A matrix acting between Fock spaces

Abstract

If $\mathcal{H}_{\nu}=(\nu _{n,k})_{n,k\geq 0}$ is the matrix with entries $\nu _{n,k}=\int _{[0,\infty )}\frac{ t^{n+k}}{n!}\,d\nu (t)$, where ν is a nonnegative Borel measure on the interval $[0,\infty )$, the matrix $\mathcal{H}_{\nu}$ acts on the space of all entire functions $f(z) =\sum_{n=0}^{\infty} a_{n} z^{n}$ and induces formally the operator in the following way:

$$ \mathcal{H}_{\nu}(f) (z)=\sum_{n=0}^{\infty} \Biggl(\sum_{k=0}^{\infty}\nu _{n,k}a_{k}\Biggr)z^{n}. $$

In this paper, for $0< p\leq \infty$, we classify for which measures the operator $\mathcal{H}_{\nu}(f)$ is well defined on $F^{p}$ and also gets an integral representation, and among them we characterize those for which $\mathcal{H}_{\nu}$ is a bounded (resp., compact) operator between $F^{p}$ and $F^{\infty }$.

1 Introduction

Throughout this paper, we write $A \lesssim B$ for nonnegative quantities A and B if there exists a constant C (independent of A and B) such that $A \leq CB$. The symbol $A \simeq B$ means that both $A \lesssim B$ and $B \lesssim A$. C denotes a finite constant that may change value from one occurrence to the next.

Let $\mathbb{C}$ be the complex plane and $H(\mathbb{C})$ be the class of all entire functions. If $0< p <\infty$, then the Fock space $F^{p}$ is the set of all $f\in H(\mathbb{C})$ such that

$$ \Vert f \Vert ^{p}_{p }:=\frac{p }{2\pi} \int _{\mathbb{C}} \bigl\vert f(z) e^{- \frac{1}{2} \vert z \vert ^{2}} \bigr\vert ^{p}\,dA(z)< \infty , $$

where $z = x+iy$ and $dA(z) = dx\,dy$ is the Lebesgue area measure on $\mathbb{C}$. Set

$$ F^{\infty }= \Bigl\{ f\in H(\mathbb{C}): \Vert f \Vert _{\infty }= \sup_{z\in \mathbb{C}} \bigl\vert f(z) \bigr\vert e^{-\frac{1}{2} \vert z \vert ^{2}}< \infty \Bigr\} . $$

In particular, $F^{2}$ is a reproducing kernel Hilbert space. The function $K_{z}(w)=e^{ z \overline{w}}$ is the reproducing kernel for $F^{2}$ and

$$ k_{z} (w)=\frac{K_{z}(w)}{\sqrt{K(z,z)}}=e^{ z \overline{w}- \frac{1}{2}|z|^{2}} $$

is the normalized kernel. Moreover, each $k_{a}$ is a unit vector in $F^{p}$, where $0< p\leq \infty$. Let $f^{\infty }$ denote the space of entire functions such that

$$ \lim_{z\to \infty}f(z)e^{-\frac{1}{2}|z|^{2}}=0. $$

Interested readers can refer to [18] for the theory of Fock spaces.

Let $0< p, q<\infty$, and μ be a nonnegative Borel measure on $\mathbb{C}$. Recall that μ is a $(p, q)$-Fock Carleson measure if the identity operator i is bounded from $F^{p}$ to $L^{q}(e^{-\frac{q}{2}|\cdot |^{2}}\,d\mu )$, i.e., there exists some constant C such that, for all $f \in F^{p}$,

$$ \biggl( \int _{\mathbb{C} } \bigl\vert f(z) e^{-\frac{1}{2} \vert z \vert ^{2}} \bigr\vert ^{q}\,d\mu (z) \biggr)^{\frac{1}{q}} \leq C \Vert f \Vert _{p } . $$

When $p=q$, μ is exactly the Fock Carleson measure for $F^{p}$ (see [11, 18]). Also, μ is called a vanishing $(p, q)$-Fock Carleson measure if

$$ \lim_{j \rightarrow \infty} \int _{\mathbb{C} } \bigl\vert f_{j}(z) e^{- \frac{1}{2} \vert z \vert ^{2}} \bigr\vert ^{q}\,d\mu (z)=0 $$

whenever $\{f_{j} \}$ is a bounded sequence in $F^{p}$ that converges to 0 uniformly on compact subsets of $\mathbb{C}$ as $j \rightarrow \infty$.

The matrix $\mathcal{H}_{\nu}= (\nu _{n,k})_{n,k\geq 0}$ induces formally an operator (which will also be denoted by $\mathcal{H}_{\nu}$) on $H(\mathbb{C})$ in the following sense. For any $f(z) =\sum_{n=0}^{\infty} a_{n} z^{n}\in H(\mathbb{C})$, by multiplication of the matrix with the sequence of Taylor coefficients of the function, we can define

$$ \mathcal{H}_{\nu}(f) (z)=\sum _{n=0}^{\infty}\Biggl(\sum_{k=0}^{\infty} \nu _{n,k}a_{k}\Biggr)z^{n}. $$

(1.1)

If the right-hand side makes sense and defines a function in $H(\mathbb{C})$, then the matrix $\mathcal{H}_{\nu}$ induces formally an operator $\mathcal{H}_{\nu}$ on $H(\mathbb{C})$.

Let $\mathbb{D}$ be the open unit disc and $H(\mathbb{D})$ be all of the analytic functions on $\mathbb{D}$. If we replace $H(\mathbb{C})$ by $H(\mathbb{D})$ in the definition of operators above, there is a rich history of these operators (which will be denoted by $H_{\nu}$) on several natural $L^{p}$ spaces of analytic functions on $\mathbb{D}$, especially those on the Hardy space and the Bergman space. For example, the Hilbert operator induced by the Hilbert matrix $(\frac{1}{n+k+1})_{n,k\geq 0}$ has been studied in Hardy spaces [5] and Bergman spaces [4]; estimates on the norms have also been obtained. More generally, another approach to the study of $H_{\nu}$ on spaces of analytic functions is developed. Galanopoulos and Peláez introduced Hankel matrix $H_{\nu}= (\int _{[0,1)}t^{n+k}\,d\nu (t) )_{n,k\geq 0}$, where ν is a finite nonnegative Borel measure on $[0,1)$, and also investigated the boundedness and compactness of the operator induced by $H_{\nu}$ on the Hardy space $H^{1}$ and the Bergman space $A^{2}$ in [6]. Chatzifountas, Girela, and Peláez [2] later generalized the operator $H_{\nu}$ on Hardy spaces $H^{p}$. In [7, 8], Girela and Merchán also studied the operator $H_{\nu}$ acting on certain möbius invariant spaces on the unit disk such as the Bloch space, BMOA, the analytic Besov spaces, etc. Recently, Ye and Zhou considered a new operator, called derivative-Hilbert operator, induced by some Hankel matrix on analytic function spaces on $\mathbb{D}$ in [16, 17]. It turns out that the derivative-Hilbert operator and the Hilbert operator $H_{\nu}$ are closely related. For more results on the operator induced by some Hankel matrix, we refer to [1, 5, 14, 15].

From now on, ν denotes a nonnegative Borel measure on $[0,\infty )$. In the Fock space setting, Zhuo et al. [19] considered a special matrix $\mathcal{H}_{\nu}= (\nu _{n,k})_{n,k\geq 0}$ with entries

$$ \nu _{n,k}=\frac{1}{n!} \int _{[0,\infty )}t^{n+k}\,d\nu (t) $$

and characterized those nonnegative Borel measures ν such that the operator $\mathcal{H}_{\nu}$ was well defined on the Fock space $F^{p}$ ($0< p<\infty$). Furthermore, $\mathcal{H}_{\nu}$ was rewritten as

$$ \mathcal{H}_{\nu}(f) (z)= \int _{[0, \infty )}f(t)e^{ t z}\,d\nu (t),\quad z \in \mathbb{C}. $$

Under this integral representation of $\mathcal{H}_{\nu}$, the measures ν for which $\mathcal{H}_{\nu}$ is a bounded (resp., compact) operator from the Fock space $F^{p}$ into $F^{q}$ ($0< p,q<\infty$) were characterized. The main purpose of this paper is to extend the operator $\mathcal{H}_{\nu}$ to be well defined on the Fock space $F^{\infty}$. With $0 < p \leq \infty$, we are going to obtain some characterizations on those measures ν such that the operator $\mathcal{H}_{\nu}$ is bounded or compact from $F^{p}$ to $F^{\infty }$ and from $F^{\infty }$ to $F^{p}$, respectively.

