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Journal of Function Spaces
Volume 2014, Article ID 787840, 11 pages
http://dx.doi.org/10.1155/2014/787840
Research Article

An Odd Rearrangement of ()

1Department of Mathematics, Zhejiang University, Hangzhou 310027, China
2Department of Mathematics, Zhejiang Normal University, Jinhua 321004, China

Received 13 November 2013; Accepted 31 December 2013; Published 26 February 2014

Academic Editor: Dumitru Motreanu

Copyright © 2014 Zheng Wang and Jiecheng Chen. 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 introduce an odd rearrangement defined by , , where is a decreasing rearrangement of the measurable function . With the help of this odd rearrangement, we show that for each , there exists a such that , where is an distribution function of . Moreover, we study the boundedness of singular integral operators when they are restricted to odd rearrangement of , and we give some results on Hilbert transform.

1. Introduction

From pages 39 to 40 in [1], we know the following relations between the Hardy space and the Lebesgue space : where does not coincide with since the elements of have integral zero. This means that if and , then . On the other hand, it is well known that both Lebesgue spaces and Lorentz spaces can be characterized by distribution functions or decreasing rearrangements. This naturally raises the following question.

Question 1. For each , does there exist a function which is an appropriate modification of satisfying ? Here is a distribution function of measurable function (see Definition 1).

In this paper, we will present a positive answer to Question 1 by introducing an odd rearrangement. We will first address this problem in the case of . Then in a similar way, motivated by the decreasing rearrangement defined on in [2, pp. 44–46] that keeps the distribution function, we introduce an odd rearrangement to solve the problem on . To obtain applications of our result, the singular integral characterization of Hardy space attracts our attention. It is known that we can characterizeby the Hilbert transform and characterize , when , by the Riesz transforms (see [1]). As it is pointed out in [2], although it is important to study the boundedness of the singular integral operators on , it may be impossible for us to give some conditions on singular integral operators to ensure they map to . In this paper, we will show that some singular integral operators are of strong type when some singular integral operators is restricted to odd rearrangements of .

The distribution function and decreasing rearrangement are two important tools for us to study Lebesgue space and Lorentz space. Below, we briefly review their definition on the Euclidean space.

Definition 1. For a measurable function on , the distribution function of is the function defined on by where is the Lebesgue measure.

Definition 2. Let be a complex-valued function defined on . The decreasing rearrangement of is the function defined on by Here, we adopt the convention ; thus whenever for all .

We easily observe that when two measurable functions have the same distribution function or decreasing rearrangement, then they have the same -norm.

The following definition of odd rearrangement is the main issue in our study.

Definition 3. Let be a complex-valued function defined on . The odd rearrangement of is the function defined on by where is a decreasing rearrangement of the measurable function and is a projection: .

In this paper, we will show that and have the same distribution function and decreasing rearrangement. This indicates that the -norm of remains unchanged under the action of projection . For convenience, we note that by the subset of consists of all odd rearrangements of -functions.

As a first step, we study the properties of the odd rearrangement in Proposition 4. Next, we give criteria to determine the boundedness of the singular integral operators when it is restricted to odd rearrangements of via Theorem 6. Thirdly, we consider three kinds of singular operators: singular integral operators with rough kernels, general singular integrals, and singular integrals of nonconvolutional type. Using the boundedness of Riesz transforms and Hilbert transform, we establish the relationships between the Hardy space and the Lebesgue space. Finally, we use the rearrangement which is an extension of the odd rearrangement to obtain some results on Hilbert transform.

This paper is organized as follows. In Section 2, we list some properties of odd rearrangement and seek sufficient conditions that ensure singular integral operators be of strong type when restricted to odd rearrangements of . In Section 3, with the help of the rearrangement, we construct an odd function in but not in and show that for each and there exists a function such that and .

Throughout, the letter always denotes (possibly different) constants which are independent of all essential variables.

2. The Odd Rearrangement and Singular Integrals

From pages 46 to 48 in [2], we know many properties of the decreasing rearrangement. We find that the odd rearrangement has some similar properties with the decreasing rearrangement. The following are some properties of the odd rearrangement .

