New real-variable characterizations of Hardy spaces associated with twisted convolution

Abstract

In this paper, we give some new real-variables characterizations of the Hardy space associated with twisted convolution, including Poisson maximal function, area integral, and Littlewood-Paley g-function.

1 Introduction

In this paper, we consider the 2n linear differential operators

$$ Z_{j}=\frac{\partial}{\partial{z_{j}}}+\frac{1}{4} \bar{z}_{j},\qquad \bar{Z}_{j}=\frac{\partial}{\partial {\bar{z}_{j}}}- \frac{1}{4}z_{j} \quad \text{on } \mathbb{C}^{n}, j=1,2,\ldots,n. $$

(1)

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

$$ [Z_{j}, \bar{Z}_{j} ]=-\frac{1}{2}I, \quad j=1,2,\ldots,n. $$

(2)

The operator L defined by

$$L=-\frac{1}{2}\sum_{j=1}^{n} (Z_{j}\bar{Z}_{j}+\bar{Z}_{j}Z_{j} ) $$

is nonnegative, self-adjoint, and elliptic. Therefore it generates a diffusion semigroup $\{T_{t}^{L}\}_{t>0}=\{e^{-tL}\}_{t>0}$. The operators in (1) generate a family of ‘twisted translations’ $\tau_{w}$ on $\mathbb{C}^{n}$ defined on measurable functions by

$$\begin{aligned} (\tau_{w}f ) (z) =&\exp \Biggl(\frac{1}{2}\sum _{j=1}^{n}(w_{j}z_{j}+ \bar{w}_{j}\bar{z}_{j}) \Biggr)f(z) \\ =&f(z+w)\exp \biggl(\frac{i}{2}\operatorname{Im}(z\cdot\bar{w}) \biggr). \end{aligned}$$

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

$$\begin{aligned} (f\times g) (z) =&\int_{\mathbb{C}^{n}}f(w)\tau_{-w}g(z)\,dw \\ =&\int_{\mathbb{C}^{n}}f(z-w)g(w)\bar{\omega}(z,w)\,dw, \end{aligned}$$

where $\omega(z,w)=\exp (\frac{i}{2}\operatorname{Im}(z\cdot\bar{w}) )$. More about twisted convolution can be found in [1–3].

In [4], the authors defined the Hardy space $H_{L}^{1}(\mathbb{C}^{n})$ associated with a twisted convolution. They gave several characterizations of $H_{L}^{1}(\mathbb{C}^{n})$ via maximal functions, the atomic decomposition, and the behavior of the local Riesz transform. As applications, the boundedness of Hömander multipliers on Hardy spaces is considered in [5]. The ‘twisted cancelation’ and Weyl multipliers were introduced for the first time in [6]. Recently, Huang and Wang [7] defined the Hardy space $H_{L}^{p}(\mathbb{C}^{n})$ associated with a twisted convolution for $\frac{2n}{2n+1}< p\leq1$. Huang gave the characterizations of the Hardy space associated with twisted convolution by the Lusin area integral function and the Littlewood-Paley function defined by the heat kernel in [8] and established the boundedness of the Weyl multiplier by these characterizations in [9]. Recently, Huang and Liu gave the molecular characterization of Hardy space associated with twisted convolution in [10]. The purpose of this paper is to give some new real-variable characterizations for $H_{L}^{p}(\mathbb {C}^{n})$, including the Poisson maximal function, the Lusin area integral, and the Littlewood-Paley g-function defined by the Poisson kernel.

We first give some basic notations concerning $H_{L}^{p}(\mathbb{C}^{n})$. Let $\mathcal{B}$ denote the class of $C^{\infty}$-functions φ on $\mathbb{C}^{n}$, supported on the ball $B(0,1)$ such that $\|\varphi\|_{\infty}\leq1$ and $\|\nabla\varphi\|_{\infty}\leq 2$. For $t>0$, let $\varphi_{t}(z)=t^{-2n}\varphi(z/t)$. Given $\sigma>0$, $0<\sigma\leq+\infty$, and a tempered distribution f, define the grand maximal function

$$M_{\sigma}f(z)=\sup_{\varphi\in \mathcal{B}}\sup_{0< t< \sigma} \bigl\vert \varphi_{t}\times f(z)\bigr\vert . $$

Then the Hardy space $H_{L}^{p}(\mathbb{C}^{n})$ can be defined by

$$H_{L}^{p}\bigl(\mathbb{C}^{n}\bigr)= \bigl\{ f\in \mathcal{S}'\bigl(\mathbb{C}^{n}\bigr): M_{\infty}f \in L^{p}\bigl(\mathbb{C}^{n}\bigr) \bigr\} . $$

For any $f\in H_{L}^{p}(\mathbb{C}^{n})$, define $\| f\|_{ H_{L}^{p}(\mathbb{C}^{n})}=\| M_{\infty}f\|_{L^{p}}$.

Definition 1

Let $\frac{2n}{2n+1}< p\le1\le q\le\infty$ and $p\neq q$. A function $a(z)$ is a $H_{L}^{p,q}$-atom for the Hardy space $H_{L}^{p}(\mathbb{C}^{n})$ associated to a ball $B(z_{0},r)$ if

$$\begin{aligned} (1)&\quad \operatorname{supp} a\subset B(z_{0},r); \\ (2)& \quad \|a\|_{q}\leq\bigl\vert B(z_{0},r)\bigr\vert ^{1/q-1/p}; \\ (3)&\quad \int_{\mathbb{C}^{n}} a(w)\bar{\omega}(z_{0},w) \,dw=0. \end{aligned}$$

We define the atomic Hardy space $H^{p,q}_{L}(\mathbb{C}^{n})$ to be the set of all tempered distributions of the form $\sum_{j} \lambda_{j}a_{j}$ (the sum converges in the topology of $\mathcal{S}'(\mathbb{C}^{n})$), where $a_{j}$ are $H_{L}^{p,q}$-atoms and $\sum_{j}|\lambda_{j}|^{p}<+\infty$.

