# Weighted Endpoint Estimates for Commutators of Riesz Transforms Associated with Schrödinger Operators

Authors: Yu Liu; Jielai Sheng; Lijuan Wang
Journal: Abstract and Applied Analysis
 (2013)

---

## Abstract

LetL=-Δ+V be a Schrödinger operator, where Δ is the laplacian on ℝn and the nonnegative potential V belongs to the reverse Hölder class Bs1 for some s1≥(n/2). Assume that ω∈A1(ℝn). Denote by HL1(ω) the weighted Hardy space related to the Schrödinger operator L=-Δ+V. Let ℛb=[b,ℛ] be the commutator generated by a function
b∈BMOθ(ℝn)
and the Riesz transform ℛ=∇(-Δ+V)-(1/2). Firstly, we show that the operator ℛ is bounded from L1(ω) into
Lweak1(ω).
Secondly, we obtain the endpoint estimates for the commutator [b,ℛ]. Namely, it is bounded from the weighted Hardy space HL1(ω) into
Lweak1(ω).


---

## Body

## 1. Introduction

LetL=-Δ+V be a Schrödinger operator, where Δ is the laplacian on ℝn and the nonnegative potential V belongs to the reverse Holder class Bs1 for some s1≥(n/2) and n≥3. In this paper, we consider the Riesz transform associated with the Schrödinger operator L as follows:
(1)ℛ=∇(-Δ+V)-1/2.Letb be a locally integrable function on ℝn and let T be a linear operator. For a suitable function f, the commutator is defined by Tbf≐[b,T]f=bT(f)-T(bf). It is well known that when T is a Calderón-Zygmund operator, Coifman et al. [1] proved that [b,T] is a bounded operator on Lp for 1<p<∞ if and only if b∈BMO(ℝn).Recently, some scholars have investigated the boundedness of the commutators generated by aBMO function b and Riesz transforms associated with the Schrödinger operator (cf. [2–8]). It follows from [9] that Riesz transform associated with the Schrödinger operator L is not a Calderón-Zygmund operator if the potential V∈Bs((n/2)<s<n). Their results imply that the boundedness of the commutators of Riesz transform associated with the Schrödinger operator L depends on the nonnegative potential V. In [5], the authors have obtained the weighted Lp, 1<p<∞, and weak LlogL the estimates for the commutator ℛb. In this paper we are interested in the weighted Hardy space estimates for ℛb, which are also the weighted endpoint estimates. It is noted that our main results generalize Theorem 2.7 and Theorem  4.1 in [3] to the weighted case and the function b that we consider belongs to a larger class than the classical BMO space.Note that a nonnegative locallyLs integrable function V on ℝn is said to belong to Bs(1<s<∞) if there exists C>0 such that the reverse Hölder inequality
(2)(1|B|∫BV(x)sdx)1/s≤C|B|∫BV(x)dx
holds for every ball B in ℝn. Obviously, Bs2⊂Bs1 if s2>s1. But it is important that the Bq class has a property of “self- improvement”; that is, if V∈Bs, then V∈Bs+ɛ for some ɛ>0.AssumeV∈Bs1 for some s1>(n/2). Then the auxiliary function ρ(x,V)=ρ(x) introduced by Shen in [9] is defined as follows:
(3)ρ(x)=1m(x,V)≐supr>0{r:1rn-2∫B(x,r)‍V(y)dy≤1},x∈ℝn.In [4], the authors define the class BMOθ(ρ) of locally integrable function b such that
