# Estimates of Fractional Integral Operators on Variable Exponent Lebesgue Spaces

Authors: Canqin Tang; Qing Wu; Jingshi Xu
Journal: Journal of Function Spaces
 (2016)

---

## Abstract

By some estimates for the variable fractional maximal operator, the authors prove that the fractional integral operator is bounded and satisfies the weak-type inequality on variable exponent Lebesgue spaces.


---

## Body

## 1. Introduction

It is well known that the boundedness of fractional integral operators is the focus of study in the classical Lebesgue spaces. Sobolev [1] showed that the fractional integral operator is bounded from the classical Lebesgue space L
p
(
R
n
) to L
q
(
R
n
). In 1999, Kenig and Stein [2] have obtained the boundedness of multilinear fractional integral operator I
α
,
k from L
r
1
(
R
n
)
×
L
r
2
(
R
n
)
×
⋯
×
L
r
k
(
R
n
) to L
s
(
R
n
). During the recent three decades, variable exponent function spaces have been studied extensively; see, for example, [3–15]. The characterization and the boundedness of the classical operators on variable exponent function spaces were systemically studied; see [5, 6, 16–21]. For example, Capone et al. [22] proved that Riesz fractional integral I
α satisfies the weak-type inequality on variable exponent Lebesgue spaces in 2007.Motivated by the aforementioned results, we will consider the variable fractional integral operatorI
α
(
·
) in this paper. Before stating our results, we need to recall some notions firstly.Given a bounded open setΩ
⊂
R
n and a measurable function p
(
·
)
:
Ω
→
[
1
,
∞
)
. The variable exponent Lebesgue space L
p
(
·
)
(
Ω
) is (1)
L
p
·
Ω
≔
f
is measurable
:
∫
Ω
f
x
λ
p
x
d
x
<
∞
for some
λ
>
0equipped with the norm(2)
f
p
·
,
Ω
≔
inf
⁡
λ
>
0
:
∫
Ω
f
x
λ
p
x
d
x
≤
1
.Denote byP
(
Ω
) the set of measurable functions p on Ω with values in [
1
,
∞
) satisfying(3)
1
<
p
-
≤
p
x
≤
p
+
<
∞
,where we use the standard notation (4)
p
-
≔
e
s
s
inf
x
∈
Ω
⁡
p
x
,
p
+
≔
e
s
s
sup
x
∈
Ω
⁡
p
x
.Definition 1.
(i) Ifp
(
·
) satisfies (5)
p
x
-
p
y
≤
C
log
⁡
e
+
1
/
x
-
y
,
x
,
y
∈
R
n
,
x
-
y
<
1
2
,then we say p
(
·
) satisfies the Log-Hölder condition.
(ii) Ifp
(
·
) satisfies(6)
p
x
-
p
y
≤
C
log
⁡
e
+
x
,
∀
y
≥
x
,then we say p
(
·
) satisfies the Log-Hölder decay condition.Givenα
(
x
), 0
<
α
(
x
)
<
n, the variable fractional operator can be defined by (7)
I
α
x
f
x
=
∫
Ω
f
y
x
-
y
n
-
α
x
d
y
.Now the first result is the following theorem.Theorem 2.
Given an open setΩ
⊂
R
n and α
(
·
)
:
0
<
α
-
≤
α
+
<
n, let p
(
·
) be a function from Ω to (
1
,
∞
) such that 1
<
p
-
≤
p
+
<
∞ and (
α
(
·
)
p
(
·
)
)
+
<
n. Suppose further that α
(
·
) satisfies condition (5) and p
(
·
) satisfies conditions (3), (5), and (6). Define q
(
·
) by (8)
1
p
x
-
1
q
x
=
α
x
n
,
x
∈
Ω
.Then, the fractional integral operator I
α
(
·
) satisfies the weak-type inequality (9)
sup
t
>
0
⁡
t
χ
I
α
·
f
x
>
t
q
·
,
Ω
≤
C
f
p
·
,
Ω
.Remark 3.