The atomic quasi-norm in $H_{L}^{p,q}(\mathbb{C}^{n})$ is defined by

$$\|f\|_{L\text{-atom}}=\inf \biggl\{ \biggl(\sum_{j}| \lambda_{j}|^{p} \biggr)^{1/p} \biggr\} , $$

where the infimum is taken over all decompositions $f=\sum_{j} \lambda_{j}a_{j}$ and $a_{j}$ are $H_{L}^{p,q}$-atoms.

The following result has been proved in [4] and [7].

Proposition 1

Let $\frac{2n}{2n+1}< p\leq1$. Then for a tempered distribution f on $\mathbb{C}^{n}$, the following are equivalent:

• (i) $M_{\infty}f \in L^{p}(\mathbb{C}^{n})$.

• (ii) For some σ, $0<\sigma<+\infty$, $M_{\sigma}f \in L^{p}(\mathbb{C}^{n})$.

• (iii) For some radial function $\varphi\in\mathcal{S}$, such that $\int_{\mathbb{C}^{n}}\varphi(z)\, dz\neq0$, we have

  $$\sup_{0< t< 1}\bigl\vert \varphi_{t}\times f(z)\bigr\vert \in L^{p}\bigl(\mathbb{C}^{n}\bigr). $$

• (iv) f can be decomposed as $f=\sum_{j} \lambda_{j}a_{j}$, where $a_{j}$ are $H_{L}^{p,q}$-atoms and $\sum_{j}|\lambda_{j}|^{p}<+\infty$.

Corollary 1

Let $\frac{2n}{2n+1}< p\leq1$ and $1< q\le\infty$. Then $H_{L}^{p,q}(\mathbb{C}^{n})=H^{p}_{L}(\mathbb{C}^{n})$ with equivalent norms.

Let $\{P_{t}^{L}\}_{t>0}$ be the Poisson semigroup generated by the operator L. Then, for $f\in L^{2}(\mathbb{C}^{n})$, the function $e^{-t\sqrt{L}}f$ has the special Hermite expansion (cf. [11])

$$e^{-t\sqrt{L}}f(z)= (2\pi )^{-n}\sum_{k=0}^{\infty}e^{-\sqrt {2k+n}t}f\times \varphi_{k}(z), $$

where $\varphi_{k}$ are Laguerre functions. Therefore $e^{-t\sqrt{L}}f$ is given by the twisted convolution with the kernel

$$ P_{t}(z)= (2\pi )^{-n}\sum _{k=0}^{\infty}e^{-\sqrt{2k+n}t} \varphi_{k}(z). $$

(3)

The Poisson maximal function is defined by

$$M_{P}(f) (z)=\sup_{t>0}\bigl\vert P_{t}\times f(z)\bigr\vert . $$

We can characterize the Hardy space $H_{L}^{1}(\mathbb{C}^{n})$ as follows.

Theorem 1

$f\in H_{L}^{1}(\mathbb{C}^{n})$ if and only if $f\in L^{1}(\mathbb{C}^{n})$ and $M_{P}(f)\in L^{1}(\mathbb{C}^{n})$. Moreover, we have

$$\|f\|_{H_{L}^{1}}\sim\bigl\Vert M_{P}(f)\bigr\Vert _{L^{1}}. $$

We define the area integral associated to $\{P_{t}^{L}\}_{t>0}$ by

$$\bigl(S_{L}^{k}f\bigr) (z)= \biggl(\int _{0}^{+\infty}\int_{|z-w|< t} \bigl\vert D_{t}^{k}f(w)\bigr\vert ^{2} \frac{dw\,dt}{t^{2n+1}} \biggr)^{1/2}, $$

the Littlewood-Paley g-function by

$$\mathcal{G}_{L}^{k}(f) (z)= \biggl(\int_{0}^{\infty} \bigl\vert D_{t}^{k}f(z)\bigr\vert ^{2} \frac{dt}{t} \biggr)^{1/2}, $$

and we consider the $g_{\lambda}^{*}$-function associated with L defined by

$$g_{\lambda,k}^{*}f(z)= \biggl(\int_{0}^{\infty}\int _{\mathbb {C}^{n}} \biggl(\frac{t}{t+|z-w|} \biggr)^{2\lambda n}\bigl\vert D_{t}^{k}f(w)\bigr\vert ^{2} \frac{dw\,dt}{t^{2n+1}} \biggr)^{1/2}, $$

where $D_{t}^{k}f(z)=t^{k} (\partial_{t}^{k}P_{t}^{L}f )(z)$.

Now we can prove the main result of this paper.

