1 Introduction

Let \(dm\) denote the usual Lebesgue measure on the complex plane \({\mathbb {C}}\). For each \(p, 0 < p < \infty \hbox { and }\alpha > 0\), the Fock space \(\mathcal {F}_{\alpha }^p\) is the space of all entire functions \(f\) on \(\mathbb {C}\) for which

$$\begin{aligned} \Vert f\Vert _{p, \alpha }^p = \frac{p\alpha }{2\pi } \int \limits _{\mathbb {C}} |f(z)|^p e^{-\frac{p\alpha }{2}|z|^2} dm(z) < \infty . \end{aligned}$$

When \(p=2\), the space \(\mathcal {F}_{\alpha }^2\) is a functional Hilbert space with the inner product

$$\begin{aligned} \langle f, g \rangle = \frac{\alpha }{\pi } \int \limits _{\mathbb {C}}f (z)\overline{g(z)} e^{-\alpha |z|^2} dm(z). \end{aligned}$$

The reproducing kernel function for \(\mathcal {F}_{\alpha }^2\) is given by \(K(z,w) = \exp \{\alpha z\overline{w}\}\) (see [6, Chapter 7]). We also use the normalized kernel function \(k_{w}(z) = \exp \{\alpha z\overline{w} -\frac{\alpha }{2}|w|^2\}\). Furthermore we note that \(\Vert k_{w}\Vert _{p, \alpha }^p = 1\) for each \(p > 0\). In Janson et al. [6], considered the space \(\mathcal {F}_{\alpha }^{\infty }\) which is defined by

$$\begin{aligned} \mathcal {F}_{\alpha }^{\infty } = \left\{ f \text { is an entire function } : \Vert f\Vert _{\infty , \alpha } = \sup _{z \in \mathbb {C}}|f(z)|e^{-\frac{\alpha }{2}|z|^2}< \infty \right\} . \end{aligned}$$

In recent progresses of researches on the space \(\mathcal {F}_{\alpha }^{p}\), linear operators induced by entire functions have been studied by many authors. By Carswell et al. [1], boundedness and compactness of composition operators were characterized completely. Guo and Izuchi [5] investigate some operator theoretic properties of them. Choe et al. [2] study on a linear sum of two composition operators. Stroethoff [13] investigates the compactness of multiplication operator by using the Berezin symbol. For weighted composition operators acting on these spaces, we can refer to papers [11, 14, 15]. Recently Constantin [3] studies the Volterra-type operator on \(\mathcal {F}_{1}^p ~ (p >0)\). We consider a generalization of these type operators acting on \(\mathcal {F}_{\alpha }^{\infty }\).

Now let \(g\) and \(\varphi \) be entire functions on \(\mathbb {C}\). We define operators \(C_{\varphi }^{g}\hbox { and }T_{\varphi }^{g}\) by

$$\begin{aligned} C_{\varphi }^{g} f(z) := \int \limits _{0}^{z} f^{\prime }(\varphi (\zeta )) g(\zeta )\, d\zeta \quad \text {and} \quad T_{\varphi }^{g} f(z) := \int \limits _{0}^{z} f(\varphi (\zeta )) g(\zeta )\, d\zeta , \end{aligned}$$

for \(f\) is an entire function and \(z \in \mathbb {C}\). When \(g = {\varphi }^{\prime }\), the operator \(C_{\varphi }^{{\varphi }^{\prime }}\) is the composition operator \(C_{\varphi }\) up to a constant. If \(\varphi (z) = z\) and we replace \(g\) with \(g^{\prime }\), then \(T_{z}^{g^{\prime }}\) becomes the Volterra-type operator. The above operators \(C_{\varphi }^{g}\) and \(T_{\varphi }^{g}\) on some analytic function spaces over the unit disk of \(\mathbb {C}\) were introduced by Li and Stević [7, 8] and they investigated characterizations for boundedness and compactness of these operators. Lindström and Sanatpour [9] also considered boundedness and compactness of \(C_{\varphi }^{g}\) acting between Zygmund type spaces on the unit disk. Recently, Mengestie [10] studied the operator \(T_{\varphi }^{g^{\prime }} : \mathcal {F}_{\alpha }^{p} \rightarrow \mathcal {F}_{\alpha }^{q} \left( 0 < p, \, q < \infty \right) \).

Motivated by these studies, we will investigate the boundedness of \(C_{\varphi }^{g}\hbox { and }T_{\varphi }^{g}\) on \(\mathcal {F}_{\alpha }^{\infty }\). For the compactness of these operators, we give an estimate for the essential norm of them. To do our investigation, we need a characterization for the space \(\mathcal {F}_{\alpha }^{\infty }\) in terms of the first order derivative of entire functions. So our first purpose of this paper is to establish a characterization for \(\mathcal {F}_{\alpha }^{\infty }\) by higher order derivatives. By using this result, we will characterize the boundedness of \(C_{\varphi }^{g}\hbox { and }T_{\varphi }^{g}\), and estimate essential norms of them. Furthermore, as an application of our results, we describe a concrete form of \(g\) which induces the bounded or compact Volterra-type operators on \(\mathcal {F}_{\alpha }^{\infty }\).