Ifα
(
x
)
≡
α, Theorem 2 is consistent with Theorem 1.9 in [22].Givenα
(
·
) with 0
<
α
-
≤
α
+
<
k
n, the multilinear variable fractional integral operators can be defined by(10)
I
α
·
,
k
f
1
,
…
,
f
k
x
=
∫
Ω
k
f
1
x
-
y
1
f
2
x
-
y
2
⋯
f
k
x
-
y
k
y
1
,
y
2
,
…
,
y
k
k
n
-
α
x
d
y
1
⋯
d
y
k
.The following is our second result.Theorem 4.
Letk
∈
N and α
(
·
) satisfy the Log-Hölder condition, 0
<
α
-
≤
α
+
<
k
n
. If r
i
(
·
) satisfies the Log-Hölder condition and 1
≤
(
r
i
)
-
≤
(
r
i
)
+
≤
∞, for i
=
1
,
…
,
k, and (11)
1
s
x
=
1
r
1
x
+
1
r
2
x
+
⋯
+
1
r
k
x
-
α
x
n
>
0
,then if each r
i
>
1,(12)
I
α
·
,
k
f
1
,
…
,
f
k
s
·
,
Ω
≤
C
∏
i
=
1
k
f
i
r
i
·
,
Ω
.The following sections are the proofs of Theorems2 and 4. Here we point out that C denotes a positive constant, but it may vary from line to line.


## 2. Proof of Theorem2

All cubes are assumed to have their sides parallel to the coordinate axes;Q
=
Q
(
x
0
,
r
) denotes that a cube is centered at x
0 with side length r. Q is the Lebesgue measure of Q. B
(
x
,
r
) denotes that a ball is centered at x with radius r
.Define the variable fractional maximal operator by(13)
M
α
·
f
x
≔
sup
x
∈
Q
⁡
Q
α
x
/
n
-
1
∫
Q
f
y
d
y
.If α
(
·
)
=
0, M
α
(
·
)
f
(
x
) becomes (14)
M
f
x
=
sup
x
∈
Q
⁡
1
Q
∫
Q
f
y
d
ywhich is the usual Hardy-Littlewood maximal function.Given a functionf, denote (15)
f
p
·
,
Ω
=
∫
Ω
f
y
p
y
d
y
.Lemma 5 (see [4]).
Givenp
(
·
)
:
Ω
→
[
1
,
∞
) such that p
+
<
∞, then f
p
(
·
)
,
Ω
<
C
1 if and only if |
f
|
p
(
·
)
,
Ω
<
C
2. In particular, if either constant equals 1, we can take the other equal to 1 as well.Lemma 6 (see [4]).
Given a setΩ with finite measure and exponent functions p
(
·
)
,
q
(
·
)
:
Ω
→
[
1
,
∞
) such that p
(
x
)
≤
q
(
x
), (16)
f
p
·
,
Ω
≤
C
1
+
Ω
f
q
·
,
Ω
.The next lemma is the generalized Hölder inequality on variable exponent Lebesgue spaces.Lemma 7 (see [12]).
Ifp
(
·
)
∈
P
(
Ω
), there is a constant C such that, for all f
∈
L
p
(
·
)
(
Ω
) and all g
∈
L
p
′
(
·
)
(
Ω
), (17)
∫
Ω
f
x
g
x
d
x
≤
C
f
p
·
,
Ω
g
p
′
·
,
Ω
,where 1
/
p
′
(
·
)
+
1
/
p
(
·
)
=
1.Lemma 8 (see [23]).
Given an open setΩ and a function p
(
·
)
:
Ω
→
[
1
,
∞
) which satisfies the Log-Hölder condition, then, for any ball B such that B
∩
Ω
>
0, (18)
B
p
-
B
∩
Ω
-
p
+
B
∩
Ω
≤
C
.Lemma 9 (see [22]).
Given an open setΩ and a function p
(
·
)
:
Ω
→
[
1
,
∞
)
,
p
+
<
∞ which satisfies the Log-Hölder condition, then, for any ball B such that B
∩
Ω
>
0 and B
≤
1, (19)
χ
B
∩
Ω
p
·
,
Ω
≤
C
B
1
/
p
x
.Here and below, fort
>
0, let (20)
R
t
x
=
e
+
x
-
t
n
.Lemma 10 (see [22]).