Theorem 2

• (a) A function $f\in H_{L}^{1}(\mathbb{C}^{n})$ if and only if its Lusin area integral $S_{L}^{k}f\in L^{1}(\mathbb{C}^{n})$ and $f\in L^{1}(\mathbb{C}^{n})$. Moreover, we have

  $$\|f\|_{H_{L}^{1}}\sim\bigl\Vert S_{L}^{k}f\bigr\Vert _{L^{1}}. $$

• (b) A function $f\in H_{L}^{1}(\mathbb{C}^{n})$ if and only if its Littlewood-Paley g-function $\mathcal{G}_{L}^{k}f\in L^{1}(\mathbb{C}^{n})$ and $f\in L^{1}(\mathbb{C}^{n})$. Moreover, we have

  $$\|f\|_{H_{L}^{1}}\sim\bigl\Vert \mathcal{G}_{L}^{k}f \bigr\Vert _{L^{1}}. $$

• (c) A function $f\in H_{L}^{1}(\mathbb{C}^{n})$ if and only if its $g_{\lambda}^{*}$-function $g_{\lambda,k}^{*}f\in L^{1}(\mathbb{C}^{n})$ and $f\in L^{1}(\mathbb{C}^{n})$, where $\lambda>3$. Moreover, we have

  $$\|f\|_{H_{L}^{1}}\sim\bigl\Vert g_{\lambda,k}^{*}f\bigr\Vert _{L^{1}}. $$

Remark 1

In this paper, we just give the proofs of our results for $p=1$. In fact, we can prove the case $\frac{2n}{2n+1}< p<1$ under more conditions (such as that f vanishes weakly at infinity). The proofs of the case $\frac{2n}{2n+1}< p<1$ are quite similar to the case $p=1$, so we omit them.

Throughout the article, we will use C to denote a positive constant, which is independent of the 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 $\frac{1}{C} \leq \frac{B_{1}}{B_{2}}\leq C$.

2 Preliminaries

In this section, we give some preliminaries that we will use in the sequel.

Let $K_{t}(z)$ be the heat kernel of $\{T_{t}^{L}\}_{t>0}$. Then we can get (cf. [11])

$$ K_{t}(z)= (4\pi )^{-n} (\sinh t )^{-n}e^{-\frac{1}{4}|z|^{2}(\coth t)}. $$

(4)

It is easy to prove that the heat kernel $K_{t}(z)$ has the following estimates (cf. [8]).

Lemma 1

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

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

• (ii) $|\nabla K_{t}(z) |\leq C t^{-n-\frac{1}{2}}e^{-C\frac{|z|^{2}}{t}}$.

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

$$Q_{t}^{k}(z)=t^{2k}\partial_{s}^{k}K_{s}(z)|_{s=t^{2}}. $$

We have the following estimates [8].

Lemma 2

There exist constants $C, C_{k}>0$ such that

• (i) $|Q_{t}^{k}(z) |\leq C_{k} t^{-2n}e^{-C t^{-2}|z|^{2}}$;

• (ii) $| \nabla Q_{t}^{k}(z) |\leq C_{k} t^{-2n-1}e^{-C t^{-2}|z|^{2}}$.

By the subordination formula, we can give the following estimates as regards the Poisson kernel.

Lemma 3

There exist constants $C_{k}>0$, $A>0$ such that

• (a)
  $$ 0< P_{t}(z)\leq C_{k}\frac{t}{(t^{2}+A|z|^{2})^{(2n+1)/2}}; $$

  (5)
• (b)
  $$ \bigl\vert \nabla P_{t}(z)\bigr\vert \leq C_{k}\frac{\sqrt{t}}{(t^{2}+A|z|^{2})^{(2n+1)/2}}. $$

  (6)

Lemma 4

Let $D_{t}^{k}(z)$ be the integral kernel of the operator $D_{t}^{k}$. Then there exist constants $C_{k}>0$, $A>0$, such that

• (a)
  $$ \bigl|D_{t}^{k}(z) \bigr|\leq C_{k}\frac {t}{(t^{2}+A|z|^{2})^{(2n+1)/2}}; $$
• (b)
  $$ \bigl\vert \nabla D_{t}^{k}(z)\bigr\vert \leq C_{k}\frac{\sqrt{t}}{(t^{2}+A|z|^{2})^{(2n+1)/2}}. $$

We also need some basic properties about the tent space (cf. [12]).

Let $0< p<\infty$, and $1\leq q\leq\infty$. Then the tent space $T^{p}_{q}$ is defined as the space of functions f on $\mathbb{C}^{n}\times\mathbb{R}^{+}$, so that

$$\biggl(\int_{\Gamma(z)}\bigl\vert f(w,t)\bigr\vert ^{q}\frac {dw\,dt}{t^{2n+1}} \biggr)^{1/q}\in L^{p}\bigl( \mathbb{C}^{n}\bigr), \quad \text{when } 1\leq q< \infty $$

and

$$\sup_{(w,t)\in\Gamma(z)}\bigl\vert f(w,t)\bigr\vert \in L^{p} \bigl(\mathbb{C}^{n}\bigr),\quad \text{when } q=\infty, $$

where $\Gamma(z)$ is the standard cone whose vertex is $z\in\mathbb{C}^{n}$, i.e.,

$$\Gamma(z)=\bigl\{ (w,t):|w-z|< t\bigr\} . $$

Assume $B(z_{0},r)$ is a ball in $\mathbb{C}^{n}$, its tent $\hat{B}$ is defined by $\hat{B}=\{(w,t):|w-z_{0}|\leq r-t\}$. A function $a(z,t)$ supported in a tent $\hat{B}$, B a ball in $\mathbb{C}^{n}$, is said to be an atom in the tent space $T^{p}_{q}$ if and only if it satisfies

$$\biggl(\int_{\hat{B}}\bigl\vert a(z,t)\bigr\vert ^{2}\frac{dz\,dt}{t} \biggr)^{1/2}\leq |B|^{1/2-1/p}. $$

The atomic decomposition of $T^{p}_{q}$ is stated as follows.

Proposition 2

When $0< p\leq1$, then for any $f\in T^{p}_{2}$ can be written as $f=\sum\lambda_{k}a_{k}$, where $a_{k}$ are atoms and $\sum|\lambda_{k}|^{p}\leq C\|f\|_{T^{p}_{2}}^{p}$.

