Let 𝐿=Δ+𝑉 be a Schrödinger operator on 𝑛, where 𝑉𝐿1loc(𝑛) is a nonnegative function on 𝑛. In this article, we show that the Hardy spaces 𝐿 on product spaces can be characterized in terms of the Lusin area integral, atomic decomposition, and maximal functions.

1. Introduction

Let 𝑉𝐿1loc(𝑛) be a nonnegative function on 𝑛. The Schrödinger operator with potential 𝑉 is defined by 𝐿=Δ+𝑉on𝑛.(1.1) The operator 𝐿 is a self-adjoint positive definite operator. From the Feynman-Kac formula, it is well known (see, e.g., [1, page 195]) that the kernel 𝑡(𝑥,𝑦) of the semigroup 𝑒𝑡𝐿 satisfies the estimate0𝑡1(𝑥,𝑦)(4𝜋𝑡)𝑛/2𝑒|𝑥𝑦|2/4𝑡(1.2) for all 𝑡>0 and 𝑥,𝑦𝑛.

Let us consider the Hardy space on product domains. We note that the usual Hardy space 𝐻1(𝑛×𝑛) on the product domain is now well understood (see, e.g., [24]). In this paper we will be concerned with the space 𝐻1𝐿(𝑛×𝑛) associated to 𝐿 as introduced in [5] (see [6, 7] for one-parameter theory). Firstly, we set𝐻2(𝑛×𝑛)=(𝐿𝐿)=(𝐿𝐿)𝑢𝐿2(𝑛×𝑛)𝑢𝐿2(𝑛×𝑛)(1.3) and note that𝐿2(𝑛×𝑛)=(𝐿𝐿)𝒩(𝐿𝐿)=𝐻2(𝑛×𝑛)𝒩(𝐿𝐿),(1.4) where (𝐿𝐿) (resp., 𝒩(𝐿𝐿)) stands for the range (resp., the nullspace) of 𝐿𝐿, and the sum is orthogonal. For a function 𝑓𝐿2(𝑛×𝑛), define𝑆𝐿𝑥(𝑓)1,𝑥2=111||𝑥22||2|𝑥𝑦|<𝑡𝑦<𝑡|||𝑄𝑡21𝑡22𝑓𝑦1,𝑦2|||2𝑑𝑦1𝑑𝑦2𝑑𝑡1𝑡1𝑛+1𝑑𝑡2𝑡2𝑛+11/2,(1.5) where 𝑄𝑡1𝑡2𝑓=𝑡1𝑑𝑑𝑡1𝑡2𝑑𝑑𝑡2𝐻𝑡1𝑡2𝐻𝑓,𝑡1𝑡2𝑓𝑦1,𝑦2=𝑛×𝑛𝑡1𝑦1,𝑧1𝑡2𝑦2,𝑧2𝑓𝑧1,𝑧2𝑑𝑧1𝑑𝑧2.(1.6) The space 𝐻1𝐿(𝑛×𝑛) is defined as the completion of 𝐻2(𝑛×𝑛) in the norm given by𝑓𝐻1𝐿(𝑛×𝑛)=𝑆𝐿(𝑓)𝐿1(2𝑛).(1.7)

The main purpose of this article is to derive atomic characterizations and the maximal characterizations of 𝐻1𝐿(𝑛×𝑛). Before stating our results, let us recall some necessary notations (see also [8, 9]). Suppose that Ω𝑛×𝑛 is an open set with finite measure. Denote by 𝑚(Ω) the maximal dyadic subrectangles of Ω in the form of 𝑅=𝐼×𝐽, where 𝐼 and 𝐽 are cubes in 𝑛. Let 𝑅=𝐼×𝐽𝑚(Ω) and we denote by (𝐼) and (𝐽) the side lengths of 𝐼 and 𝐽, respectively. For given 𝜆>0, we will write 𝜆𝑅 for the 𝜆-fold dilation of 𝑅=𝐼×𝐽 with the same center.

An 𝐿-atom is a function 𝑎 on 2𝑛, together with an open set Ω of finite measure, which satisfies the following properties:(i)Ωsupp(𝑎) (see (2.4));(ii)𝑎(𝑥) can be further decomposed as 𝑎=𝑅𝑚(Ω)𝑎𝑅, where 𝐿supp1𝑖𝐿1𝑗𝑎𝑅10𝑅,𝑅𝑚(Ω)(𝐼)2𝐿1𝑖(𝐽)2𝐿1𝑗𝑎𝑅2𝐿2(2𝑛)||Ω||1,𝑖,𝑗=0,1;(1.8)(iii)𝑎𝐿2(2𝑛)|Ω|2.

We can define the atomic Hardy space 𝐻1atom(𝑛×𝑛) by𝑗=0𝜆𝑗𝛼𝑗𝜆𝑗𝑗=01and𝛼𝑗are𝐿-atoms(1.9) with the norm 𝑓𝐻1atom(𝑛×𝑛) given by the natural quotient norm:inf𝑗=0||𝜆𝑗||𝑓=𝑗=0𝜆𝑗𝛼𝑗𝜆,where𝑗𝑗=01and𝛼𝑗.are𝐿-atoms(1.10)

The first result of this paper is the following theorem.

