Order Bounded and 2-Complex Symmetric Weighted Superposition Operators on Fock Spaces

Abstract

One aim of the paper is to characterize some complex symmetric and 2-complex symmetric bounded weighted superposition operators on Fock spaces respect to the conjugations $\mathcal{J}$ and ${\mathcal{J}}_{r,s,t}$ defined by $\mathcal{J}f\left(z\right)=\overline{f\left(\overline{z}\right)}$ and ${\mathcal{J}}_{r,s,t}f\left(z\right)=t{e}^{sz}\overline{f\left(\overline{rz+s}\right)}$. Another aim is to characterize the order bounded weighted superposition operators from one Fock space into another Fock space.

1. Introduction

Consider H as a separable complex Hilbert space, where $T:H\to H$ represents a bounded linear operator, with $T^{*}$ denoting its adjoint operator. Complex symmetric operators have gained significant prominence in both theoretical studies and practical applications within the realm of non-Hermitian operators, amid a plethora of other challenges (see [1]). In order to introduce such operators, we need the following definition.

Definition 1. 

An operator $C:H\to H$ is said to be a conjugation if it satisfies the following conditions:

(a) conjugate-linear: $C(\alpha x+\beta y)=$ $\overline{\alpha }C\left(x\right)+\overline{\beta }C\left(y\right)$ for α, β∈$\mathbb{C}$ and x, y∈H;

(b) isometric: $\parallel C\left(x\right)\parallel =\parallel x\parallel$ for all $x\in H$;

(c) involutive: $C^2=I_d$ where $I_d$ is an identity operator.

For any conjugation C on H, there is an orthonormal basis ${\left\{e_n\right\}}_{n=1}^{\infty }$ for H such that $C{e}_n=e_n$ for all $n\in \mathbb{N}$ (see [2]). Indeed, there exist some conjugates on the holomorphic function spaces. For example, Hai et al. in [3] proved that the operator $\mathcal{J}f\left(z\right)=\overline{f\left(\overline{z}\right)}$ is a conjugation on Fock space ${\mathcal{F}}^2$, and ${\mathcal{J}}_{r,s,t}f\left(z\right)=t{e}^{sz}\overline{f\left(\overline{rz+s}\right)}$ is a conjugation on ${\mathcal{F}}^2$ if and only if

$$\begin{array}{c}\hfill \left|r\right|=1,\overline{r}s+\overline{s}=0,{\left|t\right|}^2{e}^{{\left|s\right|}^2}=1.\end{array}$$

(1)

Therefore, it follows from the condition (1) that if

$$r=\frac{1}{2}+\frac{\sqrt{3}}{2}i,s=1-\sqrt{3}i,t=\frac{1}{e^2}\left(\frac{\sqrt{3}}{2}+\frac{1}{2}i\right),$$

then ${\mathcal{J}}_{r,s,t}$ is a conjugation on ${\mathcal{F}}^2$. Since $\mathcal{J}={\mathcal{J}}_{1,0,1}$, ${\mathcal{J}}_{r,s,t}$ can be viewed as an extension of $\mathcal{J}$ on ${\mathcal{F}}^2$.

Definition 2. 

Let C be a conjugation on H. A bounded linear operator $T:H\to H$ is said to be complex symmetric with C if $CTC=T^{*}$.

Interestingly, if a bounded linear operator $T:H\to H$ is complex symmetric, then T can be written as a symmetric matrix relative to a certain orthonormal basis of H (see [2]). This shows that the complex symmetric operators can be regarded as a generalization of the symmetric matrices. The complex symmetric operators on abstract Hilbert spaces were studied by Garcia, Putinar, and Wogen in [2,4,5,6]. Furthermore, numerous studies also have been conducted about such operators on holomorphic function spaces (see [7,8,9,10,11,12,13]).

Concerning complex symmetric operators, Chō et al. introduced the concept of m-complex symmetric operators in [14], which has been the subject of ongoing investigation in subsequent studies in [15,16].

Definition 3. 

Let $m\in \mathbb{N}$ and C be a conjugation on H. A bounded linear operator $T:H\to H$ is said to be m-complex symmetric with C if

$$\begin{array}{c}\hfill \sum _{j=0}^m{(-1)}^{m-j}{C}_m^j{T^{*}}^{j}C{T}^{m-j}C=0,\end{array}$$

where $C_m^j=\frac{m(m-1)\cdots (m-j+1)}{j!}$.

By Definition 3, 1-complex symmetric operator is just the complex symmetric operator, and a bounded linear operator $T:H\to H$ is 2-complex symmetric with C if and only if

$$\begin{array}{c}\hfill C{T}^2-2T^{*}CT+{T^{*}}^{2}C=0.\end{array}$$

Most recently, Hu et al. in [17] characterized 2-complex symmetric weighted composition operators on Hardy space. Therefore, it is natural to study 2-complex symmetric weighted composition operators on some other holomorphic function spaces. However, since the proper description of the adjoint of the weighted composition operators on weighted Bergman space of the half-plane is very difficult, Xue et al. in [18] only characterized three kinds of 2-complex symmetric weighted composition operators on such space. Motivated by the next works of Le [19] and Zhao [20,21,22], Bai et al. in [23] characterized one kind of 2-complex symmetric weighted composition operators on Fock space. Le in [19] showed that the weighted composition operator $W_{u,\phi }$ is bounded on Fock space if and only if the weight function u belongs to Fock space, $\phi \left(z\right)=az+b$ with $\left|a\right|\le 1$, and

$$\begin{array}{c}\hfill M(u,\phi )=sup\left\{{\left|u\left(z\right)\right|}^2{e}^{\left(\right|\phi \left(z\right){|}^2-\left|z{|}^2\right)}:z\in \mathbb{C}\right\}<\infty ,\end{array}$$

(2)

where $\mathbb{C}$ denotes the complex plane. Zhao in [20,21,22] studied the unitary, invertible and normal weighted composition operator $W_{u,\phi }$ with the symbol $\phi \left(z\right)=az+b$, $\left|a\right|\le 1$, and the weight function $u\left(z\right)=e^{\overline{p}z}$ on Fock space.

Next, we turn to the study of the order bounded operators on holomorphic function spaces. Let $0<p<\infty$, $(X,\rho )$ be a metric space of holomorphic functions defined over a given domain $\mathsf{\Omega }$, $(\mathsf{\Omega },\mathcal{A},\mu )$ a measure space, and

$$\begin{array}{c}\hfill L^p(\mathsf{\Omega },\mathcal{A},\mu )=\left\{f|f:\mathsf{\Omega }\to \mathbb{C}\mathrm{is}\mathrm{measurable}\mathrm{and}{\int }_{\mathsf{\Omega }}{\left|f\right|}^{p}d\mu <\infty \right\}.\end{array}$$

Definition 4. 

An operator $T:(X,\rho )\to L^p(\mathsf{\Omega },\mathcal{A},\mu )$ is said to be order bounded if there exists a nonnegative function $g\in L^p(\mathsf{\Omega },\mathcal{A},\mu )$ such that for all $f\in X$ with $\rho (0,f)\le 1$, it holds

$$\left|T\right(f\left)\right(x\left)\right|\le g\left(x\right),a.e.\left[\mu \right].$$

The definition was introduced by Hunziker and Jarchow in [24] in the case when $(X,\rho )$ is a quasi-Banach space of holomorphic functions on $\mathsf{\Omega }$. This is an interesting property. For instance, Kwapień in [25] and Schwartz in [26] proved that if X is a Banach space, $\mu$ is any measure, $1\le p<\infty$ and $T:X\to L^p\left(\mu \right)$ is order bounded, then T is p-integral. Ueki in [27] proved that every order bounded weighted composition operator between weighted Bergman spaces is bounded. Recently, it also has been studied by experts and scholars. For example, Gao et al. in [28] proved a sufficient condition for order bounded weighted composition operators between Dirichlet spaces. Sharma in [29] proved that this sufficient condition is also necessary.

Inspired by the aforementioned researches, it is natural to desire an extension of these investigations to weighted superposition operators on Fock spaces. Hence, the primary focus of this paper is the characterization of complex symmetric, 2-complex symmetric and order bounded weighted superposition operators within Fock spaces.

2. Preliminaries

Here, denote by $\mathbb{C}$ the complex plane and by $H\left(\mathbb{C}\right)$ the set of all holomorphic functions on $\mathbb{C}$. Let $0<p\le \infty$. The goal of this paper is to study the operator on Fock spaces ${\mathcal{F}}^p$.

