# Lipschitz Estimates for One-Sided Cohen’s Commutators on Weighted One-Sided Triebel-Lizorkin Spaces

Authors: Xiao Jin Zhang; Xian Ming Hou
Journal: Journal of Function Spaces
 (2014)

---

## Abstract

We introduce one-sided Cohen’s commutators of singular integral operators and fractional integral operators, respectively. Using the extrapolation of one-sided weights, we establish the boundedness of these operators from weighted Lebesgue spaces to weighted one-sided Triebel-Lizorkin spaces.


---

## Body

## 1. Introduction

The one-sided commutators considered in this paper are related to the commutators studied by Calderón in [1]. Cohen [2] defined the Cohen type commutators of Calderón-Zygmund singular integrals (for convenience, we only consider the 1-dimensional case) by
(1)
T
A
f
x
=
p
.
v
.
∫
R
‍
A
x
-
A
y
-
A
′
y
x
-
y
x
-
y
2
Ω
(
x
-
y
)
f
(
y
)
d
y
,
where Ω satisfies certain homogeneity, smoothness, and symmetry conditions. Chen and Lu [3] proved the boundedness of the commutators T
A from Lebesgue spaces to Triebel-Lizorkin spaces for A
(
k
)
∈
Li
p
α
(
R
)
(
k
=
0,1
). A function A
∈
Li
p
α
(
R
), 0
<
α
<
1, if it satisfies
(2)
A
Li
p
α
=
sup
⁡
x
,
h
∈
R
,
h
≠
0
A
x
+
h
-
A
x
h
α
<
∞
.
Similar to an unbounded function log
⁡
|
x
|
∈
BMO, the functions in Li
p
α are not necessarily bounded either (e.g., |
x
|
α
∈
Li
p
α). Therefore, it is also nontrivial to investigate the commutators generated by operators and Lipschitz functions.In the one-sided case, we will study the weighted boundedness of the commutatorsT
A from weighted Lebesgue spaces to weighted Triebel-Lizorkin spaces. The one-sided operators were motivated as not only the generalization of the theory of both-sided ones but also the requirement in ergodic theory. Lots of results show that, for a class of smaller operators (one-sided operators) and a class of wider weights (one-sided weights), many results in harmonic analysis still hold; see [4–14]. However, for one-sided weights, classical reverse Hölder’s inequality does not hold.A functionK is called a one-sided Calderón-Zygmund kernel (OCZK) if K satisfies
(3)
∫
a
<
|
x
|
<
b
‍
K
(
x
)
d
x
≤
C
,
0
<
a
<
b
,
(4)
K
x
≤
C
x
,
x
≠
0
,
(5)
K
x
-
y
-
K
x
≤
C
y
x
2
,
x
>
2
y
,
with support in R
-
=
(
-
∞
,
0
) or R
+
=
(
0
,
+
∞
). An example of such a kernel is
(6)
K
(
x
)
=
sin
(
log
⁡
x
)
(
x
log
⁡
x
)
χ
(
-
∞
,
0
)
(
x
)
,
where χ
E denotes the characteristic function of a set E. In [15], Aimar et al. introduced the one-sided Calderón-Zygmund singular integrals which are defined by
(7)
T
+
f
x
=
lim
⁡
ɛ
→
0
+
∫
x
+
ɛ
∞
‍
K
x
-
y
f
y
d
y
,
T
-
f
(
x
)
=
lim
⁡
ɛ
→
0
+
∫
-
∞
x
-
ɛ
‍
K
(
x
-
y
)
f
(
y
)
d
y
,
where the kernels K are OCZKs.The study of weights for one-sided operators is motivated by their natural appearance in harmonic analysis, such as the one-sided Hardy-Littlewood maximal operator:(8)
M
+
f
x
=
sup
⁡
h
>
0
1
h
∫
x
x
+
h
‍
f
y
d
y
,
M
-
f
(
x
)
=
sup
⁡
h
>
0
1
h
∫
x
-
h
x
‍
f
y
d
y
.Recently, Sawyer [13] introduced the one-sided A
p classes A
p
+, A
p
-, which are defined by
(9)
A
p
+
:
A
p
+
ω
:
=
sup
⁡
a
<
b
<
c
1
c
-
a
p
∫
a
b
‍
ω
(
x
)
d
x
∫
b
c
‍
ω
(
x
)
1
-
p
′
d
x
p
-
1
<
∞
,
A
p
-
:
A
p
-
ω
:
=
sup
⁡
a
<
b
<
c
1
c
-
a
p
∫
b
c
‍
ω
(
x
)
d
x
∫
a
b
‍
ω
(
x
)
1
-
p
′
d
x
p
-
1
<
∞
,
when 1
<
p
<
∞; also, for p
=
1,
(10)
A
1
+
:
M
-
ω
(
x
)
≤
C
ω
(
x
)
,
A
1
-
:
M
+
ω
(
x
)
≤
C
ω
(
x
)
,
for some constant C.Very recently, Fu and Lu [16] introduced a class of one-sided Triebel-Lizorkin spaces and their weighted version.Definition 1.