Theorem 1.1. Let 𝐿 be the Schrödinger operator as (1.1). Then the spaces 𝐻1𝐿(𝑛×𝑛) and 𝐻1atom(𝑛×𝑛) coincide. In other words, 𝑓𝐻1𝐿(𝑛×𝑛)𝑓𝐻1atom(𝑛×𝑛),forevery𝑓𝐻1𝐿(𝑛×𝑛).(1.11)

Next we give the “maximal" characterizations of 𝐻1𝐿(𝑛×𝑛). Given a function 𝑓𝐿1(2𝑛) we define two functions:𝑓+(𝑥)=sup𝑡1,𝑡2>0||𝐻𝑡1,𝑡2||,𝒩𝑓(𝑥)𝑓(𝑥)=sup|𝑥1𝑦1|<𝑡1|𝑥2𝑦2|<𝑡2|||𝐻𝑡21,𝑡22𝑓|||.(𝑦)(1.12) Similarly, one can consider the Poisson semigroup generated by the operator 𝐿 and the operators𝑆𝑃𝑓(𝑥)=111||𝑥22||2|𝑥𝑦|<𝑡𝑦<𝑡|||𝑡1𝐿1𝑒𝑡1𝐿1𝑡2𝐿2𝑒𝑡2𝐿2|||𝑓(𝑦)2𝑑𝑦1𝑑𝑦2𝑑𝑡1𝑑𝑡2𝑡1𝑛+1𝑡2𝑛+11/2,𝑓+𝑃(𝑥)=sup𝑡1𝑡>02>0|||𝑒𝑡1𝐿1𝑒𝑡2𝐿2|||,𝒩𝑓(𝑥)𝑃𝑓(𝑥)=sup|𝑥1𝑦1|<𝑡1|𝑥2𝑦2|<𝑡2|||𝑒𝑡1𝐿1𝑒𝑡2𝐿2|||,𝑓(𝑦)(1.13) with 𝑥𝑛×𝑛 and 𝑓𝐿2(2𝑛).

Define the spaces 𝐻1max,(𝑛×𝑛), 𝐻1𝒩(𝑛×𝑛),𝐻1𝑆𝑃(𝑛×𝑛),𝐻1max,𝑃(𝑛×𝑛), and 𝐻1𝒩𝑃(𝑛×𝑛) as the completion of 𝐻2(𝑛×𝑛) in the norms given by the 𝐿1 norm of the corresponding square or maximal functions, respectively. For example,𝑓𝐻1max,(𝑛×𝑛)𝑓=+𝐿1(2𝑛).(1.14) By a similar manner, the norms of 𝐻1𝒩(𝑛×𝑛),𝐻1𝑆𝑃(𝑛×𝑛),𝐻1max,𝑃(𝑛×𝑛), and 𝐻1𝒩𝑃(𝑛×𝑛) are defined. The second result of this paper is the following.

Theorem 1.2. Let 𝐿 be the Schrödinger operator as (1.1). Then the spaces 𝐻1𝐿(𝑛×𝑛), 𝐻1max,(𝑛×𝑛), 𝐻1𝒩(𝑛×𝑛),𝐻1𝑆𝑃(𝑛×𝑛),𝐻1max,𝑃(𝑛×𝑛), and 𝐻1𝒩𝑃(𝑛×𝑛) coincide with equivalent norms.

This paper is organized as follows. In Section 2, we will give some preliminary results including the properties of Schrödinger operators and tent spaces on product spaces. The proofs of Theorems 1.1 and 1.2 will be given in Sections 3 and 4, respectively.

Throughout this paper, the letter “𝐶" or “𝑐" will denote (possibly different) constants that are independent of the essential variables.

2. Preliminaries

2.1. Tent Spaces on Product Domains

A theory of “tent spaces" was developed by Coifman et al. [10, 11]. These spaces are useful for the study of a variety of problems in harmonic analysis. In particular, we note that the tent spaces give a natural and simple approach to the atomic decomposition of functions in the classical Hardy space 𝐻𝑝(𝑛) by using the area integral functions and the theory of the Carleson measure. See also [6, 12].

Tent spaces have been studied by [13, 14] in connection with the theory of Carleson measures on product domains. Let +𝑛+1 be the usual upper half-space in 𝑛+1. For any 𝛼>0, we set Γ𝛼(𝑥)={(𝑦,𝑡)+𝑛+1×+𝑛+1,|𝑥1𝑦1|<𝛼𝑡1,and|𝑥2𝑦2|<𝛼𝑡2} as the standard cone (of aperture 𝛼) with vertex 𝑥𝑛×𝑛. In particular, we set Γ(𝑥)=Γ1(𝑥). If (𝑥,𝑡)+𝑛+1×+𝑛+1, then 𝑅𝑥,𝑡 denotes the rectangle centered at 𝑥𝑛×𝑛 whose side lengths are 𝑡1 and 𝑡2, respectively. For any open set Ω𝑛×𝑛, the tent over Ω, 𝑇(Ω), is the set(𝑥,𝑡)+𝑛+1×+𝑛+1𝑅𝑥,𝑡.Ω(2.1) For any function 𝑓(𝑦,𝑡) defined on +𝑛+1×+𝑛+1 we will write𝒜(𝑓)(𝑥)=Γ(𝑥)||||𝑓(𝑦,𝑡)2𝑑𝑦𝑑𝑡𝑡1𝑛+1𝑡2𝑛+11/2.(2.2) The tent space 𝑇𝑝2 is then defined as the space of functions 𝑓 such that 𝒜𝑓𝐿𝑝(2𝑛) and is equipped with the norm, 𝑓𝑇𝑝2=𝒜(𝑓)𝑝, 0<𝑝<.

We now introduce 𝑇12-atoms.

Definition 2.1. A function 𝐴(𝑥,𝑡)+𝑛+1×+𝑛+1 is called a 𝑇12-atom if there exists an open set Ω𝑛×𝑛 of finite measure satisfying the following properties:(i)𝐴(𝑥,𝑡) can be further decomposed as 𝐴=𝑅𝑚(Ω)𝐴𝑅, where each 𝐴𝑅 is supported in 𝑇(3𝑅), and 𝑅Ω is a maximal dyadic subrectangle of Ω in the form of 𝑅=𝐼×𝐽, where 𝐼 and 𝐽 are cubes in 𝑛;(ii)𝐴𝐿2(𝑑𝑦𝑑𝑡/𝑡1𝑡2)|Ω|1/2 and 𝑅𝑚(Ω)𝐴𝑅2𝐿2(𝑑𝑦𝑑𝑡/𝑡1𝑡2)|Ω|1.

