On Toeplitz operators between different Fock–Sobolev-type spaces

Abstract

In this paper, we characterize those positive Borel measurable symbols μ on $\mathbb{C}^{n}$ that induce the Toeplitz operators $T^{\alpha}_{\mu}$ to be bounded or compact between different Fock–Sobolev–type spaces $F^{p}_{\alpha}$ and $F^{\infty}_{\alpha}$ with $0< p\leq \infty$.

1 Introduction

Let $\mathbb{C}^{n}$ be the complex n-space and dv be the ordinary volume measure on $\mathbb{C}^{n}$ that is normalized so that $\int _{\mathbb{C}^{n}}e^{-|z|^{2}}\,dv(z)=1$. For points $z=(z_{1},z_{2},\dots ,z_{n})$ and $w=(w_{1},w_{2},\dots ,w_{n})$ in $\mathbb{C}^{n}$, we write

$$ z\overline{w}=\sum_{j=1}^{n} z_{j}\overline{w_{j}},\quad \vert z \vert = \sqrt{z \bar{z}}. $$

For every $0< p<\infty$, $\alpha \in \mathbb{R}$, we denote by $L^{p}_{\alpha}(\mathbb{C}^{n})$ the space of measurable functions f such that

$$ \Vert f \Vert _{L^{p}_{\alpha}}= \biggl( \int _{\mathbb{C}^{n}} \bigl\vert f(z)e^{- \frac{1}{2} \vert z \vert ^{2}} \bigr\vert ^{p} \frac{dv(z)}{(1+ \vert z \vert )^{\alpha}} \biggr)^{ \frac{1}{p}}< \infty . $$

For $p=\infty$, we denote by $L^{\infty}_{\alpha}(\mathbb{C}^{n})$ the spaces of the Lebesgue measurable function f on $\mathbb{C}^{n}$ such that

$$ \Vert f \Vert _{L^{\infty}_{\alpha}}=\text{esssup} \biggl\{ \frac{ \vert f(z) \vert e^{-\frac{1}{2} \vert z \vert ^{2}}}{(1+ \vert z \vert )^{\alpha}}:z\in \mathbb{C}^{n} \biggr\} < \infty . $$

Let $H(\mathbb{C}^{n})$ be the set of entire functions on $\mathbb{C}^{n}$. Then, for a given $0< p<\infty$, the Fock–Sobolev-type space $F^{p}_{\alpha}$ with the norm $\|\cdot \|_{F^{p}_{\alpha}}=\|\cdot \|_{L^{p}_{\alpha}}$ is defined as

$$ F^{p}_{\alpha}=L^{p}_{\alpha}\bigl(\mathbb{C}^{n}\bigr)\cap H\bigl(\mathbb{C}^{n}\bigr). $$

Obviously, the Fock–Sobolev-type space $F^{2}_{\alpha}$ equipped with the natural inner product defined by

$$ \langle f,g\rangle _{L^{2}_{\alpha}}= \int _{\mathbb{C}^{n}}f(z) \overline{g(z)}e^{- \vert z \vert ^{2}} \frac{dv(z)}{(1+ \vert z \vert )^{\alpha}} $$

is a reproducing kernel Hilbert space for every real α. As stated in [4], with respect to the above inner product, it is difficult to compute the reproducing kernel of $F^{2}_{\alpha}$ explicitly. Hence, we use the equivalent norm with respect to a new measure $|z|^{-\alpha}\,dv(z)$. In detail, for $\alpha \leq 0$, we will let

$$ \langle f,g\rangle _{\alpha}= \int _{\mathbb{C}^{n}}f(z)\overline{g(z)}e^{- \vert z \vert ^{2}} \frac{dv(z)}{ \vert z \vert ^{\alpha}} $$

and for $\alpha >0$ we let

$$ \langle f,g\rangle _{\alpha}= \int _{\mathbb{C}^{n}}f^{-}_{ \frac{\alpha}{2}}(z) \overline{g^{-}_{\frac{\alpha}{2}}(z)}e^{- \vert z \vert ^{2}}\,dv(z) + \int _{\mathbb{C}^{n}}f^{+}_{\frac{\alpha}{2}}(z) \overline{g^{+}_{\frac{\alpha}{2}}(z)}e^{- \vert z \vert ^{2}} \frac{dv(z)}{ \vert z \vert ^{\alpha}}, $$

where $f^{-}_{\frac{\alpha}{2}}$ is the Taylor expansion of f up to order $\frac{\alpha}{2}$ and $f^{+}_{\frac{\alpha}{2}}=f-f^{-}_{\frac{\alpha}{2}}$. Now, we can make sure that the inner product $\langle \cdot ,\cdot \rangle _{\alpha}$ generates a new Hilbert space norm on $F^{p}_{\alpha}$ that is equivalent to the $F^{p}_{\alpha}$ norm $\|\cdot \|_{F^{p}_{\alpha}}$. In particular, if we define the norm $\|\cdot \|_{\widetilde{F^{p}_{\alpha}}}$ on $F^{p}_{\alpha}$, when $\alpha \leq 0$, by

$$ \Vert f \Vert _{\widetilde{F^{p}_{\alpha}}}= \biggl( \int _{\mathbb{C}^{n}} \bigl\vert f(z) \bigr\vert ^{p}e^{- \frac{p}{2} \vert z \vert ^{2}}\frac{dv(z)}{ \vert z \vert ^{\alpha}} \biggr)^{\frac{1}{p}} $$

and when $\alpha >0$,

$$ \Vert f \Vert _{\widetilde{F^{p}_{\alpha}}}= \biggl( \int _{\mathbb{C}^{n}} \bigl\vert f^{-}_{ \frac{\alpha}{p}}(z) \bigr\vert ^{p}e^{-\frac{p}{2} \vert z \vert ^{2}}\,dv(z) \biggr)^{ \frac{1}{p}} + \biggl( \int _{\mathbb{C}^{n}} \bigl\vert f^{+}_{ \frac{\alpha}{p}}(z) \bigr\vert ^{p}e^{-\frac{p}{2} \vert z \vert ^{2}} \frac{dv(z)}{ \vert z \vert ^{\alpha}} \biggr)^{\frac{1}{p}}, $$

then we have that both $\|\cdot \|_{\widetilde{F^{p}_{\alpha}}}$ and $\|\cdot \|_{F^{p}_{\alpha}}$ are equivalent norms.

As is well known, $F^{2}_{\alpha}$ is indeed a reproducing kernel Hilbert space (see [4, Lemma 2.1] for more details). Therefore, we have

$$ K_{z}^{\alpha}(w)=\sum_{\beta} \phi _{\beta}(w) \overline{\phi _{\beta}(z)}, $$

where $\{\phi _{\beta}\}$ is any orthonormal basis for $F^{2}_{\alpha}$ with respect to $\langle \cdot ,\cdot \rangle _{\alpha}$. Note that polynomials form a dense subset of $F^{p}_{\alpha}$ (see [3, Proposition 2.3]). Also, note that monomials are mutually orthogonal, which means that $\{ \frac{z^{\beta}}{\sqrt{\langle z^{\beta},z^{\beta}\rangle _{\alpha}}} \}$ is an orthonormal basis for $F^{2}_{\alpha}$. The arguments that are identical to those in the proof of [3, Theorem 4.5] then give us that

$$ K^{\alpha}_{z}(w)= \textstyle\begin{cases} \mathcal {I}^{-\frac{\alpha}{2}}K_{z}(w), & \text{if $\alpha \leq 0$;} \\ \mathcal {I}^{-\frac{\alpha}{2}}K_{z}(w)+(K_{z})^{-}_{\frac{\alpha}{2}}(w), &\text{if $\alpha >0$.} \end{cases} $$

Here, $K_{z}(w)=e^{\bar{z} w}$ and $\mathcal {I}^{s}$ is the fractional integration operator defined as

$$ \mathcal {I}^{s}f(z)= \textstyle\begin{cases} \sum _{k=0}^{\infty}\frac{\Gamma (n+k)}{\Gamma (n+s+k)}f_{k}(z), & \text{if $s\geq 0$;} \\ \sum _{k>|s|}^{\infty}\frac{\Gamma (n+k)}{\Gamma (n+s+k)}f_{k}(z), & \text{if $s< 0$.} \end{cases} $$

Now, it is easy to see that if $\alpha \leq 0$, $(F^{2}_{\alpha},\|\cdot \|_{\widetilde{F^{2}_{\alpha}}})$ is a closed subspace of $L^{2}_{\alpha}$ with respect to $\langle \cdot ,\cdot \rangle _{\alpha}$. In this case, let $P_{\alpha}$ denote the orthogonal projection such that

$$ P_{\alpha }f(z)=\bigl\langle f,K_{z}^{\alpha}\bigr\rangle _{\alpha } $$

for any $f\in L^{2}_{\alpha}$. Unfortunately, the inner product $\langle \cdot ,\cdot \rangle _{\alpha}$ does not make sense on $L^{2}_{\alpha}$ when $\alpha >0$. That means we cannot define the Toeplitz operator on $F^{2}_{\alpha}$ in the usual way in terms of this inner product. However, according to the ideas of [4], it makes sense to define the Toeplitz operator with the complex Borel measure on $\mathbb{C}^{n}$ by the formula

$$ T^{\alpha}_{\mu }f(z)= \int _{\mathbb{C}^{n}}f(w) \overline{K_{z}^{\alpha}(w)}e^{- \vert w \vert ^{2}} \vert w \vert ^{-\alpha}\,d\mu (w) $$

if $\alpha \leq 0$ and

$$\begin{aligned} T^{\alpha}_{\mu }f(z) =& \int _{\mathbb{C}^{n}}f^{-}_{\frac{\alpha}{2}}(w) \overline{ \bigl(K^{\alpha}_{z}\bigr)^{-}_{\frac{\alpha}{2}}(w)}e^{- \vert w \vert ^{2}}\,d \mu (w) \\ &{}+ \int _{\mathbb{C}^{n}}f^{+}_{\frac{\alpha}{2}}(w) \overline{ \bigl(K^{\alpha}_{z}\bigr)^{+}_{\frac{\alpha}{2}}(w)}e^{- \vert w \vert ^{2}} \frac{d\mu (w)}{ \vert w \vert ^{\alpha}} \end{aligned}$$

if $\alpha >0$. Note that if μ satisfies

$$ \int _{\mathbb{C}^{n}} \bigl\vert K_{z}^{\alpha}(w) \bigr\vert ^{2}e^{- \vert w \vert ^{2}} \frac{d \vert \mu \vert (w)}{(1+ \vert w \vert )^{\alpha}}< \infty , \quad z\in \mathbb{C}^{n}, $$

then the Toeplitz operator $T^{\alpha}_{\mu}$ is densely defined on $F^{2}_{\alpha}$. Seqentially, we let

$$ B(z,r)=\bigl\{ w\in \mathbb{C}^{n}: \vert w-z \vert < r\bigr\} $$

for $z\in \mathbb{C}^{n}$ and $r>0$. Given a Borel measure μ on $\mathbb{C}^{n}$, the average of μ on $B(z,r)$ is $\frac{\mu (B(z,r))}{v(B(z,r))}$. Since the Lebesgue volume $v(B(z,r))\simeq r^{2n}$, we simply set the average function of μ as

$$ \widehat{\mu}_{r}(z)=\mu \bigl(B(z,r)\bigr). $$

For Borel measure μ on $\mathbb{C}^{n}$, $0< t<\infty$, the t-Berezin transform of a Toeplitz operator on $\mathbb{C}^{n}$ is defined by

$$ \widetilde{\mu}^{\alpha}_{t}(z)= \frac{1}{(1+ \vert z \vert )^{(\frac{t}{2}-1)\alpha}} \int _{\mathbb{C}^{n}} \bigl\vert k_{2,z}^{ \alpha}(w)e^{-\frac{1}{2} \vert w \vert ^{2}} \bigr\vert ^{t} \frac{d\mu (w)}{ \vert w \vert ^{\alpha}} $$

if $\alpha \leq 0$, and

$$\begin{aligned} \widetilde{\mu}^{\alpha}_{t}(z) =& \frac{1}{(1+ \vert z \vert )^{(\frac{t}{2}-1)\alpha}} \int _{\mathbb{C}^{n}} \bigl\vert \bigl(k_{2,z}^{ \alpha} \bigr)^{-}_{\frac{\alpha}{t}}(w)e^{-\frac{1}{2} \vert w \vert ^{2}} \bigr\vert ^{t}\,d \mu (w) \\ &{}+\frac{1}{(1+ \vert z \vert )^{(\frac{t}{2}-1)\alpha}} \int _{\mathbb{C}^{n}} \bigl\vert \bigl(k_{2,z}^{\alpha} \bigr)^{+}_{\frac{\alpha}{t}}(w)e^{-\frac{1}{2} \vert w \vert ^{2}} \bigr\vert ^{t}\frac{d\mu (w)}{ \vert w \vert ^{\alpha}} \end{aligned}$$

if $\alpha >0$, where $k_{2,z}^{\alpha}(w)$ is the normalization of the kernel $K_{z}^{\alpha}(w)$, and, in general, we denote by $k^{\alpha}_{p,z}=K^{\alpha}_{z}\|K^{\alpha}_{z}\|^{-1}_{F^{p}_{ \alpha}}$, for $0< p\leq \infty$, $z\in \mathbb{C}^{n}$.

Although there has been much more research on Toeplitz operators on Bergman or Hardy spaces of various domains than on Fock spaces, Toeplitz operators on Fock spaces have been studied for many years, see [10] and references therein for examples. In 2014, Mengestie had characterized the Toeplitz operators with positive measure symbols between classic Fock spaces $F^{p}_{\alpha}$ and $F^{\infty}_{\alpha}$. In more detail, the Toeplitz operator $T_{\mu}: F^{p}_{\alpha}\to F^{\infty}_{\alpha}$ is bounded (compact) if and only if μ is a (vanishing) $(1,q)$ Fock–Carleson measure, where $1\leq p\leq q<\infty$, if and only if $\widehat{\mu}_{\delta}, \widetilde{\mu}_{t}\in L^{\infty}$ for some or any $t,\delta >0$. The Toepitz operator $T_{\mu}: F^{\infty}_{\alpha}\to F^{p}_{\alpha}$ is bounded or compact if and only μ is a vanishing $(\infty ,q)$ Fock–Carleson measure, where $1\leq q<\infty$, if and only if $\widehat{\mu}_{\delta}, \widetilde{\mu}_{t}\in L^{1}$ for some or any $t,\delta >0$. Those achievements would be based on some his own results in [7]. In his other paper [8], he had researched the basic properties of Carleson-type measures of Fock–Sobolev spaces. It is worth noting that the positive measure μ is a $(\infty ,q)$ Fock–Carleson measure for $0< q<\infty$ if and only if it is also vanishing, if and only $\widehat{\mu}_{mq,r}, \widetilde{\mu}_{t,mq}\in L^{1}$ for some or any $t,r>0$.

In 2019, Lv had, in [6], successfully obtained the boundedness and compactness of Toeplitz operators with positive measure symbols on doubling the Fock space between $F^{p}_{\varphi}$ and $F^{\infty}_{\varphi}$ for $0< p\leq \infty$. She pointed out that Toeplitz operator $T_{\mu}:F^{p}_{\varphi}\to F^{\infty}_{\varphi}$ is bounded (compact) if and only if μ is a (vanishing) $\frac{p}{p+1}$ Fock–Carleson measure. Conversely, the Toeplitz operator $T_{\mu}:F^{\infty}_{\varphi}\to F^{p}_{\varphi}$ is bounded or compact if and only if $\widehat{\mu}_{\delta},\widetilde{\mu}_{t}\in L^{p}$ for some or any $t,\delta >0$. Something interesting attracts our attention, that is the so-called $\frac{p}{p+1}$ Fock–Carleson measure. As stated in [6], μ is a $(p,q)$ Fock–Carleson measure if and only if it is a $(tp,tq)$ Fock–Carleson measure for any $t>0$, because it is also equivalent to both $\widehat{\mu}_{\delta}\rho ^{2(1-\frac{q}{p})}$ and $\widetilde{\mu}_{t}\rho ^{2(1-\frac{q}{p})}$ that are bounded. That is why we call it a $\frac{p}{p+1}$ Fock–Carleson measure and we write it as $\|\mu \|_{\frac{p}{q}}=\|i\|_{F^{p/q}_{\varphi}\to L^{1}_{\varphi}( \mu )}$. Luckily, these good achievements keep being correct in the case of large Fock spaces, for which we can refer to our latest paper [1].

