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Journal of Function Spaces
Volume 2014, Article ID 765984, 8 pages
http://dx.doi.org/10.1155/2014/765984
Research Article

Marcinkiewicz Integrals on Weighted Weak Hardy Spaces

1College of Mathematics and Informatics, Henan Polytechnic University, Jiaozuo 454003, China
2Department of Mathematics, Jiaozuo University, Jiaozuo 454003, China

Received 13 January 2014; Accepted 26 February 2014; Published 3 April 2014

Academic Editor: Yongqiang Fu

Copyright © 2014 Yue Hu and Yueshan Wang. 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

We prove that, under the condition , Marcinkiewicz integral is bounded from weighted weak Hardy space to weighted weak Lebesgue space for , where belongs to the Muckenhoupt weight class. We also give weaker smoothness condition assumed on Ω to imply the boundedness of from to .

1. Introduction and Results

Suppose that is the unit sphere in equipped with Lebesgue measure . Let be homogeneous of degree zero on satisfying and where for any . Then, the Marcinkiewicz integral operator is defined by where

The operator was originally introduced by Marcinkiewicz [1] in 1938 for and sign . In 1958, Stein [2] defined the Marcinkiewicz integral of higher dimensions and proved that if , then is of strong type for and of weak type . We say that if In 1962, Benedek et al. [3] showed that if is continuously differentiable on , then is of strong type for all . In 1990, Torchinsky and Wang [4] considered the weighted case and proved that if , then, for all (Muckenhoupt weight class), is bounded on .

In 1999, Ding et al. [5] improved Torchinsky and Wang's result, by ridding of the smoothness condition, as follows.

Theorem A. Let , and satisfy (1). If , then there is independent of such that where is the dual of such that .

In 2003, Ding et al. [6] discussed the boundedness of Marcinkiewicz integral on weighted Hardy space, getting the following results.

Theorem B. Let , and . Let satisfy (1). If , then there exists a constant independent of such that .

Theorem C. Let satisfy (1) and condition (. If , then there is a independent of such that .

In 2007, C. C. Lin and Y. C. Lin [7] showed that under weaker condition assumed on , which is called -Dini type condition of order , Marcinkiewicz integral operator is bounded from to .

We say that satisfies the -Dini condition if , is homogenous of degree zero on , and where denotes the integral modulus of continuity of defined by and is a rotation in with . If , we let .

In 2002, Ding et al. [8] considered the boundedness of Marcinkiewicz integral on weak Hardy space; they obtained the following result.

Theorem D. Suppose that satisfies (1) and Then, there exists a constant such that

In this paper, we will investigate the boundedness of Marcinkiewicz integral on weighted weak Hardy space. We have the following conclusions.

Theorem 1. Suppose that , satisfies (1) and . If and , then there exists a constant independent of such that .

Theorem 2. Let , satisfy (1) and
If , then there is a independent of such that .

Remark 3. From [8, Remark 2], Condition (10) is weaker than Condition (4).

2. Some Preliminaries and Notations

We begin this section with some properties of weights which play an important role in the proofs of our main results.

A weight is a nonnegative, locally integrable function on . Let denote the ball with the center and radius . For any ball B and denotes the ball concentric with whose radius is times as long. For a given weight function and a measurable set , we also denote the Lebesgue measure of by and set weighted measure .

Definition 4. A weight is said to belong to for , if there exists a constant such that for every ball, ,
The class is defined by replacing the above inequality with
The classical weight theory was first introduced by Muckenhoupt in the study of weighted boundedness of Hardy-Littlewood maximal functions in [9].
We know that if with , then for all and for some . Therefore, we will use the notation to denote the critical index of . Obviously, if , , then we have .
We state the following results about class that will be used later on.

Lemma 5 (see [10]). Let , . Then, for any ball and any , there exists an absolute constant such that where depends neither on nor on .

Lemma 6 (see [10]). Let with . Then, for any , there exists an absolute independent of such that

We also need the following result about -Dini condition.

Lemma 7 (see [11]). Suppose that and satisfies -Dini condition. Then, for ,

Given a Muckenhoupt weight function on , for , we denote by the space of all functions satisfying When , will be taken to mean , and we set . And also the weighted weak Lebesgue spaces is defined by

Let us now turn to the weighted weak Hardy spaces. The (unweighted) weak spaces have first appeared in the work of Fefferman et al. [12]. The atomic decomposition theory of weak space on was given by Fefferman and Soria in [13]. In 2000, Quek and Yang [14] introduced the weighted weak Hardy spaces and established their atomic decompositions. Moreover, by using the atomic decomposition theory of , Quek and Yang also obtained the boundedness of Calderón-Zygmund type operators on these weighted spaces.

We write to denote the Schwartz space of all rapidly decreasing smooth functions and to denote the space of all tempered distributions, that is, the topological dual of . Let , and . Define where , and .

For the grand maximal function of is defined by Then, we can define the weighted weak Hardy space by

Moreover, we set .

Lemma 8 (see [14]). Let . For every in , there exists a sequence of bounded measurable functions such that(i) in the sense of distributions.(ii)Each can be further decomposed into , where satisfies the following:(a)each is supported in a ball with and  . Here, denotes the characteristic function of the set and ;(b), where is independent of and ;(c) for every multi-index with .
Conversely, if has a decomposition satisfying (i) and (ii), then . Moreover, we have .

3. Proof of Theorem 1

It suffices to show that there exists a constant , such that, for any ,

For any given , we choose such that . Then, by Lemma 8, for every , we can write where and satisfies (a)–(c) in Lemma 8. Then,

We claim that the following inequality holds: In fact, it follows from Minkowski's integral inequality and the bounded overlapping property of the cubes that Notice that ; then from Chebyshev's inequality and the boundedness of on we have

We set where . Then, we get Since , then, by Lemmas 5 and 8 and the fact , we get An application of Chebyshev's inequality gives us that Below, we will give the estimate of integral . We have

For and , we have . Thus, By Minkowski's inequality for integral, From the fact that and Lemma 6, we have Since , then there exists a such that . Thus, we have Hence,

For and with , we can see that . Since , we have . By the vanishing moment condition of , But Note that for any and . Thus, From the fact that and is homogeneous of degree zero, we get Then, For , we have . Then, Hence, Since , then and for some . From Lemma 6, Analogously, we have Then, Hence,

Combining the estimate of and , we get . This completes the proof of the Theorem 1.

4. Proof of Theorem 2

We follow the strategy of the proof of Theorem 1. For any given , we may choose such that . For every , and then by Lemma 8, we can write where , and satisfies (a)–(c) in Lemma 8. Then,

Similar to the estimate of in the proof of Theorem 1, we obtain

We set where . Then, we get Since , and then by Lemmas 5 and 8 and the fact , similar to the estimate in the proof of Theorem 1, we get An application of Chebyshev's inequality gives us that Below, we give the estimate of integral . We have For and , we have . Thus, Then, Minkowski's inequality for integral gives us that Since , then . From Lemmas 6 and 5, we have Hence, for , Hölder's inequality and imply that Then,

We now estimate . Since , we have and hence

Denote . Minkowski's inequality for integrals and the fact for give us

Notice that

By Hölder's inequality and Lemma 7, we have Then, But from which we get Thus,

This completes the proof of Theorem 2.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

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