If $0<p<\infty$, then ${\mathcal{F}}^p$ is the space of all functions $f\in H\left(\mathbb{C}\right)$ such that

$$\begin{array}{c}\hfill {\parallel f\parallel }_p={\left(\frac{p}{2\pi }{\int }_{\mathbb{C}}{\left|f\left(z\right)\right|}^p{e}^{-\frac{{p\left|z\right|}^2}{2}}d\nu \left(z\right)\right)}^{\frac{1}{p}}<\infty ,\end{array}$$

where $\nu \left(z\right)$ denotes the usual Lebesgue area measure on $\mathbb{C}$. In particular, if $p=2$, then the space ${\mathcal{F}}^2$ is a Hilbert space with the inner product

$$\begin{array}{c}\hfill \langle f,g\rangle =\frac{1}{\pi }{\int }_{\mathbb{C}}f\left(z\right)\overline{g\left(z\right)}{e}^{-{\left|z\right|}^2}d\nu \left(z\right).\end{array}$$

The reproducing kernel functions of ${\mathcal{F}}^2$ are given by

$$K_w\left(z\right)=e^{\overline{w}z},z\in \mathbb{C}.$$

A direct calculation shows that $\parallel K_w{\parallel }_2=e^{\frac{{\left|w\right|}^2}{2}}$. Let $k_w$ be the normalization of $K_w$, that is,

$$k_w\left(z\right)=\frac{K_w\left(z\right)}{\parallel K_w{\parallel }_2}=e^{\overline{w}z-\frac{{\left|w\right|}^2}{2}},z\in \mathbb{C}.$$

From the calculations, it follows that $k_w\in {\mathcal{F}}^p$ and $\parallel k_w{\parallel }_p=1$. By [30], there exists a positive constant C independent of $f\in {\mathcal{F}}^p$ and $z\in \mathbb{C}$ such that

$$\begin{array}{c}\hfill \left|f\left(z\right)\right|\le {\left(\frac{2\pi }{p}\right)}^{\frac{1}{p}}{e}^{\frac{{\left|z\right|}^2}{2}}{\parallel f\parallel }_p.\end{array}$$

(3)

If $p=\infty$, then ${\mathcal{F}}^{\infty }$ is a Banach space of all functions $f\in H\left(\mathbb{C}\right)$ such that

$$\begin{array}{c}\hfill {\parallel f\parallel }_{\infty }=\underset{z\in \mathbb{C}}{sup}\left|f\left(z\right)\right|e^{-\frac{{\left|z\right|}^2}{2}}<\infty .\end{array}$$

The conclusion (3) with the factor ${\left(\frac{2\pi }{p}\right)}^{\frac{1}{p}}$ replaced by 1 still holds for $p=\infty$.

Here, I would like to mention the role that Fock space ${\mathcal{F}}^2$ plays in physics. For instance, ${\mathcal{F}}^2$ is used to describe systems with varying numbers of particles such as the states of quantum harmonic oscillators, and the reproducing kernels in ${\mathcal{F}}^2$ are used to characterize the coherent states in quantum physics. For more about Fock spaces see [30].

For two given holomorphic functions $\phi$ and u defined on $\mathbb{C}$, the weighted superposition operator usually denoted by $S_{(u,\phi )}$ on or between the subspaces of $H\left(\mathbb{C}\right)$ is defined by

$$S_{(u,\phi )}f\left(z\right)=u\left(z\right)\phi \left(f\left(z\right)\right).$$

When $u=1$, it is the superposition operator usually denoted by $S_{\phi }$. While $\phi \left(z\right)=z$, it is the multiplication operator usually denoted by $M_u$. The weighted superposition operator is a typical example of nonlinear operators. Recently, it has been studied on Bloch and Bergman spaces in [31]. The questions for $S_{(u,\phi )}$ is technically more difficult than the questions for $S_{\phi }$ because of the presence of the multiplier u. We now immediately illustrate this with an example which is similar to that in [32]. Assume that $f\left(z\right)=1$ and $g\left(z\right)=z$. Then, for a nonzero holomorphic function $\phi$ on $\mathbb{C}$, we have

$$\begin{array}{c}\hfill \parallel S_{(u,\phi )}{f\parallel }_2={\left|\phi \left(1\right)\right|\parallel u\parallel }_2\end{array}$$

and

$$\begin{array}{c}\hfill \parallel S_{(u,\phi )}{g\parallel }_2={\parallel u\phi \parallel }_2,\end{array}$$

which shows that if $S_{(u,\phi )}$ maps ${\mathcal{F}}^2$ into itself, then u and $u\phi$ belong to ${\mathcal{F}}^2$. Now, consider the function $h\left(z\right)=sin\left(z^2\right)/z^2$. When $\left|z\right|=r$ gets sufficiently large, we have ${\left|h\left(z\right)\right|}^2\asymp e^{r^2}/r^4$, where the notation $a\asymp b$ means that there exist two positive constants $C_1$ and $C_2$ such that $C_{1}b\le a\le C_{2}b$. Then

$$\begin{array}{c}\hfill {\int }_0^{2\pi }{\int }_0^{\infty }{\left|h\left(r{e}^{it}\right)\right|}^{2}r{e}^{-r^2}drdt<\infty ,\end{array}$$

which implies that $h\in {\mathcal{F}}^2$. But, since

$$|S_{(z^2,z)}{h\left(z\right)|}^{2}r{e}^{-r^2}={\left|z\right|}^4{\left|h\left(z\right)\right|}^{2}r{e}^{-r^2}\asymp r$$

for sufficiently large r, we see that $S_{(z^2,z)}h$ does not belong to ${\mathcal{F}}^2$. This shows that $S_{(z^2,z)}$ does not map ${\mathcal{F}}^2$ into itself. It exhibits the existence of interplay between u and $\phi$ in defining $S_{(u,\phi )}$ on ${\mathcal{F}}^2$. It is this that makes the study of such operators so interesting. In order to identify the bounded weighted superposition operators between Fock spaces, Mengestie in [32] obtained the following result.

Theorem 1. 

Let φ and u be holomorphic functions on $\mathbb{C}$ and $0<p,q\le \infty$.

(a) If $p\le q$, then $S_{(u,\phi )}:{\mathcal{F}}^p\to {\mathcal{F}}^q$ is bounded if and only if either and u is a constant and $\phi \left(z\right)=az+b$ for some a, $b\in \mathbb{C}$ or $u\in {\mathcal{F}}^q$ and φ is a constant.

(b) If $p>q$, then $S_{(u,\phi )}:{\mathcal{F}}^p\to {\mathcal{F}}^q$ is bounded if and only if $u\in {\mathcal{F}}^q$ and φ is a constant.

Remark 1. 

(a) Theorem 1 shows except in the case when φ is a constant, every weighted superposition operator from one into another Fock space is a superposition operator since $S_{(u,\phi )}f=\alpha af+\alpha b=S_{\psi }f$, where $\psi \left(z\right)=\alpha az+\alpha b$.

(b) Theorem 1 implies that the operator $S_{(u,\phi )}$ is bounded on ${\mathcal{F}}^2$ if and only if either u is a constant and $\phi \left(z\right)=az+b$ for some a, $b\in \mathbb{C}$ or $u\in {\mathcal{F}}^2$ and φ is a constant. Therefore, in Section 3, Section 4 and Section 5 of the paper, we consider the operators $S_{(u,\phi )}$ defined by the functions $u=\alpha$ a constant and $\phi \left(z\right)=az+b$ for some a, $b\in \mathbb{C}$ or $u\in {\mathcal{F}}^2$ and $\phi =a$ a constant.

One of the main difficulties in dealing with nonlinear operator theory is the lack of reasonable definitions that can be applied to a wide range of operators. To this end, Felke et al. thought that reasonable definition is in the sense that it should reduce to the familiar property in case of linear operators and attempts to preserve or share some of the useful linear structures. This idea motivated Felke et al. in [33] to study the various spectral forms for the weighted superposition operators on Fock spaces.

Let the operator $S_{(u,\phi )}$ be bounded on ${\mathcal{F}}^2$. Also inspired by the idea of Felke et al., we define the “adjoint” operator of $S_{(u,\phi )}$ by

$$\begin{array}{c}\hfill S_{(u,\phi )}^{*}f\left(z\right)=\langle f,S_{(u,\phi )}{K}_z\rangle \end{array}$$

for every $f\in {\mathcal{F}}^2$ and $z\in \mathbb{C}$. At this moment, replacing the operator T by $S_{(u,\phi )}$ in Definitions 1–4, respectively, we obtain the corresponding definitions about the operator $S_{(u,\phi )}$, and consequently this allows us to study the complex symmetry, 2-complex symmetry and order boundedness for the operators $S_{(u,\phi )}$ on Fock spaces in this paper.