Proposition 2.2. Suppose 𝑓𝑇12. Then 𝑓=𝑘𝜆𝑘𝐴𝑘, where 𝐴𝑘 are 𝑇12-atoms, 𝜆𝑘, and 𝑘|𝜆𝑘|𝑐𝑓𝑇12 so that the sum converges in the 𝑇12 norm. Moreover, if one assumes that 𝑓𝑇22, then the sum also converges in the 𝑇22 norm.

Proof. See [5, Proposition  3.3] for the proof.

2.2. Some Results on Product Spaces

We recall that the strong maximal function is defined as follows:𝑀𝑠𝑓(𝑥)=sup𝑥𝑅=𝐼×𝐽1||𝑅||𝑅||||𝑓(𝑥)𝑑𝑥,(2.3) where 𝐼 and 𝐽 are cubes in 𝑛. It is well known that the operator 𝑀𝑠 is bounded on 𝐿𝑝(2𝑛), for 1<𝑝<.

Now for any open set Ω𝑛×𝑛 with finite measure, we setΩ=𝑥𝑛×𝑛,𝑀𝑠𝜒Ω1(𝑥)>22𝑛.(2.4) By the strong maximal theorem, |Ω|𝑐|Ω|. Denote by 𝑚1(Ω) the dyadic subrectangles 𝑅=𝐼×𝐽Ω that are maximal in the 𝑥1 direction, where 𝐼and𝐽 are dyadic cubes in 𝑛. Define 𝑚2(Ω) similarly. It is well known that Journé’s covering lemma holds (see [8, 15]).

Lemma 2.3. Let Ω𝑛×𝑛. For any 𝑅=𝐼×𝐽𝑚2(Ω), one sets 𝛾1(𝑅)=sup𝐼𝑙𝑙×𝐽Ω|𝑙|/|𝐼|. Define 𝛾2 similarly. Then for any 𝛿>0, one has 𝑅𝑚2(Ω)||𝑅||𝛾1𝛿(𝑅)𝑐𝛿||Ω||,𝑅𝑚1(Ω)||𝑅||𝛾2𝛿(𝑅)𝑐𝛿||Ω||,(2.5) where 𝑐𝛿 is a constant depending only on 𝛿, but not on Ω.

The following lemma shows that in order to prove that an operator is bounded from 𝐻1atom(𝑛×𝑛) to 𝐿1(2𝑛), we just need to check that the operator is uniformly bounded on the 𝐿-atoms.

Lemma 2.4. Assume that 𝑇 is either a linear operator or a positive sublinear operator, bounded on 𝐿2(2𝑛) and for every 𝐿-atom 𝑎, 𝑇(𝑎)𝐿1(2𝑛)𝑐(2.6) with constant 𝑐 independent on 𝑎. Then 𝑇 can extend to a bounded operator from 𝐻1𝐿(𝑛×𝑛) to 𝐿1(2𝑛), and 𝑇(𝑓)𝐿1(2𝑛)𝑐𝑓𝐻1𝐿(𝑛×𝑛).(2.7)

Proof. Its proof is similar to that of [16, Lemma  3.3] and we omit it here. See also [17].

2.3. Some Properties of the Schrödinger Operator 𝐿 on 𝑛

Let 𝐿 be the Schrödinger operator as (1.1), and let 𝑡(𝑥,𝑦) be the kernels of the operators of semigroup {𝑒𝑡𝐿}.

First we note that, for each 𝑘, there exist two positive constants 𝐶𝑘 and 𝑐𝑘 such that the time derivatives of 𝑡 satisfy||||𝜕𝑘𝜕𝑡𝑘𝑡||||(𝑥,𝑦)𝐶𝑘𝑡(𝑛+2𝑘)/2||||exp𝑥𝑦2𝑐𝑘𝑡(2.8) for all 𝑡>0 and almost all 𝑥,𝑦𝑛. For the proof, see, for example, [18, 19].

Next, for 𝑠>0, we define||||𝔽(𝑠)=𝜓measurable𝜓(𝑧)𝐶|𝑧|𝑠1+|𝑧|2𝑠.(2.9) Then for any nonzero function 𝜓𝔽(𝑠), we have that {0|𝜓(𝑡)|2𝑑𝑡/𝑡}1/2<. Denote 𝜓𝑡(𝑧)=𝜓(𝑡𝑧). It follows from the spectral theory in [20] that for any 𝑓𝐿2(𝑛),0𝜓𝑡𝐿𝑓2𝐿2(𝑋)𝑑𝑡𝑡1/2=0𝜓𝑡𝐿𝜓𝑡𝐿𝑓,𝑓𝑑𝑡𝑡1/2=0||𝜓||2𝑡𝐿𝑑𝑡𝑡𝑓,𝑓1/2=𝜅𝑓𝐿2(𝑛),(2.10) where 𝜅={0|𝜓(𝑡)|2𝑑𝑡/𝑡}1/2. As an application, we have 𝑆𝐿(𝑓)2=𝑐𝑓2 for some constant 𝑐.

Lemma 2.5. Let 𝑓𝐿2(𝑛) and 𝑢=𝑒𝑡𝐿𝑓. Then for any 𝑝>0, there exists a constant 𝐶=𝐶(𝑛,𝑝)>0 such that sup𝐵((𝑥0,𝑡0),𝑟)||||1𝑢(𝑥,𝑡)𝐶𝑟𝑛+1𝐵((𝑥0,𝑡0),2𝑟)||||𝑢(𝑥,𝑡)𝑝𝑑𝑥𝑑𝑡1/𝑝,(2.11) where 𝐵((𝑥0,𝑡0),𝑟)=𝐵𝑟(𝑥0)×[𝑡0𝑐𝑟,𝑡0] with 𝑡0>2𝑐𝑟 and 𝑐>0.

Proof. For the proof, we refer to [7, Lemma  8.4].

