# Endpoint Estimates for Generalized Commutators of Hardy Operators onH1 Space

Authors: Xiao Yu; Shanzhen Lu
Journal: Journal of Function Spaces and Applications
 (2013)

---

## Abstract

We study theH1-boundedness of the generalized commutators of Hardy operator with a homogeneous kernel as follows: ℋΩ,A,βmf(x)=(1/|x|n-β)∫|y|<|x|(‍Ω(x-y)/|x-y|m-1)Rm(A;x,y)f(y)dy, where Rm(A;x,y)=A(x)-∑|α|<m‍(1/α!)DαA(y)(x-y)α with m∈Z+, 0≤β<n and Ω∈Lip1(Sn-1). We prove that, when m≥1, ℋΩ,A,βm is not bounded from H1 to Ln/(n-β) unless ℋΩ,A,βm≡0. Finally, we prove that ℋΩ,A,βm is bounded from H1 to Ln/(n-β),∞ with m≥1.


---

## Body

## 1. Introduction

Letf∈Lp(ℝ+) with 1<p<∞; the classical Hardy operator is defined by
(1)Hf(x)=1x∫0x‍f(t)dt,x≠0.A famous result proved by Hardy [1] can be stated as follows;
(2)∥Hf∥Lp(ℝ+)≤pp-1∥f∥Lp(ℝ+).Hardy [1] also pointed out the fact that the constant p/(p-1) in (2) is the best possible. Later, Hardy operator was studied by many mathematicians; please see [2, 3] for more details.In 1995, Christ and Grafakos [4] studied the following n-dimensional Hardy operator;
(3)ℋf(x)=1νn|x|n∫|y|<|x|‍f(y)dy,x∈ℝn∖{0},
where νn is the volume of the unit ball in ℝn, and they proved the following inequality:
(4)∥ℋf∥Lp≤pp-1∥f∥Lp,1<p<∞.Furthermore, Christ and Grafakos [4] also showed that the constant p/(p-1) in (3) is the best possible.In 2007, Fu et al. [5] considered the following commutator of fractional Hardy operator:
(5)ℋβbf(x)=b(x)ℋβf(x)-ℋβ(fb)(x),
where ℋβf(x) is defined by
(6)ℋβf(x)=1|x|n-β∫|y|<|x|‍f(y)dy
with -n<β<n. When β=0, we simply denote ℋ0b by ℋb and ℋ0 is just the n-dimensional Hardy operator proposed by Christ and Grafakos in [4] (without considering the constant νn).In 2011, Fu et al. [6] studied the following n-dimensional fractional Hardy operator with a homogeneous kernel:
(7)ℋΩ,βf(x)=1|x|n-β∫|y|<|x|‍Ω(x-y)f(y)dy,
where Ω∈Lr(Sn-1). Fu et al. [6] proved that ℋΩ,βb is bounded on Herz type space and λ-central Morrey space. Here ℋΩ,βb is just the commutator of fractional Hardy operator with a homogeneous kernel.Recently, Zhao et al. [7] gave a counterexample to show that ℋb is not bounded from H1 to L1, and they proved that ℋb is bounded from H1 to weak L1 space where H1 denotes the Hardy space.On the other hand, in 1982, Cohen [8] studied the following generalized commutator:
(8)TA2f(x)=∫ℝn‍Ω(x-y)|x-y|n+1TA2f(x)=×(A(x)-A(y)-∇A(y)(x-y))f(y)dy,
where Ω∈L1(Sn-1) is homogeneous of degree zero and satisfies the moment condition
(9)∫Sn-1‍Ω(x)xαdσ(x)=0
for |α|=1. Cohen [8] proved that, if Ω∈Lip1(Sn-1) and ∇A∈BMO, then TA2 is bounded on Lp(ℝn) with 1<p<∞.Later, Cohen and Gosselin [9] considered another type of generalized commutator TAm(f) as follows:
(10)TAmf(x)=∫ℝn‍Ω(x-y)|x-y|n+m-1Rm(A;x,y)f(y)dy,
where Rm(A;x,y)(m∈Z+) is defined by Rm(A;x,y)=A(x)-∑|α|<m(1/α!)DαA(y)(x-y)α, the mth remainder of Taylor series of the function A at x about y, and Ω satisfies the following moment conditions:
(11)∫Sn-1‍Ω(x)xαdσ(x)=0
for |α|=m-1. Obviously, if we choose m=1, TAm becomes [A,T], the commutator of T generalized by A and T.Cohen and Gosselin [9] proved that TAm is bounded on Lp(ℝn) for 1<p<∞ if Ω∈Lip1(Sn-1) and the function A has derivatives of order m-1 in BMO(ℝn). Later, TAm was studied by many mathematicians; for example, see [10, 11] or [12] for more details. Particularly in [11], Lu and Wu studied the endpoint estimates of TAm on H1 space.It should be pointed out that the generalized commutators of some operators play an important role in the study of partial differential equation. Recently, by using theW1,p estimate for the elliptic equation of divergence form with partially BMO coefficients and the Lp boundedness of the Cohen-Gosselin type generalized commutators proved by Yan in [12], Wang and Zhang [13] gave a new proof of Wu’s theorem in [14]. Here we would like to point out that the method used in [13] is much simpler than that in [14].In this paper, we will consider the following generalized commutator of fractional Hardy operator with a homogeneous kernel:(12)ℋΩ,A,βmf(x)=1|x|n-β∫|y|<|x|‍Ω(x-y)|x-y|m-1f(y)Rm(A;x,y)dy,
