞∫Ck∫y≥xΩx−ybx−byfyyn−αdyp2dx=∑k=−∞+∞∫Ck∫2kn≥y≥xΩx−ybx−byfyyn−αdy+∫y>2knΩx−ybx−byfyyn−αdyp2dx≤C∑k=−∞+∞∫Ck∫x≤y≤2knΩx−ybx−byfyyn−αdyp2dx+C∑k=−∞+∞∫Ck∫y>2knΩx−ybx−byfyyn−αdyp2dx≔J1+J2.Using Hölder’s inequality, we have(29)J1≤C∑k=−∞+∞∫Ck1xn−α∫y≤2knΩx−ybx−byfydyp2dx≤C∑k=−∞+∞2−kn−αp2∫Bk∑i=−∞k∫BiΩx−ybx−byfydyp2dx≤CfLp1ℝnp2.For the termJ2, we see at once that
(30)J2≤C∑k=−∞+∞∫Ck∑i=k+∞∫CiΩx−ybx−bBkfyyn−αdyp2dx+C∑k=−∞+∞∫Ck∑i=k+∞∫CiΩx−yby−bBkfyyn−αdyp2dx≔J21+J22.The Hölder inequality, along with (24), implies
(31)J21≤C∑k=−∞+∞∫Ckbx−bBkp2∑i=k+∞∫CiΩx−yfyyn−αdyp2dx≤C∑k=−∞+∞∫Bkbx−bBkp2∑i=k+∞fLp1Bi∫CiΩx−yqdy1/q∫Ciy−n−αsdy1/sp2dx≤C∑k=−∞+∞∑i=k+∞2k−in/2fLp1Bip2∑i=k+∞2k−inp2′/2p2p2/p2′≤CfLpℝnp.The termJ22 need a further decomposition as follows:
(32)J22≤C∑k=−∞+∞∫Bk∑i=k+∞∫CiΩx−yby−bBifyyn−αdyp2dx+C∑k=−∞+∞∫Bk∑i=k+∞∫CiΩx−ybBi−bBkfyyn−αdyp2dx≔J221+J222.Applying the Hölder inequality, we deduce(33)J221≤C∑k=−∞+∞∫Bk∑i=k+∞∫Cifyp1yn−αp1dy1/p∫CiΩx−yqdy1/q∫Ciby−bBisdy1/sp2dx≤CbCBMOsℝnp∑k=−∞+∞∑i=k+∞2k−in/pfLpBip≤C∑k=−∞+∞∑i=k+∞2k−in/2fLp1Bip2∑i=k+∞2k−inp2′/2p2p2/p2′≤CfLpℝnp.From the fact that(34)bBk−bBi≤∑j=ki−1bBj−bBj+1≤∑j=ki−11Bj∫Bj+1by−bBj+1dy≤Ci−kbCBMOp2Bi,the term J222 can be estimated as follows:
(35)J222≤CbCBMOp2ℝnp2∑k=−∞+∞∫Bk∑i=k+∞i−k∫CiΩx−yfyyn−αdyp2dx≤C∑k=−∞+∞∫Bk∑i=k+∞i−kfLp1Bi∫CiΩx−yqdy1/q∫Ciy−n−αsdy1/sp2dx≤C∑k=−∞+∞∑i=k+∞2k−in/pfLpBip≤C∑i=−∞+∞fLp1Bip2∑k=−∞ii−k2k−in/2≤CfLp1ℝnp2,which is the desired result and (a) is obtained.Next, we verify (b) inspired by [18]. Namely, we need to show that there is a constant CΩ,α,p,n such that
(36)Θ≔Θb,r≔1Br∫Brby−bBrdy≤Cforanyr∈ℝ+.For abbreviation, we assume thatb,HΩ,αLp1→Lp2=1, b,HΩ,α∗Lp1→Lp2=1, and bBr=0 since b−bBr,HΩ,α=b,HΩ,α. Let fy=sgnbyχBry. It is easy to check that 1/Br∫ℝnfybydy=Θ.Applying (3) and (10), we deduce that for A and γ appearing in (10), there is a constant C1<1 such that σD>0 for D=x′∈Sn−1:Ωx′≥2A/log2/C1γ. For x′∈D and y′∈Sn−1 with x′−y′≤C1, we obtain from (10) that