For0
<
α
<
1, 1
<
p
<
∞, and an appropriate weight ω, the weighted one-sided Triebel-Lizorkin spaces F
˙
p
,
+
α
,
∞
(
ω
) and F
˙
p
,
-
α
,
∞
(
ω
) are defined by
(11)
f
F
˙
p
,
+
α
,
∞
(
ω
)
≈
sup
⁡
h
>
0
1
h
1
+
α
∫
x
x
+
h
‍
f
-
f
x
,
x
+
h
L
p
ω
<
∞
,
f
F
˙
p
,
-
α
,
∞
(
ω
)
≈
sup
⁡
h
>
0
1
h
1
+
α
∫
x
-
h
x
‍
f
-
f
x
-
h
,
x
L
p
(
ω
)
<
∞
.In [16], the authors proved the boundedness for the one-sided commutators (with symbols A
∈
Li
p
α
(
R
)) of Calderón-Zygmund singular integral, T
A
+
f, and fractional integral, I
β
,
A
+
f, respectively. T
A
+
f and I
β
,
A
+
f are defined as follows:
(12)
T
A
+
f
(
x
)
=
p
.
v
.
∫
x
∞
‍
K
(
x
-
y
)
(
A
(
x
)
-
A
(
y
)
)
f
(
y
)
d
y
,
I
β
,
A
+
f
(
x
)
=
∫
x
∞
‍
1
y
-
x
1
-
β
(
A
(
x
)
-
A
(
y
)
)
f
(
y
)
d
y
.LetA
(
x
) be locally integrable functions on R. Denote by R
m
(
A
;
x
,
y
) the mth order remainder of the Taylor series of A at x about y, precisely:
(13)
R
m
(
A
;
x
,
y
)
=
A
(
x
)
-
∑
|
k
|
≤
m
-
1
‍
1
k
!
A
(
k
)
(
y
)
x
-
y
k
,
m
≥
1
.Cohen’s commutators of one-sided singular integrals are defined by(14)
T
A
,
m
+
f
x
=
p
.
v
.
∫
x
∞
‍
K
x
-
y
y
-
x
m
-
1
R
m
A
;
x
,
y
f
y
d
y
,
T
A
,
m
-
f
(
x
)
=
p
.
v
.
∫
-
∞
x
‍
K
(
x
-
y
)
x
-
y
m
-
1
R
m
(
A
;
x
,
y
)
f
(
y
)
d
y
.Obviously, whenm
=
1, T
A
,
m
+
=
T
A
+. Therefore, the results of this paper are the extension of [16].Theorem 2.
Assume that1
<
p
<
q
<
∞ and 1
/
p
-
1
/
q
=
α. Let A
(
k
)
(
x
)
∈
L
i
p
α
(
R
), k
=
0,1
,
…
,
m
-
1. Then, one gets the following.(i)
Ifω
∈
A
(
p
,
q
)
+, there exists the constant C
>
0 such that
(15)
T
A
,
m
+
f
L
q
(
ω
q
)
≤
C
f
L
p
(
ω
p
)
.
