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Recent Advances in Function Spaces and its Applications in Fractional Differential Equations 2020

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Volume 2020 |Article ID 3543165 | https://doi.org/10.1155/2020/3543165

Lei Zhang, Shaoguang Shi, "A Characterization of Central BMO Space via the Commutator of Fractional Hardy Operator", Journal of Function Spaces, vol. 2020, Article ID 3543165, 7 pages, 2020. https://doi.org/10.1155/2020/3543165

A Characterization of Central BMO Space via the Commutator of Fractional Hardy Operator

Academic Editor: Lishan Liu
Received25 Mar 2020
Accepted23 Apr 2020
Published31 May 2020

Abstract

This paper is devoted in characterizing the central BMO space via the commutator of the fractional Hardy operator with rough kernel. Precisely, by a more explicit decomposition of the operator and the kernel function, we will show that if the symbol function belongs to the central BMO space, then the commutator are bounded on Lebesgue space. Conversely, the boundedness of the commutator implies that the symbol function belongs to the central BMO space by exploiting the center symmetry of the Hardy operator deeply.

1. Introduction

In this paper, we focus on the need for characterizing the central BMO space via the boundedness of the commutators of the following fractional Hardy operators

is the dual operator of . Here, satisfies and denotes the unit sphere in .

For a function , the commutators of and can be written as

In [1], Fu et al. considered the boundedness of and on homogeneous Herz spaces and Lebesgue spaces under the assumption that satisfies (2) and (4). We recall the results from ([1], Proposition 3.1 and Theorem 3.1) as

Suppose that

Then,

For the boundedness of the classical fractional Hardy operator, see [2]. For a ball (i.e., a ball centered at with radius ) and , is the central BMO function space, which was introduced by Lu and Yang [3] via the norm

It is easy to see that CBMO can be understood as a local version of the classical BMO space at the origin [4] and

Hence, the famous John-Nirenberg inequality for BMO space is not true for CBMO space, which reveals that they have quite different properties.

In the study of harmonic analysis, the characterization of function spaces via the boundedness of the commutators plays an important role in the field of PDEs, see, for example, [511] and the references therein. However, there are less attention paid for the commutators with rough kernels since the characterization depends heavily on the smoothness of the kernel function . Under the premise that is smooth enough, i.e., or , see, for example, [1215]. It is difficult to weaken the smoothness of , Chen and Ding [16] considered a characterization of space under the condition that satisfies the following Hölder condition of log type

It is easy to check that (10) is weaker than the Lipschitz condition and stronger than the condition (4). For and , we call satisfies the –Dini condition if where is defined as

As a useful supplement of [1], we give a characterization of the CBMO space via the boundedness of and as follows.

Theorem 1. Let . (a)If then (b)if

A part of Theorem 1 has been proven in [1], we will show the rest of Theorem 1 in Section 2. In what follows, we will denote by a positive constant which may vary from line to line. The symbol means and for the set of all integers. Last, but not least, , , , and with .

2. Proof of Theorem 1.1

We prove Theorem 1 in this section. To do so, we need one lemma about the estimates of the kernel function , which plays a key role in the proof.

Lemma 2. Let satisfy (2) and (10). Then, (a) with be given in (10) and .(b)If furthermore satisfies the –Dini condition, then, there is a such that for , , with ; we have

Proof. We give the proof by a slight modification from [17]. For , we first show that Indeed, the first inequality can be obtained immediately from ([18], Lemma 2). Since (a) can be shown by (2) and (10) as We are left to show (b). The first estimate can be obtained directly by the –Dini condition as and the estimate follows from ([19], p.65-77).
Accordingly, the second can be deduced similarly. In fact, which is the desired one.

The following is the boundedness of the fractional Hardy operators.

Theorem 3. Let with and satisfy (2) and (4). Then, both and are bounded from to .

Proof. Since the boundedness for is contained in ([1], Proposition 3.1), it is enough to check the boundedness for . Namely, the task is now left to show that there exist constants such that for any , one has To do so, we first recall a useful estimates from [1] as for , and . Since Applying Hölder’s inequality to for and (24), we have which is our desired result.

Now, we can prove Theorem 1. Without loss of generality, we can assume that in the proof of (a) since . We see at once that the boundedness of is just ([1], Theorem 3.1). To complete the proof of (a), what is left is to show is that

It is easy to check that

Using Hölder’s inequality, we have

For the term , we see at once that

The Hölder inequality, along with (24), implies

The term need a further decomposition as follows:

Applying the Hölder inequality, we deduce

From the fact that the term can be estimated as follows: which is the desired result and (a) is obtained.

Next, we verify (b) inspired by [18]. Namely, we need to show that there is a constant such that

For abbreviation, we assume that , , and since . Let It is easy to check that

Applying (3) and (10), we deduce that for and appearing in (10), there is a constant such that for For and with , we obtain from (10) that

Writing we see at once that for , We conclude from (16) that since and , and hence, , and finally, that

Furthermore,

For abbreviation, we write

This, along with the estimates for and , one has

Consequently,

This in turn implies that Thus, is proved if . If , we see immediately that

According to , , , Lemma 2 and (37), we can obtain that for and ,

We continue to choose for and get

It is easy to check that

To deal with the term , we first obtain from (44) that

Then, the estimate for consists of two cases.

Case 1. . Since for , we conclude from (47) and that

Case 2. . In this case, This in turn reveals that