3 The proofs of the main results

Let

$$M_{H}f(z)=\sup_{t>0}\bigl\vert K_{t} \times f(z)\bigr\vert ,\quad f\in L^{1}\bigl(\mathbb{C}^{n} \bigr) $$

be the heat maximal function. Then we can characterize $H_{L}^{1}(\mathbb{C}^{n})$ by the maximal function $M_{H}f$ as follows (cf. [4] or [8]).

Lemma 5

$f\in H_{L}^{1}(\mathbb{C}^{n})$ if and only if $M_{H}f\in L^{1}(\mathbb{C}^{n})$ and $f\in L^{1}(\mathbb{C}^{n})$.

Now, we give the proof of Theorem 1.

Proof of Theorem 1

If $f\in H_{L}^{1}(\mathbb{C}^{n})$, then, by Lemma 5, we get $M_{H}f\in L^{1}(\mathbb{C}^{n})$. Since

$$P_{t}(z)=\frac{1}{\sqrt{\pi}}\int_{0}^{\infty}K_{t^{2}/4\mu }(z)e^{-\mu} \mu^{-1/2}\, d\mu, $$

we have $\|M_{P}(f)\|_{L^{1}}\leq C \|M_{H}(f)\|_{L^{1}}$, i.e., $M_{P}f\in L^{1}(\mathbb{C}^{n})$.

For the reverse, there exists a function η defined on $(1,\infty )$ that is rapidly decreasing at ∞ and satisfies the moment conditions (cf. [13])

$$\int_{1}^{\infty}\eta(t)\, dt=1, \qquad \int _{1}^{\infty}t^{k}\eta(t)\, dt=0,\quad k=1,2, \ldots. $$

Let

$$ \Phi(z)=\int_{1}^{\infty}\eta(t)P_{t}(z)\, dt. $$

(7)

Since

$$\bigl(1+s^{2}\bigr)^{-(2n+1)/2}=\sum_{k< R}a_{k}s^{k}+O \bigl(s^{R}\bigr),\quad 0\leq s< \infty $$

for appropriate binomial coefficients $a_{k}$, we have

$$ \frac{t}{(t^{2}+A|z|^{2})^{(2n+1)/2}}=\sum_{k< R}a_{k}t|z|^{-1-2n} \biggl(\frac{t}{|z|} \biggr)^{k}+O\bigl(t^{R+1}|z|^{-2n-1-R} \bigr). $$

(8)

By (8) and Lemma 3, we know that Φ and any derivative of Φ are rapidly decreasing. Thus $\Phi\in\mathcal{S}$ and

$$\int_{\mathbb{C}^{n}}\Phi(z)\, dz=\int_{1}^{\infty}\eta(t)\, dt=1. $$

Therefore,

$$M_{\Phi}(f) (z)\leq M_{P}(f) (z)\dot{\int}_{1}^{\infty}\bigl\vert \eta(t)\bigr\vert \, dt\leq C M_{P}(f) (z). $$

This proves that $M_{P}(f)\in L^{1}(\mathbb{C}^{n})$ implies $f\in H_{L}^{1}(\mathbb{C}^{n})$ and the proof of Theorem 1 is complete. □

In order to get our results, we need the following lemma (cf. Lemma 5 in [8]).

Lemma 6

• (i) The operators $S_{L}^{k}$ and $\mathcal{G}_{L}^{k}$ are isometries on $L^{2}(\mathbb{C}^{n})$ up to constant factors. Exactly,

  $$ \bigl\Vert \mathcal{G}_{L}^{k} f \bigr\Vert _{L^{2}}\sim \| f \|_{L^{2}},\qquad \bigl\Vert S_{L}^{k} f \bigr\Vert _{L^{2}}\sim \| f \|_{L^{2}}. $$

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

  $$C^{-1}\|f\|_{L^{2}}\leq\bigl\Vert g_{\lambda,k}^{*}f\bigr\Vert _{L^{2}}\leq C\|f\|_{L^{2}}. $$

We define the new Lusin type area integral operator by

$$\bigl(S_{L,\alpha}^{k}f\bigr) (z)= \biggl(\int _{0}^{+\infty}\int_{|z-w|< \alpha t}\bigl\vert D_{t}^{k}f(w)\bigr\vert ^{2}\frac{dw\,dt}{t^{2n+1}} \biggr)^{1/2}, $$

where $\alpha>0$.

Lemma 7

It is easy to see that the above definition of the area integral operator is independent of α in the sense of $\|(S_{L}^{\alpha}f)\|_{L^{p}}\sim \|(S_{L}^{\beta}f)\|_{L^{p}}$, for $0<\alpha<\beta<\infty$ and $0< p<\infty$ (cf. [12]). In the following, we use $S_{L}^{k}$ to denote $S_{L,1}^{k}$.

Proof of Theorem 2

(a) By Lemma 4, we can prove that there exists a constant $C>0$ such that for any atom $a(z)$ of $H_{L}^{1}(\mathbb{C}^{n})$, we have

$$ \bigl\Vert S_{L}^{k}a\bigr\Vert _{L^{1}}\leq C. $$

(9)

In the following, we will show that $f\in H_{L}^{1}(\mathbb{C}^{n})$ when $S_{L}^{k}f\in L^{1}(\mathbb{C}^{n})$ and $f\in L^{1}(\mathbb{C}^{n})$.