where Rm(A;x,y)=A(x)-∑|α|<m(1/α!)DαA(y)(x-y)α,0≤β<n, and Ω∈Lip1(Sn-1).As the Hardy operator is controlled by the Hardy-Littlewood maximal function, we have(13)ℋΩ,A,βm≤CMΩ,A,βf(x),
where MΩ,A,βf(x) is defined by
(14)MΩ,A,βf(x)=supr>01rn-β∫|x-y|<r‍|Ω(x-y)||x-y|m-1|Rm(A;x,y)f(y)|dy.By a simple computation or from [15, pp. 221-222], we have
(15)MΩ,A,βf(x)=supr>01rn-β∫|x-y|<r‍|Ω(x-y)||x-y|m-1|Rm(A;x,y)f(y)|dy≤Csupr>01rn-β∫(r/2)<|x-y|<r‍|Ω(x-y)||x-y|m-1|Rm(A;x,y)f(y)|dy,
and thus
(16)ℋΩ,A,βmf(x)≤Csupr>01rn-β∫|x-y|<r‍|Ω(x-y)||x-y|m-1|Rm(A;x,y)f(y)|dy≤Csupr>01rn-β∫(r/2)<|x-y|<r‍|Ω(x-y)||x-y|m-1|Rm(A;x,y)f(y)|dy≤Csupr>0r-(n-β+m-1)∫(r/2)<|x-y|<r‍|Ω(x-y)|×|Rm(A;x,y)f(y)|dy.So we have(17)ℋΩ,A,βmf(x)≤CM~Ω,βAf(x),
where
(18)M~Ω,βAf(x)=supr>0r-(n-β+m-1)×∫(r/2)<|x-y|<r‍|Ω(x-y)Rm(A;x,y)f(y)|dy.From [15, p. 222], we have the following lemma.Lemma 1 (see [15]).
Let0<β<n and Ω∈Lip1(Sn-1). If 1<p,q<∞ with 1/p-1/q=β/n, and A has derivatives of order m-1 in BMO(ℝn), then
(19)∥M~Ω,βAf∥Lq≤C∑|α|=m-1‍∥DαA∥BMO∥f∥Lp,
where the constant C is independent of f and A.By checking [15, p. 222] carefully, we deduce that (19) still holds if we take β=0. So we have the following proposition.Proposition 2.
Let0≤β<n, 1/p-1/q=β/n with 1<p,q<∞. If Ω∈Lip1(Sn-1) and A has derivatives of order m-1 in BMO(ℝn), then
(20)∥ℋΩ,A,βmf∥Lq≤C∑|α|=m-1‍∥DαA∥BMO∥f∥Lp,
where the constant C is independent of f and A.Definition 3 (see [16]).
One says a functiona(x) is an H1 atom if a satisfies the following conditions:
(21)(i)supp(a)⊂B(x0,r),(ii)∥a∥L∞≤|B(x0,r)|-1,(iii)∫‍a(x)dx=0.It is well known that, if a functionf belongs to H1, then it can be written as f=∑i=1∞λiai where each ai is an H1 atom. Moreover, one has
(22)∥f∥H1~inf{∑i=-∞+∞‍|λi|},
where the infimum is taken over all decompositions of f.Definition 4 (see [17]).
A functionf is said to belong to BMO(ℝn) if the following sharp maximal function is bounded:
(23)f♯(x)=supB1|B|∫B‍|f(y)-fB|dy<∞,
where the supreme is taken over all balls B⊂ℝn and fB=(1/|B|)∫B‍f(x)dx and ∥f∥BMO=∥f♯∥L∞.Proposition 5 (see [17]).
Let1<p<∞ and f∈BMO(ℝn); then one has(a)
∥f∥BMO~supB((1/|B|)∫B‍|f(x)-fB|pdx)1/p;(b)
∥f∥BMO~supBinfa∈R(1/|B|)∫B‍|f(x)-a|dx.
Obviously, whenm=1, ℋΩ,A,βm can be written as ℋΩ,A,β1f(x):=ℋΩ,A,βf(x)=A(x)ℋΩ,βf(x)-ℋΩ,β(fA)(x), just the commutator of fractional Hardy operator with a homogeneous kernel.For the casem=2,β=0, and Ω≡1, Lu and Zhao [18] proved that ℋΩ,A,02 is bounded on Herz type spaces and Morrey-Herz type spaces.In this paper, we would like to show thatℋΩ,A,βm is not bounded from H1 to Ln/(n-β) for all m∈Z+. Furthermore, we will prove that ℋΩ,A,βm is bounded from H1 to Ln/(n-β),∞, where Ln/(n-β),∞ denotes the weak Ln/(n-β) space. Some ideas of this paper come from Zhao et al. [7].In this chapter, we would like to show that, ifA∈BMO(ℝn), ℋΩ,A,β(0≤β<n) is not bounded from H1 to Ln/(n-β).To show this, letA(x)=χ(4,∞)(x)∈BMO, Ω≡1, and f0(x)=χ(0,4)(x)-χ(-4,0)(x); then for x>8 and n=1, we have
(24)|ℋΩ,A,βf0(x)|=|1x1-β∫04‍(1-0)×1dy|=4x1-β,
and then
(25)∫R1‍|ℋΩ,A,βf0(x)|1/(1-β)dx≥∫8∞‍41/(1-β)xdx=∞,
which indicates that ℋΩ,A,β is not bounded from H1 to Ln/(n-β).