3. Auxiliary Lemmas

First, the proofs of Lemmas 1 and 2 can be obtained since the linear span of the reproducing kernel functions $\{K_w:w\in \mathbb{C}\}$ is dense in ${\mathcal{F}}^2$, and then we omit the detailed proofs.

Lemma 1. 

Let the operator $S_{(u,\phi )}$ be bounded on ${\mathcal{F}}^2$ and C be a conjugation on ${\mathcal{F}}^2$. Then the operator $S_{(u,\phi )}$ is 2-complex symmetric on ${\mathcal{F}}^2$ with the conjugation C if and only if for each w, $z\in \mathbb{C}$ it holds

$$\begin{array}{c}\hfill (C{S}_{(u,\phi )}^2-2S_{(u,\phi )}^{*}C{S}_{(u,\phi )}+S_{(u,\phi )}^{*2}C)K_w\left(z\right)=0.\end{array}$$

Lemma 2. 

Let the operator $S_{(u,\phi )}$ be bounded on ${\mathcal{F}}^2$ and C be a conjugation on ${\mathcal{F}}^2$. Then the operator $S_{(u,\phi )}$ is complex symmetric on ${\mathcal{F}}^2$ with the conjugation C if and only if for each w, $z\in \mathbb{C}$ it holds

$$\begin{array}{c}\hfill (C{S}_{(u,\phi )}-S_{(u,\phi )}^{*}C)K_w\left(z\right)=0.\end{array}$$

In order to fulfill the research objectives of the paper, we need the formula for the adjoint as follows.

Lemma 3. 

Let u be a constant α and $\phi \left(z\right)=az+b$ for some a, $b\in \mathbb{C}$. Then for each $f\in {\mathcal{F}}^2$ it follows that

$$S_{(u,\phi )}^{*}f\left(z\right)=\overline{\alpha }\left[\overline{a}f\left(z\right)+\overline{b}f\left(0\right)\right].$$

Proof. 

From Theorem 1, it follows that the operator $S_{(u,\phi )}$ is bounded on ${\mathcal{F}}^2$. For each $f\in {\mathcal{F}}^2$ and $z\in \mathbb{C}$, we have

$$\begin{array}{cc}\hfill S_{(u,\phi )}^{*}f\left(z\right)& =〈S_{(u,\phi )}^{*}f,K_z〉=〈f,S_{(u,\phi )}{K}_z〉\hfill \\ \hfill & =\frac{1}{\pi }{\int }_{\mathbb{C}}f\left(w\right)\overline{S_{(u,\phi )}{K}_z\left(w\right)}{e}^{-{\left|w\right|}^2}d\nu \left(w\right)\hfill \\ \hfill & =\frac{1}{\pi }{\int }_{\mathbb{C}}f\left(w\right)\overline{\alpha (a{K}_z\left(w\right)+b)}{e}^{-{\left|w\right|}^2}d\nu \left(w\right)\hfill \\ \hfill & =\frac{1}{\pi }{\int }_{\mathbb{C}}f\left(w\right)\overline{\alpha (a{e}^{\overline{z}w}+b)}{e}^{-{\left|w\right|}^2}d\nu \left(w\right)\hfill \\ \hfill & =\frac{1}{\pi }{\int }_{\mathbb{C}}f\left(w\right)\left(\overline{\alpha }\overline{a}{e}^{z\overline{w}}+\overline{\alpha }\overline{b}\right)e^{-{\left|w\right|}^2}d\nu \left(w\right)\hfill \\ \hfill & =\overline{\alpha }\overline{a}\frac{1}{\pi }{\int }_{\mathbb{C}}f\left(w\right)e^{z\overline{w}}{e}^{-{\left|w\right|}^2}d\nu \left(w\right)+\overline{\alpha }\overline{b}\frac{1}{\pi }{\int }_{\mathbb{C}}f\left(w\right)e^{-{\left|w\right|}^2}d\nu \left(w\right)\hfill \\ \hfill & =\overline{\alpha }\overline{a}\langle f,K_z\rangle +\overline{\alpha }\overline{b}\langle f,K_0\rangle \hfill \\ \hfill & =\overline{\alpha }\left[\overline{a}f\left(z\right)+\overline{b}f\left(0\right)\right],\hfill \end{array}$$

from which the desired result follows. The proof is completed. □

Similar to Lemma 3, we have the following result. So, we omit the proof.

Lemma 4. 

Let $u\in {\mathcal{F}}^2$ and φ be a constant a. Then for each $f\in {\mathcal{F}}^2$ it follows that

$$S_{(u,\phi )}^{*}f\left(z\right)=\frac{\overline{a}}{\pi }{\int }_{\mathbb{C}}f\left(w\right)\overline{u\left(w\right)}{e}^{-{\left|w\right|}^2}d\nu \left(w\right).$$

For any $n\in \mathbb{N}\cup \left\{0\right\}$, let

$$e_n\left(z\right)=\sqrt{\frac{1}{n!}}{z}^n.$$

By [30], the set ${\left(e_n\right)}_{n\in \mathbb{N}}$ is an orthonormal basis for ${\mathcal{F}}^2$, and then $\parallel z^n{\parallel }_2=\sqrt{n!}$.

If we consider the function $u\left(z\right)=a_k{z}^k$, $k\in \mathbb{N}\cup \left\{0\right\}$, in Lemma 4, then we obtain the following result.

Lemma 5. 

Let $u\left(z\right)=a_k{z}^k$, $k\in \mathbb{N}\cup \left\{0\right\}$, and φ be a constant a. Then

$$S_{(u,\phi )}^{*}{\mathcal{J}}_{r,s,t}{K}_w\left(z\right)=t\overline{a}{\overline{a}}_k{e}^{sw}{(s+rw)}^k.$$

Proof. 

From Lemma 4, we have

$$\begin{array}{cc}\hfill S_{(u,\phi )}^{*}{\mathcal{J}}_{r,s,t}{K}_w\left(z\right)& =S_{(u,\phi )}^{*}t{e}^{sz}{e}^{w(rz+s)}=\frac{\overline{a}}{\pi }{\int }_{\mathbb{C}}t{e}^{sz}{e}^{w(rz+s)}\overline{u\left(z\right)}{e}^{-{\left|z\right|}^2}d\nu \left(z\right)\hfill \\ \hfill & =\frac{\overline{a}{\overline{a}}_k}{\pi }{\int }_{\mathbb{C}}t{e}^{sz}{e}^{w(rz+s)}\overline{z^k}{e}^{-{\left|z\right|}^2}d\nu \left(z\right)\hfill \\ \hfill & =\frac{t\overline{a}{\overline{a}}_k}{\pi }{e}^{sw}{\int }_{\mathbb{C}}{e}^{(s+rw)z}\overline{z^k}{e}^{-{\left|z\right|}^2}d\nu \left(z\right)\hfill \\ \hfill & =\frac{t\overline{a}{\overline{a}}_k}{\pi }{e}^{sw}\sum _{n=0}^{\infty }\frac{{(s+rw)}^n}{n!}{\int }_{\mathbb{C}}{z}^n\overline{z^k}{e}^{-{\left|z\right|}^2}d\nu \left(z\right)\hfill \\ \hfill & =t\overline{a}{\overline{a}}_k{e}^{sw}\frac{{(s+rw)}^k}{k!}\frac{1}{\pi }{\int }_{\mathbb{C}}{z}^k\overline{z^k}{e}^{-{\left|z\right|}^2}d\nu \left(z\right)\hfill \\ \hfill & =t\overline{a}{\overline{a}}_k{e}^{sw}\frac{{(s+rw)}^k}{k!}{\parallel z^k\parallel }_2^2\hfill \\ \hfill & =t\overline{a}{\overline{a}}_k{e}^{sw}{(s+rw)}^k.\hfill \end{array}$$

The proof is completed. □

Since $\mathcal{J}={\mathcal{J}}_{1,0,1}$, we have

Corollary 1. 

Let $u\left(z\right)=a_k{z}^k$, $k\in \mathbb{N}\cup \left\{0\right\}$, and φ be a constant a. Then

$$S_{(u,\phi )}^{*}\mathcal{J}{K}_w\left(z\right)=\overline{a}{\overline{a}}_k{w}^k.$$

Next, we will see how the operator $S_{(u,\phi )}^{*}$ acts on the constant-valued functions.

Lemma 6. 

Let $u\left(z\right)=a_k{z}^k$, $k\in \mathbb{N}\cup \left\{0\right\}$, and φ be a constant a. Then

$$\begin{array}{c}\hfill \left(S_{(u,\phi )}^{*}b\right)\left(z\right)=\left\{\begin{array}{c}0,k\in \mathbb{N}\hfill \\ \overline{a}{\overline{a}}_{k}b,k=0.\hfill \end{array}\right.\end{array}$$

(4)

Proof. 