Unfortunately, this phenomenon will not occur in the case of Fock–Sobolev-type spaces, because the definition of the largest Fock–Soboblev–type space $F^{\infty}_{\alpha}$ differs, totally, from doubling Fock spaces $F^{\infty}_{\varphi}$. In other cases, including the classic Fock space, doubling Fock space, and large Fock space, the infinite index would be thought of as the limit of the finite index. However, This is not the case here. Moreover, $F^{\infty}_{\alpha}\neq \lim_{p\to \infty}F^{p}_{\alpha}$. In this paper, some terminologies and symbols are still similar to those in [6]. We are going to achieve some characterizations on those $\mu \geq 0$ such that Toeplitz operator $T_{\mu}$ is bounded or compact from $F^{p}_{\alpha}$ to $F^{\infty}_{\alpha}$ and the converse case, respectively, for $0< p\leq \infty$.

We will end this introduction with a comment on some notations.

Firstly, for conciseness, we will denote by $I_{(\star )}$ the integral in Line $(\star )$. Take, for example, that $I_{\text{(1)}}$ refers to the integral in Line (1).

Secondly, for positive quantities A and B (which may depend on a variety of parameters or variables), we will use the notation $A\lesssim B$ if there exists an unimportant constant C such that $A\leq CB$. The notation $A\gtrsim B$ will have a similar meaning. We write $A\simeq B$ if $A\lesssim B$ and $A\gtrsim B$ at the same time.

2 Carleosn measures and related results

In this section, we are going to characterize Fock–Carleson measures. For this purpose, we need some conclusions about the reproducing kernel $K_{z}^{\alpha}$, which are partly from [3, 4].

Lemma 2.1

The upper pointwise estimate and the properties of the Bergman kernel $K^{\alpha}_{z}$ are as follows:

• (1) Suppose that f belongs to the Fock–Sobolev-type space $F^{p}_{\alpha}$ for any real α. Then, for any $z,w\in \mathbb{C}^{n}$, $p,a,t>0$, $\alpha \in \mathbb{R}$, we have

  $$\begin{aligned} \frac{ \vert f(z) \vert ^{p}e^{-a \vert z \vert ^{2}}}{(1+ \vert z \vert )^{\alpha}}\lesssim \int _{ \vert w-z \vert < t} \bigl\vert f(w) \bigr\vert ^{p}e^{-a \vert w \vert ^{2}} \frac{dv(w)}{(1+ \vert w \vert )^{\alpha}}. \end{aligned}$$

• (2) The Bergman kernel satisfies

  $$\begin{aligned} \bigl\vert K^{\alpha}_{z}(w) \bigr\vert \lesssim \textstyle\begin{cases} (1+ \vert z \vert \vert w \vert )^{\frac{\alpha}{2}}\exp (\frac{1}{2} \vert z \vert ^{2}+\frac{1}{2} \vert w \vert ^{2}- \frac{1}{8} \vert z-w \vert ^{2}), & \textit{if $\alpha \leq 0$;} \\ (1+ \vert w\bar{z} \vert )^{\frac{\alpha}{2}}\exp (\frac{1}{2} \vert z \vert ^{2}+\frac{1}{2} \vert w \vert ^{2}- \frac{1}{8} \vert z-w \vert ^{2}), & \textit{if $\alpha >0$.} \end{cases}\displaystyle \end{aligned}$$

  More specifically,

  $$\begin{aligned} \bigl\vert K^{\alpha}_{z}(z) \bigr\vert \simeq \bigl(1+ \vert z \vert \bigr)^{\alpha }e^{ \vert z \vert ^{2}} \end{aligned}$$

  for any $z\in \mathbb{C}^{n}$, and there is a small enough $r_{0}>0$ such that

  $$\begin{aligned} \bigl\vert K^{\alpha}_{z}(w) \bigr\vert \gtrsim \bigl(1+ \vert z \vert \bigr)^{\alpha }\exp \biggl( \frac{1}{2} \vert z \vert ^{2}+ \frac{1}{2} \vert w \vert ^{2} \biggr),\quad \forall z\in B(w,r_{0}). \end{aligned}$$

• (3) Together with the above, the estimate of the norm of the reproducing kernel is

  $$ \bigl\Vert K^{\alpha}_{z} \bigr\Vert _{F^{p}_{\alpha}} \lesssim \textstyle\begin{cases} (1+ \vert z \vert )^{\alpha -\frac{\alpha}{p}}e^{\frac{ \vert z \vert ^{2}}{2}}, & 0< p< \infty ; \\ e^{\frac{ \vert z \vert ^{2}}{2}}, & p=\infty . \end{cases} $$

• (4) For $0< p\leq \infty$, the normalization of the kernel $k^{\alpha}_{p,z}\to 0$ uniformly on compact subsets of $\mathbb{C}^{n}$ as $|z|\to \infty$.

Proof

We only discuss the statement (4) in the case of $p=\infty$, as the others are the basic properties in [3, 4] like [4, Lemmas 2.1, 2.3] and [3, Proposition 4.8]. By the statement (3), to be continued, we can easily see that

$$ \frac{ \vert K^{\alpha}_{z}(w) \vert }{ \Vert K^{\alpha}_{z} \Vert _{F^{\infty}_{\alpha}}} \lesssim \bigl\vert K^{\alpha}_{z}(w) \bigr\vert e^{-\frac{ \vert z \vert ^{2}}{2}}. $$

Moreover, in terms of the statement (2),

$$ \bigl\vert K^{\alpha}_{z}(w) \bigr\vert e^{-\frac{ \vert z \vert ^{2}}{2}} \lesssim \textstyle\begin{cases} (1+ \vert z \vert \vert w \vert )^{\frac{\alpha}{2}}\exp (\frac{1}{2} \vert w \vert ^{2}-\frac{1}{8} \vert z-w \vert ^{2}), & \text{if $\alpha \leq 0$;} \\ (1+ \vert w\bar{z} \vert )^{\frac{\alpha}{2}}\exp (\frac{1}{2} \vert w \vert ^{2}-\frac{1}{8} \vert z-w \vert ^{2}), & \text{if $\alpha >0$.} \end{cases} $$

This tells us the statement (4) is true when w is fixed and $|z|\to \infty$. □

Sequentially, we will study Carleson measures on $F_{\alpha}^{p}$. If $\mu \geq 0$ is a Borel measure, then we define the norm $\|\cdot \|_{L^{p}_{\alpha ,\mu}\cap H(\mathbb{C}^{n})}$ on $L^{p}_{\alpha ,\mu}\cap H(\mathbb{C}^{n})$ by

$$ \Vert f \Vert ^{p}_{L^{p}_{\alpha ,\mu}\cap H(\mathbb{C}^{n})}= \int _{\mathbb{C}^{n}} \bigl\vert f^{-}_{\frac{\alpha}{p}}(z)e^{-\frac{1}{2} \vert z \vert ^{2}} \bigr\vert ^{p}\,d \mu (z) + \int _{\mathbb{C}^{n}} \bigl\vert f^{+}_{\frac{\alpha}{p}}(z)e^{- \frac{1}{2} \vert z \vert ^{2}} \bigr\vert ^{p}\frac{d\mu (z)}{ \vert z \vert ^{\alpha}}, $$

when $\alpha >0$ and

$$ \Vert f \Vert ^{p}_{L^{p}_{\alpha ,\mu}\cap H(\mathbb{C}^{n})}= \int _{\mathbb{C}^{n}} \bigl\vert f(z)e^{-\frac{1}{2} \vert z \vert ^{2}} \bigr\vert ^{p} \frac{d\mu (z)}{ \vert z \vert ^{\alpha}}, $$

when $\alpha \leq0$. If $0< p,q<\infty$ and μ is a weighted $(p,q)$-Fock–Carleson measure, then the inclusion

$$ \iota _{p}: F^{p}_{\alpha}\to L^{q}_{\alpha ,\mu}\cap H\bigl(\mathbb{C}^{n}\bigr) $$

is bounded. We call μ a vanishing weighted $(p,q)$-Fock–Carleson measure for $F^{p}_{\alpha}$ if

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

when $\alpha >0$ and

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

when $\alpha \leq0$, whenever $\{f_{j}\}$ is bounded in $F^{p}_{\alpha}$ and converges to 0 uniformly on a compact subset of $\mathbb{C}^{n}$ as $j\to \infty$.

The following three theorems characterize (vanishing) $(p,q)$ Fock–Carleson measures for all possible $0< p,q<\infty$. All of them have been proven in [2].

Theorem 2.2

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

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

• (2) $(1+|z|)^{(\frac{1}{p}-\frac{1}{q})q\alpha}\widetilde{\mu}^{\alpha}_{t}(z) \in L^{\infty}(dv)$ for some or any $t>0$;

• (3) $(1+|z|)^{(\frac{1}{p}-\frac{1}{q})q\alpha}\widehat{\mu}_{\delta}(z) \in L^{\infty}(dv)$ for some or any $\delta >0$;

• (4) The sequence

  $$ \bigl\{ \bigl(1+ \vert a_{k} \vert \bigr)^{(\frac{1}{p}-\frac{1}{q})q\alpha} \widehat{\mu}_{r}(a_{k}) \bigr\} _{k=1}^{\infty} \in l^{\infty } $$

  for some or any r-lattice $\{a_{k}\}_{k=1}^{\infty}$.

Furthermore,

$$\begin{aligned} \Vert \iota _{p} \Vert ^{q}_{F^{p}_{\alpha}\to L^{q}_{\alpha ,\mu}\cap H( \mathbb{C}^{n})} \simeq & \bigl\Vert \bigl(1+ \vert \cdot \vert \bigr)^{(\frac{1}{p}- \frac{1}{q})q\alpha} \widetilde{\mu}^{\alpha}_{t} \bigr\Vert _{L^{\infty}(dv)} \\ \simeq & \bigl\Vert \bigl(1+ \vert \cdot \vert \bigr)^{(\frac{1}{p}-\frac{1}{q})q\alpha} \widehat{\mu}_{\delta} \bigr\Vert _{L^{\infty}(dv)} \\ \simeq & \bigl\Vert \bigl\{ \bigl(1+ \vert a_{k} \vert \bigr)^{(\frac{1}{p}-\frac{1}{q})q \alpha}\widehat{\mu}_{r}(a_{k}) \bigr\} _{k}^{\infty} \bigr\Vert _{l^{ \infty}}. \end{aligned}$$

Theorem 2.3

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

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

• (2) $(1+|z|)^{(\frac{1}{p}-\frac{1}{q})q\alpha}\widetilde{\mu}^{\alpha}_{t}(z) \to 0$ as $|z|\to \infty$ for some or any $t>0$;

• (3) $(1+|z|)^{(\frac{1}{p}-\frac{1}{q})q\alpha}\widehat{\mu}_{\delta}(z) \to 0$ as $|z|\to \infty$ for some or any $\delta >0$;

• (4) The sequence

  $$ \lim_{k\to \infty}\bigl(1+ \vert a_{k} \vert \bigr)^{(\frac{1}{p}-\frac{1}{q})q\alpha} \widehat{\mu}_{r}(a_{k})=0 $$

  for some or any r-lattice $\{a_{k}\}_{k=1}^{\infty}$.

Theorem 2.4

Let $0< q< p<\infty$ and $\mu \geq 0$. 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) $(1+|z|)^{(\frac{1}{p}-\frac{1}{q})q\alpha}\widetilde{\mu}^{\alpha}_{t}(z) \in L^{\frac{p}{p-q}}(dv)$ for some or any $t>0$;

• (4) $(1+|z|)^{(\frac{1}{p}-\frac{1}{q})q\alpha}\widehat{\mu}_{\delta}(z) \in L^{\frac{p}{p-q}}(dv)$ for some or any $\delta >0$;

• (5) The sequence

  $$ \bigl\{ \bigl(1+ \vert a_{k} \vert \bigr)^{(\frac{1}{p}-\frac{1}{q})q\alpha} \widehat{\mu}_{r}(a_{k}) \bigr\} _{k=1}^{\infty} \in l^{\frac{p}{p-q}} $$

  for some or any r-lattice $\{a_{k}\}_{k=1}^{\infty}$.

Furthermore,

$$\begin{aligned} \Vert \iota _{p} \Vert ^{q}_{F^{p}_{\alpha}\to L^{q}_{\alpha ,\mu}\cap H( \mathbb{C}^{n})} \simeq & \bigl\Vert \bigl(1+ \vert \cdot \vert \bigr)^{(\frac{1}{p}- \frac{1}{q})q\alpha} \widetilde{\mu}^{\alpha}_{t} \bigr\Vert _{L^{ \frac{p}{p-q}}} \\ \simeq & \bigl\Vert \bigl(1+ \vert \cdot \vert \bigr)^{(\frac{1}{p}-\frac{1}{q})q\alpha} \widehat{\mu}_{\delta} \bigr\Vert _{L^{\frac{p}{p-q}}} \\ \simeq & \bigl\Vert \bigl\{ \bigl(1+ \vert a_{k} \vert \bigr)^{(\frac{1}{p}-\frac{1}{q})q \alpha}\widehat{\mu}_{r}(a_{k}) \bigr\} _{k}^{\infty} \bigr\Vert _{l^{ \frac{p}{p-q}}}. \end{aligned}$$

The well-informed reader will note that the descriptions in Theorem 2.2 are a little different from those in [5, Theorem 3.4] and [9, Theorem 25]. The reason why this situation occurred is mainly the measurement of the unit ball. For example, in the case of Fock-type spaces,

$$ C_{1}\bigl[\Phi '\bigl( \vert z \vert ^{2}\bigr)\bigr]^{-1}\bigl[\Psi '\bigl( \vert z \vert ^{2}\bigr)\bigr]^{-(n-1)}\leq \bigl\vert B(z,r) \bigr\vert \leq C_{2}\bigl[\Phi '\bigl( \vert z \vert ^{2}\bigr)\bigr]^{-1}\bigl[\Psi '\bigl( \vert z \vert ^{2}\bigr) \bigr]^{-(n-1)} $$

for some constants $C_{1}$ and $C_{2}$. See [9] for more information.

3 Bounded Toeplitz operators

In this section, we are going to achieve some characterizations on those $\mu \geq 0$ such that the Toeplitz operator $T_{\mu}$ is bounded or compact from $F^{p}_{\alpha}$ to $F^{\infty}_{\alpha}$ and the converse, respectively, for $0< p\leq \infty$. To study the compactness, we need the following lemma, part of which can been found in [2, Lemma 3.2].

Lemma 3.1

Let $0< p,q\leq \infty$ and suppose μ is a t-Fock–Carleson measure. The Toeplitz operator $T^{\alpha}_{\mu}$ is well defined on $F^{p}_{\alpha}$. Moreover, for $R>0$, the Toeplitz operator $T^{\alpha}_{\mu _{R}}$ is compact from $F^{p}_{\alpha}$ to $F^{q}_{\alpha}$, where $\mu _{R}(V)=\int _{V\cap \{z:|z|\leq R\}}\,d\mu$ for $V\subset \mathbb{C}^{n}$ measurable.

Proof

It suffices to discuss the case of an infinite index, because by [2, Lemmas 3.1, 3.2], we can conclude that $T^{\alpha}_{\mu}$ is well defined on $F^{p}_{\alpha}$ and $T^{\alpha}_{\mu _{R}}$ is compact from $F_{\alpha}^{p}$ to $F_{\alpha}^{q}$ for $0< p,q<\infty$, respectively.