Recall that if 𝐿 is a nonnegative, self-adjoint operator on 𝐿2(𝑛), and 𝐸𝐿(𝜆) denotes its spectral decomposition, then for every bounded Borel function 𝐹[0,), one defines the operator 𝐹(𝐿)𝐿2(𝑛)𝐿2(𝑛) by the formula𝐹(𝐿)=0𝐹(𝜆)𝑑𝐸𝐿(𝜆).(2.12) In particular, the operator cos(𝑡𝐿) is then well defined on 𝐿2(𝑛). Moreover, it follows from [21, Theorem  3] that the Schwartz kernel 𝐾cos(𝑡𝐿)(𝑥,𝑦) of cos(𝑡𝐿) satisfiessupp𝐾cos(𝑡𝐿)(𝑥,𝑦)(𝑥,𝑦)𝑛×𝑛||||.𝑥𝑦𝑡(2.13) See also [22]. By the Fourier inversion formula, whenever 𝐹 is an even bounded Borel function with the Fourier transform of 𝐹, 𝐹𝐿1(), we can write 𝐹(𝐿) in terms of cos(𝑡𝐿). In fact, using (2.12) we have𝐹𝐿=(2𝜋)1𝑡𝐹(𝑡)cos𝐿𝑑𝑡,(2.14) which, when combined with (2.13), gives𝐾𝐹(𝐿)(𝑥,𝑦)=(2𝜋)1|||||𝑡|𝑥𝑦𝐹(𝑡)𝐾cos(𝑡L)(𝑥,𝑦)𝑑𝑡.(2.15)

Lemma 2.6. Let 𝜙𝐶0() be even, 𝜙0, 𝜙𝐿1()=1,supp𝜙[1,1], and set 𝜑(𝑥)=2𝜙(2𝑥)𝜙(𝑥) and 𝜑𝑡(𝑥)=(1/𝑡)𝜑(𝑥/𝑡) for 𝑡>0. Let Φ and Φ𝑡 denote the Fourier transform of 𝜑 and 𝜑𝑡,respectively. Then, the kernels of the operators Φ𝑡(𝐿) and 𝑡2𝐿Φ𝑡(𝐿) have supports contained in {(𝑥,𝑦)𝑛×𝑛|𝑥𝑦|𝑡}.

Proof. For the proof, we refer the reader to [7, Lemma  3.5].

Lemma 2.7. Let Ψ(𝑥)=𝑥2Φ(𝑥) as in Lemma 2.6; then the operator 𝜋𝐿(𝑓)(𝑥)=0Ψ𝑡1𝐿𝑡Ψ2𝐿(𝑓(,𝑡))(𝑥)𝑑𝑡1𝑑𝑡2𝑡1𝑡2(2.16) is bounded from 𝑇𝑝2 to 𝐿𝑝, if 1<𝑝<.

Proof. For the proof, we refer to [5, Lemma  3.4].

3. Proof of Theorem 1.1

3.1. The Inclusion of 𝐻1𝐿(𝑛×𝑛)𝐻1atom(𝑛×𝑛)

Let 𝑓𝐻1𝐿(𝑛×𝑛)𝐻2(𝑛×𝑛). Then 𝐹=𝑡21𝐿𝑒𝑡21𝐿𝑡22𝐿𝑒𝑡22𝐿𝑓𝑇12𝑇22. We start with a suitable version of the Calderón reproducing formula. Let Φ be as in Lemma 2.6, Ψ(𝑥)=𝑥2Φ(𝑥), and let 𝑐Ψ be a constant such that 𝑐Ψ0𝑡Ψ(𝑡)𝑒𝑡2𝑑𝑡=1. By 𝐿2-functional calculus [23], one can write𝑓(𝑥)=𝑐2Ψ0Ψ𝑡1𝐿𝑡21𝐿𝑒𝑡21𝐿𝑡Ψ2𝐿𝑡22𝐿𝑒𝑡22𝐿𝑓(𝑥)𝑑𝑡1𝑑𝑡2𝑡1𝑡2,(3.1) where the integral converges in 𝐿2(2𝑛). By Proposition 2.2, 𝐹 has a 𝑇12-atomic decomposition: 𝐹=𝑘=0𝜆𝑘𝐴𝑘, where 𝑘=0|𝜆𝑘|𝐶𝐹𝑇12𝐶𝑓𝐻1𝐿(𝑛×𝑛), and 𝐴𝑘 are 𝑇12-atoms associated to an open set Ω𝑘. It is easy to see that the sum converges in 𝑇12 and 𝑇22. We have𝑓(𝑥)=𝑐2Ψ0Ψ𝑡1𝐿𝑡21𝐿𝑒𝑡21𝐿𝑡Ψ2𝐿𝑡22𝐿𝑒𝑡22𝐿𝑓(𝑥)𝑑𝑡1𝑑𝑡2𝑡1𝑡2=𝑘=0𝑐2Ψ0Ψ𝑡1𝐿𝑡Ψ2𝐿𝜆𝑘𝐴𝑘(,𝑡)𝑑𝑡1𝑑𝑡2𝑡1𝑡2=𝑘=0𝜇𝑘𝑎𝑘(𝑥),(3.2) where 𝜇𝑘=𝑐0𝜆𝑘 and 𝑎𝑘(𝑥)=𝑐010Ψ(𝑡1𝐿)Ψ(𝑡2𝐿)(𝐴𝑘(,𝑡))(𝑥)𝑑𝑡1𝑑𝑡2/𝑡1𝑡2. Here 𝑐0 is a large constant determined later. Using Lemma 2.7, we can show that the sum (3.2) converges in 𝐿2(2𝑛).

It is known that 𝑘=0|𝜇𝑘|𝐶𝑓𝐻1𝑆𝐿(𝑛×𝑛). Now we turn to check that 𝑎𝑘 are 𝐿-atoms associated to open sets Ω𝑘. Since 𝐴𝑘 are 𝑇12-atoms, then 𝐴𝑘=𝑅𝑚(Ω𝑘)𝐴𝑘,𝑅. Thus𝑎𝑘(𝑥)=Ω𝑅𝑚𝑘𝑐010Ψ𝑡1𝐿𝑡Ψ2𝐿𝐴𝑘,𝑅(,𝑡)(𝑥)𝑑𝑡1𝑑𝑡2𝑡1𝑡2=Ω𝑅𝑚𝑘𝑎𝑘,𝑅(𝑥).(3.3) Lemma 2.6 implies that for 𝑖,𝑗=0,1, supp((𝐿1)𝑖(𝐿1)𝑗𝑎𝑘Ω)𝑘 and supp((𝐿1)𝑖(𝐿1)𝑗𝑎𝑘,𝑅)10𝑅.

