${\mathcal{H}}^p$-Boundedness of Weyl multipliers

Abstract

In this paper, we will define Hardy spaces ${\mathcal{H}}^p({\mathbb{C}}^n)$ associated with twisted convolution operators and give the atomic decomposition of ${\mathcal{H}}^p({\mathbb{C}}^n)$, where $\frac{2n}{2n+1}<p\le 1$. Then we consider the boundedness of the Weyl multiplier on ${\mathcal{H}}^p({\mathbb{C}}^n)$.

MSC:42B30, 42B25, 42B35.

1 Introduction

The ‘twisted translation’ ${\tau }_w$ on ${\mathbb{C}}^n$ is defined on measurable functions by

$$\begin{array}{rcl}({\tau }_{w}f)(z)& =& exp(\frac{1}{2}\sum _{j=1}^n(w_j{z}_j+{\overline{w}}_j{\overline{z}}_j))f(z)\\ =& f(z+w)exp(\frac{i}{2}Im(z\cdot \overline{w}))\end{array}$$

and the ‘twisted convolution’ of two functions f and g on ${\mathbb{C}}^n$ can be defined as

$$\begin{array}{rcl}(f\times g)(z)& =& {\int }_{{\mathbb{C}}^n}f(w){\tau }_{-w}g(z)dw\\ =& {\int }_{{\mathbb{C}}^n}f(z-w)g(w)\overline{\omega }(z,w)dw,\end{array}$$

where $\omega (z,w)=exp(\frac{i}{2}Im(z\cdot \overline{w}))$.

In order to define the space ${\mathcal{H}}^p({\mathbb{C}}^n)$ associated with ‘twisted convolution’, we first define the following version maximal operator in terms of twisted convolutions.

Let $\mathcal{B}=\{\phi \in C^{\mathrm{\infty }}({\mathbb{C}}^n):supp\phi \subset B(0,1),{\parallel \phi \parallel }_{\mathrm{\infty }}\le 1,{\parallel \mathrm{\nabla }\phi \parallel }_{\mathrm{\infty }}\le 2\}$, and for $t>0$, ${\phi }_t(z)=t^{-2n}\phi (\frac{z}{t})$. Given $\sigma \in (0,+\mathrm{\infty }]$ and a tempered distribution f, define the grand maximal function

$$M_{\sigma }f(z)=\underset{\phi \in \mathcal{B},0<t<\sigma }{sup}|{\phi }_t\times f(z)|,$$

(1)

where $B(0,1)=\{z\in {\mathbb{C}}^n:|z|<1\}$.

Definition 1 Let f be a tempered distribution on ${\mathbb{C}}^n$ and $\frac{2n}{2n+1}<p\le 1$, we say that f belongs to the Hardy space ${\mathcal{H}}^p({\mathbb{C}}^n)$ if and only if the grand maximal function

$$M_{\sigma }f(z)=\underset{\phi \in \mathcal{B},0<t<\sigma }{sup}|{\phi }_t\times f(z)|$$

lies in $L^p({\mathbb{C}}^n)$, that is, ${\mathcal{H}}^p({\mathbb{C}}^n)=\{f\in {\mathcal{S}}^{\prime }({\mathbb{C}}^n):M_{\sigma }f\in L^p\}$, where $\mathcal{S}({\mathbb{C}}^n)$ denotes the Schwartz space. We set ${\parallel f\parallel }_{{\mathcal{H}}^p}={\parallel M_{\sigma }f\parallel }_{L^p}$.

Remark 1 From Theorem A in [1], we know that for some σ, $0<\sigma <\mathrm{\infty }$, $M_{\sigma }f\in L^p$ if and only if $M_{\mathrm{\infty }}f\in L^p$, when $\frac{2n}{2n+1}<p\le 1$, so the ${\mathcal{H}}^p$ also can be defined as $\{f\in {\mathcal{S}}^{\prime }({\mathbb{C}}^n):M_{\mathrm{\infty }}f\in L^p\}$, where $\frac{2n}{2n+1}<p\le 1$. The case of $p=1$ has been considered in [1].