For any $f\in F^{\infty}_{\alpha}$, by the definition of a Toeplitz operator, it suffices to prove that, for any $z\in \mathbb{C}^{n}$, when $\alpha >0$,

$$\begin{aligned}& \int _{\mathbb{C}^{n}} \bigl\vert f^{+}_{\frac{\alpha}{2}}(w) \bigr\vert \bigl\vert \bigl(K_{z}^{\alpha} \bigr)^{+}_{\frac{\alpha}{2}}(w) \bigr\vert e^{- \vert w \vert ^{2}} \frac{d\mu (w)}{ \vert w \vert ^{\alpha}} \\& \quad = \int _{ \vert w \vert \geq 1} \bigl\vert f^{+}_{\frac{\alpha}{2}}(w) \bigr\vert \bigl\vert \bigl(K_{z}^{ \alpha} \bigr)^{+}_{\frac{\alpha}{2}}(w) \bigr\vert e^{- \vert w \vert ^{2}} \frac{d\mu (w)}{ \vert w \vert ^{\alpha}} \end{aligned}$$

(1)

$$\begin{aligned}& \qquad {}+ \int _{ \vert w \vert < 1} \bigl\vert f^{+}_{\frac{\alpha}{2}}(w) \bigr\vert \bigl\vert \bigl(K_{z}^{ \alpha} \bigr)^{+}_{\frac{\alpha}{2}}(w) \bigr\vert e^{- \vert w \vert ^{2}} \frac{d\mu (w)}{ \vert w \vert ^{\alpha}}< \infty . \end{aligned}$$

(2)

Note that for a fixed small $0<\delta <R$, Theorems 2.2 and 2.3 show us that

$$\begin{aligned} \bigl(1+ \vert \cdot \vert \bigr)^{(t-1)\alpha}\widehat{\mu}_{\delta}(\cdot )\in L^{\infty}(dv), \end{aligned}$$

(3)

for some t, maybe different from the condition. Now, we divide the integral above into two cases. When $|w|\geq 1$, [2, Lemma 2.3] tells us that

$$\begin{aligned} \mathrm{I}_{\text{(1)}} \lesssim & \int _{ \vert w \vert \geq 1} \biggl( \frac{ \vert f^{+}_{\frac{\alpha}{2}}(w) \vert }{(1+ \vert w \vert )^{\alpha}}e^{-\frac{1}{2} \vert w \vert ^{2}} \biggr) \bigl\vert \bigl(K_{z}^{\alpha} \bigr)^{+}_{\frac{\alpha}{2}}(w) \bigr\vert e^{- \frac{1}{2} \vert w \vert ^{2}} \widehat{\mu}_{\delta}(w)\,dv(w) \\ \lesssim & \Vert f \Vert _{F^{\infty}_{\alpha}} \biggl\Vert \frac{\widehat{\mu}_{\delta}(\cdot )}{(1+ \vert \cdot \vert )^{(1-t)\alpha}} \biggr\Vert _{L^{\infty}} \underbrace{ \int _{\mathbb{C}^{n}} \bigl\vert \bigl(K_{z}^{\alpha} \bigr)^{+}_{\frac{\alpha}{2}}(w) \bigr\vert e^{-\frac{1}{2} \vert w \vert ^{2}} \frac{dv(w)}{ \vert w \vert ^{(t-1)\alpha}}}_{\mathrm{II}}. \end{aligned}$$

According to the proof in [4, Lemma 3.4], for any $z\in \mathbb{C}^{n}$,

$$\begin{aligned} \mathrm{II} \lesssim &\sum_{ \vert m \vert >\frac{\alpha}{2}} \frac{ \vert z \vert ^{ \vert m \vert }\Gamma (n+ \vert m \vert )}{m!\Gamma (n-\frac{\alpha}{2}+ \vert m \vert )} \int _{\mathbb{C}^{n}} \vert w \vert ^{ \vert m \vert + \vert t-1 \vert \alpha}e^{-\frac{1}{2} \vert w \vert ^{2}}\,dv(w) \\ \lesssim &\sum_{ \vert m \vert >\frac{\alpha}{2}} \vert z \vert ^{ \vert m \vert }\bigl( \vert m! \vert \bigr)^{-1}n^{ \vert m \vert } \vert m \vert ^{ \frac{\alpha}{2}}\Gamma \biggl(n+\frac{ \vert m \vert + \vert t-1 \vert \alpha}{2} \biggr)< \infty . \end{aligned}$$

On the other hand, the Cauchy integral formula over the unit polydisk and the maximum modulus theorem show us that

$$\begin{aligned} \sup_{ \vert w \vert < 1} \bigl\vert f_{\frac{\alpha}{2}}^{+}(w) \bigr\vert \vert w \vert ^{- \frac{\alpha}{2}} \lesssim &\sup_{ \vert w \vert < 1}\sum _{ \vert \gamma \vert > \frac{\alpha}{2}} \biggl\vert \frac{\partial ^{\gamma }f(0)}{\gamma !} \biggr\vert \vert w \vert ^{ \vert \gamma \vert -\frac{\alpha}{2}} \\ \lesssim &\sup_{ \vert w \vert < 2} \bigl\vert f(w) \bigr\vert \sum _{k>\frac{\alpha}{2}} \frac{ \vert w \vert ^{k-\frac{\alpha}{2}}}{2^{k}} \\ \lesssim &\sup_{ \vert w \vert < 2} \bigl\vert f(w) \bigr\vert \bigl(1+ \vert w \vert \bigr)^{-\alpha}. \end{aligned}$$

(4)

Also, when $|w|<1$, by Stirling’s formula,

$$\begin{aligned} \bigl\vert \bigl(K_{z}^{\alpha} \bigr)^{+}_{\frac{\alpha}{2}}(w) \bigr\vert \vert w \vert ^{- \frac{\alpha}{2}} \lesssim \vert w \vert ^{-\frac{\alpha}{2}}\sum _{k> \frac{\alpha}{2}}\frac{k^{\frac{\alpha}{2}}}{k!} \vert z\bar{w} \vert ^{k} \lesssim \sum_{k>\frac{\alpha}{2}} \frac{k^{\frac{\alpha}{2}}}{k!} \vert z \vert ^{k}. \end{aligned}$$

(5)

Therefore, we can see that

$$\begin{aligned} I_{\text{(2)}} \lesssim & \int _{ \vert w \vert < 1} \biggl(\sup_{ \vert w \vert < 2} \frac{ \vert f(w) \vert }{(1+ \vert w \vert )^{\alpha}} \biggr) \bigl( \bigl\vert \bigl(K_{z}^{\alpha} \bigr)^{+}_{ \frac{\alpha}{2}}(w) \bigr\vert \vert w \vert ^{-\frac{\alpha}{2}} \bigr)e^{- \frac{1}{2} \vert w \vert ^{2}}\,d\mu (w) \\ \lesssim & \Vert f \Vert _{F^{\infty}_{\alpha}} \int _{ \vert w \vert < 1} \bigl( \bigl\vert \bigl(K_{z}^{ \alpha} \bigr)^{+}_{\frac{\alpha}{2}}(w) \bigr\vert \vert w \vert ^{-\frac{\alpha}{2}} \bigr)e^{-\frac{1}{2} \vert w \vert ^{2}}\,d\mu (w) \\ \lesssim & \Vert f \Vert _{F^{\infty}_{\alpha}} \biggl\Vert \frac{\widehat{\mu}_{\delta}(\cdot )}{(1+ \vert \cdot \vert )^{(1-t)\alpha}} \biggr\Vert _{L^{\infty}}\sum _{k>\frac{\alpha}{2}} \frac{k^{\frac{\alpha}{2}}}{k!} \vert z \vert ^{k}. \end{aligned}$$

Sequentially, in order to prove the compactness of $T^{\alpha}_{\mu _{R}}$ for $q=\infty$, we suppose that $\{f_{k}\}_{k\geq 1}\subset F^{p}_{\alpha}(0< p\leq \infty )$ is a bounded sequence and $f_{k}$ uniformly converges to 0 on compact subset of $\mathbb{C}^{n}$ as $k\to \infty$. By Montel’s theorem, we will show that

$$ \lim_{k\to \infty} \bigl\Vert T^{\alpha}_{\mu _{R}}f_{k} \bigr\Vert _{F^{\infty}_{\alpha}}=0. $$

By the definition of a Topelitz operator, we will omit the details about the case $\alpha \leq 0$ and the other comes into play:

$$\begin{aligned}& \bigl\vert T^{\alpha}_{\mu _{R}}f_{k}(z) \bigr\vert \frac{e^{-\frac{1}{2} \vert z \vert ^{2}}}{(1+ \vert z \vert )^{\alpha}} \\& \quad \lesssim \frac{e^{-\frac{1}{2} \vert z \vert ^{2}}}{(1+ \vert z \vert )^{\alpha}} \int _{ \mathbb{C}^{n}} \bigl\vert (f_{k})_{\frac{\alpha}{2}}^{-}(w) \bigr\vert \bigl\vert \bigl(K_{z}^{\alpha} \bigr)_{\frac{\alpha}{2}}^{-}(w) \bigr\vert e^{- \vert w \vert ^{2}}\,d \mu _{R}(w) \\& \qquad {}+\frac{e^{-\frac{1}{2} \vert z \vert ^{2}}}{(1+ \vert z \vert )^{\alpha}} \int _{\mathbb{C}^{n}} \bigl\vert (f_{k})_{\frac{\alpha}{2}}^{+}(w) \bigr\vert \bigl\vert \bigl(K_{z}^{ \alpha} \bigr)_{\frac{\alpha}{2}}^{+}(w) \bigr\vert \frac{e^{- \vert w \vert ^{2}}}{ \vert w \vert ^{\alpha}}\,d\mu _{R}(w). \end{aligned}$$

(6)

Now, we pay attention to the integral $I_{\text{(6)}}$, because the first will be discussed in a similar way. For some suitable $r_{0}>0$, we divide the integral $I_{\text{(6)}}$ into

$$\begin{aligned} I_{\text{(6)}} =&\frac{e^{-\frac{1}{2} \vert z \vert ^{2}}}{(1+ \vert z \vert )^{\alpha}} \int _{ \vert w \vert \geq r_{0}} \bigl\vert (f_{k})_{\frac{\alpha}{2}}^{+}(w) \bigr\vert \bigl\vert \bigl(K_{z}^{\alpha} \bigr)_{\frac{\alpha}{2}}^{+}(w) \bigr\vert \frac{e^{- \vert w \vert ^{2}}}{ \vert w \vert ^{\alpha}}\,d\mu _{R}(w) \end{aligned}$$

(7)

$$\begin{aligned} &{}+\frac{e^{-\frac{1}{2} \vert z \vert ^{2}}}{(1+ \vert z \vert )^{\alpha}} \int _{ \vert w \vert < r_{0}} \bigl\vert (f_{k})_{\frac{\alpha}{2}}^{+}(w) \bigr\vert \bigl\vert \bigl(K_{z}^{ \alpha} \bigr)_{\frac{\alpha}{2}}^{+}(w) \bigr\vert \frac{e^{- \vert w \vert ^{2}}}{ \vert w \vert ^{\alpha}}\,d\mu _{R}(w). \end{aligned}$$

(8)

According to the proof in [4, Proposition 3.2],

$$\begin{aligned} \int _{\mathbb{C}^{n}} \bigl\vert K^{\alpha}_{w}(z) \bigr\vert ^{p} e^{-\frac{p}{2} \vert z \vert ^{2}} \frac{dv(z)}{(1+ \vert z \vert )^{\alpha}}\lesssim \bigl(1+ \vert w \vert \bigr)^{p\alpha -\alpha}e^{ \frac{p}{2} \vert w \vert ^{2}}, \quad 0< p< \infty . \end{aligned}$$

(9)

Therefore, if we combine [2, Lemma 2.3] and the condition (3), the integral $I_{\text{(7)}}$ comes into play, when $k\to \infty$,

$$\begin{aligned} I_{\text{(7)}} \lesssim & \frac{e^{-\frac{1}{2} \vert z \vert ^{2}}}{(1+ \vert z \vert )^{\alpha}} \int _{r_{0}\leq \vert w \vert \leq R} \bigl\vert (f_{k})_{\frac{\alpha}{2}}^{+}(w) \bigr\vert \frac{ \vert (K_{z}^{\alpha} )_{\frac{\alpha}{2}}^{+}(w) \vert }{ (1+ \vert w \vert )^{(t-1)\alpha}}\frac{e^{- \vert w \vert ^{2}}}{ \vert w \vert ^{\alpha}} \frac{\widehat{\mu}_{\delta}(w)\,dv(w)}{(1+ \vert w \vert )^{(1-t)\alpha}} \\ \lesssim &\frac{e^{-\frac{1}{2} \vert z \vert ^{2}}}{(1+ \vert z \vert )^{\alpha}} \biggl\Vert \frac{\widehat{\mu}_{\delta}(\cdot )}{(1+ \vert \cdot \vert )^{(1-t)\alpha}} \biggr\Vert _{L^{\infty}}\sup_{r_{0}\leq \vert w \vert \leq R} \bigl\vert (f_{k})_{ \frac{\alpha}{2}}^{+}(w) \bigr\vert \frac{ \vert (K_{z}^{\alpha} )_{\frac{\alpha}{2}}^{+}(w) \vert }{e^{ \vert w \vert ^{2}}(1+ \vert w \vert )^{t\alpha}} \\ \lesssim &\sup_{r_{0}\leq \vert w \vert \leq R} \frac{ \vert f_{k}(w) \vert e^{-\frac{1}{2} \vert w \vert ^{2}}}{(1+ \vert w \vert )^{t\alpha}} \int _{ \mathbb{C}^{n}} \frac{ \vert K^{\alpha}_{w}(z) \vert }{e^{\frac{1}{2} \vert z \vert ^{2}}e^{\frac{1}{2} \vert w \vert ^{2}}} \frac{dv(z)}{(1+ \vert z \vert )^{\alpha}}\to 0. \end{aligned}$$

On the other hand, in view of the estimates (5) and (4),

$$\begin{aligned} I_{\text{(8)}} \lesssim & \frac{e^{-\frac{1}{2} \vert z \vert ^{2}}}{(1+ \vert z \vert )^{\alpha}} \biggl(\sup _{ \vert w \vert \leq 2r_{0}} \frac{ \vert f_{k}(w) \vert e^{-\frac{1}{2} \vert w \vert ^{2}}}{(1+ \vert w \vert )^{\alpha}} \biggr) \int _{ \vert w \vert \leq r_{0}} \bigl\vert \bigl(K_{z}^{\alpha} \bigr)_{ \frac{\alpha}{2}}^{+}(w) \bigr\vert e^{-\frac{1}{2} \vert w \vert ^{2}} \frac{d\mu (w)}{ \vert w \vert ^{\frac{\alpha}{2}}} \\ \lesssim & \biggl(\frac{e^{-\frac{1}{2} \vert z \vert ^{2}}}{(1+ \vert z \vert )^{\alpha}} \sum_{s>\frac{\alpha}{2}} \frac{s^{\frac{\alpha}{2}}}{s!} \vert z \vert ^{s} \biggr) \biggl(\sup _{ \vert w \vert \leq 2r_{0}} \frac{ \vert f_{k}(w) \vert e^{-\frac{1}{2} \vert w \vert ^{2}}}{(1+ \vert w \vert )^{\alpha}} \biggr) \biggl\Vert \frac{\widehat{\mu}_{\delta}(\cdot )}{(1+ \vert \cdot \vert )^{(1-t)\alpha}} \biggr\Vert _{L^{\infty}} \\ \lesssim & \frac{ \vert z \vert e^{ \vert z \vert }e^{-\frac{1}{2} \vert z \vert ^{2}}}{(1+ \vert z \vert )^{- \vert \alpha \vert }} \biggl( \sup_{ \vert w \vert \leq 2r_{0}} \frac{ \vert f_{k}(w) \vert e^{-\frac{1}{2} \vert w \vert ^{2}}}{(1+ \vert w \vert )^{\alpha}} \biggr) \biggl\Vert \frac{\widehat{\mu}_{\delta}(\cdot )}{(1+ \vert \cdot \vert )^{(1-t)\alpha}} \biggr\Vert _{L^{\infty}} . \end{aligned}$$

Thus, when $k\to \infty$, the integral $I_{\text{(8)}}$ will tend to zero. Now, the proof is finished. □

In the following, we will characterize the boundedness and compactness of positive Toeplitz operators from $F^{p}_{\alpha}$ to $F^{\infty}_{\alpha}$ and from $F^{\infty}_{\alpha}$ to $F^{p}_{\alpha}$ with $0< p<\infty$. Now, we state the main results as follows.

Theorem 3.2

Let $0< p<\infty$, and $\mu \geq 0$. Then,

• (1) $T^{\alpha}_{\mu}:F^{p}_{\alpha}\to F^{\infty}_{\alpha}$ is bounded if and only if μ is a p-Fock–Carleson measure. Furthermore,

  $$ \bigl\Vert T^{\alpha}_{\mu} \bigr\Vert _{F^{p}_{\alpha}\to F^{\infty}_{\alpha}} \simeq \Vert \mu \Vert _{p}. $$

• (2) $T^{\alpha}_{\mu}:F^{p}_{\alpha}\to F^{\infty}_{\alpha}$ is compact if and only if μ is a vanishing p-Fock–Carleson measure.