To continue, we write𝑅𝑚(Ω𝑘)(𝐼)2𝐿1𝑖(𝐽)2𝐿1𝑗𝑎𝑘,𝑅22𝑅𝑚(Ω𝑘)sup21||||2𝑛(𝐼)2𝐿1𝑖(𝐽)2𝐿1𝑗𝑎𝑘,𝑅||||(𝑥)(𝑥)𝑑𝑥2𝑐01𝑅𝑚(Ω𝑘)sup21||||+𝑛+1×+𝑛+1𝐴𝑘,𝑅×(𝑦,𝑡)(𝐼)2𝐿1𝑖Ψ𝑡1𝐿(𝐽)2𝐿1𝑗Ψ𝑡2𝐿(𝑦)𝑑𝑦𝑑𝑡𝑡1𝑡2||||2𝑐01𝑅𝑚(Ω𝑘)𝐴𝑘,𝑅2𝐿2(𝑑𝑦𝑑𝑡/𝑡1𝑡2)||Ω𝑘||1,(3.4) if we choose 𝑐0 large enough. By a similar argument, we have(𝐼)2𝐿1𝑖(𝐽)2𝐿1𝑗𝑎𝑘2||Ω𝑘||1/2.(3.5)

For general 𝑓𝐻1𝐿(𝑛×𝑛), we just need a standard argument. The inclusion of 𝐻1𝐿(𝑛×𝑛)𝐻1atom(𝑛×𝑛) is finished.

3.2. The Inclusion of 𝐻1Atom(𝑛×𝑛)𝐻1𝐿(𝑛×𝑛)

By Lemma 2.4, it is enough to show that 𝑆𝐿(𝑎)1𝐶 for any 𝐿-atom 𝑎 associated to an open set Ω. For 𝑅=𝐼×𝐽𝑚(Ω), we let 𝑙 be the maximal dyadic cube such that 𝐼𝑙 and Ω𝑙×𝐽. Also we let 𝑆 be the maximal dyadic cube such that 𝐽𝑆 and Ω𝑙×𝑆. Let 𝑅=100𝑙×100𝑆. We can see that |𝑅𝑚(Ω)𝑅|𝐶|Ω|, since𝑅𝑚(Ω)𝑅𝑥𝑛×𝑛,𝑀𝑠𝜒Ω1(𝑥)>2002𝑛.(3.6) Due to Hölders inequality one has𝑅𝑚(Ω)𝑅𝑆𝐿|||||(𝑎)(𝑥)𝑑𝑥𝑅𝑚(Ω)𝑅|||||1/22𝑛𝑆2𝐿(𝑎)(𝑥)𝑑𝑥1/2||Ω||𝐶1/2𝑎2𝐶.(3.7) Let us prove (𝑅𝑚(Ω)𝑅)𝑐𝑆𝐿(𝑎)(𝑥)𝑑𝑥𝐶. One can write(𝑅𝑚(Ω)𝑅)𝑐𝑆𝐿(𝑎)(𝑥)𝑑𝑥𝑅𝑚(Ω)(100𝑙×100𝑆)𝑐𝑆𝐿𝑎𝑅𝑥1,𝑥2𝑑𝑥1𝑑𝑥2𝑅𝑚(Ω)𝑥1100𝑙+𝑥2100𝑆𝑆𝐿𝑎𝑅𝑥1,𝑥2𝑑𝑥1𝑑𝑥2=I+II.(3.8) We only estimate the term I since the proof of the term II is similar. Observe thatI=𝑅𝑚(Ω)𝑥1100𝑙𝑥2100𝐽+𝑥2100𝐽𝑆𝐿𝑎𝑅𝑥1,𝑥2𝑑𝑥2𝑑𝑥1=I1+I2.(3.9) Consider term I1. By Hölder’s inequality, we obtainI1𝑅𝑚(Ω)||𝐽||1/2𝑥1100𝑙𝑥2100𝐽𝑆2𝐿𝑎𝑅𝑥1,𝑥2𝑑𝑥21/2𝑑𝑥1𝐶𝑅𝑚(Ω)||𝐽||1/2𝑥1100𝑙|𝑥1𝑦1|<𝑡1𝑛|||𝑡21𝐿𝑒𝑡21𝐿𝑎𝑅,𝑥2𝑦1|||2𝑑𝑥2𝑑𝑦1𝑑𝑡1𝑡1𝑛+11/2𝑑𝑥1𝐶𝑅𝑚(Ω)||𝐽||1/2𝑥1100𝑙0(𝐼)𝐵(𝑥1,𝑡1)𝑛|||𝑡21𝐿𝑒𝑡21𝐿𝑎𝑅,𝑥2𝑦1|||2𝑑𝑥2𝑑𝑦1𝑑𝑡1𝑡1𝑛+11/2𝑑𝑥1+𝐶𝑅𝑚(Ω)||𝐽||1/2𝑥1100𝑙(𝐼)𝐵(𝑥1,𝑡1)𝑛|||𝑡21𝐿𝑒𝑡21𝐿𝑎𝑅,𝑥2𝑦1|||2𝑑𝑥2𝑑𝑦1𝑑𝑡1𝑡1𝑛+11/2𝑑𝑥1=I11+I12.(3.10) Consider the term I11. It follows from estimate (2.8) thatI11𝐶𝑅𝑚(Ω)||𝐽||1/2×𝑥1100𝑙0(𝐼)𝐵(𝑥1,𝑡1)𝑛𝑛𝑡1𝑡1+||𝑦1𝑧1||𝑛+1||𝑎𝑅𝑧1,𝑥2||𝑑𝑧12𝑑𝑥2𝑑𝑦1𝑑𝑡1𝑡1𝑛+11/2𝑑𝑥1.(3.11) Let 𝑥𝐼 denote the center of cube 𝐼. Note that 𝑥1100𝑙 and |𝑥1𝑦1|<𝑡1<𝑙(𝐼). We use Hölder’s inequality to obtainI11𝐶𝑅𝑚(Ω)||𝐽||1/2||𝐼||1/2𝑥1100𝑙0(𝐼)𝑡1||𝑥1𝑥𝐼||2𝑛+2𝑑𝑡11/2𝑑𝑥1𝑎𝑅𝐿2(2𝑛)𝐶𝑅𝑚(Ω)||𝑅||1/2𝑎𝑅𝐿2(2𝑛)𝛾1(𝑅)1/𝑛𝐶𝑅𝑚(Ω)𝑎𝑅2𝐿22𝑛1/2𝑅𝑚(Ω)||𝑅||𝛾1(𝑅)2/𝑛1/2𝐶,(3.12) where in the last inequality we use Lemma 2.3.