2 Preliminaries

In this section, we state some lemmas for the proof of our main results. The following lemma can be found in [13].

Lemma 2.1

Suppose $\alpha > 0$. For every positive integer n,

$$ \int _{0}^{\infty}r^{np}e^{-\frac{\alpha p}{2}r^{2}}r \,dr \simeq \biggl( \frac{n!}{\alpha ^{n}} \biggr)^{\frac{p}{2}}n^{-\frac{p}{4}+\frac{1}{2}}. $$

The following formula is used many times throughout this paper. See [18, Corollary 2.5] for a proof.

Lemma 2.2

Suppose $\alpha > 0$ and β is real. Then

$$ \frac{\alpha}{\pi} \int _{\mathbb{C}} \bigl\vert e^{\beta z \bar{w}} \bigr\vert e^{-\alpha \vert z \vert ^{2}}\,dA(z)= e^{-\frac{\beta ^{2} \vert w \vert ^{2}}{4\alpha}} $$

for all $w\in \mathbb{C}$.

Lemma 2.3

Let $f(z)=\sum a_{n} z^{n}$ be an entire function. Then

$$ f\in F^{\infty} \quad \Rightarrow \quad \sup_{n\in \mathbb{N}} \vert a_{n} \vert ( n! )^{\frac{1}{2}}n^{- \frac{1}{4}}< \infty . $$

Proof

By Cauchy’s integral formula and Hölder’s inequality, it is easy to see that

$$ \vert a_{n} \vert r^{n} e^{-\frac{ r^{2}}{2}}\leq \sup _{ \vert z \vert =r} \bigl\vert f\bigl(re^{i\theta}\bigr) \bigr\vert e^{- \frac{ r^{2}}{2}} $$

for every $n \in \mathbb{N}$ and $r > 0$, and hence $\|a_{n}z ^{n}\|_{\infty}\lesssim \|f\|_{\infty}$. Note that, by elementary calculations and Stirling’s formula,

$$ \bigl\Vert z^{n} \bigr\Vert _{\infty}= \biggl( \frac{n}{e} \biggr)^{\frac{n}{2}}\simeq ( n! )^{\frac{1}{2}}n^{-\frac{1}{4}}. $$

Thus, for every n,

$$ \vert a_{n} \vert ( n! )^{\frac{1}{2}}n^{-\frac{1}{4}}\lesssim \Vert f \Vert _{ \infty}. $$

 □

Let $D(z, r)$ denote the Euclidean disk centered at z with radius r. The following basic estimate for integral averages of functions in Fock spaces can be found in [18, Lemma 2.32].

Lemma 2.4

For any $0< p<\infty$ and $r\in (0,\infty )$, there exists a positive constant $C =C(p,r)$ such that, for all $z\in \mathbb{C}$,

$$\begin{aligned} \bigl\vert f(z) e^{-\frac{1}{2} \vert z \vert ^{2}} \bigr\vert ^{p} &\leq C \int _{D(z,r)} \bigl\vert f(z)e^{- \frac{1}{2} \vert z \vert ^{2}} \bigr\vert ^{p}\,dA(z) \end{aligned}$$

for all entire functions f.

We also need the following result. See [18, Corollary 2.8] for a proof.

Lemma 2.5

Let $0< p\leq \infty$ and $f\in F^{p}$. Then

$$ \bigl\vert f(z) \bigr\vert \leq \Vert f \Vert _{p }e^{\frac{1}{2} \vert z \vert ^{2}} $$

for all $z\in \mathbb{C}$.

Given $r > 0$, a sequence $\{a_{k}\}$ in $\mathbb{C}$ is called an r-lattice if $\bigcup^{\infty}_{k=1}D(a_{k},r)$ covers $\mathbb{C}$ and the disks $\{D(a_{k},r/3)\}^{\infty}_{k=1}$ are pairwise disjoint. For any $\delta > 0$, there exists a positive integer m (depending only on r and δ) such that every point in $\mathbb{C}$ belongs to at most m of the sets $D(a_{k},\delta )$, see [18, page 118]. The following three lemmas characterize the $(p,q)$-Fock Carleson measure and vanishing $(p,q)$-Fock Carleson measure for $0< p,q<\infty$, which can be found in [9]. For a Borel measure μ on $\mathbb{C}$, define $\widehat{\mu}_{r}(z)=\frac{\mu (D(z,r))}{\pi r^{2}}$. Let $t>0$, the t-Berezin transform of μ is defined by

$$ \tilde{\mu}_{t}(z)= \int _{\mathbb{C}}e^{-\frac{t}{2}|z-w|^{2}}\,d\mu (w), $$

whenever these integrals converge, see [18, page 120].

Lemma 2.6

Let $0< p\leq q<\infty$, and let $\mu \geq 0$. Then the following statements are equivalent:

• (1) μ is a $(p,q)$-Fock Carleson measure;

• (2) $\widehat{\mu}_{r}(z)$ is bounded on $\mathbb{C}$ for some (or any) $r>0$;

• (3) $\tilde{\mu}_{t}(z)$ is bounded on $\mathbb{C}$ for some (or any) $t>0$. Furthermore,

  $$ \Vert i \Vert _{ F^{p} \to L^{q} (e^{-\frac{q}{2} \vert \cdot \vert ^{2}}\,d\mu )}\simeq \bigl\Vert \widehat{ \mu}_{r}^{\frac{1}{q}} \bigr\Vert _{L^{\infty}}\simeq \bigl\Vert \tilde{\mu}_{t}^{ \frac{1}{q}} \bigr\Vert _{L^{\infty}}. $$

Lemma 2.7

Let $0< p\leq q<\infty$, and let $\mu \geq 0$. Then the following statements are equivalent:

• (1) μ is a vanishing $(p,q)$-Fock Carleson measure;

• (2) $\widehat{\mu}_{r}(z)\to 0$ as $z \to \infty$ for some (or any) $r>0$;

• (3) $\tilde{\mu}_{t}(z)\to 0$ as $z \to \infty$ for some (or any) $t>0$.

Lemma 2.8

Let $0< q< p<\infty$ and let $\mu \geq 0$. Set $s=\frac{p}{q}$ and $s'$ to be the conjugate exponent of s. Then the following statements are equivalent:

• (1) μ is a $(p,q)$-Fock Carleson measure;

• (2) μ is a vanishing $(p,q)$-Fock Carleson measure;

• (3) $\widehat{\mu}_{r}(z)\in L^{s'}(dA)$ for some (or any) $r>0$.

In the light of above three lemmas, the notion of (vanishing) $(p,q)$-Fock Carleson measures does not depend on the particular value of p, q, but depends only on the ratio $s=\frac{p}{q}$ in the case $0 < q < p < \infty$. Let $\Lambda ^{s}$ be the class of all $(p,q)$-Fock Carleson measures such that $s=\frac{p}{q}$ and $\Lambda ^{s}_{0}$ be the class of all vanishing $(p,q)$-Fock Carleson measures such that $s=\frac{p}{q}$. When $0 < s\leq 1$ (equivalently, $p \leq q$), we simply write Λ and $\Lambda _{0}$ for $\Lambda ^{s}$ and $\Lambda ^{s}_{0}$ respectively. That is,

$$ \Lambda = \bigl\{ \mu \geq 0 : \widehat{\mu}_{r} \in L^{\infty}( \mathbb{C}) \text{ for some } r > 0\bigr\} $$

and

$$ \Lambda _{0}=\Bigl\{ \mu \geq 0 : \lim_{|z|\to \infty} \widehat{\mu}_{r}(z) = 0 \text{ for some } r > 0\Bigr\} . $$

Notice that $\Lambda ^{s}\subset \Lambda$ and $\Lambda ^{s}_{0}\subset \Lambda _{0}$ for all $s > 0$.

3 Conditions such that $\mathcal{H}_{\nu}$ is well defined on $F^{p}$

In this section, we first clarify for which measures the operator $\mathcal{H}_{\nu}$ is well defined on Fock spaces and also gets an integral representation. Throughout the paper, a nonnegative Borel measure μ on $[0,\infty )$ can be seen as a Borel measure on $\mathbb{C}$ by identifying it with the measure $\mu ^{\star}$ defined by

$$ \mu ^{\star}(A)=\mu \bigl(A\cap [0,\infty \bigr)) $$

for any Borel subset A of $\mathbb{C}$. In this way a positive Borel measure μ on $[0,\infty )$ is an a $(p,q)$-Fock Carleson measure if and only if there exists a positive constant C such that

$$ \int _{(a-r,a+r)}d\mu (t) \leq Cr^{2} $$

for any $a\in [0,\infty )$ and any fixed $0 \leq r<\infty$.