Given a setG and two nonnegative functions r
(
·
) and s
(
·
), suppose that, for each y
∈
G, (21)
s
y
-
r
y
≤
C
log
⁡
e
+
y
.Then, there exists a constant C
t such that, for every function f obeying f
y
≤
1
,
y
∈
G, (22)
∫
G
f
y
r
y
d
y
≤
C
t
∫
G
f
y
s
y
d
y
+
∫
G
R
t
y
r
-
G
d
y
.Lemma 11.
Givenα
(
·
)
:
0
<
α
-
≤
α
+
<
n, fix ϵ
,
0
<
ϵ
<
m
a
x
(
α
-
,
n
-
α
+
). Then, there exists a constant C
=
C
(
α
-
,
n
,
ϵ
) such that, for all f
∈
L
l
o
c
1
(
R
n
) and all x
∈
R
n,(23)
I
α
·
f
x
≤
C
M
α
·
-
ϵ
f
x
1
/
2
M
α
·
+
ϵ
f
x
1
/
2
.Proof.
Letδ
>
0 be given. Then, (24)
I
α
·
f
x
=
∫
x
-
y
<
δ
f
y
x
-
y
α
x
-
n
d
y
+
∫
x
-
y
≥
δ
f
y
x
-
y
α
x
-
n
d
y
=
I
1
x
+
I
2
x
.Let R
i
=
y
2
-
i
-
1
δ
≤
x
-
y
<
2
-
i
δ; we have (25)
I
1
x
≤
∑
i
=
0
∞
∫
R
i
f
y
x
-
y
n
-
α
x
d
y
≤
∑
i
=
0
∞
1
2
-
i
-
1
δ
n
-
α
x
∫
R
i
f
y
d
y
≤
∑
i
=
0
∞
2
n
-
α
-
2
-
i
ϵ
δ
ϵ
1
2
-
i
δ
n
-
α
x
+
ϵ
∫
B
i
f
y
d
y
,where B
i
=
{
y
∣
x
-
y
≤
2
-
i
δ
}. Hence, I
1
(
x
)
≤
C
ϵ
δ
ϵ
(
M
α
(
·
)
-
ϵ
f
)
(
x
)
. Similarly, (26)
I
2
x
≤
∑
i
=
1
∞
2
i
-
1
δ
α
x
-
n
∫
B
-
i
f
y
d
y
≤
C
ϵ
′
δ
-
ϵ
M
α
·
+
ϵ
f
x
.By choosing δ
ϵ
=
[
M
α
(
·
)
+
ϵ
f
(
x
)
/
M
α
(
·
)
-
ϵ
f
(
x
)
]
1
/
2, one finds (27)
I
α
·
f
x
≤
C
M
α
·
-
ϵ
f
x
1
/
2
M
α
·
+
ϵ
f
x
1
/
2
.Lemma 12.
Given an open setΩ
⊂
R
n and α
(
·
)
:
0
<
α
-
≤
α
+
<
n, let p
(
·
)
:
Ω
→
[
1
,
∞
) be such that 1
=
p
-
≤
p
+, (
α
(
·
)
p
(
·
)
)
+
<
n, and such that p
(
·
) satisfies the Log-Hölder condition. Define q
(
·
) by (28)
1
p
x
-
1
q
x
=
α
x
n
,
x
∈
Ω
.Then, for all f
∈
L
p
(
·
)
(
Ω
) such that f
p
(
·
)
,
Ω
≤
1 and such that |
f
(
x
)
|
≥
1 or f
(
x
)
=
0
,
x
∈
Ω,(29)
M
α
·
f
x
≤
C
M
f
x
p
x
/
q
x
.Proof.