Proposition 4.   ,
,
  ,
  ,
  ,
  ,
  ,
  ,
  .
  ,
  .

Proof. We only need to prove that ; since other properties can be showed similar with the proof of Proposition  1.4.5 in [2], we know that , .
(i) If , then ,. So we have This means that .
(ii) If , , for convenience, we choose Since is a nonnegative decreasing function defined on , it follows that then we have So if , then we obtain that this means that And if , then we obtain that this means that Let ; we conclude that .

Firstly, we show that for each , we can find one sequence of simple functions that converge to .

Lemma 5. For each , there exists one sequence satisfying where .

Proof. We take Since is a nonnegative decreasing right continuous function defined on , it follows that is an open set of , which means that is an open ball of . We take ; then the theorem can be showed easily.

Note that is a radical function in . It is easy for us to show that is a subspace of . Now we give criteria to determine the boundedness of the singular integral operators when they are restricted to odd rearrangements of via Theorem 6, Corollary 7, and Corollary 8.

Theorem 6. Suppose that is a singular integral operator satisfying the following.(1)There exists a constant independent of such that (2) is weak type of when restricted to the functions of .Then is strong type of with norm at most .

Proof. For each , according to Lemma 5, there exists one sequence satisfying where . We can easily obtain that Due to the fact that is weak type of , we obtain that Then converges to in measure, so there exists some subsequence converges to a.e. Via Fatou’s lemma, we conclude that

Theorem 6 is a criterion to determine the boundedness of the singular integral operators when they are restricted to odd rearrangements of . By Theorem 6, we can find two simpler criteria. Corollary 7 is an immediate consequence of Theorem 6, while Corollary 8 can be showed later.

Corollary 7. Suppose that is a singular integral operator satisfying(1)for any , ;(2); (3) is weak type of when restricted to the functions of .Then is strong type of .

Corollary 8. Suppose that is a singular integral operator satisfying the following:(1) is a bounded operator from to ;(2) is weak type of .Then is strong type of .

Now let us study the boundedness of singular integral operators with the help of Theorem 6 and Corollary 7. Firstly, we consider the boundedness of singular integral operators with rough kernels. The method of rotations is used here; thus we need to consider the boundedness of Hilbert transform firstly.

Lemma 9. Consider the characteristic function of an interval ,

Proof. As in [2, pp. 251-252], for the characteristic function of an interval , So it is not difficult for us to get the above results.

Proposition 10. For each , we have, Equality can occur if .

Proof. Since Hilbert transform is bounded linear operator mapping to and satisfying Using Lemma 9, it is an immediate consequence of Corollary 7.

With the help of boundedness of Hilbert transform, we establish the relationship between andin the following corollary.

Corollary 11. For all ,

Proof. We use singular integral characterization of in [2]; following Proposition 10, we get that for all .

Since the boundedness of Hilbert transform has been known, we can use the method of rotations to study singular integral operator where is odd and integrable over .

Lemma 12. If is odd and integrable over , then

Proof. Via the method of rotation we can obtain where are the directional Hilbert transforms defined by Since the following identity is valid for all matrices, : we take ; then Since for any the following inequality holds: we obtain Then, via Minkowski’s integral inequality, we obtain

According to [3], if with , then is weak type of . With the help of Corollary 8 and Lemma 12, we easily get the following proposition.

Proposition 13. If is an odd function defined on and , then

We know that Riesz transforms satisfy the conditions in Proposition 13, so maps to . Then we establish one relationship between and in the following corollary.

Corollary 14. (i) For all , then Equality can occur if and only if a.e.
(ii) For all ,

Remark 15. Corollary 14 is the high dimensional extension of Corollary 11.

Here we can prove Corollary 8.

Proof of Corollary 8. Since is a bounded operator from to and , then via Corollary 14, we obtain that So Corollary 8 is an immediate consequence of Theorem 6 and Corollary 14.

Remark 16. When is odd and integrable over , whether is weak type of is an open problem. So we suppose in Proposition 13. As showed in [4], is weak type when restricted to radial functions in . But we can try to show that can be weak type when restricted to functions in , and then the same conclusion of Proposition 13 can be obtained. On the other hand, we try to omit the “odd” condition in the following proposition.

