# Commutators of Higher Order Riesz Transform Associated with Schrödinger Operators

Authors: Yu Liu; Lijuan Wang; Jianfeng Dong
Journal: Journal of Function Spaces and Applications
 (2013)

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## Abstract

LetL=-Δ+V be a Schrödinger operator on ℝn (n≥3), where V≢0 is a nonnegative potential belonging to certain reverse Hölder class Bs for s≥n/2. In this paper, we prove the boundedness of commutators ℛbHf=bℛHf-ℛH(bf) generated by the higher order Riesz transform ℛH=∇2(-Δ+V)-1, where b∈BMOθ(ρ), which is larger than the space BMO(ℝn). Moreover, we prove that ℛbH is bounded from the Hardy space HL1(ℝn) into weak Lweak1(ℝn).


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## Body

## 1. Introduction

LetL=-Δ+V be a Schrödinger operator on ℝn, n≥3, where V≢0 is a nonnegative potential belonging to the reverse Hölder class Bs for some s≥n/2. In this paper, we will consider the higher order Riesz transforms associated with the Schrödinger operator L defined by ℛH≐∇2L-1 and the commutator
(1)ℛbH(f)(x)=ℛH(bf)(x)-b(x)ℛHf(x),x∈ℝn.
We also consider its dual higher order transforms associated with the Schrödinger operator L defined by ℛ~H≐L-1∇2 and the commutator
(2)ℛ~bH(f)(x)=ℛ~H(bf)(x)-b(x)ℛ~Hf(x),x∈ℝn,
where b∈BMO∞(ρ), which is larger than the space BMO(ℝn).Because the investigation of commutators of singular integral operators plays an important role in Harmonic analysis and PDE, many authors concentrate on this topic. It is well known that Coifman et al. [1] proved that [b,T] is a bounded operator on Lp for 1<p<∞ if and only if b∈BMO(ℝn) when T is a Calderón-Zygmund operator. See [2, 3] for the research development of the commutator Tb on Euclidean spaces ℝn and [4–6] on spaces of homogeneous type.In recent years, singular integral operators related to Schrödinger operators and their commutators have been brought to many scholars attention. See, for example, [7–19] and their references. Especially, Guo et al. [12] investigated the boundedness of the commutators ℛbH when b∈BMO(ℝn). But their method is not valid to prove the boundedness of the commutators ℛbH when b∈BMO∞(ρ). In fact, since
(3)ℛH=∇2(-Δ)-1(-Δ)(-Δ+V)-1=∇2(-Δ)-1(I-V(-Δ+V)-1),
then ℛbH may be written as follows:
(4)ℛbH(f)=[b,T1T2]f=[b,T1](T2f)+T1[b,T2]f,
where T1=∇2(-Δ)-1 and T2=I-V(-Δ+V)-1. If b∈BMO(ℝn), by using Corollary  1 in [12], we obtain the Lp boundedness of ℛbH. But if b∈BMOθ(ρ) and b∉BMO(ℝn), it follows from [1] that [b,T1] is not bounded on Lp, and then we cannot obtain the Lp boundedness of ℛbH.Motivated by [12, 15, 17], our aim in this paper is to investigate the Lp estimates and endpoint estimates for ℛbH when b∈BMO∞(ρ). Different from the classical higher order Riesz transform, there exist some new problems for the higher order Riesz transform ℛH. We need to obtain some new estimates for ℛH when the potential V satisfies more stronger conditions.A nonnegative locallyLs-integrable function V (1<s<∞) is called to belong to Bs if there exists a constant C>0 such that the reverse Hölder inequality
(5)(1|B(x,r)|∫B(x,r)‍Vsdy)1/s≤C(1|B(x,r)|∫B(x,r)Vdy)
holds for every ball B in ℝn.Moreover, a locally bounded nonnegative functionV∈B∞, if there exists a positive constant C such that
(6)∥V∥L∞(B(x,r))≤C|B(x,r)|∫B(x,r)‍V(y)dy
holds for every B(x,r) in ℝn and 0<r<∞.Obviously,Bs2⊂Bs1, if s2>s1. But it is important that the Bs class has a property of  ‘‘self-improvement’’; that is, if V∈Bs, then V∈Bs+ε for some ε>0. Furthermore, it is easy to see that B∞⊆Bs for any 1<s<∞.Assume thatV≥0 and V∈Llocn/2(ℝn). The Schrödinger operator L=-Δ+V generates a (C0) semigroup {TtL}t>0={e-tL}t>0. The maximal function with respect to the semigroup {TtL}t>0 is given by
(7)MLf(x)=supt>0|TtLf(x)|.
The Hardy space HL1(ℝn) associated with the Schrödinger operator L is defined as follows in terms of the maximal function mentioned earlier (cf. [20]).Definition 1.
A functionf∈L1(ℝn) is said to be in HL1(ℝn) if the semigroup maximal function MLf belongs to L1(ℝn). The norm of such a function is defined by
(8)∥f∥HL1=∥MLf∥L1.We introduce the auxiliary functionρ(x,V)=ρ(x) defined by
(9)ρ(x)=1m(x,V)≐supr>0{r:1rn-2∫B(x,r)‍V(y)dy≤1},x∈ℝn.
It is known that 0<ρ(x)<∞ for any x∈ℝn (from Lemma 8 in Section 2).Definition 2.
Let1<q≤∞. A measurable function a is called a (1,q)ρ-atom associated to the ball B(x,r) if r<ρ(x) and the following conditions hold:(i)
suppa⊂B(x,r) for some x∈ℝn and r>0,(ii)
∥a∥Lq(ℝn)≤|B(x,r)|1/q-1,(iii)
whenr<ρ(x)/4, ∫ℝn‍a(x)dx=0.The spaceHL1(ℝn) admits the following atomic decompositions (cf. [21]).Proposition 3.
Letf∈L1(ℝn). Then, f∈HL1(ℝn) if and only if f can be written as f=∑j‍λjaj, where aj are (1,q)ρ-atoms and ∑j‍|λj|<∞. Moreover,
(10)∥f∥HL1~inf{∑j‍|λj|},
where the infimum is taken over all atomic decompositions of f into HL1-atoms.Following [17], the class BMOθ(ρ) of locally integrable function b is defined as follows:
(11)1|B(x,r)|∫B(x,r)|b(y)-bB|dy≤C(1+rρ(x))θ,
for all x∈ℝn and r>0, where θ>0 and bB=(1/|B|)∫B‍b(y)dy. A norm for b∈BMOθ(ρ), denoted by [b]θ, is given by the infimum of the constants satisfying (11), after identifying functions that differ upon a constant. If we let θ=0 in (11), then BMOθ(ρ) is exactly the John-Nirenberg space BMO(ℝn). Denote that BMO∞(ρ)=⋃θ>0‍BMOθ(ρ). It is easy to see that BMO(ℝn)⊂BMOθ(ρ)⊂BMOθ′(ρ) for 0<θ≤θ′. Bongioanni et al. [17] gave some examples to clarify that the space BMO(ℝn) is a subspace of BMO∞(ρ).Letρ1(x) be the auxiliary function of |∇V(x)|. Our main results are given as follows.Theorem 4.
Suppose thatV∈Bs for some s≥n, |∇V|∈Bs1 (s1≥n/2), ρ(x)≲ρ1(x), and ρ(x)≲1. Let b∈BMO∞(ρ). The commutator ℛ~bH(f) is bounded on Lp(ℝn) for s1′<p<∞, where (1/s1)+(1/s1′)=1.By duality, we immediately have the following.Corollary 5.
Suppose thatV∈Bs for some s≥n, |∇V|∈Bs1 (s1≥n/2), ρ(x)≲ρ1(x), and ρ(x)≲1. Let b∈BMO∞(ρ). The commutator ℛbH is bounded on Lp(ℝn) for 1<p<s1.Furthermore, we get the endpoint estimate for the commutatorℛbH.Theorem 6.
Suppose thatV∈Bs for some s≥n, |∇V|∈Bs1 (s1≥n/2), ρ(x)≲ρ1(x), and ρ(x)≲1. Let b∈BMO∞(ρ). Then, for any λ>0,
(12)|{x∈ℝn:|ℛbH(f)(x)|>λ}|≲[b]θλ∥f∥HL1(ℝn),mmmmmmmmmmmmmmmmm∀f∈HL1(ℝn).
Namely, the commutator ℛbH is bounded from HL1(ℝn) into Lweak1(ℝn).This paper is organized as follows. In Section2, we collect some known facts about the auxiliary function ρ(x) and some necessary estimates for the kernel of the higher order Riesz transform ℛH. In Section 3, we give the proof of Theorems 4 and 6. Section 4 gives the corresponding results when the potential V satisfies stronger conditions. In Section 5, we give some examples for the potentials V in Theorems 4 and 6.Throughout this paper, unless otherwise indicated, we always assume that0≢V∈Bs for some s>n. We will use C to denote the positive constants, which are not necessarily same at each occurrence even be different in the same line, and may depend on the dimension n and the constant in (5) or (6). By A~B and A≲B, we mean that there exist some constants C,C′ such that 1/C≤A/B≤C and A≤C′B, respectively.


