Complex Symmetry of Toeplitz Composition Operators on the Bergman Space

Abstract

In this paper, we characterize the conditions for a complex symmetric Toeplitz composition operator $T_{\psi }{C}_{\varphi }$ on the Bergman space $A^2\left(\mathbb{D}\right)$, where the symbol function $\psi$ is harmonic or $\psi \left(z\right)=A{\overline{z}}^n{z}^m+B{\overline{z}}^s{z}^t$ for $A,B\in \mathbb{C}\setminus \left\{0\right\}$ and $n-m=t-s$. We also present necessary and sufficient conditions for the commutativity of $T_{\psi }{C}_{\varphi }$ and $C_{\xi ,\theta }$, where $C_{\xi ,\theta }f\left(z\right)=e^{i\xi }\overline{f\left(e^{i\theta }\overline{z}\right)}$ for $\xi ,\theta \in \mathbb{R}$.

1. Introduction

Let $\mathcal{B}\left(\mathcal{H}\right)$ be the algebra of bounded linear operators on a separable complex Hilbert space $\mathcal{H}.$ In the operator theory of Hilbert spaces, symmetry is a central research topic. A symmetric operator, which embodies the most fundamental form of symmetry in Hilbert spaces, is defined by $\langle Tf,g\rangle =\langle f,Tg\rangle$ for all $f,g\in \mathcal{H}$. An anti-linear operator $C:\mathcal{H}\to \mathcal{H}$ is said to be a conjugation if C is involutive and isometric, that is, $C^2=I$ and $\parallel Cf\parallel =\parallel f\parallel$ for all $f\in \mathcal{H}$. A bounded linear operator T on a Hilbert space $\mathcal{H}$ is said to be complex symmetric if there exists a conjugation C on $\mathcal{H}$ such that $T=C{T}^{*}C$. We also call T a C-symmetric operator.

The complex symmetric operator class was systematically introduced by Garcia and Putinar [1,2] on Hilbert spaces. The introduction of complex symmetric operators not only unifies the symmetry of classical operators such as self-adjoint and normal operators, but also includes a large number of non-normal operators on concrete function spaces including Hardy, Bergman and Fock spaces [3,4,5,6,7]. This theoretical framework has been widely applied to the theory of Toeplitz operators in complex analysis, non-Hermitian Hamiltonians in quantum mechanics, and the invariant subspace problem in functional analysis [8,9,10].

The Bergman space $A^2\left(\mathbb{D}\right)$ is the Hilbert space of analytic functions on the unit disk that are square-integrable with respect to the area measure. It plays a fundamental role in both complex analysis and operator theory. For any function $f\in A^2\left(\mathbb{D}\right)$, its Bergman norm is defined as

$${\parallel f\parallel }_{A^2}^2={\int }_{\mathbb{D}}{\left|f\left(z\right)\right|}^{2}dA\left(z\right)<\infty .$$

This space is endowed with the inner product

$$\langle f,g\rangle ={\int }_{\mathbb{D}}f\left(z\right)\overline{g\left(z\right)}dA\left(z\right).$$

Here, $\mathbb{D}=\{z\in \mathbb{C}\mid |z|<1\}$ is the unit disk, and $dA\left(z\right)$ denotes the normalized Lebesgue area measure on $\mathbb{D}$, so that $dA\left(z\right)={\displaystyle \frac{1}{\pi }}dxdy={\displaystyle \frac{1}{\pi }}rdrd\theta$. Therefore, the Bergman space $A^2\left(\mathbb{D}\right)$ can be expressed as

$$A^2\left(\mathbb{D}\right)=\left\{f:\mathbb{D}\to \mathbb{C}\mid f\mathrm{is}\mathrm{analytic}\mathrm{and}{\int }_D{\left|f\left(z\right)\right|}^{2}dA\left(z\right)<\infty \right\}.$$

An orthonormal basis for $A^2\left(\mathbb{D}\right)$ is given by the monomials $e_n\left(z\right)=\sqrt{n+1}{z}^n$ ($n=0,1,2,\dots$), which satisfy

$${\langle e_m,e_n\rangle }_{A^2\left(\mathbb{D}\right)}={\delta }_{mn},i.e.,{\int }_{\mathbb{D}}{e}_m\left(z\right)\overline{e_n\left(z\right)}dA\left(z\right)={\delta }_{mn}.$$

The reproducing kernel on the Bergman space $A^2\left(\mathbb{D}\right)$ is defined as $K_z\left(w\right)={\displaystyle \frac{1}{{(1-\overline{z}w)}^2}}$ for all $z,w\in \mathbb{D}$. A fundamental property states that every function $f\in A^2\left(\mathbb{D}\right)$ satisfies the reproducing formula

$$f\left(z\right)=\langle f,K_z\rangle ={\int }_{\mathbb{D}}f\left(w\right)\overline{K_z\left(w\right)}dA\left(w\right).$$

The orthogonal projection $P:L^2(\mathbb{D},dA)\to A^2\left(\mathbb{D}\right)$ is called the Bergman projection. By the reproducing property, the Bergman projection has the integral representation

$$P\left(f\right)\left(z\right)={\int }_{\mathbb{D}}f\left(w\right)\overline{K_z\left(w\right)}dA\left(w\right),f\in L^2\left(\mathbb{D}\right).$$

For any bounded linear operator T on $A^2\left(\mathbb{D}\right)$, let $T^{*}$ denote its adjoint. This is the unique bounded linear operator satisfying

$$\langle Tf,g\rangle =\langle f,T^{*}g\rangle ,\forall f,g\in A^2\left(\mathbb{D}\right).$$

For a symbol $\psi \in L^{\infty }\left(\mathbb{D}\right)$, the Toeplitz operator $T_{\psi }:A^2\left(\mathbb{D}\right)\to A^2\left(\mathbb{D}\right)$ is defined by $T_{\psi }f=P\left(\psi f\right).$ For an analytic self-map $\varphi :\mathbb{D}\to \mathbb{D}$, the composition operator $C_{\varphi }$ is defined by $C_{\varphi }f=f\circ \varphi .$ Consequently, for a function $\psi \in L^{\infty }\left(\mathbb{D}\right)$ and an analytic self-map $\varphi :\mathbb{D}\to \mathbb{D}$, the Toeplitz composition operator $T_{\psi }{C}_{\varphi }:A^2\left(\mathbb{D}\right)\to A^2\left(\mathbb{D}\right)$ is defined by

$$T_{\psi }{C}_{\varphi }f=P\left(\psi (f\circ \varphi )\right),\mathrm{for}\mathrm{all}f\in A^2\left(\mathbb{D}\right).$$

See [11] for further background on these operators on Bergman spaces.

In 2014, Guo and Zhu [12] raised an interesting question concerning the characterization of complex symmetric Toeplitz operators on the Hardy space $H^2$. Subsequently, Ko and Lee [3] provided a characterization for the complex symmetry of such operators on the Hardy space $H^2\left(\mathbb{D}\right)$. Recently, this research direction has been extended to weighted Bergman spaces. Ko, Lee and Lee [6] have investigated complex symmetric Toeplitz operators with respect to a special conjugation $C_{\xi ,\theta }$ on the weighted Bergman space $A_{\alpha }^2\left(\mathbb{D}\right)$. Additionally, many scholars have also conducted continuous research on the complex symmetry of composition operators. Earlier, Jung, Kim, Ko and Lee [13] studied the complex symmetry of weighted composition operators on the Hardy space $H^2\left(\mathbb{D}\right)$. Later, Ahamed and Rahman [7] explored the same problem for weighted composition operators on the weighted Bergman space over the half-plane.

Additionally, Gupta and Singh [14] first introduced the concept of Toeplitz composition operators on the Fock space and studied its relevant properties, such as boundedness and compactness. Later, Gupta and Malhotra [15] investigated the complex symmetry of Toeplitz composition operators on the Hardy space $H^2\left(\mathbb{D}\right)$ with respect to the special conjugation $C_{\xi ,\theta }$.

Although the complex symmetry of Toeplitz composition operators with respect to $C_{\xi ,\theta }$ has been established on Hardy spaces, corresponding research on Bergman spaces has not yet been undertaken. Furthermore, the interaction between Toeplitz operators and composition operators introduces new phenomena that do not arise in isolated cases, making the study of Toeplitz composition operators particularly challenging. Our findings not only help fill this gap but also lay the groundwork for further research on more general classes of operators on Bergman spaces.

Motivated by all those works, we will investigate the complex symmetry of Toeplitz composition operators on the Bergman space $A^2\left(\mathbb{D}\right)$. In Section 2, we investigate the complex symmetry of Toeplitz composition operators $T_{\psi }{C}_{\varphi }$ on the Bergman space $A^2\left(\mathbb{D}\right)$ with respect to the special conjugation $C_{\xi ,\theta }.$ In Section 3, we investigate the commutativity of $T_{\psi }{C}_{\varphi }$ and $C_{\xi ,\theta }$, and establish the necessary and sufficient conditions for their commutativity on the Bergman space $A^2\left(\mathbb{D}\right)$.

2. Complex Symmetric Toeplitz Composition Operators on ${\mathit{A}}^{\mathbf{2}}\left(\mathbb{D}\right)$

In this section, we derive conditions for the complex symmetry of Toeplitz composition operator; we require an explicit formula for the adjoint of a composition operator. For linear fractional self-maps of $\mathbb{D}$, P. Hurst [16] extended the adjoint formula given by C. Cowen [17] on the Hardy space to the Bergman space, which is as follows:

We first recall the adjoint formula for composition operators induced by linear fractional self-maps on the Bergman space. For the linear symbols $\varphi \left(z\right)=az$ and $\varphi \left(z\right)=az+b$ considered in this paper, it follows from the general theory of linear fractional transformations (see Theorem 2 in [16] and Theorem 2 in [17]) that

$$C_{\varphi }^{*}=M_g{C}_{\sigma }{M}_h^{*}.$$

For $\varphi \left(z\right)=az+b$ with $a\ne 0$ and $\left|a\right|+\left|b\right|\le 1$ [18], we obtain

$$C_{\varphi }^{*}=M_g{C}_{\sigma },$$

where $\sigma \left(z\right)={\displaystyle \frac{\overline{a}z}{-\overline{b}z+1}}$, $g\left(z\right)={(-\overline{b}z+1)}^{-2}$ and $h\left(z\right)=1.$

In the special case $\varphi \left(z\right)=az$ (i.e., $b=0$), the above expression simplifies to

$$C_{\varphi }^{*}=C_{\sigma },$$

where $\sigma \left(z\right)=\overline{a}z$ and $g\left(z\right)=h\left(z\right)=1.$

Lemma 1

([17]). Let $\varphi \left(z\right)=az+b$ be an analytic self-map of $\mathbb{D}$ with $a\ne 0$. Then the map $\sigma \left(z\right)={\displaystyle \frac{\overline{a}z}{-\overline{b}z+1}}$ is also an analytic self-map of $\mathbb{D}$.