We first assume that $f\in L^{1}(\mathbb{C}^{n})\cap L^{2}(\mathbb{C}^{n})$. When $S_{L}^{k}f\in L^{1}(\mathbb{C}^{n})$, we know $D_{t}^{k}f\in T_{2}^{1}$. By Proposition 2, we get

$$ D_{t}^{k}f(z)=\sum _{j}\lambda_{j}a_{j}(z,t), $$

(10)

where $a_{j}(z,t)$ are atoms of $T^{1}_{2}$ and $\sum_{j}|\lambda_{j}|<\infty$. By the spectrum theorem (cf. [14]), we can prove

$$ f(z)=4\int_{0}^{\infty}D_{t}^{k} \bigl(D_{t}^{k}f(z) \bigr)\frac{dt}{t}. $$

(11)

By (10) and (11), we get

$$f(z)=4\int_{0}^{+\infty} D_{t}^{k} \biggl( \sum_{j}\lambda_{j}a_{j}(z,t) \biggr)\frac{dt}{t} =C\sum_{j} \lambda_{j}\int_{0}^{+\infty}D_{t}^{k}a_{j}(z,t) \frac{dt}{t}. $$

Therefore, it is sufficient to prove $\alpha_{j}=\int_{0}^{+\infty}D_{t}^{k}a_{j}(z,t)\frac{dt}{t}$, $i=1,2, \ldots$ , are bounded in $H_{L}^{1}(\mathbb{C}^{n})$ uniformly, i.e., there exists a constant $C>0$ such that for any atom $a(z,t)$ in $T^{1}_{2}$,

$$\|\alpha\|_{H_{L}^{1}}= \biggl\Vert \int_{0}^{+\infty }D_{t}^{k}a(z,t) \frac{dt}{t}\biggr\Vert _{H_{L}^{1}}\leq C. $$

We assume that $a(z,t)$ is supported in $\hat{B}(z_{0},r)$, where $\hat{B}(z_{0},r)$ denotes the tent of the ball $B(z_{0},r)$, then

$$\Bigl\Vert \sup_{t>0}\bigl\vert e^{-t\sqrt{L}}\alpha(z) \bigr\vert \Bigr\Vert _{L^{1}} \leq\Bigl\Vert \Bigl(\sup _{t>0}\bigl\vert e^{-t\sqrt{L}}\alpha(z)\bigr\vert \Bigr) \chi_{B^{*}}\Bigr\Vert _{L^{1}} +\Bigl\Vert \Bigl(\sup _{t>0}\bigl\vert e^{-t\sqrt{L}}\alpha(z)\bigr\vert \Bigr) \chi_{(B^{*})^{c}}\Bigr\Vert _{L^{1}} =I_{1}+I_{2}, $$

where $B^{*}=B(z_{0}, 2r)$.

By the Hölder inequality, we get

$$I_{1}\leq\bigl\vert B^{*}\bigr\vert ^{1/2} \biggl(\int_{\mathbb{C}^{n}}\Bigl(\sup_{t>0}\bigl\vert e^{-t\sqrt{L}}\alpha(z)\bigr\vert \Bigr)^{2}\, dz \biggr)^{1/2} \leq\bigl\vert B^{*}\bigr\vert ^{1/2}\|\alpha\|_{L^{2}}. $$

By the self-adjointness of $D_{t}^{k}$ and Lemma 5, we can get

$$\begin{aligned} \|\alpha\|_{L^{2}} =&\sup_{\|\beta\|_{L^{2}}\leq1}\int _{\mathbb{C}^{n}}\alpha(z)\bar{\beta}(z)\, dz \\ =&\sup_{\|\beta\|_{L^{2}}\leq1}\int_{\mathbb{C}^{n}} \biggl(\int _{0}^{+\infty}D_{t}^{k}a(z,t) \frac{dt}{t} \biggr)\bar {\beta}(z)\, dz \\ =&\sup_{\|\beta\|_{L^{2}}\leq1}\int_{0}^{+\infty}\int _{\mathbb{C}^{n}}D_{t}^{k}a(z,t)\bar{\beta}(z)\, dz \frac{dt}{t} \\ =&\sup_{\|\beta\|_{L^{2}}\leq1}\int_{0}^{+\infty}\int _{\mathbb{C}^{n}}a(z,t)D_{t}^{k}\bar{\beta}(z)\, dz \frac{dt}{t} \\ \leq&\sup_{\|\beta\|_{L^{2}}\leq1} \biggl(\int_{\mathbb{C}^{n}}\int _{0}^{+\infty}\bigl\vert a(z,t)\bigr\vert ^{2}\frac{dz\, dt}{t} \biggr)^{1/2} \\ &{} \times \biggl(\int_{\mathbb{C}^{n}} \int_{0}^{+\infty} \bigl\vert D_{t}^{k}\bar{\beta}(z)\bigr\vert ^{2}\frac{dz\, dt}{t} \biggr)^{1/2} \\ \leq&|B|^{-1/2}\|\beta\|_{L^{2}}\leq|B|^{-1/2}. \end{aligned}$$

This gives the proof of $I_{1}\leq C$.

