http://dx.doi.org/10.4236/am.2013.411A3005
Boundedness of Hyper-Singular Parametric
Marcinkiewicz Integrals with Variable Kernels
Qiquan Fang1*, Xianliang Shi2
1
Department of Mathematics, Zhejiang University of Science and Technology, Hangzhou, China
2
College of Mathematics and Computer Science, Hunan Normal University, Key Laboratory of High Performance Computing and Stochastic Information Processing (Ministry of Education of China),
Hunan Normal University, Changsha, China Email: *[email protected]
Received September 13, 2013; revised October 13, 2013; accepted October 20, 2013
Copyright © 2013 Qiquan Fang, Xianliang Shi. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
ABSTRACT
In this article, we consider the boundedness of , ,
b on Hardy type space Hbp
Rn . Where 0 <<n ,
1 2 2
, , 0 , , 2 1 2
d , t
t
f x F f x
t
b b
, ,
1
,
d
m
t x y t n i
i
x x y
.
i
F f x
,
i
b x b y
f y
x y
b
y
1, 2, ,
,
,1 , 0, 1 0 1 mn
m i i i i
b b b b R i m
b
Keywords: Hyper-Singular Marcinkiewicz Integral; Variable Kernel; Multilinear Commutator; Hardy Type Space
1. Introduction
A function
x z,
n rL R L
defined on is said to be-
long to , if it satisfies the following
three conditions:
n n
n1S
, ,
R R
1)
x,z
x z for any ,x zRn and any> 0
;
2)
1
1
1
0,
: sup , d
n r n
n n
L R L S
r r
S y R
z y z z
<3)
Sn1
x z,
d z 0, for anyn
xR .
In [1], the authors considered the hyper-singular pa- rametric Marcinkiewicz integral with variable kernel as follows:
1 2 2
, 0 , 2 1 2
d , t
t
f x F f x
t
where
,
,
d .
t x y t n
x x y
F f x f y y
x y
When 0, we set ,0
, which is the para- metric Marcinkiewicz integral with variable kernels con- sidered in [2].
For > 0, the homogenous Lipschitz space
nR
is the space of function f such that 1
, , 0
, sup
n
h x h R h
f x f
h
where kh denotes -th difference operator (see [3]). k
In 2006, Lu and Xu studied the boundedness of the
commutator of ,
m b
in [4]. They proved that:
Theorem A [4]. Suppose
n1 forS
10 < , , 0 <
2
n
b R .
m
If n 1
p
n
and 1 1
q p n
, then ,
m b
maps m,
p n b s
H R
continuously into Lq
Rn . Here , m b is defined as
follows:
1 2 2
, 0 , , 3
d ,
m m
b b t
t
f x F f x
t
(1)
where
, , 1 d .
m m
b t x y t n
x y
F f x b x b y f y y
x y
Let b
b b1, 2,,bm
,1 i m, i 0,
n , i ib R
m
i
i1
, ,
, 0 1.
In this article, we
mainly consider the commutator b defined by
1 2 2
, , 0 , , 2 1 2
d ,
t
t
f x F f x
t
b b
(2)
where
, ,
1
,
d .
t
m
i i
n x y t
i
F f x
x x y
b x b y f y y
x y
b
Given any positive integer , for all , we
denote by
m 1 j m
m j
C
2 ,the family of all finite subsets
of of dif-
ferent elements. For any
1 , ,
j
1, 2,,m
jm j
C
, we associate the com-
plementary sequence given by
1, 2,,m
\ , (see [5]).For any
1 , 2 ,,
j
Cmj
1, 2, ,
, we will
denote b b b j
.
b 1 2
and the product
j
b b b b When C0m, we have ,
by definition, we have . Similarly, when ,
we have
1
b m
m
C
and b 1. With this notation, if
1 j ,
we write
1
1 .
j j
b b
b
When 1
m i i
, we write
i m
i i b
b
i
1 .