In the following lemma, we introduce a conjugation on the Bergman space $A^2\left(\mathbb{D}\right)$, which is crucial for our study of the complex symmetry of Toeplitz composition operators.

Lemma 2

([3,6]). For every ξ and θ, let $C_{\xi ,\theta }$: $A^2\left(\mathbb{D}\right)\to A^2\left(\mathbb{D}\right)$ be defined by

$$C_{\xi ,\theta }f\left(z\right)=e^{i\xi }\overline{f\left(e^{i\theta }\overline{z}\right)}.$$

Then $C_{\xi ,\theta }$ is a conjugation on $A^2\left(\mathbb{D}\right)$. Moreover, $C_{\xi ,\theta }$ and $C_{\tilde{\xi },\tilde{\theta }}$ are unitarily equivalent where $(\tilde{\xi },\tilde{\theta })$ satisfies the equation $\tilde{\xi }-k\tilde{\theta }=-\xi +k\theta -2n\pi$ for every $k\in \mathbb{N}$ and $n\in \mathbb{Z}$.

The formula presented in the following lemma forms the foundation for the subsequent characterization of the complex symmetry of Toeplitz composition operators.

Lemma 3

([19]). Let P be the Bergman projection on $L^2\left(\mathbb{D}\right)$. Then, for nonnegative integers $n,m$, we have

$$P\left({\overline{z}}^n{z}^m\right)=\left\{\begin{array}{cc}0,\hfill & ifm<n;\hfill \\ {\displaystyle \frac{m-n+1}{m+1}}{z}^{m-n},\hfill & ifm\ge n.\hfill \end{array}\right.$$

We now investigate the conditions under which the Toeplitz composition operator $T_{\psi }{C}_{\varphi }$ on $A^2\left(\mathbb{D}\right)$ is complex symmetric with respect to the conjugation $C_{\xi ,\theta }$.

Theorem 1.

For $\psi \in L^{\infty }\left(\mathbb{D}\right)$ such that $\psi \left(z\right)={\sum }_{n=0}^{\infty }\widehat{\psi }\left(n\right)z^n+{\sum }_{n=1}^{\infty }\overline{\widehat{\psi }(-n)}{\overline{z}}^n,$ and an analytic self-map $\varphi :\mathbb{D}⟶\mathbb{D}$ given by $\varphi \left(z\right)=az$ with $a\ne 0.$ Let $T_{\psi }{C}_{\varphi }$ be a Toeplitz composition operator on $A^2\left(\mathbb{D}\right)$. Then $T_{\psi }{C}_{\varphi }$ on $A^2\left(\mathbb{D}\right)$ is complex symmetric operators with respect to the conjugation $C_{\xi ,\theta }$ if and only if for $m\in \mathbb{Z},$

$$\widehat{\psi }\left(m\right)=\overline{\widehat{\psi }(-m)}{a}^m{\lambda }^m.$$

Proof. 

By Lemma 2, let $C_{\xi ,\theta }{z}^k=\mu {\left(\lambda z\right)}^k$ where $\mu =e^{i\xi }$ and $\lambda =e^{-i\theta }$ with $\left|\lambda \right|=\left|\mu \right|=1$. Assume $T_{\psi }{C}_{\varphi }$ is complex symmetric with respect to the conjugation $C_{\xi ,\theta }$, then for all $k\in \mathbb{N}\cup \left\{0\right\},$ we have

$$\begin{array}{c}\hfill T_{\psi }{C}_{\varphi }{C}_{\xi ,\theta }{z}^k=C_{\xi ,\theta }{\left(T_{\psi }{C}_{\varphi }\right)}^{*}{z}^k.\end{array}$$

(1)

Thus, we reduce the left-hand side of (1) to get

$$\begin{array}{cc}\hfill T_{\psi }{C}_{\varphi }{C}_{\xi ,\theta }{z}^k& =T_{\psi }{C}_{\varphi }{e}^{i\xi }{\left(\overline{e^{i\theta }\overline{z}}\right)}^k=T_{\psi }{C}_{\varphi }{\left(\mu \lambda z\right)}^k\hfill \\ \hfill & =T_{\psi }\left[\mu {\lambda }^k{\left(\varphi \left(z\right)\right)}^k\right]=T_{\psi }\left(\mu {\lambda }^k{\left(az\right)}^k\right)\hfill \\ \hfill & =P\left[\left({\displaystyle \sum _{n=1}^{\infty }}\overline{\widehat{\psi }(-n)}{\overline{z}}^n+{\displaystyle \sum _{n=0}^{\infty }}\widehat{\psi }\left(n\right)z^n\right)\left(\mu {\lambda }^k{\left(az\right)}^k\right)\right]\hfill \\ \hfill & =\mu {\lambda }^k{\displaystyle \sum _{n=0}^{\infty }}\widehat{\psi }\left(n\right)a^k{z}^{n+k}+\mu {\lambda }^k{\displaystyle \sum _{n=1}^k}{\displaystyle \frac{k-n+1}{k+1}}\overline{\widehat{\psi }(-n)}{a}^k{z}^{k-n}.\hfill \end{array}$$

(2)

From the adjoint formula for $\varphi \left(z\right)=az$ given earlier, we have $C_{\varphi }^{*}=C_{\sigma }$, where $\sigma \left(z\right)=\overline{a}z$ and $g\left(z\right)=h\left(z\right)=1$. Reducing the right-hand side of (1) further yields

$$\begin{array}{cc}\hfill C_{\xi ,\theta }{\left(T_{\psi }{C}_{\varphi }\right)}^{*}{z}^k& =C_{\xi ,\theta }{C}_{\varphi }^{*}{T}_{\psi }^{*}{z}^k=C_{\xi ,\theta }{C}_{\varphi }^{*}{T}_{\overline{\psi }}{z}^k\hfill \\ \hfill & =C_{\xi ,\theta }{C}_{\varphi }^{*}P\left[\left({\displaystyle \sum _{n=0}^{\infty }}\overline{\widehat{\psi }\left(n\right)}{\overline{z}}^n+{\displaystyle \sum _{n=1}^{\infty }}\widehat{\psi }(-n)z^n\right)z^k\right]\hfill \\ \hfill & =C_{\xi ,\theta }{C}_{\sigma }\left({\displaystyle \sum _{n=0}^k}{\displaystyle \frac{k-n+1}{k+1}}\overline{\widehat{\psi }\left(n\right)}{z}^{k-n}+{\displaystyle \sum _{n=1}^{\infty }}\widehat{\psi }(-n)z^{n+k}\right)\hfill \\ & =C_{\xi ,\theta }\left({\displaystyle \sum _{n=0}^k}{\displaystyle \frac{k-n+1}{k+1}}\overline{\widehat{\psi }\left(n\right)}{\left(\overline{a}z\right)}^{k-n}+{\displaystyle \sum _{n=1}^{\infty }}\widehat{\psi }(-n){\left(\overline{a}z\right)}^{n+k}\right)\hfill \\ & =\mu {\displaystyle \sum _{n=0}^k}{\displaystyle \frac{k-n+1}{k+1}}\widehat{\psi }\left(n\right)a^{k-n}{\lambda }^{k-n}{z}^{k-n}+\mu {\displaystyle \sum _{n=1}^{\infty }}\overline{\widehat{\psi }(-n)}{a}^{n+k}{\lambda }^{n+k}{z}^{n+k}.\hfill \end{array}$$

(3)

It follows from (1) that

$$\begin{array}{cc}\hfill & \mu {\displaystyle \sum _{n=0}^{\infty }}\widehat{\psi }\left(n\right)a^k{\lambda }^k{z}^{n+k}+\mu {\displaystyle \sum _{n=1}^k}{\displaystyle \frac{k-n+1}{k+1}}\overline{\widehat{\psi }(-n)}{a}^k{\lambda }^k{z}^{k-n}\hfill \\ & =\mu {\displaystyle \sum _{n=0}^k}{\displaystyle \frac{k-n+1}{k+1}}\widehat{\psi }\left(n\right)a^{k-n}{\lambda }^{k-n}{z}^{k-n}+\mu {\displaystyle \sum _{n=1}^{\infty }}\overline{\widehat{\psi }(-n)}{a}^{n+k}{\lambda }^{n+k}{z}^{n+k}\hfill \end{array}$$

(4)

for all $k\in \mathbb{N}\cup \left\{0\right\}$.

Hence, by the uniqueness of power series expansions on the unit disk, the coefficients of $z^p$ for all $p\in \mathbb{N}\cup \left\{0\right\}$ must be equal on both sides of (4). By comparing the coefficients of $z^p$, we derive the following key relation through case analysis based on the relationship between p and k. We consider the following two cases:

(1) If $p\ge k$: (4) reduces to

$$\mu \widehat{\psi }(p-k)a^k{\lambda }^k=\mu \overline{\widehat{\psi }\left(-\left(p-k\right)\right)}{a}^p{\lambda }^p.$$

Let $m=p-k\ge 0$, we get

$$\begin{array}{c}\hfill \widehat{\psi }\left(m\right)=\overline{\widehat{\psi }(-m)}{a}^m{\lambda }^m.\end{array}$$

(5)

(2) If $0\le p\le k-1$: (4) reduces to

$$\mu {\displaystyle \frac{p+1}{k+1}}\overline{\widehat{\psi }\left(-\left(k-p\right)\right)}{a}^k{\lambda }^k=\mu {\displaystyle \frac{p+1}{k+1}}\widehat{\psi }(k-p)a^p{\lambda }^p.$$

In this case, for $0\le p\le k-1$, we obtain the same condition (5) for $m=k-p\ge 1$, so the above relation holds for all $m\ge 0$. Hence, $\widehat{\psi }\left(m\right)=\overline{\widehat{\psi }(-m)}{a}^m{\lambda }^m$ holds for all $m\in \mathbb{N}$.

Conversely, suppose that for all $m\in \mathbb{N}$, $\widehat{\psi }\left(m\right)=\overline{\widehat{\psi }(-m)}{a}^m{\lambda }^m.$ Under this assumption, it suffices to prove that (1) holds for all $z\in \mathbb{D}.$ From (2) and (3), we compare the coefficients of $z^p$ for all $p\in \mathbb{N}\cup \left\{0\right\}.$ We distinguish two cases based on the range of p:

(1) If $p\ge k$, (2) reduces to $\mu \widehat{\psi }(p-k)a^k{\lambda }^k$ and (3) simplifies to $\mu \overline{\widehat{\psi }(-(p-k))}{a}^p{\lambda }^p.$ Setting $m=p-k$, (5) turns into

$$\widehat{\psi }(p-k)=\overline{\widehat{\psi }(-(p-k))}{a}^{p-k}{\lambda }^{p-k}.$$

Substituting it into $\mu \overline{\widehat{\psi }(-(p-k))}{a}^p{\lambda }^p$ yields

$$\mu \widehat{\psi }(p-k)a^{-(p-k)}{\lambda }^{-(p-k)}·a^p{\lambda }^p=\mu \widehat{\psi }(p-k)a^k{\lambda }^k.$$

Thus, the contributions match when $p\ge k$.