By Lemma 2, we can prove

$$\begin{aligned}& \sup_{s>0} \biggl\vert e^{-s\sqrt{L}}\int _{0}^{+\infty }D_{t}^{k}a(z,t) \frac{dt}{t} \biggr\vert \\& \quad = \sup_{s>0} \biggl\vert e^{-s\sqrt{L}}\int _{0}^{+\infty}(-t\sqrt {L})^{k}e^{-t\sqrt{L}}a(z,t) \frac{dt}{t} \biggr\vert \\& \quad = \sup_{s>0} \biggl\vert \int_{0}^{+\infty}(-t \sqrt{L})^{k}e^{-(s+t)\sqrt{L}}a(z,t)\frac{dt}{t} \biggr\vert \\& \quad = \sup_{s>0} \biggl\vert \int_{0}^{+\infty} \biggl(\frac{t}{s+t} \biggr)^{k} \bigl(-(s+t) \sqrt{L} \bigr)^{k}e^{-(s+t)\sqrt{L}}a(z,t)\frac{dt}{t} \biggr\vert \\& \quad = \sup_{s>0} \biggl\vert \int_{0}^{+\infty} \biggl(\frac{t}{s+t} \biggr)^{k} \int_{\mathbb{C}^{n}}D_{s+t}^{k}(z-w)a(w,t) \frac{dw\, dt}{t} \biggr\vert \\& \quad \leq \sup_{s>0}\int_{0}^{+\infty} \frac{t}{s+t} \int_{\mathbb{C}^{n}}\frac{s+t}{((s+t)^{2}+A|z-w|^{2})^{(2n+1)/2}}\bigl\vert a(w,t)\bigr\vert \frac{dw \, dt}{t} \\& \quad \leq \sup_{s>0} \biggl(\int_{0}^{r} \int_{B}(s+t)^{-4n} \biggl(1+A\frac{|z-w|^{2}}{(s+t)^{2}} \biggr)^{-(2n+1)} \biggl(\frac{t}{s+t} \biggr)^{2} \frac{dw \, dt}{t} \biggr)^{1/2} \\& \qquad {} \times \biggl(\int_{0}^{r}\int _{B}\bigl\vert a(w,t)\bigr\vert ^{2} \frac{dw\, dt}{t} \biggr)^{1/2} \\& \quad \leq |B|^{-1/2}|z-z_{0}|^{-(2n+1)} \biggl(\int _{0}^{r}\int_{B}t\, dw\, dt \biggr)^{1/2} \\& \quad \leq Cr|z-z_{0}|^{-(2n+1)}. \end{aligned}$$

Then we get

$$I_{2}\leq C r\int_{(B^{*})^{c}}|z-z_{0}|^{-(2n+1)} \, dz\leq C. $$

When $f\in L^{1}(\mathbb{C}^{n})$, we can proceed similarly to Proposition 14 in [15]. In fact, we let $f_{s}=T_{2^{-s}}^{L}f$, $s\geq0$. Then, by $f\in L^{1}(\mathbb{C}^{n})$ and Lemma 3, we know $f_{s}\in L^{2}(\mathbb{C}^{n})$ and $\|S_{L}^{k}f_{s}\|_{1}\leq\|S_{L}^{k}f\|_{1}$. By the above proof, we get

$$\|f_{s} \|_{H_{L}^{1}(\mathbb{C}^{n})}\lesssim \bigl\Vert S_{L}^{k}f_{s} \bigr\Vert _{L^{1}}\leq \bigl\Vert S_{L}^{k}f \bigr\Vert _{L^{1}}. $$

By the monotone convergence theorem, we have

$$\|f_{s}-f_{n} \|_{H_{L}^{1}}\leq \bigl\Vert S_{L}^{k} (f_{s}-f_{n}) \bigr\Vert _{L^{1}}\rightarrow0,\quad \text{when } s,n\rightarrow +\infty. $$

Therefore, $\{f_{s}\}$ is a Cauchy sequence in $H^{1}_{L}(\mathbb{C}^{n})$ and there exists $g\in H^{1}_{L}(\mathbb{C}^{n})$ such that

$$\lim_{s\rightarrow+\infty}f_{s}=g \quad \text{in } H^{1}_{L}\bigl(\mathbb{C}^{n}\bigr). $$

As

$$\lim_{s\rightarrow+\infty}f_{s}=f \quad \text{in } ( \mathit{BMO}_{L})^{*}, $$

we know $f=g\in H^{1}_{L}(\mathbb{C}^{n})$ and $\|f\|_{H_{L}^{1}(\mathbb{C}^{n})}\lesssim\|S_{L}^{k}f\|_{L^{1}}$.

This gives the proof of Theorem 2(a).

(b) Firstly, by Lemma 4, we can prove that there exists a positive constant C such that for any atom $a(z)$ of $H_{L}^{1}(\mathbb{C}^{n})$, we have

$$\bigl\Vert \mathcal{G}_{L}^{k}a\bigr\Vert _{L^{1}}\leq C. $$

For the reverse, by (a), it is sufficient to prove

$$ \bigl\Vert S_{L}^{k+1}f\bigr\Vert _{L^{1}}\leq C \bigl\Vert \mathcal{G}_{L}^{k}f\bigr\Vert _{L^{1}}. $$

(12)

Our proof is motivated by [16]. Let

$$F(z) (t)= \bigl(\partial_{t}^{k}e^{-t\sqrt{L}}f \bigr) (z),\qquad V(z,s)=e^{-s\sqrt{L}}F(z). $$

Then

$$V(z,s) (t)=e^{-s\sqrt{L}} \bigl(\partial_{t}^{k}e^{-t\sqrt{L}}f \bigr) (z)= \bigl(\partial_{t}^{k}e^{-(s+t)\sqrt{L}}f \bigr) (z). $$

Therefore

$$\begin{aligned} \int_{0}^{+\infty} \bigl\vert V(z,s) (t) \bigr\vert ^{2}t^{2k-1} \, dt =&\int_{0}^{+\infty} \bigl\vert \bigl(\partial_{t}^{k}e^{-(s+t)\sqrt {L}}f \bigr) (z) \bigr\vert ^{2}t^{2k-1}\, dt \\ =&\int_{s}^{+\infty} \bigl\vert \bigl( \partial_{t}^{k}e^{-t\sqrt{L}}f \bigr) (z) \bigr\vert ^{2}(t-s)^{2k-1} \, dt. \end{aligned}$$