Proposition 17. Let with and suppose for some that Then and there exists a constant , such that for all ,

Proof. Since , according to [2, pp. 274-275], the singular integral operator is bounded on . Then we obtain For each , this means that According to [3], is weak type of , so Proposition 17 is an immediate consequence of Corollary 7.

Now let us apply Corollary 8 to the general singular integral operators and show their boundedness on when restricted to odd rearrangement.

The following is the definition of the general singular operators.

The function defined that satisfies size estimates, the smooth estimate, expressed in terms of Hömander’s condition, and cancelation condition, Condition implies that there exists a sequence as such that the following limit exists: The tempered distribution has the form

Proposition 18. Let the function K be defined on that satisfies (45), (46), and (47). Suppose that the operator is given by convolution with . Then .

Proof. By the results of [1, pp. 93-94], admits an extension that is bounded for and is of weak type . Also, there exists a constant such that for all , Thus Proposition 18 is an immediate consequence of Corollary 8.

Finally, we give results of operators. Corollary 8 is used again.

Proposition 19. Assume that is in , and is an element of that is associated with the kernel . Then for all , where is a constant that depends on the dimension.

Proof. By the results of [1, pp. 171–190], admits an extension that is bounded for and is of weak type . Also, there exists a constant such that for all ,
Thus Proposition 19 is an immediate consequence of Corollary 8 too.

3. Some Results on Hilbert Transform

In Section 2, we find that if , by odd rearrangement, we obtain and . In this section, we use the rearrangement, which is like the odd rearrangement, to show that for given and , there exists a function satisfying and .

Proposition 20. For given and , there exists a function satisfying and .

Proof. Since , we discuss it in two cases.
Case 1. Given that and , then there exists a function satisfying and .
Case 2. Given that and , then there exists a function satisfying and .
Proof of Case 1. We take , , and . Then is a continuous function defined on . And we consider the function Then we obtain . We take then Here, we use three steps to prove .
Step 1. Consider
Proof. We take ; then . Consider
Step 2. . Consider
Step 3. is decreasing on the interval . Consider So we obtain that Thus is decreasing on the interval . Since is continuous on the interval , it is easy to conclude that . Now let us discuss Case 1.
, then there exists satisfying Taking , then we obtain that At the same time, we can easily verify that .
Proof of Case 2. We know that / = C1 in Section 2. If we take , then ; we note that Here, we prove that .
We know that Hilbert transform satisfies Thus we obtain Then Since , so we just need an estimate of : In order to estimate the limit of , we consider the following function: We obtain that thus This means that . Now we construct the following functions:
When , we have; and when, we have
Using the continuity of on the interval , we obtain that for each , there exists a constant satisfying So Case 2 is true.

Here, we can get some corollaries.

Corollary 21. Suppose that , are two disjoint open intervals with given same length; note that is the distance between and ; then depends only on . Precisely(i) when ;(ii); (iii) and are disjoint open intervals.

Proof. Since Hilbert transform, , commutes with translations and dilations, we can suppose that Because Hilbert transform, , commutes with dilations, we obtain that We take ; then we obtain that where , and is defined in the proof of Proposition 20. According to the proof of Proposition 20, we conclude the following:(i);(ii);(iii) is decreasing and continuous on the interval .And the corollary is consequence of the three items above.

As in [1], we know that the elements of have integral zero. In Exercise 6.4.3 of [1], we learn that, for some , if is compactly supported with integral zero, then . In the following example, with the help of rearrangement, we construct an odd function in while not in . Thus “compact" condition cannot be removed without modifying other conditions properly.

Example 22. According to Corollary 21, we construct two sequences of intervals and satisfying the following:(i) and and are disjoint intervals;(ii), ;(iii) and ,;(iv),.

We take and , since ; we obtain that Since in as , we obtain that in as . This means that there is a sequence satisfying By noting that , then we obtain that Since , we obtain that if and only if . And we easily obtain that Since , we obtain that Then