(2) If $0\le p\le k-1$, (2) reduces to $\mu {\displaystyle \frac{p+1}{k+1}}\overline{\widehat{\psi }\left(-\left(k-p\right)\right)}{a}^k{\lambda }^k$ and (3) simplifies to $\mu {\displaystyle \frac{p+1}{k+1}}\widehat{\psi }(k-p)a^p{\lambda }^p.$ Setting $m=k-p$, (5) turns into

$$\widehat{\psi }(k-p)=\overline{\widehat{\psi }\left(-\left(k-p\right)\right)}{a}^{k-p}{\lambda }^{k-p}.$$

Substituting it into $\mu {\displaystyle \frac{p+1}{k+1}}\widehat{\psi }(k-p)a^p{\lambda }^p$ yields

$$\mu {\displaystyle \frac{p+1}{k+1}}\overline{\widehat{\psi }(-(k-p))}{a}^{k-p}{\lambda }^{k-p}·a^p{\lambda }^p=\mu {\displaystyle \frac{p+1}{k+1}}\overline{\widehat{\psi }(-(k-p))}{a}^k{\lambda }^k.$$

Hence, the contributions are equal.

From the above two cases, under the assumption that (5) holds, we obtain $T_{\psi }{C}_{\varphi }{C}_{\xi ,\theta }{z}^k=C_{\xi ,\theta }{\left(T_{\psi }{C}_{\varphi }\right)}^{*}{z}^k$. Hence the Toeplitz composition operator $T_{\psi }{C}_{\varphi }$ is complex symmetric with respect to the conjugation $C_{\xi ,\theta }$. □

Example 1.

Let $\psi \left(z\right)=z+\overline{z}\in L^{\infty }\left(\mathbb{D}\right)$ and $\varphi \left(z\right)=iz$. Then $\varphi \left(z\right)$ is an analytic self-map of $\mathbb{D}$, and the Fourier coefficients of ψ are given by

$$\widehat{\psi }\left(1\right)=1,\overline{\widehat{\psi }(-1)}=1,$$

all other coefficients are zero. Consider the conjugation $C_{\xi ,\theta }$ with $\theta =\pi /2$ where $\lambda =e^{-i\theta }=-i$ and $\xi \in \mathbb{R}$. For $m=1$, we have $\widehat{\psi }\left(1\right)=1$ and $\overline{\widehat{\psi }(-1)}a\lambda =1·i·(-i)=1.$ For $m\ne 1,$ both sides are zero. The conditions given above satisfy

$$\widehat{\psi }\left(m\right)=\overline{\widehat{\psi }(-m)}{a}^m{\lambda }^m,m\in \mathbb{Z}.$$

This still satisfies $T_{\psi }{C}_{\varphi }{C}_{\xi ,\theta }=C_{\xi ,\theta }{\left(T_{\psi }{C}_{\varphi }\right)}^{*}$. Therefore, $T_{\psi }{C}_{\varphi }$ is complex symmetric with respect to the conjugation $C_{\xi ,\pi /2}$.

Theorem 2.

For $\psi \in L^{\infty }\left(\mathbb{D}\right)$ such that $\psi \left(z\right)={\sum }_{n=0}^{\infty }\widehat{\psi }\left(n\right)z^n+{\sum }_{n=1}^{\infty }\overline{\widehat{\psi }(-n)}{\overline{z}}^n,$ and an analytic self-map $\varphi :\mathbb{D}⟶\mathbb{D}$ given by $\varphi \left(z\right)=az+b$ with $a\ne 0.$ Let $T_{\psi }{C}_{\varphi }$ be a Toeplitz composition operator on $A^2\left(\mathbb{D}\right)$. Then $T_{\psi }{C}_{\varphi }$ on $A^2\left(\mathbb{D}\right)$ is a complex symmetric operator with respect to the conjugation $C_{\xi ,\theta }$ if and only if for $k,p\in \mathbb{N}\cup \left\{0\right\},$

$$\begin{array}{cc}\hfill & {\displaystyle \sum _{n=max(0,p-k)}^p}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{p-n}}\right)\widehat{\psi }\left(n\right)a^{p-n}{b}^{k-p+n}{\lambda }^k+{\displaystyle \sum _{n=1}^{k-p}}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{p+n}}\right){\displaystyle \frac{p+1}{p+n+1}}\overline{\widehat{\psi }(-n)}{a}^{p+n}{b}^{k-p-n}{\lambda }^k\hfill \\ \hfill & ={\displaystyle \sum _{n=max(0,k-p)}^k}\left({\displaystyle \genfrac{}{}{0pt}{}{p+1}{p-k+n}}\right){\displaystyle \frac{k-n+1}{k+1}}\widehat{\psi }\left(n\right)a^{k-n}{b}^{p-k+n}{\lambda }^p\hfill \\ \hfill & +{\displaystyle \sum _{n=1}^{p-k}}\left({\displaystyle \genfrac{}{}{0pt}{}{p-1}{p-n-k}}\right)\overline{\widehat{\psi }(-n)}{a}^{k+n}{b}^{p-k-n}{\lambda }^p.\hfill \end{array}$$

Proof. 

We still follow the proof method of Theorem 1. Assume $T_{\psi }{C}_{\varphi }$ is complex symmetric with respect to the conjugation $C_{\xi ,\theta }$, we have $T_{\psi }{C}_{\varphi }{C}_{\xi ,\theta }{z}^k=C_{\xi ,\theta }{\left(T_{\psi }{C}_{\varphi }\right)}^{*}{z}^k$ for all $k\in \mathbb{N}\cup \left\{0\right\}.$ Hence, we obtain

$$\begin{array}{cc}\hfill & T_{\psi }{C}_{\varphi }{C}_{\xi ,\theta }{z}^k=T_{\psi }{C}_{\varphi }\mu {\left(\lambda z\right)}^k=T_{\psi }\left(\mu {\lambda }^k{(az+b)}^k\right)\hfill \\ \hfill & =P\left[\left({\displaystyle \sum _{n=0}^{\infty }}\widehat{\psi }\left(n\right)z^n+{\displaystyle \sum _{n=1}^{\infty }}\overline{\widehat{\psi }(-n)}{\overline{z}}^n\right)\left(\mu {\lambda }^k{(az+b)}^k\right)\right]\hfill \\ \hfill & =\mu {\lambda }^k{\displaystyle \sum _{n=0}^{\infty }}\widehat{\psi }\left(n\right)z^n{\displaystyle \sum _{m=0}^k}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{m}}\right)a^m{b}^{k-m}{z}^m\hfill \\ \hfill & +\mu {\lambda }^k{\displaystyle \sum _{n=1}^{\infty }}\overline{\widehat{\psi }(-n)}{\overline{z}}^n{\displaystyle \sum _{m=0}^k}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{m}}\right)a^m{b}^{k-m}{z}^m\hfill \\ & =\mu {\lambda }^k{\displaystyle \sum _{m=0}^k}{\displaystyle \sum _{n=0}^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{m}}\right)\widehat{\psi }\left(n\right)a^m{b}^{k-m}{z}^{m+n}\hfill \\ & +\mu {\lambda }^k{\displaystyle \sum _{m=0}^k}{\displaystyle \sum _{n=1}^m}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{m}}\right){\displaystyle \frac{m-n+1}{m+1}}\overline{\widehat{\psi }(-n)}{a}^m{b}^{k-m}{z}^{m-n}\hfill \end{array}$$

(6)

and

$$\begin{array}{cc}\hfill C_{\xi ,\theta }{\left(T_{\psi }{C}_{\varphi }\right)}^{*}{z}^k& =C_{\xi ,\theta }{C}_{\varphi }^{*}{T}_{\psi }^{*}{z}^k=C_{\xi ,\theta }{C}_{\varphi }^{*}{T}_{\overline{\psi }}{z}^k\hfill \\ \hfill & =C_{\xi ,\theta }{C}_{\varphi }^{*}P\left[\left({\displaystyle \sum _{n=0}^{\infty }}\overline{\widehat{\psi }\left(n\right)}{\overline{z}}^n+{\displaystyle \sum _{n=1}^{\infty }}\widehat{\psi }(-n)z^n\right)z^k\right]\hfill \\ \hfill & =C_{\xi ,\theta }{C}_{\varphi }^{*}\left({\displaystyle \sum _{n=0}^k}{\displaystyle \frac{k-n+1}{k+1}}\overline{\widehat{\psi }\left(n\right)}{z}^{k-n}+{\displaystyle \sum _{n=1}^{\infty }}\widehat{\psi }(-n)z^{n+k}\right).\hfill \end{array}$$

As recalled earlier, for a linear analytic self-map $\varphi \left(z\right)=az+b$, the adjoint of $C_{\varphi }$ is given by $C_{\varphi }^{*}=M_g{C}_{\sigma }$, where $g\left(z\right)={(1-\overline{b}z)}^{-2}$ and $\sigma \left(z\right)={\displaystyle \frac{\overline{a}z}{1-\overline{b}z}}$. Moreover, since $\varphi$ maps $\mathbb{D}$ into itself, we have $\left|a\right|+\left|b\right|\le 1$ by [18], which implies $\left|b\right|<1$. Consequently, we have ${\displaystyle \frac{1}{{(1-\overline{b}z)}^i}}={\sum }_{j=0}^{\infty }\left({\displaystyle \genfrac{}{}{0pt}{}{i+j-1}{j}}\right){\left(\overline{b}z\right)}^j$ for $z\in \mathbb{D}$. Therefore,

$$\begin{array}{cc}\hfill & C_{\xi ,\theta }{\left(T_{\psi }{C}_{\varphi }\right)}^{*}{z}^k=C_{\xi ,\theta }{M}_g{C}_{\sigma }\left({\displaystyle \sum _{n=0}^k}{\displaystyle \frac{k-n+1}{k+1}}\overline{\widehat{\psi }\left(n\right)}{z}^{k-n}+{\displaystyle \sum _{n=1}^{\infty }}\widehat{\psi }(-n)z^{n+k}\right)\hfill \\ \hfill & =C_{\xi ,\theta }\left({\displaystyle \sum _{n=0}^k}{\displaystyle \frac{k-n+1}{k+1}}\overline{\widehat{\psi }\left(n\right)}{\overline{a}}^{k-n}{\left({\displaystyle \frac{1}{1-\overline{b}z}}\right)}^{k-n+2}{z}^{k-n}+{\displaystyle \sum _{n=1}^{\infty }}\widehat{\psi }(-n){\overline{a}}^{n+k}{\left({\displaystyle \frac{1}{1-\overline{b}z}}\right)}^{n+k+2}{z}^{n+k}\right)\hfill \\ & =C_{\xi ,\theta }\left({\displaystyle \sum _{j=0}^{\infty }}{\displaystyle \sum _{n=0}^k}\left({\displaystyle \genfrac{}{}{0pt}{}{k-n+j+1}{j}}\right){\displaystyle \frac{k-n+1}{k+1}}\overline{\widehat{\psi }\left(n\right)}{\overline{a}}^{k-n}{\overline{b}}^j{z}^{k-n+j}\phantom{{\displaystyle \sum _{j=0}^{\infty }}{\displaystyle \sum _{n=1}^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{n+k+j-1}{j}}\right)}\right.\hfill \\ \hfill & \left.+{\displaystyle \sum _{j=0}^{\infty }}{\displaystyle \sum _{n=1}^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{n+k+j-1}{j}}\right)\widehat{\psi }(-n){\overline{a}}^{n+k}{\overline{b}}^j{z}^{n+k+j}\right).\hfill \end{array}$$