## 2. Endpoint Estimates forℋΩ,A,βm from H1 to Ln/(n-β)

In Section1, we know that, when m=1, ℋΩ,A,β1 is not bounded from H1 to Ln/(n-β). In this section, we will prove that, when m≥2, ℋΩ,A,βm is also not bounded from H1 to Ln/(n-β) unless ℋΩ,A,βm≡0. We have the following conclusions.Theorem 6.
Letm≥2,0≤β<n, and Ω∈Lip1(Sn-1). Assume that A has derivatives of order m-1 in BMO(ℝn); then the following two statements are equivalent;(i)
ℋΩ,A,βm maps H1(ℝn) continuously into Ln/(n-β);(ii)
for anyH1 atom supported on certain ball B and u∈3B∖2B, there is
(26)∫(4B)c‍|∑|α|=m-1‍1α!Kα(x,u)1|x|n-β∫B‍DαA(y)a(y)dy|n/(n-β)×dx≤C,where Kα(x,u)=Ω(x-u)(x-u)α/|x-u|m-1 with |α|=m-1.In order to prove Theorem6, we need the following lemma.Lemma 7 (see [9]).
Letb be a function on ℝn with mth order derivatives in Llocq(ℝn) for some q>n; then
(27)|Rm(b;x,y)|≤Cm,n|x-y|m×∑|α|=m‍(1|Q~(x,y)|∫Q~(x,y)‍|Dαb(z)|qdz)1/q,
where Q~(x,y) is the cube centered at x and having diameter 5n|x-y|.Proof of Theorem6.
Suppose thata(x) is an H1 atom supported on B(x0,r) and satisfies (21). Now we take a~(x)=a(x+x0); then a~ is also an H1 atom and satisfies
(28)(i′)supp(a~)⊂B(0,r),(ii′)∥a~∥L∞≤|B(0,r)|-1,(iii′)∫ℝn‍a~(x)dx=0.
Thus by the main results in [19] and the atomic decomposition of the space H1(ℝn), it suffices to show that, for any H1 atom a~, we have ∥ℋΩ,A,βma~∥Ln/(n-β)≤C.
LetB=B(0,r) and A~(x)=A(x)-∑|α|=m-1(1/α!)mB(DαA)xα; then Rm(A;x,y)=Rm(A~;x,y). For each H1 atom a~, we split each ℋΩ,A,βma~(x) as
(29)ℋΩ,A,βma~(x)=χ4B(x)ℋΩ,A,βma~(x)+χ(4B)c(x)ℋΩ,A,βma~(x):=μ1(x)+μ2(x).
Forμ1(x), taking n/(n-β)<q<∞ and p so that 1/p-1/q=β/n, it follows from Proposition 2 that
(30)∥μ1∥Ln/(n-β)≤|4B|(n-β)/n-1/q∥ℋΩ,A,βma~∥Lq≤C∑|α|=m-1‍∥DαA∥BMO|B|1-1/p∥a~∥Lp≤C∑|α|=m-1‍∥DαA∥BMO.
Forμ2(x), as x∈(4B)c and |y|<|x|, we can deduce {y:|y|<|x|}∩{y:y∈B(0,r)}={y:y∈B(0,r)}; thus we have
(31)μ2(x)=χ(4B)c1|x|n-β×∫B(0,r)‍Ω(x-y)|x-y|m-1Rm(A~;x,y)a~(y)dy.
Next we denoteB(0,r)=B and B(0,kr)=kB; then by the vanishing condition of a~, we decompose μ2 as follows:
(32)μ2(x)=χ(4B)c(x)1|x|n-β∫B(Ω(x-y)|x-y|m-1Rm-1(A~;x,y)‍=-Ω(x-u)|x-u|m-1Rm-1(A~;x,u)Ω(x-y)|x-y|m-1)×a~(y)dy-χ(4B)c(x)1|x|n-β×∑|α|=m-1‍1α!∫B‍[Kα(x,y)-Kα(x,u)]DαA~(y)a~(y)dy-χ(4B)c(x)1|x|n-β×∑|α|=m-1‍1α!∫B‍Kα(x,u)DαA~(y)a~(y)dy=μ21(x,u)-μ22(x,u)-μ23(x,u),
where u∈3B∖2B.
Forμ21(x,u), as u∈3B∖2B, y∈B, and x∈(4B)c, we have
(33)|x-y|~|x-u|~|x|.