Theorem 3.1

Suppose $0 < p\leq \infty$ and ν is a nonnegative Borel measure on $[0, \infty )$. If $e^{\epsilon |\cdot |^{2}}\nu \in \Lambda$ for any fixed $\epsilon >\frac{1}{2}$, then the power series in (1.1) is a well defined entire function for every $f\in F^{p}$. Furthermore,

$$ \mathcal{H}_{\nu}(f) (z)= \int _{[0, \infty )}f(t)e^{ t z}\,d\nu (t),\quad z \in \mathbb{C}, f\in F^{p} . $$

(3.1)

Proof

The case when $0 < p <\infty$ has been proved in [19]. We now consider the case $p =\infty$. Suppose that $e^{\epsilon |\cdot |^{2}}\nu \in \Lambda$ with any fixed $\epsilon >\frac{1}{2}$, it might as well assume that $e^{\epsilon |\cdot |^{2}}\nu$ is $(1,1)$-Fock Carleson measure for any $0< s<\infty$ by Lemma 2.5. For any $0 < r < \infty$, fix $f(z)=\sum_{n=0}^{\infty} a_{n} z^{n} \in F^{\infty}$ and $|z| \leq r$. We deduce that, by Lemma 2.2,

$$\begin{aligned} \int _{[0, \infty )} \bigl\vert f(t)e^{ t z} \bigr\vert \,d \nu (t)&\leq \Vert f \Vert _{\infty} \int _{[0, \infty )} \bigl\vert e^{ t z} \bigr\vert e^{\frac{1}{2} \vert t \vert ^{2}}\,d\nu (t) \\ &= \Vert f \Vert _{\infty} \int _{[0, \infty )} \bigl\vert e^{ tz} \bigr\vert e^{(\frac{1}{2}- \epsilon ) \vert t \vert ^{2}}e^{\epsilon \vert t \vert ^{2}}\,d\nu (t) \\ &\lesssim \Vert f \Vert _{\infty } \int _{\mathbb{C}} \bigl\vert e^{ t z} \bigr\vert e^{(\frac{1}{2}- \epsilon ) \vert t \vert ^{2}}\,dA(t) \\ &\simeq \Vert f \Vert _{\infty }e^{ \frac{ \vert z \vert ^{2}}{4\epsilon -2}} \leq \Vert f \Vert _{\infty }e^{ \frac{ \vert r \vert ^{2}}{4\epsilon -2}}. \end{aligned}$$

So the integral in (3.1) uniformly converges for every domain $\{z:|z| \leq r\}$, the resulting function is analytic in $\mathbb{C}$ and, for every $z \in \mathbb{C}$,

$$\begin{aligned} \int _{[0, \infty )}f(t)e^{ t z}\,d\nu (t)&=\sum _{n=0}^{\infty} \int _{[0, \infty )}f(t)\frac{t^{n}}{n!}\,d\nu (t) z^{n}= \sum_{n=0}^{ \infty} \frac{1}{n!} \int _{[0, \infty )}\sum_{k=0}^{\infty}a_{k} t ^{n+k}\,d\nu (t) z^{n}. \end{aligned}$$

(3.2)

By Lemma 2.6, we may assume that $e^{\epsilon |\cdot |^{2}}\nu$ is $(2,2)$-Fock Carleson measure without loss of generality. This together with Hölder’s inequality and Lemma 2.1 shows that

$$\begin{aligned} \vert \nu _{n,k} \vert &\leq \frac{1}{n!} \int _{[0,\infty )} \vert t \vert ^{n+k}\,d\nu (t) \\ &\leq \frac{1}{n!} \biggl( \int _{[0,\infty )} \bigl\vert t^{k}e^{-\frac{1}{2} \vert t \vert ^{2}} \bigr\vert ^{2}e^{ \epsilon \vert t \vert ^{2}}\,d\nu (t) \biggr)^{1/2} \biggl( \int _{[0,\infty )} \bigl\vert t^{n}e^{( \frac{1}{2}-\epsilon ) \vert t \vert ^{2}} \bigr\vert ^{2}e^{\epsilon \vert t \vert ^{2}}\,d\nu (t) \biggr)^{1/2} \\ &\leq \frac{1}{n!} \biggl( \int _{\mathbb{C}} \bigl\vert t^{k}e^{-\frac{1}{2} \vert t \vert ^{2}} \bigr\vert ^{2}\,dA(t) \biggr)^{1/2} \biggl( \int _{\mathbb{C}} \bigl\vert t^{n}e^{(\frac{1}{2}- \epsilon ) \vert t \vert ^{2}} \bigr\vert ^{2}\,dA(t) \biggr)^{1/2} \\ &\lesssim \frac{1}{n!} ( k! )^{\frac{1}{2}} \biggl( \frac{n!}{(\epsilon -\frac{1}{2}) ^{n}} \biggr)^{\frac{1}{2}}. \end{aligned}$$

Combining this with Lemma 2.3, we conclude that for every n,

$$\begin{aligned} \Biggl\vert \sum_{k=0}^{\infty}\nu _{n,k}a_{k} \Biggr\vert \leq{}& \vert \nu _{n,0}a_{0} \vert + \vert \nu _{n,1}a_{1} \vert + \vert \nu _{n,2}a_{2} \vert +\sum _{k=3}^{\infty} \vert \nu _{n,k}a_{k} \vert \\ ={}& \vert \nu _{n,0}a_{0} \vert + \vert \nu _{n,1}a_{1} \vert + \vert \nu _{n,2}a_{2} \vert +\sum_{k=3}^{ \infty} \vert \nu _{n+3,k-3}a_{k} \vert \\ \lesssim{}& \bigl( \vert a_{0} \vert + \vert a_{1} \vert + \vert a_{2} \vert \bigr)\frac{1}{n!} \biggl( \frac{n!}{(\epsilon -\frac{1}{2}) ^{n}} \biggr)^{\frac{1}{2}} \\ &{}+ \frac{1}{n!} \biggl( \frac{(n+3)!}{(\epsilon -\frac{1}{2}) ^{n+3}} \biggr)^{ \frac{1}{2}}\sum_{k=3}^{\infty} \vert a_{k} \vert ( k! )^{\frac{1}{2}}k^{- \frac{3}{2}} \\ \lesssim{}& \bigl( \vert a_{0} \vert + \vert a_{1} \vert + \vert a_{2} \vert \bigr)\frac{1}{n!} \biggl( \frac{n!}{(\epsilon -\frac{1}{2}) ^{n}} \biggr)^{\frac{1}{2}} \\ &{}+ \frac{1}{n!} \biggl( \frac{(n+3)!}{(\epsilon -\frac{1}{2}) ^{n+3}} \biggr)^{ \frac{1}{2}} \sum_{k=3}^{\infty}k^{-\frac{3}{2}+\frac{1}{4}} \sup_{k \in \mathbb{N}} \vert a_{k} \vert ( k! )^{\frac{1}{2}}k^{-\frac{1}{4}} \\ \lesssim{}& \frac{1}{n!} \biggl( \frac{(n+3)!}{(\epsilon -\frac{1}{2}) ^{n+3}} \biggr)^{\frac{1}{2}}. \end{aligned}$$

Therefore, the series in (1.1) is well defined for all $z \in \mathbb{C}$, and

$$ \sum_{k=0}^{\infty}a_{k}\nu _{n,k}=\frac{1}{n!} \int _{[0,\infty )}f(t)t^{n}\,d\nu (t). $$

By (3.2), we obtain

$$ \mathcal{H}_{\nu}(f) (z)= \int _{[0, \infty )}f(t)e^{ t z}\,d\nu (t),\quad z \in \mathbb{C}. $$

This proves the desired result. □

Theorem 3.2

Suppose that ν is a nonnegative Borel measure on $[0, \infty )$ such that

$$ \int _{[0,\infty )} \int _{[0,t)}\bigl[1+(ts)^{2}\bigr] e^{2 ts}\,d \nu (s)\,d\nu (t)< \infty . $$

(3.3)

Then the power series in (1.1) is well defined for every $f\in F^{\infty}$ and (3.1) holds.