By Lemma 2, we obtain that

$$\begin{array}{cc}\hfill C_{\xi ,\theta }{\left(T_{\psi }{C}_{\varphi }\right)}^{*}{z}^k& =\mu {\displaystyle \sum _{j=0}^{\infty }}{\displaystyle \sum _{n=0}^k}\left({\displaystyle \genfrac{}{}{0pt}{}{k-n+j+1}{j}}\right){\displaystyle \frac{k-n+1}{k+1}}\widehat{\psi }\left(n\right)a^{k-n}{b}^j{\lambda }^{k-n+j}{z}^{k-n+j}\hfill \\ \hfill & +\mu {\displaystyle \sum _{j=0}^{\infty }}{\displaystyle \sum _{n=1}^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{n+k+j-1}{j}}\right)\overline{\widehat{\psi }(-n)}{a}^{n+k}{b}^j{\lambda }^{n+k+j}{z}^{n+k+j}\hfill \end{array}$$

(7)

for all $k\in \mathbb{N}\cup \left\{0\right\}$.

It follows from (1) that

$$\begin{array}{cc}\hfill & {\displaystyle \sum _{m=0}^k}{\displaystyle \sum _{n=0}^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{m}}\right)\widehat{\psi }\left(n\right)a^m{b}^{k-m}{\lambda }^k{z}^{m+n}+{\displaystyle \sum _{m=0}^k}{\displaystyle \sum _{n=1}^m}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{m}}\right){\displaystyle \frac{m-n+1}{m+1}}\overline{\widehat{\psi }(-n)}{a}^m{b}^{k-m}{\lambda }^k{z}^{m-n}\hfill \\ \hfill & ={\displaystyle \sum _{j=0}^{\infty }}{\displaystyle \sum _{n=0}^k}\left({\displaystyle \genfrac{}{}{0pt}{}{k-n+j+1}{j}}\right){\displaystyle \frac{k-n+1}{k+1}}\widehat{\psi }\left(n\right)a^{k-n}{b}^j{\lambda }^{k-n+j}{z}^{k-n+j}\hfill \\ \hfill & +{\displaystyle \sum _{j=0}^{\infty }}{\displaystyle \sum _{n=1}^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{n+k+j-1}{j}}\right)\overline{\widehat{\psi }(-n)}{a}^{n+k}{b}^j{\lambda }^{n+k+j}{z}^{n+k+j}\hfill \end{array}$$

(8)

for all $n,k\in \mathbb{N}\cup \left\{0\right\}$ with $\left|\lambda \right|=1$.

Thus, the coefficients of $z^p$ for $p\in \mathbb{N}\cup \left\{0\right\}$ must be equal on both sides of (8). By comparing the coefficients of $z^p$, we observe that

$$\begin{array}{cc}\hfill & {\displaystyle \sum _{n=max(0,p-k)}^p}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{p-n}}\right)\widehat{\psi }\left(n\right)a^{p-n}{b}^{k-p+n}{\lambda }^k+{\displaystyle \sum _{n=1}^{k-p}}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{p+n}}\right){\displaystyle \frac{p+1}{p+n+1}}\overline{\widehat{\psi }(-n)}{a}^{p+n}{b}^{k-p-n}{\lambda }^k\hfill \\ \hfill & ={\displaystyle \sum _{n=max(0,k-p)}^k}\left({\displaystyle \genfrac{}{}{0pt}{}{p+1}{p-k+n}}\right){\displaystyle \frac{k-n+1}{k+1}}\widehat{\psi }\left(n\right)a^{k-n}{b}^{p-k+n}{\lambda }^p\hfill \\ \hfill & +{\displaystyle \sum _{n=1}^{p-k}}\left({\displaystyle \genfrac{}{}{0pt}{}{p-1}{p-k-n}}\right)\overline{\widehat{\psi }(-n)}{a}^{k+n}{b}^{p-k-n}{\lambda }^p\hfill \end{array}$$

(9)

for each $k,p\in \mathbb{N}\cup \left\{0\right\}.$

Conversely, assume that (9) holds for all $n,k$ and $p\in \mathbb{N}\cup \left\{0\right\}$. From (6) and (7), we obtain

$$\begin{array}{cc}\hfill (T_{\psi }{C}_{\varphi }{C}_{\xi ,\theta }-C_{\xi ,\theta }{\left(T_{\psi }{C}_{\varphi }\right)}^{*})z^k& =\mu {\displaystyle \sum _{m=0}^k}{\displaystyle \sum _{n=0}^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{m}}\right)\widehat{\psi }\left(n\right)a^m{b}^{k-m}{\lambda }^k{z}^{m+n}\hfill \\ \hfill & +\mu {\displaystyle \sum _{m=0}^k}{\displaystyle \sum _{n=1}^m}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{m}}\right){\displaystyle \frac{m-n+1}{m+1}}\overline{\widehat{\psi }(-n)}{a}^m{b}^{k-m}{\lambda }^k{z}^{m-n}\hfill \\ \hfill & -\mu \left({\displaystyle \sum _{j=0}^{\infty }}{\displaystyle \sum _{n=0}^k}\left({\displaystyle \genfrac{}{}{0pt}{}{k-n+j+1}{j}}\right){\displaystyle \frac{k-n+1}{k+1}}\widehat{\psi }\left(n\right)a^{k-n}{b}^j{\lambda }^{k-n+j}{z}^{k-n+j}\phantom{{\displaystyle \sum _{j=0}^{\infty }}{\displaystyle \sum _{n=1}^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{n+k+j-1}{j}}\right)}\right.\hfill \\ \hfill & \left.+\mu {\displaystyle \sum _{j=0}^{\infty }}{\displaystyle \sum _{n=1}^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{n+k+j-1}{j}}\right)\overline{\widehat{\psi }(-n)}{a}^{n+k}{b}^j{\lambda }^{n+k+j}{z}^{n+k+j}\right)\hfill \\ \hfill & =0.\hfill \end{array}$$

Thus, the Toeplitz composition operator $T_{\psi }{C}_{\varphi }$ is complex symmetric with respect to the conjugation $C_{\xi ,\theta }$ on $A^2\left(\mathbb{D}\right).$ □

Example 2

(applying the algebraic condition). Although the conditions in Theorem 2 are algebraically rather complex, each term reflects the case where the Bergman projection acts on a specific Fourier component. In fact, it suffices to examine just a few representative pairs of indices $(k,p)$ to determine the complex symmetry. We will illustrate this below using specific operators.

Let $\varphi \left(z\right)={\displaystyle \frac{1}{3}}z+{\displaystyle \frac{1}{6}}$, which is an analytic self-map of $\mathbb{D}$ with coefficients $a={\displaystyle \frac{1}{3}}$ and $b={\displaystyle \frac{1}{6}}$. Let $\psi \left(z\right)={\displaystyle \frac{13}{12}}+z+\overline{z}\in L^{\infty }$; its Fourier coefficients are

$$\widehat{\psi }\left(0\right)={\displaystyle \frac{13}{12}},\widehat{\psi }\left(1\right)=1,\overline{\widehat{\psi }(-1)}=1,$$

and $\widehat{\psi }\left(n\right)=0$ for all $\left|n\right|\ge 2$. Consider the conjugation $C_{\xi ,\theta }$ where we choose $\theta =0$ and $\xi \in \mathbb{R}$. Then $C_{\xi ,0}f\left(z\right)=e^{i\xi }\overline{f\left(\overline{z}\right)}$ be the conjugation with $\lambda =e^{-i·0}=1$.

We verify the complex symmetry condition from Theorem 2 for $(k,p)=(2,1)$; the condition becomes:

$$\begin{array}{cc}\hfill & {\displaystyle \sum _{n=max(0,-1)}^1}\left({\displaystyle \genfrac{}{}{0pt}{}{2}{1-n}}\right)\widehat{\psi }\left(n\right)a^{1-n}{b}^{1+n}{\lambda }^2+{\displaystyle \sum _{n=1}^1}\left({\displaystyle \genfrac{}{}{0pt}{}{2}{1+n}}\right){\displaystyle \frac{2}{2+n}}\overline{\widehat{\psi }(-n)}{a}^{1+n}{b}^{1-n}·{\lambda }^2\hfill \\ \hfill =& {\displaystyle \sum _{n=max(0,1)}^2}\left({\displaystyle \genfrac{}{}{0pt}{}{2}{-1+n}}\right){\displaystyle \frac{3-n}{3}}\widehat{\psi }\left(n\right)·a^{2-n}{b}^{-1+n}\lambda +{\displaystyle \sum _{n=1}^{-1}}\left({\displaystyle \genfrac{}{}{0pt}{}{0}{-1-n}}\right)\overline{\widehat{\psi }(-n)}{a}^{n-2}{b}^{-1-n}\lambda .\hfill \end{array}$$

Therefore, substituting $a={\displaystyle \frac{1}{3}},b=16$ into the above equation, we obtain that the left-hand side totals ${\displaystyle \frac{2}{9}}$, and the right-hand side totals ${\displaystyle \frac{2}{9}}$. The condition is satisfied for $k=2$, $p=1$.

Therefore, it is sufficient to test just a few representative $(k,p)$ pairs to determine the complex symmetry.

Next, we will study the differences between complex symmetric Toeplitz composition operators $T_{\psi }{C}_{\varphi }$ on the Bergman space $A^2\left(\mathbb{D}\right)$ and those on the Hardy space $H^2\left(\mathbb{D}\right)$. We consider finding symbols $\psi$ such that $T_{\psi }{C}_{\varphi }$ is complex symmetric on $A^2\left(\mathbb{D}\right)$ yet not on $H^2\left(\mathbb{D}\right)$.