Let $\frac{2n}{2n+1}<p\le 1$, an atom for ${\mathcal{H}}^p({\mathbb{C}}^n)$ centered at $z\in {\mathbb{C}}^n$ is a function $a(z)$ which satisfies:

• $$\begin{array}{rl}(1)& suppa?B(z,r),\\ (2)& {?a?}_{\mathrm{8}}={(2r)}^{-\frac{2n}{p}},\\ (3)& ?a(w)\overline{?}(z,w)dw=0.\end{array}$$

The atomic norm of f can be defined as

$${\parallel f\parallel }_{\mathrm{atom}}=inf\{{\left(\sum _{j=1}^{\mathrm{\infty }}{|{\lambda }_j|}^p\right)}^{\frac{1}{p}}:f=\sum {\lambda }_j{a}_j\},$$

where the infimum is taken over all decompositions of $f=\sum {\lambda }_j{a}_j$ in the sense of ${\mathcal{S}}^{\prime }({\mathbb{C}}^n)$ and $a_j$ are atoms.

In this paper, we will first give the atomic decomposition of ${\mathcal{H}}^p({\mathbb{C}}^n)$ as follows.

Theorem 1 Let $f\in {\mathcal{S}}^{\prime }({\mathbb{C}}^n)$ and $\frac{2n}{2n+1}<p\le 1$, then $f\in {\mathcal{H}}^p({\mathbb{C}}^n)$ if and only if $f=\sum {\lambda }_j{a}_j$, where $a_j$ are atoms. Moreover, ${\parallel f\parallel }_{{\mathcal{H}}^p}\sim {\parallel f\parallel }_{\mathrm{atom}}$.

Remark 2 The case $p=1$ has been proved in [1], so we will consider the case $\frac{2n}{2n+1}<p<1$ in this paper.

The boundedness of the Weyl multiplier has been considered by many authors (cf. [2] and [3]). In this paper, we will consider the boundedness of the Weyl multiplier on ${\mathcal{H}}^p({\mathbb{C}}^n)$. We first give some notations for Weyl multipliers. On ${\mathbb{C}}^n$ consider the 2n linear differential operators

$$Z_j=\frac{\partial }{{\partial }_{z_j}}+\frac{1}{4}{\overline{z}}_j,{\overline{Z}}_j=\frac{\partial }{{\partial }_{{\overline{z}}_j}}-\frac{1}{4}{z}_j,j=1,2,\dots ,n.$$

(2)

Together with the identity they generate a Lie algebra $h^n$ which is isomorphic to the $2n+1$ dimensional Heisenberg algebra. The only nontrivial commutation relations are

$$[Z_j,{\overline{Z}}_j]=-\frac{1}{2}I,j=1,2,\dots ,n.$$

(3)

The operator L defined by

$$L=-\frac{1}{2}\sum _{j=1}^n(Z_j{\overline{Z}}_j+{\overline{Z}}_j{Z}_j)$$

is nonnegative, self-adjoint, and elliptic. Therefore it generates a diffusion semigroup ${\{T_t^L\}}_{t>0}={\{e^{-tL}\}}_{t>0}$. There exists an irreducible projective representation W of ${\mathbb{C}}^n$ into a separable Hilbert space $H_W$ such that

$$W(z+v)=\omega (z,v)W(z)W(v).$$

Given a function f in $L^1({\mathbb{C}}^n)$ its Weyl transform $\tau (f)$ is a bounded operator on $H_W$ defined by

$$\tau (f)={\int }_{{\mathbb{C}}^n}f(z)W(z)dz.$$

(4)

Let $H=-\mathrm{\Delta }+{|x|}^2$ be the Hermite operator, then we have $\tau (Lf)=\tau (f)H$ or more generally

$$\tau (\varphi (L)f)=\tau (f)\varphi (H).$$

(5)

We say that a bounded operator M on $L^2({\mathbb{R}}^n)$ is a Weyl multiplier on $L^p({\mathbb{R}}^n)$ if the operator $T_M$ initially defined on $L^1\cup L^p$ by

$$\tau (T_{M}f)=\tau (f)M$$

extends to a bounded operator on $L^p({\mathbb{C}}^n)$. In [4], the author considered multipliers of the form $\varphi (H)$ and proved the $L^p$-boundedness of $\varphi (H)$.