Throughout this paper, the notation \(a {\lesssim }_{s,t} b\) means that there exists a positive constant \(C_{s,t}\) depending only on some parameters \(s, t\) such that \(a \le C_{s,t}b\). Moreover, if both \(a {\lesssim }_{s,t} b\hbox { and } b {\lesssim }_{s,t} a\) hold, then one says that \(a {\approx }_{s,t} b\).

2 Higher Order Derivatives Criterion for \(\mathcal {F}_{\alpha }^{\infty }\)

In this section, we shall obtain the \(n\)-th derivative criterion for \({\mathcal {F}}_{\alpha }^{\infty }.\) For the case of the usual Bloch space \(\mathcal {B}\) on the unit disk, a corresponding result have been obtained by Stroethoff [12].

Theorem 1

Suppose that \(n\) is a positive integer and \(f\) is an entire function. Then \(f \in \mathcal {F}_{\alpha }^{\infty }\) if and only if \(\frac{|f^{(n)}(z)|}{(1+|z|)^n}e^{-\frac{\alpha }{2}|z|^2} \in L^{\infty }(\mathbb {C})\).

Proof

First suppose that \(f \in \mathcal {F}_{\alpha }^{\infty }\). Now we consider the following operator:

$$\begin{aligned} P_{\alpha }f(z) = \frac{\alpha }{\pi } \int \limits _{\mathbb {C}} e^{\alpha z\overline{w}}f(w)e^{-\alpha |w|^2}dm(w). \end{aligned}$$

By [6, Theorem 7.1], this \(P_{\alpha }\) is a bounded projection from \(\mathcal {L}_{\alpha }^{\infty }\) onto \(\mathcal {F}_{\alpha }^{\infty }\). Here \(\mathcal {L}_{\alpha }^{\infty }\) denotes the space of measurable functions \(g\) such that \(\mathrm{esssup}_{z \in \mathbb {C}}|g(z)|e^{-\frac{\alpha }{2}|z|^2} < \infty .\) Since \(\mathcal {F}_{\alpha }^{\infty } \subset \mathcal {L}_{\alpha }^{\infty }\), hence, we obtain

$$\begin{aligned} f(z) = \frac{\alpha }{\pi } \int \limits _{\mathbb {C}} e^{\alpha z\overline{w}}f(w)e^{-\alpha |w|^2}dm(w). \end{aligned}$$

Differentiating under the integral sign gives

$$\begin{aligned} |f^{(n)}(z)|&\le \frac{{\alpha }^{n+1}}{\pi } \int \limits _{\mathbb {C}} |w|^n e^{\alpha \mathrm{Re}(z\overline{w})} |f(w)|e^{-\alpha |w|^2}dm(w)\\&\le \Vert f\Vert _{\infty , \alpha } \frac{{\alpha }^{n+1}}{\pi } \int \limits _{\mathbb {C}} |w|^n e^{\alpha \mathrm{Re}(z\overline{w})} e^{-\frac{\alpha }{2}|w|^2}dm(w)\\&\le \Vert f\Vert _{\infty , \alpha }e^{\frac{\alpha }{2}|z|^2} \frac{{\alpha }^{n+1}}{\pi } \int \limits _{\mathbb {C}} (1+|w|)^n e^{-\frac{\alpha }{2}|w-z|^2}dm(w)\\&\le \Vert f\Vert _{\infty , \alpha }e^{\frac{\alpha }{2}|z|^2}(1+|z|)^n \frac{{\alpha }^{n+1}}{\pi } \int \limits _{\mathbb {C}} (1+|w-z|)^n e^{-\frac{\alpha }{2}|w-z|^2}dm(w). \end{aligned}$$

In the last line, we use an elementary inequality: \(1+|w| \le (1+|z|)(1+|w-z|)\) \((z, w \in \mathbb {C})\) which is verified by an application of the triangle inequality. Since

$$\begin{aligned} \int \limits _{\mathbb {C}} |w-z|^n e^{-\frac{\alpha }{2}|w-z|^2}dm(w) = \pi \left( \frac{2}{\alpha }\right) ^{\frac{n+2}{2}}\cdot \Gamma \left( \frac{n}{2}+1\right) , \end{aligned}$$

and so we obtain that

$$\begin{aligned} \sup _{z \in \mathbb {C}} \frac{|f^{(n)}(z)|}{(1+|z|)^n}e^{-\frac{\alpha }{2}|z|^2} {\lesssim }_{\alpha , n} \Vert f\Vert _{\infty , \alpha }, \end{aligned}$$
(1)

for all \(n \ge 1\).