## 2. Some Lemmas

In this section, we collect some known results about auxiliary functionρ(x) and some necessary estimates for the kernel of the higher order Riesz transform in the paper.Lemma 7.
V
∈
B
s
(
s
≥
n
/
2
) is a doubling measure; that is, there exists a constant C>0 such that
(13)∫B(x,2r)‍V(y)dy≤C∫B(x,r)‍V(y)dy.
Especially, there exist constants μ≥1 and C such that
(14)∫B(x,tr)‍V(y)dy≤Ctnμ∫B(x,r)‍V(y)dy
holds for every ball B(x,r) and t>1.Lemma 8.
There exist constantsC,k0>0 such that
(15)1C(1+|x-y|ρ(x))-k0≤ρ(y)ρ(x)≤C(1+|x-y|ρ(x))k0/(k0+1).
In particular, ρ(y)~ρ(x) if |x-y|<Cρ(x).Using the Hölder inequality andBs condition, we have the following.Lemma 9.
Let(16)∫B(x,R)‍V(y)dy|x-y|n-2≤CRn-2∫B(x,R)‍V(y)dy.
Moreover, ifV∈Bn, then there exists C>0 such that
(17)∫B(x,R)‍V(y)dy|x-y|n-1≤CRn-1∫B(x,R)‍V(y)dy.Lemma 10.
(
1
) For 0<r<R<∞,
(18)1rn-2∫B(x,r)‍V(y)dy≤C(rR)2-n/s1Rn-2∫B(x,R)‍V(y)dy,1rn-2∫B(x,r)‍V(y)dy~1iffr~ρ(x).
(
2
) There exist C>0 and k0′>0 such that
(19)1Rn-2∫B(x,R)‍V(y)dy≤C(1+Rρ(x))k0′.LetΓ(x,y) be the fundamental solution of L. Then, there exists Cl>0 such that for each l>0,
(20)|Γ(x,y)|≤Cl(1+|x-y|/ρ(x))l1|x-y|n-2.
In particular, Γ(x,y)=Γ(x,y,0)=Γ(y,x,0) is the fundamental solution of the Schrödinger operator L. If V∈Bn, then there exists Cl>0 such that for each l>0,
(21)|∇Γ(x,y)|≤Cl(1+|x-y|/ρ(x))l1|x-y|n-1.
The previous facts had been obtained by Shen in [8].We denote the fundamental solution of-Δ by Γ0(x,y), which satisfies the following.(i)
There existsC>0 such that
(22)|Γ0(x,y)|≤C|x-y|n-2.(ii)
There existsC>0 such that
(23)|∇Γ0(x,y)|≤C|x-y|n-1.Lemma 11.
Suppose thatV∈Bs for some s>n and |∇V|∈Bs1 for some s1>n/2. Assume that (-Δ+V)u=0 in B(x0,2R). Then,
(24)|∇2u(x)|≲(1+Rρ(x0))k0′×supB(x0,2R)|u(y)|(∫B(x0,2R)‍|∇V(y)||x-y|n-1dy+1R2)+1R(1+Rρ(x0))k0′supB(x0,2R)|∇u(y)|.Proof.
Letϕ∈Cc∞(B(x0,2R)) such that ϕ≡1 on B(x0,(3/2)R), 0<ϕ≤1, |∇ϕ|≤CR-1, |∇2ϕ|≤CR-2, and |∇3ϕ|≤CR-3. Since
(25)u(x)ϕ(x)=∫ℝn‍Γ0(x,y)(-Δ)(uϕ)(y)dy,
then, for x∈B(x0,R),
(26)∇2u(x)=∫ℝn‍∇Γ0(x,y)(-Δ)(∇(uϕ))(y)dy=∫ℝn‍∇Γ0(x,y)(-Δ)(∇uϕ+u∇ϕ)(y)dy=∫ℝn‍∇Γ0(x,y)(-V(y)u(y)∇ϕ(y)-2∇u(y)·∇2ϕ(y)-u(y)Δ∇ϕ(y)∇2ϕ(y))dy=+∫ℝn‍∇Γ0(x,y)(-∇Vuϕ-V∇uϕ)dy.
Therefore, we have, for x∈B(x0,R),
(27)|∇2u(x)|≲∫B(x0,2R)‍|∇V(y)||u(y)||x-y|n-1dy+∫B(x0,2R)‍V(y)|∇u(y)||x-y|n-1dy≲+1Rn∫B(x0,2R)‍|V(y)u(y)|dy+1Rn+1∫B(x0,2R)‍|∇u(y)|dy≲+1Rn+2∫B(x0,2R)‍|u(y)|dy≲supB(x0,2R)|u(y)|(∫B(x0,2R)‍|∇V(y)||x-y|n-1dy≲supB(x0,2R)|u(y)|mll+1Rn∫B(x0,2R)‍|V(y)|dy+1R2∫B(x0,2R)‍|∇V(y)||x-y|n-1)≲+supB(x0,2R)|∇u(y)|(∫B(x0,2R)‍|V(y)||x-y|n-1dy+1R)≲(1+Rρ(x0))k0′supB(x0,2R)|u(y)|≲×(∫B(x0,2R)‍|∇V(y)||x-y|n-1dy+1R2)≲+1R(1+Rρ(x0))k0′supB(x0,2R)|∇u(y)|,
where we use Lemma 9 and (2) in Lemma 10 in the last step.
Therefore, we complete the proof of the lemma.Furthermore, we get the following corollary via the proof of Lemma11.Corollary 12.
Suppose thatV∈Bs for some s>n and |∇V|∈Bs1 for some s1>n/2. There exists a constant Cl>0 such that for each l>0,
(28)|∇2Γ(x,y)|≤Cl(1+|x-y|/ρ(x))l1|x-y|n-2×(∫B(x,2|x-y|)‍|∇V(z)||x-y|n-1dz+1|x-y|2).Lemma 13.
Suppose thatV∈Bs for some s>n and |∇V|∈Bs1 for some s1>n/2. There exists a constant Cl>0 such that for each l>0,
(29)|∇y2Γ(x+h,y)-∇y2Γ(x,y)|≤Cl(1+|x-y|/ρ(x))l|h|δ|x-y|n-2+δ≤×(∫B(x,2|x-y|)‍|∇V(z)||x-z|n-1dz+1|x-y|2),
where δ=1-n/t>0 for t>n.Proof.
LetR=|x-y|/4. Assume that |h|<R/2. It follows from the embedding theorem of Morrey, Corollary  1, and Remark  4.9 in [8] that
(30)|∇y2Γ(x+h,y)-∇y2Γ(x,y)|≤C|h|1-n/t(∫B(x,R)‍|∇x∇y2Γ(z,y)|tdz)1/t≤C(|h|R)1-n/t(1+Rm(x,V))k0supB(x,2R)|∇y2Γ(z,y)|≤Cl(1+|x-y|/ρ(x))l|h|δ|x-y|n-2+δ≤×(∫B(x,2|x-y|)‍|∇V(z)||x-z|n-1dz+1|x-y|2),
where δ=1-n/t>0.Similarly, we have the following.Lemma 14.
Suppose thatV∈Bs for some s>n and |∇V|∈Bs1 for some s1>n/2. There exists a constant Cl>0 such that for each l>0,
(31)|∇x2Γ(x,y+h)-∇x2Γ(x,y)|≤Cl(1+|x-y|/ρ(x))l|h|δ|x-y|n-2+δ≤×(∫B(x,2|x-y|)‍|∇V(z)||x-z|n-1dz+1|x-y|2),
where δ=1-n/t>0 for t>n.Corollary 15.
Suppose thatV∈Bs for some s>n, |∇V|∈Bs1 for some s1>n/2, and there exists a constant C such that
(32)|∇V(x)|≤Cm(x,V)3.
There exists a constant Cl>0 such that for each l>0,
(33)|∇2Γ(x,y)|≤Cl(1+|x-y|/ρ(x))l1|x-y|n.Proof.
Since|∇V(z)|≤Cm(z,V)3 for z∈B(x,2|x-y|), then by using Lemma 8,
(34)m(z,V)≲(1+|x-z|m(x,V))k0m(x,V)≲(1+|x-y|m(x,V))k0m(x,V).
Therefore, by Corollary 12,
(35)|∇2Γ(x,y)|≤Cl(1+|x-y|/ρ(x))l1|x-y|n-2≤×(∫B(x,2|x-y|)‍|∇V(z)||x-z|n-1dz+1|x-y|2)≲Cl(1+|x-y|/ρ(x))l(1+|x-y|m(x,V))3k0m(x,V)3|x-y|n-2≤×(∫B(x,2|x-y|)‍1|x-z|n-1dz+1|x-y|2)≲Cl(1+|x-y|/ρ(x))l(1+|x-y|m(x,V))3k0+3|x-y|n+1≤×(∫B(x,2|x-y|)‍1|x-z|n-1dz+1|x-y|2)≲Cl(1+|x-y|/ρ(x))l-3k0-31|x-y|n.Furthermore, we obtain the following corollary by using Corollary12 and Lemma 14.Corollary 16.
Suppose thatV∈Bs for some s>n, |∇V|∈Bs1 for some s1>n/2 and satisfies (32). There exists a constant Cl>0 such that for each l>0,
(36)|∇x2Γ(x,y+h)-∇x2Γ(x,y)|≤Cl(1+|x-y|/ρ(x))l|h|δ|x-y|n+δ,|∇x2Γ(x+h,y)-∇x2Γ(x,y)|≤Cl(1+|x-y|/ρ(x))l|h|δ|x-y|n+δ,
where δ=1-n/t>0 for t>n.Remark 17.
Following Remark 5 in [22], we know that if V is a nonnegative polynomial, condition (32) holds true. Therefore, Corollaries 15 and 16 also hold true.Lemma 18.
Suppose thatV∈Bs for some s>n, and (1/s)+(1/s′)=1.(1)
L-1(-Δ) and L-1V are bounded on the space Lp(ℝn), where 1≤p≤s.(2)
ℛH is bounded on the space Lp(ℝn) for 1<p≤s.(3)
ℛ~H is bounded on the space Lp(ℝn) for s′<p<∞.SinceℛH=∇2(-Δ)-1(-Δ)(-Δ+V)-1, by using (1) in Lemma 18, we obtain the following.Lemma 19.
Suppose thatV∈Bs for some s>n. Then, for any λ>0,
(37)|{x∈ℝn:|ℛH(f)(x)|>λ}|≤Cλ∥f∥L1(ℝn),mmmmmmmmmmlmmmmm∀f∈L1(ℝn).