It is clear that the constant-valued function $f\left(z\right)\equiv b$ belongs to ${\mathcal{F}}^2$. Then, for each $z\in \mathbb{C}$, we have

$$\begin{array}{c}\hfill \left(S_{(u,\phi )}^{*}b\right)\left(z\right)=\langle S_{(u,\phi )}^{*}b,K_z\rangle =\langle b,S_{(u,\phi )}{K}_z\rangle =\langle b,a{a}_k{w}^k\rangle =\overline{a}{\overline{a}}_{k}b\langle 1,w^k\rangle ,\end{array}$$

from which the desired result follows. □

Let X be an inner product space with the inner product ${\langle x,y\rangle }_X$. Here, we say that X is also a normed space with the norm ${\parallel x\parallel }_X={\langle x,x\rangle }_X^{\frac{1}{2}}$, $x\in X$. Next, we have the following interesting result, although it is not used in the paper.

Lemma 7. 

Let the operator $T_n$ be complex symmetric on H with the conjugation C for every $n\in \mathbb{N}$. If the sequence ${\left(T_n\right)}_{n\in \mathbb{N}}$ converges to the operator T in the operator norm topology on H, then T is complex symmetric on H with the conjugation C.

Proof. 

Since ${\left(T_n\right)}_{n\in \mathbb{N}}$ converges to the operator T in H, for arbitrary $\epsilon >0$ there exists some $N\in \mathbb{N}$ such that for all $n\ge N$,

$$\parallel T_n{-T\parallel }_{H\to H}<\frac{\epsilon }{2},$$

where ${\parallel S\parallel }_{H\to H}$ denotes the norm of the operator S on H. Because

$$\parallel T_n{-T\parallel }_{H\to H}={\parallel T_n^{*}-T^{*}\parallel }_{H\to H},$$

it holds

$$\parallel T_n^{*}-T^{*}{\parallel }_{H\to H}<\frac{\epsilon }{2}$$

for all $n\ge N$. Since each $T_n$ is complex symmetric on H with C, we have $(C{T}_n-T_n^{*}C)f=0$ for all $f\in H$. Then, for $f\in H$ with ${\parallel f\parallel }_H=1$ and for $n\ge N$, it follows that

$$\begin{array}{cc}\hfill \parallel (CT-T^{*}C){f\parallel }_H& =\parallel (C{T}_n-T_n^{*}C)f-(CT-T^{*}C){f\parallel }_H\hfill \\ \hfill & =\parallel (C{T}_n-CT)f-(T_n^{*}C-T^{*}C){f\parallel }_H\hfill \\ \hfill & \le \parallel (C{T}_n-CT){f\parallel }_H+{\parallel (T_n^{*}C-T^{*}C)f\parallel }_H\hfill \\ \hfill & \le \parallel C{T}_n{-CT\parallel }_{H\to H}{\parallel f\parallel }_H+\parallel T_n^{*}C-T^{*}{C\parallel }_{H\to H}{\parallel f\parallel }_H\hfill \\ \hfill & =\parallel C{T}_n{-CT\parallel }_{H\to H}+{\parallel T_n^{*}C-T^{*}C\parallel }_{H\to H}\hfill \\ \hfill & \le \parallel T_n{-T\parallel }_{H\to H}+{\parallel T_n^{*}-T^{*}\parallel }_{H\to H}\hfill \\ \hfill & <\epsilon ,\hfill \end{array}$$

which shows

$$\begin{array}{c}\hfill \underset{{\parallel f\parallel }_H=1}{sup}{\parallel (CT-T^{*}C)f\parallel }_H\le \epsilon ,\end{array}$$

that is,

$$\begin{array}{c}\hfill \parallel CT-T^{*}{C\parallel }_{H\to H}\le \epsilon .\end{array}$$

From the arbitrariness of $\epsilon$, we obtain $\parallel CT-T^{*}{C\parallel }_{H\to H}=0$. That is, $CT=T^{*}C$, which shows that T is complex symmetric on H with C. The proof is completed. □

4. Complex Symmetric Operators $S_{(\alpha ,az+b)}$ on ${\mathcal{F}}^2$

Theorem 1 tells us that the operator $S_{(u,\phi )}$ is bounded on ${\mathcal{F}}^2$ if and only if one of the conditions holds: (i) u is a constant and $\phi \left(z\right)=az+b$ for some a, $b\in \mathbb{C}$; (ii) $u\in {\mathcal{F}}^2$ and $\phi$ is a constant. Therefore, in this section we study the operator $S_{(u,\phi )}$ on ${\mathcal{F}}^2$ defined by the functions $\phi \left(z\right)=az+b$ and $u\left(z\right)=\alpha$.

The first result is following.

Theorem 2. 

Let u be a constant α and $\phi \left(z\right)=az+b$ for some a, $b\in \mathbb{C}$. Then the operator $S_{(u,\phi )}$ is complex symmetric on ${\mathcal{F}}^2$ with the conjugation $\mathcal{J}$.

Proof. 

First, it follows from Theorem 1 that the operator $S_{(u,\phi )}$ is bounded on ${\mathcal{F}}^2$. For each w, $z\in \mathbb{C}$, from Lemma 3 we have

$$S_{(u,\phi )}^{*}\mathcal{J}{K}_w\left(z\right)=S_{(u,\phi )}^{*}{e}^{wz}=\overline{\alpha }\left(\overline{a}{e}^{wz}+\overline{b}\right)=\overline{\alpha }\overline{a}{e}^{wz}+\overline{\alpha }\overline{b}.$$

From the calculations, we also have

$$\mathcal{J}{S}_{(u,\phi )}{K}_w\left(z\right)=\mathcal{J}\left[\alpha (a{K}_w\left(z\right)+b)\right]=\overline{\alpha }\overline{a}{e}^{wz}+\overline{\alpha }\overline{b}$$

for each w, $z\in \mathbb{C}$. So, we get that

$$\mathcal{J}{S}_{(u,\phi )}{K}_w\left(z\right)=S_{(u,\phi )}^{*}\mathcal{J}{K}_w\left(z\right)$$

for each w, $z\in \mathbb{C}$. By Lemma 3.2, the operator $S_{(u,\phi )}$ is complex symmetric on ${\mathcal{F}}^2$ with the conjugation $\mathcal{J}$. The proof is completed. □

Theorem 3. 

Let u be a constant α and $\phi \left(z\right)=az+b$ for some a, $b\in \mathbb{C}$. Then the operator $S_{(u,\phi )}$ is complex symmetric on ${\mathcal{F}}^2$ with the conjugation ${\mathcal{J}}_{r,s,t}$ if and only if one of the conditions holds: (i) $\alpha b=0$; (ii) $s=0$.

Proof. 

For each w, $z\in \mathbb{C}$, from Lemma 3 we have

$$S_{(u,\phi )}^{*}{\mathcal{J}}_{r,s,t}{K}_w\left(z\right)=S_{(u,\phi )}^{*}t{e}^{sz}{e}^{w(rz+s)}=t\overline{\alpha }\overline{a}{e}^{sz}{e}^{w(rz+s)}+t\overline{\alpha }\overline{b}{e}^{sw}.$$

On the other hand, we have

$${\mathcal{J}}_{r,s,t}{S}_{(u,\phi )}{K}_w\left(z\right)={\mathcal{J}}_{r,s,t}\left[\alpha (a{K}_w\left(z\right)+b)\right]=t\overline{\alpha }\overline{a}{e}^{sz}{e}^{w(rz+s)}+t\overline{\alpha }\overline{b}{e}^{sz}$$

for each w, $z\in \mathbb{C}$. Therefore, by Lemma 3.2, the operator $S_{(u,\phi )}$ is complex symmetric on ${\mathcal{F}}^2$ with the conjugation ${\mathcal{J}}_{r,s,t}$ if and only if $\overline{\alpha }\overline{b}{e}^{sz}=\overline{\alpha }\overline{b}{e}^{sw}$ for all w, $z\in \mathbb{C}$. This is equivalent to either $\alpha b=0$ or $s=0$. The proof is completed. □

In the next two results, we assume that $\alpha ,a\ne 0$. Otherwise, if $\alpha =0$, then $S_{(u,\phi )}$ is zero operator on ${\mathcal{F}}^2$; if $a=0$, then $S_{(u,\phi )}f=\alpha b$ for all $f\in {\mathcal{F}}^2$, and then the study makes little sense.

Theorem 4. 