Definition 1.1. Let 0 <p1, b be defined as above
such that
,1 , 0 < < 1, 1i
m n
i i i
b R i m
n
.
A function a x
on R is called a -atom if
p q b, ,
1) suppaB x0,r :
xRn: xx0 r
, for some
0
n
x R and r> 0;
2)
1 1
0, q p
q
a B x r
3) n
d for anyRa x b x x
, 1, , \
m j
C m
0 <
and j1, 2,, .m
1
Definition 1.2. Let , we say that a distribu-
tion
p
f on Rn belongs to Hbp
Rn if and only if fcan be written as i i
i
f a
in the distributional sense,where each ai is a
p q, ,b
-atom and. p i i
Moreover,1
inf
p
p p i H
i
f
b
with the infimum taken over all the above decomposi- tions of f as above
Definition 1.3. A function
1
, n r n 1
x z L R L S r
,
r
L
is said to satisfy the
-Dini condition, if
1 1
0 d < , 0 1,
r
(3)where r
r
denotes the integral modulus of continu- ity of order of defined by
1 1
, 0 ,
1
,
sup sup
, d .
n
n n
r S
x R y S y z
r r
x z y
x z z z
We will denote simply Lr-Dini condition for Lr,- Dini condition when 0.
2. Main Theorem
Now let us formulate our main results as follows.
Theorem 2.1.Suppose that , ,
b is the commutator
(2), and let
1 1
0 1,1 p n ,
q p n .
1, n r n
x z L R L S
, then , ,
b is bounded
from Lp
Rn into Lq
Rn . That is,
, , Lq Rn C f LpRn .
b b
Theorem 2.2. Suppose that , ,
b is the commutator
(2), and let n < p 1,1 1
n q p n
.
If
x z, satisfies the following two conditions: 1)
x z, satisfies Lr,-Dini condition (3); 2) there exists> max , , 1 2
n n n
r
n n n
n n
p p
such
that
x z, L
Rn L Sr n1 then
1
0 0
< 2
2 < 4
, ,
d
d . r r
n n
K x K n
n r
r
y K y K
x x x y x x x
x
x y x
y CK
K
, ,
b is boun-
ded from Hbp
Rn
q n
L R
into
. That is where the constant is independent of and.
> 0
C K
y
, , f Lq Rn C f HpRn .
b
b b
lemma 2.2. [7] Let 0 <<n, 1 < p< n,1 1
q p n
and T , be defined as
Remark Obviously, , ,
b is the commutator of the
operator ,
in [1]. At the same time, we change the
course of the statement in [4]. , x Rn f y
d .yx y
, nx x y
T f If there exists
In order to prove our Theorems, we need several
pre-liminary lemmas. r> p, such that
1, n r n ,
x z L R L S
then
Lemma 2.1. [6] Let 0<n
, r1
and suppose
If there exists a
n r n1x z L R L S
.
,
T is bounded from Lp
Rn into Lq
Rn . That is
1
, Lq Rn L Rn L Sr n LpRn .
T f C f
constant 0 < < 1 2
R such that y <RK, then for any
3. Proofs
0
n
x R ,
3.1. Proof of Theorem 2.1.
Applying the Minkowski’ inequality, we can get
1 2 2
, , 0 , 2 1 2
1
2 2
2 1 2 0
1
1 2 2 1 2 1
1 ,
d
, d
d
, d
d
sup
n
n i n
t
m
i i
n x y t
i
m
i i
n
R x y
i
m
i i
R i x y R
x y
t
f x F f x
t
x x y t
b x b y f y y
t
x y
x x y t
b x b y f y y
t
x y
b x b y
C
x y
b b
,
, 1
d
,
d .
n n n
n
n
R R R
x x y
x y f y
x y x y
x x y
C f y y T f
x y
b b
y
x
3.2. Proof of Theorem 2.2. By Lemma 2.2 , we have
1
, ,
,
.