(ii)
Ifω
∈
A
(
p
,
q
)
-, there exists the constant C
>
0 such that
(16)
T
A
,
m
-
f
L
q
(
ω
q
)
≤
C
f
L
p
(
ω
p
)
.Theorem 3.
Assume that1
<
p
<
∞ and 0
<
α
<
1. Let A
(
k
)
(
x
)
∈
L
i
p
α
(
R
), k
=
0,1
,
…
,
m
-
1. Then one gets the following.(i)
Ifω
∈
A
p
+, there exists the constant C
>
0 such that
(17)
T
A
,
m
+
f
F
˙
p
,
+
α
,
∞
(
ω
)
≤
C
f
L
p
(
ω
)
.
(ii)
Ifω
∈
A
p
-, there exists the constant C
>
0 such that
(18)
T
A
,
m
-
f
F
˙
p
,
-
α
,
∞
(
ω
)
≤
C
f
L
p
(
ω
)
.The other main objects in this paper are one-sided Cohen’s commutators of fractional integral operators, which are defined by(19)
I
β
,
A
+
,
m
f
x
=
∫
x
∞
‍
1
y
-
x
m
-
β
R
m
A
;
x
,
y
f
y
d
y
,
I
β
,
A
-
,
m
f
(
x
)
=
∫
-
∞
x
‍
1
x
-
y
m
-
β
R
m
(
A
;
x
,
y
)
f
(
y
)
d
y
.
Obviously, when m
=
1, I
β
,
A
+
,
m
=
I
β
,
A
+.Theorem 4.
Assume that1
<
p
<
q
<
∞ and 1
/
p
-
1
/
q
=
α
+
β. Let A
(
k
)
(
x
)
∈
L
i
p
α
(
R
), k
=
0,1
,
…
,
m
-
1. Then one gets the following.(i)
Ifω
∈
A
(
p
,
q
)
+, there exists the constant C
>
0 such that
(20)
I
β
,
A
+
,
m
f
L
q
(
ω
q
)
≤
C
f
L
p
(
ω
p
)
.
(ii)
Ifω
∈
A
(
p
,
q
)
-, there exists the constant C
>
0 such that
(21)
I
β
,
A
-
,
m
f
L
q
(
ω
q
)
≤
C
f
L
p
(
ω
p
)
.Theorem 5.
Assume that1
<
p
,
q
<
∞
,
0
<
β
<
1, and 1
/
p
-
1
/
q
=
β. Let A
(
k
)
(
x
)
∈
L
i
p
α
(
R
), k
=
0,1
,
…
,
m
-
1. Then one gets the following.(i)
Ifω
∈
A
(
p
,
q
)
+, there exists the constant C
>
0 such that
(22)
I
β
,
A
+
,
m
f
F
˙
q
,
+
α
,
∞
(
ω
q
)
≤
C
f
L
p
(
ω
p
)
.
(ii)
Ifω
∈
A
(
p
,
q
)
-, there exists the constant C
>
0 such that
(23)
I
β
,
A
-
,
m
f
F
˙
q
,
-
α
,
∞
(
ω
q
)
≤
C
f
L
p
(
ω
p
)
.Throughout this paper the letterC will denote a positive constant that may vary from line to line.


## 2. Estimates for the One-Sided Cohen Type Commutators of Singular Integrals

This section begins with some necessary lemmas.Lemma 6 (see [17]).