Thus we have(34)|Ω(x-y)|x-y|m-1Rm-1(A~;x,y)-Ω(x-u)|x-u|m-1Rm-1(A~;x,u)|≤|Ω(x-y)|x-y|m-1Rm-1(A~;x,y)-Ω(x-y)|x-u|m-1Rm-1(A~;x,y)|+|Ω(x-y)|x-u|m-1Rm-1(A~;x,y)-Ω(x-u)|x-u|m-1Rm-1(A~;x,y)|+|Ω(x-u)|x-u|m-1||Rm-1(A~;x,y)-Rm-1(A~;x,u)|:=I+II+III.
For the termI, by Lemma 7, we have
(35)I≤C|Ω(x-y)||x-y|m|y-u||Rm-1(A~;x,y)|≤C|y-u||x-y|m|x-y|m-1k∑|α|=m-1‍∥DαA∥BMO≤Ck2-k∑|α|=m-1‍∥DαA∥BMO.
For the termII, by the similar estimates of I, we obtain
(36)II≤Ck2-k∑|α|=m-1‍∥DαA∥BMO.
For the termIII, by the following formula (see [20]):
(37)Rm(A;x,y)-Rm(A;x,z)=∑|α|<m‍1β!Rm-|α|(DαA;z,y)(x-z)α,
and then together with Lemma 7, we have
(38)III≤C∑|α|=m-1‍∥DαA∥BMO∑i=0m-2‍|x-y|-(m-1)+l|y-u|m-1-i≤C∑|α|=m-1‍∥DαA∥BMOk2-k.
Thus forμ21(x,u), by the size condition of a~, we get the following estimates:
(39)∥μ21(·,u)∥Ln/(n-β)≤C∑|α|=m-1‍∥DαA∥BMO×∑k=2∞‍k2-k[∫2k+1B∖2kB‍(1|x|n-β×∫B‍|a~(y)|dy1|x|n-β)n/(n-β)dx](n-β)/n≤C∑|α|=m-1‍∥DαA∥BMO∑k=2∞‍k2-k≤C∑|α|=m-1‍∥DαA∥BMO.
Forμ22(x,u), since Ω∈Lip1(Sn-1), we have the following estimates of |Kα(x-y)-Kα(x-u)|:
(40)|Kα(x-y)-Kα(x-u)|≤|Ω(x-y)(x-y)α|x-y|m-1-Ω(x-y)(x-u)α|x-y|m-1|+|Ω(x-y)(x-u)α|x-y|m-1-Ω(x-u)(x-u)α|x-y|m-1|+|Ω(x-u)(x-u)α|x-y|m-1-Ω(x-u)(x-u)α|x-u|m-1|≤C|y-u||x-y|.
Thus by the size condition ofa~ and Lemma 7, we have
(41)∥μ22(·,u)∥Ln/(n-β)≤C∑|α|=m-1‍∑k=2∞‍[∫2k+1B∖2kB‍×(∫B‍|y-u||x-y||x|n-β|a~(y)|[∫2k+1B∖2kB‍=×|Dα(A~)(y)|dy∫B‍|y-u||x-y||x|n-β)n/(n-β)dx](n-β)/n≤C∑|α|=m-1∥DαA∥BMO‍×∑k=2∞|2kB|(β-n)/n‍|B|-1|B||2kB|(n-β)/n2-k≤C∑|α|=m-1‍∥DαA∥BMO.
Now we can deduce that∥ℋΩ,A,βma~∥Ln/(n-β)≤C is equivalent to ∥μ23(·,u)∥Ln/(n-β)≤C. By the vanishing condition of a~, we can easily get
(42)∫(4B)c‍|∑|α|=m-1‍1α!Kα(x,u)1|x|n-β∫B‍DαA(y)a~(y)dy|n/(n-β)×dx≤C.
Consequently, we have finished the proof of Theorem6.Next we would like to show thatℋΩ,A,βm is not bounded from H1 to Ln/(n-β) unless ℋΩ,A,βm≡0. We have the following theorem.Theorem 8.
Letm≥2,0≤β<n, and Ω∈Lip1(Sn-1), and assume that A has derivatives of order m-1 in BMO(ℝn). Then the following two statements are equivalent:(i)
ℋΩ,A,βm maps H1 continuously into Ln/(n-β),(ii)
A is a polynomial of degree no more than m-1 or Ω≡0.Remark 9.
From Theorem8, we can draw the conclusion that, when m≥2, ℋΩ,A,βm is not bounded from H1 to Ln/(n-β) unless ℋΩ,A,βm≡0.Proof of Theorem8.
It is clear that (ii)⇒ (i) is obvious. We only need to prove (i) ⇒ (ii).
Leta~ be an H1 atom supported on the ball B=B(0,r), and denote Cα=(1/α!)∫B‍DαA(y)a~(y)dy with |α|=m-1. By Theorem 6, for any u∈3B∖2B and N>8 with N∈Z+, we have
(43)C≥∫(4B)c‍|1|x|n-β∑|α|=m-1‍CαΩ(x-u)(x-u)α|x-u|m-1|n/(n-β)dx≥C1∫(4B)c‍|1|x-u|n-β∑|α|=m-1‍CαΩ(x-u)(x-u)α|x-u|m-1|n/(n-β)dx≥C1∫8r≤|x|≤Nr‍|1|x|n-β∑|α|=m-1‍CαΩ(x)xα|x|m-1|n/(n-β)dx=C1log(N8)∫Sn-1‍|∑|α|=m-1‍CαΩ(x′)(x′)α|n/(n-β)dσ(x′).