To prove another direction, we may establish the following inequality:

$$\begin{aligned} \Vert f\Vert _{\infty , \alpha } {\lesssim }_{\alpha } \sup _{z \in \mathbb {C}} \frac{|f^{(n)}(z)|}{(1+|z|)^n}e^{-\frac{\alpha }{2}|z|^2} + \sum _{j=0}^{n-1}|f^{(j)}(0)| \end{aligned}$$
(2)

for \(n \ge 1\). The above inequality is verified by induction on \(n\). First we consider the case \(n = 1\). For \(z \in \mathbb {C}\) we have that

$$\begin{aligned} |f(z) \!-\! f(0)| \le \int _{0}^{1}\!|z||f^{\prime }(tz)|dt \le \left( \sup _{z \in \mathbb {C}} \frac{|f^{\prime }(z)|}{1+|z|}e^{-\frac{\alpha }{2}|z|^2} \right) \cdot \int \limits _{0}^{1}\!|z|(1\!+\!|tz|)e^{\frac{\alpha }{2}|tz|^2}dt.\nonumber \\ \end{aligned}$$
(3)

The last integral in the above line is dominated by \(\frac{3}{\alpha }e^{\frac{\alpha }{2}|z|^2}\) when \(|z| > 1\) or \(2e^{\frac{\alpha }{2}|z|^2}\) when \(|z| \le 1\). Thus (3) gives that

$$\begin{aligned} \Vert f - f(0)\Vert _{\infty , \alpha } {\lesssim }_{\alpha } \sup _{z \in \mathbb {C}} \frac{|f^{\prime }(z)|}{1+|z|}e^{-\frac{\alpha }{2}|z|^2}, \end{aligned}$$

and so (2) is true when \(n =1\). Next we assume that (2) holds when \(n = k\). By this assumption, we may prove that

$$\begin{aligned} \sup _{z \in \mathbb {C}} \frac{|f^{(k)}(z)|}{(1+|z|)^k}e^{-\frac{\alpha }{2}|z|^2} {\lesssim }_{\alpha } \sup _{z \in \mathbb {C}} \frac{|f^{(k+1)}(z)|}{(1+|z|)^{k+1}}e^{-\frac{\alpha }{2}|z|^2} + |f^{(k)}(0)|. \end{aligned}$$
(4)

As in (3), we have that

$$\begin{aligned} |f^{(k)}(z) - f^{(k)}(0)| \le \left( \sup _{z \in \mathbb {C}} \frac{|f^{(k+1)}(z)|}{(1+|z|)^{k+1}}e^{-\frac{\alpha }{2}|z|^2} \right) \cdot \int \limits _{0}^{1}|z|(1+|tz|)^{k+1}e^{\frac{\alpha }{2}|tz|^2}dt. \end{aligned}$$

For \(z \in \mathbb {C}\) with \(|z| >1\),

$$\begin{aligned} \int \limits _{0}^{1}|z|(1+|tz|)^{k+1}e^{\frac{\alpha }{2}|tz|^2}dt&\le \int \limits _{0}^{1}|z|(1+|z|)^{k+1}e^{\frac{\alpha }{2}|z|^2 t}dt\\&< (1+|z|)^{k+1} \cdot \frac{2}{\alpha |z|}e^{\frac{\alpha }{2}|z|^2} < \frac{4}{\alpha }(1+|z|)^k e^{\frac{\alpha }{2}|z|^2}. \end{aligned}$$

On the other hand, we also obtain that

$$\begin{aligned} \int \limits _{0}^{1}|z|(1+|tz|)^{k+1}e^{\frac{\alpha }{2}|tz|^2}dt \le 2(1+|z|)^k e^{\frac{\alpha }{2}|z|^2} \end{aligned}$$

for \(z \in \mathbb {C}\) with \(|z| \le 1\). Hence we have that

$$\begin{aligned} \frac{|f^{(k)}(z) - f^{(k)}(0)|}{(1+|z|)^k}e^{-\frac{\alpha }{2}|z|^2} {\lesssim }_{\alpha } \sup _{z \in \mathbb {C}} \frac{|f^{(k+1)}(z)|}{(1+|z|)^{k+1}}e^{-\frac{\alpha }{2}|z|^2} \end{aligned}$$

for all \(z \in \mathbb {C}\), and so the desired estimate (4) is obtained. By induction on \(n\), we see that (2) is true for all \(n \ge 1\). \(\square \)

Remark

Inequality (1) also gives \(|f^{(j)}(0)| {\lesssim }_{\alpha , j} \Vert f\Vert _{\infty , \alpha }\) for all \(j \ge 1\). Combining this with (1) and (2), we obtain

$$\begin{aligned} \Vert f\Vert _{\infty , \alpha } {\approx }_{\alpha , n} \sup _{z \in \mathbb {C}} \frac{|f^{(n)}(z)|}{(1+|z|)^n}e^{-\frac{\alpha }{2}|z|^2} + \sum _{j=0}^{n-1}|f^{(j)}(0)|. \end{aligned}$$
(5)

In [6], Janson et al. have also mentioned the subspace \(\mathcal {F}_{\alpha ,0}^{\infty }\) of \(\mathcal {F}_{\alpha }^{\infty }.\) Namely, \(\mathcal {F}_{\alpha ,0}^{\infty }\) consists of all entire functions \(f\) which satisfy \(|f(z)|e^{-\frac{\alpha }{2}|z|^2} \rightarrow 0\) as \(|z| \rightarrow \infty \). It is expected that the \(n\)-th derivative criterion in Theorem 1 carries over to the space \(\mathcal {F}_{\alpha ,0}^{\infty }\).