q n
n q n
n r n n p n
L R
R L R
L R L S R L R
f
C T f
C f
b
b
b
Noting that r> n
n , we can choose l1 such that 1
< < n
r l
. It is easy to see that . Next , we
choose such that
1
l
>
r
2
l
2 1
1 1
.
l l n
It follows from
Theorem 2.1 that , ,b is bounded from Ll1
Rninto Ll2
Rn . That is
1
2
, ,f Ll Rn C Rn f Ll Rn .
b b
(4)
By the atomic decomposition theory on Hardy type space, it suffices to prove that there is a constant
such that for all -atom the following holds
> 0
C
p l b, ,1
, , q n Rn .
L R
a C
b b
Without loss of generality we may assume that
0
suppa B B x d, . We write
0
8 ,8 :
B nBB x nd B x d0,
. We split
, ,a Lq Rn
b into two parts as follows:
1 1
, , q n , , d , , d 1 2
q q
q q
C
B B
L R
a a x x a x x
b b b : I I .
We can easily see that l2>q. By (4) and the size condition of atom a, we have
2
2
2 2 1 2 1
1
1 1 1 1 1 1 1 1
1 , , d n l n n .
l l
q l q l q l l p
R L R R
B
I a x x B C B a C B B C
b b b Rn
b
Next we estimate I2. Let us consider , , a x
b :
00
1 2 2
2
, , 0 2 1 2
1
1 2 2
2 1 2 2
1 21 22
, d
d
, d
d
: .
m x x d
i i
n x y t
i
m
i i
n x x d x y t
i
x x y t
a x b x b y a y y
t
x y
x x y t
b x b y a y y
t
x y
I I
b
0 0 2
x y x x x x d for
0,
,
C
yB x d x B . By the mean value theorem, we
have
2 22 2 2 1 2
0 2 .
xy xx d Cd xy
Thus, by the Minkowski’s inequality for integrals,
0
1 2 2
21 2 1 2
1
1 2
2 2 2 2
0 1
1 1
2
2 2
0 1 2
, d
d
, 1 1
d 2
,
d ,
n
n
n n
m x x d
i i
n
R x y
i
n
R B
n n
R B R B
x x y t
I b x b y a y y
t
x y
x x y
C x y a y y
x y x y x x d
x x y d
C a y y C d x x
x y x y
b
b b x xy a
y d .yApplying the Hölder inequality and the size condition of a, we have
1
1 1
0 0
0 1
1
1 1
0
1
1 1 1
1 1 1
2 2
1
2 1 1 1 1 1 1
1 0
2
, d
, d d , d d
, d d l n n r n .
n B
r
r r l
r r r l r l
B B x x x y x x B
r
x x n r r l n r r p
L R L R L S
x x S
x x y a y y
x x y y a y y x x y y a y y B
t x z z t a B C x x B
So we can get
11 2 1 2
21 n n r n 0 .
n r n n r n p n
R L R L S
I C d x x
b
> n ,
r
n we have Noting that
1
1 1
1 2 1 2
21: C B
J
21 01 1 2
1 2 1 1 2 1 2
d
d .
n n r n
n n n
q q
n r n q
q n r n p n
C
R L R L S B
q n r n q
n r n p n n n r n p n n r p n
R d R R
I C d x x x
C d C d d C
b
b b n b
For I22, we write
,
1
,
: d
m
t x y t n i
i
x x y
G a x b x b y a y y
x y
b i .