IfA is a function with derivatives of order m
-
1 in L
i
p
α
(
R
)
(
0
<
α
≤
1
), then, for the mth remainder of A, there is a constant C
>
0 such that
(24)
R
m
A
;
x
,
y
≤
C
A
(
m
-
1
)
L
i
p
α
x
-
y
m
-
1
+
α
,
R
m
A
;
x
,
y
-
R
m
A
;
x
,
z
≤
C
A
(
m
-
1
)
L
i
p
α
∑
l
=
0
m
-
1
‍
x
-
z
l
z
-
y
m
-
1
-
l
+
α
,
R
m
A
;
x
,
y
-
R
m
A
;
z
,
y
≤
C
A
m
-
1
L
i
p
α
×
∑
l
=
1
m
-
1
‍
x
-
z
l
z
-
y
m
-
1
-
l
+
α
+
x
-
z
m
-
1
+
α
.The primary tool in the proof of Theorem3 is an extrapolation theorem that appeared in [18].Lemma 7.
LetT be a sublinear operator defined in C
c
∞
(
R
) satisfying
(25)
ω
T
f
∞
≤
C
ω
f
∞
,
for all ω
-
1
∈
A
1
-. Then, for 1
<
p
<
∞,
(26)
T
f
L
p
(
w
)
≤
C
f
L
p
(
w
)
holds whenever w
∈
A
p
+.Lemma 8 (see [9]).
Suppose thatω
∈
A
1
-; then there exists ε
1
>
0 such that, for all 1
<
r
≤
1
+
ε
1, w
r
∈
A
1
-.Proof of Theorem2.
For convenience, we only prove case(
i
). By Lemma 6 and assumption (4), it is easy to prove that
(27)
T
A
,
m
+
f
L
q
w
q
≤
∫
x
∞
‍
K
x
-
y
y
-
x
m
-
1
R
m
A
;
x
,
y
f
y
d
y
L
q
w
q
≤
C
∫
x
∞
‍
f
y
y
-
x
1
-
α
d
y
L
q
w
q
≤
C
I
α
+
f
L
q
w
q
≤
C
f
L
p
w
p
,
where 1
/
p
-
1
/
q
=
α. In the last inequality, we use the boundedness of I
α
+ that appeared in [19].Proof of Theorem3.
Without loss of generality, we only prove case (i). Letx
∈
R, h
>
0, and J
=
[
x
,
x
+
8
h
]. Write f
=
f
1
+
f
2, where f
1
=
f
χ
J. Then
(28)
1
h
1
+
α
∫
x
x
+
2
h
‍
T
A
,
m
+
f
(
y
)
-
T
A
,
m
+
f
x
,
x
+
2
h
d
y
≤
2
h
1
+
α
∫
x
x
+
2
h
‍
T
A
,
m
+
f
y
-
T
A
,
m
+
f
2
x
+
2
h
d
y
≤
2
h
1
+
α
∫
x
x
+
2
h
‍
T
A
,
m
+
f
1
y
d
y
+
2
h
1
+
α
∫
x
x
+
2
h
‍
T
A
,
m
+
f
2
y
-
T
A
,
m
+
f
2
x
+
2
h
d
y
≤
2
h
1
+
α
∫
x
x
+
2
h
‍
T
A
,
m
+
f
1
y
d
y
+
1
h
1
+
α
×
∫
x
x
+
2
h
‍
∫
x
+
8
h
∞
‍
K
y
-
z
z
-
y
m
-
1
R
m
(
A
;
y
,
z
)
-
K
x
+
2
h
-
z
z
-
x
+
2
h
m
-
1
K
y
-
z
z
-
y
m
-
1
×
R
m
A
;
x
+
2
h
,
z
f
z
d
z
d
y
:
=
I
(
x
)
+
I
I
(
x
)
.
To estimateI
I
(
x
), we have
(29)
K
y
-
z
z
-
y
m
-
1
R
m
A
;
y
,
z
-
K
x
+
2
h
-
z
z
-
x
+
2
h
m
-
1
R
m
A
;
x
+
2
h
,
z
≤
K
y
-
z
z
-
y
m
-
1
R
m
A
;
y
,
z
-
R
m
A
;
x
+
2
h
,
z
+
1
z
-
y
m
-
1
R
m
A
;
x
+
2
h
,
z
×
K
y
-
z
-
K
x
+
2
h
-
z
1
z
-
y
m
-
1
+
1
z
-
x
+
2
h
m
-
1
R
m
A
;
x
+
2
h
,
z
1
z
-
y
m
-
1
-
1
z
-
x
+
2
h
m
-
1
K
x
+
2
h
-
z
×
1
z
-
y
m
-
1
-
1
z
-
x
+
2
h
m
-
1
K
x
+
2
h
-
z
:
=
I
I
1
(
x
)
+
I
I
2
(
x
)
+
I
I
3
(
x
)
.