For each positive integer \(n\), we denote

$$\begin{aligned} \mathcal {F}^{(n)} = \left\{ \text {f is entire} \,:\, \lim _{|z| \rightarrow \infty }\frac{|f^{(n)}(z)|}{(1+|z|)^n}e^{-\frac{\alpha }{2}|z|^2} = 0 \right\} . \end{aligned}$$

Now we will show that \(\mathcal {F}_{\alpha ,0}^{\infty }\) is equal to \(\mathcal {F}^{(n)}\).

Lemma 1

If \(f \in \mathcal {F}^{(n)}\), then \(\lim _{r \rightarrow 1^{-}}\Vert f-f_r\Vert _{\infty , \alpha } = 0\) where \(f_r(z) = f(rz)\) for \(0 \le r <1\).

Proof

Suppose that \(f \in \mathcal {F}^{(n)}\) and take \(\epsilon >0\) arbitrarily. Then there exists \(R>0\) such that

$$\begin{aligned} \frac{|f^{(n)}(z)|}{(1+|z|)^n}e^{-\frac{\alpha }{2}|z|^2} < \frac{\epsilon }{4} \quad \text {for |z| > R.} \end{aligned}$$

If \(1/2 < r < 1\) and \(|z| > 2R\), then we have that

$$\begin{aligned} \frac{|f^{(n)}(z) -(f_r)^{(n)}(z)|}{(1+|z|)^n}e^{-\frac{\alpha }{2}|z|^2} \le \frac{|f^{(n)}(z)|}{(1+|z|)^n}e^{-\frac{\alpha }{2}|z|^2} + \frac{r^n|f^{(n)}(rz)|}{(1+|rz|)^n}e^{-\frac{\alpha }{2}|rz|^2} < \frac{\epsilon }{2}. \end{aligned}$$

The uniform continuity of \(f^{(n)}\) on the compact set \(\{z \, : \, |z| \le 2R\}\) implies that

$$\begin{aligned} \max _{|z| \le 2R}|f^{(n)}(z) - f^{(n)}(rz)| < \frac{\epsilon }{2}, \end{aligned}$$

and so we have that

$$\begin{aligned} |f^{(n)}(z) - (f_r)^{(n)}(z)|&\le |f^{(n)}(z) - f^{(n)}(rz)| + |f^{(n)}(rz) - (f_r)^{(n)}(z)|\\&< \frac{\epsilon }{2} + (1-r^n)\max _{|z| \le 2R}|f^{(n)}(z)| \end{aligned}$$

for \(|z| \le 2R\) and \(r\) is close to \(1\) sufficiently. Combining these estimates with (5), we obtain that

$$\begin{aligned} \Vert f - f_r\Vert _{\infty , \alpha } {\lesssim }_{\alpha } \epsilon + (1-r^n)\max _{|z| \le 2R}|f^{(n)}(z)| + \sum _{j = 0}^{n-1}(1-r^j)|f^{(j)}(0)|. \end{aligned}$$

By letting \(r \rightarrow 1^{-}\), we can get the result. \(\square \)

Proposition 1

\(\mathcal {F}^{(n)}\) is the closure of the polynomials in \(\mathcal {F}_{\alpha }^{\infty }\) under the norm \(\Vert \cdot \Vert _{\infty , \alpha }.\)

Proof

For \(f (z) = \sum _{j= 0}^{\infty }a_j z^j, 0<r<1\) and each positive integer \(N\), we put \(P_N (z) = \sum _{j= 0}^{N}a_j r^j z^j\). Cauchy’s integral formula implies that

$$\begin{aligned} |a_j| \le \frac{1}{t^j}e^{\frac{\alpha }{2}t^2}\Vert f\Vert _{\infty , \alpha }. \end{aligned}$$

Thus if we put \(t = \sqrt{\frac{j}{\alpha }}\), then we have that

$$\begin{aligned} |a_j| \le \left( \frac{\alpha e}{j}\right) ^{\frac{j}{2}}\Vert f\Vert _{\infty , \alpha }. \end{aligned}$$

Since \(\Vert z^j\Vert _{\infty , \alpha } = \left( \frac{j}{\alpha e}\right) ^{\frac{j}{2}}\), we see that

$$\begin{aligned} \Vert f_r- P_N\Vert _{\infty , \alpha } \le \Vert f\Vert _{\infty , \alpha } \cdot \sum _{j = N+1}^{\infty }r^{j}. \end{aligned}$$

Now we assume that \(f \in \mathcal {F}^{(n)}\) and \(\epsilon >0\). By Lemma 1, there exists \(0< r_0 < 1\) such that \(\Vert f - f_{r_0}\Vert _{\infty , \alpha } < \epsilon \). By using the above argument, we also have that

$$\begin{aligned} \Vert f- P_N\Vert _{\infty , \alpha } < \epsilon + \Vert f\Vert _{\infty , \alpha } \cdot \sum _{j = N+1}^{\infty }{r_{0}}^{j}. \end{aligned}$$