Proof

Assume that ν satisfies (3.3) and fix $f(z)=\sum_{n=0}^{\infty} a_{n} z^{n} \in F^{\infty}$. From the definition of $\nu _{n,k}$, it is elementary to check that

$$\begin{aligned} \sum_{n=0}^{\infty} \bigl\vert \nu _{n,0} a_{0}z^{n} \bigr\vert &= \vert a_{0} \vert \sum_{n=0}^{ \infty} \biggl(\frac{1}{n!} \biggr)^{\frac{1}{2}} \vert z \vert ^{n} \biggl(\frac{1}{n!} \biggr)^{\frac{1}{2}} \int _{[0,\infty )} \vert t \vert ^{n}\,d\nu (t) \\ &\leq \vert a_{0} \vert \Biggl(\sum _{n=0}^{\infty}\frac{1}{n!} \vert z \vert ^{2n} \Biggr)^{ \frac{1}{2}} \Biggl( \sum_{n=0}^{\infty} \frac{1}{n!} \int _{[0,\infty )} \int _{[0,\infty )} \vert t \vert ^{n} \vert s \vert ^{n}\,d\nu (s)\,d\nu (t) \Biggr)^{ \frac{1}{2}} \\ &= \vert a_{0} \vert \Biggl(\sum_{n=0}^{\infty} \frac{1}{n!} \vert z \vert ^{2n} \Biggr)^{ \frac{1}{2}} \Biggl( \sum_{n=0}^{\infty}\frac{2}{n!} \int _{[0,\infty )} \int _{[0,t)}(st)^{n}\,d\nu (s)\,d\nu (t) \Biggr)^{\frac{1}{2}} \\ &\lesssim \vert a_{0} \vert \Biggl(\sum _{n=0}^{\infty}\frac{1}{n!} \vert z \vert ^{2n} \Biggr)^{ \frac{1}{2}} \biggl( \int _{[0,\infty )} \int _{[0,t)}e^{st}\,d\nu (s)\,d\nu (t) \biggr)^{\frac{1}{2}}. \end{aligned}$$

The same arguments show that

$$\begin{aligned} \sum_{n=0}^{\infty} \bigl\vert \nu _{n,1}a_{1}z^{n} \bigr\vert \lesssim \vert a_{1} \vert \Biggl(\sum_{n=0}^{ \infty} \frac{1}{n!} \vert z \vert ^{2n} \Biggr)^{\frac{1}{2}} \biggl( \int _{[0, \infty )} \int _{[0,t)}(st)e^{st}\,d\nu (s)\,d\nu (t) \biggr)^{\frac{1}{2}}. \end{aligned}$$

On the other hand, for any $n \in \mathbb{N}$,

$$\begin{aligned} \Biggl\vert \sum_{k=2}^{\infty}\nu _{n,k}a_{k} \Biggr\vert &\leq \sum _{k=2}^{\infty} \vert a_{k} \vert ( k! )^{\frac{1}{2}}k^{-\frac{1}{4}}k^{\frac{1}{4}} \biggl( \frac{1}{k!} \biggr)^{\frac{1}{2}} \vert \nu _{n,k} \vert \\ &\leq \sup_{k\in \mathbb{N}} \vert a_{k} \vert ( k! )^{\frac{1}{2}}k^{- \frac{1}{4}}\sum_{k=2}^{\infty}k^{\frac{1}{4}} \biggl(\frac{1}{k!} \biggr)^{ \frac{1}{2}} \vert \nu _{n,k} \vert . \end{aligned}$$

However, by Hölder’s inequality, we have

$$\begin{aligned} \sum_{k=2}^{\infty} \biggl(\frac{1}{k!} \biggr)^{\frac{1}{2}}k^{ \frac{1}{4}} \vert \nu _{n,k} \vert ={}& \frac{1}{n!}\sum_{k=2}^{\infty} \biggl( \frac{1}{k!} \biggr)^{\frac{1}{2}}k^{\frac{1}{4}} \biggl( \int _{[0,\infty )}t^{n+k}\,d\nu (t) \biggr)^{2\cdot \frac{1}{2}} \\ \lesssim {}&\frac{1}{n!}\sum_{k=2}^{\infty}k^{\frac{1}{4}-1} \biggl( \frac{1}{(k-2)!} \int _{[0,\infty )} \int _{[0,t)}(st)^{k}(st)^{n}\,d\nu (s)\,d \nu (t) \biggr)^{\frac{1}{2}} \\ \lesssim {}&\frac{1}{n!} \Biggl(\sum_{k=2}^{\infty}k^{-\frac{3}{2}} \Biggr)^{\frac{1}{2}} \Biggl(\sum_{k=2}^{\infty} \frac{1}{(k-2)!} \int _{[0, \infty )} \int _{[0,t)}(st)^{k}(st)^{n}\,d\nu (s)\,d \nu (t) \Biggr)^{ \frac{1}{2}} \\ \lesssim {}&\frac{1}{n!} \biggl( \int _{[0,\infty )} \int _{[0,t)}(st)^{2}e^{ st}(st)^{n} \,d\nu (s)\,d\nu (t) \biggr)^{\frac{1}{2}}. \end{aligned}$$

It follows from Lemma 2.3 and (3.3) that the series $\sum_{k=0}^{\infty}a_{k}\nu _{n,k}$ is absolutely convergent, and

$$ \sum_{k=0}^{\infty}a_{k}\nu _{n,k}=\frac{1}{n!} \int _{[0,\infty )}f(t)t^{n}\,d\nu (t). $$

Furthermore,

$$\begin{aligned} \Biggl\vert \sum_{n=0}^{\infty}\Biggl(\sum _{k=0}^{\infty}\nu _{n,k}a_{k} \Biggr)z^{n} \Biggr\vert \lesssim{}& \sum_{n=0}^{\infty} \bigl\vert \nu _{n,0}a_{0}z^{n} \bigr\vert + \sum_{n=0}^{ \infty} \bigl\vert \nu _{n,1}a_{1}z^{n} \bigr\vert \\ &{}+\sum_{n=0}^{\infty} \biggl( \biggl( \frac{1}{n!} \biggr)^{2} \int _{[0, \infty )} \int _{[0,t)}e^{ st}(st)^{n+2}\,d\nu (s)\,d\nu (t) \biggr)^{ \frac{1}{2}} \vert z \vert ^{n} \\ \leq {}& \Biggl(\sum_{n=0}^{\infty} \frac{1}{n!} \vert z \vert ^{2n} \Biggr)^{ \frac{1}{2}} \biggl( \int _{[0,\infty )} \int _{[0,t)}e^{2st}\,d\nu (s)\,d\nu (t) \biggr)^{\frac{1}{2}} \\ &{}+ \Biggl(\sum_{n=0}^{\infty}\frac{1}{n!} \vert z \vert ^{2n} \Biggr)^{\frac{1}{2}} \Biggl(\sum _{n=0}^{\infty}\frac{1}{n!} \int _{[0,\infty )} \int _{[0,t)}e^{ st}(st)^{n+2}\,d\nu (s)\,d\nu (t) \Biggr)^{\frac{1}{2}} \\ = {}& \biggl( \int _{[0,\infty )} \int _{[0,t)}\bigl[1+(ts)^{2}\bigr]e^{2 st}\,d \nu (s)\,d\nu (t) \biggr)^{\frac{1}{2}} \Biggl(\sum_{n=0}^{\infty} \frac{1}{n!} \vert z \vert ^{2n} \Biggr)^{\frac{1}{2}} \end{aligned}$$

for each $z\in \mathbb{C}$. This shows that the power series in (1.1) represents an entire function and

$$ \mathcal{H}_{\nu}(f) (z)= \int _{[0, \infty )}f(t)e^{ t z}\,d\nu (t),\quad z \in \mathbb{C}. $$

The proof is completed. □

We also obtain the following necessary condition for the operator $\mathcal{H}_{\nu}$ to be well defined on Fock spaces.

Theorem 3.3

Let ν be a nonnegative Borel measure on $[0, \infty )$. If the integral

$$ \mathcal{H}_{\nu}(f) (z)= \int _{[0, \infty )}f(t)e^{ t z}\,d\nu (t),\quad z \in \mathbb{C}$$

converges absolutely for every $f\in F^{\infty}$, then $e^{\frac{1}{2}|\cdot |^{2}}\,d\nu$ is a $(\infty ,1)$-Fock Carleson measure.

