#### Abstract

We found that the classical Calderón-Zygmund singular integral operators are bounded on both the classical Hardy spaces and the product Hardy spaces. The purpose of this paper is to extend this result to a more general class. More precisely, we introduce a class of singular integral operators including the classical Calderón-Zygmund singular integral operators and show that they are bounded on both the classical Hardy spaces and the product Hardy spaces.

#### 1. Introduction

The classical Hardy spaces and the product Hardy spaces play important roles in Harmonic analysis, which are due to the original work of Fefferman and Stein [1] and Gundy and Stein [2], respectively. It is well known that these two Hardy spaces are essentially different. For instance, see [35]. It has been known that the classical Calderón-Zygmund singular integral operators are bounded on the classical Hardy spaces and the product singular integral operators are bounded on the product Hardy spaces. Surprisingly, in [6], we found that the classical Calderón-Zygmund singular integral operators are also bounded on the product Hardy spaces. More precisely, if is a bounded operator on with , where the kernel and satisfies for and , then is bounded on both the classical Hardy spaces and the product Hardy space .

A natural question arises: weather there exist a more general class of operators that are bounded both on the classical Hardy spaces and the product Hardy spaces. The purpose of this paper is to answer this question. Now we first recall the definitions of the classical Hardy spaces (see [1] for more details) and the product Hardy spaces (see [2, 7] for more details).

We let be the set including all that satisfy and for all . And let be the set including all that satisfy and for all .

Given an , the Littlewood-Paley-Stein square function of is defined by , where , . And the discrete square function is defined by , where are dyadic cubes in with the side length and the center and is the characteristic function. It is well known that if , then and, for different , .

The classical Hardy space is then defined by where denotes the space of distributions modulo polynomials. The norm is defined by .

Similarly, given a , the product square function of is defined by , where , . And the discrete square function is defined by , where are dyadic intervals in with the side length , and the center , respectively. Also, for , and, for different ,  .

The product Hardy space is then defined by The norm is defined by .

The following theorem is our main result.

Theorem 1. If , and is an operator bounded on with , where the kernel and satisfies for all , then is bounded on both and .

Remark 2. (I) In [8], we have shown that the operator is bounded on for all .
(II) It is easy to verify that the classical Calderón-Zygmund singular integral operators are contained in our class. Moreover, some more operators will be in our class. For example, the operator with .

Throughout this paper, we do the following conventions.(a)The notation means that for some positive constants .(b)If is a cube or interval, then we denote by its center and by its side length.(c)For and a large positive integer , we denote the set , where are dyadic cubes in with side length and .(d)For and a large positive integer , we denote the set , where and are dyadic intervals in with side length and , respectively} and .(e) means the minimum of and .

#### 2. Proof of Theorem 1

The first crucial tool in the proof of Theorem 1 is to apply the following discrete Calderón identity (see [9] for more details).

Lemma 3. If , then consider the following.
() Suppose that with . Then for all , there exist a large positive integer (depending only on ) and a function such that , where the series converges in . Moreover, and .
() Suppose that with . Then for all , there exist a large positive integer (depending only on ) and a function such that , where the series converges in . Moreover, and .

For the proof, we refer readers to [9].

The second crucial tool in the proof of Theorem 1 is the following orthogonal estimates.

Lemma 4. Suppose that and is the kernel as in Theorem 1; then
() for with , one has , for all and , where is a constant independent of and ;
() for with , one has , for all and , where is a constant independent of , , and .

Proof. (a) Since is single-parameter dilation invariant, that is, for each , satisfies the same hypotheses, with the same bounds as . We just need to show that , for all . Without loss of generality, we may assume that . To get the required estimate, we will discuss it in the following three cases: (I) ; (II) or ; (III) .
For case (I), , by the moment condition of , we have
For case (II), or , we have
For case (III), , we let with and when and when . We have
(b) Without loss of generality, we may assume that . The required estimate will be discussed in the following four cases: (I) ; (II) ; (III) ; (IV) .
For case (I), , by the moment condition of , we have
For case (II), , similarly, we have
The cases (II) and (III) are symmetric, so case (III) follows.
For case (IV), , we let with and when and when . Then
This completes the proof of Lemma 4.

As a consequence of Lemma 4, we have the following.

Lemma 5. (a) Under hypothesis (a) of Lemma 4, one has for all and .
(b) Under hypothesis (b) of Lemma 4, one has for all , , and and .

The proof of Lemma 5 is based on the following two observations: convolution operation is commutative; that is, (or );    (or ) satisfies the same estimate as (or ) in Lemma 4 with the bound (or ). The details are left to the readers.

The last crucial tool in the proof of Theorem 1 is the following strongly maximal function estimates.

Lemma 6. Suppose that , , , , , and , and . Then consider the following.
() If , , and , then one has where is the strongly maximal operator.
() If , , and , then one has

For the proof, we refer readers to [10].

Proof of Theorem 1. Firstly we show that is bounded on the classical Hardy space . Since is dense in , we just need to show that, for all , we have ; that is, for a fixed ,
Note that
For , , and , applying (a) of Lemma 3, we have
By (a) of Lemma 5, we have
Applying (a) of Lemma 6 with and , we have
Therefore,
Applying Fefferman-Stein’s vector-valued strong maximal inequality (see [11] for more details) on , we have
The proof of ’s boundedness on is almost the same as above; that is, we just need to replace (a) of the required lemmas to (b). Here we omit the details.

#### Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.

#### Acknowledgment

Chaoqiang Tan is supported by SRF for the Doctoral Program of Higher Education (Grant no. 20104402120002).