### 2.1. Some Lemmas Related to BMO SpacesBMOθ(ρ)

In this section, we recall some propositions and lemmas for the BMO spacesBMOθ(ρ) in [17].A ballB(x,ρ(x)) is called critical. In [20], Dziubański and Zienkiewicz gave the following covering lemma on ℝn.Proposition 20.
There exists a sequence of points{xk}k=1∞ in ℝn, such that the family of critical balls Qk=B(xk,ρ(xk)), k≥1, satisfies the following:(i)
⋃k‍Qk=ℝn;(ii)
there existsN=N(ρ)such that for everyk∈ℕ,card{j:4Qj⋂‍4Qk≠∅}≤N.Lemma 21.
Letθ>0 and 1≤p<∞. If b∈BMOθ(ρ), then
(38)(1|B(x,r)|∫B(x,r)‍|b(y)-bB|pdy)1/p≤C[b]θ(1+rρ(x))θ′,
for all B=B(x,r), with x∈ℝn and r>0, where θ′=(1+k0)θ and k0 is the constant appearing in Lemma 8.Lemma 22.
Letb∈BMOθ(ρ), B=B(x0,r), and p≥1. Then,
(39)(1|2kB|∫2kB‍|b(y)-bB|pdy)1/p≤C[b]θk(1+2krρ(x0))θ′,
for all k∈ℕ, with θ′ as in (38).Given thatα>0, we define the following maximal functions for g∈Lloc1(ℝn) and x∈ℝn:
(40)Mρ,αg(x)=supx∈B∈ℬρ,α1|B|∫B‍|g|,Mρ,α♯g(x)=supx∈B∈ℬρ,α1|B|∫B‍|g-gB|,
where ℬρ,α={B(y,r):y∈ℝn,r≤αρ(y)}.Also, given a ballQ⊂ℝn, for g∈Lloc1(Q) and x∈Q, we define
(41)MQg(x)=supx∈B∈ℱ(Q)1|B⋂‍Q|∫B⋂‍Q‍|g|,MQ♯g(x)=supx∈B∈ℱ(Q)1|B⋂‍Q|∫B⋂‍Q‍|g-gB|,
where ℱ(Q)={B(y,r):y∈Q,r>0}.Lemma 23.
For1<p<∞, there exist β and γ such that if {Qk}k=1∞ is a sequence of balls as in Proposition 20, then
(42)∫ℝn‍|Mρ,β(g)|p≤C(∑k‍|Qk|(1|Qk|∫2Qk‍|g|)∫ℝn‍|Mρ,γ♯(g)|p+∑k‍|Qk|(1|Qk|∫2Qk‍|g|))p,
for all g∈Lloc1(ℝn).