For the term I12, we apply the definition of 𝐿-atom to obtainI12=𝐶𝑅𝑚(Ω)||𝐽||1/2(𝐼)2×𝑥1100𝑙(𝐼)𝐵(𝑥1,𝑡1)𝑛|||𝑡41𝐿2𝑒𝑡21𝐿(𝐼)2𝐿1𝑎𝑅,𝑥2𝑦1|||2𝑑𝑥2𝑑𝑦1𝑑𝑡1𝑡1𝑛+51/2𝑑𝑥1𝐶𝑅𝑚(Ω)||𝐽||1/2(𝐼)2×𝑥1100𝑙(𝐼)𝐵(𝑥1,𝑡1)𝑛𝑛𝑡1𝑡1+||𝑦1𝑧1||𝑛+1||(𝐼)2𝐿1𝑎𝑅,𝑥2𝑧1||𝑑𝑧12𝑑𝑥2×𝑑𝑦1𝑑𝑡1𝑡1𝑛+51/2𝑑𝑥1𝐶𝑅𝑚(Ω)||𝐽||1/2||𝐼||1/2(𝐼)2×𝑥1100𝑙(𝐼)𝑡13𝑡1+||𝑥1𝑥𝐼||2𝑛+2𝑑𝑡11/2𝑑𝑥1(𝐼)2𝐿1𝑎𝐼𝑅𝐿2(2𝑛)𝐶𝑅𝑚(Ω)||𝑅||1/2𝑥1100𝑙(𝐼)||𝑥1𝑥𝐼||𝑛+1𝑑𝑥1(𝐼)2𝐿1𝑎𝐼𝑅𝐿2(2𝑛)𝐶𝑅𝑚(Ω)||𝑅||1/2𝛾1(𝑅)1/𝑛(𝐼)2𝐿1𝑎𝐼𝑅𝐿2(2𝑛)𝐶,(3.13) which, together with estimate of I11, show that I1𝐶. By a similar argument as mentioned previously, we can show I2𝐶. We have obtained the required estimate 𝑆𝐿(𝑎)1𝐶. This completes the proof of Theorem 1.1.

4. Proof of Theorem 1.2

In this section, we will give the proof of Theorem 1.2 in the following routine: 𝐻1atom(𝑛×𝑛)𝐻1𝒩(𝑛×𝑛)𝐻1max,(𝑛×𝑛)𝐻1max,𝑃(𝑛×𝑛)𝐻1𝒩𝑃(𝑛×𝑛)𝐻1𝑆𝑃(𝑛×𝑛𝐻)1atom(𝑛×𝑛). The main idea comes from [9, 16].

Step I. 𝐻1atom(𝑛×𝑛)𝐻1𝒩(𝑛×𝑛). It is similar to the proof of 𝐻1atom(𝑛×𝑛)𝐻1𝐿(𝑛×𝑛). Here we omit the details.

Step II. 𝐻1𝒩(𝑛×𝑛)𝐻1max,(𝑛×𝑛) because 𝑓+(𝑥)𝒩(𝑓)(𝑥).

Step III. 𝐻1max,(𝑛×𝑛)𝐻1max,𝑃(𝑛×𝑛). By 𝐿2-functional calculus ([23]), we have𝑒𝑡1𝐿𝑒𝑡2𝐿𝑓𝑥1,𝑥2=1𝜋1𝜋0𝑒𝑢1𝑢1𝑒𝑢2𝑢2𝑒(𝑡21/4𝑢1)𝐿𝑒(𝑡22/4𝑢2)𝐿𝑓𝑥1,𝑥2𝑑𝑢1𝑑𝑢2.(4.1) Therefore,𝑓+𝑃11𝐶𝜋1𝜋0𝑒𝑢1𝑢1𝑒𝑢2𝑢2𝑓+1𝑑𝑢1𝑑𝑢2𝑓𝐶+1,(4.2) which completes the proof of 𝐻1max,(𝑛×𝑛)𝐻1max,𝑃(𝑛×𝑛).