Hence

$$\sup_{s>0} \int_{0}^{+\infty} \bigl\vert V(z,s) (t) \bigr\vert ^{2}t^{2k-1}\, dt \leq\int _{0}^{+\infty} \bigl\vert \bigl(t^{k} \partial_{t}^{k}e^{-t\sqrt {L}}f \bigr) (z) \bigr\vert ^{2}\frac{dt}{t} = \bigl(\mathcal{G}_{L}^{k}f(z) \bigr)^{2}. $$

Let $\mathbf{X}=L^{2} ((0,\infty), t^{2k-1}\, dt )$. Then

$$\sup_{s>0} \bigl\Vert e^{-s\sqrt{L}}F(z) \bigr\Vert _{\mathbf{X}}=\mathcal {G}_{L}^{k}f(z)\in L^{1} \bigl(\mathbb{C}^{n}\bigr). $$

Therefore $F\in H_{\mathbf{X}}^{1}(\mathbb{C}^{n})$, here $H_{\mathbf{X}}^{1}(\mathbb{C}^{n})$ can be seen as a vector-valued Hardy space. This shows that $\widetilde{S_{L}^{1}}F(z)\in L^{1}(\mathbb{C}^{n})$, where

$$\widetilde{ S_{L}^{1}}F(z)= \biggl(\int _{0}^{+\infty}\int_{|z-w|< 2t} \bigl\Vert D_{t}^{1}F(w) \bigr\Vert _{\mathbf{X}}^{2} \frac{dw\, dt}{t^{2n+1}} \biggr)^{1/2}. $$

By

$$\begin{aligned} \bigl(S_{L}^{1}F(z) \bigr)^{2} =&\int _{0}^{+\infty}\int_{|z-w|< 2t} \bigl\Vert D_{t}^{1}(z) \bigr\Vert _{\mathbf{X}}^{2} \frac{dw\, dt}{t^{2n+1}} \\ =&\int_{0}^{+\infty}\int_{\vert z-w\vert < 2t} \int_{0}^{+\infty} \bigl\vert (-t\sqrt{L})e^{-t\sqrt{L}}F(w) (s) \bigr\vert ^{2}s^{2k-1}\, ds\frac{dw\, dt}{t^{2n+1}} \\ =&\int_{0}^{+\infty} \int_{0}^{+\infty} \int_{\vert z-w\vert < 2t} \bigl\vert (-\sqrt{L})^{k+1}e^{-(s+t)\sqrt{L}}f(w) \bigr\vert ^{2}t^{1-2n}s^{2k-1}\, dw\, dt\, ds \\ =&\int_{0}^{+\infty}\int_{s}^{+\infty} \int_{\vert z-w\vert < 2(t-s)} \bigl\vert (-\sqrt{L})^{k+1}e^{-t\sqrt{L}}f(w) \bigr\vert ^{2}(t-s)^{1-2n}s^{2k-1}\, dw\,dt\,ds \\ =&\int_{0}^{+\infty}\int_{0}^{t} \int_{\vert z-w\vert < 2(t-s)} \bigl\vert (-\sqrt{L})^{k+1}e^{-t\sqrt{L}}f(w) \bigr\vert ^{2}(t-s)^{1-2n}s^{2k-1}\, dw\,ds\,dt \\ \geq&\int_{0}^{+\infty}\int_{0}^{t/2} \int_{\vert z-w\vert < 2(t-s)} \bigl\vert (-\sqrt{L})^{k+1}e^{-t\sqrt{L}}f(w) \bigr\vert ^{2}(t-s)^{1-2n}s^{2k-1}\, dw\,ds\,dt \\ \geq&\int_{0}^{+\infty}\int_{0}^{t/2} \int_{\vert z-w\vert < t} \bigl\vert (-\sqrt{L})^{k+1}e^{-t\sqrt{L}}f(w) \bigr\vert ^{2}t^{1-2n}s^{2k-1}\, dw\,ds\,dt \\ =&\frac{1}{2k2^{2k}}\int_{0}^{+\infty}\int _{\vert z-w\vert < t} \bigl\vert (-t\sqrt {L})^{k+1}e^{-t\sqrt{L}}f(w) \bigr\vert ^{2} t^{-1-2n}\, dw\,dt \\ =&\frac{1}{2k2^{2k}}\int_{0}^{+\infty}\int _{\vert z-w\vert < t} \bigl\vert D_{t}^{k+1}f(w) \bigr\vert ^{2}\frac{dw\, dt}{t^{2n+1}}=\frac{1}{2k2^{2k}} \bigl(S_{L}^{k+1}f(z) \bigr)^{2}, \end{aligned}$$

we get $S_{L}^{k+1}f\in L^{1}(\mathbb{C}^{n})$. Then $f\in H_{L}^{1}(\mathbb{C}^{n})$ follows from (a).

This completes the proof of Theorem 2(b).