Since \(f \in \mathcal {F}_{\alpha }^{\infty }\) by Theorem 1, this implies that \( \limsup _{N \rightarrow \infty }\Vert f- P_N\Vert _{\infty , \alpha } \le \epsilon \). This completes the proof. \(\square \)

Corollary 1

For each positive integer \(n\), an entire function \(f\) belongs to \(\mathcal {F}_{\alpha , 0}^{\infty }\) if and only if \(\frac{|f^{(n)}(z)|}{(1+|z|)^n}e^{-\frac{\alpha }{2}|z|^2} \rightarrow 0\) as \(|z| \rightarrow \infty \).

Proof

By noting that the polynomial set is dense in \(\mathcal {F}_{\alpha }^1\) (see [4, Proposition 5]) and \(\mathcal {F}_{\alpha }^{1}\) is dense in \(\mathcal {F}_{\alpha , 0}^{\infty }\) (see [6, Theorem 7.2]), we see that \(\mathcal {F}_{\alpha , 0}^{\infty }\) is also the closure of the polynomials in \(\mathcal {F}_{\alpha }^{\infty }\) under the norm \(\Vert \cdot \Vert _{\infty , \alpha }\). Hence Proposition 1 implies \(\mathcal {F}_{\alpha , 0}^{\infty } = \mathcal {F}^{(n)}\). We obtain the desired characterization for the space \(\mathcal {F}_{\alpha , 0}^{\infty }\). \(\square \)

3 Applications to Operators on \(\mathcal {F}_{\alpha }^{\infty }\)

In this section, we investigate linear operators induced by entire functions \(g\) and \(\varphi \). First we consider the boundedness of \(C_{\varphi }^{g}\) on \(\mathcal {F}_{\alpha }^{\infty }\) and the estimate for its essential norm.

Theorem 2

The operator \(C_{\varphi }^{g}\) is bounded on \(\mathcal {F}_{\alpha }^{\infty }\) if and only if \(\varphi \) and \(g\) satisfy

$$\begin{aligned} \sup _{z \in \mathbb {C}} \frac{|g(z)|(1+|\varphi (z)|)}{1+|z|} \exp \left\{ \frac{\alpha }{2}\left( |\varphi (z)|^2 - |z|^2\right) \right\} < \infty . \end{aligned}$$
(6)

If \(C_{\varphi }^{g}\) is bounded on \(\mathcal {F}_{\alpha }^{\infty }\), then it holds that

$$\begin{aligned} \Vert C_{\varphi }^{g}\Vert _{e} {\approx }_{\alpha } \limsup _{|\varphi (z)| \rightarrow \infty } \frac{|g(z)|(1+|\varphi (z)|)}{1+|z|} \exp \left\{ \frac{\alpha }{2}\left( |\varphi (z)|^2 - |z|^2\right) \right\} . \end{aligned}$$
(7)

Here \(\Vert C_{\varphi }^{g}\Vert _{e}\) denotes the essential norm of \(C_{\varphi }^{g}\). Hence \(C_{\varphi }^{g}\) is compact if and only if

$$\begin{aligned} \lim _{|\varphi (z)| \rightarrow \infty } \frac{|g(z)|(1+|\varphi (z)|)}{1+|z|} \exp \left\{ \frac{\alpha }{2}\left( |\varphi (z)|^2 - |z|^2\right) \right\} = 0. \end{aligned}$$

Proof

By applying Theorem 1 to \(\Vert C_{\varphi }^{g}f\Vert _{\infty , \alpha }\), we see that condition (6) is a sufficient condition for the boundedness of \(C_{\varphi }^{g}\).

To prove the other direction, we consider the function \( f_{z}(w) = k_{\varphi (z)}(w) \) for fixed \(z \in \mathbb {C}\) with \(|\varphi (z)|>1\). Since \(f_z \in \mathcal {F}_{\alpha }^{\infty }\) and \(\Vert f_z\Vert _{\infty , \alpha } \le 1\), it follows from Theorem 1 that

$$\begin{aligned} \frac{|g(z)|(1+|\varphi (z)|)}{1+|z|} \exp \left\{ \frac{\alpha }{2}\left( |\varphi (z)|^2 - |z|^2\right) \right\}&\le \frac{2}{\alpha } \frac{|f_{z}^{\prime }(\varphi (z))g(z)|}{1+|z|}e^{-\frac{\alpha }{2}|z|^2} \nonumber \\&{\lesssim }_{\alpha } \Vert C_{\varphi }^{g}f_z\Vert _{\infty , \alpha } \le \Vert C_{\varphi }^{g}\Vert . \end{aligned}$$
(8)