(37)Ωy′=Ωx′−Ωx′−Ωy′≥Ωx′−Ωx′−Ωy′≥Alog2/C1γ.WritingE=x∈ℝn:x>C2rand&x′∈DwithC2=3C1−1+1>4, we see at once that for x∈E, b,HΩ,αfx≥HΩ,αbfx−bxHΩ,αfx≔K1x−K2x. We conclude from (16) that x−y′−x′≤3y/x≤C1 since y<r and x>C2y>4y, and hence, Ωx−y′≥A/log2/C1γ, and finally, that
(38)K1x≥Alog2/C1γ1xn−α∫y<xbyfydy=Alog2/C1γxn−α∫Brbyfydy=C3rnxn−αΘ.Furthermore,(39)K2x≤bxxn−α∫BrfyΩx−y′−Ωx′dy≤Cbxxn−α∫Brfylogx/yγdy≤Cbxxn−α∫Brfylogx/rγdy≤C4rnbxxn−αlogx/rγ.For abbreviation, we write(40)F=x∈E:bx>C3Θ2C4logxrγ,x<Θp2/nr;F1=E\F∩x<Θp2/nr;F2=E∩C5F+C2rn1/n<x<Θp2/nr.This, along with the estimates forK1 and K2, one has
(41)χFLp1ℝnfLp2′ℝn≥χFLp1ℝnb,HΩ,αfLp1′ℝn≥∫F1b,HΩ,αfxdx≥∫F1C3rnΘxn−α−C3rnΘ2xn−αpdx≥C3rnΘ2∫C3F+C2rn1/nΘp2/nrρα−1dρ∫Ddσx′≥σDC3rnΘαΘαp2/nrα−C5αF+C2rnα/n.Consequently,(42)F+C2rnα/n≥C5−αΘαp2/nrα−C5−ααC3σDχFLp1ℝnfLp2′ℝnΘrn≥C6Θαp2/nrα.This in turn implies thatC6Θp2rn≤∣F∣+C2rn. Thus, b is proved if Θ≤2C6−1C2n1/p2. If Θ>2C6−1C2n1/p2, we see immediately that
(43)F≥C6Θp2rn−C2nrn≥C6Θp2rn2.According toF⊂E, C2>4, y>C2x, Lemma 2 and (37), we can obtain that for x∈Br and y∈F,
(44)x−y≈y,Ωx−y′≥Alog2/C1γ.We continue to choosef∗y=sgnbyχFy for x∈Br and get
(45)b,HΩ,α∗f∗x≥∫FΩx−ybyyn−αdy−bx∫FΩx−yf∗yyn−αdy≔L1x+L2x.It is easy to check that(46)L2x≤C7bx∫C2rΘp2/nrρα−1dρ∫Sn−1dσy′≤C8bxΘαp2/nrα−C2αrα.To deal with the termL1x, we first obtain from (44) that
(47)L1x≥AC3Θ2C4log1/C1γ∫Flogy/rγyn−αdy.Then, the estimate forL1x consists of two cases.Case 1.
γ≥n. Since 4r<C2r<y<Θp2/nr for y∈F, we conclude from (47) and F≥C6Θp2rn/2 that
(48)L1x≥C9Θrn∫Flogy/ry/rnlogy/rγ−nyαdy≥C9Θ1+αp2/nrn−αlogp2/nΘp2/nnlogC2γ−nF≥C10Θ1+αp2/nlogΘnrα.Case 2.
1<γ<n. In this case,
(49)L1x≥C9Θ1+αp2/nrn−α∫Flogyrγy/rndy≥C11Θ1+αp2/nlogΘγrα.
This in turn reveals that(50)b,HΩ,α∗f∗x≥C11Θ1+αp2/nlogΘγrα−C9bxrαΘαp2/n,∀x∈Br.Having disposed of the above estimates, we can now obtain(51)C12rn+αΘ1+αp2/n≥∫x<Θp2/nrdx1/p1rn/p2′≥rn/p2′f∗Lp1ℝn≥χBrLp2′ℝnb,HΩ,α∗f∗Lp2ℝn≥∫Brb,HΩ,α∗f∗dx≥C11ΘlogΘγrα∫Brdx−C9Θαp2/nrα∫Brbxdx≥C11Θ1+αp2/nrn+αlogΘγ−C9Θ1+αp2/nrn+α.Namely,C11logΘγ−C9≤C12. This in turn shows that there is a constant C≔CΩ,α,n,p,γ such that Θ≤C, whence reaching the desired fact.



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