# Hardy Spaces Associated to Schrödinger Operators on Product Spaces

Authors: Liang Song; Chaoqiang Tan
Journal: Journal of Function Spaces and Applications
 (2012)

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## Abstract

LetL=−Δ+V be a Schrödinger operator on ℝn, where V∈Lloc1(ℝn) is a nonnegative function on ℝn. 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.


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## Body

## 1. Introduction

LetV∈Lloc1(ℝn) be a nonnegative function on ℝn. The Schrödinger operator with potential V is defined by (1.1)L=-Δ+Von⁡Rn.
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 ht(x,y) of the semigroup e-tL satisfies the estimate(1.2)0≤ht(x,y)≤1(4πt)n/2e-|x-y|2/4t
for all t>0 and x,y∈ℝn.Let us consider the Hardy space on product domains. We note that the usual Hardy spaceH1(ℝn×ℝn) on the product domain is now well understood (see, e.g., [2–4]). In this paper we will be concerned with the space HL1(ℝn×ℝn) associated to L as introduced in [5] (see [6, 7] for one-parameter theory). Firstly, we set(1.3)H2(Rn×Rn)=R(L⊗L)¯={(L⊗L)u∈L2(Rn×Rn):u∈L2(Rn×Rn)}¯
and note that(1.4)L2(Rn×Rn)=R(L⊗L)¯⊕N(L⊗L)=H2(Rn×Rn)⊕N(L⊗L),
where ℛ(L⊗L) (resp., 𝒩(L⊗L)) stands for the range (resp., the nullspace) of L⊗L, and the sum is orthogonal. For a function f∈L2(ℝn×ℝn), define(1.5)SL(f)(x1,x2)=(∬|x2-y2|<t2|x1-y1|<t1|Qt12t22f(y1,y2)|2dy1dy2dt1t1n+1dt2t2n+1)1/2,
where (1.6)Qt1t2f=(-t1ddt1)(-t2ddt2)Ht1t2f,Ht1t2f(y1,y2)=∬Rn×Rnht1(y1,z1)ht2(y2,z2)f(z1,z2)dz1dz2.
The space HL1(ℝn×ℝn) is defined as the completion of H2(ℝn×ℝn) in the norm given by(1.7)‖f‖HL1(Rn×Rn)=‖SL(f)‖L1(R2n).The main purpose of this article is to derive atomic characterizations and the maximal characterizations ofHL1(ℝn×ℝn). Before stating our results, let us recall some necessary notations (see also [8, 9]). Suppose that Ω⊂ℝn×ℝn 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 ℝn. Let R=I×J∈m(Ω) and we denote by ℓ(I) and ℓ(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.AnL-atom is a function a on ℝ2n, together with an open set Ω of finite measure, which satisfies the following properties:(i)
supp(a)⊂Ω̃ (see (2.4));(ii)
a(x) can be further decomposed as a=∑R∈m(Ω)aR, where
(1.8)supp⁡((L-1)i⊗(L-1)jaR)⊂10R,∑R∈m(Ω)‖(l(I)-2L-1)i⊗(l(J)-2L-1)jaR‖L2(R2n)2≤|Ω|-1,i,j=0,1;(iii)
∥a∥L2(ℝ2n)≤|Ω|-2.We can define the atomic Hardy spaceHatom1(ℝn×ℝn) by(1.9){∑j=0∞λjαj:{λj}j=0∞∈l1andαjareL-atoms}
with the norm ∥f∥Hatom1(ℝn×ℝn) given by the natural quotient norm:(1.10)inf⁡{∑j=0∞|λj|:f=∑j=0∞λjαj,where{λj}j=0∞∈l1andαjareL-atoms}.The first result of this paper is the following theorem.Theorem 1.1.
LetL be the Schrödinger operator as (1.1). Then the spaces HL1(ℝn×ℝn) and Hatom1(ℝn×ℝn) coincide. In other words,
(1.11)‖f‖HL1(Rn×Rn)≈‖f‖Hatom1(Rn×Rn),foreveryf∈HL1(Rn×Rn).Next we give the “maximal" characterizations ofHL1(ℝn×ℝn). Given a function f∈L1(ℝ2n) we define two functions:(1.12)fh+(x):=sup⁡t1,t2>0|Ht1,t2f(x)|,Nhf(x):=sup⁡|x1-y1|<t1|x2-y2|<t2|Ht12,t22f(y)|.
Similarly, one can consider the Poisson semigroup generated by the operator L and the operators(1.13)SPf(x):=(∬|x2-y2|<t2|x1-y1|<t1|t1L1e-t1L1⊗t2L2e-t2L2f(y)|2dy1dy2dt1dt2t1n+1t2n+1)1/2,fP+(x):=sup⁡t1>0t2>0|e-t1L1⊗e-t2L2f(x)|,NPf(x):=sup⁡|x1-y1|<t1|x2-y2|<t2|e-t1L1⊗e-t2L2f(y)|,
with x∈ℝn×ℝn and f∈L2(ℝ2n).Define the spacesHmax,h1(ℝn×ℝn), H𝒩h1(ℝn×ℝn),HSP1(ℝn×ℝn),Hmax,P1(ℝn×ℝn), and H𝒩P1(ℝn×ℝn) as the completion of H2(ℝn×ℝn) in the norms given by the L1 norm of the corresponding square or maximal functions, respectively. For example,(1.14)‖f‖Hmax⁡,h1(Rn×Rn):=‖fh+‖L1(R2n).
By a similar manner, the norms of H𝒩h1(ℝn×ℝn),HSP1(ℝn×ℝn),Hmax,P1(ℝn×ℝn), and H𝒩P1(ℝn×ℝn) are defined. The second result of this paper is the following.Theorem 1.2.
LetL be the Schrödinger operator as (1.1). Then the spaces HL1(ℝn×ℝn), Hmax,h1(ℝn×ℝn), H𝒩h1(ℝn×ℝn),HSP1(ℝn×ℝn),Hmax,P1(ℝn×ℝn), and H𝒩P1(ℝn×ℝn) coincide with equivalent norms.This paper is organized as follows. In Section2, 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(ℝn) 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 ℝ+n+1 be the usual upper half-space in ℝn+1. For any α>0, we set Γα(x)={(y,t)∈ℝ+n+1×ℝ+n+1,|x1-y1|<αt1,and|x2-y2|<αt2} as the standard cone (of aperture α) with vertex x∈ℝn×ℝn. In particular, we set Γ(x)=Γ1(x). If (x,t)∈ℝ+n+1×ℝ+n+1, then Rx,t denotes the rectangle centered at x∈ℝn×ℝn whose side lengths are t1 and t2, respectively. For any open set Ω⊂ℝn×ℝn, the tent over Ω, T(Ω), is the set(2.1){(x,t)∈R+n+1×R+n+1:Rx,t⊂Ω}.
For any function f(y,t) defined on ℝ+n+1×ℝ+n+1 we will write(2.2)A(f)(x)=(∬Γ(x)|f(y,t)|2dydtt1n+1t2n+1)1/2.
The tent space T2p is then defined as the space of functions f such that 𝒜f∈Lp(ℝ2n) and is equipped with the norm, ∥f∥T2p=∥𝒜(f)∥p, 0<p<∞.We now introduceT21-atoms.Definition 2.1.
A functionA(x,t):ℝ+n+1×ℝ+n+1→ℂ is called a T21-atom if there exists an open set Ω⊂ℝn×ℝn of finite measure satisfying the following properties:(i)
A(x,t) can be further decomposed as A=∑R∈m(Ω)AR, where each AR is supported in T(3R), and R⊂Ω is a maximal dyadic subrectangle of Ω in the form of R=I×J, where I and J are cubes in ℝn;(ii)
∥A∥L2(dydt/t1t2)≤|Ω|-1/2 and ∑R∈m(Ω)∥AR∥L2(dydt/t1t2)2≤|Ω|-1.Proposition 2.2.
Supposef∈T21. Then f=∑kλkAk, where Ak are T21-atoms, λk∈ℂ, and ∑k|λk|≤c∥f∥T21 so that the sum converges in the T21 norm. Moreover, if one assumes that f∈T22, 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:(2.3)Msf(x)=sup⁡x∈R=I×J1|R|∫R|f(x)|dx,
where I and J are cubes in ℝn. It is well known that the operator Ms is bounded on Lp(ℝ2n), for 1<p<∞.Now for any open setΩ⊂ℝn×ℝn with finite measure, we set(2.4)Ω̃={x∣∈Rn×Rn,Ms(χΩ)(x)>122n}.
By the strong maximal theorem, |Ω̃|≤c|Ω|. Denote by m1(Ω) the dyadic subrectangles R=I×J⊂Ω that are maximal in the x1 direction, where IandJ are dyadic cubes in ℝn. Define m2(Ω) similarly. It is well known that Journé’s covering lemma holds (see [8, 15]).Lemma 2.3.
LetΩ⊂ℝn×ℝn. For any R=I×J∈m2(Ω), one sets γ1(R)=supI⊂ll×J⊂Ω̃|l|/|I|. Define γ2 similarly. Then for any δ>0, one has
(2.5)∑R∈m2(Ω)|R|γ1-δ(R)≤cδ|Ω|,∑R∈m1(Ω)|R|γ2-δ(R)≤cδ|Ω|,
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 fromHatom1(ℝn×ℝn) to L1(ℝ2n), we just need to check that the operator is uniformly bounded on the L-atoms.Lemma 2.4.
Assume thatT is either a linear operator or a positive sublinear operator, bounded on L2(ℝ2n) and for every L-atom a,
(2.6)‖T(a)‖L1(R2n)≤c
with constant c independent on a. Then T can extend to a bounded operator from HL1(ℝn×ℝn) to L1(ℝ2n), and
(2.7)‖T(f)‖L1(R2n)≤c‖f‖HL1(Rn×Rn).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 OperatorL on ℝn