Proof

(1) First, we assume that $T^{\alpha}_{\mu}:F^{p}_{\alpha}\to F^{\infty}_{\alpha}$ is bounded. From the proof in [2, Theorem 3.3], it follows that

$$\begin{aligned} \bigl\vert \widetilde{\mu}^{\alpha}_{2}(z) \bigr\vert \lesssim &\bigl(1+ \vert z \vert \bigr)^{\alpha - \frac{\alpha}{p}} \biggl\vert T^{\alpha}_{\mu} \biggl( \frac{k_{2,z}^{\alpha}}{ \Vert k_{2,z}^{\alpha} \Vert _{F^{p}_{\alpha}}} \biggr) (z) \biggr\vert \frac{e^{-\frac{1}{2} \vert z \vert ^{2}}}{(1+ \vert z \vert )^{\alpha}} \\ \lesssim &\bigl(1+ \vert z \vert \bigr)^{\alpha -\frac{\alpha}{p}} \bigl\Vert T^{\alpha}_{\mu} \bigr\Vert _{F^{p}_{ \alpha}\to F^{\infty}_{\alpha}} \biggl\Vert \frac{k_{2,z}^{\alpha}}{ \Vert k_{2,z}^{\alpha} \Vert _{F^{p}_{\alpha}}} \biggr\Vert _{F^{p}_{\alpha}}. \end{aligned}$$

(10)

Together with Theorem 2.2, μ is a p-Fock–Carleson measure. Furthermore,

$$ \Vert \mu \Vert _{p}\simeq \sup_{z\in \mathbb{C}^{n}} \bigl\vert \widetilde{\mu}^{\alpha}_{2}(z) \bigr\vert \bigl(1+ \vert z \vert \bigr)^{ \frac{\alpha}{p}-\alpha}\lesssim \bigl\Vert T^{\alpha}_{\mu} \bigr\Vert _{F^{p}_{\alpha} \to F^{\infty}_{\alpha}}. $$

On the other hand, if μ is a p-Fock–Carleson measure, then $\widehat{\mu}_{\delta}(z)(1+|z|)^{\frac{\alpha}{p}-\alpha}$ is bounded for $\delta >0$. Given any $f\in F^{p}_{\alpha}$, we only want to obtain that, when $\alpha >0$,

$$\begin{aligned}& \sup_{z\in \mathbb{C}^{n}} \bigl\vert T^{\alpha}_{\mu}f(z) \bigr\vert \frac{e^{-\frac{1}{2} \vert z \vert ^{2}}}{(1+ \vert z \vert )^{\alpha}} \\& \quad\lesssim \sup_{z\in \mathbb{C}^{n}} \frac{e^{-\frac{1}{2} \vert z \vert ^{2}}}{(1+ \vert z \vert )^{\alpha}} \int _{\mathbb{C}^{n}} \bigl\vert f_{\frac{\alpha}{2}}^{-}(w) \bigr\vert \bigl\vert \bigl(K_{z}^{\alpha} \bigr)_{\frac{\alpha}{2}}^{-}(w) \bigr\vert e^{- \vert w \vert ^{2}}\,d\mu (w) \\& \qquad {}+\sup_{z\in \mathbb{C}^{n}} \frac{e^{-\frac{1}{2} \vert z \vert ^{2}}}{(1+ \vert z \vert )^{\alpha}} \int _{\mathbb{C}^{n}} \bigl\vert f_{\frac{\alpha}{2}}^{+}(w) \bigr\vert \bigl\vert \bigl(K_{z}^{\alpha} \bigr)_{\frac{\alpha}{2}}^{+}(w) \bigr\vert \frac{e^{- \vert w \vert ^{2}}}{ \vert w \vert ^{\alpha}}\,d\mu (w)< \infty . \end{aligned}$$

(11)

Now, we pay attention to the integral $I_{\text{(11)}}$, because the first will be discussed in a similar way. For some suitable $r_{0}>0$, we divide the integral $I_{\text{(11)}}$ into

$$\begin{aligned} I_{\text{(11)}} =&\frac{e^{-\frac{1}{2} \vert z \vert ^{2}}}{(1+ \vert z \vert )^{\alpha}} \int _{ \vert w \vert \geq r_{0}} \bigl\vert f_{\frac{\alpha}{2}}^{+}(w) \bigr\vert \bigl\vert \bigl(K_{z}^{\alpha} \bigr)_{\frac{\alpha}{2}}^{+}(w) \bigr\vert \frac{e^{- \vert w \vert ^{2}}}{ \vert w \vert ^{\alpha}}\,d\mu (w) \end{aligned}$$

(12)

$$\begin{aligned} &{}+\frac{e^{-\frac{1}{2} \vert z \vert ^{2}}}{(1+ \vert z \vert )^{\alpha}} \int _{ \vert w \vert < r_{0}} \bigl\vert f_{\frac{\alpha}{2}}^{+}(w) \bigr\vert \bigl\vert \bigl(K_{z}^{\alpha} \bigr)_{\frac{\alpha}{2}}^{+}(w) \bigr\vert \frac{e^{- \vert w \vert ^{2}}}{ \vert w \vert ^{\alpha}}\,d\mu (w). \end{aligned}$$

(13)

If choosing an r-lattice $\{a_{k}\}_{k\geq 1}$, then we can see that, by [2, Lemma 2.3] and the upper pointwise estimate in Lemma 2.1,

$$\begin{aligned} I_{\text{(12)}} \lesssim & \frac{e^{-\frac{1}{2} \vert z \vert ^{2}}}{(1+ \vert z \vert )^{\alpha}} \int _{ \vert w \vert \geq r_{0}} \bigl\vert \bigl(K_{z}^{\alpha} \bigr)_{\frac{\alpha}{2}}^{+}(w) \bigr\vert \frac{ \vert f_{\frac{\alpha}{2}}^{+}(w) \vert e^{-\frac{1}{2} \vert w \vert ^{2}}}{(1+ \vert w \vert )^{\frac{\alpha}{p}}} \frac{e^{-\frac{1}{2} \vert w \vert ^{2}}}{(1+ \vert w \vert )^{\alpha}} \frac{\widehat{\mu}^{\alpha}_{\delta}(w)\,dv(w)}{(1+ \vert w \vert )^{-\frac{\alpha}{p}}} \\ \lesssim & \Vert f \Vert _{F^{p}_{\alpha}}\sum _{k: \vert a_{k} \vert \geq r+r_{0}} \int _{B(a_{k},r)} \frac{e^{-\frac{1}{2} \vert z \vert ^{2}}}{(1+ \vert z \vert )^{\alpha}} \bigl\vert \bigl(K_{z}^{ \alpha} \bigr)_{\frac{\alpha}{2}}^{+}(w) \bigr\vert e^{-\frac{1}{2} \vert w \vert ^{2}} \frac{\widehat{\mu}^{\alpha}_{\delta}(w)\,dv(w)}{(1+ \vert w \vert )^{\alpha -\frac{\alpha}{p}}} \\ \lesssim & \Vert f \Vert _{F^{p}_{\alpha}} \biggl\Vert \frac{\widehat{\mu}_{\delta}(\cdot )}{(1+ \vert \cdot \vert )^{\alpha -\frac{\alpha}{p}}} \biggr\Vert _{L^{\infty}}\sum _{k: \vert a_{k} \vert \geq r+r_{0}}\sup_{w\in B(a_{k},r)} \frac{ \vert (K_{z}^{\alpha} )_{\frac{\alpha}{2}}^{+}(w) \vert }{(1+ \vert z \vert )^{\alpha}} \frac{e^{-\frac{1}{2} \vert w \vert ^{2}}}{e^{\frac{1}{2} \vert z \vert ^{2}}} \\ \lesssim & \Vert f \Vert _{F^{p}_{\alpha}} \biggl\Vert \frac{\widehat{\mu}_{\delta}(\cdot )}{(1+ \vert \cdot \vert )^{\alpha -\frac{\alpha}{p}}} \biggr\Vert _{L^{\infty}}\sup_{ \vert w \vert \geq r_{0}} \int _{\mathbb{C}^{n}}e^{- \frac{1}{2} \vert z \vert ^{2}} \bigl\vert K^{\alpha}_{w}(z) \bigr\vert e^{-\frac{1}{2} \vert w \vert ^{2}} \frac{dv(z)}{(1+ \vert z \vert )^{\alpha}}. \end{aligned}$$

Next, the integral $I_{\text{(13)}}$ comes into play in a similar way as in the estimate $I_{\text{(8)}}$,

$$\begin{aligned} I_{\text{(13)}} \lesssim & \frac{e^{-\frac{1}{2} \vert z \vert ^{2}}}{(1+ \vert z \vert )^{\alpha}} \int _{ \vert w \vert < r_{0}} \frac{ \vert (K_{z}^{\alpha} )_{\frac{\alpha}{2}}^{+}(w) \vert }{ \vert w \vert ^{\frac{\alpha}{2}}e^{\frac{1}{2} \vert w \vert ^{2}}} \biggl(\sup _{ \vert w \vert < 2r_{0}} \frac{ \vert f(w) \vert e^{-\frac{1}{2} \vert w \vert ^{2}}}{(1+ \vert w \vert )^{\frac{\alpha}{p}}} \biggr)\frac{d\mu (w)}{(1+ \vert w \vert )^{\alpha -\frac{\alpha}{p}}} \\ \lesssim & \Vert f \Vert _{F^{p}_{\alpha}} \biggl\Vert \frac{\widehat{\mu}_{\delta}(\cdot )}{(1+ \vert \cdot \vert )^{\alpha -\frac{\alpha}{p}}} \biggr\Vert _{L^{\infty}} \biggl(\sum _{k>\frac{\alpha}{2}} \frac{k^{\frac{\alpha}{2}}}{k!} \vert z \vert ^{k} \biggr) \frac{e^{-\frac{1}{2} \vert z \vert ^{2}}}{(1+ \vert z \vert )^{\alpha}}. \end{aligned}$$

All in all, $T^{\alpha}_{\mu}$ is bounded from $F^{p}_{\alpha}$ to $F^{\infty}_{\alpha}$ and $\|T^{\alpha}_{\mu}\|_{F^{p}_{\alpha}\to F^{\infty}_{\alpha}}\simeq \|\mu \|_{p}$.

(2) We suppose μ is a vanishing p-Fock–Carleson measure. By Theorem 2.3,

$$ \widehat{\mu}_{\delta}(z) \bigl(1+ \vert z \vert \bigr)^{\frac{\alpha}{p}-\alpha}\to 0, \quad \text{as } \vert z \vert \to \infty . $$

As $T^{\alpha}_{\mu _{R}}$ is compact from $F^{p}_{\alpha}$ to $F^{\infty}_{\alpha}$ by Lemma 3.1, if we note that $\mu -\mu _{R}\geq 0$, $T^{\alpha}_{\mu -\mu _{R}}$ is also bounded from $F^{p}_{\alpha}$ to $F^{\infty}_{\alpha}$. For $r>0$,

$$ \lim_{R\to \infty}\sup_{z\in \mathbb{C}^{n}} \widehat{(\mu - \mu _{R})}_{r}(z) \bigl(1+ \vert z \vert \bigr)^{\frac{\alpha}{p}-\alpha}=0. $$

Therefore, the condition (1) tells us that, when $R\to \infty$,

$$\begin{aligned} \bigl\Vert T^{\alpha}_{\mu}-T^{\alpha}_{\mu _{R}} \bigr\Vert _{F^{p}_{\varphi}\to F^{ \infty}_{\varphi}} =& \bigl\Vert T^{\alpha}_{\mu -\mu _{R}} \bigr\Vert _{F^{p}_{\varphi} \to F^{\infty}_{\varphi}}\simeq \Vert \mu -\mu _{R} \Vert _{p} \\ \simeq &\sup_{z\in \mathbb{C}^{n}}\widehat{(\mu -\mu _{R})}_{r}(z) \bigl(1+ \vert z \vert \bigr)^{ \frac{\alpha}{p}-\alpha}\to 0. \end{aligned}$$

Hence, we can see that $T^{\alpha}_{\mu}:F^{p}_{\alpha}\to F^{\infty}_{\alpha}$ is compact.

Conversely, we assume that $T^{\alpha}_{\mu}:F^{p}_{\alpha}\to F^{\infty}_{\alpha}$ is compact. Then, $\widehat{\mu}_{\delta}(z)(1+|z|)^{\frac{\alpha}{p}-\alpha}$ is bounded for $\delta >0$. Obviously, $\{ \frac{k^{\alpha}_{2,z}}{\|k^{\alpha}_{2,z}\|_{F^{p}_{\alpha}}}: z\in \mathbb{C}^{n} \}$ is bounded in $F^{p}_{\alpha}$. Therefore, $\{T^{\alpha}_{\mu} ( \frac{k^{\alpha}_{2,z}}{\|k^{\alpha}_{2,z}\|_{F^{p}_{\alpha}}} ): z\in \mathbb{C}^{n} \}$ is relatively compact in $F^{\infty}_{\alpha}$. For any sequence $\{z_{j}\}_{j\geq 1}$ with $|z_{j}|\to \infty$, there exists a subsequence of $\{T^{\alpha}_{\mu} ( \frac{k^{\alpha}_{2,z_{j}}}{\|k^{\alpha}_{2,z_{j}}\|_{F^{p}_{\alpha}}} ) \}_{j\geq 1}$ converging to some h in $F^{\infty}_{\alpha}$. Without loss of generality, we may assume that

$$ \lim_{ \vert z_{j} \vert \to \infty} \biggl\Vert T^{\alpha}_{\mu } \biggl( \frac{k^{\alpha}_{2,z_{j}}}{ \Vert k^{\alpha}_{2,z_{j}} \Vert _{F^{p}_{\alpha}}} \biggr)-h \biggr\Vert _{F^{\infty}_{\alpha}}=0. $$

Our goal is to show $h=0$. We will only prove the case $\alpha >0$. For any $w\in \mathbb{C}^{n}$, the definition of a Toeplitz operator implies that

$$\begin{aligned}& \frac{e^{-\frac{1}{2} \vert w \vert ^{2}}}{(1+ \vert w \vert )^{\alpha}} \biggl\vert T^{\alpha}_{ \mu} \biggl( \frac{k^{\alpha}_{2,z_{j}}}{ \Vert k^{\alpha}_{2,z_{j}} \Vert _{F^{p}_{\alpha}}} \biggr) (w) \biggr\vert \\& \quad\lesssim \frac{e^{-\frac{1}{2} \vert w \vert ^{2}}}{(1+ \vert w \vert )^{\alpha}} \int _{ \mathbb{C}^{n}} \bigl\vert \bigl(K_{w}^{\alpha} \bigr)_{\frac{\alpha}{2}}^{-}( \xi ) \bigr\vert \frac{ \vert (k^{\alpha}_{2,z_{j}})_{\frac{\alpha}{2}}^{-}(\xi ) \vert }{ \Vert k^{\alpha}_{2,z_{j}} \Vert _{F^{p}_{\alpha}}}e^{- \vert \xi \vert ^{2}}\,d\mu (\xi ) \\& \qquad {}+\frac{e^{-\frac{1}{2} \vert w \vert ^{2}}}{(1+ \vert w \vert )^{\alpha}} \int _{\mathbb{C}^{n}} \bigl\vert \bigl(K_{w}^{\alpha} \bigr)_{\frac{\alpha}{2}}^{+}(\xi ) \bigr\vert \frac{ \vert (k^{\alpha}_{2,z_{j}})_{\frac{\alpha}{2}}^{+}(\xi ) \vert }{ \Vert k^{\alpha}_{2,z_{j}} \Vert _{F^{p}_{\alpha}}} \frac{e^{- \vert \xi \vert ^{2}}}{ \vert \xi \vert ^{\alpha}}\,d\mu (\xi ). \end{aligned}$$

(14)

Now, we pay attention to the integral $I_{\text{(14)}}$ and we divide it into two integrals for some suitable $R>0$

$$\begin{aligned} I_{\text{(14)}} \lesssim & \frac{e^{-\frac{1}{2} \vert w \vert ^{2}}}{(1+ \vert w \vert )^{\alpha}} \int _{ \vert \xi \vert > R} \bigl\vert \bigl(K_{w}^{\alpha} \bigr)_{\frac{\alpha}{2}}^{+}(\xi ) \bigr\vert \frac{ \vert (k^{\alpha}_{2,z_{j}})_{\frac{\alpha}{2}}^{+}(\xi ) \vert }{ \Vert k^{\alpha}_{2,z_{j}} \Vert _{F^{p}_{\alpha}}} \frac{e^{- \vert \xi \vert ^{2}}}{ \vert \xi \vert ^{\alpha}}\,d\mu (\xi ) \end{aligned}$$

(15)

$$\begin{aligned} &{}+\frac{e^{-\frac{1}{2} \vert w \vert ^{2}}}{(1+ \vert w \vert )^{\alpha}} \int _{ \vert \xi \vert \leq R} \bigl\vert \bigl(K_{w}^{\alpha} \bigr)_{\frac{\alpha}{2}}^{+}( \xi ) \bigr\vert \frac{ \vert (k^{\alpha}_{2,z_{j}})_{\frac{\alpha}{2}}^{+}(\xi ) \vert }{ \Vert k^{\alpha}_{2,z_{j}} \Vert _{F^{p}_{\alpha}}} \frac{e^{- \vert \xi \vert ^{2}}}{ \vert \xi \vert ^{\alpha}}\,d\mu (\xi ). \end{aligned}$$

(16)