By considering the identity function \(\pi (z) = z\), we see that the boundedness of \(C_{\varphi }^{g}\) implies that

$$\begin{aligned} \sup _{z \in \mathbb {C}} \frac{|g(z)|}{1+|z|}e^{-\frac{\alpha }{2}|z|^2} < \infty , \end{aligned}$$
(9)

and so we have that

$$\begin{aligned} \frac{|g(z)|(1+|\varphi (z)|)}{1+|z|} \exp \left\{ \frac{\alpha }{2}\left( |\varphi (z)|^2 - |z|^2\right) \right\} \le 2 e^{\frac{\alpha }{2}}\sup _{z \in \mathbb {C}} \frac{|g(z)|}{1+|z|}e^{-\frac{\alpha }{2}|z|^2}, \end{aligned}$$

for any \(z \in \mathbb {C}\) with \(|\varphi (z)| \le 1\). Hence (6) is verified by combining this with (8) and (9).

Next we will prove the estimate (7). To prove the lower estimate in (7), we take a sequence \(\{z_j\}\) in \(\mathbb {C}\) with \(|\varphi (z_j)| \rightarrow \infty \) as \(j \rightarrow \infty \). Put \(f_j(z) = k_{\varphi (z_j)}(z)\). Then \(\{f_j\}\) is a bounded sequence in \(\mathcal {F}_{\alpha }^{\infty }\) and \(f_j \rightarrow 0 ~ (\text {as} \, j \rightarrow \infty )\) uniformly on compact subsets in \(\mathbb {C}\). More precisely, we see that \(f_j \in \mathcal {F}_{\alpha , 0}^{\infty }\) for each \(j\). By using the duality \((\mathcal {F}_{\alpha , 0}^{\infty })^{*} \cong \mathcal {F}_{\alpha }^1\) (see [6, Theorem 7.4]), a standard argument shows that \(\{f_j\}\) converges to \(0\) weakly in \(\mathcal {F}_{\alpha , 0}^{\infty }\). Since \((\mathcal {F}_{\alpha }^{\infty })^{*} \subset (\mathcal {F}_{\alpha , 0}^{\infty })^{*}\), this shows that \(\{f_j\}\) also converges to \(0\) weakly in \(\mathcal {F}_{\alpha }^{\infty }\). Hence we see that \(\Vert \mathcal {K}f_{j}\Vert _{\infty , \alpha } \rightarrow 0\) as \(j \rightarrow \infty \) for any compact operators \(\mathcal {K}\) on \(\mathcal {F}_{\alpha }^{\infty }\) and

$$\begin{aligned} \Vert C_{\varphi }^{g}\Vert _{e} \ge \limsup _{j \rightarrow \infty }\Vert C_{\varphi }^{g}f_{j}\Vert _{\infty , \alpha }. \end{aligned}$$

As in the argument of (8), we also obtain that

$$\begin{aligned} \frac{|g(z_j)|(1+|\varphi (z_j)|)}{1+|z_j|} \exp \left\{ \frac{\alpha }{2}\left( |\varphi (z_j)|^2 - |z_j|^2\right) \right\} {\lesssim }_{\alpha } \Vert C_{\varphi }^{g}f_j\Vert _{\infty , \alpha }, \end{aligned}$$

for a sufficiently large \(j\). Thus we get

$$\begin{aligned} \limsup _{j \rightarrow \infty } \frac{|g(z_j)|(1+|\varphi (z_j)|)}{1+|z_j|} \exp \left\{ \frac{\alpha }{2}\left( |\varphi (z_j)|^2 - |z_j|^2\right) \right\} {\lesssim }_{\alpha } \Vert C_{\varphi }^{g}\Vert _{e}, \end{aligned}$$

and so this implies the desired lower estimate for \(\Vert C_{\varphi }^{g}\Vert _{e}\).

Now we prove the upper estimate in (7). For each positive integer \(k\) we put \({\phi }_{k}(z) = \frac{k}{k+1}z.\) Then it follows from [15, Theorem 4 (b)] (also see [1]) that the composition operator \(C_{{\phi }_k}\) is compact on \(\mathcal {F}_{\alpha }^{\infty }\). Hence we have that

$$\begin{aligned} \left\| C_{\varphi }^{g}\right\| _{e} \le \left\| C_{\varphi }^{g} - C_{\varphi }^{g}C_{{\phi }_k}\right\| = \sup _{\Vert f\Vert _{\infty , \alpha } \le 1}\left\| C_{\varphi }^{g}(I-C_{{\phi }_k})f\right\| _{\infty , \alpha }, \end{aligned}$$
(10)

where \(I\) denotes the identity operator on \(\mathcal {F}_{\alpha }^{\infty }\). By Theorem 1 the last term in (10) is comparable with

$$\begin{aligned}&\sup _{\Vert f\Vert _{\infty , \alpha } \le 1} \sup _{|\varphi (z)| \le r} \frac{\left| \left( C_{\varphi }^{g}(I-C_{{\phi }_k})f\right) ^{\prime }(z)\right| }{1+|z|}e^{-\frac{\alpha }{2}|z|^2} \nonumber \\&\quad \quad \quad + \sup _{\Vert f\Vert _{\infty , \alpha } \le 1} \sup _{|\varphi (z)| > r} \frac{\left| \left( C_{\varphi }^{g}(I-C_{{\phi }_k})f\right) ^{\prime }(z)\right| }{1+|z|}e^{-\frac{\alpha }{2}|z|^2}, \end{aligned}$$
(11)

for any fixed \(r \in (0, \infty )\). By an application of Theorem 1 once again, we see that the second term in (11) is dominated by

$$\begin{aligned} \sup _{|\varphi (z)|> r} \frac{|g(z)|(1+|\varphi (z)|)}{1+|z|} \exp \left\{ \frac{\alpha }{2}\left( |\varphi (z)|^2 - |z|^2\right) \right\} . \end{aligned}$$
(12)