Fixx
∈
Ω, and fix a ball B containing x. Then, by the definition of q
(
·
), (30)
1
B
1
-
α
x
/
n
∫
B
∩
Ω
f
y
d
y
=
B
α
x
/
n
1
B
∫
B
∩
Ω
f
y
d
y
α
x
p
x
/
n
1
B
∫
B
∩
Ω
f
y
d
y
1
-
α
x
p
x
/
n
≤
B
α
x
/
n
1
B
∫
B
∩
Ω
f
y
d
y
α
x
p
x
/
n
M
f
x
p
x
/
q
x
.To complete the proof we will show that (31)
B
α
x
/
n
1
B
∫
B
∩
Ω
f
y
d
y
α
x
p
x
/
n
≤
C
.We consider it in two cases depending on the size of B
. Suppose first that |
B
|
>
1. Let Ω
f
=
supp
⁡
f. By Chebyschev’s inequality and Lemma 5, (32)
Ω
f
≤
f
p
·
,
Ω
≤
f
p
·
,
Ω
<
1
.Therefore, since p
(
x
)
≥
1, by Lemma 6, (33)
B
α
x
/
n
1
B
∫
B
∩
Ω
f
y
d
y
α
x
p
x
/
n
≤
∫
B
∩
Ω
f
y
d
y
α
x
p
x
/
n
≤
f
1
,
Ω
f
α
x
p
x
/
n
≤
C
1
+
Ω
f
α
p
+
/
n
f
p
·
,
Ω
f
α
p
-
/
n
≤
C
.Now suppose |
B
|
≤
1. If p
-
(
B
∩
Ω
)
>
1, then (
p
′
)
+
(
B
∩
Ω
)
<
∞ and p
′
(
·
) satisfies the Log-Hölder condition, so by Lemmas 7 and 9
(34)
B
α
x
/
n
1
B
∫
B
∩
Ω
f
y
d
y
α
x
p
x
/
n
≤
B
α
x
/
n
-
α
x
p
x
/
n
χ
B
∩
Ω
p
′
·
,
Ω
α
x
p
x
/
n
f
p
·
,
Ω
α
x
p
x
/
n
≤
B
α
x
/
n
-
α
x
p
x
/
n
χ
B
∩
Ω
p
′
·
,
Ω
α
x
p
x
/
n
≤
C
B
α
x
/
n
-
α
x
p
x
/
n
B
α
x
p
x
/
n
p
′
x
≤
C
.The argument is the same when p
-
(
B
∩
Ω
)
=
1 except that instead of applying Lemma 9 we note that, by Lemma 6, (35)
χ
B
∩
Ω
p
′
·
,
Ω
α
x
p
x
/
n
=
χ
B
∩
Ω
p
′
·
,
B
∩
Ω
α
x
p
x
/
n
≤
C
1
+
B
χ
B
∩
Ω
∞
,
B
∩
Ω
α
x
p
x
/
n
≤
C
.Then, by Lemma 8, (36)
B
α
x
/
n
-
α
x
p
x
/
n
≤
B
1
-
p
x
α
+
/
n
≤
B
p
-
B
∩
Ω
-
p
+
B
∩
Ω
α
+
/
n
≤
C
.This completes the proof.Lemma 13.
Given an open setΩ
⊂
R
n and α
(
·
)
:
0
<
α
-
≤
α
+
<
n, let p
(
·
)
∈
P
(
Ω
) be such that 1
<
p
-
≤
p
+, (
α
(
·
)
p
(
·
)
)
+
<
n, and satisfy the Log-Hölder decay condition. Define q
(
·
) by (37)
1
p
x
-
1
q
x
=
α
x
n
,
x
∈
Ω
.Then, for all f
∈
L
p
(
·
)
(
Ω
) such that f
p
(
·
)
,
Ω
≤
1 and f
x
≤
1, x
∈
Ω,(38)
M
α
·
f
x
≤
C
M
f
x
p
x
/
I
q
x
,where I
r
(
x
)
=
s
u
p
|
y
|
>
|
x
|
r
(
y
).Proof.
Fixx
∈
Ω and let B be any ball containing x. It will suffice to show that (39)
1
B
1
-
α
x
/
n
∫
B
∩
Ω
f
y
d
y
≤
C
M
f
x
p
x
/
I
q
x
,where C is independent of B.