(4)1|B(x,r)|∫B(x,r)‍|b(y)-bB|dy≤C(1+rρ(x))θ,for allx∈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 (4), after identifying functions that differ upon a constant. If we let θ=0 in (4), then BMOθ(ρ) is exactly the John-Nirenberg space BMO. Denote BMO∞(ρ)=⋃θ>0‍BMOθ(ρ). It is easy to see that BMO⊂BMOθ(ρ)⊂BMOθ′(ρ) for 0<θ≤θ′. Hence, BMO⊂BMO∞(ρ).Throughout this paper, we setω(E)=∫E‍ω(x)dx for any subset E⊆ℝn. Assume that the nonnegative function ω∈Lloc1(ℝn). We say that ω∈Ap(ℝn) for 1<p<∞ if there is a constant C>0 such that
(5)supB(1|B|∫B‍ω(x)dx)(1|B|∫B‍(ω(x))1-p′dx)p-1≤C,
for all balls B in ℝn, where (1/p)+(1/p′)=1.We say thatω∈A1(ℝn), if there exists a positive constant C>0, such that
(6)Mω(x)≤Cω(x),a.e.x∈ℝn,
where M is a Hardy-Littlewood operator.Given a weight functionω∈Ap(ℝn) for 1≤p<∞, as usual we denote by Lq(ω) the space of all measurable functions satisfying
(7)∥f∥Lq(ω)q=∫ℝn‍|f(x)|qω(x)dx<∞.
When q=∞, L∞(ω) will be taken to mean L∞ and ∥f∥L∞(ω)=∥f∥L∞. Moreover, denote by Lweak1(ω) the space of all measurable functions satisfying
(8)supλ>0[λω{x∈ℝn:|f(x)|≥λ}]<∞.In the rest of this paper, we always assume thatω∈A1(ℝn).BecauseV≥0 and V∈Llocn/2(ℝn), the Schrödinger operator L generates a (C0) contraction semigroup {TsL:s>0}={e-sL:s>0}. The maximal function associated with {TsL:s>0} is defined by MLf(x)=sups>0|TsLf(x)|.The Hardy spaceHL1(ℝn) associated with the Schrödinger operator L is defined in terms of the maximal function mentioned above (cf. [10]). Recently, the weighted Hardy space HL1(ω) has been established by Liu et al. [11] and it is defined as follows.Definition 1.
A functionf∈L1(ω) is said to be in HL1(ω) if the maximal function MLf belongs to L1(ω). The norm of such a function is defined by ∥f∥HL1(ω)=∥MLf∥L1(ω).Definition 2.
Let1≤q≤∞. A function a∈Lq(ω) is called an HL1,q(ω)-atom if r<ρ(x0) and the following conditions hold:(i)
suppa⊂B(x0,r),(ii)
∥a∥Lq(ω)≤ω(B(x0,r))(1/q)-1,(iii)
ifr<(ρ(x0)/4), then ∫B(x0,r)a(x)dx=0.
In [11], Liu et al. gave the following atomic decomposition for the space HL1(ω).Proposition 3.
AssumeV∈Bs1 for some s1>(n/2). Let f∈L1(ω). Then f∈HL1(ω) if and only if f can be written as f=∑jλjaj, where aj are HL1,∞(ω)-atoms, ∑j‍|λj|<∞, and the sum converges in the HL1(ω) quasinorm. Moreover,
(9)∥f∥HL1(ω)~inf{∑j|λj|},
where the infimum is taken over all atomic decompositions of f into HL1,∞-atoms.Following the above definition of atoms and the above atomic decomposition, we know that the weighted Hardy spaceHL1(ω) is not the special case of Hardy spaces established by Yang and Zhou in [12].Now we are in a position to give the main results in this paper.Theorem 4.
Letω∈A1(ℝn). Suppose V∈Bs for s>(n/2). Then,
(10)ω{x∈ℝn:|∇(-Δ+V)1/2f(x)|>α}≤Cα∥f∥L1(ω),foreveryα>0.Theorem 5.
Letb∈BMO∞(ρ). Then, for any λ>0,
(11)ω(x∈ℝn:|ℛbf(x)|>λ)≤C[b]θλ∥f∥HL1(ω),∀f∈HL1(ω),
where
θ
>
0
. Namely, the commutator
ℛ
b
is bounded from
H
L
1
(
ω
)
into
L
w
e
a
k
1
(
ω
)
.This paper is organized as follows. In Section2, we recall some basic facts to prove main results in this paper. Section 3 gives the proof of weighted estimates of Riesz transform associated with the Schrödinger operator. In Section 4, we prove Theorem 5.Throughout this paper, the letterC stands for a constant and is not necessarily the same at each occurrence. By B1~B2, we mean that there exists a constant C>1 such that (1/C)≤(B1/B2)≤C. Moreover, for the ball B=B(x,r), we denote the ball MB by MB=B(x,Mr), where M is a positive constant.