Let u be a constant α and $\phi \left(z\right)=az+b$ for some a, $b\in \mathbb{C}$. Then the operator $S_{(u,\phi )}$ is 2-complex symmetric on ${\mathcal{F}}^2$ with the conjugation $\mathcal{J}$ if and only if one of the conditions holds: (i) $b=0$; (ii) $b\ne 0$ and $\alpha (a+b)=1$.

Proof. 

For each w, $z\in \mathbb{C}$, from Lemma 3 and a direct calculation we have

$$\begin{array}{cc}\hfill \mathcal{J}{S}_{(u,\phi )}^2{K}_w\left(z\right)& =\mathcal{J}{S}_{(u,\phi )}\left[\alpha (a{K}_w\left(z\right)+b)\right]\hfill \\ \hfill & =\mathcal{J}\alpha \left[\alpha a^2{K}_w\left(z\right)+\alpha ab+b\right]\hfill \\ \hfill & =\mathcal{J}\left[{\alpha }^2{a}^2{K}_w\left(z\right)+{\alpha }^{2}ab+\alpha b\right]\hfill \\ \hfill & ={\overline{\alpha }}^2{\overline{a}}^2{e}^{wz}+{\overline{\alpha }}^2\overline{a}\overline{b}+\overline{\alpha }\overline{b},\hfill \end{array}$$

(5)

$$\begin{array}{cc}\hfill S_{(u,\phi )}^{*}\mathcal{J}{S}_{(u,\phi )}{K}_w\left(z\right)& =S_{(u,\phi )}^{*}\mathcal{J}\left[\alpha (a{K}_w\left(z\right)+b)\right]=S_{(u,\phi )}^{*}\left(\overline{\alpha }\overline{a}{e}^{wz}+\overline{\alpha }\overline{b}\right)\hfill \\ \hfill & =\overline{\alpha }\left[\overline{a}(\overline{\alpha }\overline{a}{e}^{wz}+\overline{\alpha }\overline{b})+\overline{b}(\overline{\alpha }\overline{a}+\overline{\alpha }\overline{b})\right]\hfill \\ \hfill & =\overline{\alpha }\left(\overline{\alpha }{\overline{a}}^2{e}^{wz}+2\overline{\alpha }\overline{a}\overline{b}+\overline{\alpha }{\overline{b}}^2\right)\hfill \\ \hfill & ={\overline{\alpha }}^2{\overline{a}}^2{e}^{wz}+2{\overline{\alpha }}^2\overline{a}\overline{b}+{\overline{\alpha }}^2{\overline{b}}^2,\hfill \end{array}$$

(6)

and

$$\begin{array}{cc}\hfill S_{(u,\phi )}^{*2}\mathcal{J}{K}_w\left(z\right)& =S_{(u,\phi )}^{*2}{e}^{wz}=S_{(u,\phi )}^{*}(\overline{\alpha }\overline{a}{e}^{wz}+\overline{\alpha }\overline{b})\hfill \\ \hfill & ={\overline{\alpha }}^2{\overline{a}}^2{e}^{wz}+2{\overline{\alpha }}^2\overline{a}\overline{b}+{\overline{\alpha }}^2{\overline{b}}^2.\hfill \end{array}$$

(7)

From (5)–(7) and Lemma 1, it follows that the operator $S_{(u,\phi )}$ is 2-complex symmetric on ${\mathcal{F}}^2$ with the conjugation $\mathcal{J}$ if and only if

$$\begin{array}{c}\hfill \overline{\alpha }\overline{b}(1-\overline{\alpha }\overline{b})={\overline{\alpha }}^2\overline{a}\overline{b}.\end{array}$$

(8)

Assume that the operator $S_{(u,\phi )}$ is 2-complex symmetric on ${\mathcal{F}}^2$ with the conjugation $\mathcal{J}$. Then, from (8), we obtain that either $b=0$ or $b\ne 0$ and $\alpha (a+b)=1$.

Conversely, assume that either $b=0$ or $b\ne 0$ and $\alpha (a+b)=1$. It is clear that if $b=0$, then (8) holds, which shows that the operator $S_{(u,\phi )}$ is 2-complex symmetric on ${\mathcal{F}}^2$ with the conjugation $\mathcal{J}$. If $\alpha (a+b)=1$, that is, $\alpha a=1-\alpha b$, then

$$\begin{array}{c}\hfill {\overline{\alpha }}^2\overline{a}\overline{b}=\left(\overline{\alpha }\overline{a}\right)\left(\overline{\alpha }\overline{b}\right)=\overline{\alpha }\overline{b}(1-\overline{\alpha }\overline{b}).\end{array}$$

That is, (8) holds, which shows that the operator $S_{(u,\phi )}$ is 2-complex symmetric on ${\mathcal{F}}^2$ with the conjugation $\mathcal{J}$. The proof is completed. □

By Theorem 4, we can give some examples of 2-complex symmetric operators $S_{(u,\phi )}$ on ${\mathcal{F}}^2$ with the conjugation $\mathcal{J}$.

Example 1. 

(a) Let $\phi \left(z\right)=\frac{1}{2}z+i$ and $u\left(z\right)=\frac{5}{8}-\frac{5}{4}i$ for each $z\in \mathbb{C}$. Then the operator $S_{(u,\phi )}$ is 2-complex symmetric on ${\mathcal{F}}^2$ with the conjugation $\mathcal{J}$.

(b) Let $\phi \left(z\right)=iz-\frac{\sqrt{2}}{2}i$ and $u\left(z\right)=\frac{1}{1-\sqrt{2}}i$ for each $z\in \mathbb{C}$. Then the operator $S_{(u,\phi )}$ is 2-complex symmetric on ${\mathcal{F}}^2$ with the conjugation $\mathcal{J}$.

(c) Let $\phi \left(z\right)=(1+i)z$ and $u\left(z\right)=\pi +ei$ for each $z\in \mathbb{C}$. Then the operator $S_{(u,\phi )}$ is 2-complex symmetric on ${\mathcal{F}}^2$ with the conjugation $\mathcal{J}$.

Proof. 

(a) From a direct calculation, it follows that $\alpha (a+b)=1$. By Theorem 4, the operator $S_{(u,\phi )}$ is 2-complex symmetric on ${\mathcal{F}}^2$ with the conjugation $\mathcal{J}$.

(b) It can be similarly proved. So, the details are omitted.

(c) It is clear that $b=0$ and u is a constant. By Theorem 4, the operator $S_{(u,\phi )}$ is 2-complex symmetric on ${\mathcal{F}}^2$ with the conjugation $\mathcal{J}$. □

Now, we begin to characterize 2-complex symmetric operator $S_{(u,\phi )}$ on ${\mathcal{F}}^2$ with the conjugation ${\mathcal{J}}_{r,s,t}$.

Theorem 5. 

Let u be a constant α and $\phi \left(z\right)=az+b$ for some a, $b\in \mathbb{C}$. Then the operator $S_{(u,\phi )}$ is 2-complex symmetric on ${\mathcal{F}}^2$ with the conjugation ${\mathcal{J}}_{r,s,t}$ if and only if one of the conditions holds: (i) $b=0$; (ii) $b\ne 0$, $s=0$ and $\alpha (a+b)=1$.

Proof. 

For each w, $z\in \mathbb{C}$, from Lemma 3 we have

$$\begin{array}{cc}\hfill {\mathcal{J}}_{r,s,t}{S}_{(u,\phi )}^2{K}_w\left(z\right)& ={\mathcal{J}}_{r,s,t}{S}_{(u,\phi )}\left[\alpha (a{K}_w\left(z\right)+b)\right]\hfill \\ \hfill & ={\mathcal{J}}_{r,s,t}\left[{\alpha }^2{a}^2{K}_w\left(z\right)+{\alpha }^{2}ab+\alpha b\right]\hfill \\ \hfill & =t{e}^{sz}\left({\overline{\alpha }}^2{\overline{a}}^2{e}^{w(rz+s)}+{\overline{\alpha }}^2\overline{a}\overline{b}+\overline{\alpha }\overline{b}\right)\hfill \\ \hfill & ={\overline{\alpha }}^2{\overline{a}}^{2}t{e}^{w(rz+s)+sz}+\left({\overline{\alpha }}^2\overline{a}\overline{b}+\overline{\alpha }\overline{b}\right)t{e}^{sz},\hfill \end{array}$$