Proof

For every $f\in F^{\infty}$, the integral $\int _{[0, \infty )}f(t)e^{ t z}\,d\nu (t)$ converges absolutely for $z=0$, we have

$$ \biggl\vert \int _{[0, \infty )}f(t)\,d\nu (t) \biggr\vert = \biggl\vert \int _{[0, \infty )}f(t)e^{- \frac{1}{2} \vert t \vert ^{2}}e^{\frac{1}{2} \vert t \vert ^{2}}\,d\nu (t) \biggr\vert \leq \int _{[0, \infty )} \bigl\vert f(t) \bigr\vert e^{-\frac{1}{2} \vert t \vert ^{2}}e^{\frac{1}{2} \vert t \vert ^{2}} \,d\nu (t)< \infty . $$

By the closed graph theorem, the identity mapping is bounded from $F^{\infty}$ into $L^{1}(e^{\frac{1}{2}|\cdot |^{2}}\,d\nu )$, which implies the desired estimate. □

Specializing to the space $f^{\infty}$, we use duality theorem for $f^{\infty}$ to obtain the following result. Recall that $(f^{\infty})^{*}= F^{1}$ under the pairing

$$ \langle f,g\rangle =\frac{1}{\pi} \int _{\mathbb{C}}f(w) \overline{g(w)}e^{- |z|^{2}}\,dA(w). $$

We refer the interested reader to [18, Theorem 2.26 and page 39] for details.

Theorem 3.4

Let dν be a nonnegative Borel measure on $[0,\infty )$. If for any $f\in f^{\infty}$ the integral

$$ \mathcal{H}_{\nu}(f) (z)= \int _{[0, \infty )}f(t)e^{ t z}\,d\nu (t),\quad z \in \mathbb{C}$$

converges absolutely, then

$$\begin{aligned} \int _{[0, \infty )} \int _{[0, t)}e^{ s t}\,d\nu (s)\,d\nu (t)< \infty . \end{aligned}$$

Proof

By assumption, taking $z = 0$, there is $C>0$ such that

$$ \biggl\vert \int _{[0, r)}f(t)\,d\nu (t) \biggr\vert \leq \int _{[0, r)} \bigl\vert f(t) \bigr\vert \,d\nu (t)< \int _{[0, \infty )} \bigl\vert f(t) \bigr\vert \,d\nu (t)< C $$

for all $r \in (0,\infty )$. More specifically, choosing $f=1$, we have $\int _{[0, \infty )}\,d\nu (t)<\infty$, which means dν is a finite Borel measure. On the other hand, using Hölder’s inequality, we obtain that

$$\begin{aligned} \int _{[0, r)} \int _{\mathbb{C}} \bigl\vert f(z)e^{ t \bar{z}} \bigr\vert e^{- \vert z \vert ^{2}}\,dA(z)\,d\nu (t)\leq \Vert f \Vert _{\infty} \int _{[0, r)}e^{\frac{1}{2} \vert t \vert ^{2}}\,d\nu (t)< \infty . \end{aligned}$$

Since the set of finite linear combinations of kernel functions is dense in $f^{\infty}$ [18, Lemma 2.11] and all kernel functions belong to $F^{2}$, the reproducing property and Fubini’s theorem imply that for any $f\in f^{\infty}$,

$$\begin{aligned} \int _{[0, r)}f(t)\,d\nu (t)&= \int _{[0, r)} \int _{\mathbb{C}}f(z)e^{ t \bar{z}}e^{- |z|^{2}}\,dA(z)\,d\nu (t) \\ &= \int _{\mathbb{C}}f(z) \overline{ \int _{[0, r)}e^{ t z}\,d\nu (t)}e^{- |z|^{2}}\,dA(z) = \langle f,g_{r}\rangle , \end{aligned}$$

(3.4)

where $g_{r}(z)=\int _{[0, r)}e^{ t z}\,d\nu (t)$. By the duality relation $(f^{\infty})^{*}= F^{1}$ and the uniform boundedness principle, we get $g_{r}\in F^{1}$ and $\sup_{r}\|g_{r}\|_{1}< C$. Since $F^{1}\subset f^{\infty}$, we replace f by $g_{r}$ in (3.4) and obtain that

$$\begin{aligned} \int _{\mathbb{C}} \bigl\vert g_{r}(z) \bigr\vert ^{2}e^{- \vert z \vert ^{2}}\,dA(z) &\geq \biggl\vert \int _{ \mathbb{C}}g_{r}(z) \overline{g_{r}(z)}e^{- \vert z \vert ^{2}} \,dA(z) \biggr\vert \\ &= \biggl\vert \int _{[0, r)}g_{r}(t)\,d\nu (t) \biggr\vert = \int _{[0, r)} \int _{[0, r)}e^{ s t}\,d\nu (s)\,d\nu (t). \end{aligned}$$

However, Lemma 2.5 implies that

$$\begin{aligned} \int _{\mathbb{C}} \bigl\vert g_{r}(z) \bigr\vert ^{2}e^{- \vert z \vert ^{2}}\,dA(z) &\lesssim \Vert g_{r} \Vert _{1 } \int _{\mathbb{C}} \bigl\vert g_{r}(z)e^{-\frac{1}{2} \vert z \vert ^{2}} \bigr\vert \,dA(z) \lesssim \Vert g_{r} \Vert ^{2}_{1 }< C. \end{aligned}$$

Combining this with the previous inequality and letting $r\to \infty$, we have

$$\begin{aligned} \int _{[0, \infty )} \int _{[0, \infty )}e^{ s t}\,d\nu (s)\,d\nu (t)< C. \end{aligned}$$

This proves the desired result. □

4 The operator $\mathcal{H}_{\nu}$ acting between Fock spaces

In this section, for $0 < p\leq \infty$, we are going to characterize those measures ν for which $\mathcal{H}_{\nu}$ is a bounded (resp., compact) operator from $F^{p}$ to $F^{\infty }$ or from $F^{\infty }$ to $F^{p}$ for some $0< p\leq \infty$. Now, we state the main results as follows.

Theorem 4.1

Suppose $0 < p\leq \infty$. Let ν be a nonnegative Borel measure on $[0, \infty )$ such that $e^{\epsilon |\cdot |^{2}}\nu \in \Lambda$ with any fixed $\epsilon >\frac{1}{2}$. Then $\mathcal{H}_{\nu}$ is bounded from $F^{p}$ into $F^{\infty}$ if and only if $e^{ |\cdot |^{2}}\nu \in \Lambda$.

Proof

By Theorem 3.1, $\mathcal{H}_{\nu}$ has an integral representation (3.1). Suppose that $\mathcal{H}_{\nu}$ is a bounded operator from $F^{p}$ into $F^{\infty }$. Fixed $r>0$, Lemmas 2.4 and 2.2 show that there is $C > 0$ such that

$$\begin{aligned} C&> \bigl\Vert \mathcal{H}_{\nu}( k_{a}) \bigr\Vert _{\infty} \gtrsim \bigl\vert \mathcal{H}_{\nu }(k_{a} ) (a)e^{-\frac{1}{2} \vert a \vert ^{2}} \bigr\vert = \int _{[0, \infty )}e^{ t \overline{a}+ta- \vert a \vert ^{2}}\,d\nu (t) \\ &= \int _{[0, \infty )} e^{- \vert t -a \vert ^{2}}e^{ \vert t \vert ^{2}}\,d\nu (t) \geq \int _{(a-r,a+r)}e^{- \vert t -a \vert ^{2}}e^{ \vert t \vert ^{2}}\,d\nu (t)\gtrsim \int _{(a-r,a+r)}e^{ \vert t \vert ^{2}}\,d\nu (t) \end{aligned}$$

(4.1)

for any $a\in [0, \infty )$. This proves that $e^{ |\cdot |^{2}}\nu \in \Lambda$ by Lemma 2.6.