First, the integral $I_{\text{(15)}}$ is estimated by, after denoting by $\{a_{k}\}_{k\geq 1}$ an r-lattice and combing with [4, Proposition 3.2],

$$\begin{aligned} I_{\text{(15)}} \lesssim & \frac{e^{-\frac{1}{2} \vert w \vert ^{2}}}{(1+ \vert w \vert )^{\alpha}} \int _{ \vert \xi \vert > R} \bigl\vert \bigl(K_{w}^{\alpha} \bigr)_{\frac{\alpha}{2}}^{+}(\xi ) \bigr\vert \frac{ \vert (k^{\alpha}_{2,z_{j}})_{\frac{\alpha}{2}}^{+}(\xi ) \vert }{ \Vert k^{\alpha}_{2,z_{j}} \Vert _{F^{p}_{\alpha}}} \frac{e^{- \vert \xi \vert ^{2}}}{(1+ \vert \xi \vert )^{\frac{\alpha}{p}}} \frac{\widehat{\mu}_{\delta}(\xi )\,dv(\xi )}{(1+ \vert \xi \vert )^{\alpha -\frac{\alpha}{p}}} \\ \lesssim & \biggl\Vert \frac{\widehat{\mu}_{\delta}(\cdot )}{(1+ \vert \cdot \vert )^{\alpha -\frac{\alpha}{p}}} \biggr\Vert _{L^{\infty}} \sum_{k: \vert a_{k} \vert > R+r} \biggl[ \biggl(\sup_{\xi \in B(a_{k},r)} \bigl\vert \bigl(K_{w}^{\alpha} \bigr)_{\frac{\alpha}{2}}^{+}( \xi ) \bigr\vert e^{-\frac{1}{2} \vert \xi \vert ^{2}} \frac{e^{-\frac{1}{2} \vert w \vert ^{2}}}{(1+ \vert w \vert )^{\alpha}} \biggr) \\ &{}\times \int _{B(a_{k},r)} \frac{(1+ \vert \xi \vert )^{(\frac{1}{2}-\frac{1}{p})\alpha}}{(1+ \vert z_{j} \vert )^{(\frac{1}{2}-\frac{1}{p})\alpha}} \bigl\vert k^{\alpha}_{2,z_{j}}(\xi ) \bigr\vert \frac{e^{-\frac{1}{2} \vert \xi \vert ^{2}}}{ \vert \xi \vert ^{\frac{\alpha}{2}}}\,dv( \xi ) \biggr] \\ \lesssim & \biggl\Vert \frac{\widehat{\mu}_{\delta}(\cdot )}{(1+ \vert \cdot \vert )^{\alpha -\frac{\alpha}{p}}} \biggr\Vert _{L^{\infty}} \biggl(\sup_{ \vert \xi \vert > R} \int _{\mathbb{C}^{n}}e^{- \frac{1}{2} \vert w \vert ^{2}} \bigl\vert \bigl(K_{\xi}^{\alpha} \bigr)_{ \frac{\alpha}{2}}^{+}(w) \bigr\vert e^{-\frac{1}{2} \vert \xi \vert ^{2}} \frac{dv(w)}{(1+ \vert w \vert )^{\alpha}} \biggr) \\ &{}\times \sum_{k: \vert a_{k} \vert > R+r} \int _{B(a_{k},r)} \frac{(1+ \vert \xi \vert )^{(\frac{1}{2}-\frac{1}{p})\alpha}}{(1+ \vert z_{j} \vert )^{(\frac{1}{2}-\frac{1}{p})\alpha}} \bigl\vert k^{\alpha}_{2,z_{j}}(\xi ) \bigr\vert \frac{e^{-\frac{1}{2} \vert \xi \vert ^{2}}}{ \vert \xi \vert ^{\frac{\alpha}{2}}}\,dv( \xi ) \\ \lesssim & \biggl\Vert \frac{\widehat{\mu}_{\delta}(\cdot )}{(1+ \vert \cdot \vert )^{\alpha -\frac{\alpha}{p}}} \biggr\Vert _{L^{\infty}} \int _{ \vert \xi \vert > R}\bigl(1+ \vert z_{j}-\xi \vert \bigr)^{ \vert \frac{1}{2}-\frac{1}{p} \vert \vert \alpha \vert }e^{-\frac{1}{8} \vert z_{j}-\xi \vert ^{2}}\,dv( \xi ). \end{aligned}$$

This implies the integral $I_{\text{(15)}}$ will tend to zero as $R\to \infty$. Next, the integral $I_{\text{(16)}}$ will be divided into two subcases.

$$\begin{aligned} I_{\text{(16)}} \lesssim & \frac{e^{-\frac{1}{2} \vert w \vert ^{2}}}{(1+ \vert w \vert )^{\alpha}} \int _{ \vert \xi \vert < r_{0}} \bigl\vert \bigl(K_{w}^{\alpha} \bigr)_{\frac{\alpha}{2}}^{+}(\xi ) \bigr\vert \frac{ \vert (k^{\alpha}_{2,z_{j}})_{\frac{\alpha}{2}}^{+}(\xi ) \vert }{ \Vert k^{\alpha}_{2,z_{j}} \Vert _{F^{p}_{\alpha}}} \frac{e^{- \vert \xi \vert ^{2}}}{ \vert \xi \vert ^{\alpha}}\,d\mu (\xi ) \\ &{}+\frac{e^{-\frac{1}{2} \vert w \vert ^{2}}}{(1+ \vert w \vert )^{\alpha}} \int _{r_{0}\leq \vert \xi \vert \leq R} \bigl\vert \bigl(K_{w}^{\alpha} \bigr)_{\frac{\alpha}{2}}^{+}( \xi ) \bigr\vert \frac{ \vert (k^{\alpha}_{2,z_{j}})_{\frac{\alpha}{2}}^{+}(\xi ) \vert }{ \Vert k^{\alpha}_{2,z_{j}} \Vert _{F^{p}_{\alpha}}} \frac{e^{- \vert \xi \vert ^{2}}}{ \vert \xi \vert ^{\alpha}}\,d\mu (\xi ), \end{aligned}$$

(17)

for a small enough $0< r_{0}< R$. Since $\frac{k^{\alpha}_{2,z}}{\|k^{\alpha}_{2,z}\|_{F^{p}_{\alpha}}}\to 0$ uniformly on a compact subset of $\mathbb{C}^{n}$ when $|z|\to \infty$, it is easy to see that

$$\begin{aligned}& \lim_{ \vert z_{j} \vert \to \infty} \int _{r_{0}\leq \vert \xi \vert \leq R} \bigl\vert \bigl(K_{w}^{\alpha} \bigr)_{\frac{\alpha}{2}}^{+}(\xi ) \bigr\vert \frac{ \vert (k^{\alpha}_{2,z_{j}})_{\frac{\alpha}{2}}^{+}(\xi ) \vert }{ \Vert k^{\alpha}_{2,z_{j}} \Vert _{F^{p}_{\alpha}}} \frac{e^{- \vert \xi \vert ^{2}}}{ \vert \xi \vert ^{\alpha}}\,d\mu (\xi ) \\ & \quad \lesssim \biggl\Vert \frac{\widehat{\mu}_{\delta}(\cdot )}{(1+ \vert \cdot \vert )^{\alpha -\frac{\alpha}{p}}} \biggr\Vert _{L^{\infty}} \lim_{ \vert z_{j} \vert \to \infty} \int _{r_{0}\leq \vert \xi \vert \leq R} \bigl\vert \bigl(K_{z}^{\alpha} \bigr)_{\frac{\alpha}{2}}^{+}( \xi ) \bigr\vert \frac{ \vert (k^{\alpha}_{2,z_{j}})_{\frac{\alpha}{2}}^{+}(\xi ) \vert }{ \Vert k^{\alpha}_{2,z_{j}} \Vert _{F^{p}_{\alpha}}} \frac{e^{- \vert \xi \vert ^{2}}\,dv(\xi )}{(1+ \vert \xi \vert )^{\frac{\alpha}{p}}}=0. \end{aligned}$$

The other is, as follows, if we use a similar way as in the estimate $I_{\text{(8)}}$,

$$\begin{aligned} I_{\text{(17)}} \lesssim & \frac{e^{-\frac{1}{2} \vert w \vert ^{2}}}{(1+ \vert w \vert )^{\alpha}} \int _{ \vert \xi \vert < r_{0}} \frac{ \vert (K_{w}^{\alpha} )_{\frac{\alpha}{2}}^{+}(\xi ) \vert }{ \vert w \vert ^{\frac{1}{2}\alpha} \Vert k^{\alpha}_{2,z_{j}} \Vert _{F^{p}_{\alpha}}} \biggl(\sup _{ \vert \xi \vert \leq 2r_{0}} \frac{ \vert (k^{\alpha}_{2,z_{j}})_{\frac{\alpha}{2}}^{+}(\xi ) \vert }{(1+ \vert \xi \vert )^{\frac{\alpha}{p}}} \biggr) \frac{e^{- \vert \xi \vert ^{2}}\,d\mu (\xi )}{(1+ \vert \xi \vert )^{\alpha -\frac{\alpha}{p}}} \\ \lesssim &\frac{e^{-\frac{1}{2} \vert w \vert ^{2}}}{(1+ \vert w \vert )^{\alpha}} \int _{ \vert \xi \vert < r_{0}} \bigl( \bigl\vert \bigl(K_{w}^{\alpha} \bigr)^{+}_{\frac{\alpha}{2}}( \xi ) \bigr\vert \vert \xi \vert ^{-\frac{\alpha}{2}} \bigr) \frac{e^{-\frac{1}{2} \vert \xi \vert ^{2}}}{(1+ \vert \xi \vert )^{\alpha -\frac{\alpha}{p}}}\,d \mu (\xi ) \\ \lesssim & \biggl\Vert \frac{\widehat{\mu}_{\delta}(\cdot )}{(1+ \vert \cdot \vert )^{\alpha -\frac{\alpha}{p}}} \biggr\Vert _{L^{\infty}} \biggl(\sum_{k>\frac{\alpha}{2}} \frac{k^{\frac{\alpha}{2}}}{k!} \vert w \vert ^{k} \biggr) \frac{e^{-\frac{1}{2} \vert w \vert ^{2}}}{(1+ \vert w \vert )^{\alpha}} \\ \lesssim & \biggl\Vert \frac{\widehat{\mu}_{\delta}(\cdot )}{(1+ \vert \cdot \vert )^{\alpha -\frac{\alpha}{p}}} \biggr\Vert _{L^{\infty}} \frac{ \vert w \vert e^{ \vert w \vert }e^{-\frac{1}{2} \vert w \vert ^{2}}}{(1+ \vert w \vert )^{- \vert \alpha \vert }}. \end{aligned}$$

This implies that the integral $I_{\text{(17)}}$ tends to zero if we let $r_{0}$ go to zero simultaneously. All in all, we know that $\lim_{|z_{j}|\to \infty}T^{\alpha}_{\mu } ( \frac{k^{\alpha}_{2,z_{j}}}{\|k^{\alpha}_{2,z_{j}}\|_{F^{p}_{\alpha}}} )(w)=0$ for any $w\in \mathbb{C}^{n}$. Do not forget that $\lim_{|z_{j}|\to \infty}T^{\alpha}_{\mu } ( \frac{k^{\alpha}_{2,z_{j}}}{\|k^{\alpha}_{2,z_{j}}\|_{F^{p}_{\alpha}}} )(w)=h(w)$ for any $w\in \mathbb{C}^{n}$ under the assumption. Hence, $h=0$, that is

$$ \lim_{ \vert z_{j} \vert \to \infty} \biggl\Vert T^{\alpha}_{\mu } \biggl( \frac{k^{\alpha}_{2,z_{j}}}{ \Vert k^{\alpha}_{2,z_{j}} \Vert _{F^{p}_{\alpha}}} \biggr) \biggr\Vert _{F^{\infty}_{\varphi}}=0. $$

This, together with the estimate (10), yields that, when $|z_{j}|\to \infty$,

$$\begin{aligned} \bigl\vert \widetilde{\mu}^{\alpha}_{2}(z) \bigr\vert \lesssim \bigl(1+ \vert z \vert \bigr)^{\alpha - \frac{\alpha}{p}} \biggl\Vert T^{\alpha}_{\mu } \biggl( \frac{k^{\alpha}_{2,z_{j}}}{ \Vert k^{\alpha}_{2,z_{j}} \Vert _{F^{p}_{\alpha}}} \biggr) \biggr\Vert _{F^{\infty}_{\varphi}}\to 0. \end{aligned}$$

Hence, we can conclude that μ is a vanishing p-Fock–Carleson measure. □

Theorem 3.3

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

• (1) $T^{\alpha}_{\mu}:F^{\infty}_{\alpha}\to F^{p}_{\alpha}$ is bounded;

• (2) $T^{\alpha}_{\mu}:F^{\infty}_{\alpha}\to F^{p}_{\alpha}$ is compact;

• (3) $(1+|z|)^{\alpha -\frac{\alpha}{p}}\widetilde{\mu}^{\alpha}_{t}(z)\in L^{p}$ for some or any $t>0$;

• (4) $(1+|z|)^{\alpha -\frac{\alpha}{p}}\widehat{\mu}_{\delta}(z)\in L^{p}$ for some or any $\delta >0$;

• (5) The sequence $\{(1+|a_{k}|)^{\alpha -\frac{\alpha}{p}}\widehat{\mu}_{r}(a_{k})\}_{k=1}^{ \infty}\in l^{p}$ for some or any r-lattice $\{a_{k}\}_{k=1}^{\infty}$.

Furthermore,

$$\begin{aligned} \bigl\Vert T^{\alpha}_{\mu} \bigr\Vert _{F^{p}_{\alpha}\to F^{q}_{\alpha}} \simeq & \bigl\Vert \bigl(1+ \vert \cdot \vert \bigr)^{\alpha -\frac{\alpha}{p}} \widetilde{\mu}^{\alpha}_{t} \bigr\Vert _{L^{p}} \simeq \bigl\Vert \bigl(1+ \vert \cdot \vert \bigr)^{\alpha -\frac{\alpha}{p}} \widehat{\mu}_{ \delta} \bigr\Vert _{L^{p}} \\ \simeq & \bigl\Vert \bigl\{ \bigl(1+ \vert a_{k} \vert \bigr)^{\alpha -\frac{\alpha}{p}} \widehat{\mu}_{r}(a_{k}) \bigr\} _{k}^{\infty} \bigr\Vert _{l^{p}}. \end{aligned}$$

Proof

In view of [2, Lemma 2.4], the statements (3), (4), and (5) are equivalent.

(1) ⇒ (5): Suppose that $T^{\alpha}_{\mu}$ is bounded from $F^{\infty}_{\alpha}$ to $F^{p}_{\alpha}$. Let $\{a_{k}\}_{k=1}^{\infty}$ be an r-lattice and $\{\lambda _{k}\}_{k\geq 1}\in l^{\infty}$, and set

$$ f(z)=\sum_{k=1}^{\infty}\lambda _{k} \frac{k_{2,a_{k}}^{\alpha}(z)}{ \Vert k_{2,a_{k}}^{\alpha} \Vert _{F^{\infty}_{\alpha}}}= \sum_{k=1}^{\infty} \lambda _{k} \frac{k_{2,a_{k}}^{\alpha}(z)}{(1+ \vert a_{k} \vert )^{-\frac{\alpha}{2}}}, \quad z\in \mathbb{C}^{n}. $$

We can claim that

$$ f\in F^{\infty}_{\alpha }\quad \text{and}\quad \Vert f \Vert _{F^{\infty}_{ \alpha}}\lesssim \sup_{k}\{\lambda _{k}\}. $$

To that end, by the estimate (3) in Lemma 2.1, we can follow the norm $\|k_{2,a_{k}}^{\alpha}\|_{F^{\infty}_{\alpha}}$ from that

$$\begin{aligned} \frac{ \vert k_{2,a_{k}}^{\alpha}(z) \vert e^{-\frac{1}{2} \vert z \vert ^{2}}}{(1+ \vert z \vert )^{\alpha}} =\frac{ \vert K_{a_{k}}^{\alpha}(z) \vert }{\sqrt{K_{a_{k}}^{\alpha}(a_{k})}} \frac{e^{-\frac{1}{2} \vert z \vert ^{2}}}{(1+ \vert z \vert )^{\alpha}} \lesssim \frac{ \Vert K_{a_{k}}^{\alpha} \Vert _{F^{\infty}_{\alpha}}e^{-\frac{1}{2} \vert a_{k} \vert ^{2}}}{(1+ \vert a_{k} \vert )^{\frac{\alpha}{2}}}. \end{aligned}$$