(9)

$$\begin{array}{cc}\hfill S_{(u,\phi )}^{*}{\mathcal{J}}_{r,s,t}{S}_{(u,\phi )}{K}_w\left(z\right)& =S_{(u,\phi )}^{*}{\mathcal{J}}_{r,s,t}\left[\alpha (a{K}_w\left(z\right)+b)\right]\hfill \\ \hfill & =S_{(u,\phi )}^{*}t{e}^{sz}\left[\overline{\alpha }(\overline{a}{e}^{w(rz+s)}+\overline{b})\right]\hfill \\ \hfill & =S_{(u,\phi )}^{*}\left[\overline{\alpha }(\overline{a}t{e}^{w(rz+s)+sz}+\overline{b}t{e}^{sz})\right]\hfill \\ \hfill & ={\overline{\alpha }}^2{\overline{a}}^{2}t{e}^{w(rz+s)+sz}+{\overline{\alpha }}^2\overline{a}\overline{b}t{e}^{sz}+{\overline{\alpha }}^2\overline{a}\overline{b}t{e}^{ws}+{\overline{\alpha }}^2{\overline{b}}^{2}t,\hfill \end{array}$$

(10)

and

$$\begin{array}{cc}\hfill S_{(u,\phi )}^{*2}{\mathcal{J}}_{r,s,t}{K}_w\left(z\right)& =S_{(u,\phi )}^{*2}t{e}^{sz}{e}^{w(rz+s)}\hfill \\ \hfill & =S_{(u,\phi )}^{*}\left[\overline{\alpha }\overline{a}t{e}^{w(rz+s)+sz}+\overline{\alpha }\overline{b}t{e}^{sw}\right]\hfill \\ \hfill & ={\overline{\alpha }}^2{\overline{a}}^{2}t{e}^{w(rz+s)+sz}+\left(2{\overline{\alpha }}^2\overline{a}\overline{b}+{\overline{\alpha }}^2{\overline{b}}^2\right)t{e}^{sw}.\hfill \end{array}$$

(11)

Since $t\ne 0$ and $\alpha \ne 0$, it follows from (9)–(11) that the operator $S_{(u,\phi )}$ is 2-complex symmetric on ${\mathcal{F}}^2$ with the conjugation ${\mathcal{J}}_{r,s,t}$ if and only if

$$\begin{array}{c}\hfill \left(\overline{\alpha }\overline{a}\overline{b}+\overline{b}\right)e^{sz}+\left(2\overline{\alpha }\overline{a}\overline{b}+\overline{\alpha }{\overline{b}}^2\right)e^{sw}=2\left(\overline{\alpha }\overline{a}\overline{b}{e}^{sz}+\overline{\alpha }\overline{a}\overline{b}{e}^{sw}+\overline{\alpha }{\overline{b}}^2\right).\end{array}$$

(12)

Assume that the operator $S_{(u,\phi )}$ is 2-complex symmetric on ${\mathcal{F}}^2$ with the conjugation ${\mathcal{J}}_{r,s,t}$. Then, by letting $w=z=0$ in (12) we obtain

$$\begin{array}{c}\hfill b\left[\alpha (a+b)-1\right]=0.\end{array}$$

(13)

From (13), we obtain that $b=0$, or $\alpha (a+b)=1$. Letting $w=z$ in (12), we have

$$\begin{array}{c}\hfill \left(\overline{b}+\overline{\alpha }{\overline{b}}^2-\overline{\alpha }\overline{a}\overline{b}\right)e^{sz}=2\overline{\alpha }{\overline{b}}^2.\end{array}$$

(14)

Clearly, if $b=0$, then (14) holds. Now, assume that $b\ne 0$. At this moment, (14) becomes

$$\begin{array}{c}\hfill \left(1+\overline{\alpha }\overline{b}-\overline{\alpha }\overline{a}\right)e^{sz}=2\overline{\alpha }\overline{b}.\end{array}$$

(15)

We have obtained that $\alpha (a+b)=1$. Bringing this into (15), we obtain $e^{sz}=1$, which shows that $s=0$. Combining these, we obtain that if $b\ne 0$, then $\alpha (a+b)=1$ and $s=0$.

Conversely, we will prove that if $b=0$, then (12) holds; if $b\ne 0$, $\alpha (a+b)=1$ and $s=0$, then (12) holds.

First, it is clear that if $b=0$, then (12) holds. Therefore, by Lemma 1, the operator $S_{(u,\phi )}$ is 2-complex symmetric on ${\mathcal{F}}^2$ with the conjugation ${\mathcal{J}}_{r,s,t}$.

Now, we assume that $b\ne 0$, $\alpha (a+b)=1$ and $s=0$. At this moment, by calculating the left of (12), we obtain

$$\begin{array}{cc}\hfill \left(\overline{\alpha }\overline{a}\overline{b}+\overline{b}\right)e^{sz}+\left(2\overline{\alpha }\overline{a}\overline{b}+\overline{\alpha }{\overline{b}}^2\right)e^{sw}& =\overline{\alpha }\overline{a}\overline{b}+\overline{b}+2\overline{\alpha }\overline{a}\overline{b}+\overline{\alpha }{\overline{b}}^2=2\overline{b}+2\overline{\alpha }\overline{a}\overline{b}.\hfill \end{array}$$

By calculating the right of (12), we also obtain

$$\begin{array}{c}\hfill 2\left(\overline{\alpha }\overline{a}\overline{b}{e}^{sz}+\overline{\alpha }\overline{a}\overline{b}{e}^{sw}+\overline{\alpha }{\overline{b}}^2\right)=2\left(\overline{\alpha }\overline{a}\overline{b}+\overline{\alpha }\overline{a}\overline{b}+\overline{\alpha }{\overline{b}}^2\right)=2\overline{b}+2\overline{\alpha }\overline{a}\overline{b}.\end{array}$$

Thus, for this case, (12) holds. By Lemma 1, the operator $S_{(u,\phi )}$ is 2-complex symmetric on ${\mathcal{F}}^2$ with the conjugation ${\mathcal{J}}_{r,s,t}$. The proof is completed. □

By Theorem 5, we can give the example as follows.

Example 2. 

(a) Let $s=0$, $\phi \left(z\right)=\frac{1}{3}z+\frac{2}{3}$ and $u\left(z\right)=1$. Then the operator $S_{(u,\phi )}$ is 2-complex symmetric on ${\mathcal{F}}^2$ with the conjugation ${\mathcal{J}}_{r,s,t}$.

(b) Let $s=0$, $\phi \left(z\right)=-2iz+i$ and $u\left(z\right)=i$. Then the operator $S_{(u,\phi )}$ is 2-complex symmetric on ${\mathcal{F}}^2$ with the conjugation ${\mathcal{J}}_{r,s,t}$.

(c) Let $s=0$, $\phi \left(z\right)=\frac{1}{5}-\frac{2}{5}i$ and $u\left(z\right)=1+2i$. Then the operator $S_{(u,\phi )}$ is 2-complex symmetric on ${\mathcal{F}}^2$ with the conjugation ${\mathcal{J}}_{r,s,t}$.

Proof. 

(a) From a direct calculation, we obtain that $\alpha (a+b)=1$. By Theorem 5, the operator $S_{(u,\phi )}$ is 2-complex symmetric on ${\mathcal{F}}^2$ with the conjugation ${\mathcal{J}}_{r,s,t}$.

(b) and (c) can be similarly proved. So, the details are omitted. □

5. Operators ${\mathit{S}}_{({\mathit{a}}_{\mathbf{0}}+{\mathit{a}}_{\mathbf{1}}\mathit{z}+\cdots +{\mathit{a}}_{\mathit{k}}{\mathit{z}}^{\mathit{k}},\mathit{a})}$ on ${\mathcal{F}}^{\mathbf{2}}$

In Section 4, we consider the operator $S_{(u,\phi )}$ defined by the functions $\phi \left(z\right)=az+b$ and u a constant. From Theorem 1, we know that the case when $\phi$ is a constant and $u\in {\mathcal{F}}^2$ also induces a bounded operator $S_{(u,\phi )}$ on ${\mathcal{F}}^2$. Here, we select some special u to consider. To be precise, u is the polynomial function.

In the first result, we assume that u is not zero function. Otherwise, $S_{(u,\phi )}$ is the zero operator on ${\mathcal{F}}^2$. Our first result is following, which shows that $S_{(u,\phi )}$ still can not escape the fate of the zero operator on ${\mathcal{F}}^2$.

Theorem 6. 

Let φ be a constant a and $u\left(z\right)=a_0+a_{1}z+\cdots +a_k{z}^k$. Then the operator $S_{(u,\phi )}$ is complex symmetric on ${\mathcal{F}}^2$ with the conjugation ${\mathcal{J}}_{r,s,t}$ if and only if one of the conditions holds: (i) $s=0$, $a=0$, $a_j\ne 0$ for some $j\in \{1,2,\cdots ,k\}$; (ii) $s=0$, $a_0\ne 0$, $a_j=0$ for all $j\in \{1,2,\cdots ,k\}$; (iii) $s\ne 0$, $a=0$.

Proof. 