## 2.1. Some Lemmas Related to BMO SpacesBMOθ(ρ)

In this section, we recall some propositions and lemmas for the BMO spacesBMOθ(ρ) in [17].A ballB(x,ρ(x)) is called critical. In [20], Dziubański and Zienkiewicz gave the following covering lemma on ℝn.Proposition 20.
There exists a sequence of points{xk}k=1∞ in ℝn, such that the family of critical balls Qk=B(xk,ρ(xk)), k≥1, satisfies the following:(i)
⋃k‍Qk=ℝn;(ii)
there existsN=N(ρ)such that for everyk∈ℕ,card{j:4Qj⋂‍4Qk≠∅}≤N.Lemma 21.
Letθ>0 and 1≤p<∞. If b∈BMOθ(ρ), then
(38)(1|B(x,r)|∫B(x,r)‍|b(y)-bB|pdy)1/p≤C[b]θ(1+rρ(x))θ′,
for all B=B(x,r), with x∈ℝn and r>0, where θ′=(1+k0)θ and k0 is the constant appearing in Lemma 8.Lemma 22.
Letb∈BMOθ(ρ), B=B(x0,r), and p≥1. Then,
(39)(1|2kB|∫2kB‍|b(y)-bB|pdy)1/p≤C[b]θk(1+2krρ(x0))θ′,
for all k∈ℕ, with θ′ as in (38).Given thatα>0, we define the following maximal functions for g∈Lloc1(ℝn) and x∈ℝn:
(40)Mρ,αg(x)=supx∈B∈ℬρ,α1|B|∫B‍|g|,Mρ,α♯g(x)=supx∈B∈ℬρ,α1|B|∫B‍|g-gB|,
where ℬρ,α={B(y,r):y∈ℝn,r≤αρ(y)}.Also, given a ballQ⊂ℝn, for g∈Lloc1(Q) and x∈Q, we define
(41)MQg(x)=supx∈B∈ℱ(Q)1|B⋂‍Q|∫B⋂‍Q‍|g|,MQ♯g(x)=supx∈B∈ℱ(Q)1|B⋂‍Q|∫B⋂‍Q‍|g-gB|,
where ℱ(Q)={B(y,r):y∈Q,r>0}.Lemma 23.
For1<p<∞, there exist β and γ such that if {Qk}k=1∞ is a sequence of balls as in Proposition 20, then
(42)∫ℝn‍|Mρ,β(g)|p≤C(∑k‍|Qk|(1|Qk|∫2Qk‍|g|)∫ℝn‍|Mρ,γ♯(g)|p+∑k‍|Qk|(1|Qk|∫2Qk‍|g|))p,
for all g∈Lloc1(ℝn).