Hence, we have $T^{\alpha}_{\mu }f\in F^{p}_{\alpha}$ by the condition. Khinchine’s inequality and Fubini’s Theorem show that

$$\begin{aligned}& \int _{\mathbb{C}^{n}} \Biggl(\sum_{k=1}^{\infty} \biggl\vert \lambda _{k} \frac{T^{\alpha}_{\mu}(k_{2,a_{k}}^{\alpha})(z)}{(1+ \vert a_{k} \vert )^{-\frac{\alpha}{2}}} \biggr\vert ^{2} \Biggr)^{\frac{p}{2}} \frac{e^{-\frac{p}{2} \vert z \vert ^{2}}}{(1+ \vert z \vert )^{\alpha}}\,dv(z) \\& \quad\lesssim \int _{\mathbb{C}^{n}} \int _{0}^{1} \Biggl\vert \sum _{k=1}^{ \infty}\Psi _{k}(t)\lambda _{j} \frac{T^{\alpha}_{\mu}(k_{2,a_{k}}^{\alpha})(z)}{(1+ \vert a_{k} \vert )^{-\frac{\alpha}{2}}} \Biggr\vert ^{p}\,dt \frac{e^{-\frac{p}{2} \vert z \vert ^{2}}}{(1+ \vert z \vert )^{\alpha}}\,dv(z) \\& \quad\lesssim \int _{0}^{1} \Biggl\Vert T^{\alpha}_{\mu} \Biggl(\sum_{k=1}^{ \infty} \Psi _{k}(t)\lambda _{j} \frac{k_{2,a_{k}}^{\alpha}(z)}{(1+ \vert a_{k} \vert )^{-\frac{\alpha}{2}}} \Biggr) \Biggr\Vert ^{p}_{F^{p}_{\alpha}}\,dt, \end{aligned}$$

where $\Psi _{k}$ is the kth Rademacher function on $[0,1]$. By the bounedness of $T^{\alpha}_{\mu}$,

$$ \int _{0}^{1} \Biggl\Vert T^{\alpha}_{\mu} \Biggl(\sum_{k=1}^{\infty} \Psi _{k}(t) \lambda _{j} \frac{k_{2,a_{k}}^{\alpha}(z)}{(1+ \vert a_{k} \vert )^{-\frac{\alpha}{2}}} \Biggr) \Biggr\Vert ^{p}_{F^{p}_{\alpha}}\,dt \lesssim \bigl\Vert T^{\alpha}_{\mu} \bigr\Vert ^{p}_{F^{\infty}_{\alpha}\to F^{p}_{\alpha}} \sup_{k\geq 1} \vert \lambda _{k} \vert ^{p}. $$

Using the property of an r-lattice, we will have the estimate in other directions,

$$\begin{aligned}& \int _{\mathbb{C}^{n}} \Biggl(\sum_{k=1}^{\infty} \biggl\vert \lambda _{k} \frac{T^{\alpha}_{\mu}(k_{2,a_{k}}^{\alpha})(z)}{(1+ \vert a_{k} \vert )^{-\frac{\alpha}{2}}} \biggr\vert ^{2} \Biggr)^{\frac{p}{2}} \frac{e^{-\frac{p}{2} \vert z \vert ^{2}}}{(1+ \vert z \vert )^{\alpha}}\,dv(z) \\& \quad \gtrsim \sum_{j=1}^{\infty} \int _{B(a_{j},r)} \Biggl(\sum_{k=1}^{ \infty} \biggl\vert \lambda _{k} \frac{T^{\alpha}_{\mu}(k_{2,a_{k}}^{\alpha})(z)}{(1+ \vert a_{k} \vert )^{-\frac{\alpha}{2}}} \biggr\vert ^{2} \Biggr)^{\frac{p}{2}} \frac{e^{-\frac{p}{2} \vert z \vert ^{2}}}{(1+ \vert z \vert )^{\alpha}}\,dv(z) \\& \quad \gtrsim \sum_{j=1}^{\infty} \int _{B(a_{j},r)} \biggl\vert \lambda _{j} \frac{T^{\alpha}_{\mu}(k_{2,a_{j}}^{\alpha})(z)}{(1+ \vert a_{j} \vert )^{-\frac{\alpha}{2}}} \biggr\vert ^{p} \frac{e^{-\frac{p}{2} \vert z \vert ^{2}}}{(1+ \vert z \vert )^{\alpha}}\,dv(z). \end{aligned}$$

By the upper pointwise estimate in Lemma 2.1, it is easy to see that

$$ \biggl\vert \frac{T^{\alpha}_{\mu}(k_{2,a_{j}}^{\alpha})(a_{j})}{(1+ \vert a_{j} \vert )^{-\frac{\alpha}{2}}} \biggr\vert ^{p} \frac{e^{-\frac{p}{2} \vert a_{j} \vert ^{2}}}{(1+ \vert a_{j} \vert )^{\alpha}} \lesssim \int _{B(a_{j},r)} \biggl\vert \lambda _{j} \frac{T^{\alpha}_{\mu}(k_{2,a_{j}}^{\alpha})(z)}{(1+ \vert a_{j} \vert )^{-\frac{\alpha}{2}}} \biggr\vert ^{p}\frac{e^{-\frac{p}{2} \vert z \vert ^{2}}}{(1+ \vert z \vert )^{\alpha}}\,dv(z). $$

The above analysis implies that

$$\begin{aligned} \int _{\mathbb{C}^{n}} \Biggl(\sum_{k=1}^{\infty} \biggl\vert \lambda _{k} \frac{T^{\alpha}_{\mu}(k_{2,a_{k}}^{\alpha})(z)}{(1+ \vert a_{k} \vert )^{-\frac{\alpha}{2}}} \biggr\vert ^{2} \Biggr)^{\frac{p}{2}} \frac{e^{-\frac{p}{2} \vert z \vert ^{2}}\,dv(z)}{(1+ \vert z \vert )^{\alpha}} \gtrsim \sum _{j=1}^{ \infty} \biggl\vert \lambda _{j} \frac{T^{\alpha}_{\mu}(k_{2,a_{j}}^{\alpha})(a_{j})}{(1+ \vert a_{j} \vert )^{-\frac{\alpha}{2}}} \biggr\vert ^{p} \frac{e^{-\frac{p}{2} \vert a_{j} \vert ^{2}}}{(1+ \vert a_{j} \vert )^{\alpha}}. \end{aligned}$$

Using the definition of the Toeplitz operator $T^{\alpha}_{\mu}$, and similar calculations as in [2, Page 42], we can have the details as follows, when $\alpha \leq 0$,

$$\begin{aligned}& \sum_{j=1}^{\infty} \biggl\vert \lambda _{j} \frac{T^{\alpha}_{\mu}(k_{2,a_{j}}^{\alpha})(a_{j})}{(1+ \vert a_{j} \vert )^{-\frac{\alpha}{2}}} \biggr\vert ^{p} \frac{e^{-\frac{p}{2} \vert a_{j} \vert ^{2}}}{(1+ \vert a_{j} \vert )^{\alpha}} \\& \quad \gtrsim \sum_{j=1}^{\infty} \frac{ \vert \lambda _{j} \vert ^{p}}{(1+ \vert a_{j} \vert )^{\alpha -p\alpha}} \biggl( \int _{B(a_{j},r)} \bigl\vert k_{2,a_{j}}^{ \alpha}(w) \bigr\vert ^{2}e^{- \vert w \vert ^{2}}\frac{d\mu (w)}{ \vert w \vert ^{\alpha}} \biggr)^{p} \\& \quad \gtrsim \sum_{j=1}^{\infty} \frac{ \vert \lambda _{j} \vert ^{p}}{(1+ \vert a_{j} \vert )^{\alpha -p\alpha}} \biggl( \int _{B(a_{j},r)} \frac{ \vert w \vert ^{-\alpha}}{(1+ \vert a_{j} \vert )^{-\alpha}}\,d\mu (w) \biggr)^{p} \\& \quad \gtrsim \sum_{j=1}^{\infty} \vert \lambda _{j} \vert ^{p} \frac{\mu (B(a_{j},r))^{p}}{(1+ \vert a_{j} \vert )^{\alpha -p\alpha}}. \end{aligned}$$

In the case $\alpha >0$, because $\max_{z\in \mathbb{C}^{n}}\{1,|z|^{\alpha}\}\lesssim (1+|z|)^{ \alpha}$ for any $z\in \mathbb{C}^{n}$,

$$\begin{aligned}& \sum_{j=1}^{\infty} \biggl\vert \lambda _{j} \frac{T^{\alpha}_{\mu}(k_{2,a_{j}}^{\alpha})(a_{j})}{(1+ \vert a_{j} \vert )^{-\frac{\alpha}{2}}} \biggr\vert ^{p} \frac{e^{-\frac{p}{2} \vert a_{j} \vert ^{2}}}{(1+ \vert a_{j} \vert )^{\alpha}} \\& \quad \gtrsim \sum_{j=1}^{\infty} \frac{ \vert \lambda _{j} \vert ^{p}}{(1+ \vert a_{j} \vert )^{\alpha -p\alpha}} \biggl( \int _{B(a_{j},r)} \bigl\vert \bigl(k_{2,a_{j}}^{\alpha} \bigr)^{-}_{\frac{\alpha}{2}}(w) \bigr\vert ^{2}e^{- \vert w \vert ^{2}}\,d \mu (w) \\& \qquad {}+ \int _{B(a_{j},r)} \bigl\vert \bigl(k_{2,a_{j}}^{\alpha} \bigr)^{+}_{ \frac{\alpha}{2}}(w) \bigr\vert ^{2}e^{- \vert w \vert ^{2}} \frac{d\mu (w)}{ \vert w \vert ^{\alpha}} \biggr)^{p} \\& \quad \gtrsim \sum_{j=1}^{\infty} \frac{ \vert \lambda _{j} \vert ^{p}}{(1+ \vert a_{j} \vert )^{\alpha -p\alpha}} \biggl( \int _{B(a_{j},r)} \bigl\vert k_{2,a_{j}}^{ \alpha}(w) \bigr\vert ^{2}e^{- \vert w \vert ^{2}}\frac{d\mu (w)}{(1+ \vert w \vert )^{\alpha}} \biggr)^{p} \\& \quad \gtrsim \sum_{j=1}^{\infty} \frac{ \vert \lambda _{j} \vert ^{p}}{(1+ \vert a_{j} \vert )^{\alpha -p\alpha}} \biggl( \int _{B(a_{j},r)} \frac{(1+ \vert a_{j} \vert )^{\alpha}}{(1+ \vert w \vert )^{\alpha}}\,d\mu (w) \biggr)^{p} \\& \quad \gtrsim \sum_{j=1}^{\infty} \vert \lambda _{j} \vert ^{p} \frac{\mu (B(a_{j},r))^{p}}{(1+ \vert a_{j} \vert )^{\alpha -p\alpha}}. \end{aligned}$$

If we set $\beta _{j}=|\lambda _{j}|^{p}$, we then know that $\{\beta _{j}\}_{j=1}^{\infty}\in l^{\infty}$. Thus, we further have

$$ \sum_{j=1}^{\infty}\beta _{j} \frac{\mu (B(a_{j},r))^{p}}{(1+ \vert a_{j} \vert )^{\alpha -p\alpha}}\lesssim \bigl\Vert T^{ \alpha}_{\mu} \bigr\Vert ^{p}_{F^{\infty}_{\alpha}\to F^{p}_{\alpha}}\sup_{k \geq 1} \vert \lambda _{k} \vert ^{p}. $$

This implies that

$$ \biggl\Vert \biggl\{ \frac{\mu (B(a_{j},r))}{(1+ \vert a_{j} \vert )^{\frac{\alpha}{p}-\alpha}} \biggr\} _{j\geq 1} \biggr\Vert _{l^{p}}\lesssim \bigl\Vert T^{\alpha}_{\mu} \bigr\Vert _{F^{\infty}_{ \alpha}\to F^{p}_{\alpha}}. $$

In order to finish the proof, we only need to prove $(4)\Rightarrow (2)$ when $\alpha >0$. The basic inequality $(a+b)^{p}\leq 2^{p-1}(a^{p}+b^{p})$ when $p>1$, $(a+b)^{p}\leq a^{p}+b^{p}$ when $0< p\leq 1$ for arbitrary positive numbers a, b tells us that, for any $f\in F^{\infty}_{\alpha}$,

$$\begin{aligned}& \int _{\mathbb{C}^{n}} \bigl\vert T^{\alpha}_{\mu}f(z) \bigr\vert ^{p}e^{- \frac{p}{2} \vert z \vert ^{2}}\frac{dv(z)}{(1+ \vert z \vert )^{\alpha}} \\& \quad \lesssim \int _{\mathbb{C}^{n}} \biggl( \int _{\mathbb{C}^{n}} \bigl\vert f_{ \frac{\alpha}{2}}^{-}(w) \bigr\vert \bigl\vert \bigl(K_{z}^{\alpha} \bigr)_{ \frac{\alpha}{2}}^{-}(w) \bigr\vert e^{- \vert w \vert ^{2}}e^{-\frac{1}{2} \vert z \vert ^{2}}\,d \mu (w) \biggr)^{p}\frac{dv(z)}{(1+ \vert z \vert )^{\alpha}} \\& \qquad {}+ \int _{\mathbb{C}^{n}} \underbrace{ \biggl( \int _{\mathbb{C}^{n}} \bigl\vert f_{\frac{\alpha}{2}}^{+}(w) \bigr\vert \bigl\vert \bigl(K_{z}^{\alpha} \bigr)_{\frac{\alpha}{2}}^{+}(w) \bigr\vert e^{- \vert w \vert ^{2}}e^{-\frac{1}{2} \vert z \vert ^{2}} \frac{d\mu (w)}{ \vert w \vert ^{\alpha}} \biggr)^{p}}_{\mathrm{III}} \frac{dv(z)}{(1+ \vert z \vert )^{\alpha}}. \end{aligned}$$

(18)

Now, we pay attention to the integral $I_{\text{(18)}}$, because the first will be discussed in a similar way. For some suitable $r_{0}>0$, we use that basic inequality again to divide the integral III into

$$\begin{aligned} \mathrm{III} \lesssim & \biggl( \int _{ \vert w \vert \geq r_{0}} \bigl\vert \bigl(K_{z}^{\alpha} \bigr)_{\frac{\alpha}{2}}^{+}(w) \bigr\vert \bigl\vert f_{\frac{\alpha}{2}}^{+}(w) \bigr\vert e^{- \vert w \vert ^{2}}e^{-\frac{1}{2} \vert z \vert ^{2}} \frac{d\mu (w)}{ \vert w \vert ^{\alpha}} \biggr)^{p} \end{aligned}$$

(19)

$$\begin{aligned} &{}+ \biggl( \int _{ \vert w \vert < r_{0}} \bigl\vert \bigl(K_{z}^{\alpha} \bigr)_{ \frac{\alpha}{2}}^{+}(w) \bigr\vert \bigl\vert f_{\frac{\alpha}{2}}^{+}(w) \bigr\vert e^{- \vert w \vert ^{2}}e^{-\frac{1}{2} \vert z \vert ^{2}} \frac{d\mu (w)}{ \vert w \vert ^{\alpha}} \biggr)^{p}. \end{aligned}$$

(20)

After we put the integrals $I_{\text{(19)}}$ and $I_{\text{(20)}}$ into the integral $I_{\text{(18)}}$, we find that

$$\begin{aligned} I_{\text{(18)}} \lesssim & \int _{\mathbb{C}^{n}} \biggl( \int _{ \vert w \vert \geq r_{0}} \bigl\vert f_{\frac{\alpha}{2}}^{+}(w) \bigr\vert \frac{ \vert (K_{z}^{\alpha} )_{\frac{\alpha}{2}}^{+}(w) \vert }{e^{ \vert w \vert ^{2}}e^{\frac{1}{2} \vert z \vert ^{2}}} \frac{d\mu (w)}{ \vert w \vert ^{\alpha}} \biggr)^{p} \frac{dv(z)}{(1+ \vert z \vert )^{\alpha}} \end{aligned}$$

(21)

$$\begin{aligned} &{}+ \int _{\mathbb{C}^{n}} \biggl( \int _{ \vert w \vert < r_{0}} \bigl\vert f_{ \frac{\alpha}{2}}^{+}(w) \bigr\vert \frac{ \vert (K_{z}^{\alpha} )_{\frac{\alpha}{2}}^{+}(w) \vert }{e^{ \vert w \vert ^{2}}e^{\frac{1}{2} \vert z \vert ^{2}}} \frac{d\mu (w)}{ \vert w \vert ^{\alpha}} \biggr)^{p} \frac{dv(z)}{(1+ \vert z \vert )^{\alpha}}. \end{aligned}$$

(22)