Therefore, we consider Toeplitz composition operators $T_{\psi }{C}_{\varphi }$ with non-harmonic symbols. On Hardy spaces, the product ${\overline{z}}^n{z}^m$ simplifies to $z^{m-n}$. However, this relationship does not hold on Bergman spaces, as ${\overline{z}}^n{z}^m\ne z^{m-n}$ since $z\in \mathbb{D}$. The following theorem establishes the necessary and sufficient conditions for a Toeplitz composition operator $T_{\psi }{C}_{\varphi }$ with a non-harmonic symbol to be complex symmetric.

Symbols of the form $\psi \left(z\right)=A{\overline{z}}^n{z}^m+B{\overline{z}}^s{z}^t$ have previously been studied in [6] in the context of Toeplitz operators. In contrast, we investigate Toeplitz composition operators with such symbols, where the interaction between the two operators introduces additional complexity.

The following theorem gives necessary and sufficient conditions for $T_{\psi }{C}_{\varphi }$ with such a symbol to be complex symmetric.

Theorem 3.

Let $\psi \left(z\right)=A{\overline{z}}^n{z}^m+B{\overline{z}}^s{z}^t$ where $A,B\in \mathbb{C}\setminus \left\{0\right\}$ and $n-m=t-s$. For an analytic self-map $\varphi :\mathbb{D}⟶\mathbb{D}$ given by $\varphi \left(z\right)=az$ with $a\ne 0.$ Toeplitz composition operator $T_{\psi }{C}_{\varphi }$ on $A^2\left(\mathbb{D}\right)$ is complex symmetric with respect to the conjugation $C_{\xi ,\theta }$ if and only if $s=m$, $t=n$, and $B=A{\lambda }^{n-m}{a}^{n-m}.$

Proof. 

Assume $n>m$. Suppose $T_{\psi }{C}_{\varphi }$ is complex symmetric with respect to the conjugation $C_{\xi ,\theta }$, where $\mu =e^{i\xi }$ and $\lambda =e^{-i\theta }$. If $k\ge n-m$, then

$$\begin{array}{cc}\hfill T_{\psi }{C}_{\varphi }{C}_{\xi ,\theta }{z}^k& =T_{\psi }{C}_{\varphi }\left(\mu {\lambda }^k{z}^k\right)=T_{\psi }\left(\mu {\lambda }^k{\left(az\right)}^k\right)\hfill \\ \hfill & =P\left[\left(A{\overline{z}}^n{z}^m+B{\overline{z}}^s{z}^t\right)\mu {\lambda }^k{a}^k{z}^k\right]\hfill \\ \hfill & ={\displaystyle \frac{m+k-n+1}{m+k+1}}A\mu {\lambda }^k{a}^k{z}^{m+k-n}+{\displaystyle \frac{t+k-s+1}{t+k+1}}B\mu {\lambda }^k{a}^k{z}^{t+k-s}\hfill \end{array}$$

(10)

and

$$\begin{array}{cc}\hfill & C_{\xi ,\theta }{\left(T_{\psi }{C}_{\varphi }\right)}^{*}{z}^k=C_{\xi ,\theta }{C}_{\varphi }^{*}{T}_{\psi }^{*}{z}^k=C_{\xi ,\theta }{C}_{\sigma }{T}_{\overline{\psi }}{z}^k\hfill \\ \hfill & =C_{\xi ,\theta }{C}_{\sigma }P\left[\left(\overline{A}{z}^n{\overline{z}}^m+\overline{B}{z}^s{\overline{z}}^t\right)z^k\right]\hfill \\ \hfill & =C_{\xi ,\theta }{C}_{\sigma }\left({\displaystyle \frac{n+k-m+1}{n+k+1}}\overline{A}{z}^{n+k-m}+{\displaystyle \frac{s+k-t+1}{s+k+1}}\overline{B}{z}^{s+k-t}\right)\hfill \\ \hfill & =C_{\xi ,\theta }\left({\displaystyle \frac{n+k-m+1}{n+k+1}}\overline{A}{\overline{a}}^{n+k-m}{z}^{n+k-m}+{\displaystyle \frac{s+k-t+1}{s+k+1}}\overline{B}{\overline{a}}^{s+k-t}{z}^{s+k-t}\right)\hfill \\ \hfill & ={\displaystyle \frac{n+k-m+1}{n+k+1}}A\mu {\lambda }^{n+k-m}{a}^{n+k-m}{z}^{n+k-m}+{\displaystyle \frac{s+k-t+1}{s+k+1}}B\mu {\lambda }^{s+k-t}{a}^{s+k-t}{z}^{s+k-t}.\hfill \end{array}$$

(11)

Therefore, $T_{\psi }{C}_{\varphi }$ is a complex symmetric operator if and only if

$${\displaystyle \frac{m+k-n+1}{m+k+1}}A\mu {\lambda }^k{a}^k{z}^{m+k-n}={\displaystyle \frac{s+k-t+1}{s+k+1}}B\mu {\lambda }^{s+k-t}{a}^{s+k-t}{z}^{s+k-t}$$

and

$${\displaystyle \frac{t+k-s+1}{t+k+1}}B\mu {\lambda }^k{a}^k{z}^{t+k-s}={\displaystyle \frac{n+k-m+1}{n+k+1}}A\mu {\lambda }^{n+k-m}{a}^{n+k-m}{z}^{n+k-m},$$

or equivalently,

$$\begin{array}{c}\hfill {\displaystyle \frac{m+k-n+1}{m+k+1}}A={\displaystyle \frac{s+k-t+1}{s+k+1}}B{\lambda }^{s-t}{a}^{s-t}\end{array}$$

(12)

and

$$\begin{array}{c}\hfill {\displaystyle \frac{t+k-s+1}{t+k+1}}B={\displaystyle \frac{n+k-m+1}{n+k+1}}A{\lambda }^{n-m}{a}^{n-m}.\end{array}$$

(13)

Since $n-m=t-s$, therefore (12) and (13) can be equivalent to

$${\displaystyle \frac{1}{(m+k+1)(t+k+1)}}={\displaystyle \frac{1}{(s+k+1)(n+k+1)}}.$$

Hence, $s=m$ and $t=n$. From (13), we obtain $B=A{\lambda }^{n-m}{a}^{n-m}.$

On the other hand, if $k<n-m$, then

$$\begin{array}{cc}\hfill T_{\psi }{C}_{\varphi }{C}_{\xi ,\theta }{z}^k& =T_{\psi }{C}_{\varphi }\left(\mu {\lambda }^k{z}^k\right)\hfill \\ \hfill & =P\left[\left(A{\overline{z}}^n{z}^m+B{\overline{z}}^s{z}^t\right)\mu {\lambda }^k{a}^k{z}^k\right]\hfill \\ \hfill & ={\displaystyle \frac{t+k-s+1}{k+t+1}}B\mu {\lambda }^k{a}^k{z}^{t+k-s}\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill C_{\xi ,\theta }{\left(T_{\psi }{C}_{\varphi }\right)}^{*}{z}^k& =C_{\xi ,\theta }{C}_{\varphi }^{*}{T}_{\psi }^{*}{z}^k=C_{\xi ,\theta }{C}_{\sigma }{T}_{\overline{\psi }}{z}^k\hfill \\ \hfill & =C_{\xi ,\theta }{C}_{\sigma }P\left[\left(\overline{A}{z}^n{\overline{z}}^m+\overline{B}{z}^s{\overline{z}}^t\right)z^k\right]\hfill \\ \hfill & ={\displaystyle \frac{n+k-m+1}{n+k+1}}A\mu {\lambda }^{n+k-m}{a}^{n+k-m}{z}^{n+k-m}.\hfill \end{array}$$

We can also obtain that $T_{\psi }{C}_{\varphi }$ is a complex symmetric operator if and only if

$$\begin{array}{c}\hfill {\displaystyle \frac{t+k-s+1}{t+k+1}}B={\displaystyle \frac{n+k-m+1}{n+k+1}}A{\lambda }^{n-m}{a}^{n-m}\end{array}$$

or equivalently,

$$B=A{\lambda }^{n-m}{a}^{n-m}.$$

Hence $T_{\psi }{C}_{\varphi }$ is complex symmetric with respect to the conjugation $C_{\xi ,\theta }$ if and only if $s=m$, $t=n$, and $B=A{\lambda }^{n-m}{a}^{n-m}.$ □

Remark 1.

Let $\psi \left(z\right)=A{\overline{z}}^n{z}^m+B{\overline{z}}^s{z}^t$ where $A,B\in \mathbb{C}\setminus \left\{0\right\}$ and $n-m=t-s$. For an analytic self-map $\varphi :\mathbb{D}⟶\mathbb{D}$ given by $\varphi \left(z\right)=az+b$ with $a\ne 0.$ A Toeplitz composition operator $T_{\psi }{C}_{\varphi }$ on $A^2\left(\mathbb{D}\right)$ is never complex symmetric with respect to the conjugation $C_{\xi ,\theta }$.

Proof. 

We proceed by contradiction. Suppose that $T_{\psi }{C}_{\varphi }$ is complex symmetric with respect to $C_{\xi ,\theta }$. We obtain

$$T_{\psi }{C}_{\varphi }{C}_{\xi ,\theta }{z}^k=C_{\xi ,\theta }{\left(T_{\psi }{C}_{\varphi }\right)}^{*}{z}^k$$

for each $k\in \mathbb{N}\cup \left\{0\right\}$. Therefore, to obtain a contradiction, it suffices to find some k for which this equality fails. Next, we consider the case where $k=0$.