## 2. Preliminaries

Firstly, we recall some lemmas of the auxiliary functionρ(x) which have been proved by Shen in [9]. Throughout this section we always assume V∈Bs1 for some s1>(n/2).Lemma 6.
The measure
V
(
x
)
d
x
satisfies the doubling condition; that is, there exists
C
>
0
such that
(12)
∫
B
(
x
,
2
r
)
V
(
y
)
d
y
≤
C
∫
B
(
x
,
r
)
V
(
y
)
d
y
holds for all balls
B
(
x
,
r
)
in
ℝ
n
.Lemma 7.
For
0
<
r
<
R
<
∞
,
(13)
1
/
r
n
-
2
∫
B
(
x
,
r
)
V
(
y
)
d
y
≤
C
(
R
r
)
(
n
/
s
1
)
-
2
1
/
R
n
-
2
∫
B
(
x
,
R
)
V
(
y
)
d
y
.Lemma 8.
If
r
=
ρ
(
x
)
, then
(14)
1
/
r
n
-
2
∫
B
(
x
,
r
)
V
(
y
)
d
y
=
1
.
Moreover,
(15)
1
/
r
n
-
2
∫
B
(
x
,
r
)
V
(
y
)
d
y
~
1
i
f
a
n
d
o
n
l
y
i
f
r
~
ρ
(
x
)
.Lemma 9.
There exists
l
0
>
0
such that, for any
x
and
y
in
ℝ
n
,
(16)
1
C
(
1
+
|
x
-
y
|
ρ
(
x
)
)
-
l
0
≤
ρ
(
y
)
ρ
(
x
)
≤
C
(
1
+
|
x
-
y
|
ρ
(
x
)
)
l
0
/
(
l
0
+
1
)
.
In particular,
ρ
(
x
)
~
ρ
(
y
)
if
|
x
-
y
|
<
C
ρ
(
x
)
.Lemma 10.
There exists
l
1
>
0
such that
(17)
∫
B
(
x
,
R
)
V
(
y
)
|
x
-
y
|
n
-
2
d
y
≤
C
R
n
-
2
∫
B
(
x
,
R
)
V
(
y
)
d
y
≤
C
(
1
+
R
ρ
(
x
)
)
l
1
.For the proofs of Lemma6 to Lemma 10, readers can refer to [9].A ballB(x,ρ(x)) is called critical. Due to Lemma  2.3 in [10], we have Lemma 12 on ℝn.Proposition 11.
There exists a sequence of points
{
x
k
}
k
=
1
∞
in
ℝ
n
, such that the family of critical balls
Q
k
=
B
(
x
k
,
ρ
(
x
k
)
)
,
k
≥
1
,satisfiesthe following(i)
⋃k‍Qk=ℝn.(ii)
There existsN=N(ρ)such that, for everyk∈ℕ, card{j:4Qj⋂‍4Qk≠∅}≤N.Secondly, we recall the estimates of the fundamental solution of the operator-Δ+V+iτ and give the estimates of kernels of Riesz transform ℛ. Let Γ(x,y,τ) denote the fundamental solution for the operator -Δ+V+iτ, where τ∈ℝ. Clearly, Γ(x,y,τ)=Γ(y,x,-τ).Lemma 12.
Let
N
>
0
be an integer.
(1)
SupposeV∈Bs1 for s1>(n/2). Then there exists CN>0 such that, for x≠y,
(18)|Γ(x,y,τ)|≤CN(1+|x-y||τ|(1/2))N(1+|x-y|ρ(x)-1)N1|x-y|n-2.
(2)
SupposeV∈Bs1 for s1>(n/2). Then there exists CN>0 such that, for x≠y,
(19)|∇xΓ(x,y,τ)|≤CN(1+|x-y||τ|(1/2))N(1+|x-y|ρ(x)-1)N×(1|x-y|n-2∫B(y,|x-y|)V(z)dz|z-y|n-1+1|x-y|n-1).By the functional calculus, we may write(20)(-Δ+V)-(1/2)=12π∫ℝ(-iτ)-(1/2)(-Δ+V+τ)-1dτ.Letf∈C0∞(ℝn). From (-Δ+V+τ)-1f(x)=∫ℝnΓ(x,y,τ)f(y)dy, it follows that
(21)ℛf(x)=∫ℝnK(x,y)f(y)dy,
where
(22)K(x,y)=12π∫ℝ(-iτ)-(1/2)∇xΓ(x,y,τ)dτ.And the adjoint operator ofℛ is given by
(23)ℛ~f(x)=∫ℝnK~(x,y)f(y)dy,
where
(24)K~(x,y)=12π∫ℝ‍(-iτ)-(1/2)∇yΓ(y,x,τ)dτ¯.Lemma 13.
Suppose
V
∈
B
s
1
for some
s
1
>
(
n
/
2
)
. Then there exists
δ
>
0
, for any integer
N
>
0
, and
0
<
h
<
(
|
x
-
y
|
/
16
)
,
(25)
|
K
(
x
,
y
)
|
≤
C
N
(
1
+
|
x
-
y
|
ρ
(
x
)
-
1
)
N
×
(
1
|
x
-
y
|
n
-
1
∫
B
(
y
,
|
x
-
y
|
)
V
(
z
)
d
z
|
z
-
y
|
n
-
1
+
1
|
x
-
y
|
n
)
,
|
K
(
x
,
y
+
h
)
-
K
(
x
,
y
)
|
≤
C
N
(
1
+
|
x
-
y
|
ρ
(
x
)
-
1
)
N
|
h
|
δ
|
x
-
y
|
n
-
1
+
δ
×
(
∫
B
(
y
,
|
x
-
y
|
)
‍
V
(
z
)
d
z
|
z
-
y
|
n
-
1
+
1
|
x
-
y
|
)
.Lemmas12 and 13 have been proved in [9] and in [2], respectively.Thirdly, we recall some important and useful properties ofAp weights (cf. Chapter V in [13]).Proposition 14.
(i) Letω∈Ap, 1≤p<∞. Then there exist a constant C>0 and r>1 depending only on p and the Ap constant of ω, such that
(26)(1|B|∫Bωrdx)1/r≤C|B|∫Bωdx,
for any ball B.
(ii) Letω∈A1. If 1≤p<∞, then ω1-p∈Ap.At last, we review some basic facts about theBMO space BMOθ(ρ), which have been proved in [4].Proposition 15.
Let
θ
>
0
and let
1
≤
p
<
∞
. If
b
∈
B
M
O
θ
(
ρ
)
, then
(27)
(
1
|
B
(
x
,
r
)
|
∫
B
(
x
,
r
)
|
b
(
y
)
-
b
B
|
p
d
y
)
(
1
/
p
)
≤
C
[
b
]
θ
(
1
+
r
ρ
(
x
)
)
θ
′
,
for all
B
=
B
(
x
,
r
)
, with
x
∈
ℝ
n
and
r
>
0
, where
θ
′
=
(
1
+
l
0
)
θ
and
l
0
is the constant appearing in Lemma9.Lemma 16.
Letb∈BMOθ(ρ), let B=B(x0,r), and let s≥1; then
(28)(1|2kB|∫2kB|b(y)-bB|pdy)(1/p)≤C[b]θk(1+2krρ(x0))θ′,
for all
k
∈
ℕ
, with
θ
′
as in (27).