If choosing an r-lattice $\{a_{k}\}_{k\geq 1}$, we can see that, when $0< p\leq 1$,

$$\begin{aligned} I_{\text{(19)}} \lesssim & \biggl[\sum_{k: \vert a_{k} \vert \geq r_{0}+r} \int _{B(a_{k},r)} \bigl\vert \bigl(K_{z}^{\alpha} \bigr)_{\frac{\alpha}{2}}^{+}(w) \bigr\vert \bigl\vert f_{\frac{\alpha}{2}}^{+}(w) \bigr\vert e^{-\frac{1}{2} \vert z \vert ^{2}} \frac{e^{- \vert w \vert ^{2}}\widehat{\mu}_{\delta}(w)}{(1+ \vert w \vert )^{\alpha}}\,dv(w) \biggr]^{p} \\ \lesssim &\sum_{k: \vert a_{k} \vert \geq r_{0}+r}\sup_{w\in B(a_{k},r)} \bigl\vert \bigl(K_{z}^{\alpha} \bigr)_{\frac{\alpha}{2}}^{+}(w) \bigr\vert ^{p} \bigl\vert f_{\frac{\alpha}{2}}^{+}(w) \bigr\vert ^{p}e^{-\frac{p}{2} \vert z \vert ^{2}}e^{-p \vert w \vert ^{2}} \frac{\widehat{\mu}^{p}_{\delta}(w)}{(1+ \vert w \vert )^{p\alpha}} \\ \lesssim & \Vert f \Vert ^{p}_{F^{\infty}_{\alpha}}\sum _{k: \vert a_{k} \vert \geq r_{0}+r} \frac{\widehat{\mu}^{p}_{\delta}(a_{k})}{(1+ \vert a_{k} \vert )^{-\alpha}} \sup_{w \in B(a_{k},r)} \bigl\vert \bigl(K_{z}^{\alpha} \bigr)_{\frac{\alpha}{2}}^{+}(w) \bigr\vert ^{p} \frac{e^{-\frac{p}{2} \vert z \vert ^{2}}e^{-\frac{p}{2} \vert w \vert ^{2}}}{(1+ \vert w \vert )^{\alpha}} \\ \lesssim & \Vert f \Vert ^{p}_{F^{\infty}_{\alpha}}\sum _{k: \vert a_{k} \vert \geq r_{0}+r} \frac{\widehat{\mu}^{p}_{\delta}(a_{k})}{(1+ \vert a_{k} \vert )^{-\alpha}} \int _{B(a_{k},r)} \frac{ \vert (K_{z}^{\alpha} )_{\frac{\alpha}{2}}^{+}(w) \vert ^{p}}{e^{\frac{p}{2} \vert z \vert ^{2}}e^{\frac{p}{2} \vert w \vert ^{2}}} \frac{dv(w)}{(1+ \vert w \vert )^{\alpha}} \\ \lesssim & \Vert f \Vert ^{p}_{F^{\infty}_{\alpha}} \int _{\mathbb{C}^{n}}e^{- \frac{p}{2} \vert z \vert ^{2}} \bigl\vert \bigl(K_{z}^{\alpha} \bigr)_{ \frac{\alpha}{2}}^{+}(w) \bigr\vert ^{p}e^{-\frac{p}{2} \vert w \vert ^{2}} \widehat{\mu}^{p}_{\delta}(w)\,dv(w). \end{aligned}$$

Next, Fubini’s theorem and the estimate (9) show us, in detail, that

$$\begin{aligned} I_{\text{(21)}} \lesssim & \Vert f \Vert ^{p}_{F^{\infty}_{\alpha}} \int _{ \mathbb{C}^{n}} \biggl( \int _{\mathbb{C}^{n}}e^{-\frac{p}{2} \vert z \vert ^{2}} \bigl\vert \bigl(K_{z}^{\alpha} \bigr)_{\frac{\alpha}{2}}^{+}(w) \bigr\vert ^{p}e^{- \frac{p}{2} \vert w \vert ^{2}}\widehat{\mu}^{p}_{\delta}(w)\,dv(w) \biggr) \frac{dv(z)}{(1+ \vert z \vert )^{\alpha}} \\ \lesssim & \Vert f \Vert ^{p}_{F^{\infty}_{\alpha}} \int _{\mathbb{C}^{n}} \biggl( \int _{\mathbb{C}^{n}}e^{-\frac{p}{2} \vert z \vert ^{2}} \bigl\vert \bigl(K_{w}^{ \alpha} \bigr)_{\frac{\alpha}{2}}^{+}(z) \bigr\vert ^{p} e^{-\frac{p}{2} \vert w \vert ^{2}} \frac{dv(z)}{(1+ \vert z \vert )^{\alpha}} \biggr) \widehat{\mu}^{p}_{\delta}(w)\,dv(w) \\ \lesssim & \Vert f \Vert ^{p}_{F^{\infty}_{\alpha}} \int _{\mathbb{C}^{n}} \frac{\widehat{\mu}^{p}_{\delta}(w)}{(1+ \vert w \vert )^{\alpha -p\alpha}}\,dv(w). \end{aligned}$$

Combining [2, Lemma 2.3], the estimate (9), and Hölder’s inequality, we continue to estimate the case $p>1$ as follows,

$$\begin{aligned} I_{\text{(19)}} \lesssim & \biggl( \int _{ \vert w \vert \geq r_{0}} \bigl\vert f_{ \frac{\alpha}{2}}^{+}(w) \bigr\vert ^{p} \bigl\vert \bigl(K_{z}^{\alpha} \bigr)_{\frac{\alpha}{2}}^{+}(w) \bigr\vert e^{-\frac{p+1}{2} \vert w \vert ^{2}}e^{- \frac{1}{2} \vert z \vert ^{2}} \widehat{\mu}^{p}_{\delta}(w) \frac{dv(w)}{(1+ \vert w \vert )^{\alpha}} \biggr) \\ &{}\times \biggl( \int _{\mathbb{C}^{n}} \bigl\vert \bigl(K_{z}^{\alpha} \bigr)_{\frac{\alpha}{2}}^{+}(w) \bigr\vert e^{-\frac{1}{2} \vert w \vert ^{2}}e^{- \frac{1}{2} \vert z \vert ^{2}} \frac{dv(w)}{(1+ \vert w \vert )^{\alpha}} \biggr)^{p-1} \\ \lesssim & \int _{ \vert w \vert \geq r_{0}} \frac{ \vert f_{\frac{\alpha}{2}}^{+}(w) \vert ^{p} e^{-\frac{p}{2} \vert w \vert ^{2}}}{(1+ \vert w \vert )^{p\alpha}} \bigl\vert \bigl(K_{z}^{\alpha} \bigr)_{\frac{\alpha}{2}}^{+}(w) \bigr\vert e^{-\frac{1}{2} \vert w \vert ^{2}}e^{-\frac{1}{2} \vert z \vert ^{2}} \frac{\widehat{\mu}^{p}_{\delta}(w)\,dv(w)}{(1+ \vert w \vert )^{\alpha -p\alpha}} \\ \lesssim & \Vert f \Vert ^{p}_{F^{\infty}_{\alpha}} \int _{\mathbb{C}^{n}}e^{- \frac{1}{2} \vert z \vert ^{2}} \bigl\vert \bigl(K_{z}^{\alpha} \bigr)_{ \frac{\alpha}{2}}^{+}(w) \bigr\vert e^{-\frac{1}{2} \vert w \vert ^{2}} \frac{\widehat{\mu}^{p}_{\delta}(w)}{(1+ \vert w \vert )^{\alpha -p\alpha}}\,dv(w). \end{aligned}$$

Again, Fubini’s theorem and the estimate (9) show us, in detail, that

$$\begin{aligned} I_{\text{(21)}} \lesssim & \Vert f \Vert ^{p}_{F^{\infty}_{\alpha}} \int _{ \mathbb{C}^{n}} \biggl( \int _{\mathbb{C}^{n}} \frac{ \vert (K_{z}^{\alpha} )_{\frac{\alpha}{2}}^{+}(w) \vert }{e^{\frac{1}{2} \vert z \vert ^{2}}e^{\frac{1}{2} \vert w \vert ^{2}}} \frac{\widehat{\mu}^{p}_{\delta}(w)\,dv(w)}{(1+ \vert w \vert )^{\alpha -p\alpha}} \biggr)\frac{dv(z)}{(1+ \vert z \vert )^{\alpha}} \\ \lesssim & \Vert f \Vert ^{p}_{F^{\infty}_{\alpha}} \int _{\mathbb{C}^{n}} \biggl( \int _{\mathbb{C}^{n}} \frac{ \vert (K_{z}^{\alpha} )_{\frac{\alpha}{2}}^{+}(w) \vert }{e^{\frac{1}{2} \vert z \vert ^{2}}e^{\frac{1}{2} \vert w \vert ^{2}}} \frac{dv(z)}{(1+ \vert z \vert )^{\alpha}} \biggr) \frac{\widehat{\mu}^{p}_{\delta}(w)}{(1+ \vert w \vert )^{\alpha -p\alpha}}\,dv(w) \\ \lesssim & \Vert f \Vert ^{p}_{F^{\infty}_{\alpha}} \int _{\mathbb{C}^{n}} \frac{\widehat{\mu}^{p}_{\delta}(w)}{(1+ \vert w \vert )^{\alpha -p\alpha}}\,dv(w). \end{aligned}$$

Then, the integral $I_{\text{(20)}}$ comes into that, after we use the estimate (4) twice,

$$\begin{aligned} I_{\text{(20)}} \lesssim & \biggl[ \int _{ \vert w \vert < r_{0}} \frac{ \vert (K_{z}^{\alpha} )_{\frac{\alpha}{2}}^{+}(w) \vert }{e^{\frac{1}{2} \vert z \vert ^{2}}e^{\frac{1}{2} \vert w \vert ^{2}}} \biggl(\sup _{ \vert w \vert < 2r_{0}} \frac{ \vert f(w) \vert e^{-\frac{1}{2} \vert w \vert ^{2}}}{(1+ \vert w \vert )^{\alpha}} \biggr) \frac{d\mu (w)}{ \vert w \vert ^{\frac{\alpha}{2}}} \biggr]^{p} \\ \lesssim & \Vert f \Vert ^{p}_{F^{\infty}_{\alpha}}\mu ^{p}\bigl(B(0,r_{0})\bigr)\sup_{ \vert w \vert < 2r_{0}} \bigl\vert K_{z}^{\alpha}(w) \bigr\vert ^{p} \frac{e^{-\frac{p}{2} \vert z \vert ^{2}}e^{-\frac{p}{2} \vert w \vert ^{2}}}{(1+ \vert w \vert )^{p\alpha}} \\ \lesssim & \Vert f \Vert ^{p}_{F^{\infty}_{\alpha}} \frac{\mu ^{p}(B(0,r_{0}))}{(1+ \vert w \vert )^{-\alpha}}\sup_{ \vert w \vert < 2r_{0}} \bigl\vert \bigl(K_{z}^{\alpha} \bigr)_{\frac{\alpha}{2}}^{+}(w) \bigr\vert ^{p} \frac{e^{-\frac{p}{2} \vert z \vert ^{2}}e^{-\frac{p}{2} \vert w \vert ^{2}}}{(1+ \vert w \vert )^{\alpha}} \\ \lesssim & \Vert f \Vert ^{p}_{F^{\infty}_{\alpha}} \int _{\mathbb{C}^{n}}e^{- \frac{p}{2} \vert z \vert ^{2}} \bigl\vert \bigl(K_{z}^{\alpha} \bigr)_{ \frac{\alpha}{2}}^{+}(w) \bigr\vert ^{p}e^{-\frac{p}{2} \vert w \vert ^{2}} \widehat{\mu}^{p}_{\delta}(w)\,dv(w). \end{aligned}$$

We can estimate the integral $I_{\text{(22)}}$ successfully using Fubini’s theorem like the integral $I_{\text{(21)}}$.

All in all, the analysis above tells us that $\|T^{\alpha}_{\mu}\|_{F^{\infty}_{\alpha}\to F^{p}_{\alpha}} \lesssim \|(1+|\cdot |)^{\alpha -\frac{\alpha}{p}}\widehat{\mu}_{ \delta}\|_{L^{p}}$. Finally, taking $\mu _{R}$ as in Lemma 3.1, then we have $\mu -\mu _{R}\geq 0$ and, moreover,

$$ \bigl\Vert T^{\alpha}_{\mu}-T^{\alpha}_{\mu _{R}} \bigr\Vert _{F^{\infty}_{\alpha}\to F^{p}_{ \alpha}}\lesssim \bigl\Vert \bigl(1+ \vert \cdot \vert \bigr)^{\alpha -\frac{\alpha}{p}} \widehat{(\mu -\mu _{R})}_{\delta} \bigr\Vert _{L^{p}}\to 0, $$

when $R\to \infty$. Lemma 3.1 shows us that $T_{\mu _{R}}$ is compact from $F^{\infty}_{\alpha}$ to $F^{p}_{\alpha}$, and so is $T_{\mu}$. The condition (2) holds. □

The following theorem about the infinity case is not merged into Theorem 3.2, and the reasons for this can be found in the Introduction section. However, its proof is similar to that in Theorem 3.2. We can, however, give the details for completeness.

Theorem 3.4

Let $\mu \geq 0$. Then,

• (1) $T^{\alpha}_{\mu}:F^{\infty}_{\alpha}\to F^{\infty}_{\alpha}$ is bounded if and only if μ is a 1-Fock–Carleson measure. Furthermore,

  $$ \bigl\Vert T^{\alpha}_{\mu} \bigr\Vert _{F^{\infty}_{\alpha}\to F^{\infty}_{\alpha}} \simeq \Vert \mu \Vert _{1}. $$

• (2) $T^{\alpha}_{\mu}:F^{\infty}_{\alpha}\to F^{\infty}_{\alpha}$ is compact if and only if μ is a vanishing 1-Fock–Carleson measure.

Proof

(1) Suppose μ is a 1-Fock–Carleson measure. It suffices to discuss

$$ \sup_{z\in \mathbb{C}^{n}} \underbrace{\frac{e^{-\frac{1}{2} \vert z \vert ^{2}}}{(1+ \vert z \vert )^{\alpha}} \int _{\mathbb{C}^{n}} \bigl\vert \bigl(K_{z}^{\alpha} \bigr)_{\frac{\alpha}{2}}^{+}(w) \bigr\vert \bigl\vert f_{\frac{\alpha}{2}}^{+}(w) \bigr\vert \frac{e^{- \vert w \vert ^{2}}}{ \vert w \vert ^{\alpha}}\,d\mu (w)}_{\mathrm{IV}} \lesssim \Vert f \Vert _{F^{\infty}_{\alpha}} \Vert \widehat{\mu}_{\delta} \Vert _{L^{ \infty}}, $$

when $\alpha >0$, for any $f\in F^{\infty}_{\alpha}$. Therefore, fix a small enough $r_{0}>0$, and we will divide the integral IV into two cases.

$$\begin{aligned} \mathrm{IV} =&\frac{e^{-\frac{1}{2} \vert z \vert ^{2}}}{(1+ \vert z \vert )^{\alpha}} \int _{ \vert w \vert \geq r_{0}} \bigl\vert \bigl(K_{z}^{\alpha} \bigr)_{\frac{\alpha}{2}}^{+}(w) \bigr\vert \bigl\vert f_{\frac{\alpha}{2}}^{+}(w) \bigr\vert \frac{e^{- \vert w \vert ^{2}}}{ \vert w \vert ^{\alpha}}\,d\mu (w) \end{aligned}$$

(23)

$$\begin{aligned} &{}+\frac{e^{-\frac{1}{2} \vert z \vert ^{2}}}{(1+ \vert z \vert )^{\alpha}} \int _{ \vert w \vert < r_{0}} \bigl\vert \bigl(K_{z}^{\alpha} \bigr)_{\frac{\alpha}{2}}^{+}(w) \bigr\vert \bigl\vert f_{\frac{\alpha}{2}}^{+}(w) \bigr\vert \frac{e^{- \vert w \vert ^{2}}}{ \vert w \vert ^{\alpha}}\,d\mu (w). \end{aligned}$$

(24)

First, by [2, Lemma 2.3] and the estimate (9), the integral $I_{\text{(23)}}$ is estimated by

$$\begin{aligned} I_{\text{(23)}} \lesssim & \frac{e^{-\frac{1}{2} \vert z \vert ^{2}}}{(1+ \vert z \vert )^{\alpha}} \int _{ \vert w \vert \geq r_{0}} \bigl\vert \bigl(K_{z}^{\alpha} \bigr)_{\frac{\alpha}{2}}^{+}(w) \bigr\vert \frac{ \vert f_{\frac{\alpha}{2}}^{+}(w) \vert e^{-\frac{1}{2} \vert w \vert ^{2}}}{(1+ \vert w \vert )^{\alpha}}e^{- \frac{1}{2} \vert w \vert ^{2}} \widehat{\mu}_{\delta}(w)\,dv(w) \\ \lesssim & \Vert f \Vert _{F^{\infty}_{\alpha}} \Vert \widehat{\mu}_{\delta} \Vert _{L^{ \infty}}\sum_{k: \vert a_{k} \vert >r+r_{0}} \sup_{w\in B(a_{k},r)} \bigl\vert \bigl(K_{z}^{ \alpha} \bigr)_{\frac{\alpha}{2}}^{+}(w) \bigr\vert e^{-\frac{1}{2} \vert w \vert ^{2}} \frac{e^{-\frac{1}{2} \vert z \vert ^{2}}}{(1+ \vert z \vert )^{\alpha}} \\ \lesssim & \Vert f \Vert _{F^{\infty}_{\alpha}} \Vert \widehat{\mu}_{\delta} \Vert _{L^{ \infty}}\sup_{w\in \mathbb{C}^{n}} \int _{\mathbb{C}^{n}}e^{- \frac{1}{2} \vert z \vert ^{2}} \bigl\vert K^{\alpha}_{w}(z) \bigr\vert e^{-\frac{1}{2} \vert w \vert ^{2}} \frac{dv(z)}{(1+ \vert z \vert )^{\alpha}}. \end{aligned}$$

Using a similar way as in the estimate $I_{\text{(8)}}$, the integral $I_{\text{(24)}}$ is estimated by

$$\begin{aligned} I_{\text{(24)}} \lesssim & \Vert \widehat{\mu}_{\delta} \Vert _{L^{\infty}} \biggl(\frac{e^{-\frac{1}{2} \vert z \vert ^{2}}}{(1+ \vert z \vert )^{\alpha}}\sum_{k> \frac{\alpha}{2}} \frac{k^{\frac{\alpha}{2}}}{k!} \vert z \vert ^{k} \biggr) \biggl( \sup _{ \vert w \vert \leq 2r_{0}} \frac{ \vert f(w) \vert e^{-\frac{1}{2} \vert w \vert ^{2}}}{(1+ \vert w \vert )^{\alpha}} \biggr) \\ \lesssim & \Vert \widehat{\mu}_{\delta} \Vert _{L^{\infty}} \frac{ \vert z \vert e^{ \vert z \vert }e^{-\frac{1}{2} \vert z \vert ^{2}}}{(1+ \vert z \vert )^{- \vert \alpha \vert }} \biggl( \sup_{ \vert w \vert \leq 2r_{0}} \frac{ \vert f_{k}(w) \vert e^{-\frac{1}{2} \vert w \vert ^{2}}}{(1+ \vert w \vert )^{\alpha}} \biggr). \end{aligned}$$

Therefore, our desired goal is obtained.