Noting the fact that|
x
+
2
h
-
z
|
~
|
y
-
z
| and using Lemma 6 and (4) of kernel K
(
x
), we consider I
I
1
(
x
), I
I
2
(
x
), and I
I
3
(
x
), respectively:
(30)
I
I
1
x
≤
C
z
-
y
m
×
∑
l
=
1
m
-
1
‍
x
+
2
h
-
y
l
z
-
x
+
2
h
m
-
1
-
l
+
α
∑
l
=
1
m
-
1
‍
+
x
+
2
h
-
y
m
-
1
+
α
≤
C
x
+
2
h
-
y
z
-
y
2
-
α
,
I
I
2
x
≤
C
(
x
+
2
h
-
y
)
z
-
x
+
2
h
m
-
3
+
α
z
-
y
m
-
1
≤
C
x
+
2
h
-
y
z
-
y
2
-
α
,
I
I
3
x
≤
C
(
x
+
2
h
-
y
)
z
-
x
+
2
h
m
-
2
+
α
z
-
y
m
≤
C
x
+
2
h
-
y
z
-
y
2
-
α
,
where we use the differential mean value theorem for I
I
3.
Combining the above estimate, we have(31)
I
I
x
≤
C
sup
⁡
h
>
0
1
h
1
+
α
∫
x
x
+
2
h
‍
∫
x
+
8
h
∞
‍
x
+
2
h
-
y
z
-
y
2
-
α
f
z
d
z
d
y
.
Consider the following two sublinear operators defined onC
c
∞
(
R
):
(32)
M
1
+
f
(
x
)
=
sup
⁡
h
>
0
1
h
1
+
α
∫
x
x
+
2
h
‍
T
A
+
f
1
y
d
y
,
M
2
+
f
(
x
)
=
sup
⁡
h
>
0
1
h
1
+
α
∫
x
x
+
2
h
‍
∫
x
+
8
h
∞
‍
x
+
2
h
-
y
z
-
y
2
-
α
f
z
d
z
d
y
.
The above inequalities imply that
(33)
1
h
1
+
α
∫
x
x
+
2
h
‍
T
A
,
m
+
f
y
-
T
A
,
m
+
f
x
,
x
+
2
h
d
y
≤
I
x
+
I
I
x
≤
C
M
1
+
f
(
x
)
+
M
2
+
f
(
x
)
.
Thus, we will discuss the boundedness of these two operators.
ForI
(
x
), by Hölder’s inequality and Theorem 2, we get
(34)
1
h
1
+
α
∫
x
x
+
2
h
‍
T
A
,
m
+
f
1
y
d
y
≤
C
h
α
1
h
∫
x
x
+
2
h
‍
T
A
,
m
+
f
1
y
r
d
y
1
/
r
≤
C
h
α
+
1
/
r
∫
x
x
+
8
h
‍
f
y
s
d
y
1
/
s
≤
C
1
8
h
∫
x
x
+
8
h
‍
f
y
s
ω
s
ω
-
s
d
y
1
/
s
≤
C
f
ω
∞
1
8
h
∫
x
x
+
8
h
‍
ω
-
s
d
y
1
/
s
≤
C
f
ω
∞
ω
-
1
(
x
)
,
where 1
/
s
-
1
/
r
=
α, ω
-
s
∈
A
1
- for ω
-
1
∈
A
1
-.
Next, we have(35)
ω
M
1
+
f
∞
≤
C
f
ω
∞
.