LetL be the Schrödinger operator as (1.1), and let ht(x,y) be the kernels of the operators of semigroup {e-tL}.First we note that, for eachk∈ℕ, there exist two positive constants Ck and ck such that the time derivatives of ht satisfy(2.8)|∂k∂tkht(x,y)|≤Ckt-(n+2k)/2exp⁡(-|x-y|2ckt)
for all t>0 and almost all x,y∈ℝn. For the proof, see, for example, [18, 19].Next, fors>0, we define(2.9)F(s):={ψ:C→Cmeasurable:|ψ(z)|≤C|z|s(1+|z|2s)}.
Then for any nonzero function ψ∈𝔽(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 f∈L2(ℝn),(2.10){∫0∞‖ψ(tL)f‖L2(X)2dtt}1/2={∫0∞〈ψ¯(tL)ψ(tL)f,f〉dtt}1/2={〈∫0∞|ψ|2(tL)dttf,f〉}1/2=κ‖f‖L2(Rn),
where κ={∫0∞|ψ(t)|2dt/t}1/2. As an application, we have ∥SL(f)∥2=c∥f∥2 for some constant c.Lemma 2.5.
Letf∈L2(ℝn) and u=e-tLf. Then for any p>0, there exists a constant C=C(n,p)>0 such that
(2.11)sup⁡B((x0,t0),r)|u(x,t)|≤C(1rn+1∬B((x0,t0),2r)|u(x,t)|pdxdt)1/p,
where B((x0,t0),r)=Br(x0)×[t0-cr,t0] with t0>2cr and c>0.Proof.
For the proof, we refer to [7, Lemma  8.4].Recall that ifL is a nonnegative, self-adjoint operator on L2(ℝn), and EL(λ) denotes its spectral decomposition, then for every bounded Borel function F:[0,∞)→ℂ, one defines the operator F(L):L2(ℝn)→L2(ℝn) by the formula(2.12)F(L):=∫0∞F(λ)dEL(λ).
In particular, the operator cos(tL) is then well defined on L2(ℝn). Moreover, it follows from [21, Theorem  3] that the Schwartz kernel Kcos(tL)(x,y) of cos(tL) satisfies(2.13)supp⁡Kcos⁡(tL)(x,y)⊂{(x,y)∈Rn×Rn:|x-y|≤t}.
See also [22]. By the Fourier inversion formula, whenever F is an even bounded Borel function with the Fourier transform of F, F̂∈L1(ℝ), we can write F(L) in terms of cos(tL). In fact, using (2.12) we have(2.14)F(L)=(2π)-1∫-∞∞F̂(t)cos⁡(tL)dt,
which, when combined with (2.13), gives(2.15)KF(L)(x,y)=(2π)-1∫|t|≥|x-y|F̂(t)Kcos⁡⁡(tL)(x,y)dt.Lemma 2.6.
Letϕ∈C0∞(ℝ) be even, ϕ≥0, ∥ϕ∥L1(ℝ)=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)∈ℝn×ℝn:|x-y|≤t}.Proof.
For the proof, we refer the reader to [7, Lemma  3.5].Lemma 2.7.
LetΨ(x)=x2Φ(x) as in Lemma 2.6; then the operator
(2.16)πL(f)(x)=∬0∞Ψ(t1L)⊗Ψ(t2L)(f(⋅,t))(x)dt1dt2t1t2
is bounded from T2p to Lp, if 1<p<∞.Proof.
For the proof, we refer to [5, Lemma  3.4].