Conversely, suppose $e^{ |\cdot |^{2}}\nu \in \Lambda$. Given $f\in F^{p}$, by Lemmas 2.5, 2.2, and 2.6, we have

$$\begin{aligned} \bigl\vert \mathcal{H}_{\nu}(f) (z)e^{-\frac{1}{2} \vert z \vert ^{2}} \bigr\vert & \leq \int _{[0, \infty )} \bigl\vert f(t)e^{ t z}e^{-\frac{1}{2} \vert z \vert ^{2}} \bigr\vert \,d\nu (t) \\ &\leq \Vert f \Vert _{p} \int _{[0, \infty )} \bigl\vert e^{ t z} \bigr\vert e^{-\frac{1}{2} \vert z \vert ^{2}}e^{- \frac{1}{2} \vert t \vert ^{2}} e^{ \vert t \vert ^{2}}\,d\nu (t) \\ &\leq \Vert i \Vert _{ F^{1} \to L^{1} (e^{\frac{1}{2} \vert \cdot \vert ^{2}}\,d\nu )} \Vert f \Vert _{p} \int _{\mathbb{C}} \bigl\vert e^{ w z} \bigr\vert e^{-\frac{1}{2} \vert z \vert ^{2}}e^{- \frac{1}{2} \vert w \vert ^{2}}\,dA(w) \\ &\lesssim \Vert i \Vert _{ F^{1} \to L^{1} (e^{\frac{1}{2} \vert \cdot \vert ^{2}}\,d\nu )} \Vert f \Vert _{p} \end{aligned}$$

for any $z\in \mathbb{C}$. Therefore, $\mathcal{H}_{\nu}$ is bounded from $F^{p}$ into $F^{\infty }$. This completes the proof of the theorem. □

Lemma 4.1

Suppose that $0 < p,q \leq \infty$ and $\mathcal{H}_{\nu}$ is bounded from $F^{p}$ into $F^{q}$. Then $\mathcal{H}_{\nu}$ is a compact operator if and only if for any bounded sequence $\{f_{n}\}$ in $F^{p}$, which converges uniformly to 0 on every compact subset of $\mathbb{C}$, we have $\mathcal{H}_{\nu}(f_{n})\to 0$ as $n\to 0$ in $F^{q}$.

The proof of Lemma 4.1 is similar to that of Proposition 3.11 in [3]. We omit the details.

Theorem 4.2

Suppose $0 < p\leq \infty$. Let ν be a nonnegative Borel measure on $[0, \infty )$ such that $e^{\epsilon |\cdot |^{2}}\nu \in \Lambda$ with any fixed $\epsilon >\frac{1}{2}$. Then $\mathcal{H}_{\nu}$ is a compact operator from $F^{p}$ into $F^{\infty }$ if and only if $e^{ |\cdot |^{2}}\nu \in \Lambda _{0}$.

Proof

Assume that $\mathcal{H}_{\nu}$ is a compact operator from $F^{p}$ into $F^{\infty }$. It is easy to see that the function $k_{a}(z)\to 0$ uniformly on compact sets as $a \to \infty$. Using Lemma 4.1, we obtain that $\{\mathcal{H}_{\nu}(k_{a})\}$ converges to 0 in $F^{\infty }$ when $a \to \infty$. Hence, by (4.1) we deduce that

$$ \int _{|t-a|< r}e^{ |t|^{2}}\,d\nu (t)\to 0,\quad a\to \infty . $$

This proves $e^{ |\cdot |^{2}}\nu \in \Lambda _{0}$.

Conversely, assume that $\{f_{n}\}$ is a bounded sequence in $F^{p}$, and $f_{n}$ uniformly converges to 0 on compact subsets of $\mathbb{C}$ as $n\to \infty$. It follows from Lemma 2.5 that

$$\begin{aligned} \bigl\vert \mathcal{H}_{\nu}(f_{n}) (z)e^{-\frac{1}{2} \vert z \vert ^{2}} \bigr\vert &\leq \int _{[0, \infty )} \bigl\vert f_{n}(t)e^{ t z}e^{-\frac{1}{2} \vert z \vert ^{2}} \bigr\vert \,d\nu (t) \\ &\leq \Vert f_{n} \Vert _{\infty} \int _{[0, \infty )}e^{-\frac{1}{2} \vert t-z \vert ^{2}} e^{ \vert t \vert ^{2}}\,d\nu (t) \\ &\leq \Vert f_{n} \Vert _{p}\widetilde{ \bigl(e^{ \vert \cdot \vert ^{2}}\nu \bigr)}_{1}(z). \end{aligned}$$

Since $e^{ |\cdot |^{2}}\nu \in \Lambda _{0}$ by Lemma 2.7, $\widetilde{(e^{ |\cdot |^{2}}\nu )}_{1}(z)\to 0$ as $z \to \infty$. So, for any $\varepsilon > 0$, there is some $R>0$ such that when $|z|\geq R$

$$\begin{aligned} \bigl\vert \mathcal{H}_{\nu}(f_{n}) (z)e^{-\frac{1}{2} \vert z \vert ^{2}} \bigr\vert \leq \Vert f_{n} \Vert _{p} \widetilde{ \bigl(e^{ \vert \cdot \vert ^{2}}\nu \bigr)}_{1}(z)\leq \varepsilon \Vert f_{n} \Vert _{p}. \end{aligned}$$

When $|z|< R$, for the above ε, there is some $R_{1}>0$ such that

$$\begin{aligned} \int _{ \vert w \vert \geq R_{1}} \bigl\vert e^{ w z} \bigr\vert e^{-\frac{1}{2} \vert z \vert ^{2}}e^{- \frac{1}{2} \vert w \vert ^{2}}\,dA(w)&= \int _{ \vert w \vert \geq R_{1}}e^{-\frac{1}{2} \vert w-z \vert ^{2}}\,dA(w) \\ &\leq e^{-\frac{1}{4} (R_{1}-R)^{2}} \int _{\mathbb{C}}e^{-\frac{1}{4} \vert w-z \vert ^{2}}\,dA(w) \\ &= e^{-\frac{1}{4} (R_{1}-R)^{2}} \int _{\mathbb{C}}e^{-\frac{1}{4} \vert w \vert ^{2}}\,dA(w)\leq \varepsilon . \end{aligned}$$

Hence, by Lemma 2.5,

$$\begin{aligned} &\bigl\vert \mathcal{H}_{\nu}(f_{n}) (z)e^{-\frac{1}{2} \vert z \vert ^{2}} \bigr\vert \\ &\quad \leq e^{-\frac{1}{2} \vert z \vert ^{2}} \int _{[0, \infty )} \bigl\vert f_{n}(t)e^{ t z} \bigr\vert e^{- \vert t \vert ^{2}} e^{ \vert t \vert ^{2}}\,d\nu (t) \\ &\quad \lesssim e^{-\frac{1}{2} \vert z \vert ^{2}} \int _{\mathbb{C}} \bigl\vert f_{n}(w)e^{ w z} \bigr\vert e^{- \vert w \vert ^{2}}\,dA(w) \\ &\quad \leq e^{-\frac{1}{2} \vert z \vert ^{2}} \int _{ \vert w \vert < R_{1}} \bigl\vert f_{n}(w)e^{ w z} \bigr\vert e^{- \vert w \vert ^{2}}\,dA(w)+e^{-\frac{1}{2} \vert z \vert ^{2}} \int _{ \vert w \vert \geq R_{1}} \bigl\vert f_{n}(w)e^{ w z} \bigr\vert e^{- \vert w \vert ^{2}}\,dA(w) \\ &\quad \leq \int _{ \vert w \vert < R_{1}} \bigl\vert f_{n}(w) \bigr\vert e^{-\frac{1}{2} \vert w-z \vert ^{2}}e^{- \vert w \vert ^{2}}\,dA(w)+e^{-\frac{1}{2} \vert z \vert ^{2}} \int _{ \vert w \vert \geq R_{1}} \Vert f_{n} \Vert _{p }e^{ \frac{1}{2} \vert w \vert ^{2}} \bigl\vert e^{ w z} \bigr\vert e^{- \vert w \vert ^{2}}\,dA(w) \\ &\quad \leq \int _{ \vert w \vert < R_{1}} \bigl\vert f_{n}(w) \bigr\vert e^{-\frac{1}{2} \vert w \vert ^{2}}\,dA(w)+ \Vert f_{n} \Vert _{p} \varepsilon \leq \bigl(1+ \Vert f_{n} \Vert _{p}\bigr) \varepsilon , \end{aligned}$$

where the last inequality comes from the assumption that $\{f_{n}\}$ converges to 0 uniformly on compact subsets of $\mathbb{C}$ as $n\to \infty$. Therefore, by the arbitrariness of ε and Lemma 4.1, we see that $\mathcal{H}_{\nu}:F^{p} \to F^{\infty }$ is compact. □

Define the Rademacher functions $\psi _{n}(t)$ on $[0,1]$ by

$$\begin{aligned}& \psi _{0}(t)= \textstyle\begin{cases} 1, & 0\leq t-[t]< \frac{1}{2}, \\ -1, & \frac{1}{2}\leq t-[t]< 1; \end{cases}\displaystyle \\& \psi _{n}(t)=\psi _{0}\bigl(2^{n}t\bigr),\quad n>0. \end{aligned}$$

Then Khinchine’s inequality is the following, which can be found in [12].