By Lemma 7, we obtain
(36)
M
1
+
f
L
p
(
ω
)
≤
C
f
L
p
(
ω
)
for all ω
∈
A
p
+.
ForM
2
+, set I
j
=
[
x
,
x
+
2
j
h
], j
∈
Z
+. Then(37)
1
h
1
+
α
∫
x
x
+
2
h
‍
∫
x
+
8
h
∞
‍
x
+
2
h
-
y
z
-
y
2
-
α
f
z
d
z
d
y
≤
C
h
α
∫
x
x
+
2
h
‍
∑
j
=
3
∞
‍
∫
x
+
2
j
h
x
+
2
j
+
1
h
‍
1
z
-
y
2
-
α
f
z
d
z
d
y
≤
C
h
α
-
1
∑
j
=
3
∞
‍
∫
x
+
2
j
h
x
+
2
j
+
1
h
‍
1
z
-
x
2
-
α
f
z
d
z
≤
C
h
∑
j
=
3
∞
‍
1
2
j
2
-
α
∫
x
+
2
j
h
x
+
2
j
+
1
h
‍
f
z
d
z
≤
∑
j
=
3
∞
‍
2
j
+
1
2
j
2
-
α
1
2
j
+
1
h
∫
x
x
+
2
j
+
1
h
‍
f
z
d
z
≤
C
∑
j
=
3
∞
‍
2
j
+
1
2
j
2
-
α
1
2
j
+
1
h
∫
x
x
+
2
j
+
1
h
‍
f
z
t
d
z
1
/
t
=
C
∑
j
=
3
∞
‍
2
j
+
1
2
j
(
2
-
α
)
1
2
j
+
1
h
∫
x
x
+
2
j
+
1
h
‍
f
z
t
ω
t
ω
-
t
d
z
1
/
t
≤
C
f
ω
∞
∑
j
=
3
∞
‍
2
j
+
1
2
j
2
-
α
1
2
j
+
1
h
∫
x
x
+
2
j
+
1
h
‍
ω
-
t
d
z
1
/
t
≤
C
f
ω
∞
ω
-
1
x
,
where 0
<
α
<
1, ω
-
t
∈
A
1
- for ω
-
1
∈
A
1
-. Then
(38)
ω
M
2
+
f
∞
≤
C
f
ω
∞
.
By Lemma 7, we have
(39)
M
2
+
f
L
p
(
ω
)
≤
C
f
L
p
(
ω
)
for all ω
∈
A
p
+.
Combining estimates (36) and (39), the proof is completed.


## 3. Estimates for the One-Sided Commutators of Cohen Type of Fractional Integrals

In order to prove Theorems4 and 5, we will introduce the one-sided extrapolation lemma.Lemma 9 (see [18]).
Let1
<
p
0
<
∞ and let T be sublinear operator defined in C
c
∞
(
R
) satisfying
(40)
ω
T
f
∞
≤
C
f
ω
p
0
,
for every x
∈
R and ω
∈
A
(
p
0
,
∞
)
+; then, for every 1
<
p
<
p
0, 1
/
p
-
1
/
q
=
1
/
p
0, and ω
∈
A
(
p
,
q
)
+, the inequality
(41)
ω
T
f
q
≤
C
f
ω
p
holds.Lemma 10 (see [20]).
Suppose thatω
∈
A
(
p
,
q
)
+; then ω
q
∈
A
q
+ and ω
p
∈
A
p
+  for all 1
<
p
<
q
≤
∞.Proof of Theorem4.
For convenience, we only give the proof of case (i). Using Lemma6 and the boundedness of one-sided fractional integral operators I
α
+
β
+ that appeared in [19], we get
(42)
I
β
,
A
+
,
m
f
L
q
w
q
≤
∫
x
∞
‍
1
y
-
x
m
-
β
R
m
A
;
x
,
y
f
y
d
y
L
q
w
q
≤
C
∫
x
∞
‍
1
y
-
x
1
-
α
-
β
f
y
d
y
L
q
w
q
≤
C
f
L
P
w
P
,
where 1
/
p
-
1
/
q
=
α
+
β.Proof of Theorem5.