Assume $n>m.$ Suppose $T_{\psi }{C}_{\varphi }$ is complex symmetric with respect to the conjugation $C_{\xi ,\theta }.$ For $k⩾n-m,$ following a similar argument as in the proof of Theorem 3, we conclude that

$$\begin{array}{cc}\hfill T_{\psi }{C}_{\varphi }{C}_{\xi ,\theta }{z}^k& =\mu {\lambda }^k{\displaystyle \sum _{p=0}^k}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{p}}\right){\displaystyle \frac{m+p-n+1}{m+p+1}}A{a}^p{b}^{k-p}{z}^{m-n+p}\hfill \\ \hfill & +\mu {\lambda }^k{\displaystyle \sum _{p=0}^k}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{p}}\right){\displaystyle \frac{t+p-s+1}{t+p+1}}B{a}^p{b}^{k-p}{z}^{t-s+p}\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill C_{\xi ,\theta }{\left(T_{\psi }{C}_{\varphi }\right)}^{*}{z}^k& =\mu {\lambda }^{n+k-m+q}{\displaystyle \sum _{q=0}^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{n+k-m+q+1}{q}}\right){\displaystyle \frac{n+k-m+1}{n+k+1}}A{a}^{n+k-m}{b}^q{z}^{n+k-m+q}\hfill \\ \hfill & +\mu {\lambda }^{s+k-t+q}{\displaystyle \sum _{q=0}^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{s+k-t+q+1}{q}}\right){\displaystyle \frac{s+k-t+1}{s+k+1}}B{a}^{s+k-t}{b}^q{z}^{s+k-t+q}.\hfill \end{array}$$

Therefore, $T_{\psi }{C}_{\varphi }$ is a complex symmetric operator if and only if for all $k\ge 0$,

$$\begin{array}{cc}\hfill & \mu {\lambda }^k{\displaystyle \sum _{p=0}^k}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{p}}\right){\displaystyle \frac{m+p-n+1}{m+p+1}}A{a}^p{b}^{k-p}{z}^{m-n+p}\hfill \\ \hfill & +\mu {\lambda }^k{\displaystyle \sum _{p=0}^k}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{p}}\right){\displaystyle \frac{t+p-s+1}{t+p+1}}B{a}^p{b}^{k-p}{z}^{t-s+p}\hfill \\ \hfill & =\mu {\lambda }^{n+k-m+q}{\displaystyle \sum _{q=0}^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{n+k-m+q+1}{q}}\right){\displaystyle \frac{n+k-m+1}{n+k+1}}A{a}^{n+k-m}{b}^q{z}^{n+k-m+q}\hfill \\ \hfill & +\mu {\lambda }^{s+k-t+q}{\displaystyle \sum _{q=0}^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{s+k-t+q+1}{q}}\right){\displaystyle \frac{s+k-t+1}{s+k+1}}B{a}^{s+k-t}{b}^q{z}^{s+k-t+q}.\hfill \end{array}$$

(14)

We now consider whether (14) still holds in the case $k=0$ under the given assumptions. Thus, for $k=0$, (14) becomes

$$\begin{array}{cc}\hfill & {\displaystyle \frac{m-n+1}{m+1}}A{z}^{m-n}+{\displaystyle \frac{t-s+1}{t+1}}B{z}^{t-s}\hfill \\ \hfill & ={\lambda }^{n-m+q}{\displaystyle \sum _{q=0}^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{n-m+q+1}{q}}\right){\displaystyle \frac{n-m+1}{n+1}}A{a}^{n-m}{b}^q{z}^{n-m+q}\hfill \\ \hfill & +{\lambda }^{s-t+q}{\displaystyle \sum _{q=0}^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{s-t+q+1}{q}}\right){\displaystyle \frac{s-t+1}{s+1}}B{a}^{s-t}{b}^q{z}^{s-t+q}.\hfill \end{array}$$

(15)

We now combine like terms in (15) to get

$$\begin{array}{cc}\hfill & {\lambda }^{n-m+q}{\displaystyle \sum _{q\ne 2m-2n,q=0}^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{n-m+q+1}{q}}\right){\displaystyle \frac{n-m+1}{m+1}}A{a}^{n-m}{b}^q{z}^{n-m+q}\hfill \\ \hfill & +\left({\lambda }^{m-n}\left({\displaystyle \genfrac{}{}{0pt}{}{m-n+1}{2m-2n}}\right){\displaystyle \frac{n-m+1}{m+1}}A{a}^{n-m}{b}^{2m-2n}-{\displaystyle \frac{m-n+1}{m+1}}A\right)z^{m-n}\hfill \\ \hfill & +{\lambda }^{s-t+q}{\displaystyle \sum _{q\ne 2t-2s,q=0}^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{s-t+q+1}{q}}\right){\displaystyle \frac{s-t+1}{s+1}}B{a}^{s-t}{b}^q{z}^{s-t+q}\hfill \\ \hfill & +\left({\lambda }^{t-s}\left({\displaystyle \genfrac{}{}{0pt}{}{t-s+1}{2t-2s}}\right){\displaystyle \frac{s-t+1}{s+1}}B{a}^{s-t}{b}^{2t-2s}-{\displaystyle \frac{t-s+1}{t+1}}B\right)z^{t-s}\hfill \\ \hfill & =0.\hfill \end{array}$$

(16)

By the equation (16), we can let $n-m=t-s=d>0$. Hence, we obtain

$$\begin{array}{cc}\hfill & {\lambda }^{d+q}{\displaystyle \sum _{q\ne -2d,q=0}^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{d+q+1}{q}}\right){\displaystyle \frac{d+1}{m+1}}A{a}^d{b}^q{z}^{d+q}\hfill \\ \hfill & +\left({\lambda }^{-d}\left({\displaystyle \genfrac{}{}{0pt}{}{-d+1}{-2d}}\right){\displaystyle \frac{d+1}{m+1}}A{a}^d{b}^{-2d}-{\displaystyle \frac{-d+1}{m+1}}A\right)z^{-d}\hfill \\ \hfill & +{\lambda }^{-d+q}{\displaystyle \sum _{q\ne 2d,q=0}^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{-d+q+1}{q}}\right){\displaystyle \frac{-d+1}{s+1}}B{a}^{-d}{b}^q{z}^{-d+q}\hfill \\ \hfill & +\left({\lambda }^d\left({\displaystyle \genfrac{}{}{0pt}{}{d+1}{2d}}\right){\displaystyle \frac{-d+1}{s+1}}B{a}^{-d}{b}^{2d}-{\displaystyle \frac{d+1}{t+1}}B\right)z^d\hfill \\ \hfill & =0.\hfill \end{array}$$

(17)

By (17), it can be expressed in the form ${\sum }_{r=-d}^{\infty }{C}_r{z}^r=0$, where $C_r$ depends on the range of $r.$ Moreover, the coefficient of $z^r$ for all $r\in [-d,d+q]$ must be equal on both sides of (17). We can consider the following cases:

(1) $r=-d$: $\left({\displaystyle \genfrac{}{}{0pt}{}{-d+1}{-2d}}\right)=0$ and $\left({\displaystyle \genfrac{}{}{0pt}{}{-d+1}{0}}\right)=1$; therefore

$$\begin{array}{c}\hfill C_{-d}=-{\displaystyle \frac{-d+1}{m+1}}A+{\lambda }^{-d}{\displaystyle \frac{-d+1}{s+1}}B{a}^{-d}=0;\end{array}$$

(18)

(2) $r=d$:

$$\begin{array}{c}\hfill C_d={\lambda }^d\left({\displaystyle \genfrac{}{}{0pt}{}{d+1}{m+1}}\right)A{a}^d+{\lambda }^d\left({\displaystyle \genfrac{}{}{0pt}{}{d+1}{s+1}}\right){\displaystyle \frac{-d+1}{2d}}B{a}^{-d}{b}^{2d}-{\displaystyle \frac{d+1}{t+1}}B=0;\end{array}$$

(19)

(3) $-d<r<d$:

$$\begin{array}{c}\hfill C_{r_1}={\lambda }^r\left({\displaystyle \genfrac{}{}{0pt}{}{r+1}{r+d}}\right){\displaystyle \frac{-d+1}{s+1}}B{a}^{-d}{b}^{r+d}=0;\end{array}$$

(20)

(4) $r\ge d+1$:

$$\begin{array}{c}\hfill C_{r_2}={\lambda }^r\left({\displaystyle \genfrac{}{}{0pt}{}{r+1}{r-d}}\right){\displaystyle \frac{d+1}{m+1}}A{a}^d{b}^{r-d}+{\lambda }^r\left({\displaystyle \genfrac{}{}{0pt}{}{r+1}{r+d}}\right){\displaystyle \frac{-d+1}{s+1}}B{a}^{-d}{b}^{r+d}=0.\end{array}$$

(21)

Since A is a global constant, we first consider the case $r\ge d+1$:

(1) If $d=1$, (21) reduces to

$${\lambda }^r\left({\displaystyle \genfrac{}{}{0pt}{}{r+1}{r-1}}\right){\displaystyle \frac{1+1}{m+1}}A{a}^d{b}^{r-1}+{\lambda }^r\left({\displaystyle \genfrac{}{}{0pt}{}{r+1}{r+1}}\right){\displaystyle \frac{-1+1}{s+1}}B{a}^{-1}{b}^{r+1}=0,$$

which yields $A=0;$

(2) If $d\ge 2$, $\left({\displaystyle \genfrac{}{}{0pt}{}{r+1}{r+d}}\right)=0$, thus (21) reduces to

$${\lambda }^r\left({\displaystyle \genfrac{}{}{0pt}{}{r+1}{r-d}}\right){\displaystyle \frac{d+1}{m+1}}A{a}^d{b}^{r-d}=0,$$

which also leads to $A=0$.

Then, we get $A=0$; substituting it into (18) and (20), we find that $B=0$ is the only condition for the equalities to hold.

Next, we substitute $A=0$ into (19):

$${\lambda }^d\left({\displaystyle \genfrac{}{}{0pt}{}{d+1}{2d}}\right){\displaystyle \frac{-d+1}{2d}}B{a}^{-d}{b}^{2d}-{\displaystyle \frac{d+1}{t+1}}B=0.$$

This can be further divided into two cases:

(1) $d=1$: the equation reduces to $-{\displaystyle \frac{2}{t+1}}B=0$, which yields $B=0$;

(2) $d\ge 2$: we have $\left({\displaystyle \genfrac{}{}{0pt}{}{d+1}{2d}}\right)=0$. Substituting this into the original equation simplifies it to $-{\displaystyle \frac{d+1}{t+1}}B=0$, which also yields $B=0.$

Thus, we conclude that $A=0$ and $B=0$. When $k=0$, only $A=B=0$ satisfies all four cases of r, which yields a contradiction to our assumption that $A,B\in \mathbb{C}\setminus \left\{0\right\}$.

This is a contradiction. Therefore, (14) does not hold in the case $k=0$, so we conclude that the Toeplitz composition operator $T_{\psi }{C}_{\varphi }$ on $A^2\left(\mathbb{D}\right)$ is never complex symmetric with respect to the conjugation $C_{\xi ,\theta }$. □

3. The Toeplitz Composition Operator Commutes with the Conjugation Operator

In this section, we now characterize the commutativity between Toeplitz composition operator $T_{\psi }{C}_{\varphi }$ and conjugation operator $C_{\xi ,\theta }$ on $A^2\left(\mathbb{D}\right)$. We also obtain that $T_{\psi }{C}_{\varphi }$ is normal if and only if it is both complex symmetric with respect to the conjugation $C_{\xi ,\theta }$ and commutes with the conjugation $C_{\xi ,\theta }$.

Theorem 4.