It follows at once from the definition ofq
(
·
) that 1
/
I
p
(
x
)
-
1
/
I
q
(
x
)
=
α
(
x
)
/
n. Therefore, by Hölder’s inequality, (40)
1
B
1
-
α
x
/
n
∫
B
∩
Ω
f
y
d
y
=
B
α
x
/
n
1
B
∫
B
∩
Ω
f
y
d
y
α
x
I
p
x
/
n
1
B
∫
B
∩
Ω
f
y
d
y
1
-
α
x
I
p
x
/
n
≤
∫
B
∩
Ω
f
y
I
p
x
d
y
α
x
/
n
M
f
x
I
p
x
/
I
q
x
.Since f
x
≤
1, M
f
(
x
)
≤
1, so M
f
(
x
)
I
p
(
x
)
/
I
q
(
x
)
≤
M
f
(
x
)
p
(
x
)
/
I
q
(
x
). Therefore, to complete the proof we only need to show that (41)
∫
B
∩
Ω
f
y
I
p
x
d
y
α
x
/
n
≤
C
.For(42)
min
⁡
∫
B
∩
Ω
f
y
I
p
x
d
y
α
-
/
n
,
∫
B
∩
Ω
f
y
I
p
x
d
y
α
+
/
n
≤
∫
B
∩
Ω
f
y
I
p
x
d
y
α
x
/
n
≤
max
⁡
∫
B
∩
Ω
f
y
I
p
x
d
y
α
-
/
n
,
∫
B
∩
Ω
f
y
I
p
x
d
y
α
+
/
n
,
0
<
α
-
≤
α
+
<
n
,it suffices to prove (43)
∫
B
∩
Ω
f
y
I
p
x
d
y
≤
C
.Define the sets(44)
D
x
≔
y
∈
B
∩
Ω
,
y
≤
x
,
D
x
≔
y
∈
B
∩
Ω
,
y
>
x
.Then, (45)
∫
B
∩
Ω
f
y
I
p
x
d
y
=
∫
D
x
f
y
I
p
x
d
y
+
∫
D
x
f
y
I
p
x
d
y
.The integral in the right-hand side is easy to estimate. If y
∈
D
x, then I
p
(
x
)
≥
p
(
y
). Thus, since |
f
(
x
)
|
≤
1, by Lemma 5, (46)
∫
D
x
f
y
I
p
x
d
y
≤
∫
D
x
f
y
p
y
d
y
≤
f
p
·
,
Ω
≤
1
.To estimate the first integral, we will apply Lemma 10. If y
∈
D
x, then, by the Log-Hölder decay condition, (47)
p
y
-
I
p
x
≤
C
log
⁡
e
+
y
.Therefore, for any t
>
1 and (
I
p
)
-
≥
1, R
t
(
y
)
(
I
p
)
- is integrable, and(48)
∫
D
x
f
y
I
p
x
d
y
≤
C
t
∫
D
x
f
y
p
y
d
y
+
∫
D
x
R
t
y
I
p
-
d
y
≤
C
f
p
·
,
Ω
≤
C
.This completes the proof.Lemma 14 (see [22]).
Given an open setΩ
⊂
R
n, let p
(
·
)
:
Ω
→
[
1
,
∞
) satisfy the Log-Hölder decay condition. Suppose f
∈
L
p
(
·
)
(
Ω
) such that f
p
(
·
)
,
Ω
≤
1. Then, for every ball B and every x
∈
B, (49)
1
B
∫
B
∩
Ω
f
y
d
y
p
x
≤
C
1
B
∫
B
∩
Ω
f
y
p
y
/
p
-
d
y
p
-
+
S
x
,where S
(
·
)
∈
L
1
(
R
n
) and C
=
C
(
n
,
p
(
·
)
)
.Proof of Theorem2.