## 3. Proofs of the Main Results

Firstly, in order to prove Theorem4, we need the following lemmas. As usual, for f∈Lloc1(ℝn), we denote by Mp the p-maximal function which is defined as
(43)Mpf(x)=supr>0(1|B(x,r)|∫B(x,r)‍|f(y)|pdy)1/p.Lemma 24.
Suppose thatV∈Bs for some s≥n, |∇V|∈Bs1 (s1≥n/2), ρ(x)≲ρ1(x), and ρ(x)≲1. Let b∈BMOθ(ρ). Then, there exists a constant C such that
(44)1|Q|∫Q‍|ℛ~bHf(y)|dy≤C[b]θinfy∈QMpf(y),
for all f∈Llocp(ℝn) for p>s1′ and every ball Q=B(x0,ρ(x0)).Proof.
Letf∈Lp(ℝn) and Q=B(x0,ρ(x0)). Write ℛ~bHf as
(45)ℛ~bHf=(b-bQ)ℛ~Hf-ℛ~H(f(b-bQ)).
Firstly, by the Hölder inequality with p>s1′ and Lemma 21,
(46)1|Q|∫Q‍|(b-bQ)ℛ~Hf(y)|dy≤(1|Q|∫Q‍|b-bQ|p′dy)1/p′(1|Q|∫Q‍|ℛ~Hf(y)|pdy)1/p≤C[b]θ(1|Q|∫Q‍|ℛ~Hf(y)|pdy)1/p.
If we write f=f1+f2 with f1=fχ2Q, then
(47)(1|Q|∫Q‍|ℛ~Hf1(y)|pdy)1/p≤C(1|Q|∫2Q‍|f(y)|pdy)1/p≤Cinfy∈QMpf(y),
where we use the fact that ℛ~H is bounded on Lp(ℝn) with s′<s1′<p<∞.
By Corollary12 and the Hölder inequality, we have
(48)|ℛ~Hf2(x)|=|∫|x0-z|>2ρ(x0)‍∇z2Γ(x,z)f(z)dz|≲I1(x)+I2(x),
where
(49)I1(x)=∫|x0-z|>2ρ(x0)‍|f(z)|(1+|x-z|/ρ(x))l|x-z|ndz,I2(x)=∫|x0-z|>2ρ(x0)‍|f(z)|(1+|x-z|/ρ(x))l|x-z|n-2mmmmml×∫B(z,|x-z|/4)‍|∇V(u)||u-z|n-1dudz.
For x∈Q, note that ρ(x)~ρ(x0) by using Lemma 8. We also note that |x-z|~|x0-z|. Then,
(50)|I1(x)|≤C∑k≥1‍2-lk(2kρ(x0))n∫2kρ(x0)≤|x0-z|<2k+1ρ(x0)‍|f(z)|dz≤C∑k≥1‍2-lk(2kρ(x0))n∫|x0-z|<2k+1ρ(x0)‍|f(z)|dz≤C∑k≥1‍2-lk(1|B(x0,2kρ(x0))|∫|x0-z|<2k+1ρ(x0)|f(z)|pdz)1/p≤Cinfy∈QMpf(y).
Sincex∈Q, then
(51)I2(x)≲∫|x0-z|>2ρ(x0)‍|f(z)|(1+|x0-z|/ρ(x0))l|x0-z|n-2mmmmmmmll×∫B(x0,4|x0-z|)‍|∇V(u)||u-z|n-1dudz≲∑k≥1‍2-lk(2kρ(x0))n-2∫|x0-z|<2k+1ρ(x0)‍|f(z)|dz===×∫B(x0,2k+3ρ(x0))‍|∇V(u)||u-z|n-1≲∑k≥1‍2-lk(2kρ(x0))n-2===×∫|x0-z|<2kρ(x0)‍|f(z)|ℐ1(|∇V|χB(x0,2k+3ρ(x0)))dz.
Using the Hölder inequality and the boundedness of the fractional integral ℐ1 with 1/s1=1/p′+1/n, we have
(52)∫|x0-z|<2kρ(x0)‍|f(z)|ℐ1(|∇V|χB(x0,2k+3ρ(x0)))dz×∥fχB(x0,2k+3ρ(x0))∥p∥ℐ1(|∇V|χB(x0,2k+3ρ(x0)))∥p′≤∥fχB(x0,2k+3ρ(x0))∥p∥|∇V|χB(x0,2k+3ρ(x0))∥s1.
Since |∇V|∈Bs1, we obtain
(53)∥|∇V|χB(x0,2k+3ρ(x0))∥s1≲(2kρ(x0))-n/s1′∫B(x0,2kρ(x0))‍|∇V(z)|dz≲(2kρ(x0))n-2-n/s1′1(2kρ(x0))n-2∫B(x0,2kρ(x0))‍|∇V(z)|dz≲(2kρ(x0))n-2-n/s1′(2kρ(x0)m(x0,|∇V|))k0′≲(2kρ(x0))n-2-n/s1′(2k)k0′,
where we use the assumption that m(x0,|∇V|)≲m(x0,V) and (2) in Lemma 10.
We also have(54)∥fχB(x0,2k+3ρ(x0))∥p≲(2kρ(x0))n/pinfy∈QMpf(y).
Therefore, using the fact thatn/p-n/s1′=1, we obtain
(55)I2(x)≲∑k≥1‍2-lk(2kρ(x0))n-2(2kρ(x0))n-2-n/s1′×(2k)k0′(2kρ(x0))n/pinfy∈QMpf(y)≲∑k≥1‍2-lk+kk0′+1ρ(x0)infy∈QMpf(y)≲infy∈QMpf(y),
where we choose l large enough such that the previous series converges and we use the fact that ρ(x0)≲1.
To deal with the second term of (45), we split again f=f1+f2 with f1=fχ2Q.
Firstly, using the Hölder inequality and the boundedness ofℛ~H on Lp(ℝn),
(56)1|Q|∫Q‍|ℛ~H((b-bQ)f1)(y)|dy≤(1|Q|∫Q‍|ℛ~H((b-bQ)f1)(y)|p1dy)1/p1≤(1|Q|∫2Q|((b-bQ)f)(y)|p1dy)1/p1≤(1|Q|∫Q‍|b-bQ|vdy)1/v(1|Q|∫2Q‍|f(y)|pdy)1/p≤C[b]θinfy∈QMpf(y),
where p1/p+p1/v=1, p>p1, and we have used Lemma 21 in the last inequality.
For the remaining term, we firstly see the fact thatρ(x)~ρ(x0) and |x-z|~|x0-z|. Then, we deal with
(57)|ℛ~H[f2(b-bQ)](x)|=|∫|x0-z|>2ρ(x0)‍∇z2Γ(x,z)[f2(b-bQ)](z)dz|≲I~1(x)+I~2(x),
where
(58)I~1(x)=∫|x0-z|>2ρ(x0)‍|f2(z)b-bQ||x0-z|n(1+|x0-z|/ρ(x0))ldz,I~2(x)=∫|x0-z|>2ρ(x0)‍|f2(z)(b-bQ)|(1+|x0-z|/ρ(x0))l|x0-z|n-2mmmmmmk×∫B(z,|x0-z|/4)‍|∇V(u)||u-z|n-1dudz.
By the Hölder inequality and Lemma22, we have
(59)I~1(x)≲∑k≥1‍2-lk(2kρ(x0))n∫2kρ(x0)≤|x0-z|<2k+1ρ(x0)‍|[f2(b-bQ)](z)|dz≤C∑k≥1‍2-lk(2kρ(x0))n∫|x0-z|<2k+1ρ(x0)‍|[f2(b-bQ)](z)|dz≤C∑k≥1‍2-lk(1|B(x0,2kρ(x0))|∫|x0-z|<2k+1ρ(x0)‍|f(z)|pdz)1/pmmmmk×(1|B(x0,2kρ(x0))|∫|x0-z|<2k+1ρ(x0)‍|b-bQ|p′dz)1/p′≤C∑k≥1‍2-lk+θ′kk[b]θinfy∈QMpf(y)≤Cinfy∈QMpf(y),
where 1/p+1/p′=1, and we choose l large enough. The following estimate is similar to the estimate of I2(x). We repeat the previous method.
Then,(60)I~2(x)≲∫|x0-z|>2ρ(x0)‍|f2(z)(b-bQ)|(1+|x0-z|/ρ(x0))l|x0-z|n-2mmmmmmll×∫B(x0,4|x0-z|)‍|∇V(u)||u-z|n-1dudz≲∑k≥1‍2-lk(2kρ(x0))n-2==×∫|x0-z|<2k+1ρ(x0)‍|f2(z)(b-bQ)|dz==×∫B(x0,2k+3ρ(x0))‍|∇V(u)||u-z|n-1≲∑k≥1‍2-lk(2kρ(x0))n-2mm×∫|x0-z|<2kρ(x0)‍|f2(z)(b-bQ)|mmmmmmmmm×ℐ1(|∇V|χB(x0,2k+3ρ(x0)))dz.
Using the Hölder inequality and the boundedness of the fractional integral ℐ1 with 1/s1=1/p~′+1/n, we have
(61)∫|x0-z|<2kρ(x0)‍|f2(z)(b-bQ)|ℐ1(|∇V|χB(x0,2k+3ρ(x0)))dz×∥f2(z)(b-bQ)χB(x0,2k+3ρ(x0))∥p~×∥ℐ1(|∇V|χB(x0,2k+3ρ(x0)))∥p~′≤∥f2(z)(b-bQ)χB(x0,2k+3ρ(x0))∥p~∥|∇V|χB(x0,2k+3ρ(x0))∥s1.
Since |∇V|∈Bs1, we have already obtained
(62)∥|∇V|χB(x0,2k+3ρ(x0))∥s1≲(2kρ(x0))n-2-n/s1′(2k)k0′.
Also,(63)∥f2(z)(b-bQ)χB(x0,2k+3ρ(x0))∥p~≲(2kρ(x0))n/p~≲×(1|B(x0,2kρ(x0))|∫|x0-z|<2kρ(x0)|f(z)|pdz)1/p≲×(1|B(x0,2kρ(x0))|∫|x0-z|<2kρ(x0)|b-bQ|νdz)1/ν≲2kθ′k(2kρ(x0))n/p~[b]θinfy∈QMpf(y),
where p~/p+p~/ν=1.