Step IV. 𝐻1max,𝑃(𝑛×𝑛)𝐻1𝒩𝑃(𝑛×𝑛), for any (𝑦1,𝑦2,𝑡1,𝑡2)Γ(𝑥), and 𝑥𝑛×𝑛. Applying Lemma 2.5 with 0<𝑝<1, we obtain|||𝑒𝑡1𝐿𝑒𝑡2𝐿𝑓𝑦1,𝑦2|||𝑝𝐶𝑡1𝑛+1𝑡2𝑛+1𝑡1𝑡1/2𝐵(𝑥1,2𝑡1)𝑡2𝑡2/2𝐵(𝑥2,2𝑡2)|||𝑒𝑠1𝐿𝑒𝑠2𝐿𝑓𝑧1,𝑧2|||𝑝𝑑𝑧2𝑑𝑠2𝑑𝑧1𝑑𝑠1𝐶𝑡𝑛1𝑡𝑛2𝐵(𝑥1,2𝑡1)𝐵(𝑥2,2𝑡2)𝑓+𝑃𝑝(𝑧)𝑑𝑧𝐶𝑀𝑠||𝑓+𝑃||𝑝(𝑥).(4.3) Therefore,𝒩𝑓𝐿1(2𝑛)𝑀𝑠||𝑓+𝑃||𝑝1/𝑝𝐿1(2𝑛)𝑓𝐶+𝑃𝐿1(2𝑛).(4.4) This completes the proof of 𝐻1max,𝑃(𝑛×𝑛)𝐻1𝒩𝑃(𝑛×𝑛).

Lemma 4.1. Let 𝑓 and 𝑔 be the functions of 𝐿2(𝑛), and suppose that 𝜙𝐶0(𝑛) is radial and supp(𝜙)𝐵(0,1). Let 𝑢(𝑥,𝑡)=𝑒𝑡𝐿𝑓(𝑥). Then one has +𝑛+1|||𝜕𝑢|||𝜕𝑡(𝑥,𝑡)2||𝑔𝜙𝑡||(𝑥)2𝑡𝑑𝑥𝑑𝑡𝐶𝑛||||𝑓(𝑥)2||||𝑔(𝑥)2𝑑𝑥++𝑛+1||||𝑢(𝑥,𝑡)2||𝑔Ψ𝑡||(𝑥)2𝑑𝑥𝑑𝑡𝑡,(4.5) where Ψ𝐶0(𝑛) with Ψ=0 is a vector-value function independent on 𝑓 and 𝑔 and supp(Ψ)𝐵(0,1).

Proof. We note that 𝜕Δ+2𝜕𝑡2|𝑢|2=𝜕Δ+2𝜕𝑡2𝑢𝜕𝑢+Δ+2𝜕𝑡2||||𝑢𝑢+2𝑢2|||+2𝜕𝑢|||𝜕𝑡2=2𝑉|𝑢|2||||+2𝑢2|||+2𝜕𝑢|||𝜕𝑡2|||2𝜕𝑢|||𝜕𝑡2.(4.6) Following the steps in [24], we obtain +𝑛+1𝜕Δ+2𝜕𝑡2|𝑢|2||(𝑥,𝑡)𝑔𝜙𝑡||(𝑥)2𝑡𝑑𝑥𝑑𝑡𝐶𝑛||||𝑓(𝑥)2||||𝑔(𝑥)2𝑑𝑥++𝑛+1||||𝑢(𝑥,𝑡)2||𝑔Ψ𝑡||(𝑥)2𝑑𝑥𝑑𝑡𝑡.(4.7) Then the lemma follows readily.

Applying Lemma 4.1, we can obtain the following lemma.

Lemma 4.2. Suppose that 𝑓,𝑔𝐿2(𝑛×𝑛). Let 𝜙 and Ψ be functions as in Lemma 4.1. Then one has +𝑛+1×+𝑛+1|||𝑡1𝐿𝑒𝑡1𝐿𝑡2𝐿𝑒𝑡2𝐿|||𝑓(𝑥)2||𝜙𝑔𝑡1𝜙𝑡2||(𝑥)2𝑑𝑥𝑑𝑡𝑡1𝑡2𝐶+𝑛+1×+𝑛+1|||𝑒𝑡1𝐿𝑒𝑡2𝐿|||𝑓(𝑥)2||Ψ𝑔𝑡1Ψ𝑡2||(𝑥)2𝑑𝑥𝑑𝑡𝑡1𝑡2+𝑛+𝑛+1|||𝐼𝑒𝑡2𝐿|||𝑓(𝑥)2||𝑔𝑥1,Ψ𝑡2𝑥2||2𝑑𝑥2𝑑𝑡2𝑡2𝑑𝑥1+𝑛+𝑛+1|||𝑒𝑡1𝐿|||𝐼𝑓(𝑥)2||𝑔,𝑥2Ψ𝑡1𝑥1||2𝑑𝑥1𝑑𝑡1𝑡1𝑑𝑥2+2𝑛||||𝑓(𝑥)2||||𝑔(𝑥)2.𝑑𝑥(4.8)

Proof. The proof of Lemma 4.2 can be obtained by iterating Lemma 4.1.

We begin to show 𝐻1𝒩𝑃(𝑛×𝑛)𝐻1𝑆𝑃(𝑛×𝑛) by following the idea of [9] (see also [25, pages 107–109]).

Set 𝑢(𝑥,𝑡)=𝑒𝑡1𝐿𝑒𝑡2𝐿𝑓(𝑥) and 𝑢(𝑥)=sup(𝑦,𝑡)Γ(𝑥)|𝑢(𝑦,𝑡)|.

We claim that||𝑥𝑆𝑃||||(𝑓)(𝑥)>𝛼𝐶𝑥𝑢||>𝛼+𝛼2𝑢𝛼||𝑢||(𝑥)2.𝑑𝑥(4.9) Once the claim holds, we integrate 𝛼 from 0 to to complete the proof of 𝐻1𝒩𝑃(𝑛×𝑛)𝐻1𝑆𝑃(𝑛×𝑛). Now we turn to prove the claim.