## 3. Weighted Estimates of Riesz Transforms Associated with the Schrödinger Operators

In this section, we need to prove the weighted estimates of the Riesz transform associated with the Schrödinger operator, which will be used in the proof of Theorem5.Theorem 17.
Let
ω
∈
A
1
(
ℝ
n
)
. Suppose
V
∈
B
s
for some
s
≥
(
n
/
2
)
. Then, for
p
0
′
≤
p
<
∞
,
(29)
∥
(
-
Δ
+
V
)
-
(
1
/
2
)
∇
f
∥
L
p
(
ω
)
≤
C
∥
f
∥
L
p
(
ω
)
,
where
p
0
′
=
p
0
/
(
p
0
-
1
)
and
(
1
/
p
0
)
=
(
1
/
s
)
-
(
1
/
n
)
.Proof.
By the improvement of “Bs”, V∈Bs1 for s<s1<n. It follows from the proof of Theorem  0.5 in [9] that
(30)|(-Δ+V)1/2∇f(x)|≤C{M(|f|p1′)(x)}1/p1′|(-Δ+V)1/2∇f(x)|≤+2supɛ>0|∫|y-x|>ɛK10(y,x)f(y)dy|,
where (1/p1′)=1-(1/p1)=1-(1/s1)+(1/n) and K10(x,y) is the kernel for the operator ∇(-Δ)-1/2.
Sinceω∈A1(ℝn), then it follows from Proposition  7.2 in [14] that ω∈Ap(ℝn) for any p>1.
By [13, Theorem 1, page 201] and [13, Corollary 2, page 205]
(31)∥(-Δ+V)-1/2∇f∥Lp(ω)≤C∥f∥Lp(ω),fors1′<p<∞.
Sincep1′<p0′, then
(32)∥(-Δ+V)-1/2∇f∥Lp(ω)≤C∥f∥Lp(ω),forp0′<p<∞.By duality, we have the following.Corollary 18.
Let
ω
∈
A
1
(
ℝ
n
)
. Suppose
V
∈
B
s
for some
s
≥
(
n
/
2
)
. Then, for
1
<
p
≤
p
0
,
(33)
∥
∇
(
-
Δ
+
V
)
-
1
/
2
f
∥
L
p
(
ω
)
≤
C
∥
f
∥
L
p
(
ω
)
,
where
(
1
/
p
0
)
=
(
1
/
s
)
-
(
1
/
n
)
.Proof.
It follows from (ii) in Proposition14 that ω1-p′∈Ap′ for p0′<p′<∞, where (1/p)+(1/p′)=1. By using Theorem 17, we know that
(34)∥(-Δ+V)-1/2∇f∥Lp′(ω1-p′)≤C∥f∥Lp′(ω1-p′).
Therefore, by duality and (34),
(35)∥∇(-Δ+V)-1/2f∥Lp(ω)=sup∥g∥Lp′(ω)=1|∫ℝn∇(-Δ+V)-1/2f(x)g(x)ω(x)dx|=sup∥g∥Lp′(ω)=1|∫ℝnf(y)((-Δ+V)-1/2∇)(gω)(y)dy|≤sup∥g∥Lp′(ω)=1(∫ℝn‍|f(y)|pω(y)dy)1/p×(∫ℝn|((-Δ+V)-1/2∇)(gω)(y)|p′×(ω(y))-(p′/p)dy∫ℝn|((-Δ+V)-1/2∇)(gω)(y)|p′)1/p′=sup∥g∥Lp′(ω)=1(∫ℝn‍|f(y)|pω(y)dy)1/p×(∫ℝn|((-Δ+V)-1/2∇)(gω)(y)|p′×(ω(y))1-p′dy∫ℝn|((-Δ+V))|p′)1/p′≤Csup∥g∥Lp′(ω)=1(∫ℝn‍|f(y)|pω(y)dy)1/p≤Csup∥g∥Lp′(ω)=1×(∫ℝn|g(y)|p′ω(y)dy)1/p′≤C∥f∥Lp(ω).