Let $\psi \in L^{\infty }\left(\mathbb{D}\right)$ such that $\psi \left(z\right)={\sum }_{n=0}^{\infty }\widehat{\psi }\left(n\right)z^n+{\sum }_{n=1}^{\infty }\overline{\widehat{\psi }(-n)}{\overline{z}}^n$ and for a self-analytic linear transformation $\varphi \left(z\right)=az+b(a\ne 0)$, $\varphi :\mathbb{D}⟶\mathbb{D}$. Let $T_{\psi }{C}_{\varphi }$ be a Toeplitz composition operator on $A^2\left(\mathbb{D}\right)$. Then $T_{\psi }{C}_{\varphi }$ on $A^2\left(\mathbb{D}\right)$ commutes with respect to the conjugation $C_{\xi ,\theta }$ if and only if for each $m,k\in \mathbb{N}\cup \left\{0\right\},$ we have

(i)

$b=0$: $\widehat{\psi }\left(m\right)=\overline{\widehat{\psi }\left(m\right)}\alpha {\lambda }^m,$ where $\alpha ={\left({\displaystyle \frac{\overline{a}}{a}}\right)}^k;$

(ii)

$b\ne 0$:

$$\begin{array}{cc}\hfill & {\lambda }^k{\displaystyle \sum _{m=0}^{min(k,p)}}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{m}}\right)\left(\widehat{\psi }(p-m)a^m{b}^{k-m}-\overline{\widehat{\psi }(p-m)}{\overline{a}}^m{\overline{b}}^{k-m}\right)\hfill \\ & ={\lambda }^p{\displaystyle \sum _{m=p+1}^k}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{m}}\right){\displaystyle \frac{p+1}{m+1}}\left(\overline{\widehat{\psi }(p-m)}{a}^m{b}^{k-m}-\widehat{\psi }(p-m){\overline{a}}^m{\overline{b}}^{k-m}\right).\hfill \end{array}$$

Proof. 

If the Toeplitz composition operator $T_{\psi }{C}_{\varphi }$ commutes with the conjugation $C_{\xi ,\theta }$, we have

$$\begin{array}{c}\hfill T_{\psi }{C}_{\varphi }{C}_{\xi ,\theta }{z}^k=C_{\xi ,\theta }{T}_{\psi }{C}_{\varphi }{z}^k,\mathrm{for}\mathrm{each}k\in \mathbb{N}\cup \left\{0\right\}.\end{array}$$

(22)

We consider the following two cases:

Case ($\mathrm{i}$): Suppose $b=0$. We simplify the left-hand side of (22) to get

$$\begin{array}{cc}\hfill T_{\psi }{C}_{\varphi }{C}_{\xi ,\theta }{z}^k& =T_{\psi }{C}_{\varphi }{\left(\mu \lambda z\right)}^k=T_{\psi }\left(\mu {\lambda }^k{\left(az\right)}^k\right)\hfill \\ \hfill & =P\left[\left({\displaystyle \sum _{n=1}^{\infty }}\overline{\widehat{\psi }(-n)}{\overline{z}}^n+{\displaystyle \sum _{n=0}^{\infty }}\widehat{\psi }\left(n\right)z^n\right)\left(\mu {\lambda }^k{\left(az\right)}^k\right)\right]\hfill \\ & =\mu {\lambda }^k{\displaystyle \sum _{n=0}^{\infty }}\widehat{\psi }\left(n\right)a^k{z}^{n+k}+\mu {\lambda }^k{\displaystyle \sum _{n=1}^k}{\displaystyle \frac{k-n+1}{k+1}}\overline{\widehat{\psi }(-n)}{a}^k{z}^{k-n}\hfill \end{array}$$

and simplify the right-hand side of (22) to get

$$\begin{array}{cc}\hfill C_{\xi ,\theta }{T}_{\psi }{C}_{\varphi }{z}^k& =C_{\xi ,\theta }{T}_{\psi }\left(a^k{z}^k\right)\hfill \\ \hfill & =C_{\xi ,\theta }P\left[\left({\displaystyle \sum _{n=0}^{\infty }}\widehat{\psi }\left(n\right)z^n+{\displaystyle \sum _{n=1}^{\infty }}\overline{\widehat{\psi }(-n)}{\overline{z}}^n\right)a^k{z}^k\right]\hfill \\ \hfill & =C_{\xi ,\theta }\left({\displaystyle \sum _{n=0}^{\infty }}\widehat{\psi }\left(n\right)a^k{z}^{k+n}+{\displaystyle \sum _{n=1}^k}{\displaystyle \frac{k-n+1}{k+1}}\overline{\widehat{\psi }(-n)}{a}^k{z}^{k-n}\right)\hfill \\ & =\mu {\displaystyle \sum _{n=0}^{\infty }}\overline{\widehat{\psi }\left(n\right)}{\overline{a}}^k{\lambda }^{k+n}{z}^{k+n}+\mu {\displaystyle \sum _{n=1}^k}{\displaystyle \frac{k-n+1}{k+1}}\widehat{\psi }(-n){\overline{a}}^k{\lambda }^{k-n}{z}^{k-n}.\hfill \end{array}$$

Hence, it follows from (22) that

$$\begin{array}{cc}\hfill & \mu {\lambda }^k{\displaystyle \sum _{n=0}^{\infty }}\widehat{\psi }\left(n\right)a^k{z}^{n+k}+\mu {\lambda }^k{\displaystyle \sum _{n=1}^k}{\displaystyle \frac{k-n+1}{k+1}}\overline{\widehat{\psi }(-n)}{a}^k{z}^{k-n}\hfill \\ & =\mu {\displaystyle \sum _{n=0}^{\infty }}\overline{\widehat{\psi }\left(n\right)}{\overline{a}}^k{\lambda }^{k+n}{z}^{k+n}+\mu {\displaystyle \sum _{n=1}^k}{\displaystyle \frac{k-n+1}{k+1}}\widehat{\psi }(-n){\overline{a}}^k{\lambda }^{k-n}{z}^{k-n}\hfill \end{array}$$

(23)

for all $k\in \mathbb{N}\cup \left\{0\right\}.$

By comparing the coefficients of the same power $z^p$ of (23) for $p\in \mathbb{N}\cup \left\{0\right\}$, we obtain the following key relation through case analysis based on the relationship between p and k. There are two cases to consider:

(1) If $p\ge k$, (23) reduces to

$$\widehat{\psi }(p-k)a^k{\lambda }^k=\overline{\widehat{\psi }(p-k)}{\overline{a}}^k{\lambda }^p.$$

Then, by letting $m=p-k\ge 0$, we obtain

$$\widehat{\psi }\left(m\right)=\overline{\widehat{\psi }\left(m\right)}{\left({\displaystyle \frac{\overline{a}}{a}}\right)}^k{\lambda }^m.$$

(2) If $0\le p\le k-1$, (23) reduces to

$${\displaystyle \frac{p+1}{k+1}}\overline{\widehat{\psi }(p-k)}{a}^k{\lambda }^k={\displaystyle \frac{p+1}{k+1}}\widehat{\psi }(p-k){\overline{a}}^k{\lambda }^p.$$

Let $m=p-k\le -1$. Then

$$\begin{array}{c}\hfill \overline{\widehat{\psi }\left(m\right)}=\widehat{\psi }\left(m\right){\left({\displaystyle \frac{\overline{a}}{a}}\right)}^k{\lambda }^m.\end{array}$$

(24)

Further, let $m=-j$, then we can express(24) as

$$\overline{\widehat{\psi }(-j)}=\widehat{\psi }(-j){\left({\displaystyle \frac{\overline{a}}{a}}\right)}^k{\lambda }^{-j}.$$

Thus, we can conclude that for $m\in \mathbb{Z}$,

$$\widehat{\psi }\left(m\right)=\overline{\widehat{\psi }\left(m\right)}\alpha {\lambda }^m,\mathrm{where}\alpha ={\left({\displaystyle \frac{\overline{a}}{a}}\right)}^k.$$

Conversely, if $\widehat{\psi }\left(m\right)=\overline{\widehat{\psi }\left(m\right)}\alpha {\lambda }^m$ with $\alpha ={\left({\displaystyle \frac{\overline{a}}{a}}\right)}^k.$ Then, by the derivation above, we obtain $(T_{\psi }{C}_{\varphi }{C}_{\xi ,\theta }-C_{\xi ,\theta }{T}_{\psi }{C}_{\varphi })z^k=0.$ Thus, $T_{\psi }{C}_{\varphi }$ commutes with the conjugation $C_{\xi ,\theta }.$

Case ($\mathrm{ii}$): If $b\ne 0.$ Then, we obtain

$$\begin{array}{cc}\hfill & T_{\psi }{C}_{\varphi }{C}_{\xi ,\theta }{z}^k=T_{\psi }{C}_{\varphi }{e}^{i\xi }{\left(\overline{e^{i\theta }\overline{z}}\right)}^k=T_{\psi }\left(\mu {\lambda }^k{(az+b)}^k\right)\hfill \\ \hfill & =P\left[\left({\displaystyle \sum _{n=1}^{\infty }}\overline{\widehat{\psi }(-n)}{\overline{z}}^n+{\displaystyle \sum _{n=0}^{\infty }}\widehat{\psi }\left(n\right)z^n\right)\left(\mu {\lambda }^k{(az+b)}^k\right)\right]\hfill \\ \hfill & =\mu {\lambda }^k{\displaystyle \sum _{m=0}^k}{\displaystyle \sum _{n=0}^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{m}}\right)\widehat{\psi }\left(n\right)a^m{b}^{k-m}{z}^{m+n}+\mu {\lambda }^k{\displaystyle \sum _{m=0}^k}{\displaystyle \sum _{n=1}^m}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{m}}\right){\displaystyle \frac{m-n+1}{m+1}}\overline{\widehat{\psi }(-n)}{a}^m{b}^{k-m}{z}^{m-n}\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill & C_{\xi ,\theta }{T}_{\psi }{C}_{\varphi }{z}^k=C_{\xi ,\theta }{T}_{\psi }{(az+b)}^k\hfill \\ & =C_{\xi ,\theta }P\left[\left({\displaystyle \sum _{n=0}^{\infty }}\widehat{\psi }\left(n\right)z^n+{\displaystyle \sum _{n=1}^{\infty }}\overline{\widehat{\psi }(-n)}{\overline{z}}^n\right){\displaystyle \sum _{m=0}^k}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{m}}\right)a^m{b}^{k-m}{z}^m\right]\hfill \\ & =C_{\xi ,\theta }\left({\displaystyle \sum _{m=0}^k}{\displaystyle \sum _{n=0}^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{m}}\right)\widehat{\psi }\left(n\right)z^n{a}^m{b}^{k-m}{z}^{m+n}\phantom{{\displaystyle \sum _{m=0}^k}{\displaystyle \sum _{n=1}^m}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{m}}\right){\displaystyle \frac{m-n+1}{m+1}}}\right.\hfill \\ \hfill & \left.+{\displaystyle \sum _{m=0}^k}{\displaystyle \sum _{n=1}^m}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{m}}\right){\displaystyle \frac{m-n+1}{m+1}}\overline{\widehat{\psi }(-n)}{a}^m{b}^{k-m}{z}^m\right)\hfill \\ \hfill & =\mu {\displaystyle \sum _{m=0}^k}{\displaystyle \sum _{n=0}^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{m}}\right)\overline{\widehat{\psi }\left(n\right)}{\overline{a}}^m{\overline{b}}^{k-m}{\lambda }^{m+n}{z}^{m+n}\hfill \\ \hfill & +\mu {\displaystyle \sum _{m=0}^k}{\displaystyle \sum _{n=1}^m}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{m}}\right){\displaystyle \frac{m-n+1}{m+1}}\widehat{\psi }(-n){\overline{a}}^m{\overline{b}}^{k-m}{\lambda }^{m-n}{z}^{m-n}.\hfill \end{array}$$