Note that this estimate does not depend on \(k\).

Next assertion is

$$\begin{aligned} \lim _{k \rightarrow \infty }\sup _{\Vert f\Vert _{\infty , \alpha } \le 1} \sup _{|\varphi (z)| \le r} \frac{\left| \left( C_{\varphi }^{g}(I-C_{{\phi }_k})f\right) ^{\prime }(z)\right| }{1+|z|}e^{-\frac{\alpha }{2}|z|^2} = 0. \end{aligned}$$
(13)

Put \(w = \varphi (z)\) and denote \(\left[ \frac{k}{k+1}w, w\right] \) the radial segment. By integrating \(f^{\prime }\) along \(\left[ \frac{k}{k+1}w, w\right] \), we obtain that

$$\begin{aligned} \left| f^{\prime }(w) - f^{\prime }\left( \frac{k}{k+1}w\right) \right| \le \frac{1}{k+1}|w||f^{\prime \prime }(\xi (w))|, \end{aligned}$$

for some \(\xi (w) \in \left[ \frac{k}{k+1}w, w\right] \). Furthermore an application of Cauchy’s estimate to \(f^{\prime \prime }\) on the circle with center at \(\xi (w)\) and radius \(r\) gives that

$$\begin{aligned} |f^{\prime \prime }(\xi (w))| \le \frac{2}{r^2}\max _{|\zeta |=2r}|f(\zeta )|. \end{aligned}$$

Hence we have that

$$\begin{aligned} \left| \left( C_{\varphi }^{g}(I\!-\!C_{{\phi }_k})f\right) ^{\prime }(z)\right|&\le |g(z)|\left\{ \left| f^{\prime }(w) \!- \!f^{\prime } \left( \frac{k}{k+1}w\right) \right| + \frac{1}{k+1}\left| f^{\prime }\left( \frac{k}{k+1}w\right) \right| \right\} \\&{\lesssim }_{\alpha } \frac{|g(z)| \cdot \Vert f\Vert _{\infty , \alpha }}{k+1} \left\{ \frac{2e^{2\alpha r^2}}{r} + (1+r)e^{\frac{\alpha }{2}r^2} \right\} , \end{aligned}$$

for fixed \(r \in (0, \infty )\). Combinig this with (9) prove (13). By (10)–(13), we obtain that

$$\begin{aligned} \Vert C_{\varphi }^{g}\Vert _{e} {\lesssim }_{\alpha } \sup _{|\varphi (z)| > r} \frac{|g(z)|(1+|\varphi (z)|)}{1+|z|} \exp \left\{ \frac{\alpha }{2}\left( |\varphi (z)|^2 - |z|^2\right) \right\} . \end{aligned}$$

Since \(r \in (0, \infty )\) was arbitrary, we get the desired upper estimate in (7). \(\square \)

For the operator \(T_{\varphi }^{g}\), on the other hand, Theorem 1 gives that

$$\begin{aligned} \Vert T_{\varphi }^{g}f\Vert _{\infty , \alpha }&{\approx }_{\alpha } \sup _{z \in \mathbb {C}} \frac{|f(\varphi (z))g(z)|}{1+|z|}e^{-\frac{\alpha }{2}|z|^2}\\&\le \Vert f\Vert _{\infty , \alpha } \sup _{z \in \mathbb {C}} \frac{|g(z)|}{1+|z|} \exp \left\{ \frac{\alpha }{2}\left( |\varphi (z)|^2 - |z|^2\right) \right\} , \end{aligned}$$

and so the condition

$$\begin{aligned} \sup _{z \in \mathbb {C}} \frac{|g(z)|}{1+|z|} \exp \left\{ \frac{\alpha }{2}\left( |\varphi (z)|^2 - |z|^2\right) \right\} < \infty \end{aligned}$$

is a sufficient condition for the boundedness of \(T_{\varphi }^{g}\) on \(\mathcal {F}_{\alpha }^{\infty }\). Furthermore the test function \(k_{\varphi (z)}(w)\) implies that this condition is also a necessary condition. By the same argument in Theorem 2, we can obtain an estimate for the essential norm \(\Vert T_{\varphi }^{g}\Vert _{e}\). Hence we obtain the following result.