This completes the proof of the above corollary.In order to prove Theorem4, we need the following lemmas.Lemma 19 (see [11, Lemma 3.12]).
Letr=ρ(x). Then there exists a constant C>0, such that
(36)∫|x-y|>r|K(y,x)|ω(y)dy≤Cω(x).
By using the proof of Lemma  3.15 in [11], we immediately have the following.Lemma 20.
Let
r
=
ρ
(
x
)
. Then there exists a constant
C
>
0
, such that
(37)
∫
|
x
-
y
|
≤
r
|
K
(
y
,
x
)
-
K
0
(
y
,
x
)
|
ω
(
y
)
d
y
≤
C
ω
(
x
)
,
where
K
0
(
y
,
x
)
is the kernel of the operator
∇
(
-
Δ
)
-
(
1
/
2
)
.Now we give the proof of Theorem4.Proof of Theorem4.
We show Theorem4 by a method similar to the one used in the proof of Theorem 2 in [15]. By the Calderón-Zygmund decomposition in the proof of Theorem 3.5 on page  413 in [16], given f∈L1(ω) and α>0, we have f=f1+f2, with f2=∑k‍bk, such that the following hold.(1)
|f1(x)|≤Cα, for a.e. x∈ℝn.(2)
Eachbk is supported in a ball Bk,
(38)∫Bk|bk(x)|ω(x)dx≤Cαω(Bk),∫Bkbk(x)dx=0.(3)
∑k‍ω(Bk)≤(C/α)∥f∥L1(ω).
Becauseω∈A1(ℝn)⊆A2(ℝn), by Corollary 18, we know that ℛ is bounded on L2(ω); it is clear that
(39)ω{x∈ℝn:|ℛf1(x)|>α2}≤Cα2∥f1∥L2(ω)2≤Cα∥f∥L1(ω).
LetBk=B(xk,rk) and Ω=⋃k‍B(xk,2rk). Then
(40)ω(Ω)≤C∑kω(Bk)≤Cα∥f∥L1(ω).
We only need to considerℛf2(x) for x∈Ωc. If rk≥(1/(m(xk,V))), then (1/(m(y,V)))≤Crk for any y∈Bk. By Lemma 19, we get
(41)∫|x-xk|≥2rk|ℛbk(x)|dx≤∫|x-xk|≥2rk∫Bk|K(x,y)||bk(y)|ω(x)dydx≤C∥bk∥L1(ω).
Ifrk<1/(m(xk,V)), then 1/(m(xk,V))~1/(m(y,V)) for any y∈Bk. Since K0(x,y) is a Calderón-Zygmund kernel, by Lemma 19, Lemma 20, and [16, Lemma 3.3, page 413], we obtain
(42)∫|x-xk|≥2rk|ℛbk(x)|ω(x)dx≤∫2rk≤|x-xk|<2/(m(xk,V))|ℛbk(x)|ω(x)dx+∫|x-xk|≥2/(m(xk,V))|ℛbk(x)|ω(x)dx≤∫2rk≤|x-xk|<2/(m(xk,V))×∫Bk|K(x,y)-K0(x,y)||bk(y)|dyω(x)dx+∫2rk≤|x-xk|<2/(m(xk,V))×∫Bk|K0(x,y)-K0(x,xk)||bk(y)|dyω(x)dx+∫|x-xk|≥2/(m(xk,V))‍∫Bk|K(x,y)||bk(y)|dyω(x)dx≤C∥bk∥L1(ω).
Then(43)∫Ωc|ℛf2(x)|dx≤∑k∫(B(x,2rk))c|ℛbk(x)|ω(x)dx≤C∑k∥bk∥L1(ω)≤Cα∑kω(Bk)≤C∥f∥L1(ω).
Therefore
(44)ω{x∈Ωc:|ℛf2(x)|>α2}≤Cα∥f∥L1.
Theorem4 is proved by combination of (39), (40), and (44).