In this paper, we will prove the following.

Theorem 2 Let

$${\mathrm{\Delta }}_{+}\varphi (N)=\varphi (N+1)-\varphi (N)\mathit{\text{and}}{\mathrm{\Delta }}_{-}\varphi (N)=\varphi (N)-\varphi (N+1).$$

Suppose that the function ϕ satisfies

$$|{\mathrm{\Delta }}_{-}^k{\mathrm{\Delta }}_{+}^m\varphi (N)|\le C{N}^{-(k+m)}$$

(6)

with k, m positive integers such that $k+m=0,1,\dots ,\nu$, where $\nu =n+1$ when n is odd and $\nu =n+2$ when n is even. Then $\varphi (H)$ is a Weyl multiplier on ${\mathcal{H}}^p({\mathbb{C}}^n)$, where $\frac{2n}{2n+1}<p\le 1$.

Remark 3 The case $p=1$ has been proved in [3].

Throughout the article, we will use A and C to denote the positive constants, which are independent of main parameters and may be different at each occurrence. By $B_1\sim B_2$, we mean that there exists a constant $C>1$ such that $\frac{1}{C}\le \frac{B_1}{B_2}\le C$.

The paper is organized as follows. In Section 2, we will give the proof of Theorem 1. Theorem 2 will be proved in Section 3.

2 Atomic decomposition for ${\mathcal{H}}^p({\mathbb{C}}^n)$

The local Hardy space $h^p({\mathbb{C}}^n)$ has been defined in [5]; let $f\in {\mathcal{S}}^{\prime }({\mathbb{C}}^n)$ and write

$$f_{\sigma }^{\ast }(z)=\underset{\phi \in \mathcal{B},0<t<\sigma }{sup}|{\phi }_t\ast f(z)|.$$

Proposition 1 When $\frac{2n}{2n+1}<p<1$, the following conditions are equivalent:

• (I₁) $f\in h^p({\mathbb{C}}^n)$;

• (I₂) $f_{\sigma }^{\ast }(z)\in L^p({\mathbb{C}}^n)$;

• (I₃) $f=\sum {\lambda }_j{a}_j$, where ${\sum }_{j=1}^n{|{\lambda }_j|}^p<\mathrm{\infty }$, $supp{a}_j\subset B(z_j,r_j)$, ${\parallel a_j\parallel }_{\mathrm{\infty }}\le {(2r_j)}^{-\frac{2n}{p}}$, and $\int a_j(z)dz=0$, whenever $r_j<\sigma$.

Consider a partition of ${\mathbb{C}}^n$ into a mesh of balls $B_j=B(z_j,\frac{\sigma }{2})$, $j=1,2,\dots$ , and construct a ${\mathbb{C}}^{\mathrm{\infty }}$ partition of unity ${\phi }_j$ such that $supp{\phi }_j\subset B(z_j,\sigma )$. The proof of the following lemma is quite similar to Theorem 2.2 in [1], so we omit it.