(c) By $S_{L}^{k}f(z)\leq (\frac{1}{2} )^{2\lambda n}g_{\lambda,k}^{*}f(z)$, we know $f\in H_{L}^{1}(\mathbb{C}^{n})$ when $g_{\lambda,k}^{*}f\in L^{1}(\mathbb{C}^{n})$ and $f\in L^{1}(\mathbb{C}^{n})$. In the following, we show there exists a constant $C>0$ such that for any atom $a(z)$ of $H^{1}_{L}(\mathbb{C}^{n})$, we have

$$\bigl\Vert g_{\lambda,k}^{*}a \bigr\Vert _{L^{1}}\leq C. $$

Without loss of generality, we may assume $a(z)$ is supported in $B(0,r)$, then

$$\begin{aligned} g_{\lambda,k}^{*}a(z)^{2} =&\int_{0}^{\infty} \int_{\mathbb{C}^{n}} \biggl(\frac{t}{t+|z-w|} \biggr)^{2\lambda n} \bigl\vert D_{t}^{k}a(w) \bigr\vert ^{2} \frac{dw \, dt}{t^{2n+1}} \\ =&\int_{0}^{\infty} \int_{|z-w|< t} \biggl(\frac{t}{t+|z-w|} \biggr)^{2\lambda n} \bigl\vert D_{t}^{k}a(w) \bigr\vert ^{2}\frac{dw\, dt}{t^{2n+1}} \\ &{}+\sum_{i=1}^{\infty}\int _{0}^{\infty} \int_{2^{i-1}t\leq|z-w|< 2^{i}t} \biggl( \frac{t}{t+|z-w|} \biggr)^{2\lambda n} \bigl\vert D_{t}^{k}a(w) \bigr\vert ^{2}\frac{dw\, dt}{t^{2n+1}} \\ \leq& C S^{1}_{L}a(z)^{2}+\sum _{i=1}^{\infty}2^{-2i\lambda n}S^{k}_{L,2^{i}}a(z)^{2}. \end{aligned}$$

Therefore,

$$\bigl\Vert g_{\lambda,k}^{*}a \bigr\Vert _{L^{1}}\leq C \bigl\Vert S_{L}^{1}a \bigr\Vert _{L^{1}}+ \sum _{i=1}^{\infty}2^{-i\lambda n} \bigl\Vert S^{k}_{L,2^{i}}a \bigr\Vert _{L^{1}}. $$

By part (a), we have $\|S_{L}^{k}a\|_{L^{1}}\leq C$. In the following, we will prove that

$$ \bigl\Vert S^{k}_{L,2^{i}}a\bigr\Vert _{L^{1}}\leq C 2^{3in}. $$

(13)

First, by Lemma 5, we can obtain

$$ \bigl\Vert S^{k}_{L,2^{i}}a \bigr\Vert _{L^{1}(B(0, 2^{i+2}r))}\leq \bigl\vert B\bigl(0, 2^{i+2}r\bigr) \bigr\vert ^{1/2} \bigl\Vert S^{k}_{L,2^{i}}a \bigr\Vert _{L^{2}} \leq C 2^{2i n}. $$

(14)

Let $z\notin B(0, 2^{i+2}r)$. We have

$$\begin{aligned} S^{k}_{L,2^{i}} a(z)^{2} \leq& \int _{0}^{\infty}\int_{|z-w|< 2^{i}t} \biggl( \int_{B(0,r)} \bigl\vert D^{k}_{t}(w-v)-D^{k}_{t}(w) \bigr\vert \bigl\vert a(v) \bigr\vert \, dv \biggr)^{2} \frac{dw \, dt}{t^{2n+1}} \\ \leq& \int^{\frac{|z|}{2^{i+1}}}_{0} \int_{|z-w|< 2^{i}t} ( \cdots )^{2} \frac{dw\, dt}{t^{2n+1}} + \int^{\infty}_{\frac{|z|}{2^{i+1}}} \int_{|z-w|< 2^{i}t} ( \cdots )^{2} \frac{dw\, dt}{t^{2n+1}} \\ =&I_{1}+I_{2}. \end{aligned}$$

For $z \notin B(0,2^{i+2}r)$, when $|z-w|< 2^{i}t \leq\frac{|z|}{2}$, we have $|w| \sim|z|$. By Lemma 4, we get

$$\begin{aligned} I_{1} \leq& C \int^{\frac{|z|}{2^{i+1}}}_{0} \int _{|z-w|< 2^{i} t} \biggl( \int_{B(0,r)} \frac{\sqrt{t}}{(t^{2}+A|w|^{2})^{(2n+1)/2}}|v| \bigl\vert a(v) \bigr\vert \, dv \biggr)^{2} \frac{dw\, dt}{t^{2n+1}} \\ \leq& C 2^{2i n} \int^{\frac{|z|}{2^{i+1}}}_{0} t^{-4n} \biggl( \frac{|z|}{t} \biggr)^{-(4n+3)} \biggl( \frac{r}{t} \biggr)^{2} \frac{dt}{t} \leq C 2^{2i n-i}\frac{ r^{2}}{|z|^{4n+2}}. \end{aligned}$$

By Lemma 4 again, we get

$$\begin{aligned} I_{2} \leq& C \int^{\infty}_{\frac{|z|}{2^{i+1}}} \int _{|z-w|< 2^{i}t} \biggl( \int_{B(0,r)} t^{-2n} \biggl( \frac{r}{t} \biggr) \bigl\vert a(v) \bigr\vert \, dv \biggr)^{2} \frac{dw\, dt}{t^{2n+1}} \\ \leq& C 2^{2i n}\int^{\infty}_{\frac{|z|}{2^{i+1}}} t^{-4n} \biggl( \frac{r}{t} \biggr)^{2} \frac{dt}{t} \leq C 2^{i(6n+2)} \frac{ r^{2}}{|z|^{2(2n+1)}}. \end{aligned}$$

Thus,

$$ \int_{|z| \geq2^{i+2}r} \bigl\vert S^{k}_{L,2^{i}} a(z)\bigr\vert \, dz \leq C 2^{3i n+i} \int_{|z| \geq 2^{i+2}r} \frac{r}{|z|^{2n+1}}\, dz\leq C 2^{3i n}. $$

(15)

Therefore, when $\lambda>3$, we prove $\|g_{\lambda,k}^{*}a \| _{L^{1}}\leq C$. Then Theorem 2(c) is proved. □