## 4. Proof of Theorem5

Forf∈HL1(ω) we can write f=∑j=-∞∞‍λjaj, where each aj is a HL1,∞(ω) atom and ∑j=-∞∞‍|λj|≤2∥f∥HL1(ω). Suppose that suppaj⊆Bj=B(xj,rj) with rj<ρ(xj). Write
(45)ℛbf(x)=∑j=-∞∞λj(b(x)-bBj)ℛaj(x)χ8Bj(x)+∑j:rj≥(ρ(xj)/4)λj(b(x)-bBj)+∑∑×ℛaj(x)χ(8Bj)c(x)+∑j:rj<(ρ(xj)/4)λj(b(x)-bBj)+∑∑×ℛaj(x)χ(8Bj)c(x)-ℛ(∑j=-∞∞λj(b-bBj)aj)(x)=A1(x)+A2(x)+A3(x)+A4(x).Using Hölder inequality, the weightedLq boundedness of ℛ, Propositions 14 and 15, for (1/q)+(1/t)+(1/ν)=1 and (1-(1/q))t=r with 1<q<p0, we have
(46)∥(b(x)-bB)ℛaj(x)χ8Bj(x)∥L1(ω)≤(∫8Bj‍|b(x)-bB|νdx)1/ν×∥ℛaj∥Lq(ω)(∫8Bj‍ω(x)rdx)1/t≤C|8Bj|1/ν|8Bj|1/t(1|8Bj|∫8Bj‍|b(x)-bB|νdx)1/ν×∥aj∥Lq(ω)(1|8Bj|∫8Bj‍ω(x)rdx)1/t≤C[b]θ|8Bj|1/ν|8Bj|1/tω(8Bj)(1/q)-1×(1|8Bj|∫8Bj‍ω(x)dx)r/t≤C[b]θ,sincerj<ρ(xj).When we consider the termA2(x), we note that ρ(xj)>rj≥ρ(xj)/4 as following:
(47)∥(b(x)-bBj)ℛaj(x)χ(8Bj)c(x)∥L1(ω)≲∫Bj|aj(y)|dy×{∫|x-xj|≥8rj|K(x,y)||b(x)-bBj|ω(x)dx}≲∫Bj|aj(y)|dy×{∫|x-xj|≥8rjCl(1+|x-y|/ρ(x))l1|x-y|n-1≲×(∫B(x,2|x-xj|)V(z)|x-z|n-1dz+1|x-y|)1|x-y|n-1×|b(x)-bBj|ω(x)dx}≲∫Bj|aj(y)|{I1(y)+I2(y)}dy.Note that|x-xj|~|x-y| and
(48)(1+|x-y|ρ(x))≥C(1+|x-xj|ρ(x))≥C(1+|x-xj|ρ(xj))1/(k0+1).Then by Lemma16 and Proposition 14, for any y∈Bj,
(49)I2(y)≲∑k=1∞∫2k+3rj≤|x-xj|<2k+4rj1(1+|x-xj|ρ(xj)-1)N/(k0+1)∑∫∑∫×1|x-xj|n|b(x)-bBj|ω(x)dx≲∑k=1∞2-((k+1)N)/(k0+1)1|B(x,2k+3rj)|≲×∫|x-xj|<2k+4rj|b-bBj|ω(x)dx≲∑k=1∞2-(((k+1)N)/(k0+1))(1|2k+4Bj|∫2k+4Bj|b-bBj|r′dx)1/r′≲×(1|2k+4Bj|∫2k+4Bjω(x)rdx)1/r≲∑k=1∞2-((k+1)N)/(k0+1)[b]θk(1+2k+4rjρ(xj))(k0+1)θ≲×(1|2k+4Bj|∫2k+4Bj‍ω(x)dx)≲∑k=1∞2-((k+1)N)/(k0+1)[b]θk(1+2k+4rjρ(xj))(k0+1)θ≲×(1|2k+5Bj|∫B(y,2k+5rj)ω(x)dx)≲∑k=1∞2-((k+1)N)/(k0+1)[b]θk(1+2k+4rjρ(xj))(k0+1)θM(w)(y)≤C[b]θM(ω)(y),
where we have chosen N large enough.Similarly, for(1/p)+(1/q)+(1/r)=1 and (1/s1)=(1/p)+(1/n), using Hölder inequality, Proposition 14, and the boundedness of the fractional integral ℐ1, we obtain, for any y∈Bj,