Lemma 1 Let $\frac{2n}{2n+1}<p<1$, assume $M_{\sigma }f(z)\in L^p$, then $g_j(z)=f(z){\phi }_j(z)\overline{\omega }(z_j,z)\in h^p$, $j=1,2,\dots$ , moreover, there exists $C>0$ such that

$$\sum _{j=1}^{\mathrm{\infty }}{\parallel g_j\parallel }_{h^p}\le C{\parallel M_{\sigma }f(z)\parallel }_p.$$

By Proposition 1 and Lemma 1, we know that every element in ${\mathcal{H}}^p({\mathbb{C}}^n)$ can be written as $f=\sum {\lambda }_j{a}_j$, where

• (I) ${({\sum }_{j=1}^{\mathrm{\infty }}{|{\lambda }_j|}^p)}^{\frac{1}{p}}\le C{\parallel f\parallel }_{{\mathcal{H}}^p}$;

• (II) $a_j$ is supported in $B(z_j,r_j)$, and ${\parallel a_j\parallel }_{\mathrm{\infty }}\le {(2r_j)}^{-\frac{2n}{p}}$;

• (III) whenever $r_j<\sigma$, there exists ${\xi }_j$ such that $|{\xi }_j-z_j|\le 2\sigma$ and $\int a_j(z)\overline{\omega }({\xi }_j,z)dz=0$.

This is not yet the atomic decomposition for ${\mathcal{H}}^p$. In order to obtain it we must first replace condition (III) with a centered cancellation property.

Lemma 2 Let $\frac{2n}{2n+1}<p<1$ and $a(z)$ be a function supported on $B=B(z_0,r)$, $r<\sigma$ such that

(I₁) ${\parallel a\parallel }_{\mathrm{\infty }}\le {(2r)}^{-\frac{2n}{p}}$;

(I₂) $\int a_j(z)\overline{\omega }(\xi ,z)dz=0$ for some ξ, $|\xi -z_0|\le 2\sigma$. If σ is sufficiently small, $a(z)$ can be decomposed as $a(z)=\sum {\lambda }_j{\alpha }_j(z)$, where

• (a) ${\sum }_{j=1}^{\mathrm{\infty }}{|{\lambda }_j|}^p\le C$;

• (b) $supp{\alpha }_j\subset B(z_j,r_j)$, ${\parallel {\alpha }_j\parallel }_{\mathrm{\infty }}\le {(2r_j)}^{-\frac{2n}{p}}$;

• (c) $\int {\alpha }_j(z)\overline{\omega }(z_j,z)dz=0$, whenever $r_j<\sigma$.

Proof Write $a(z)=g^{(1)}(z)+b^{(1)}(z)$, where

$$b^{(1)}(z)=(\frac{1}{|B|}{\int }_{B}a(w)\overline{\omega }(z_0,w)dw){\chi }_B(z)\omega (z_0,z).$$

Then the function $\frac{1}{2}{g}^{(1)}$ satisfies (b) and (c); on the other hand

$$\begin{array}{rcl}|b^{(1)}(z)|& =& \frac{1}{|B|}|{\int }_{B}a(w)(\overline{\omega }(z_0,w-z_0)-\overline{\omega }(\xi ,w-z_0))dw|\\ \le & \frac{1}{|B|}{\int }_B|a(w)|\left|(\overline{\omega }(z_0,w-z_0)-\overline{\omega }(\xi ,w-z_0))\right|dw\\ \le & \frac{{\parallel a\parallel }_{\mathrm{\infty }}}{|B|}{\int }_B\left|(\overline{\omega }(z_0,w-z_0)-\overline{\omega }(\xi ,w-z_0))\right|dw\\ \le & C\cdot r^{1-\frac{2n}{p}}\cdot \sigma .\end{array}$$

Let $q=\frac{2np}{2n-p}$, since $\frac{2n}{2n+1}<p<1$, we have $q>1$, hence

$${\parallel b^{(1)}(z)\parallel }_q\le C{\left({\int }_B\right|r^{1-\frac{2n}{p}}\cdot \sigma {|}^{q}dz)}^{\frac{1}{q}}\le C\sigma .$$