Fixf
∈
L
p
(
·
)
(
Ω
); without loss of generality we may assume that f
p
(
·
)
,
Ω
=
1. Since q
+
<
∞, by Lemma 5 it will suffice to prove that (50)
t
χ
I
α
·
f
>
t
q
·
,
Ω
=
∫
x
∈
Ω
:
I
α
·
f
>
t
t
q
x
d
x
≤
C
.Fix ϵ
,
0
<
ϵ
<
m
a
x
(
α
-
,
n
-
α
+
), such that (51)
p
-
ϵ
q
+
/
n
+
1
>
1
.Define r
(
·
)
:
Ω
→
(
1
,
∞
) by (52)
r
x
=
2
ϵ
q
x
/
n
+
1
.Since r
-
>
1 and (
r
′
)
+
<
∞, by (23) and Young’s inequality, for each x, (53)
I
α
·
f
x
≤
C
M
α
·
-
ϵ
f
x
1
/
2
M
α
·
+
ϵ
f
x
1
/
2
≤
C
M
α
·
-
ϵ
f
x
r
x
/
2
r
x
+
C
M
α
·
+
ϵ
f
x
r
′
x
/
2
r
′
x
≤
C
r
+
M
α
·
-
ϵ
f
x
r
x
/
2
+
C
r
′
+
M
α
·
+
ϵ
f
x
r
′
x
/
2
.Therefore,(54)
∫
I
α
·
f
>
t
t
q
x
d
x
≤
∫
M
α
·
-
ϵ
f
x
r
x
/
2
>
t
/
2
C
t
q
x
d
x
+
∫
M
α
·
+
ϵ
f
x
r
′
x
/
2
>
t
/
2
C
t
q
x
d
x
≕
I
1
+
I
2
.
Denotef
=
f
1
+
f
2, where f
1
=
f
χ
{
|
f
|
≥
1
} and f
2
=
f
χ
{
|
f
|
<
1
}.
Therefore, by (29) and (38), (55)
M
α
·
-
ϵ
f
x
r
x
/
2
≤
2
r
+
M
α
·
-
ϵ
f
1
x
r
x
/
2
+
M
α
·
-
ϵ
f
2
x
r
x
/
2
≤
C
M
f
1
x
r
x
p
x
/
2
q
x
+
M
f
2
x
r
x
p
x
/
2
I
q
x
,
M
α
·
+
ϵ
f
x
r
′
x
/
2
≤
C
M
f
1
x
r
′
x
p
x
/
2
q
x
+
M
f
2
x
r
′
x
p
x
/
2
I
q
x
.Therefore, (56)
I
1
≤
C
∫
M
f
1
x
r
x
p
x
/
2
q
x
>
t
/
2
C
t
q
x
d
x
+
C
∫
M
f
2
x
r
x
p
x
/
2
I
q
x
>
t
/
2
C
t
q
x
d
x
.We estimate each integral in turn.
For eachN
>
0, define f
N
=
f
χ
B
N
(
0
), where B
N
(
0
)
=
{
x
∈
R
n
:
x
≤
N
}. Then, for each N the set {
M
f
N
(
x
)
r
(
x
)
p
(
x
)
/
2
q
(
x
)
>
t
/
2
C
} is bounded, and by the monotone convergence theorem, the first integral is immediate: (57)
∫
M
f
1
x
r
x
p
x
/
2
q
x
>
t
/
2
C
t
q
x
d
x
=
lim
N
→
∞
⁡
∫
M
f
N
x
r
x
p
x
/
2
q
x
>
t
/
2
C
t
q
x
d
x
.So, it will suffice to prove that (58)
∫
M
f
N
x
r
x
p
x
/
2
q
x
>
t
/
2
C
t
q
x
d
x
≤
C
,where C is independent of N
. Fix N
>
0; for each x
∈
{
M
f
N
(
x
)
r
(
x
)
p
(
x
)
/
2
q
(
x
)
>
t
/
2
C