## 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(ℝn) 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 ℝ+n+1 be the usual upper half-space in ℝn+1. For any α>0, we set Γα(x)={(y,t)∈ℝ+n+1×ℝ+n+1,|x1-y1|<αt1,and|x2-y2|<αt2} as the standard cone (of aperture α) with vertex x∈ℝn×ℝn. In particular, we set Γ(x)=Γ1(x). If (x,t)∈ℝ+n+1×ℝ+n+1, then Rx,t denotes the rectangle centered at x∈ℝn×ℝn whose side lengths are t1 and t2, respectively. For any open set Ω⊂ℝn×ℝn, the tent over Ω, T(Ω), is the set(2.1){(x,t)∈R+n+1×R+n+1:Rx,t⊂Ω}.
For any function f(y,t) defined on ℝ+n+1×ℝ+n+1 we will write(2.2)A(f)(x)=(∬Γ(x)|f(y,t)|2dydtt1n+1t2n+1)1/2.
The tent space T2p is then defined as the space of functions f such that 𝒜f∈Lp(ℝ2n) and is equipped with the norm, ∥f∥T2p=∥𝒜(f)∥p, 0<p<∞.We now introduceT21-atoms.Definition 2.1.
A functionA(x,t):ℝ+n+1×ℝ+n+1→ℂ is called a T21-atom if there exists an open set Ω⊂ℝn×ℝn of finite measure satisfying the following properties:(i)
A(x,t) can be further decomposed as A=∑R∈m(Ω)AR, where each AR is supported in T(3R), and R⊂Ω is a maximal dyadic subrectangle of Ω in the form of R=I×J, where I and J are cubes in ℝn;(ii)
∥A∥L2(dydt/t1t2)≤|Ω|-1/2 and ∑R∈m(Ω)∥AR∥L2(dydt/t1t2)2≤|Ω|-1.Proposition 2.2.
Supposef∈T21. Then f=∑kλkAk, where Ak are T21-atoms, λk∈ℂ, and ∑k|λk|≤c∥f∥T21 so that the sum converges in the T21 norm. Moreover, if one assumes that f∈T22, 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:(2.3)Msf(x)=sup⁡x∈R=I×J1|R|∫R|f(x)|dx,
where I and J are cubes in ℝn. It is well known that the operator Ms is bounded on Lp(ℝ2n), for 1<p<∞.Now for any open setΩ⊂ℝn×ℝn with finite measure, we set(2.4)Ω̃={x∣∈Rn×Rn,Ms(χΩ)(x)>122n}.
By the strong maximal theorem, |Ω̃|≤c|Ω|. Denote by m1(Ω) the dyadic subrectangles R=I×J⊂Ω that are maximal in the x1 direction, where IandJ are dyadic cubes in ℝn. Define m2(Ω) similarly. It is well known that Journé’s covering lemma holds (see [8, 15]).Lemma 2.3.
LetΩ⊂ℝn×ℝn. For any R=I×J∈m2(Ω), one sets γ1(R)=supI⊂ll×J⊂Ω̃|l|/|I|. Define γ2 similarly. Then for any δ>0, one has
(2.5)∑R∈m2(Ω)|R|γ1-δ(R)≤cδ|Ω|,∑R∈m1(Ω)|R|γ2-δ(R)≤cδ|Ω|,
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 fromHatom1(ℝn×ℝn) to L1(ℝ2n), we just need to check that the operator is uniformly bounded on the L-atoms.Lemma 2.4.
Assume thatT is either a linear operator or a positive sublinear operator, bounded on L2(ℝ2n) and for every L-atom a,
(2.6)‖T(a)‖L1(R2n)≤c
with constant c independent on a. Then T can extend to a bounded operator from HL1(ℝn×ℝn) to L1(ℝ2n), and
(2.7)‖T(f)‖L1(R2n)≤c‖f‖HL1(Rn×Rn).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 OperatorL on ℝn