For each w, $z\in \mathbb{C}$, from Lemma 5 we have

$${\mathcal{J}}_{r,s,t}{S}_{(u,\phi )}{K}_w\left(z\right)=t\overline{a}{e}^{sz}\left[{\overline{a}}_0+{\overline{a}}_1(rz+s)+\cdots +{\overline{a}}_k{(rz+s)}^k\right]$$

and

$$S_{(u,\phi )}^{*}{\mathcal{J}}_{r,s,t}{K}_w\left(z\right)=t\overline{a}{e}^{sw}\left[{\overline{a}}_0+{\overline{a}}_1(s+rw)+\cdots +{\overline{a}}_k{(s+rw)}^k\right].$$

Therefore, by Lemma 3.2, the operator $S_{(u,\phi )}$ is complex symmetric on ${\mathcal{F}}^2$ with the conjugation ${\mathcal{J}}_{r,s,t}$ if and only if

$$\begin{array}{c}\hfill \overline{a}{e}^{sz}\left[{\overline{a}}_0+{\overline{a}}_1(s+rz)+\cdots +{\overline{a}}_k{(s+rz)}^k\right]=\overline{a}{e}^{sw}\left[{\overline{a}}_0+{\overline{a}}_1(s+rw)+\cdots +{\overline{a}}_k{(s+rw)}^k\right]\end{array}$$

(16)

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

Assume that $S_{(u,\phi )}$ is complex symmetric on ${\mathcal{F}}^2$ with the conjugation ${\mathcal{J}}_{r,s,t}$. Letting $w=0$ in (16), we have

$$\begin{array}{c}\hfill \overline{a}{e}^{sz}\left[{\overline{a}}_0+{\overline{a}}_1(s+rz)+\cdots +{\overline{a}}_k{(s+rz)}^k\right]=\overline{a}\left({\overline{a}}_0+{\overline{a}}_{1}s+\cdots +{\overline{a}}_k{s}^k\right)\end{array}$$

(17)

for all $z\in \mathbb{C}$. Let $s+rz=W$ and $d=\frac{s}{r}$. Then W can fetch all the complex numbers and (17) becomes

$$\begin{array}{c}\hfill \overline{a}{e}^{d(W-s)}\left({\overline{a}}_0+{\overline{a}}_{1}W+\cdots +{\overline{a}}_k{W}^k\right)=\overline{a}\left({\overline{a}}_{1}s+\cdots +{\overline{a}}_k{s}^k\right)\end{array}$$

(18)

for all $W\in \mathbb{C}$. Taking the derivative of W in (18), we obtain

$$\begin{array}{c}\hfill d\overline{a}\left({\overline{a}}_0+{\overline{a}}_{1}W+\cdots +{\overline{a}}_k{W}^k\right)+\overline{a}\left({\overline{a}}_1+2{\overline{a}}_{2}W+\cdots +k{\overline{a}}_k{W}^{k-1}\right)=0\end{array}$$

(19)

for all $W\in \mathbb{C}$. Then, from (19) we obtain

$$\begin{array}{c}\hfill \left\{\begin{array}{c}d\overline{a}{\overline{a}}_0+\overline{a}{\overline{a}}_1=0\hfill \\ d\overline{a}{\overline{a}}_1+2\overline{a}{\overline{a}}_2=0\hfill \\ \cdots \cdots \hfill \\ d\overline{a}{\overline{a}}_{k-1}+k\overline{a}{\overline{a}}_k=0\hfill \\ d\overline{a}{\overline{a}}_k=0.\hfill \end{array}\right.\end{array}$$

(20)

If $d\ne 0$, then it follows from (20) that

$$\begin{array}{c}\hfill \overline{a}{\overline{a}}_k=\overline{a}{\overline{a}}_{k-1}=\cdots =\overline{a}{\overline{a}}_0=0.\end{array}$$

(21)

Since u is a nonzero function, there exists some $a_j$ such that $a_j\ne 0$. Then, from (21) we obtain $a=0$.

If $d=0$, then it follows from (20) that

$$\begin{array}{c}\hfill \overline{a}{\overline{a}}_k=\overline{a}{\overline{a}}_{k-1}=\cdots =\overline{a}{\overline{a}}_1=0.\end{array}$$

(22)

From (22), we deduce that either $a_j\ne 0$ for some $j\in \{1,2,\cdots ,k\}$, $a=0$ or $a_j=0$ for each $j\in \{1,2,\cdots ,k\}$, $a_0\ne 0$.

Conversely, if one of the conditions (i), (ii) and (iii) holds, then (16) clearly holds, which shows that the operator $S_{(u,\phi )}$ is complex symmetric on ${\mathcal{F}}^2$ with the conjugation ${\mathcal{J}}_{r,s,t}$. The proof is completed. □

Since $\mathcal{J}={\mathcal{J}}_{1,0,1}$, we have the following corollary.

Corollary 2. 

Let φ be a constant a and $u\left(z\right)=a_0+a_{1}z+\cdots +a_k{z}^k$. Then the operator $S_{(u,\phi )}$ is complex symmetric on ${\mathcal{F}}^2$ with the conjugation $\mathcal{J}$ if and only if one of the conditions holds: (i) $a=0$, $a_j\ne 0$ for some $j\in \{1,2,\cdots ,k\}$; (ii) $a_0\ne 0$, $a_j=0$ for all $j\in \{1,2,\cdots ,k\}$.

Next, we turn to characterize the 2-complex symmetry. Our result is following.

Theorem 7. 

Let φ be a constant a and $u\left(z\right)=a_k{z}^k$, $k\in \mathbb{N}$. Then the operator $S_{(u,\phi )}$ is 2-complex symmetric on ${\mathcal{F}}^2$ with the conjugation ${\mathcal{J}}_{r,s,t}$ if and only if $a=0$.

Proof. 

For each w, $z\in \mathbb{C}$, from Lemmas 5 and 6 we have

$${\mathcal{J}}_{r,s,t}{S}_{(u,\phi )}^2{K}_w\left(z\right)=\overline{a}{\overline{a}}_{k}t{e}^{sz}{(s+rz)}^k,$$

$$\begin{array}{cc}\hfill {S^{*}}_{(u,\phi )}^2{\mathcal{J}}_{r,s,t}{K}_w\left(z\right)& =S_{(u,\phi )}^{*}\overline{a}{\overline{a}}_{k}t{e}^{sw}{(s+rw)}^k=0,\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill S_{(u,\phi )}^{*}{\mathcal{J}}_{r,s,t}{S}_{(u,\phi )}{K}_w\left(z\right)& =S_{(u,\phi )}^{*}\overline{a}{\overline{a}}_{k}t{e}^{sz}{(s+rz)}^k\hfill \\ \hfill & =\frac{{\overline{a}}^2{\overline{a}}_k^{2}t}{\pi }{\int }_{\mathbb{C}}{(s+rz)}^k{e}^{sz}{\overline{z}}^k{e}^{-{\left|z\right|}^2}dA\left(z\right)\hfill \\ \hfill & ={\overline{a}}^2{\overline{a}}_k^{2}t\sum _{l=0}^k{C}_k^l{s}^{2(k-l)}{r}^l\frac{k!}{(k-l)!},\hfill \end{array}$$

Therefore, by Lemma 1, the operator $S_{(u,\phi )}$ is 2-complex symmetric on ${\mathcal{F}}^2$ with the conjugation ${\mathcal{J}}_{r,s,t}$ if and only if

$$\begin{array}{c}\hfill \overline{a}{\overline{a}}_k{e}^{sz}{(s+rz)}^k=2{\overline{a}}^2{\overline{a}}_k^2\sum _{l=0}^k{C}_k^l{s}^{2(k-l)}{r}^l\frac{k!}{(k-l)!}\end{array}$$

(23)

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

Assume that the operator $S_{(u,\phi )}$ is 2-complex symmetric on ${\mathcal{F}}^2$ with the conjugation ${\mathcal{J}}_{r,s,t}$. We will divide into two cases for considerations.

Case 1. Assume that $s=0$. Since $r\ne 0$, (23) becomes

$$\begin{array}{c}\hfill \overline{a}{\overline{a}}_k{z}^k-2{\overline{a}}^2{\overline{a}}_k^{2}k!=0\end{array}$$

(24)

for all $z\in \mathbb{C}$. From (24), it follows that $a{a}_k=0$. By the assumption of $u\ne 0$, $a=0$.