LetN→+∞, we know log(N/8)→+∞. Thus we have
(44)∫Sn-1‍|∑|α|=m-1‍CαΩ(x′)(x′)α|n/(n-β)dσ(x′)=0.
From (44) we can deduce
(45)∑|α|=m-1‍CαΩ(x′)(x′)α=0.
IfΩ≡0, (45) is obviously true. Otherwise, we can easily obtain
(46)Cα=1α!∫B‍DαA(y)a~(y)dy=0.
Sincea~ is arbitrary, DαA must be a constant for each α with |α|=m-1. So we can deduce that A is a polynomial with degree no more than m-1.
Consequently, we have finished the proof of Theorem8.


## 3. Boundedness ofℋΩ,A,βm from H1 to Ln/(n-β),∞

In Section2, we prove that, when m≥2, ℋΩ,A,βm is not bounded from H1 to Ln/(n-β) unless ℋΩ,A,βm≡0. In this section, we will prove that ℋΩ,A,βm is bounded from H1 to Ln/(n-β),∞ with m≥1. Here Ln/(n-β),∞ is defined by
(47)∥f∥Ln/(n-β),∞=supλ>0λ|{x∈ℝn:|f(x)|>λ|(n-β)/n<∞.Our results can be stated as follows.Theorem 10.
Suppose thatm≥2,0≤β<n, and Ω∈Lip1(Sn-1). If A has derivatives of order m-1 in BMO(ℝn), then there exists a constant C independent of f and A, such that
(48)|{x∈ℝn:|ℋΩ,A,βmf(x)|>λ}|(n-β)/n≤C∑|α|=m-1‍∥DαA∥BMO∥f∥H1λ
for any λ>0.For the casem=1, we have the following.Theorem 11.
Let0≤β<n, A∈BMO(ℝn), and Ω∈Lip1(Sn-1); then there exists a constant C independent of f and A, such that
(49)|{x∈ℝn:|ℋΩ,A,β1f(x)|>λ}|(n-β)/n≤C∑|α|=m-1‍∥DαA∥BMO∥f∥H1λ
for any λ>0.Before the proof of Theorems10 and 11, we need the following lemma.Lemma 12.
Letℋ~βα be defined by
(50)ℋ~βαh(x)=1|x|n-β∫|y|<|x|‍h(y)(x-y)αΩ(x-y)|x-y|m-1dy,
where 0≤β<n,Ω∈Lip1(Sn-1), and |α|=m-1; then ℋ~βα is bounded from L1 to Ln/(n-β),∞.Proof.
For anyλ>0, we have
(51)λ|{x∈ℝn:|ℋ~βαh(x)|>λ}|(n-β)/n=λ|{x∈ℝn:|1|x|n-β∫|y|<|x|‍h(y)(x-y)α×Ω(x-y)|x-y|m-1dy|>λ}|(n-β)/n≤λ|{x∈ℝn:|1|x|n-β∫|y|<|x|‍|h(y)|1|x-y|m-1-|α|dy|>λ}|(n-β)/n≤λ|{x∈ℝn:|1|x|n-β∫|y|<|x|‍|h(y)|dy|>λ}|(n-β)/n≤λ|{x∈ℝn:|x|<∥h∥L11/(n-β)λ1/(n-β)}|(n-β)/n=λ(∫Sn-1‍∫0(∥h∥L11/(n-β))/(λ1/(n-β))‍rn-1drdσ(x′))(n-β)/n≤Cλ(∥h∥L1n/(n-β)λn/(n-β))(n-β)/n=C∥h∥L1.Proposition 13.
By the proof of Lemma12 with minor changes, one can draw the conclusion that ℋΩ,β is bounded from L1 to Ln/(n-β),∞ with Ω∈Lip1(Sn-1).Proof of Theorem10.
Before giving the proof of Theorem10, we introduce some notations that are very useful in this section.
For multi-indicesα=(α1,…,αn),β=(β1,…,βn), we denote α-β=(α1-β1,…,αn-βn). Furthermore, β<α means that for each i, we have βi<αi. Finally, we denote Cαβ=∏j=1mCαjβj.
From [19] and by the atomic decomposition of H1, it suffices to show
(52)|{x∈ℝn:|ℋΩ,A,βma~(x)|>λ}|(n-β)/n≤Cλ∑|α|=m-1‍∥DαA∥BMO
for any H1 atom a~, where a~ is defined in Section 2. First we have the following decomposition:
(53)|{x∈ℝn:ℋΩ,A,βma~(x)>λ}|(n-β)/n≤|{x∈B(0,2r):ℋΩ,A,βma~(x)>λ}|(n-β)/n+|{x∈ℝn∖B(0,2r):ℋΩ,A,βma~(x)>λ}|(n-β)/n:=L1+L2.