Theorem 3

The operator \(T_{\varphi }^{g}\) is bounded on \(\mathcal {F}_{\alpha }^{\infty }\) if and only if \(\varphi \) and \(g\) satisfy

$$\begin{aligned} \sup _{z \in \mathbb {C}} \frac{|g(z)|}{1+|z|} \exp \left\{ \frac{\alpha }{2}\left( |\varphi (z)|^2 - |z|^2\right) \right\} < \infty . \end{aligned}$$

If \(T_{\varphi }^{g}\) is bounded on \(\mathcal {F}_{\alpha }^{\infty }\), then the essential norm \(\Vert T_{\varphi }^{g}\Vert _{e}\) is comparable with

$$\begin{aligned} \limsup _{|\varphi (z)| \rightarrow \infty } \frac{|g(z)|}{1+|z|} \exp \left\{ \frac{\alpha }{2}\left( |\varphi (z)|^2 - |z|^2\right) \right\} . \end{aligned}$$

For each entire functions \(g\), integral type operators \(J_{g}\) and \(I_{g}\) are defined by

$$\begin{aligned} J_{g}f(z) = \int \limits _{0}^{z}f(\zeta )g^{\prime }(\zeta ) \, d\zeta \quad \text {and} \quad I_{g}f(z) = \int \limits _{0}^{z}f^{\prime }(\zeta )g(\zeta ) \, d\zeta , \end{aligned}$$

respectively. These operators are the special cases of \(T_{\varphi }^{g}\) and \(C_{\varphi }^{g}\). In fact, if we replace \(g\) with \(g^{\prime }\) and put \(\varphi (z) = z\) in the definition of \(T_{\varphi }^{g}\), then we have that \(T_{z}^{g^{\prime }}f = J_{g}f\). On the other hand, by the definition of \(C_{\varphi }^{g}\), we see that \(C_{z}^{g}f = I_{g}f\). Hence we can describe the characterizations for the boundedness and compactness of \(J_{g}\) and \(I_{g}\) as an application of Theorems 2 and 3.

Corollary 2

For each entire function \(g\), the following hold.

(a):

\(J_{g}\) is bounded on \(\mathcal {F}_{\alpha }^{\infty }\) if and only if \(g\) is a polynomial of degree \(\le 2\).

(b):

\(J_{g}\) is compact on \(\mathcal {F}_{\alpha }^{\infty }\) if and only if \(g\) is a polynomial of degree \(\le 1\).

(c):

\(I_{g}\) is bounded on \(\mathcal {F}_{\alpha }^{\infty }\) if and only if \(g\) is a constant function.

(d):

\(I_{g}\) is compact on \(\mathcal {F}_{\alpha }^{\infty }\) if and only if \(g \equiv 0\).

Proof

Claims (c) and (d) are an immediate consequence of Theorem 2. So we may only prove (a) and (b). By Theorem 3 we see that \(J_{g}\) is bounded if and only if \(g\) satisfies

$$\begin{aligned} \sup _{z \in \mathbb {C}}\frac{|g^{\prime }(z)|}{1+|z|} < \infty . \end{aligned}$$

This implies that \(g^{\prime }\) must be a polynomial of degree \(\le 1\), and so the degree of \(g\) is less than or equal to \(2\). Theorem 3 also says that

$$\begin{aligned} \lim _{|z| \rightarrow \infty }\frac{|g^{\prime }(z)|}{1+|z|} = 0 \end{aligned}$$

is equivalent to the compactness of \(J_{g}\), and so \(g^{\prime }\) is constant. Hence \(J_{g}\) is compact if and only if the degree of \(g\) must be less that or equal to \(1\). \(\square \)

Remark

The claim (a) and (b) for the Fock space \(\mathcal {F}^p (= \mathcal {F}_{1}^p) ~ (0<p< \infty )\) is proved by Constantin [3, Theorem 1].

Finally we will mention the action of differential and integral operators on \(\mathcal {F}_{\alpha }^{\infty }\). We put

$$\begin{aligned} Df(z) = f^{\prime }(z) \quad \text {and} \quad If(z) = \int \limits _{0}^{z}\! f(\zeta ) \, d\zeta , \end{aligned}$$

for each entire function \(f\). Take a sequence \(\{z_j\}\) in \(\mathbb {C}\) with \(|z_j| \rightarrow \infty \) as \(j \rightarrow \infty \) and \(f \in \mathcal {F}_{\alpha }^{\infty }\). Then we have

$$\begin{aligned} \Vert Df\Vert _{\infty , \alpha } \ge |f^{\prime }(z_j)|e^{-\frac{\alpha }{2}|z_j|^2} = \frac{|f^{\prime }(z_j)|}{1+|z_j|}e^{-\frac{\alpha }{2}|z_j|^2}(1+|z_j|). \end{aligned}$$

By applying Theorem 1 to the above, we see that \(D\) is not bounded on \(\mathcal {F}_{\alpha }^{\infty }\). Since \(If(z) = J_{z}f(z)\), on the other hand, Corollary 2 implies that \(I\) is always bounded (and compact) on \(\mathcal {F}_{\alpha }^{\infty }\).