The boundedness of the strong maximal operator 𝑀𝑠 on 𝐿2(2𝑛) implies||||𝑥𝑀𝑠𝜒{𝑢>𝛼}1(𝑥)102𝑛||||𝐶2𝑛𝑀2𝑠𝜒{𝑢>𝛼}(𝑥)𝑑𝑥𝐶2𝑛𝜒2{𝑢>𝛼}||𝑢(𝑥)𝑑𝑥=𝐶||.>𝛼(4.10) To prove (4.9), one just need to estimate||||𝑥𝑆𝑃(𝑓)(𝑥)>𝛼,𝑀𝑠𝜒{𝑢>𝛼}<1102𝑛||||.(4.11) Observe that𝑀𝑠(𝜒{𝑢>𝛼})<1/102𝑛𝑆2𝑃(𝑓)(𝑥)𝑑𝑥+𝑛+1×+𝑛+1|||𝑡1𝐿𝑒𝑡1𝐿𝑡2𝐿𝑒𝑡2𝐿|||𝑓(𝑦)2||||𝑀𝑅(𝑦,𝑡)𝑠𝜒{u>𝛼}<1102𝑛||||𝑑𝑦𝑑𝑡𝑡1𝑛+1𝑡2𝑛+1𝑅|||𝑡1𝐿𝑒𝑡1𝐿𝑡2𝐿𝑒𝑡2𝐿|||𝑓(𝑦)2𝑑𝑦𝑑𝑡𝑡1𝑡2,(4.12) where 𝑅={(𝑦,𝑡)|+𝑛+1×+𝑛+1,|𝑅(𝑦,𝑡){𝑢>𝛼}|<|𝑅(𝑦,𝑡)|/102𝑛} and 𝑅(𝑦,𝑡)={(𝑥1,𝑥2)|𝑛×𝑛|𝑥1𝑦1|<𝑡1,|𝑥2𝑦2|<𝑡2}. Now we choose a radial nonnegative function 𝜙𝐶0(𝑛) such that 𝜙(𝑥)=1 if |𝑥|1/3 and 𝜙(𝑥)=0 if |𝑥|>1. Also set 𝑔(𝑥)=𝜒{𝑢𝛼}(𝑥). We can easily check that if (𝑦,𝑡)𝑅, then 𝑔(𝜙𝑡1𝜙𝑡2)(𝑦)𝐶, for some positive constant 𝐶. Applying Lemma 4.2, we have𝑀𝑠(𝜒𝑢>𝛼)<1/102𝑛𝑆2𝑃(𝑓)(𝑥)𝑑𝑥𝐶+𝑛+1×+𝑛+1|||𝑒𝑡1𝐿𝑒𝑡2𝐿|||𝑓(𝑦)2||Ψ𝑔𝑡1Ψ𝑡2(||𝑦)2𝑑𝑦𝑑𝑡𝑡1𝑡2+𝑛+𝑛+1|||𝑒𝑡2𝐿𝑓𝑥1𝑦,2|||2||𝑔𝑥1,Ψ𝑡2𝑦2||2𝑑𝑦2𝑑𝑡2𝑡2𝑑𝑥1+𝑛+𝑛+1|||𝑒𝑡1𝐿𝑓,𝑥2𝑦1|||2||𝑔,𝑥2Ψ𝑡1𝑦1||2𝑑𝑦1𝑑𝑡1𝑡1𝑑𝑥2+2𝑛||||𝑓(𝑥)2||||𝑔(𝑥)2𝑑𝑥=I+II+III+IV.(4.13)

We firstly consider the term I. If 𝑔(Ψ𝑡1Ψ𝑡2)(𝑦)0, then 𝑢(𝑥)𝛼, for some 𝑥𝑅(𝑦,𝑡). As a consequence, |𝑢(𝑦,𝑡)|𝛼. ThusI𝐶𝛼2+𝑛+1×+𝑛+1||Ψ𝑔𝑡1Ψ𝑡2(||𝑦)2𝑑𝑦𝑑𝑡𝑡1𝑡2=𝐶𝛼2+𝑛+1×+𝑛+1||Ψ(1𝑔)𝑡1Ψ𝑡2||(𝑦)2𝑑𝑦𝑑𝑡𝑡1𝑡2𝐶𝛼21𝑔22𝐶𝛼2||𝑢||.>𝛼(4.14) Consider the term II. If 𝑔(𝑥1,)Ψ𝑡2(𝑦2)0, then there exists 𝑥2, such that 𝑢(𝑥1,𝑥2)𝛼 and |𝑥2𝑦2|<𝑡2. Thus, 𝑒𝑡2𝐿𝑓(𝑥1,)(𝑦2)𝛼. We obtainII𝐶𝛼2𝑛+𝑛+1||𝑔𝑥1,Ψ𝑡2𝑦2||2𝑑𝑦2𝑑𝑡2𝑡2𝑑𝑥1=𝐶𝛼2𝑛+𝑛+1||𝑥1𝑔1,Ψ𝑡2𝑦2||2𝑑𝑦2𝑑𝑡2𝑡2𝑑𝑥1𝐶𝛼21𝑔22𝐶𝛼2||𝑢||.>𝛼(4.15) Similarly, III𝐶𝛼2|{𝑢>𝛼}|. For the term IV, it follows from the fact 𝑓(𝑥)𝑢(𝑥) thatIV𝑢(𝑥)𝛼||||𝑓(𝑥)2𝑑𝑥𝑢(𝑥)𝛼||𝑢(||𝑥)2𝑑𝑥.(4.16)

Summarizing the estimates aforementioned, we have𝑀𝑠(𝜒𝑢>𝛼)<1/102𝑛𝑆2𝑃𝛼(𝑓)(𝑥)𝑑𝑥𝐶2||𝑢||+>𝛼𝑢𝛼||𝑢||(𝑥)2,(4.17) which completes the proof of 𝐻1𝒩𝑃(𝑛×𝑛)𝐻1𝑆𝑃(𝑛×𝑛).

Step V. 𝐻1𝑆𝑃(𝑛×𝑛)𝐻1atom(𝑛×𝑛). It is similar to the proof of 𝐻1𝐿(𝑛×𝑛)𝐻1atom(𝑛×𝑛). We omit the details.


The authors thank Y. S. Han and L. X. Yan for helpful suggestions. L. Song is supported by NNSF of China (no. 11001276) and the Fundamental Research Funds for the Central Universities (no. 11lgpy79). C. Q. Tan is supported by Specialized Research Fund for the Doctoral Program of Higher Education (no. 20104402120002) and NSF of Guangdong (no. 10451503101006384).