On the other hand, we assume that $T^{\alpha}_{\mu}:F^{\infty}_{\alpha}\to F^{\infty}_{\alpha}$ is bounded. Similar to the proof in [2, Theorem 3.3], we can calculate that, by the definitions of a Toeplitz operator and the Berezin transform, that

$$\begin{aligned} \bigl\vert \widetilde{\mu}^{\alpha}_{2}(z) \bigr\vert \lesssim &\bigl(1+ \vert z \vert \bigr)^{-\alpha} \biggl\vert T^{ \alpha}_{\mu} \biggl( \frac{k_{2,z}^{\alpha}}{(1+ \vert z \vert )^{-\frac{\alpha}{2}}} \biggr) (z) \biggr\vert e^{-\frac{1}{2} \vert z \vert ^{2}} \\ \lesssim & \bigl\Vert T^{\alpha}_{\mu} \bigr\Vert _{F^{\infty}_{\alpha}\to F^{\infty}_{ \alpha}} \biggl\Vert \frac{k_{2,z}^{\alpha}}{(1+ \vert z \vert )^{-\frac{\alpha}{2}}} \biggr\Vert _{F^{\infty}_{\alpha}}. \end{aligned}$$

(25)

Together with Theorem 2.2, μ is a 1-Fock–Carleson measure. Furthermore,

$$ \Vert \mu \Vert _{1}\simeq \sup_{z\in \mathbb{C}^{n}} \bigl\vert \widetilde{\mu}_{2}(z) \bigr\vert \lesssim \bigl\Vert T^{\alpha}_{\mu} \bigr\Vert _{F^{\infty}_{\alpha}\to F^{\infty}_{ \alpha}}. $$

(2) Suppose that μ is a vanishing 1-Fock–Carleson measure. By Theorem 2.3,

$$ \widehat{\mu}_{\delta}(z)\to 0, \quad \text{as } \vert z \vert \to \infty . $$

As $T^{\alpha}_{\mu _{R}}$ is compact from $F^{\infty}_{\alpha}$ to $F^{\infty}_{\alpha}$ by Lemma 3.1, if we note that $\mu -\mu _{R}\geq 0$, $T^{\alpha}_{\mu -\mu _{R}}$ is also bounded from $F^{\infty}_{\alpha}$ to $F^{\infty}_{\alpha}$. For $r>0$,

$$ \lim_{R\to \infty}\sup_{z\in \mathbb{C}^{n}} \widehat{(\mu - \mu _{R})}_{r}(z)=0. $$

Therefore, the condition (1) tells us that, when $R\to \infty$,

$$\begin{aligned} \bigl\Vert T^{\alpha}_{\mu}-T^{\alpha}_{\mu _{R}} \bigr\Vert _{F^{\infty}_{\varphi}\to F^{ \infty}_{\varphi}} = \bigl\Vert T^{\alpha}_{\mu -\mu _{R}} \bigr\Vert _{F^{\infty}_{ \varphi}\to F^{\infty}_{\varphi}} \simeq \sup_{z\in \mathbb{C}^{n}} \widehat{( \mu -\mu _{R})}_{r}(z)\to 0. \end{aligned}$$

Hence, we can see that $T^{\alpha}_{\mu}:F^{\infty}_{\alpha}\to F^{\infty}_{\alpha}$ is compact.

On the other hand, we know that $\widehat{\mu}_{\delta}(z)$ is bounded for $\delta >0$. Obviously, $\{\frac{k^{\alpha}_{2,z}}{(1+|z|)^{-\frac{\alpha}{2}}}: z\in \mathbb{C}^{n} \}$ is bounded in $F^{\infty}_{\alpha}$. Therefore, $\{T^{\alpha}_{\mu} ( \frac{k^{\alpha}_{2,z}}{(1+|z|)^{-\frac{\alpha}{2}}} ): z\in \mathbb{C}^{n} \}$ is relatively compact in $F^{\infty}_{\alpha}$. Next, our goal will be to obtain that, as in Theorem 3.2,

$$ \lim_{j\to \infty}T^{\alpha}_{\mu } \biggl( \frac{k^{\alpha}_{2,z_{j}}}{(1+ \vert z_{j} \vert )^{-\frac{\alpha}{2}}} \biggr) (w)=0, $$

for any $w\in \mathbb{C}^{n}$. By the definition of a Topelitz operator, we will omit the details about the case $\alpha \leq 0$ and the other comes into play,

$$\begin{aligned}& T^{\alpha}_{\mu } \biggl( \frac{k^{\alpha}_{2,z_{j}}}{(1+ \vert z_{j} \vert )^{-\frac{\alpha}{2}}} \biggr) (w) \frac{e^{-\frac{1}{2} \vert w \vert ^{2}}}{(1+ \vert w \vert )^{\alpha}} \\& \quad \lesssim \frac{e^{-\frac{1}{2} \vert w \vert ^{2}}}{(1+ \vert w \vert )^{\alpha}} \int _{ \mathbb{C}^{n}} \bigl\vert \bigl(K_{w}^{\alpha} \bigr)_{\frac{\alpha}{2}}^{-}( \xi ) \bigr\vert \frac{ \vert (k^{\alpha}_{2,z_{j}})_{\frac{\alpha}{2}}^{-}(\xi ) \vert }{(1+ \vert z_{j} \vert )^{-\frac{\alpha}{2}}}e^{- \vert \xi \vert ^{2}}\,d\mu (\xi ) \\& \qquad {}+\frac{e^{-\frac{1}{2} \vert w \vert ^{2}}}{(1+ \vert w \vert )^{\alpha}} \int _{\mathbb{C}^{n}} \bigl\vert \bigl(K_{w}^{\alpha} \bigr)_{\frac{\alpha}{2}}^{+}(\xi ) \bigr\vert \frac{ \vert (k^{\alpha}_{2,z_{j}})_{\frac{\alpha}{2}}^{+}(\xi ) \vert }{(1+ \vert z_{j} \vert )^{-\frac{\alpha}{2}}} \frac{e^{- \vert \xi \vert ^{2}}}{ \vert \xi \vert ^{\alpha}}\,d\mu (\xi ). \end{aligned}$$

(26)

Now, we pay attention to the integral $I_{\text{(26)}}$ and we divide it into two integrals for suitable $0< r_{0}< R<\infty$,

$$\begin{aligned} I_{\text{(26)}} =&\frac{e^{-\frac{1}{2} \vert w \vert ^{2}}}{(1+ \vert w \vert )^{\alpha}} \int _{ \vert \xi \vert >R} \bigl\vert \bigl(K_{w}^{\alpha} \bigr)_{ \frac{\alpha}{2}}^{+}(\xi ) \bigr\vert \frac{ \vert (k^{\alpha}_{2,z_{j}})_{\frac{\alpha}{2}}^{+}(\xi ) \vert }{(1+ \vert z_{j} \vert )^{-\frac{\alpha}{2}}} \frac{e^{- \vert \xi \vert ^{2}}}{ \vert \xi \vert ^{\alpha}}\,d\mu (\xi ) \end{aligned}$$

(27)

$$\begin{aligned} &{}+\frac{e^{-\frac{1}{2} \vert w \vert ^{2}}}{(1+ \vert w \vert )^{\alpha}} \int _{r_{0}| \leq \xi |\leq R} \bigl\vert \bigl(K_{w}^{\alpha} \bigr)_{ \frac{\alpha}{2}}^{+}(\xi ) \bigr\vert \frac{ \vert (k^{\alpha}_{2,z_{j}})_{\frac{\alpha}{2}}^{+}(\xi ) \vert }{(1+ \vert z_{j} \vert )^{-\frac{\alpha}{2}}} \frac{e^{- \vert \xi \vert ^{2}}}{ \vert \xi \vert ^{\alpha}}\,d\mu (\xi ) \\ &{}+\frac{e^{-\frac{1}{2} \vert w \vert ^{2}}}{(1+ \vert w \vert )^{\alpha}} \int _{ \vert \xi \vert < r_{0}} \bigl\vert \bigl(K_{w}^{\alpha} \bigr)_{\frac{\alpha}{2}}^{+}(\xi ) \bigr\vert \frac{ \vert (k^{\alpha}_{2,z_{j}})_{\frac{\alpha}{2}}^{+}(\xi ) \vert }{(1+ \vert z_{j} \vert )^{-\frac{\alpha}{2}}} \frac{e^{- \vert \xi \vert ^{2}}}{ \vert \xi \vert ^{\alpha}}\,d\mu (\xi ). \end{aligned}$$

(28)

In view of [4, Proposition 3.2], the integral $I_{\text{(27)}}$ is estimated by

$$\begin{aligned} I_{\text{(27)}} \lesssim & \frac{e^{-\frac{1}{2} \vert w \vert ^{2}}}{(1+ \vert w \vert )^{\alpha}} \int _{ \vert \xi \vert >R} \bigl\vert \bigl(K_{w}^{\alpha} \bigr)_{\frac{\alpha}{2}}^{+}(\xi ) \bigr\vert \frac{ \vert (k^{\alpha}_{2,z_{j}})_{\frac{\alpha}{2}}^{+}(\xi ) \vert }{(1+ \vert z_{j} \vert )^{-\frac{\alpha}{2}}} \frac{e^{- \vert \xi \vert ^{2}}\widehat{\mu}_{\delta}(\xi )}{(1+ \vert \xi \vert )^{\alpha}}\,dv( \xi ) \\ \lesssim & \Vert \widehat{\mu}_{\delta} \Vert _{L^{\infty}} \frac{e^{-\frac{1}{2} \vert w \vert ^{2}}}{(1+ \vert w \vert )^{\alpha}} \int _{ \vert \xi \vert >R} \bigl\vert \bigl(K_{w}^{\alpha} \bigr)_{\frac{\alpha}{2}}^{+}(\xi ) \bigr\vert \bigl\vert \bigl(K_{z_{j}}^{\alpha} \bigr)_{\frac{\alpha}{2}}^{+}( \xi ) \bigr\vert e^{- \vert \xi \vert ^{2}} \frac{e^{-\frac{1}{2} \vert z_{j} \vert ^{2}}\,dv(\xi )}{(1+ \vert \xi \vert )^{\alpha}} \\ \lesssim & \Vert \widehat{\mu}_{\delta} \Vert _{L^{\infty}} \biggl(\sup_{ \vert \xi \vert > R} \int _{\mathbb{C}^{n}}e^{-\frac{1}{2} \vert w \vert ^{2}} \bigl\vert \bigl(K_{\xi}^{ \alpha} \bigr)_{\frac{\alpha}{2}}^{+}(w) \bigr\vert e^{-\frac{1}{2} \vert \xi \vert ^{2}} \frac{dv(w)}{(1+ \vert w \vert )^{\alpha}} \biggr) \\ &{}\times \sum_{k: \vert a_{k} \vert >R+r} \int _{B(a_{k},r)}e^{-\frac{1}{2} \vert z_{j} \vert ^{2}} \bigl\vert \bigl(K_{z_{j}}^{\alpha} \bigr)_{\frac{\alpha}{2}}^{+}( \xi ) \bigr\vert e^{-\frac{1}{2} \vert \xi \vert ^{2}} \frac{dv(\xi )}{(1+ \vert \xi \vert )^{\alpha}} \\ \lesssim & \Vert \widehat{\mu}_{\delta} \Vert _{L^{\infty}} \int _{ \vert \xi \vert >R}e^{- \frac{1}{2} \vert z_{j} \vert ^{2}} \bigl\vert \bigl(K_{z_{j}}^{\alpha} \bigr)_{ \frac{\alpha}{2}}^{+}( \xi ) \bigr\vert e^{-\frac{1}{2} \vert \xi \vert ^{2}} \frac{dv(\xi )}{(1+ \vert \xi \vert )^{\alpha}}. \end{aligned}$$

This implies the integral $I_{\text{(27)}}$ will tend to zero as $R\to \infty$. Since $\frac{|(k^{\alpha}_{2,z_{j}})_{\frac{\alpha}{2}}^{+}(\xi )|}{(1+|z_{j}|)^{-\frac{\alpha}{2}}} \to 0$ uniformly on a compact subset of $\mathbb{C}^{n}$ when $|z_{j}|\to \infty$, it is easy to see that

$$ \lim_{ \vert z_{j} \vert \to \infty} \int _{r_{0}\leq \vert \xi \vert \leq R} \bigl\vert \bigl(K_{w}^{ \alpha} \bigr)_{\frac{\alpha}{2}}^{+}(\xi ) \bigr\vert \frac{ \vert (k^{\alpha}_{2,z_{j}})_{\frac{\alpha}{2}}^{+}(\xi ) \vert }{(1+ \vert z_{j} \vert )^{-\frac{\alpha}{2}}} \frac{e^{- \vert \xi \vert ^{2}}}{ \vert \xi \vert ^{\alpha}}\,d\mu (\xi )=0. $$

Next, the integral $I_{\text{(28)}}$ will be estimated by, in view of estimate (5),

$$\begin{aligned} I_{\text{(28)}} \lesssim & \frac{e^{-\frac{1}{2} \vert w \vert ^{2}}}{(1+ \vert w \vert )^{\alpha}} \int _{ \vert \xi \vert < r_{0}} \frac{ \vert (K_{w}^{\alpha})^{+}_{\frac{\alpha}{2}}(\xi ) \vert }{ \vert \xi \vert ^{\frac{\alpha}{2}}e^{\frac{1}{2} \vert \xi \vert ^{2}}} \frac{ \Vert k^{\alpha}_{2,z_{j}} \Vert _{F^{\infty}_{\alpha}}}{(1+ \vert z_{j} \vert )^{-\frac{\alpha}{2}}}\,d \mu (\xi ) \\ \lesssim &\frac{e^{-\frac{1}{2} \vert w \vert ^{2}}}{(1+ \vert w \vert )^{\alpha}} \int _{ \vert \xi \vert < r_{0}} \bigl( \bigl\vert \bigl(K_{w}^{\alpha} \bigr)^{+}_{\frac{\alpha}{2}}( \xi ) \bigr\vert \vert \xi \vert ^{-\frac{\alpha}{2}} \bigr)e^{-\frac{1}{2} \vert \xi \vert ^{2}}\,d \mu (\xi ) \\ \lesssim & \Vert \widehat{\mu}_{\delta} \Vert _{L^{\infty}} \sum_{k> \frac{\alpha}{2}}\frac{k^{\frac{\alpha}{2}}}{k!} \vert w \vert ^{k} \frac{e^{-\frac{1}{2} \vert w \vert ^{2}}}{(1+ \vert w \vert )^{\alpha}}. \end{aligned}$$

This implies the integral $I_{\text{(28)}}$ tends to zero if $r_{0}$ goes to zero simultaneously. Thus, our aims have been achieved. This, together with the estimate (25), yields that

$$\begin{aligned} \bigl\vert \widetilde{\mu}^{\alpha}_{2}(z) \bigr\vert \lesssim \bigl\Vert T^{\alpha}_{\mu} \bigr\Vert _{F^{ \infty}_{\alpha}\to F^{\infty}_{\alpha}} \biggl\Vert \frac{k_{2,z_{j}}^{\alpha}}{(1+ \vert z \vert )^{-\frac{\alpha}{2}}} \biggr\Vert _{F^{ \infty}_{\alpha}}\to 0, \quad \vert z_{j} \vert \to \infty . \end{aligned}$$

Hence, we can conclude that μ is a vanishing 1-Fock–Carleson measure. □