Since $supp{b}^{(1)}(z)\subset B(z_0,\sigma )$ and $\overline{\omega }(z_0,z)b^{(1)}(z)\in h^p$, we have

$$\begin{array}{rcl}{\parallel \overline{\omega }(z_0,z)b^{(1)}(z)\parallel }_{h^p}& \le & C(\sigma ){\parallel {(\overline{\omega }(z_0,z)b^{(1)}(z))}_1^{\ast }\parallel }_p\\ =& C(\sigma ){\left({\int }_B\right|{(\overline{\omega }(z_0,z)b^{(1)}(z))}_1^{\ast }{|}^{p}dz)}^{\frac{1}{p}}.\end{array}$$

Let $l=\frac{q}{p}$ and $I={\{{\int }_B{|{(\overline{\omega }(z_0,z)b^{(1)}(z))}_1^{\ast }|}^{p}dz\}}^{\frac{1}{p}}$, then

$$\begin{array}{rcl}{I}^p& \le & {\left\{{\int }_B\right|{(\overline{\omega }(z_0,z)b^{(1)}(z))}_1^{\ast }{|}^{q}dz\}}^{\frac{1}{l}}\cdot {(2r)}^{\frac{2n}{l^{\prime }}}\\ =& {\parallel {(\overline{\omega }(z_0,z)b^{(1)}(z))}_1^{\ast }\parallel }_q^{\frac{q}{l}}\cdot {(2r)}^{\frac{2n}{l^{\prime }}},\end{array}$$

thus

$$I\le {\parallel {(\overline{\omega }(z_0,z)b^{(1)}(z))}_1^{\ast }\parallel }_q\cdot {(2r)}^{\frac{2n(q-p)}{pq}}\le C\cdot \sigma \cdot {(1+\sigma )}^{\frac{2n+p}{p}}.$$

Then we have ${\parallel \overline{\omega }(z_0,z)b^{(1)}(z)\parallel }_{h^p}\le C\cdot \sigma \cdot {(1+\sigma )}^{\frac{2n+p}{p}}$. Let σ be small enough such that $C\cdot \sigma \cdot {(1+\sigma )}^{\frac{2n+p}{p}}<\frac{1}{2}$. By Proposition 1, we have

$$b^{(1)}(z)=\sum _{j=1}^{\mathrm{\infty }}{\eta }_j^{(1)}{a}_j^{(1)}(z),\text{where }\sum _{j=1}^{\mathrm{\infty }}{\left|{\eta }_j^{(1)}\right|}^p\le \frac{1}{2}.$$

The functions $a_j^{(1)}(z)$ are as in (b) and (c). We can now decompose the function $a_j^{(1)}(z)$ whose support is contained in a ball $B(z_j,r_j)$, with $r_j<\sigma$, as we did for $a(z)$, thus

$$b^{(1)}(z)=g^{(2)}(z)+b^{(2)}(z),$$

where $g^{(2)}(z)=\sum {\lambda }_j^{(2)}{\alpha }_j^{(2)}(z)$, ${\alpha }_j^{(2)}(z)$ satisfy (b) and (c), and ${\sum }_{j=1}^{\mathrm{\infty }}{|{\lambda }_j^{(2)}|}^p\le \frac{1}{2}$. Moreover,

$$b^{(2)}(z)=\sum {\eta }_j^{(2)}{a}_j^{(2)}(z),$$

where the $a_j^{(2)}(z)$ are as in (b) and (c) and

$$\sum _{j=1}^{\mathrm{\infty }}|{\eta }_j^{(2)}{|}^p\le \frac{1}{4}.$$

So we can construct sequences $b^{(k)}$ and $g^{(k)}$ such that

$$b^{(k)}=g^{(k+1)}+b^{(k+1)},$$

$g^{(k+1)}=\sum {\lambda }_j^{(k+1)}{\alpha }_j^{(k+1)}(z)$, where the ${\alpha }_j^{(k+1)}(z)$ satisfy (b) and (c), and also

$$\sum _{j=1}^{\mathrm{\infty }}|{\lambda }_j^{(k+1)}{|}^p\le \frac{1}{2^{k-1}},$$