Therefore, using thatn/p-n/s1′=1, we obtain
(64)I2(x)≲∑k≥1‍2-lkk(2kρ(x0))n-2(2kρ(x0))n-2-n/s1′×(2k)k0′+θ′(2kρ(x0))n/p~[b]θinfy∈QMpf(y)≲∑k≥1‍k2-lk+kk0′+k+kθ′[b]θρ(x0)infy∈QMpf(y)≲[b]infy∈QMpf(y),
where we choose l large enough such that the previous series converges and we use the fact that ρ(x0)≲1.
Therefore, this completes the proof.Remark 25.
Similarly, we can conclude that the previous lemma also holds if the critical ballQ is replaced by 2Q.Lemma 26.
Suppose thatV∈Bs for some s≥n, |∇V|∈Bs1 (s1≥n/2), ρ(x)≲ρ1(x), and ρ(x)≲1. Let b∈BMOθ(ρ). Then, there exists a constant C such that
(65)∫(2B)c‍|∇z2Γ(x,z)-∇z2Γ(y,z)||b(z)-bB||f(z)|dz≤C[b]θinfy∈QMpf(y),
for all f∈Llocp(ℝn) for p>s1′ and x,y∈B=B(x0,r), with r<γρ(x0), where γ≥1.Proof.
Denote thatQ=B(x0,γρ(x0)). Note that ρ(x)~ρ(x0) and |x-z|~|x0-z|. By the estimate (29), we have
(66)∫(2B)c‍|∇z2Γ(x,z)-∇z2Γ(y,z)||b(z)-bB||f(z)|dz≲rδ∫Q∖2B‍|f(z)||b(z)-bB||x0-z|n+δdz=+rδρ(x0)l∫Qc‍|f(z)||b(z)-bB||x0-z|n+δ+ldz=+rδ∫Q∖2B‍|f(z)(b-bQ)||x0-z|n-2+δ∫B(x0,|x0-z|/4)‍|∇V(u)||u-z|n-1dudz=+rδρ(x0)l∫Qc‍|f(z)(b-bQ)||x0-z|n-2+l+δmmmmmmmmlll×∫B(x0,|x0-z|/4)‍|∇V(u)||u-z|n-1dudz=I1+I2+I3+I4.
For I1, by using the Hölder inequality and Lemma 22, we have
(67)I1≲∑j=2j0‍2-jδ|B(x0,2jr)|∫2jB‍|f(z)||b(z)-bB|dz≲∑j=2j0‍2-jδj[b]θ(1+2jrρ(x0))θ′infy∈BMpf(y)≲∑j=2∞‍2-jδj[b]θinfy∈BMpf(y)≲[b]θinfy∈BMpf(y),
where j0 is the least integer such that 2j0≥(γρ(x0))/r.
To deal withI2, using Lemma 22 and choosing l>θ′, we have
(68)I2≲ρ(x0)lrl∑j=j0-1∞‍2-j(δ+l)|B(x0,2jr)|∫2jB‍|f(z)||b(z)-bB|dz≲ρ(x0)lrl∑j=j0∞‍2-j(δ+l)j[b]θ(1+2jrρ(x0))θ′infy∈BMpf(y)≲∑j=j0∞‍j2-jδ(ρ(x0)2jr)l-θ′[b]θinfy∈BMpf(y)≲[b]θinfy∈BMpf(y),
where we use the fact that ρ(x0)/2jr≤1/γ when j>j0.
To deal withI3, by using Lemma 22 and j≤j0,
(69)I3≲1rn-2∑j=2j0‍2-j(n-2+δ)mmmlm×∫2jB‍|f(z)||b(z)-bB|ℐ1(∇Vχ2j+2B)(z)dz≲1rn-2∑j=2j0‍2-j(n-2+δ)∥fχ2jB∥pmmmlm×∥(b-bB)χ2jB∥ν∥ℐ1(∇Vχ2j+2B)∥p~′≲1rn-2∑j=2j0‍2-j(n-2+δ)j|2jB|1/p~mmmlm×[b]θinfy∈BMpf(y)∥∇Vχ2j+2B∥s1,
where 1/s1=1/p~′+1/n and 1/p+1/ν+1/p~′=1.
Since|∇V|∈Bs1, then
(70)∥∇Vχ2j+2B∥s1≲∥∇VχQ∥s1≲ρ(x0)-n/s1′∫Q‍|∇V(z)|dz≲ρ(x0)n/s1-2,
for all j≤j0, where we use the fact that m(x0,|∇V|)≲m(x0,V).
Therefore,(71)I3≲[b]θinfy∈BMpf(y)rn/p~-n+2ρ(x0)2-n/s1∑j=2j0‍j2-j(n-2+δ-n/p~)≲[b]θinfy∈BMpf(y)r(rρ(x0))2-n/s1∑j=2j0‍j2-j(n/-3+δ)≲[b]θinfy∈BMpf(y)(rρ(x0))2-n/s12j0(3-n/s1)∑j=2j0‍j2-jδ≲[b]θinfy∈BMpf(y),
where we use that r≤γρ(x0)≲1.
At last, forI4 we have, for j>j0,
(72)I4≲ρ(x0)lrn-2+l∑j=j0-1∞‍2-j(n-2+δ+l)ρ(x0)lrn-2+lmmmm×∫2jB‍|f(z)||b(z)-bB|ℐ1(∇Vχ2j+2B)(z)dz≲ρ(x0)lrn-2+l∑j=j0-1∞‍2-j(n-2+δ+l)∥fχ2jB∥pρ(x0)lrn-2+lmmmm×∥(b-bB)χ2jB∥ν∥ℐ1(∇Vχ2j+2B)∥p~′≲ρ(x0)lrn-2+l∑j=j0-1∞‍2-j(n-2+δ+l)j(2jr)θ′+n/p~ρ(x0)θ′ρ(x0)lrn-2+lmmmm×[b]θinfy∈BMpf(y)∥∇Vχ2j+2B∥s1,
where 1/s1=1/p~′+1/n and 1/p+1/ν+1/p~′=1.
Furthermore, by using Lemma7,
(73)∥∇Vχ2j+2B∥s1≲(2jr)-n/s1′∫2jB‍|∇V(z)|dz=(2jr)-n/s1′∫B(x0,(2jr/ρ(x0))ρ(x0))‍|∇V(z)|dz≲(2jr)nμ-n/s1′1ρ(x0)nμ∫Q‍|∇V(z)|dz≲(2jr)nμ-n/s1′1ρ(x0)nμρ(x0)n-2ρ(x0)n-2∫Q‍|∇V(z)|dz≲(2jr)nμ-n/s1′ρ(x0)n-2ρ(x0)nμ(1+ρ(x0)m(x0,|∇V|))k0′≲(2jr)nμ-n/s1′ρ(x0)n-2ρ(x0)nμ,
where we use the fact that m(x0,|∇V|)≲m(x0,V). Consider
(74)I4≲ρ(x0)lrn-2+l∑j=j0-1∞‍2-j(n-2+δ+l)j(2jr)θ′+n/p~ρ(x0)θ′mmmmmmmm×[b]θinfy∈BMpf(y)(2jr)nμ-n/s1′ρ(x0)n-2ρ(x0)nμ≲ρ(x0)lrn-2+l∑j=j0-1∞‍2-j(n-2+δ+l)j(2jr)θ′+1ρ(x0)θ′mmmmmmmm×[b]θinfy∈BMpf(y)(2jr)nμρ(x0)n-2ρ(x0)nμ≲ρ(x0)lrn-2+l∑j=j0-1∞‍2-j(n-2+δ+l)j(2jr)θ′+1ρ(x0)θ′mmmmmmmm×[b]θinfy∈BMpf(y)(2jr)nμρ(x0)n-2ρ(x0)nμ≲[b]θinfy∈BMpf(y)(ρ(x0)r)n-2+l-θ′-1-nμ×ρ(x0)∑j=j0-1∞‍2-j(δ+n-2+l-nμ-θ′-1)j≲[b]θinfy∈BMpf(y),
where we choose l large enough such that the previous series converges and we use the fact that ρ(x0)≲1.Proof of Theorem4.
We start with a functionf∈Lp(ℝn) for s1′<p<∞. By Lemmas 24 and 26 and Remark 25, we have
(75)∥ℛ~bHf∥Lqq≤∫ℝn‍|Mρ,β(ℛ~bHf)(x)|qdx≲∫ℝn‍|Mρ,γ♯(ℛ~bHf)(x)|qdx+C∑k‍|Qk|(1|Qk|∫2Qk|ℛ~bHf(x)|dx)q≤C∫ℝn‍|Mρ,γ♯(ℛ~bHf)(x)|qdx+C[b]θp∑k‍∫2Qk‍|Mpf(x)|qdx≤C∫ℝn‍|Mρ,γ♯(ℛ~bHf)(x)|qdx+C[b]θq∥f∥Lqq,
where we use the finite overlapping property given by Proposition 20 and the boundedness of Mp in Lq(ℝn) for p<q.
Next, we consider the term∫ℝn‍|Mρ,γ♯(ℛ~bHf)(x)|qdx. Our goal is to find a pointwise estimate of Mρ,γ♯(ℛ~bHf)(x). Let x∈ℝn and B=B(x0,r), with r<γρ(x0) such that x∈B. If f=f1+f2, with f1=fχ2B, then we write
(76)ℛ~bHf=(b-bB)ℛ~Hf-ℛ~H(f1(b-bB))-ℛ~H(f2(b-bB)).
Therefore, we need to control the mean oscillation on B of each term that we call J1, J2, and J3. By using the Hölder inequality and Lemma 21, we obtain
(77)J1≤2|B|∫B‍|(b-bB)ℛ~Hf(x)|dx≲(2|B|∫B‍|b-bB|p′dx)1/p′(1|B|∫B|ℛ~Hf(x)|pdx)1/p≤C[b]θMp(ℛ~Hf)(x),
since r/ρ(x0)<γ.
To estimateJ2, let 1<p~<p. Then,
(78)J2≤2|B|∫B‍|ℛ~H[(b-bB)f1](x)|dx≲(1|B|∫B‍|ℛ~H[(b-bB)f1](x)|p~dx)1/p~≲(1|B|∈ℛbH|(b-bB)f1(x)|s~dx)1/s~≲(1|B|∫2B|b-bB|vdx)1/v(1|B|∫2B|f(x)|pdx)1/p≲[b]θMp(f)(x),
where v=pp~/(p-p~).