Khinchine’s inequality

For $0< p<\infty$ there exist constants $0< a_{p}\leq B_{p}<\infty$ such that, for all natural numbers m and all complex numbers $c_{1}, c_{2},\ldots,c_{m}$, we have

$$ a_{p}\Biggl(\sum^{m}_{n=1} \vert c_{n} \vert ^{2}\Biggr)^{p/2}\leq \int _{0}^{1} \Biggl\vert \sum ^{m}_{n=1}c_{n} \psi _{n}(t) \Biggr\vert ^{p}\,dt\leq B_{p}\Biggl(\sum ^{m}_{n=1} \vert c_{n} \vert ^{2}\Biggr)^{p/2}. $$

The next lemma is a partial result about atomic decomposition for Fock spaces, which can be found as Theorem 2.34 in [18].

Lemma 4.2

Let $0< p\leq \infty$. For $\lambda = \{\lambda _{j}\}^{\infty}_{j=1}\in l^{p}$, set

$$ S(\lambda ) (z) =\sum_{j=1}^{\infty}\lambda _{j} k_{a_{j}}(z),\quad z \in \mathbb{C}, $$

then S is a bounded operator from $l^{p}$ to $F^{p}$.

Theorem 4.3

Suppose $0 < p <\infty$. Let ν be a nonnegative Borel measure on $[0, \infty )$ such that $e^{\epsilon |\cdot |^{2}}\nu \in \Lambda$ with any fixed $\epsilon >\frac{1}{2}$. Then the following statements are equivalent:

• (i) $\mathcal{H}_{\nu}$ is a bounded operator from $F^{\infty }$ into $F^{p}$;

• (ii) $\mathcal{H}_{\nu}$ is a compact operator from $F^{\infty }$ into $F^{p}$;

• (iii) $(\widetilde{e^{ |\cdot |^{2}}\nu})_{t}(z)\in L^{p}(dA)$ for some (or any) $t>0$;

• (iv) $(\widehat{e^{ |\cdot |^{2}}\nu})_{r}(z)\in L^{p}(dA)$ for some (or any) $r>0$;

• (v) The sequence $\{(\widehat{e^{ |\cdot |^{2}}\nu})_{r}(a_{k})\}_{k=1}^{\infty}\in l^{p}$ for some (or any) r-lattice $\{a_{k}\}_{k=1}^{\infty}$.

Proof

By [10, Lemma 2.3], we get the equivalence of (iii), (iv), and (v). The implication (ii) ⇒ (i) is trivial.

(i) ⇒ (v). Given any bounded sequence $\{\lambda _{j}\}^{ \infty}_{j=1}$ and r-lattice $\{a_{j}\}^{\infty}_{j=1}$, Lemma 4.2 shows that $f(z) =\sum_{j=1}^{\infty}\lambda _{j} k_{a_{j}}(z) \in F^{\infty }$ with $\|f\|_{\infty }\lesssim \|\{\lambda _{j}\}_{j}\|_{l^{\infty}}$. Note that $\mathcal{H}_{\nu}: F^{\infty }\to F^{p}$ is bounded. By Khinchine’s inequality and Fubini’s theorem, we have

$$\begin{aligned} & \int _{\mathbb{C}} \Biggl(\sum_{j=1}^{\infty} \bigl\vert \lambda _{j}\mathcal{H}_{ \nu}( k_{a_{j}}) (z) \bigr\vert ^{2} \Biggr)^{\frac{p}{2}}e^{-\frac{p}{2} \vert z \vert ^{2}} \,dA(z) \\ &\quad \lesssim \int _{\mathbb{C}} \int _{0}^{1} \Biggl\vert \sum _{j=1}^{\infty} \psi _{j}(t)\lambda _{j}\mathcal{H}_{\nu}( k_{a_{j}}) (z) \Biggr\vert ^{p}\,dt e^{- \frac{p}{2} \vert z \vert ^{2}}\,dA(z) \\ &\quad = \int _{0}^{1} \int _{\mathbb{C}} \Biggl\vert \mathcal{H}_{\nu}\Biggl( \sum_{j=1}^{ \infty}\psi _{j}(t) \lambda _{j} k_{a_{j}}\Biggr) (z) \Biggr\vert ^{p} e^{- \frac{p}{2} \vert z \vert ^{2}}\,dA(z)\,dt \\ &\quad \lesssim \int _{0}^{1} \Vert \mathcal{H}_{\nu} \Vert ^{p}_{F^{\infty }\to F^{p} } \Biggl\Vert \sum _{j=1}^{\infty}\psi _{j}(t)\lambda _{j} k_{a_{j}} \Biggr\Vert ^{p}_{ \infty} \,dt \\ &\quad \lesssim \Vert \mathcal{H}_{\nu} \Vert ^{p}_{F^{\infty }\to F^{p} } \bigl\Vert \{ \lambda _{j}\}_{j} \bigr\Vert _{l^{\infty}}^{p}, \end{aligned}$$

where $\psi _{j}(t)$ is the jth Rademacher function on $[0,1]$. In addition, it follows from Lemma 2.4 and (4.1) that

$$\begin{aligned} & \int _{\mathbb{C}} \Biggl(\sum_{j=1}^{\infty} \bigl\vert \lambda _{j}\mathcal{H}_{ \nu}( k_{a_{j}}) (z) \bigr\vert ^{2} \Biggr)^{\frac{p}{2}}e^{-\frac{p}{2} \vert z \vert ^{2}} \,dA(z) \\ &\quad \gtrsim \sum_{k=1}^{\infty} \int _{D(a_{k},r)} \Biggl(\sum_{j=1}^{ \infty} \bigl\vert \lambda _{j}\mathcal{H}_{\nu}( k_{a_{j}}) (z) \bigr\vert ^{2} \Biggr)^{ \frac{p}{2}}e^{-\frac{p }{2} \vert z \vert ^{2}} \,dA(z) \\ &\quad \gtrsim\sum_{k=1}^{\infty} \int _{D(a_{k},r)} \bigl\vert \lambda _{k} \mathcal{H}_{\nu}( k_{a_{k}}) (z) \bigr\vert ^{p}e^{-\frac{p }{2} \vert z \vert ^{2}}\,dA(z) \\ &\quad \gtrsim\sum_{k=1}^{\infty} \vert \lambda _{k} \vert ^{p} \bigl\vert \mathcal{H}_{\nu}( k_{a_{k}}) (a_{k} ) \bigr\vert ^{p}e^{-\frac{p }{2} \vert a_{k} \vert ^{2}} \gtrsim \sum_{k=1}^{\infty} \vert \lambda _{k} \vert ^{p}\widehat{\bigl(e^{ \vert \cdot \vert ^{2}}\nu \bigr)}_{r}(a_{k})^{p}. \end{aligned}$$

Setting $\beta _{k}=|\lambda _{k}|^{p}$, then $\{\beta _{k}\}^{\infty}_{k=1}\in l^{\infty}$. Consequently,

$$\begin{aligned} \sum_{k=1}^{\infty}\beta _{k} \widehat{\bigl(e^{ \vert \cdot \vert ^{2}}\nu \bigr)}_{r}(a_{k})^{p} \lesssim \Vert \mathcal{H}_{\nu} \Vert ^{p}_{F^{\infty }\to F^{p} } \bigl\Vert \{\beta _{j} \}_{j} \bigr\Vert _{l^{\infty}}. \end{aligned}$$

The duality argument shows that $\{\widehat{(e^{ |\cdot |^{2}}\nu )}_{r}(a_{k})^{p}\}^{\infty}_{k=1} \in l^{1}$, which means $\{\widehat{(e^{ |\cdot |^{2}}\nu )}_{r}(a_{k})\}^{\infty}_{k=1} \in l^{p}$.