Hence, it follows from (22) that

$$\begin{array}{cc}\hfill & \mu {\lambda }^k{\displaystyle \sum _{m=0}^k}{\displaystyle \sum _{n=0}^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{m}}\right)\widehat{\psi }\left(n\right)a^m{b}^{k-m}{z}^{m+n}\hfill \\ \hfill & +\mu {\lambda }^k{\displaystyle \sum _{m=0}^k}{\displaystyle \sum _{n=1}^m}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{m}}\right){\displaystyle \frac{m-n+1}{m+1}}\overline{\widehat{\psi }(-n)}{a}^m{b}^{k-m}{z}^{m-n}\hfill \\ \hfill & =\mu {\displaystyle \sum _{m=0}^k}{\displaystyle \sum _{n=0}^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{m}}\right)\overline{\widehat{\psi }\left(n\right)}{\overline{a}}^m{\overline{b}}^{k-m}{\lambda }^{m+n}{z}^{m+n}\hfill \\ \hfill & +\mu {\displaystyle \sum _{m=0}^k}{\displaystyle \sum _{n=1}^m}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{m}}\right){\displaystyle \frac{m-n+1}{m+1}}\widehat{\psi }(-n){\overline{a}}^m{\overline{b}}^{k-m}{\lambda }^{m-n}{z}^{m-n}\hfill \end{array}$$

(25)

for all $k\in \mathbb{N}\cup \left\{0\right\}.$

Thus, the coefficients of $z^p$ must be equal on both sides of (25) for $p\in \mathbb{N}\cup \left\{0\right\}$. By comparing the coefficients of $z^p$, we observe that

$$\begin{array}{cc}\hfill & {\lambda }^k{\displaystyle \sum _{m=0}^{min(k,p)}}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{m}}\right)\widehat{\psi }(p-m)a^m{b}^{k-m}+{\lambda }^k{\displaystyle \sum _{m=p+1}^k}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{m}}\right){\displaystyle \frac{p+1}{m+1}}\overline{\widehat{\psi }(p-m)}{a}^m{b}^{k-m}\hfill \\ \hfill & ={\lambda }^p{\displaystyle \sum _{m=0}^{min(k,p)}}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{m}}\right)\overline{\widehat{\psi }(p-m)}{\overline{a}}^m{\overline{b}}^{k-m}+{\lambda }^p{\displaystyle \sum _{m=p+1}^k}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{m}}\right){\displaystyle \frac{p+1}{m+1}}\widehat{\psi }(p-m){\overline{a}}^m{\overline{b}}^{k-m}.\hfill \end{array}$$

Further, it can be written as

$$\begin{array}{cc}\hfill & {\lambda }^k{\displaystyle \sum _{m=0}^{min(k,p)}}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{m}}\right)\left(\widehat{\psi }(p-m)a^m{b}^{k-m}-\overline{\widehat{\psi }(p-m)}{\overline{a}}^m{\overline{b}}^{k-m}\right)\hfill \\ & ={\lambda }^p{\displaystyle \sum _{m=p+1}^k}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{m}}\right){\displaystyle \frac{p+1}{m+1}}\left(\overline{\widehat{\psi }(p-m)}{a}^m{b}^{k-m}-\widehat{\psi }(p-m){\overline{a}}^m{\overline{b}}^{k-m}\right).\hfill \end{array}$$

Conversely, if the above formula holds for all $k,p\in \mathbb{N}\cup \left\{0\right\}$, then through the derivation above, we can obtain $(T_{\psi }{C}_{\varphi }{C}_{\xi ,\theta }-C_{\xi ,\theta }{T}_{\psi }{C}_{\varphi })z^k=0.$ Thus, $T_{\psi }{C}_{\varphi }$ commutes with the conjugation $C_{\xi ,\theta }$. □

Example 3

(applying the commutativity condition). Although the condition in Theorem 4 is algebraically involved, testing a few representative pairs $(k,p)$ suffices to determine commutativity, as we illustrate below.

Let $\varphi \left(z\right)={\displaystyle \frac{1}{3}}z+{\displaystyle \frac{1}{6}}$, which is an analytic self-map of $\mathbb{D}$ with coefficients $a={\displaystyle \frac{1}{3}}$ and $b={\displaystyle \frac{1}{6}}$. Let $\psi \left(z\right)={\displaystyle \frac{13}{12}}+z+\overline{z}\in L^{\infty }$; its Fourier coefficients are

$$\widehat{\psi }\left(0\right)={\displaystyle \frac{13}{12}},\widehat{\psi }\left(1\right)=1,\overline{\widehat{\psi }(-1)}=1,$$

and $\widehat{\psi }\left(n\right)=0$ for all $\left|n\right|\ge 2$. Take the conjugation $C_{\xi ,\theta }$ where we choose $\theta =0$ and $\xi \in \mathbb{R}$, so $\lambda =1$.

We verify the commutativity condition from Theorem 4 for two cases.

Case 1: $(k,p)=(0,1)$. We have $k=0,p=1,and\lambda =1$. So the condition becomes

$$\begin{array}{cc}\hfill & {\lambda }^0{\displaystyle \sum _{m=0}^0}\left({\displaystyle \genfrac{}{}{0pt}{}{0}{m}}\right)\left(\widehat{\psi }(1-m)a^m{b}^{-m}-\overline{\widehat{\psi }(1-m)}{\overline{a}}^m{\overline{b}}^{-m}\right)\hfill \\ \hfill & ={\lambda }^1{\displaystyle \sum _{m=2}^0}\left({\displaystyle \genfrac{}{}{0pt}{}{0}{m}}\right){\displaystyle \frac{2}{m+1}}\left(\overline{\widehat{\psi }(1-m)}{a}^m{b}^{-m}-\widehat{\psi }(1-m){\overline{a}}^m{\overline{b}}^{-m}\right).\hfill \end{array}$$

We can obtain the left-hand side totals $\widehat{\psi }\left(1\right)a^0{b}^0-\overline{\widehat{\psi }\left(1\right)}{\overline{a}}^0{\overline{b}}^0=\widehat{\psi }\left(1\right)-\overline{\widehat{\psi }\left(1\right)}=1-1=0$, and for the right-hand side, the summation range is $m=2$ to $k=0$; this is an empty sum and equals 0.

Thus both sides equal 0, and the identity holds for $(k,p)=(0,1)$.

Case 2: $(k,p)=(1,0)$. We have $k=1,p=0,and\lambda =1$. So the condition becomes:

$$\begin{array}{cc}\hfill & {\lambda }^1{\displaystyle \sum _{m=0}^0}\left({\displaystyle \genfrac{}{}{0pt}{}{1}{m}}\right)\left(\widehat{\psi }(0-m)a^m{b}^{1-m}-\overline{\widehat{\psi }(0-m)}{\overline{a}}^m{\overline{b}}^{1-m}\right)\hfill \\ \hfill & ={\lambda }^0{\displaystyle \sum _{m=1}^1}\left({\displaystyle \genfrac{}{}{0pt}{}{1}{m}}\right){\displaystyle \frac{0+1}{m+1}}\left(\overline{\widehat{\psi }(0-m)}{a}^m{b}^{1-m}-\widehat{\psi }(0-m){\overline{a}}^m{\overline{b}}^{1-m}\right).\hfill \end{array}$$

Therefore, we can obtain the left-hand side totals $\widehat{\psi }\left(0\right)b-\overline{\widehat{\psi }\left(0\right)}\overline{b}=0$ and the right-hand side totals ${\displaystyle \frac{1}{2}}\left(\overline{\widehat{\psi }(-1)}·a^1{b}^0-\widehat{\psi }(-1)·{\overline{a}}^1{\overline{b}}^0\right)={\displaystyle \frac{1}{2}}\left(1·a-1·a\right)={\displaystyle \frac{1}{2}}\left({\displaystyle \frac{1}{3}}-{\displaystyle \frac{1}{3}}\right)=0.$ Both sides coincide, so the condition holds for $(k,p)=(1,0)$.

Thus, testing a few representative pairs $(k,p)$ suffices to determine commutativity without verifying the condition for all $(k,p)$. Thus, a few representative pairs $(k,p)$ suffice to decide commutativity.

Corollary 1.

Let $\psi \in L^{\infty }\left(\mathbb{D}\right)$ and let ϕ denote an analytic self-map from $\mathbb{D}$ into itself. Given that the Toeplitz composition operator $T_{\psi }{C}_{\varphi }$ is complex symmetric operators with respect to the conjugation $C_{\xi ,\theta }$ on $A^2\left(\mathbb{D}\right)$ and further, assume that $T_{\psi }{C}_{\varphi }$ commutes with $C_{\xi ,\theta }$. Then $T_{\psi }{C}_{\varphi }$ is a normal operator on $A^2\left(\mathbb{D}\right)$.

Proof. 

By assumption, $T_{\psi }{C}_{\varphi }$ is a complex symmetric operator with respect to the conjugation $C_{\xi ,\theta }$ and commutes with $C_{\xi ,\theta }$, which implies that $T_{\psi }{C}_{\varphi }$ is a self-adjoint. That is to say,

$${\left(T_{\psi }{C}_{\varphi }\right)}^{*}=C_{\xi ,\theta }{T}_{\psi }{C}_{\varphi }{C}_{\xi ,\theta }=C_{\xi ,\theta }{C}_{\xi ,\theta }{T}_{\psi }{C}_{\varphi }=T_{\psi }{C}_{\varphi }.$$

Therefore, $T_{\psi }{C}_{\varphi }$ is a normal operator on $A^2\left(\mathbb{D}\right).$ □

4. Conclusions

Toeplitz composition operators, which are formed by composing a Toeplitz operator and a composition operator, have been investigated extensively on Hardy spaces, yet the corresponding theory on Bergman spaces remains relatively underdeveloped. In this paper, we study the complex symmetry and commutativity of Toeplitz composition operators on the Bergman space $A^2\left(\mathbb{D}\right)$. We first characterize the complex symmetry of Toeplitz composition operators $T_{\psi }{C}_{\varphi }$ with respect to the conjugation $C_{\xi ,\theta }$. We consider both the harmonic and the polynomial cases for the symbol $\psi$ and obtain explicit necessary and sufficient conditions on the analytic self-map $\varphi$ and the symbol $\psi$. We also investigate the commutativity between Toeplitz composition operators and the conjugation $C_{\xi ,\theta }$, and prove that such commutativity imposes strong restrictions on the structure of the symbol $\psi$. Our results extend known results for individual Toeplitz operators and composition operators to the broader framework of Toeplitz composition operators.