ForJ3, by Lemma 26, we obtain
(79)J3≤C1|B|2∬B|(ℛ~Hf2(b-bB))ℛ~H(f2(b-bB))(u)C1|B|2mkm-(ℛ~Hf2(b-bB))(y)|dudy≤C1|B|2∬B∫(2B)c|∇2Γ(u,z)-∇2Γ(y,z)|C1|B|2mmmk×|b(z)-bB||f(z)|dzdudy≤C[b]θMpf(x).
Therefore, we have proved that
(80)|Mρ,γ♯(ℛ~bHf)(x)|≤C[b]θ(Mpℛ~Hf(x)+Mpf(x)).
Then, we have obtained the desired result.Proof of Theorem6.
Forf∈HL1(ℝn), we can write f=∑j=-∞∞‍λjaj, where each aj is a (1,q)ρ atom and ∑j=-∞∞‍|λj|≤2∥f∥HL1. Suppose that suppaj⊆Bj=B(xj,rj) with rj<ρ(xj). Write
(81)ℛbHf(x)=∑j=-∞∞‍λj(b(x)-bBj)ℛHaj(x)χ8Bj(x)+∑j:rj≥ρ(xj)/4‍λj(b(x)-bBj)ℛHaj(x)χ(8Bj)c(x)+∑j:rj<ρ(xj)/4‍λj(b(x)-bBj)ℛHaj(x)χ(8Bj)c(x)-ℛH(∑j=-∞∞‍λj(b-bBj)aj)(x)=A1(x)+A2(x)+A3(x)+A4(x).
Using the Hölder inequality, the(Lq,Lq) boundedness of ℛH with 1<q<s, and Lemma 21,
(82)∥(b(x)-bB)ℛHaj(x)χ8Bj(x)∥L1(ℝn)≤(∫8Bj‍|b(x)-bB|q′dx)1/q′∥ℛHaj∥Lq≤(∫8Bj‍|b(x)-bB|q′dx)1/q′∥aj∥Lq≲(1|Bj|∫8Bj‍|b(x)-bB|q′dx)1/q′≲C[b]θ,
since rj<ρ(xj).
When we consider the termA2(x), we note that ρ(xj)>rj≥ρ(xj)/4. Consider
(83)∥(b(x)-bBj)ℛHaj(x)χ(8Bj)c(x)∥L1(ℝn)≲∫Bj‍|aj(y)|dy{∫|x-xj|≥8rj‍|∇2Γ(x,y)||b(x)-bBj|dx}≲∫Bj‍|aj(y)|dy×{∫|x-xj|≥8rj‍Cl(1+|x-y|/ρ(x))l1|x-y|n-2mmmmmmmm×(∫B(x,2|x-y|)‍|∇V(z)||x-z|n-1dz+1|x-y|2)mmmmmmmm×|b(x)-bBj|dx∫|x-xj|≥8rj‍Cl(1+(|x-y|/ρ(x)))l}≲∫Bj‍|aj(y)|dy{I1+I2}.
Note that|x-xj|~|x-y| and
(84)(1+|x-y|ρ(x))≥C(1+|x-xj|ρ(x))≥C(1+|x-xj|ρ(xj))1/(k0+1).
Then, by Lemma 22,
(85)I2≲∑k=1∞‍∫2k+3rj≤|x-xj|<2k+4rj‍1(1+|x-xj|ρ(xj)-1)l/(k0+1)mmmmmmmmmmmmk×1|x-xj|n|b(x)-bBj|dx≲∑k=1∞‍2-(k+1)l/(k0+1)1|B(x,2k+3rj)|mkkm×∫|x-xj|<2k+4rj‍|b-bBj|dx≲∑k=1∞‍2-(k+1)l/(k0+1)1|B(x,2k+4rj)|mkkm×∫|x-xj|<2k+4rj‍|b-bBj|dx≲∑k=1∞‍2-(k+1)l/(k0+1)[b]θk(1+2k+4rjρ(xj))(k0+1)θ≤C[b]θ,
where we choose l large enough.
Similarly,(86)I1≲∑k=1∞‍∫2k+3rj≤|x-xj|<2k+4rj‍1(1+|x-xj|ρ(xj)-1)l/(k0+1)mmmmmmmmmmmmmk×1|x-xj|n-2ℐ1(∇Vχ2k+4Bj)mmmmmmmmmmmmmk×(x)|b(x)-bBj|dx≲∑k=1∞‍2-(k+1)l/(k0+1)1(2k+3rj)n-2mmmm×∫|x-xj|<2k+4rj‍ℐ1(∇Vχ2k+4Bj)(x)|b-bBj|dx≲∑k=1∞‍2-(k+1)l/(k0+1)(2k+3rj)2mmmm×(1(2k+3rj)n∫|x-xj|<2k+4rj‍|b-bBj|qdx)1/qmmmm×(2k+3rj)-n/q′∥ℐ1(∇Vχ2k+4Bj)∥q′≲∑k=1∞‍2-(k+1)l/(k0+1)(2k+3rj)2mmmm×(1(2k+3rj)n∫|x-xj|<2k+4rj|b-bBj|qdx)1/qmmmm×(2k+3rj)-n/q′∥∇Vχ2k+4Bj∥s1≲∑k=1∞‍2-(k+1)l/(k0+1)[b]θk(1+2k+4rjρ(xj))(k0+1)θmmmm×(2kρ(xj))n-n/s1′-n/q′(2k)k0′≲∑k=1∞‍2-(k+1)l/(k0+1)[b]θk(1+2k+4rjρ(xj))(k0+1)θmmmm×(2kρ(xj))(2k)k0′≤C[b]θ,
where we choose l large enough and we use the fact that ρ(xj)≲1.
Therefore, ifρ(xj)>rj≥ρ(xj)/4, then
(87)∥(b(x)-bBj)ℛHaj(x)χ(8Bj)c(x)∥L1(ℝn)≤C[b]θ.
ForA3, by using the vanishing condition of aj and Lemma 14, then
(88)∥(b(x)-bBj)ℛHaj(x)χ(8Bj)c(x)∥L1(ℝn)≲∫Bj‍|aj(y)|dy{∫|x-xj|≥8rj‍|∇2Γ(x,y)-∇2Γ(x,xj)|mmmmmmmmmmmmmm×|b(x)-bBj|dx∫|x-xj|≥8rj‍}≲∫Bj‍|aj(y)|dy×{∫|x-xj|≥8rj‍Cl(1+|x-y|/ρ(x))l|y-xj|δ|x-y|n-2+δmmmmmmmm×(∫B(x,2|x-y|)‍|∇V(z)||x-z|n-1dz+1|x-y|2)mmmmmmmm×|b(x)-bBj|dx|y-xj|δ|x-y|n-2+δ}=∫Bj‍|aj(y)|dy{I~1+I~2}.
First of all, we need to obtain the following new estimate:
(89)∥|∇V|χB(xj,2k+3rj)∥s1≲(2krj)-n/s1′∫B(xj,2krj)‍|∇V(z)|dz≲(2krj)n-2-n/s1′1(2krj)n-2∫B(xj,2krj)‍|∇V(z)|dz≲(2krj)n-2-n/s1′(1+2krjm(xj,|∇V|))k0′≲(2krj)n-2-n/s1′(1+2krjm(xj,V))k0′,
where we use the assumption that m(x0,|∇V|)≲m(x0,V) and (2) in Lemma 10. Consider
(90)I~1≲∫Bj‍|aj(y)|dy×∑k=1∞‍∫2k+3rj≤|x-xj|<2k+4rj‍ℐ1(∇Vχ2k+4Bj)(x)(1+|x-xj|ρ(xj)-1)l/(l0+1)mmmmmmmmmmmmmmm×|xj-y|δ|b(x)-bBj||x-xj|n-2|x-xj|δdx≲∑k=1∞‍2-(k+3)δ(2k+3rj)2-n/q′(1+2krjρ(xj)-1)l/(l0+1)==m×(1|B(x,2k+3rj)|∫|x-xj|<2k+4rj|b-bBj|qdx)1/q==m×∥ℐ1(∇Vχ2k+4Bj)∥q′≲∑k=1∞‍2-(k+3)δ(2k+3rj)2-n/q′(1+2krjρ(xj)-1)l/(l0+1)==m×(1|B(x,2k+3rj)|∫|x-xj|<2k+4rj|b-bBj|qdx)1/q==m×∥∇Vχ2k+4Bj∥s1≲∑k=1∞‍2-(k+3)δ[b]θk(1+2k+4rjρ(xj))(k0+1)θ==m×(2krj)n-2-n/s1′+2-n/q′(1+2krjm(xj,V))k0′-l/(l0+1)≲∑k=1∞‍2-(k+3)δ+k(k0+1)θ[b]θk(2krj)==m×(1+2krjm(xj,V))k0′-l/(l0+1)+1(1+2krjm(xj,V))-1≲∑k=1∞‍2-(k+3)δ+k(k0+1)θ[b]θk(2krj)(2krjm(xj,V))-1=∑k=1∞‍2-(k+3)δ+k(k0+1)θ[b]θkρ(xj)≲[b]θ,
where we use the fact that δ>(k0+1)θ, ρ(xj)≲1, and we choose l>(k0′+1)/(l0+1).
Secondly,(91)I~2≲∫Bj‍|aj(y)|dy×∑k=1∞‍∫2k+3rj≤|x-xj|<2k+4rj‍Cl(1+|x-xj|ρ(xj)-1)l/(l0+1)mmmmmmmmmmmllmmm×|xj-y|δ|b(x)-bBj||x-xj|n|x-xj|δdx≲∑k=1∞‍2-(k+3)δ1|B(x,2k+3rj)|∫|x-xj|<2k+4rj‍|b-bBj|dx≲∑k=1∞‍2-(k+3)δ1|B(x,2k+4rj)|∫|x-xj|<2k+4rj‍|b-bBj|dx≲∑k=1∞‍2-(k+3)δ[b]θk(1+2k+4rjρ(xj))(k0+1)θ≲∑k=1∞‍2-(k+3)δ+k(k0+1)θ[b]θk≲[b]θ,
where we use the fact that δ>(k0+1)θ.
Therefore, ifrj≤ρ(xj)/4, then
(92)∥(b(x)-bBj)ℛHaj(x)χ(8Bj)c(x)∥L1(ℝn)≤C[b]θ.
Thus, we have(93)|{x∈ℝn:|Ai(x)|>λ4}|≤Cλ∥Ai(x)∥L1≤C[b]θλ∑j=-∞∞‍|λj|,i=1,2,3.
Note that
(94)∥(b-bBj)aj∥L1≤(∫Bj‍|b(x)-bB|q′dx)1/q′∥aj∥Lq≤(1μ(Bj)∫Bj‍|b(x)-bB|q′dx)1/q′≤C[b]θ(1+rjρ(xj))θ′≤C[b]θ,
where rj<ρ(xj).
By the weak(1,1) boundedness of ℛH (cf. Lemma 19), we get
(95)|{x∈ℝn:|A4(x)|>λ4}|≤Cλ∥∑j=-∞∞‍λj(b-b(xj))aj∥L1≤C[b]θλ∑j=-∞∞‍|λj|.
Therefore,
(96)|{x∈ℝn:|[b,T]f(x)|>λ4}|≤C∑i=14‍|{x∈ℝn:|Ai(x)|>λ4}|≤C[b]θλ∑j=-∞∞‍|λj|≤C[b]θλ∥f∥HL1.
This completes the proof of Theorem 6.