(iii) ⇒ (ii). Assume that $\{f_{n}\}$ is a bounded sequence in $F^{\infty}$, and $f_{n}$ uniformly converges to 0 on compact subsets of $\mathbb{C}$ as $n\to \infty$. It follows from Lemma 2.5 that

$$\begin{aligned} \int _{\mathbb{C}} \bigl\vert \mathcal{H}_{\nu}(f_{n}) (z)e^{-\frac{1}{2} \vert z \vert ^{2}} \bigr\vert ^{p}\,dA(z) &\leq \int _{\mathbb{C}} \Vert f_{n} \Vert ^{p}_{\infty} \biggl\vert \int _{[0, \infty )}e^{-\frac{1}{2} \vert t-z \vert ^{2}} e^{ \vert t \vert ^{2}}\,d\nu (t) \biggr\vert ^{p}\,dA(z) \\ &\leq \Vert f_{n} \Vert ^{p}_{\infty } \int _{\mathbb{C}} \bigl\vert \widetilde{\bigl(e^{ \vert \cdot \vert ^{2}}\nu \bigr)}_{1}(z) \bigr\vert ^{p}\,dA(z). \end{aligned}$$

Since $(\widetilde{e^{ |\cdot |^{2}}\nu})_{1}(z)\in L^{p}(dA)$, for any $\varepsilon > 0$, there is some $R>0$ such that

$$\begin{aligned} \int _{ \vert z \vert \geq R} \bigl\vert \mathcal{H}_{\nu}(f_{n}) (z)e^{-\frac{1}{2} \vert z \vert ^{2}} \bigr\vert ^{p}\,dA(z) \leq \Vert f_{n} \Vert ^{p}_{\infty } \int _{ \vert z \vert \geq R} \bigl\vert \widetilde{\bigl(e^{ \vert \cdot \vert ^{2}}\nu \bigr)}_{1}(z) \bigr\vert ^{p}\,dA(z)\leq \varepsilon \Vert f_{n} \Vert ^{p}_{\infty}. \end{aligned}$$

Note that $(\widetilde{e^{ |\cdot |^{2}}\nu})_{1}(z)\in L^{p}(dA)$ means $(\widehat{e^{ |\cdot |^{2}}\nu})_{r}(z)\in L^{p}(dA)$, hence $(\widehat{e^{ |\cdot |^{2}}\nu})_{r}(z)\in L^{\infty}(\mathbb{C})$. Lemma 2.6 yields that

$$\begin{aligned} \int _{[0,\infty )} \bigl\vert e^{\frac{1}{2}t\bar{z}} \bigr\vert e^{-\frac{1}{4} \vert t \vert ^{2}} e^{ \vert t \vert ^{2}}\,d\nu (t) &\leq \int _{\mathbb{C}} \bigl\vert e^{\frac{1}{2}w \bar{z}} \bigr\vert e^{-\frac{1}{4} \vert w \vert ^{2}}\,dA(w). \end{aligned}$$

Hence, for the above ε and R, there is some $R_{1}>0$ such that

$$\begin{aligned} &\int _{ \vert z \vert < R} \biggl\vert e^{-\frac{1}{2} \vert z \vert ^{2}} \int _{[ R_{1},\infty )} \bigr\vert e^{t z} \bigl\vert e^{-\frac{1}{2} \vert t \vert ^{2}} e^{ \vert t \vert ^{2}}\,d\nu (t) \biggr\vert ^{p}\,dA(z) \\ &\quad = \int _{ \vert z \vert < R} \biggl\vert \int _{[ R_{1},\infty )}e^{-\frac{1}{2} \vert t-z \vert ^{2}} e^{ \vert t \vert ^{2}}\,d\nu (t) \biggr\vert ^{p}\,dA(z) \\ &\quad = \int _{ \vert z \vert < R} e^{-\frac{1}{4} \vert R_{1}-R \vert ^{2}} \biggl\vert \int _{[ R_{1}, \infty )}e^{-\frac{1}{4} \vert t-z \vert ^{2}} e^{ \vert t \vert ^{2}}\,d\nu (t) \biggr\vert ^{p}\,dA(z) \\ &\quad \leq \int _{ \vert z \vert < R} e^{-\frac{1}{4} \vert R_{1}-R \vert ^{2}} \biggl\vert \int _{[0,\infty )}e^{- \frac{1}{4} \vert t-z \vert ^{2}} e^{ \vert t \vert ^{2}}\,d\nu (t) \biggr\vert ^{p}\,dA(z) \\ &\quad \leq \int _{ \vert z \vert < R}e^{-\frac{1}{4} \vert R_{1}-R \vert ^{2}} \biggl\vert \int _{\mathbb{C}}e^{- \frac{1}{4} \vert w-z \vert ^{2}}\,dA(w) \biggr\vert ^{p}\,dA(z) \\ &\quad \leq C \int _{ \vert z \vert < R}e^{-\frac{1}{4} \vert R_{1}-R \vert ^{2}}\,dA(z)\leq \varepsilon . \end{aligned}$$

Together with the fact that $f_{n}$ uniformly converges to 0 on compact subsets of $\mathbb{C}$ as $n\to \infty$, we obtain, by Lemma 2.5,

$$\begin{aligned} &\int _{ \vert z \vert < R} \bigl\vert \mathcal{H}_{\nu}(f_{n}) (z)e^{-\frac{1}{2} \vert z \vert ^{2}} \bigr\vert ^{p}\,dA(z) \\ &\quad \leq \int _{ \vert z \vert < R} \biggl\vert e^{-\frac{1}{2} \vert z \vert ^{2}} \int _{[0, \infty )} \bigr\vert f_{n}(t)e^{ t z} \bigl\vert e^{- \vert t \vert ^{2}} e^{ \vert t \vert ^{2}}\,d\nu (t) \biggr\vert ^{p}\,dA(z) \\ &\quad \leq \int _{ \vert z \vert < R} \biggl\vert e^{-\frac{1}{2} \vert z \vert ^{2}} \int _{[0, R_{1})} \bigr\vert f_{n}(t)e^{ t z} \bigl\vert e^{- \vert t \vert ^{2}} e^{ \vert t \vert ^{2}}\,d\nu (t) \biggr\vert ^{p}\,dA(z) \\ &\qquad {} + \int _{ \vert z \vert < R} \biggl\vert e^{-\frac{1}{2} \vert z \vert ^{2}} \int _{[ R_{1}, \infty )} \bigr\vert f_{n}(t)e^{ t z} \bigl\vert e^{- \vert t \vert ^{2}} e^{ \vert t \vert ^{2}}\,d\nu (t) \biggr\vert ^{p}\,dA(z) \\ &\quad \leq \int _{ \vert z \vert < R} \biggl\vert e^{-\frac{1}{2} \vert z \vert ^{2}} \int _{[0, R_{1})} \bigr\vert f_{n}(t)e^{ t z} \bigl\vert e^{- \vert t \vert ^{2}} e^{ \vert t \vert ^{2}}\,d\nu (t) \biggr\vert ^{p}\,dA(z) \\ &\qquad {} + \Vert f_{n} \Vert ^{p}_{\infty } \int _{ \vert z \vert < R} \biggl\vert e^{-\frac{1}{2} \vert z \vert ^{2}} \int _{[ R_{1},\infty )} \bigr\vert e^{t z} \bigl\vert e^{-\frac{1}{2} \vert t \vert ^{2}} e^{ \vert t \vert ^{2}}\,d\nu (t) \biggr\vert ^{p}\,dA(z) \\ &\quad \leq \int _{ \vert z \vert < R} \biggl\vert \int _{[0, R_{1})} \bigl\vert f_{n}(t) \bigr\vert e^{- \frac{1}{2} \vert t-z \vert ^{2}} e^{ \vert t \vert ^{2}}\,d\nu (t) \biggr\vert ^{p}\,dA(z)+ \Vert f_{n} \Vert ^{p}_{\infty}\varepsilon \\ &\quad \leq \bigl(1+ \Vert f_{n} \Vert ^{p}_{\infty} \bigr)\varepsilon , \end{aligned}$$

while n is large enough. Therefore, by the arbitrariness of ε and Lemma 4.1, we see that $\mathcal{H}_{\nu}:F^{\infty }\to F^{p}$ is compact. □

The following result is a direct consequence of Theorem 4.3 and Lemma 2.8.

Corollary 4.1

Suppose $1< p <\infty$. Let ν be a nonnegative Borel measure on $[0, \infty )$ that satisfies the condition in Theorem 3.1. Then the following statements are equivalent:

• (i) $\mathcal{H}_{\nu}$ is a bounded operator from $F^{\infty }$ into $F^{p}$;

• (ii) $\mathcal{H}_{\nu}$ is a compact operator from $F^{\infty }$ into $F^{p}$;

• (iii) $e^{ |\cdot |^{2}}\nu \in \Lambda ^{\frac{p}{p-1}}$.