LetL be the Schrödinger operator as (1.1), and let ht(x,y) be the kernels of the operators of semigroup {e-tL}.First we note that, for eachk∈ℕ, there exist two positive constants Ck and ck such that the time derivatives of ht satisfy(2.8)|∂k∂tkht(x,y)|≤Ckt-(n+2k)/2exp⁡(-|x-y|2ckt)
for all t>0 and almost all x,y∈ℝn. For the proof, see, for example, [18, 19].Next, fors>0, we define(2.9)F(s):={ψ:C→Cmeasurable:|ψ(z)|≤C|z|s(1+|z|2s)}.
Then for any nonzero function ψ∈𝔽(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 f∈L2(ℝn),(2.10){∫0∞‖ψ(tL)f‖L2(X)2dtt}1/2={∫0∞〈ψ¯(tL)ψ(tL)f,f〉dtt}1/2={〈∫0∞|ψ|2(tL)dttf,f〉}1/2=κ‖f‖L2(Rn),
where κ={∫0∞|ψ(t)|2dt/t}1/2. As an application, we have ∥SL(f)∥2=c∥f∥2 for some constant c.Lemma 2.5.
Letf∈L2(ℝn) and u=e-tLf. Then for any p>0, there exists a constant C=C(n,p)>0 such that
(2.11)sup⁡B((x0,t0),r)|u(x,t)|≤C(1rn+1∬B((x0,t0),2r)|u(x,t)|pdxdt)1/p,
where B((x0,t0),r)=Br(x0)×[t0-cr,t0] with t0>2cr and c>0.Proof.
For the proof, we refer to [7, Lemma  8.4].Recall that ifL is a nonnegative, self-adjoint operator on L2(ℝn), and EL(λ) denotes its spectral decomposition, then for every bounded Borel function F:[0,∞)→ℂ, one defines the operator F(L):L2(ℝn)→L2(ℝn) by the formula(2.12)F(L):=∫0∞F(λ)dEL(λ).
In particular, the operator cos(tL) is then well defined on L2(ℝn). Moreover, it follows from [21, Theorem  3] that the Schwartz kernel Kcos(tL)(x,y) of cos(tL) satisfies(2.13)supp⁡Kcos⁡(tL)(x,y)⊂{(x,y)∈Rn×Rn:|x-y|≤t}.
See also [22]. By the Fourier inversion formula, whenever F is an even bounded Borel function with the Fourier transform of F, F̂∈L1(ℝ), we can write F(L) in terms of cos(tL). In fact, using (2.12) we have(2.14)F(L)=(2π)-1∫-∞∞F̂(t)cos⁡(tL)dt,
which, when combined with (2.13), gives(2.15)KF(L)(x,y)=(2π)-1∫|t|≥|x-y|F̂(t)Kcos⁡⁡(tL)(x,y)dt.Lemma 2.6.
Letϕ∈C0∞(ℝ) be even, ϕ≥0, ∥ϕ∥L1(ℝ)=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)∈ℝn×ℝn:|x-y|≤t}.Proof.
For the proof, we refer the reader to [7, Lemma  3.5].Lemma 2.7.
LetΨ(x)=x2Φ(x) as in Lemma 2.6; then the operator
(2.16)πL(f)(x)=∬0∞Ψ(t1L)⊗Ψ(t2L)(f(⋅,t))(x)dt1dt2t1t2
is bounded from T2p to Lp, if 1<p<∞.Proof.
For the proof, we refer to [5, Lemma  3.4].


## 3. Proof of Theorem1.1

### 3.1. The Inclusion ofHL1(ℝn×ℝn)⊂Hatom1(ℝn×ℝn)

Letf∈HL1(ℝn×ℝn)∩H2(ℝn×ℝn). Then F=t12Le-t12L⊗t22Le-t22Lf∈T21∩T22. 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 cΨ∫0∞tΨ(t)e-t2dt=1. By L2-functional calculus [23], one can write(3.1)f(x)=cΨ2∬0∞Ψ(t1L)t12Le-t12L⊗Ψ(t2L)t22Le-t22Lf(x)dt1dt2t1t2,
where the integral converges in L2(ℝ2n). By Proposition 2.2, F has a T21-atomic decomposition: F=∑k=0∞λkAk, where ∑k=0∞|λk|≤C∥F∥T21≤C∥f∥HL1(ℝn×ℝn), and Ak are T21-atoms associated to an open set Ωk. It is easy to see that the sum converges in T21 and T22. We have(3.2)f(x)=cΨ2∬0∞Ψ(t1L)t12Le-t12L⊗Ψ(t2L)t22Le-t22Lf(x)dt1dt2t1t2=∑k=0∞cΨ2∬0∞Ψ(t1L)⊗Ψ(t2L)(λkAk(⋅,t))dt1dt2t1t2=:∑k=0∞μkak(x),
where μk=c0λk and ak(x)=c0-1∬0∞Ψ(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(ℝ2n).It is known that∑k=0∞|μk|≤C∥f∥HSL1(ℝn×ℝn). Now we turn to check that ak are L-atoms associated to open sets Ωk. Since Ak are T21-atoms, then Ak=∑R∈m(Ωk)Ak,R. Thus(3.3)ak(x)=∑R∈m(Ωk)c0-1∬0∞Ψ(t1L)⊗Ψ(t2L)Ak,R(⋅,t)(x)dt1dt2t1t2=:∑R∈m(Ωk)ak,R(x).
Lemma 2.6 implies that for i,j=0,1, supp((L-1)i⊗(L-1)jak)⊂Ω̃k and supp((L-1)i⊗(L-1)jak,R)⊂10R.To continue, we write(3.4)∑R∈m(Ωk)‖(l(I)-2L-1)i⊗(l(J)-2L-1)jak,R‖22≤∑R∈m(Ωk)sup⁡‖h‖2≤1|∫R2n(l(I)-2L-1)i⊗(l(J)-2L-1)jak,R(x)h(x)dx|2≤c0-1∑R∈m(Ωk)sup⁡‖h‖2≤1|∬R+n+1×R+n+1Ak,R(y,t)×(l(I)-2L-1)iΨt1(L)⊗(l(J)-2L-1)jΨt2(L)h(y)dydtt1t2∬R+n+1×R+n+1Ak,R(y,t)|2≤c0-1∑R∈m(Ωk)‖Ak,R‖L2(dydt/t1t2)2≤|Ωk|-1,
if we choose c0 large enough. By a similar argument, we have(3.5)‖(l(I)-2L-1)i⊗(l(J)-2L-1)jak‖2≤|Ωk|-1/2.For generalf∈HL1(ℝn×ℝn), we just need a standard argument. The inclusion of HL1(ℝn×ℝn)⊂Hatom1(ℝn×ℝn) is finished.