, 0
0
0 1
1
0 0
0
,
1 d
,
d ,
1 d
m j
m j
m
m j
t x y t n
j C
0
.
i i x y t n
i m
m j
n x y t j C
x x y
G a x x x y x a y y
x y
x x y
b x b x a y y
x y
x x y
m
x x y x a y y
x y
b b b b b
b b b b
So
22
I is dominated by
00 2 0
x x d j
1
2 2
22 2 0 2 1 2
1
1
2 2
1
0 0 2 1 2
1 2
, d
d
, d
1 d
: .
m j
m
i i n
x x d x y t
i m
m j
n x y t C
x x y t
I b x b x a y y
t
x y
x x y t
x x y x a y y
t
x y
K K
b b
b bNow let us estimate K1. By the vanishing condition of a, we have
0
0
1
2 2
1 2 0 0 2 1 2
1
1 2
0 0 2 2 1 2
1
i
0 0
0
0 0
d
, , d
d
, , d
1
, , d
2
, , d .
n
n
n
m
j j
x x d x y t
i m
j j R x x d
R B
R B
t
K b x b x D x x y a y y
t
t
b x b x D x x y a y y
t
C x x D x x y a y y
x x d
C x x D x x y a y y
bb
where
0
0
0
, ,
, , .
n n
x x x
x x y
D x x y
x y x x
1 < < 0,
n n
n n
p p
Hölder’s inequality and Lemma 2.1,
1 0 1 0 1 1 1 0 0 1 0 0 1 0 0 2 2 0 0 2 2 0, , d d
, , d d
, , d d
2 , ,
j j j j j q q C B q q C B B
q q q
C
B B
q q q
B d x x d
j
B d x x d
j
K
C D x x y x x a y y x
C a y D x x y x x x y
C a y D x x y x x x y
C a y d D x x y
b b b b
1 0 0 2 0 1 0 1 1 0 2 2 0 2 0 2 0 1 0 2 d d 2 , 1 2 2 2 1 2 2 2 j j j j j q q n n r rj q r
B d x x d
j
y x n n n
n r
j q r j r d
y x j B j d n n j p y x j j B j x y
C a y d D x x y x y
C a y d d
y x
C a y d
d
, d d
d dy
b b b
0 2 1 0 2 0 1 1 1 1 1 2 1 1 2 1 0 2 1 1 1 1 0 d d2 2 d d
1 d . j j j l n y x r d d y x n n n
j n j n n
p p r
p d y x B j d n n r p l L R n n n n
n n p l p l
y
C d a y y
C d a B
C d C
b b b b Now we estimate K2. Applying Minkowski's inequality, the size condition of a, we obtain
0 0 0 1 1 22 0 0 2 2 1 2
0 1 0 0 0 0 1 0 0 0 0 , d 1 d , 1 d 2 sup s n m j m j i m j m m j n
R x x d
j C m n B j C m i i
x x x x
j C i i
x x y t
K C a y x x y x y
t
x y
x x y
C a y x x y x y
x y x x d
b x b x
C x x
x x
b b b b
b b b b
0 0 0 1 0 0 0 1 0 0 , d up , d, d .
i n m j n m j i i n B m n R B j C m n R B j C
b y b x x x y
y x a
x y
y x
x x y
C x x y x a y y
x y
C x x d x x y a y y
b b y ySo we have
11
2 0
0
.
n n r n
m j
n n
m n n
n r p
r
R L R L S
j C
K C x x d
Thus
1
1 2
1 1
0 =0
1 =0
d
.
n n r n
m j
n
m j
n q q C B
n n n
r p
R L R L S
m n n q q
r C
B
j C
n nm n n
n n
r p r q
R
j C
R
K
C d
d x x
C d d
C
b
b
b
x
, we have So when |xx0|>Cd
1
22: 22 n ,
q q
C R
B
J I C
b
2 21 22 Rn .
I J J C
b
Combining the estimates for I1 and I2, we have
, ,a Lq Rn C Rn .
b b
This completes the proof of Theorem 2.2.
4. Acknowledgements
The authors would like to thank anonymous rev for their comments and suggestions. The authors are par-tially supported by project 11226108, 11071065, 11171306 funded by NSF of China, project 20094306110004 fu ed by RFDP of high education of China.
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