} there exists a ball B
x containing x such that (59)
2
n
2
B
x
∫
2
B
x
∩
Ω
f
y
d
y
r
x
p
x
/
2
q
x
≥
1
B
x
∫
B
x
∩
Ω
f
N
y
d
y
r
x
p
x
/
2
q
x
>
t
2
C
.Therefore, by the Besicovitch covering lemma (see [24]), there exists a sequence {
B
k
} of {
B
x
} such that (60)
M
f
N
x
r
x
p
x
/
2
q
x
>
t
2
C
⊂
⋃
k
2
B
kand such that the balls {
2
B
k
} have finite overlap uniformly bounded by a constant depending only on the dimension. Hence, (61)
∫
M
f
N
x
r
x
p
x
/
2
q
x
>
t
/
2
C
t
q
x
d
x
≤
∑
k
∫
2
B
k
∩
Ω
t
q
x
d
x
≤
2
n
r
+
p
+
∑
k
∫
2
B
k
∩
Ω
1
2
B
k
∫
2
B
k
∩
Ω
f
y
d
y
r
x
p
x
/
2
d
x
.For(62)
r
x
p
x
2
-
r
y
p
y
2
≤
r
x
2
p
x
-
p
y
+
p
y
2
r
x
-
r
y
≤
r
+
2
p
x
-
p
y
+
ϵ
/
n
P
+
1
-
α
+
-
ϵ
/
n
p
+
2
p
x
-
p
y
≤
C
p
x
-
p
y
,by Lemma 14, we have (63)
∫
M
f
N
x
r
x
p
x
/
2
q
x
>
t
/
2
C
t
q
x
d
x
≤
C
∑
k
∫
2
B
k
∩
Ω
1
2
B
k
∫
2
B
k
∩
Ω
f
y
r
y
p
y
/
r
p
-
d
y
r
p
-
/
2
+
S
x
d
x
≤
C
∑
k
∫
2
B
k
∩
Ω
M
f
·
r
·
p
·
/
r
p
-
x
r
p
-
/
2
d
x
+
C
∫
2
B
k
∩
Ω
S
x
d
x
≤
C
∑
k
∫
2
B
k
∩
Ω
f
x
r
x
p
x
/
r
p
-
r
p
-
/
2
d
x
+
C
∫
2
B
k
∩
Ω
S
x
d
x
≤
C
∫
Ω
f
x
r
x
p
x
/
2
d
x
+
C
∫
Ω
S
x
d
x
,where, in the third inequality, since (
r
p
)
-
/
2
>
1, we used the boundedness of maximal operator on L
(
r
p
)
-
/
2
(
Ω
), and in the last inequality, we used the fact that the balls have uniformly bounded finite overlap.
ForS
(
·
)
∈
L
1
(
R
n
), ∫
Ω
‍
S
x
d
x
≤
C; then, it will suffice to prove (64)
∫
Ω
f
x
r
x
p
x
/
2
d
x
≤
C
.We consider the two cases f
x
≥
1 and f
x
<
1 to estimate it.
If|
f
(
x
)
|
≥
1, since r
(
x
)
/
2
=
1
/
ϵ
q
+
/
n
+
1
<
1, therefore (65)
∫
Ω
f
x
r
x
p
x
/
2
d
x
<
∫
Ω
f
x
p
x
d
x
≤
C
.If |
f
(
x
)
|
<
1, since 2
/
r
-
=
ϵ
/
n
q
+
+
1
>
1, by Hölder’s inequality (66)
∫
Ω
f
x
r
x
p
x
/
2
d
x
≤
C
∫
Ω
f
x
r
x
p
x
/
2
·
2
/
r
-
d
x
r
-
/
2
Ω
1
-
r
-
/
2
≤
C
∫
Ω
f
x
p
x
d
x
r
-
/
2
Ω
1
-
r
-
/
2
≤
C
Ω
≤
C
.