### 3.2. The Inclusion ofHAtom1(ℝn×ℝn)⊂HL1(ℝn×ℝn)

By Lemma2.4, it is enough to show that ∥SL(a)∥1≤C for any L-atom a associated to an open set Ω. For R=I×J∈m(Ω), we let l be the maximal dyadic cube such that I⊆l and l×J⊆Ω̃. Also we let S be the maximal dyadic cube such that J⊆S and l×S⊆Ω̃̃. Let R′=100l×100S. We can see that |∪R∈m(Ω)R′|≤C|Ω|, since(3.6)⋃R∈m(Ω)R′⊆{x∣∈Rn×Rn,Ms(χΩ̃̃)(x)>12002n}.
Due to Hölders inequality one has(3.7)∫⋃R∈m(Ω)R′SL(a)(x)dx≤|⋃R∈m(Ω)R′|1/2(∫R2nSL2(a)(x)dx)1/2≤C|Ω|1/2‖a‖2≤C.
Let us prove ∫(⋃R∈m(Ω)R′)cSL(a)(x)dx≤C. One can write(3.8)∫(⋃R∈m(Ω)R′)cSL(a)(x)dx≤∑R∈m(Ω)∬(100l×100S)cSL(aR)(x1,x2)dx1dx2≤∑R∈m(Ω)(∬x1∉100l+∬x2∉100S)SL(aR)(x1,x2)dx1dx2=:I+II.
We only estimate the term I since the proof of the term II is similar. Observe that(3.9)I=∑R∈m(Ω)∫x1∉100l(∫x2∈100J+∫x2∉100J)SL(aR)(x1,x2)dx2dx1=:I1+I2.
Consider term I1. By Hölder’s inequality, we obtain(3.10)I1≤∑R∈m(Ω)|J|1/2∫x1∉100l(∫x2∈100JSL2(aR)(x1,x2)dx2)1/2dx1≤C∑R∈m(Ω)|J|1/2∫x1∉100l(∫|x1-y1|<t1∫Rn|t12Le-t12LaR(⋅,x2)(y1)|2dx2dy1dt1t1n+1)1/2dx1≤C∑R∈m(Ω)|J|1/2∫x1∉100l(∫0l(I)∫B(x1,t1)∫Rn|t12Le-t12LaR(⋅,x2)(y1)|2dx2dy1dt1t1n+1)1/2dx1+C∑R∈m(Ω)|J|1/2∫x1∉100l(∫l(I)∞∫B(x1,t1)∫Rn|t12Le-t12LaR(⋅,x2)(y1)|2dx2dy1dt1t1n+1)1/2dx1=:I11+I12.
Consider the term I11. It follows from estimate (2.8) that(3.11)I11≤C∑R∈m(Ω)|J|1/2×∫x1∉100l{∫0l(I)∫B(x1,t1)∫Rn(∫Rnt1(t1+|y1-z1|)n+1|aR(z1,x2)|dz1)2dx2dy1dt1t1n+1}1/2dx1.
Let xI denote the center of cube I. Note that x1∉100l and |x1-y1|<t1<l(I). We use Hölder’s inequality to obtain(3.12)I11≤C∑R∈m(Ω)|J|1/2|I|1/2∫x1∉100l(∫0l(I)t1|x1-xI|2n+2dt1)1/2dx1⋅‖aR‖L2(R2n)≤C∑R∈m(Ω)|R|1/2‖aR‖L2(R2n)γ1(R)-1/n≤C(∑R∈m(Ω)‖aR‖L2(R2n)2)1/2(∑R∈m(Ω)|R|γ1(R)-2/n)1/2≤C,
where in the last inequality we use Lemma 2.3.For the term I12, we apply the definition of L-atom to obtain(3.13)I12=C∑R∈m(Ω)|J|1/2l(I)2×∫x1∉100l(∫l(I)∞∫B(x1,t1)∫Rn|t14L2e-t12L(l(I)-2L-1)aR(⋅,x2)(y1)|2dx2dy1dt1t1n+5)1/2dx1≤C∑R∈m(Ω)|J|1/2l(I)2×∫x1∉100l{∫l(I)∞∫B(x1,t1)∫Rn(∫Rnt1(t1+|y1-z1|)n+1|(l(I)-2L-1)aR(⋅,x2)(z1)|dz1)2dx2×dy1dt1t1n+5}1/2dx1≤C∑R∈m(Ω)|J|1/2|I|1/2l(I)2×∫x1∉100l(∫l(I)∞t1-3(t1+|x1-xI|)2n+2dt1)1/2dx1‖(l(I)-2L-1⊗I)aR‖L2(R2n)≤C∑R∈m(Ω)|R|1/2∫x1∉100ll(I)|x1-xI|n+1dx1‖(l(I)-2L-1⊗I)aR‖L2(R2n)≤C∑R∈m(Ω)|R|1/2γ1(R)-1/n‖(l(I)-2L-1⊗I)aR‖L2(R2n)≤C,
which, together with estimate of I11, show that I1≤C. By a similar argument as mentioned previously, we can show I2≤C. We have obtained the required estimate ∥SL(a)∥1≤C. This completes the proof of Theorem 1.1.