Letx
∈
R, h
>
0. Write f
=
f
1
+
f
2, where f
1
=
f
χ
J, J
=
[
x
,
x
+
4
h
]. Then
(43)
1
h
1
+
α
∫
x
x
+
2
h
‍
I
β
,
A
+
,
m
f
y
-
I
β
,
A
+
,
m
f
x
,
x
+
2
h
d
y
≤
2
h
1
+
α
∫
x
x
+
2
h
‍
I
β
,
A
+
,
m
f
y
-
I
β
,
A
+
,
m
f
2
x
+
2
h
d
y
≤
2
h
1
+
α
∫
x
x
+
2
h
‍
I
β
,
A
+
,
m
f
1
y
d
y
+
2
h
1
+
α
∫
x
x
+
2
h
‍
I
β
,
A
+
,
m
f
2
y
-
I
β
,
A
+
,
m
f
2
x
+
2
h
d
y
.
:
=
I
I
I
(
x
)
+
I
V
(
x
)
.
By Lemma 6, we have
(44)
I
β
,
A
+
,
m
f
2
y
-
I
β
,
A
+
,
m
f
2
x
+
2
h
≤
∫
x
+
4
h
∞
‍
1
z
-
y
m
-
β
R
m
(
A
;
y
,
z
)
-
1
z
-
x
+
2
h
m
-
β
R
m
A
;
x
+
2
h
,
z
f
z
d
z
≤
C
∫
x
+
4
h
∞
‍
1
z
-
y
m
-
β
×
R
m
A
;
y
,
z
-
R
m
A
;
x
+
2
h
,
z
f
(
z
)
d
z
+
C
∫
x
+
4
h
∞
‍
R
m
A
;
x
+
2
h
,
z
×
1
z
-
y
m
-
β
-
1
z
-
x
+
2
h
m
-
β
f
z
d
z
≤
C
∫
x
+
4
h
∞
‍
x
+
2
h
-
y
z
-
y
2
-
α
-
β
f
z
d
z
.
Consider the following two sublinear operators defined onC
c
∞
(
R
):
(45)
M
3
+
f
x
=
sup
⁡
h
>
0
1
h
1
+
α
∫
x
x
+
2
h
‍
I
β
,
A
+
,
m
f
1
y
d
y
,
M
4
+
f
x
=
sup
⁡
h
>
0
1
h
1
+
α
×
∫
x
x
+
2
h
‍
∫
x
+
4
h
∞
‍
x
+
2
h
-
y
z
-
y
2
-
α
-
β
f
z
d
z
d
y
.
We conclude from (43) and (45) that
(46)
1
h
1
+
α
∫
x
x
+
2
h
‍
I
β
,
A
+
,
m
f
y
-
I
β
,
A
+
,
m
f
x
,
x
+
2
h
d
y
≤
I
I
I
(
x
)
+
I
V
(
x
)
≤
C
M
3
+
f
(
x
)
+
M
4
+
f
(
x
)
.
Ifω
∈
A
(
1
/
β
,
∞
)
+, then ω
-
1
/
1
-
β
∈
A
1
-; see [16]. By Lemma 8, there exists t
>
1, such that ω
-
t
/
(
1
-
β
)
∈
A
1
-. Let r
=
t
/
(
1
-
β
) and 1
/
s
-
1
/
r
=
β. Using the boundedness of I
β
+ from L
s to L
r, together with Lemma 6, we have
(47)
1
h
1
+
α
∫
x
x
+
2
h
‍
I
β
,
A
+
,
m
f
1
y
d
y
≤
C
h
α
1
h
∫
x
x
+
2
h
‍
I
β
,
A
+
,
m
f
1
y
r
d
y
1
/
r
≤
C
h
α
+
1
/
r
∫
x
x
+
2
h
‍
∫
y
∞
‍
R
m
A
;
y
,
z
z
-
y
m
-
β
f
1
z
d
z
r
d
y
1
/
r
≤
C
h
α
+
1
/
r
∫
x
x
+
2
h
‍
∫
y
∞
‍
h
α
z
-
y
1
-
β
f
1
z
d
z
r
d
y
1
/
r
≤
C
h
β
1
h
∫
x
x
+
4
h
‍
f
y
s
d
y
1
/
s
≤
C
h
β
1
h
∫
x
x
+
4
h
‍
f
y
s
ω
s
ω
-
s
d
y
1
/
s
≤
C
h
β
1
h
∫
x
x
+
4
h
‍
f
y
1
/
β
ω
1
/
β
d
y
β
×
1
h
∫
x
x
+
4
h
‍
ω
-
r
d
y
1
/
r
≤
C
f
ω
L
1
/
β
ω
-
1
(
x
)
.