$b^{(k+1)}=\sum {\eta }_j^{(k+1)}{a}_j^{(k+1)}(z)$, where the $a_j^{(k+1)}(z)$ satisfy (b) and (c), and also

$$\sum _{j=1}^{\mathrm{\infty }}|{\eta }_j^{(k+1)}{|}^p\le \frac{1}{2^{k+1}}.$$

This shows that $a(z)={\sum }_k{g}^{(k)}(z)$ and gives the proof of Lemma 2. □

Lemma 3 There exists $C>0$ such that for any ${\mathcal{H}}^p$-atom, we have ${\parallel M_{\mathrm{\infty }}a(z)\parallel }_p<C$, where $\frac{2n}{2n+1}<p<1$.

Proof Without loss of generality, we can assume that $a(z)$ is an atom supported on $B(0,r)$. Let $\phi \in \mathcal{B}$, $t>0$, then

$${\phi }_t\times a(z)=\int [{\phi }_t(z-w)-{\phi }_t(z)]a(w)\overline{\omega }(z,w)dw+{\phi }_t(z)\hat{a}(-iz).$$

If $|z|>2r$ and ${\phi }_t\times a(z)\ne 0$, we have $t>|z|-r>\frac{1}{2}z$, so that

$$|{\phi }_t\times a(z)|\le C_{1}r\frac{r^{\frac{2n(p-1)}{p}}}{{|z|}^{2n+1}}+C_2\frac{|\hat{a}(-iz)|}{{|z|}^{2n}}.$$

Let $I_1=r\frac{r^{\frac{2n(p-1)}{p}}}{{|z|}^{2n+1}}$, and $I_2=\frac{|\hat{a}(-iz)|}{{|z|}^{2n}}$, then $|{\phi }_t\times a(z)|\le C(I_1+I_2)$. Therefore

$${\int }_{|z|>2r}|M_{\mathrm{\infty }}a(z){|}^{p}dz\le {\int }_{|z|>2r}{|I_1+I_2|}^{p}dz\le C({\int }_{|z|>2r}{|I_1|}^{p}dz+{\int }_{|z|>2r}{|I_2|}^{p}dz).$$

First we have

$${\int }_{|z|>2r}{|I_1|}^{p}dz\le C_1{\int }_{|z|>2r}\frac{r^{2n(p-1)+p}}{{|z|}^{(2n+1)p}}dz\le C_1^{\prime }.$$

By Hardy’s inequality (cf. [[6], Theorem 7.22, p.341]), we get

$${\int }_{|z|>2r}{|I_2|}^{p}dz\le C_2{\int }_{|z|>2r}\frac{{|\hat{a}(-iz)|}^p}{{|z|}^{2np}}dz\le C_2^{\prime }.$$

We also have

$$\begin{array}{rcl}{\int }_{|z|\le 2r}|M_{\mathrm{\infty }}a(z){|}^{p}dz& \le & {\left({\int }_{|z|\le 2r}\right|M_{\mathrm{\infty }}a(z)\left|dz\right)}^{\frac{1}{p}}\cdot {(2r)}^{2n(1-p)}\\ \le & C{\parallel a\parallel }_{\mathrm{\infty }}^p{(4r)}^{2np}{(4r)}^{2n(1-p)}\le C.\end{array}$$

This completes the proof of Lemma 3. □

Proof of Theorem 1 By Lemma 2 and Lemma 3, we can obtain the proof of Theorem 1. □

3 The boundedness of the Weyl multiplier on ${\mathcal{H}}^p({\mathbb{C}}^n)$

In order to prove Theorem 2, we need to give some characterizations for ${\mathcal{H}}^p({\mathbb{C}}^n)$. Let $K_t^L(z)$ be the heat kernel of ${\{T_t^L\}}_{t>0}$, then we can get (cf. [4])

$$K_t(z)={(4\pi )}^{-n}{(sinht)}^{-n}{e}^{-\frac{1}{4}{|z|}^2(cotht)}.$$

(7)

It is easy to prove that the heat kernel $K_t(z)$ has the following estimates (cf. [3]).