(50)I1(y)≲∑k=1∞∫2k+3rj≤|x-xj|<2k+4rj1(1+|x-xj|ρ(xj)-1)N/(k0+1)×|b(x)-bBj||x-xj|n-1ℐ1(Vχ2k+4Bj)(x)ω(x)dx≲∑k=1∞2-((k+1)N)/(k0+1)1(2k+3rj)n-1≲×∫|x-xj|<2k+4rjℐ1(Vχ2k+4Bj)(x)|b-bBj|ω(x)dx≲∑k=1∞2-((k+1)N)/(k0+1)(2k+3rj)1-n+(n/q)+(n/r)≲×(1(2k+3rj)n∫|x-xj|<2k+4rj|b-bBj|qdx)1/q×(1|2k+3Bj|∫2k+3Bjω(x)rdx)1/r×∥ℐ1(Vχ2k+4Bj)∥p≲∑k=1∞2-((k+1)N)/(k0+1)(2k+3rj)2-(n/s1)≲×(1(2k+3rj)n∫|x-xj|<2k+4rj|b-bBj|qdx)1/q×(1|2k+3Bj|∫2k+3Bj‍ω(x)dx)∥Vχ2k+4Bj∥s1≲∑k=1∞2-((k+1)N)/(k0+1)[b]θk(1+2k+4rjρ(xj))(k0+1)θ≲×(1|2k+5Bj|∫B(y,2k+5rj)ω(x)dx)≲×(2k+3rj)2-n/s1∥Vχ2k+4Bj∥s1≲∑k=1∞2-((k+1)N)/(k0+1)[b]θk(1+2k+4rjρ(xj))(k0+1)θ≲×M(ω)(y)1(2k+3rj)n-2∫2k+3BjVdx≲∑k=1∞2-((k+1)N)/(k0+1)[b]θk(1+2k+4rjρ(xj))(k0+1)θ≲×(1+2k+4rjρ(xj))l1M(ω)(y)≤C[b]θM(ω)(y),
where we have chosen N large enough and we have used Lemma 10 in the sixth inequality above.Therefore,(51)∫Bj|aj(y)|{I1(y)+I2(y)}dy≤C[b]θ∫Bj|aj(y)|M(ω)(y)dy≤C[b]θ∫Bj|aj(y)|ω(y)dy≤C[b]θ,
where we have used the fact that ω∈A1.Thus, ifρ(xj)>rj≥(ρ(xj))/4, then
(52)∥(b(x)-bBj)ℛaj(x)χ(8Bj)c(x)∥L1(ω)≤C[b]θ.ForA3, by using the vanishing condition of aj and Lemma 13, then
(53)∥(b(x)-bBj)ℛaj(x)χ(8Bj)c(x)∥L1(ω)≲∫Bj|aj(y)|dy×{∫|x-xj|≥8rj|K(x,y)-K(x,xj)|∫×|b(x)-bBj|ω(x)dx}≲∫Bj|aj(y)|dy×{∫|x-xj|≥8rjCl(1+|x-y|/ρ(x))N|y-xj|δ|x-y|n-1+δ×(∫B(x,2|x-y|)|V(z)||x-z|n-1dz+1|x-y|)∫×|b(x)-bBj|ω(x)dx∫|x-xj|≥8rjCl(1+|x-y|/ρ(x))N|y-xj|δ|x-y|n-1+δ}=∫Bj|aj(y)|{I~1(y)+I~2(y)}dy.
First of all, we need to obtain the following estimate:
(54)∥VχB(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|))l1,
where we have used Lemma 10 in the last inequality above.Similarly, for(1/p)+(1/q)+(1/r)=1 and (1/s1)=(1/p)+(1/n), using Hölder inequality, Proposition 14, and the boundedness of the fractional integral ℐ1, we obtain, for any y∈Bj,
(55)I~1(y)≲∑k=1∞∫2k+3rj≤|x-xj|<2k+4rj‍ℐ1(Vχ2k+4Bj)(x)(1+|x-xj|ρ(xj)-1)N/(l0+1)×|xj-y|δ|b(x)-bBj||x-xj|n-1|x-xj|δω(x)dx≲∑k=1∞2-(k+3)δ(2k+3rj)1-n+n/q+n/r(1+2krjρ(xj)-1)N/(l0+1)×(1|B(x,2k+3rj)|∫|x-xj|<2k+4rj‍|b-bBj|qdx)1/q×(1|B(x,2k+3rj)|∫|x-xj|<2k+4rj‍ω(x)rdx)1/r×∥ℐ1(Vχ2k+4Bj)∥p≲∑k=1∞2-(k+3)δ(2k+3rj)1-n/p(1+2krjρ(xj)-1)N/(l0+1)×(1|B(x,2k+3rj)|∫|x-xj|<2k+4rj‍|b-bBj|qdx)1/q×(1|B(x,2