Case 2. Assume that $s\ne 0$. (23) is equivalent to

$$\begin{array}{c}\hfill \overline{a}{\overline{a}}_k\sum _{j=0}^{\infty }\frac{s^j}{j!}{z}^j{(s+rz)}^k=2{\overline{a}}^2{\overline{a}}_k^2\sum _{l=0}^k{C}_k^l{s}^{2(k-l)}{r}^l\frac{k!}{(k-l)!}\end{array}$$

(25)

for all $z\in \mathbb{C}$. From (25), it follows that the coefficients of z and $z^2$ in (25) equal zero. That is,

$$\begin{array}{c}\hfill \overline{a}{\overline{a}}_k(s^2+kr)=0\end{array}$$

(26)

and

$$\begin{array}{c}\hfill \overline{a}{\overline{a}}_k\left(\frac{s^{k+2}}{2}+kr{s}^k+\frac{k(k-1)}{2}{s}^{k-2}{r}^2\right)=0.\end{array}$$

(27)

If $a{a}_k\ne 0$, then from (26) and (27) we deduce that $k^2=k(k-1)$, which is a contraction. So, we obtain that $a{a}_k=0$. Also, by the assumption of $u\ne 0$, $a=0$.

Conversely, if $a=0$, then (23) clearly holds. The proof is completed. □

Similar to Theorem 7, we obtain the following result, whose proof is omitted.

Theorem 8. 

Let φ be a constant a and u be a constant $a_k$. Then the operator $S_{(u,\phi )}$ is 2-complex symmetric on ${\mathcal{F}}^2$ with the conjugation ${\mathcal{J}}_{r,s,t}$ if and only if one of the conditions holds: (i) $a=0$; (ii) $s=0$ and $a{a}_k=1$.

Remark 2. 

In Theorem 5.2, readers also can consider the general situation of u such as $u\left(z\right)=a_0+a_1+\cdots +a_n{z}^n$ the polynomial function.

Corollary 3. 

Let φ be a constant a and u be a constant $a_k$. Then the operator $S_{(u,\phi )}$ is 2-complex symmetric on ${\mathcal{F}}^2$ with the conjugation $\mathcal{J}$ if and only if one of the conditions holds: (i) $a=0$; (ii) $a{a}_k=1$.

6. Order Bounded Operators $S_{(u,\phi )}:{\mathcal{F}}^p\to {\mathcal{F}}^q$

In this section, we characterize the order bounded weighted superposition operators from one into another Fock space. By Definition 4, the operator $S_{(u,\phi )}:{\mathcal{F}}^p\to {\mathcal{F}}^q$ is order bounded if there exists a nonnegative function $g\in L^q(\mathbb{C},e^{-\frac{{q\left|z\right|}^2}{2}}d\nu )$ such that for all $f\in {\mathcal{F}}^p$ with ${\parallel f\parallel }_p\le 1$, it holds that

$$\begin{array}{c}\hfill |S_{(u,\phi )}\left(f\right)\left(z\right)|\le g\left(z\right),\mathrm{a.e.}[e^{-\frac{{q\left|z\right|}^2}{2}}d\nu ].\end{array}$$

(28)

Theorem 9. 

Let φ and u be holomorphic functions on $\mathbb{C}$, and $0<p,q\le \infty$. Then the operator $S_{(u,\phi )}:{\mathcal{F}}^p\to {\mathcal{F}}^q$ is order bounded if and only if φ is a constant and $u\in {\mathcal{F}}^q$.

Proof. 

We divide into two cases to complete the proof of necessity.

Case 1. Assume that $p\le q$. If the operator $S_{(u,\phi )}:{\mathcal{F}}^p\to {\mathcal{F}}^q$ is order bounded, it follows from Definition 4 that the operator $S_{(u,\phi )}:{\mathcal{F}}^p\to {\mathcal{F}}^q$ is bounded. Thus, by Theorem 1 (a), either $\phi \left(z\right)=az+b$ for some a, $b\in \mathbb{C}$ and $u=\alpha$ is a constant or $\phi$ is a constant and $u\in {\mathcal{F}}^q$. First, we assume that $\phi \left(z\right)=az+b$ for some a, $b\in \mathbb{C}$ and $u=\alpha$. By applying Definition 4 to this case, there exists a nonnegative function $g\in L^q(\mathbb{C},e^{-\frac{{q\left|z\right|}^2}{2}}d\nu )$ such that for all $f\in {\mathcal{F}}^p$ with ${\parallel f\parallel }_p\le 1$, it holds that

$$\begin{array}{c}\hfill |S_{(u,\phi )}\left(f\right)\left(z\right)|=|\alpha \left|\right|af\left(z\right)+b|\le g\left(z\right),\mathrm{a.e.}[e^{-\frac{{q\left|z\right|}^2}{2}}d\nu ].\end{array}$$

(29)

Particularly, letting $f\left(z\right)=k_w\left(z\right)$ in (29), we obtain

$$\begin{array}{c}\hfill |S_{(u,\phi )}\left(k_w\right)\left(z\right)|=|\alpha \left|\right|a{e}^{\overline{w}z-\frac{{\left|w\right|}^2}{2}}+b|\le g\left(z\right),\mathrm{a.e.}[e^{-\frac{{q\left|z\right|}^2}{2}}d\nu ].\end{array}$$

(30)

Letting $z=w$ in (30), we have

$$\begin{array}{c}\hfill |S_{(u,\phi )}\left(k_w\right)\left(w\right)|=|\alpha \left|\right|a{e}^{\frac{{\left|w\right|}^2}{2}}+b|\le g\left(w\right),\mathrm{a.e.}[e^{-\frac{{q\left|z\right|}^2}{2}}d\nu ],\end{array}$$

which shows that

$$\begin{array}{c}\hfill {\int }_{\mathbb{C}}|a{e}^{\frac{{\left|w\right|}^2}{2}}+b{|}^q{e}^{-\frac{{q\left|w\right|}^2}{2}}d\nu \left(w\right)<\infty .\end{array}$$

(31)

If $a\ne 0$, then for sufficiently large $\left|w\right|$ we have

$$\begin{array}{c}\hfill |a{e}^{\frac{{\left|w\right|}^2}{2}}+b|\ge |a|e^{\frac{{\left|w\right|}^2}{2}}-\left|b\right|>0.\end{array}$$

Thus,

$$\begin{array}{cc}\hfill {\int }_{\mathbb{C}}|a{e}^{\frac{{\left|w\right|}^2}{2}}+b{|}^q{e}^{-\frac{{q\left|w\right|}^2}{2}}d\nu \left(w\right)& \ge {\int }_{\mathbb{C}}{\left(\left|a\right|e^{\frac{{\left|w\right|}^2}{2}}-\left|b\right|\right)}^q{e}^{-\frac{{q\left|w\right|}^2}{2}}d\nu \left(w\right)\hfill \\ \hfill & ={\int }_{\mathbb{C}}{\left(\left|a\right|-\left|b\right|e^{-\frac{{\left|w\right|}^2}{2}}\right)}^{q}d\nu \left(w\right)=\infty ,\hfill \end{array}$$

which contradicts (31) and hence $a=0$. That is, $\phi$ is a constant and u is a constant. We know that the constant functions belong to ${\mathcal{F}}^q$, and then $u\in {\mathcal{F}}^q$.

Case 2. Assume that $p>q$. If the operator $S_{(u,\phi )}:{\mathcal{F}}^p\to {\mathcal{F}}^q$ is order bounded, then it is bounded. Thus, from Theorem 1 (b), it follows that $\phi$ is a constant and $u\in {\mathcal{F}}^q$.

Now, we prove sufficiency. Assume that $\phi \left(z\right)=a$ is a constant and $u\in {\mathcal{F}}^q$. Then, for $f\in {\mathcal{F}}^p$, we have

$$|S_{(u,\phi )}f\left(z\right)|=|au\left(z\right)|$$

for all $z\in \mathbb{C}$. Since $u\in {\mathcal{F}}^q$, $\left|au\right|\in L^q(\mathbb{C},e^{-\frac{{q\left|z\right|}^2}{2}}d\nu )$. Thus, the operator $S_{(u,\phi )}:{\mathcal{F}}^p\to {\mathcal{F}}^q$ is order bounded. The proof is completed. □

Remark 3. 

From Theorem 9 and Theorem 1, it follows that the operators $S_{(u,\phi )}$ with $u=\alpha$, $\phi \left(z\right)=az+b$ with $a\ne 0$ are not order bounded on ${\mathcal{F}}^2$, but they are bounded on ${\mathcal{F}}^2$.

Theorem 9 gives the following case.

Corollary 4. 

Let φ and u be holomorphic functions on $\mathbb{C}$, and $0<p<\infty$. Then the operator $S_{(u,\phi )}:{\mathcal{F}}^{\infty }\to {\mathcal{F}}^p$ is order bounded if and only if φ is a constant and $u\in {\mathcal{F}}^p$.