Lemma 4 There exists a positive constant $C>0$ such that

• (i) $|K_t(z)|\le C{t}^{-n}{e}^{-C\frac{{|z|}^2}{t}}$;

• (ii) $|\mathrm{\nabla }{K}_t(z)|\le C{t}^{-n-\frac{1}{2}}{e}^{-C\frac{{|z|}^2}{t}}$.

Let $Q_t^k(z)$ be the twisted convolution kernel of $Q_t^k=t^{2k}{\partial }_s^k{T}_s^L{|}_{s=t^2}$, then

$$Q_t^k(z)=t^{2k}{\partial }_s^k{K}_s(z){|}_{s=t^2}.$$

Lemma 5 There exist constants $C,C_k>0$ such that

• (i) $|Q_t^k(z)|\le C_k{t}^{-2n}{e}^{-C{t}^{-2}{|z|}^2}$;

• (ii) $|\mathrm{\nabla }{Q}_t^k(z)-Q_t^k(w)|\le C_k{t}^{-2n-1}{e}^{-C{t}^{-2}{|z|}^2}|z-w|$.

In the following, we define the Lusin area integral operator by

$$\left(S_L^{k}f\right)(z)={\left({\int }_0^{+\mathrm{\infty }}{\int }_{|z-w|<t}\right|Q_t^{k}f(w){|}^2\frac{dwdt}{t^{2n+1}})}^{1/2}$$

and the Littlewood-Paley g-function

$${\mathcal{G}}_L^k(f)(z)={\left({\int }_0^{\mathrm{\infty }}\right|Q_t^{k}f(z){|}^2\frac{dt}{t})}^{1/2}.$$

We also consider the $g_{\lambda }^{\ast }$-function associated with L defined by

$$g_{\lambda ,k}^{\ast }f(x)={\left({\int }_0^{\mathrm{\infty }}{\int }_{{\mathbb{C}}^n}{\left(\frac{t}{t+|z-w|}\right)}^{2\lambda n}\right|Q_t^{k}f(w){|}^2\frac{dwdt}{t^{2n+1}})}^{1/2}.$$

We have the following lemma, whose proof is standard (cf. [3]).

Lemma 6

• (i) The operators $S_L^k$ and ${\mathcal{G}}_L^k$ are isometries on $L^2({\mathbb{C}}^n)$.

• (ii) When $\lambda >1$, there exists a constant $C>0$, such that

  $$C^{-1}{\parallel f\parallel }_{L^2}\le {\parallel g_{\lambda ,k}^{\ast }f\parallel }_{L^2}\le C{\parallel f\parallel }_{L^2}.$$

Now we can prove the following lemma.

Lemma 7 Let $\frac{2n}{2n+1}<p<1$ and $f\in {\mathcal{S}}^{\prime }({\mathbb{C}}^n)$, then we have:

• (1) $f\in {\mathcal{H}}^p({\mathbb{C}}^n)$ if and only if its Lusin area integral $S_L^{k}f\in L^p({\mathbb{C}}^n)$. Moreover, we have

  $${\parallel f\parallel }_{{\mathcal{H}}^p}\sim {\parallel S_L^{k}f\parallel }_{L^p}.$$

• (2) $f\in {\mathcal{H}}^p({\mathbb{C}}^n)$ if and only if its Littlewood-Paley g-function ${\mathcal{G}}_L^{k}f\in L^p({\mathbb{C}}^n)$. Moreover, we have

  $${\parallel f\parallel }_{{\mathcal{H}}^p}\sim {\parallel {\mathcal{G}}_L^{k}f\parallel }_{L^p}.$$