In the last inequality, we use the fact ω
-
r
=
ω
-
t
/
(
1
-
β
)
∈
A
1
- for all ω
∈
A
(
1
/
β
,
∞
)
+.
By Lemma9, there exists p
0
=
1
/
β such that
(48)
ω
M
3
+
f
∞
≤
C
f
ω
L
1
/
β
;
thus
(49)
M
3
+
f
L
q
(
ω
q
)
≤
C
f
L
p
(
ω
p
)
for all ω
∈
A
(
p
,
q
)
+.
ForM
4
+, let I
j
=
[
x
,
x
+
2
j
h
], j
∈
Z
+. Then
(50)
1
h
1
+
α
∫
x
x
+
2
h
‍
∫
x
+
4
h
∞
‍
x
+
2
h
-
y
z
-
y
2
-
α
-
β
f
z
d
z
d
y
≤
C
h
α
∫
x
x
+
2
h
‍
∑
j
=
2
∞
‍
∫
x
+
2
j
h
x
+
2
j
+
1
h
‍
1
z
-
y
2
-
α
-
β
f
z
d
z
d
y
≤
C
h
α
-
1
∑
j
=
2
∞
‍
∫
x
+
2
j
h
x
+
2
j
+
1
h
‍
1
z
-
x
2
-
α
-
β
f
z
d
z
≤
C
h
β
∑
j
=
2
∞
‍
2
j
+
1
2
j
2
-
α
-
β
1
2
j
+
1
h
∫
x
x
+
2
j
+
1
h
‍
f
z
d
z
≤
C
h
β
∑
j
=
2
∞
‍
2
j
+
1
2
j
2
-
α
-
β
1
2
j
+
1
h
∫
x
x
+
2
j
+
1
h
‍
f
z
s
d
z
1
/
s
=
C
h
β
∑
j
=
2
∞
‍
2
j
+
1
2
j
2
-
α
-
β
×
1
2
j
+
1
h
∫
x
x
+
2
j
+
1
h
‍
f
z
s
ω
s
ω
-
s
d
z
1
/
s
≤
C
f
ω
L
1
/
β
∑
j
=
2
∞
‍
2
j
+
1
1
-
β
2
j
2
-
α
-
β
×
1
2
j
+
1
h
∫
x
x
+
2
j
+
1
h
‍
ω
-
r
d
z
1
/
r
≤
C
f
ω
L
1
/
β
ω
-
1
(
x
)
,
where ω
-
r
=
ω
-
t
/
(
1
-
β
)
∈
A
1
- (t
>
1) for all ω
∈
A
(
1
/
β
,
∞
)
+. Then
(51)
ω
M
4
+
f
∞
≤
C
f
ω
L
1
/
β
.
By Lemma 9, the inequlity
(52)
M
4
+
f
L
q
(
ω
q
)
≤
C
f
L
p
(
ω
p
)
holds for ω
∈
A
(
p
,
q
)
+.
This completes the proof of case (i). For case (ii), we omit the details since they are similar to those of the proof of (i) withω
∈
A
(
p
,
q
)
- instead of ω
∈
A
(
p
,
q
)
+.



---

*Source: 720875-2014-12-08.xml*