## 3.1. The Inclusion ofHL1(ℝn×ℝn)⊂Hatom1(ℝn×ℝn)

Letf∈HL1(ℝn×ℝn)∩H2(ℝn×ℝn). Then F=t12Le-t12L⊗t22Le-t22Lf∈T21∩T22. 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 cΨ∫0∞tΨ(t)e-t2dt=1. By L2-functional calculus [23], one can write(3.1)f(x)=cΨ2∬0∞Ψ(t1L)t12Le-t12L⊗Ψ(t2L)t22Le-t22Lf(x)dt1dt2t1t2,
where the integral converges in L2(ℝ2n). By Proposition 2.2, F has a T21-atomic decomposition: F=∑k=0∞λkAk, where ∑k=0∞|λk|≤C∥F∥T21≤C∥f∥HL1(ℝn×ℝn), and Ak are T21-atoms associated to an open set Ωk. It is easy to see that the sum converges in T21 and T22. We have(3.2)f(x)=cΨ2∬0∞Ψ(t1L)t12Le-t12L⊗Ψ(t2L)t22Le-t22Lf(x)dt1dt2t1t2=∑k=0∞cΨ2∬0∞Ψ(t1L)⊗Ψ(t2L)(λkAk(⋅,t))dt1dt2t1t2=:∑k=0∞μkak(x),
where μk=c0λk and ak(x)=c0-1∬0∞Ψ(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(ℝ2n).It is known that∑k=0∞|μk|≤C∥f∥HSL1(ℝn×ℝn). Now we turn to check that ak are L-atoms associated to open sets Ωk. Since Ak are T21-atoms, then Ak=∑R∈m(Ωk)Ak,R. Thus(3.3)ak(x)=∑R∈m(Ωk)c0-1∬0∞Ψ(t1L)⊗Ψ(t2L)Ak,R(⋅,t)(x)dt1dt2t1t2=:∑R∈m(Ωk)ak,R(x).
Lemma 2.6 implies that for i,j=0,1, supp((L-1)i⊗(L-1)jak)⊂Ω̃k and supp((L-1)i⊗(L-1)jak,R)⊂10R.To continue, we write(3.4)∑R∈m(Ωk)‖(l(I)-2L-1)i⊗(l(J)-2L-1)jak,R‖22≤∑R∈m(Ωk)sup⁡‖h‖2≤1|∫R2n(l(I)-2L-1)i⊗(l(J)-2L-1)jak,R(x)h(x)dx|2≤c0-1∑R∈m(Ωk)sup⁡‖h‖2≤1|∬R+n+1×R+n+1Ak,R(y,t)×(l(I)-2L-1)iΨt1(L)⊗(l(J)-2L-1)jΨt2(L)h(y)dydtt1t2∬R+n+1×R+n+1Ak,R(y,t)|2≤c0-1∑R∈m(Ωk)‖Ak,R‖L2(dydt/t1t2)2≤|Ωk|-1,
if we choose c0 large enough. By a similar argument, we have(3.5)‖(l(I)-2L-1)i⊗(l(J)-2L-1)jak‖2≤|Ωk|-1/2.For generalf∈HL1(ℝn×ℝn), we just need a standard argument. The inclusion of HL1(ℝn×ℝn)⊂Hatom1(ℝn×ℝn) is finished.


## 3.2. The Inclusion ofHAtom1(ℝn×ℝn)⊂HL1(ℝn×ℝn)

By Lemma2.4, it is enough to show that ∥SL(a)∥1≤C for any L-atom a associated to an open set Ω. For R=I×J∈m(Ω), we let l be the maximal dyadic cube such that I⊆l and l×J⊆Ω̃. Also we let S be the maximal dyadic cube such that J⊆S and l×S⊆Ω̃̃. Let R′=100l×100S. We can see that |∪R∈m(Ω)R′|≤C|Ω|, since(3.6)⋃R∈m(Ω)R′⊆{x∣∈Rn×Rn,Ms(χΩ̃̃)(x)>12002n}.
Due to Hölders inequality one has(3.7)∫⋃R∈m(Ω)R′SL(a)(x)dx≤|⋃R∈m(Ω)R′|1/2(∫R2nSL2(a)(x)dx)1/2≤C|Ω|1/2‖a‖2≤C.
Let us prove ∫(⋃R∈m(Ω)R′)cSL(a)(x)dx≤C. One can write(3.8)∫(⋃R∈m(Ω)R′)cSL(a)(x)dx≤∑R∈m(Ω)∬(100l×100S)cSL(aR)(x1,x2)dx1dx2≤∑R∈m(Ω)(∬x1∉100l+∬x2∉100S)SL(aR)(x1,x2)dx1dx2=:I+II.
We only estimate the term I since the proof of the term II is similar. Observe that(3.9)I=∑R∈m(Ω)∫x1∉100l(∫x2∈100J+∫x2∉100J)SL(aR)(x1,x2)dx2dx1=:I1+I2.
Consider term I1. By Hölder’s inequality, we obtain(3.10)I1≤∑R∈m(Ω)|J|1/2∫x1∉100l(∫x2∈100JSL2(aR)(x1,x2)dx2)1/2dx1≤C∑R∈m(Ω)|J|1/2∫x1∉100l(∫|x1-y1|<t1∫Rn|t12Le-t12LaR(⋅,x2)(y1)|2dx2dy1dt1t1n+1)1/2dx1≤C∑R∈m(Ω)|J|1/2∫x1∉100l(∫0l(I)∫B(x1,t1)∫Rn|t12Le-t12LaR(⋅,x2)(y1)|2dx2dy1dt1t1n+1)1/2dx1+C∑R∈m(Ω)|J|1/2∫x1∉100l(∫l(I)∞∫B(x1,t1)∫Rn|t12Le-t12LaR(⋅,x2)(y1)|2dx2dy1dt1t1n+1)1/2dx1=:I11+I12.
Consider the term I11. It follows from estimate (2.8) that(3.11)I11≤C∑R∈m(Ω)|J|1/2×∫x1∉100l{∫0l(I)∫B(x1,t1)∫Rn(∫Rnt1(t1+|y1-z1|)n+1|aR(z1,x2)|dz1)2dx2dy1dt1t1n+1}1/2dx1.
Let xI denote the center of cube I. Note that x1∉100l and |x1-y1|<t1<l(I). We use Hölder’s inequality to obtain(3.12)I11≤C∑R∈m(Ω)|J|1/2|I|1/2∫x1∉100l(∫0l(I)t1|x1-xI|2n+2dt1)1/2dx1⋅‖aR‖L2(R2n)≤C∑R∈m(Ω)|R|1/2‖aR‖L2(R2n)γ1(R)-1/n≤C(∑R∈m(Ω)‖aR‖L2(R2n)2)1/2(∑R∈m(Ω)|R|γ1(R)-2/n)1/2≤C,
where in the last inequality we use Lemma 2.3.For the term I12, we apply the definition of L-atom to obtain(3.13)I12=C∑R∈m(Ω)|J|1/2l(I)2×∫x1∉100l(∫l(I)∞∫B(x1,t1)∫Rn|t14L2e-t12L(l(I)-2L-1)aR(⋅,x2)(y1)|2dx2dy1dt1t1n+5)1/2dx1≤C∑R∈m(Ω)|J|1/2l(I)2×∫x1∉100l{∫l(I)∞∫B(x1,t1)∫Rn(∫Rnt1(t1+|y1-z1|)n+1|(l(I)-2L-1)aR(⋅,x2)(z1)|dz1)2dx2×dy1dt1t1n+5}1/2dx1≤C∑R∈m(Ω)|J|1/2|I|1/2l(I)2×∫x1∉100l(∫l(I)∞t1-3(t1+|x1-xI|)2n+2dt1)1/2dx1‖(l(I)-2L-1⊗I)aR‖L2(R2n)≤C∑R∈m(Ω)|R|1/2∫x1∉100ll(I)|x1-xI|n+1dx1‖(l(I)-2L-1⊗I)aR‖L2(R2n)≤C∑R∈m(Ω)|R|1/2γ1(R)-1/n‖(l(I)-2L-1⊗I)aR‖L2(R2n)≤C,
which, together with estimate of I11, show that I1≤C. By a similar argument as mentioned previously, we can show I2≤C. We have obtained the required estimate ∥SL(a)∥1≤C. This completes the proof of Theorem 1.1.