## 4. Another Case

In this section, we obtain same results for the commutatorℛH if we impose another condition on V. Via Corollaries 15 and 16 in Section 2 and Theorems  1.2 and  1.3 in [23], we obtain the following theorems.Theorem 27.
Suppose thatV∈Bs for some s>n, |∇V|∈Bs1 for some s1>n/2, and |∇V| satisfies (32). Let b∈BMO∞(ρ). The commutator ℛbH is bounded on Lp(ℝn) for 1<p<s1.Theorem 28.
Suppose thatV∈Bs for some s>n, |∇V|∈Bs1 for some s1>n/2, and |∇V| satisfies (32). Let b∈BMO∞(ρ). Then, for any λ>0,
(97)|{x∈ℝn:|ℛbH(f)(x)|>λ}|≲[b]θλ∥f∥HL1(ℝn),mmmmmmmmmmmmmmmmmm∀f∈HL1(ℝn).
Namely, the commutator ℛbH is bounded from HL1(ℝn) into Lweak1(ℝn).Remark 29.
Following Remark 5 in [22], we know that if V is a non-negative polynomial, condition (32) holds true. Furthermore, we know that if V(x)=|p(x)|α, where p(x) is a polynomial and α>0, condition (32) holds true (see Remark 6 in [24]).


## 5. Examples

In this section, we give some examples for the potentials which can satisfy the assumption in Theorems4 and 6. We always assume that n=3 throughout this section. Denote the norm of ℝ3 by |x|=(x12+x22+x32)1/2.Example  1.  Let
(98)V(x)=1+|x|=1+(x12+x22+x32)1/2.Following [25], we know that if P(x) is a polynomial of degree k and α>0, then V(x)=|P(x)|α belongs to B∞(ℝ3). For 1<p<∞, it is easy to see that B∞(ℝ3)⊆Bp(ℝ3). Moreover, it follows from (0.14) in [26] that
(99)m(x,V)~∑|β|≤k‍|∂xβP(x)|α/(α|β|+2).Therefore,V(x)=1+|x|=1+(x12+x22+x32)1/2∈B∞(ℝ3).IfV1(x)≲V2(x), then
(100)1m(x,V2)=˙supr>0{r:1r∫B(x,r)‍V2(y)dy≤1}≲1m(x,V1)=˙supr>0{r:1r∫B(x,r)‍V1(y)dy≤1}.
Thus,
(101)∇V(x)=(x1(x12+x22+x32)1/2,x2(x12+x22+x32)1/2,mmmmmmmx3(x12+x22+x32)1/2).Therefore,|∇V(x)|=1. Clearly, |∇V(x)|∈B∞(ℝ3). So, V(x)≥|∇V(x)|. Therefore, ρ(x)≲ρ1(x). Also, since V(x)≥1, then ρ(x)≲1. Then, the potential V(x)=1+|x|=1+(x12+x22+x32)1/2 satisfies the assumption of Theorems 4 and 6.Example  2.  Let V(x)=1+|x|3=1+(x12+x22+x32)3/2. By the previous argument, we conclude that V∈B∞(ℝn).Then,(102)∇V(x)=(3(x12+x22+x32)1/2x1,3(x12+x22+x32)1/2x2,3(x12+x22+x32)1/2x3).Thus,|∇V(x)|=3|x|2=3(x12+x22+x32). Clearly, |∇V(x)|∈B∞(ℝ3). From (99), we know that m(x,V)~1+|x|3 and m(x,|∇V|)~1+|x|. Therefore, ρ(x)≲ρ1(x). Also, since V(x)≥1, then ρ(x)≲1. Then, the potential V(x)=1+|x|=1+(x12+x22+x32)3/2 satisfies the assumption of Theorems 4 and 6.Example  3.  Let V(x)=1+(x14+x24+x34)1/2. By the previous argument, we conclude that V∈B∞(ℝn).Then,(103)∇V(x)=(2(x14+x24+x34)-1/2x13,2(x14+x24+x34)-1/2x23,2(x14+x24+x34)-1/2x33).Thus,(104)|∇V(x)|=2(x16+x26+x36x14+x24+x34)1/2.
From (99), we know that m(x,V)~1+|x|. Since
(105)(x16+x26+x36)~(x12+x22+x32)3,(x14+x24+x34)~(x12+x22+x32)2,
then
(106)|∇V(x)|~(x12+x22+x32)1/2.
Thus, |∇V(x)|∈B∞(ℝ3). From (99), we know that m(x,|∇V|)~1+|x|1/2. Therefore, ρ(x)≲ρ1(x). Also, since V(x)≥1, then ρ(x)≲1. Then, the potential V(x)=1+(x14+x24+x34)1/2 satisfies the assumption of Theorems 4 and 6.



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*Source: 842375-2013-04-27.xml*