For the termL1, choosing B(0,r)=B and B(0,kr)=kB with k∈Z+, then by Proposition 2, Hölder inequality, and the size condition of a~, we have
(54)λ|{x∈B(0,2r):ℋΩ,A,βma~(x)>λ}|(n-β)/n≤C(∫B(0,2r)‍|ℋΩ,A,βma~(x)|n/(n-β)dx)(n-β)/n≤C|2B|(n-β)/n-1/q(∫B(0,2r)|ℋΩ,A,βma~(x)|q‍dx)1/q≤C|2B|(n-β)/n-1/q∑|α|=m-1‍∥DαA∥BMO∥a~∥Lp≤C∑|α|=m-1‍∥DαA∥BMO,
where 1/p-1/q=β/n and p>1.
For the termL2, as x∈ℝn∖B(0,2r), we have
(55){y:|y|<|x|}∩{y:|y|<r}={y:|y|<r}.
Then for a fixedB=B(0,r), we set AB(x)=A(x)-∑|α|=m-1(1/α!)(DαA)Bxα. It is easy to check Rm(A;x,y)=Rm(AB;x,y).
Thus by the vanishing condition ofa~ and the fact supp(a~)⊂B(0,r), we can decompose ℋΩ,A,βma~(x) as follows:
(56)ℋΩ,A,βma~(x)=1|x|n-β∫ℝn‍{Rm(AB;x,y)Ω(x-y)|x-y|m-1-[Rm-1(AB;x,0)-∑|α|=m-1‍1α!DαAB(x)xα]×Ω(x)|x|m-1Ω(x-y)|x-y|m-1}a~(y)dy=-1|x|n-β∫ℝn‍[Rm-1(AB;x,0)-∑|α|=m-1‍1α!DαAB(x)xα]×[Ω(x)|x|m-1-Ω(x-y)|x-y|m-1]a~(y)dy+1|x|n-β×∫ℝn[∑|α|<m-1‍1α!DαAB(y)(x-y)α‍-∑|α|<m-1‍DαAB(0)xα]Ω(x-y)|x-y|m-1a~(y)dy+1|x|n-β×∫ℝn[∑|α|=m-1‍1α!DαAB(y)xα‍-∑|α|=m-1‍1α!DαAB(x)xα]Ω(x-y)|x-y|m-1a~(y)dy+1|x|n-β∫ℝn‍∑|α|=m-1‍1α!DαAB(y)×∑γ<α‍Cαγxγ(-y)α-γΩ(x-y)|x-y|m-1a~(y)dy:=I1+I2+I3+I4.
Here we can simply denote eachIi by
(57)Ii=1|x|n-β∫ℝn‍Ki(AB;x,y)a~(y)dy.
ForK1, by the fact that Ω∈Lip1(Sn-1) and |x-y|~|x|, we have
(58)|Ω(x)|x|m-1-Ω(x-y)|x-y|m-1|≤|Ω(x)|x|m-1-Ω(x-y)|x|m-1|+|Ω(x-y)|x|m-1-Ω(x-y)|x-y|m-1|=|Ω((x/|x|)(|x|/|x-y|))|x|m-1-Ω((x-y)/|x-y|)|x|m-1|+|Ω(x-y)|x|m-1-Ω(x-y)|x-y|m-1|≤C|y||x|m.
As|x|>r for x∈ℝn∖2B, Lemma 7 in this paper and Lemma  2.2 in [8] tell us that
(59)|K1(AB;x,y)|≤C|x|m-1[∑|α|=m-1‍(1|Q~|∫Q~‍|DαA(z)-(DαA)B|qdz)q+∑|α|=m-1‍|DαA(x)-(DαA)B|]|y||x|m≤C|y||x|(∑|α|=m-1‍∥DαA∥BMO(C1+log|x|r)+∑|α|=m-1‍|DαA(x)-(DαA)B|),
where C1 is a constant only depending on n, and Q~ is a cube centered at x and having diameter 5n|x|.
Thus we obtain(60)λ|{x∈ℝn∖B(0,2r):|I1|>λ}|(n-β)/n≤(∫ℝn∖B(0,2r)|I1|n/(n-β)‍dx)(n-β)/n≤C(×a~(y)dy∫Rn‍|y||x|)n/(n-β)dx∫ℝn∖B(0,2r)|x|(β-n)(n/(n-β))×(∫ℝn‍|y||x|(∑|α|=m-1‍∥DαA∥BMO(C1+log|x|r)+∑|α|=m-1‍(DαA(x)-(DαA)B))×a~(y)dy∫ℝn‍|y||x|(∑|α|=m-1‍∥DαA∥BMO(C1+log|x|r))n/(n-β)dx)(n-β)/n≤C({+∑|α|=m-1‍(DαA(x)-(DαA)B))n/(n-β)}∑k=1+∞‍∫2k+1B∖2kB‍|x|-n|B|1/(n-β)|x|n/(β-n)×(∑|α|=m-1‍∥DαA∥BMO(C1+log|x|r)+∑|α|=m-1‍(DαA(x)-(DαA)B))n/(n-β)dx)(n-β)/n≤C({+∑|α|=m-1‍(DαA(x)-(DαA)B))n/(n-β)}∑k=1+∞|B|1/(n-β)‍|2kB|(1/(β-n))-1∫2k+1B∖2kB‍×(∑|α|=m-1‍∥DαA∥BMO(C1+log|x|r)+∑|α|=m-1‍(DαA(x)-(DαA)B))n/(n-β)dx)(n-β)/n≤C(∑|α|=m-1‍∥DαA∥BMOn/(n-β)∑k=1+∞‍(C1+log|2k+1r|r)n/(n-β)×|B|1/(n-β)|2k+1B|(1/(β-n))-1|2k+1B|∑|α|=m-1‍)(n-β)/n+(∑k=1+∞|B|1/(n-β)‍|2kB|(1/(n-β))-1∫2k+1B∖2kB‍×∑|α|=m-1‍(DαA(x)-(DαA)B)n/(n-β)dx)(n-β)/n≤C(∑|α|=m-1‍∥DαA∥BMOn/(n-β)∑k=1∞‍(k+C1)n/(n-β)|B|1/(n-β)×|2k+1B|(1/(β-n))-1|2k+1B|∑|α|=m-1‍)(n-β)/n≤C∑|α|=m-1‍∥DαA∥BMO.