## 4. Proof of Theorem1.2

In this section, we will give the proof of Theorem1.2 in the following routine: Hatom1(ℝn×ℝn)⊆H𝒩h1(ℝn×ℝn)⊆Hmax,h1(ℝn×ℝn)⊆Hmax,P1(ℝn×ℝn)⊆H𝒩P1(ℝn×ℝn)⊆HSP1(ℝn×ℝn)⊆Hatom1(ℝn×ℝn). The main idea comes from [9, 16].Step I. Hatom1(ℝn×ℝn)⊆H𝒩h1(ℝn×ℝn). It is similar to the proof of Hatom1(ℝn×ℝn)⊂HL1(ℝn×ℝn). Here we omit the details.Step II. H𝒩h1(ℝn×ℝn)⊆Hmax,h1(ℝn×ℝn) because fh+(x)≤𝒩h(f)(x).Step III. Hmax,h1(ℝn×ℝn)⊆Hmax,P1(ℝn×ℝn). By L2-functional calculus ([23]), we have(4.1)e-t1L⊗e-t2Lf(x1,x2)=1π1π∬0∞e-u1u1e-u2u2e-(t12/4u1)L⊗e-(t22/4u2)Lf(x1,x2)du1du2.
Therefore,(4.2)‖fP+‖1≤C1π1π∬0∞e-u1u1e-u2u2‖fh+‖1du1du2≤C‖fh+‖1,
which completes the proof of Hmax,h1(ℝn×ℝn)⊆Hmax,P1(ℝn×ℝn).Step IV. Hmax,P1(ℝn×ℝn)⊆H𝒩P1(ℝn×ℝn), for any (y1,y2,t1,t2)∈Γ(x), and x∈ℝn×ℝn. Applying Lemma 2.5 with 0<p<1, we obtain(4.3)|e-t1L⊗e-t2Lf(y1,y2)|p≤Ct1n+1t2n+1∫t1/2t1∫B(x1,2t1)∫t2/2t2∫B(x2,2t2)|e-s1L⊗e-s2Lf(z1,z2)|pdz2ds2dz1ds1≤Ct1nt2n∫B(x1,2t1)∫B(x2,2t2)(fP+)p(z)dz≤CMs(|fP+|p)(x).
Therefore,(4.4)‖Nhf‖L1(R2n)≤‖(Ms(|fP+|p))1/p‖L1(R2n)≤C‖fP+‖L1(R2n).
This completes the proof of Hmax,P1(ℝn×ℝn)⊆H𝒩P1(ℝn×ℝn).Lemma 4.1.
Letf and g be the functions of L2(ℝn), and suppose that ϕ∈C0∞(ℝn) is radial and supp(ϕ)⊆B(0,1). Let u(x,t)=e-tLf(x). Then one has
(4.5)∬R+n+1|∂u∂t(x,t)|2|g*ϕt(x)|2tdxdt≤C(∫Rn|f(x)|2|g(x)|2dx+∬R+n+1|u(x,t)|2|g*Ψt(x)|2dxdtt),
where Ψ∈C0∞(ℝn) with ∫Ψ=0 is a vector-value function independent on f and g and supp(Ψ)⊆B(0,1).Proof.
We note that(4.6)(Δ+∂2∂t2)(|u|2)=(Δ+∂2∂t2)u⋅u¯+(Δ+∂2∂t2)u¯⋅u+2|∇u|2+2|∂u∂t|2=2V|u|2+2|∇u|2+2|∂u∂t|2≥2|∂u∂t|2.
Following the steps in [24], we obtain
(4.7)∬R+n+1(Δ+∂2∂t2)(|u|2)(x,t)|g*ϕt(x)|2tdxdt≤C(∫Rn|f(x)|2|g(x)|2dx+∬R+n+1|u(x,t)|2|g*Ψt(x)|2dxdtt).
Then the lemma follows readily.Applying Lemma4.1, we can obtain the following lemma.Lemma 4.2.
Suppose thatf,g∈L2(ℝn×ℝn). Let ϕ and Ψ be functions as in Lemma 4.1. Then one has
(4.8)∬R+n+1×R+n+1|t1Le-t1L⊗t2Le-t2Lf(x)|2|g*(ϕt1⊗ϕt2)(x)|2dxdtt1t2≤C{∬R+n+1×R+n+1|e-t1L⊗e-t2Lf(x)|2|g*(Ψt1⊗Ψt2)(x)|2dxdtt1t2+∫Rn∫R+n+1|(I⊗e-t2L)f(x)|2|g(x1,⋅)*Ψt2(x2)|2dx2dt2t2dx1+∫Rn∫R+n+1|(e-t1L⊗I)f(x)|2|g(⋅,x2)*Ψt1(x1)|2dx1dt1t1dx2+∫R2n|f(x)|2|g(x)|2dx∬R+n+1×R+n+1|e-t1L⊗e-t2Lf(x)|2|g*(Ψt1⊗Ψt2)(x)|2dxdtt1t2}.Proof.