To estimate the second term, we need to show that(67)
∫
M
f
2
x
r
x
p
x
/
2
I
q
x
>
t
/
2
C
t
q
x
d
x
≤
C
.We first need to control the quantity q
(
x
)
/
I
q
(
x
)
. Fix ϵ
>
0; then, there exists y
,
|
y
|
>
|
x
|, such that I
q
(
x
)
≤
(
1
+
ϵ
)
q
(
y
)
. Then, by the Log-Hölder decay condition, we have (68)
0
≤
I
q
x
-
q
x
≤
1
+
ϵ
q
y
-
q
x
≤
q
y
-
q
x
+
ϵ
q
+
≤
C
log
⁡
e
+
x
+
ϵ
q
+
.Since ϵ
>
0 is arbitrary and q
-
≥
1, it follows that (69)
0
≤
I
q
x
q
x
-
1
≤
C
log
⁡
e
+
x
.For R
t
(
x
)
=
(
e
+
|
x
|
)
-
t
n, R
2
q
+
/
q
-
(
x
) is integrable, and by Lemma 10
(70)
∫
M
f
2
x
r
x
p
x
/
2
I
q
x
>
t
/
2
C
t
q
x
d
x
≤
C
∫
M
f
2
x
r
x
p
x
/
2
I
q
x
>
t
/
2
C
t
q
x
I
q
x
/
q
x
d
x
+
C
∫
M
f
2
x
r
x
p
x
/
2
I
q
x
>
t
/
2
C
R
2
q
+
/
q
-
x
d
x
≤
C
∫
M
f
2
x
r
x
p
x
/
2
I
q
x
>
t
/
2
C
t
I
q
x
d
x
.The proof now proceeds as it does for the first term.
We estimateI
2 as we did before; here we omit the detail. Then, we finish the proof of Theorem 2.


## 3. Proof of Theorem4

To prove Theorem4, we need the following lemma.Lemma 15 (see [16]).
LetΩ be bounded. Suppose that p
(
·
) satisfies conditions (3) and (5) and α
(
·
) also satisfies the Log-Hölder condition: (71)
inf
x
∈
Ω
⁡
α
x
>
0
,
sup
x
∈
Ω
⁡
α
x
p
x
<
n
.If 1
/
q
(
x
)
=
1
/
p
(
x
)
-
α
(
x
)
/
n, then there exists a positive constant C such that for all f
∈
L
q
(
·
)
(
Ω
)
(72)
I
α
·
f
p
·
,
Ω
≤
C
f
q
·
,
Ω
.Proof of Theorem4.
Sinceα
(
x
)
>
0, some r
i
<
∞. If r
j
+
1
x
=
⋯
=
r
k
(
x
)
=
∞, 1
≤
j
<
k, since α
(
x
)
/
n
<
1
/
r
1
(
x
)
+
1
/
r
2
(
x
)
+
⋯
+
1
/
r
j
(
x
)
≤
j, we have k
n
-
α
(
x
)
>
(
k
-
j
)
n. Then, integration in y
j
+
1
,
…
,
y
k reduces matters to the case when all r
i
(
x
) are finite (and k
=
j). Thus, we can assume that all r
i
(
x
)
<
∞.
Now, observe that0
<
α
x
<
∑
i
=
1
k
‍
n
/
r
i
x; we can find 0
<
α
i
(
x
)
<
n
/
r
i
(
x
) such that α
(
x
)
=
∑
i
=
1
k
‍
α
i
(
x
). Let 1
/
s
i
(
x
)
=
1
/
r
i
(
x
)
-
α
i
(
x
)
/
n. Since 0
<
α
i
(
x
)
/
n
<
1
/
r
i
(
x
)
≤
1 and (73)
y
1
n
-
α
1
x
y
2
n
-
α
2
x
⋯
y
k
n
-
α
k
x
≤
y
1
,
y
2
,
…
,
y
k
k
n
-
α
x
,it follows that (74)
I
α
·
,
k
f
1
,
…
,
f
k
x
≤
∏
i
=
1
k
I
α
i
·
f
i
x
,where I
α
i
(
·
) is the variable exponent fractional integral operator.
Because∑
i
=
1
k
‍
1
/
s
i
(
x
)
=
1
/
s
(
x
), by Hölder’s inequality and (72), we obtain (75)
I
α
·
,
k
f
1
,
…
,
f
k
x
s
·
,
Ω
≤
∏
i
=
1
k
I
α
i
·
f
i
x
s
i
·
,
Ω
≤
∏
i
=
1
k
f
i
x
r
i
·
,
Ω
.
This finishes the proof.



---

*Source: 2438157-2016-06-27.xml*