Next we will give the estimates ofI2. First by a cumbersome but straightforward computation, we have
(61)|K2(AB,x,y)|≤∑k=0m-2‍∑|α|=k‍|Rm-k-1(DαAB;0,y)||x|-m+k+1.
Also noting the fact that(62)|Rm-k-1(DαAB;0,y)|≤C|y|m-k-1×∑|α|=m-1‍(1|Q~0|∫Q~0|DαA(z)-(DαA)B|q‍dz)1/q≤Crm-k-1∑|α|=m-1‍∥DαA∥BMO,
where q>n, and Q~0 is a cube centered at 0 and having diameter 5n|y|.
Thus we obtain(63)|K2(AB,x,y)|≤C∑|α|=m-1‍∥DαA∥BMO∑k=0m-2‍rm-k-1|x|-m+k+1.
So we have the following estimates ofI2:
(64)I2≤C1|x|n-β∫ℝn‍∑k=0m-2‍rm-k-1|x|-m+k+1|a~(y)|dy=C∑k=0m-2‍rm-k-1|x|-m+k+1-n+β∫B‍|a~(y)|dy≤C∑k=0m-2‍rm-k-1|x|-m+k+1-n+β.
Now we get(65)λ|{x∈ℝn∖B(0,2r):|I2|>λ}|(n-β)/n≤C∑|α|=m-1‍∥DαA∥BMO(∫ℝn∖B(0,2r)‍|I2|n/(n-β)dx)(n-β)/n≤C∑|α|=m-1‍∥DαA∥BMO×(∑j=1+∞‍∫2j+1B∖2jB‍|x|(n(1+k-m)/(n-β))-ndx∑k=0m-2‍r(m-k-1)(n/(n-β))×∑j=1+∞‍∫2j+1B∖2jB‍|x|(n(1+k-m)/(n-β))-ndx)(n-β)/n≤C∑|α|=m-1‍∥DαA∥BMO×(∑j=1∞‍|2j+1B|((1+k-m)/(n-β))-1|2j+1B|∑k=0m-2‍|B|(m-k-1)/(n-β)×∑j=1∞‍|2j+1B|((1+k-m)/(n-β))-1|2j+1B|)(n-β)/n≤C∑|α|=m-1∥DαA∥BMO‍×(∑j=1∞‍2jn(((1+k-m)/(n-β))-1+1)∑k=0m-2‍|B|((m-k-1)/(n-β))+((1+k-m)/(n-β))-1+1×∑j=1∞‍2jn(((1+k-m)/(n-β))-1+1))(n-β)/n≤C∑|α|=m-1‍∥DαA∥BMO.
ForI3, by the vanishing condition of a~(y), we can split I3 as follows:
(66)I3=1|x|n-β∫ℝn‍[∑|α|=m-1‍1α!DαAB(y)xα-∑|α|=m-1‍1α!DαAB(x)xα]×Ω(x-y)|x-y|m-1a~(y)dy≤1|x|n-β∑|α|=m-1‍1α!∫ℝn‍[DαA(y)-(DαA)B]×(x-y)αΩ(x-y)|x-y|m-1a~(y)dy+1|x|n-β∑|α|=m-1‍1α!∫ℝn‍DαAB(y)×∑γ<α‍Cαγ(x-y)γyα-γ×Ω(x-y)|x-y|m-1a~(y)dy+1|x|n-β∑|α|=m-1‍1α!∫ℝn‍xα[DαA(x)-(DαA)B]×Ω(x-y)|x-y|m-1a~(y)dy=I31+I32+I33.
For the termI31, by Lemma 12 and the size condition of a~, we obtain
(67)λ|{x∈ℝn∖B(0,2r):|I31|>λ}|≤∑|α|=m-1‍∥(DαA-(DαA)B)a~∥L1≤∑|α|=m-1‍∥DαA∥BMO.