The proof of Lemma4.2 can be obtained by iterating Lemma 4.1.We begin to showH𝒩P1(ℝn×ℝn)⊆HSP1(ℝn×ℝn) by following the idea of [9] (see also [25, pages 107–109]).Setu(x,t)=e-t1L⊗e-t2Lf(x) and u*(x)=sup(y,t)∈Γ(x)|u(y,t)|.We claim that(4.9)|{x∣SP(f)(x)>α}|≤C(|{x∣u*>α}|+α-2∫u*≤α|u*(x)|2dx).
Once the claim holds, we integrate α from 0 to ∞ to complete the proof of H𝒩P1(ℝn×ℝn)⊆HSP1(ℝn×ℝn). Now we turn to prove the claim.The boundedness of the strong maximal operatorMs on L2(ℝ2n) implies(4.10)|{x∣Ms(χ{u*>α})(x)≥1102n}|≤C∫R2nMs2(χ{u*>α})(x)dx≤C∫R2nχ{u*>α}2(x)dx=C|{u*>α}|.
To prove (4.9), one just need to estimate(4.11)|{x∣SP(f)(x)>α,Ms(χ{u*>α})<1102n}|.
Observe that(4.12)∫Ms(χ{u*>α})<1/102nSP2(f)(x)dx≤∬R+n+1×R+n+1|t1Le-t1L⊗t2Le-t2Lf(y)|2|R(y,t)∩{Ms(χ{u*>α})<1102n}|dydtt1n+1t2n+1≤∬R*|t1Le-t1L⊗t2Le-t2Lf(y)|2dydtt1t2,
where R*={(y,t)|∈ℝ+n+1×ℝ+n+1,|R(y,t)∩{u*>α}|<|R(y,t)|/102n} and R(y,t)={(x1,x2)|∈ℝn×ℝn:|x1-y1|<t1,|x2-y2|<t2}. Now we choose a radial nonnegative function ϕ∈C0∞(ℝn) such that ϕ(x)=1 if |x|≤1/3 and ϕ(x)=0 if |x|>1. Also set g(x)=χ{u*≤α}(x). We can easily check that if (y,t)∈R*, then g*(ϕt1⊗ϕt2)(y)≥C, for some positive constant C. Applying Lemma 4.2, we have(4.13)∫Ms(χu*>α)<1/102nSP2(f)(x)dx≤C{∬R+n+1×R+n+1|e-t1L⊗e-t2Lf(y)|2|g*(Ψt1⊗Ψt2)(y)|2dydtt1t2+∫Rn∫R+n+1|e-t2Lf(x1,⋅)(y2)|2|g(x1,⋅)*Ψt2(y2)|2dy2dt2t2dx1+∫Rn∫R+n+1|e-t1Lf(⋅,x2)(y1)|2|g(⋅,x2)*Ψt1(y1)|2dy1dt1t1dx2+∬R2n|f(x)|2|g(x)|2dx∬R+n+1×R+n+1|e-t1L⊗e-t2Lf(y)|2|g*(Ψt1⊗Ψt2)(y)|2dydtt1t2}=:I+II+III+IV.We firstly consider the term I. Ifg*(Ψt1⊗Ψt2)(y)≠0, then u*(x)≤α, for some x∈R(y,t). As a consequence, |u(y,t)|≤α. Thus(4.14)I≤Cα2∬R+n+1×R+n+1|g*(Ψt1⊗Ψt2)(y)|2dydtt1t2=Cα2∬R+n+1×R+n+1|(1-g)*(Ψt1⊗Ψt2)(y)|2dydtt1t2≤Cα2‖1-g‖22≤Cα2|{u*>α}|.
Consider the term II. If g(x1,·)*Ψt2(y2)≠0, then there exists x2, such that u*(x1,x2)≤α and |x2-y2|<t2. Thus, e-t2Lf(x1,·)(y2)≤α. We obtain(4.15)II≤Cα2∫Rn∫R+n+1|g(x1,⋅)*Ψt2(y2)|2dy2dt2t2dx1=Cα2∫Rn∫R+n+1|(1-g(x1,⋅))*Ψt2(y2)|2dy2dt2t2dx1≤Cα2‖1-g‖22≤Cα2|{u*>α}|.
Similarly, III≤Cα2|{u*>α}|. For the term IV, it follows from the fact f(x)≤u*(x) that(4.16)IV≤∫u*(x)≤α|f(x)|2dx≤∫u*(x)≤α|u*(x)|2dx.Summarizing the estimates aforementioned, we have(4.17)∫Ms(χu*>α)<1/102nSP2(f)(x)dx≤C(α2|{u*>α}|+∫u*≤α|u*(x)|2),
which completes the proof of H𝒩P1(ℝn×ℝn)⊆HSP1(ℝn×ℝn).Step V. HSP1(ℝn×ℝn)⊆Hatom1(ℝn×ℝn). It is similar to the proof of HL1(ℝn×ℝn)⊂Hatom1(ℝn×ℝn). We omit the details.



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*Source: 179015-2012-02-06.xml*