• (3) $f\in {\mathcal{H}}^p({\mathbb{C}}^n)$ if and only if its ${\mathcal{G}}_{\lambda }^{\ast }$-function ${\mathcal{G}}_{\lambda ,k}^{\ast }f\in L^p({\mathbb{C}}^n)$, where $\lambda >4$. Moreover, we have

  $${\parallel f\parallel }_{{\mathcal{H}}^p}\sim {\parallel {\mathcal{G}}_{\lambda ,k}^{\ast }f\parallel }_{L^p}.$$

Proof (1) By Lemma 6, we know there exists a constant $C>0$ such that, for any atom $a(x)$ of ${\mathcal{H}}^p({\mathbb{C}}^n)$, we have

$${\parallel S_L^{k}a\parallel }_{L^p}\le C.$$

For the reverse, by Theorem 1, we can prove similarly to Proposition 4.1 in [7].

• (2) Firstly, we can prove ${\mathcal{G}}_L^k$ is uniformly bounded on atoms of ${\mathcal{H}}^p({\mathbb{C}}^n)$. For the reverse, we can prove the following inequality (cf. Theorem 7. 28 in [8]):

  $${\parallel S_L^{k+1}f\parallel }_{L^p}\le C{\parallel {\mathcal{G}}_L^{k}f\parallel }_{L^p}.$$

  (8)

Then (2) follows from part (1) and (8).

• (3) By $S_L^{k}f(z)\le {(\frac{1}{2})}^{2\lambda n}{g}_{\lambda ,k}^{\ast }f(z)$, we know $f\in {\mathcal{H}}^p({\mathbb{C}}^n)$ when $g_{\lambda ,k}^{\ast }f\in L^p({\mathbb{C}}^n)$. In the following, we show there exists a constant $C>0$ such that for any atom $a(z)$ of ${\mathcal{H}}^p({\mathbb{C}}^n)$, we have

  $${\parallel g_{\lambda ,k}^{\ast }a\parallel }_{L^p}\le C.$$

We assume $a(z)$ is supported in $B(z_0,r)$, then

$$g_{\lambda ,k}^{\ast }a{(z)}^2\le C{S}_L^{k}a{(z)}^2+\sum _{k=1}^{\mathrm{\infty }}{2}^{-2k\lambda n}{S}_L^{2^k}a{(z)}^2.$$

Then

$${\parallel g_{\lambda ,k}^{\ast }a\parallel }_{L^p}\le C_1{\parallel S_L^{k}a\parallel }_{L^p}+C_2\sum _{k=1}^{\mathrm{\infty }}{2}^{-k\lambda n}{\parallel S_L^{2^k}a\parallel }_{L^p}.$$

By part (1), we have ${\parallel S_L^{k}a\parallel }_{L^p}\le C$. We can prove (cf. [3])

$${\parallel S_L^{2^k}a\parallel }_{L^p}\le C{2}^{4kn}.$$

(9)

Therefore, when $\lambda >4$, we have ${\parallel g_{\lambda ,k}^{\ast }a\parallel }_{L^p}\le C$. Then Lemma 7 is proved. □

In the following, we give the proof of Theorem 2.

Proof of Theorem 2 Firstly, by Lemma 4.1 in [2], we get

$${\mathcal{G}}_L^{k+1}(F)(z)\le C{\mathcal{G}}_{\frac{k}{n},1}^{\ast }(f)(z),$$

where $F(z)=T_{\varphi }f(z)$, then, by Lemma 7, when $k>4n$,

$${\parallel \varphi (L)f\parallel }_{{\mathcal{H}}^p}\le C{\parallel {\mathcal{G}}_L^{k+1}(F)(z)\parallel }_{L^p}\le C{\parallel {\mathcal{G}}_{\frac{k}{n},1}^{\ast }(f)(z)\parallel }_{L^p}\le C{\parallel f\parallel }_{{\mathcal{H}}^p}.$$

This completes the proof of Theorem 2. □