For the termI32, by the vanishing condition of a~ and the fact Ω∈Lip1(Sn-1), we have
(68)I32=1|x|n-β∑|α|=m-1‍1α!∫ℝn‍[DαAB(y)-(DαA)B]×∑|γ|<α‍Cαγ(x-y)γyα-γ×Ω(x-y)|x-y|m-1×a~(y)dy≤C1|x|n-β∑|α|=m-1‍1α!∫ℝn‍|DαAB(y)-(DαA)B|×∑|γ|<m-1‍rm-1-|γ||x-y|1-m+|γ|×|a~(y)|dy≤C∑|γ|<m-1‍1|x|n-β|B|(m-1-|γ|)/n×∫ℝn‍∑|α|=m-1‍|DαAB(y)-(DαA)B|×|x-y|1-m+|γ||a~(y)|dy.
Thus we get(69)λ|{x∈ℝn∖B(0,2r):|I32|>λ}|(n-β)/n≤(∫ℝn∖B(0,2r)‍|I32|n/(n-β)dx)(n-β)/n≤C(∑|γ|<m-1|B|((m-1-|γ|)/n)(n/(n-β))‍∑k=1∞‍∫2k+1B∖2kB‍×(1|x|n-β∫ℝn‍∑|α|=m-1‍|DαAB(y)-(DαA)B|×|x-y|1-m+|γ||a~(y)|dy∑|α|=m-1‍)n/(n-β)dx)(n-β)/n≤C(∑|γ|<m-1‍|B|(m-1-|γ|)/(n-β)∑k=1∞|2kB|(1-m+|γ|+β-n)/(n-β)‍|2kB|×(∫B‍∑|α|=m-1‍|DαAB(y)-(DαA)B||a~(y)|dy)n/(n-β))(n-β)/n≤C∑|α|=m-1‍∥DαA∥BMO(∑|γ|<m-1‍|B|(m-1-|γ|)/(n-β)|B|(1-m+|γ|+β-n)/(n-β)×|B|1-1+1∑k=1∞‍k2kn((1-m+|γ|+β-n)/(n-β))∑|γ|<m-1‍|B|(m-1-|γ|)/(n-β)|B|(1-m+|γ|+β-n)/(n-β))(n-β)/n≤C∑|α|=m-1‍∥DαA∥BMO.
For the termI33, by the vanishing condition of a~, Ω∈Lip1(Sn-1), and (58), we have
(70)I33=1|x|n-β∑|α|=m-1‍1α!xα[DαA(x)-(DαA)B]×∫ℝn‍(Ω(x-y)|x-y|m-1-Ω(x)|x|m-1)a~(y)dy≤1|x|n-β∑|α|=m-1‍1α!xα[DαA(x)-(DαA)B]×∫ℝn‍|Ω(x-y)|x-y|m-1-Ω(x)|x-y|m-1||a~(y)|dy+1|x|n-β∑|α|=m-1‍1α!xα[DαA(x)-(DαA)B]×∫ℝn‍|Ω(x)|x-y|m-1-Ω(x)|x|m-1||a~(y)|dy≤C1|x|n-β∑|α|=m-1‍1α!|x|m-1|DαA(x)-(DαA)B|×∫B‍|y||x-y|m|a~(y)|dy.
Thus by Lemma  2.2 in [8], we obtain
(71)λ|{x∈ℝn∖B(0,2r):|I33|>λ}|(n-β)/n≤(∫ℝn∖B(0,2r)‍|I33|n/(n-β)dx)(n-β)/n≤C(∫B‍|y||x-y|m|a~(y)|dy|n/(n-β)dx∫ℝn∖B(0,2r)‍×|1|x|n-β∑|α|=m-1‍1α!|x|m-1|DαA(x)-(DαA)B|×∫B‍|y||x-y|m|a~(y)|dy|n/(n-β)dx)(n-β)/n≤C(∑|α|=m-1‍|DαA(x)-(DαA)B|n/(n-β)dx∑k=1∞‍∫2k+1∖2kB‍|x|(β-n-1)(n/(n-β))|B|1/(n-β)×∑|α|=m-1‍|DαA(x)-(DαA)B|n/(n-β)dx)(n-β)/n≤C∑|α|=m-1‍∥DαA∥BMO×(∑k=1∞‍(k+1)|2kB|(β-n-1)/(n-β)|2kB|×|B|1/(n-β)∑k=1∞‍(k+1)|2kB|(β-n-1)/(n-β)|2kB|)(n-β)/n≤C∑|α|=m-1‍∥DαA∥BMO.
Finally, we will give the estimates ofI4. By a similar argument as in the estimates of I32, we can easily get
(72)λ|{x∈ℝn∖B(0,2r):|I4|>λ}|(n-β)/n≤C∑|α|=m-1‍∥DαA∥BMO.
Combining the estimates ofI1,I2,I3, and I4, we finish the proof of Theorem 10.Proof of Theorem11.
Theorem11 was proved in [7] in the case β=0 and Ω≡1. For the case 0≤β<n and Ω≡1, we can easily prove Theorem 11 by the proof of Theorem  5.3 in [7] with minor changes. Then for the case 0≤β<n and Ω∈Lip1(Sn-1), by the main idea used in the proof of Theorem 10, we can prove Theorem 11 easily and we omit the details here.



---

*Source: 410305-2013-06-10.xml*