k+3rj)|∫|x-xj|<2k+4rj‍ω(x)rdx)1/r×∥Vχ2k+4Bj∥s1≲∑k=1∞2-(k+3)δ(2k+3rj)2-n/s1(1+2krjρ(xj)-1)N/(l0+1)×(1|B(x,2k+3rj)|∫|x-xj|<2k+4rj‍|b-bBj|qdx)1/q×(1|B(x,2k+3rj)|∫|x-xj|<2k+4rj‍ω(x)dx)×∥Vχ2k+4Bj∥s1≲∑k=1∞2-(k+3)δ(2k+3rj)2(1+2krjρ(xj)-1)N/(l0+1)×(1|B(x,2k+3rj)|∫|x-xj|<2k+4rj‍|b-bBj|qdx)1/q×(1|B(x,2k+3rj)|∫|x-xj|<2k+4rj‍ω(x)dx)×(1|B(x,2k+3rj)|∫|x-xj|<2k+4rj‍V(z)dz)≲∑k=1∞2-(k+3)δ(2k+3rj)2(1+2krjρ(xj)-1)N/(l0+1)×(1|B(x,2k+3rj)|∫|x-xj|<2k+4rj‍|b-bBj|qdx)1/q×(1|B(y,2k+5rj)|∫|x-y|<2k+5rj‍ω(x)dx)×(1|B(x,2k+3rj)|∫|x-xj|<2k+4rj‍V(z)dz)≲∑k=1∞2-(k+3)δ[b]θk(1+2k+4rjρ(xj))(k0+1)θM(ω)(y)×(1+2krjm(xj,V))l1-(N/(l0+1))≲∑k=1∞2-(k+3)δ[b]θkM(ω)(y)≲[b]θM(ω)(y),
where we have chosen N large enough.Secondly, for anyy∈Bj,
(56)I~2(y)≲∫Bj|aj(y)|dy×∑k=1∞∫2k+3rj≤|x-xj|<2k+4rj‍Cl(1+|x-xj|ρ(xj)-1)N/(l0+1)∑×|xj-y|δ|b(x)-bBj||x-xj|n|x-xj|δω(x)dx≲∑k=1∞2-(k+3)δ(1+2k+3rjρ(xj))-N/(l0+1)1|B(x,2k+3rj)|×∫|x-xj|<2k+4rj‍|b-bBj|ω(x)dx≲∑k=1∞2-(k+3)δ(1+2k+3rjρ(xj))-N/(l0+1)×(1|2k+4Bj|∫2k+4Bj|b-bBj|r′dx)1/r′×(1|2k+4Bj|∫2k+4Bj‍ω(x)rdx)1/r≲∑k=1∞2-(k+3)δ[b]θk(1+2k+4rjρ(xj))(k0+1)θ-(N/(l0+1))×(1|2k+4Bj|∫2k+4Bj‍ω(x)dx)≲∑k=1∞2-(k+3)δ[b]θk(1|2k+5Bj|∫B(y,2k+5rj)‍ω(x)dx)≲∑k=1∞2-(k+3)δ[b]θkM(ω)(y)≲[b]θM(ω)(y),
where we have also chosen N large enough.Therefore,(57)∫Bj‍|aj(y)|{I~1(y)+I~2(y)}dy≤C[b]θ∫Bj‍|aj(y)|M(ω)(y)dy≤C[b]θ∫Bj‍|aj(y)|ω(y)dy≤C[b]θ.Therefore, ifrj≤ρ(xj)/4, then
(58)∥(b(x)-bBj)ℛaj(x)χ(8Bj)c(x)∥L1(ω)≤C[b]θ.Thus, we have(59)ω{x∈ℝn:|Ai(x)|>λ4}≤Cλ∥Ai(x)∥L1(ω)≤C[b]θλ∑j=-∞∞|λj|,i=1,2,3.
Note that, for (1/r′)+(1/r)=1,
(60)∥(b-bBj)aj∥L1(ω)≤|8Bj|(1|8Bj|∫8Bj‍|b(x)-bB|r′dx)1/r′∥aj∥L∞(ω)×(1|8Bj|∫8Bj‍ω(x)rdx)1/r≤|8Bj|(1|8Bj|∫Bj‍|b(x)-bB|r′dx)1/r′ω(Bj)-1×(1|8Bj|∫8Bj‍ω(x)dx)≤C[b]θ(1+rjρ(xj))θ′≤C[b]θ,
where rj<ρ(xj).By Theorem4, we get
(61)ω{x∈ℝn:|A4(x)|>λ4}≤Cλ∥∑j=-∞∞λj(b-b(xj)aj∥L1(ω)≤C[b]θλ∑j=-∞∞‍|λj|.
Therefore,
(62)ω{x∈ℝn:|[b,ℛ]f(x)|>λ4}≤C∑i=14ω{x∈ℝn:|Ai(x)|>λ4}≤C[b]θλ∑j=-∞∞|λj|≤C[b]θλ∥f∥HL1(ω).This completes the proof of Theorem5.



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*Source: 281562-2013-12-09.xml*