2-Complex Symmetric Weighted Composition Operators on the Weighted Bergman Space of the Unit Ball

Abstract

There are two aims in this paper. One is to completely characterize complex symmetric and 2-complex symmetric weighted composition operators induced by some special symbols on the weighted Bergman space of the unit ball, and the other is to fully characterize the complex symmetry of the difference of such operators on the space.

1. Introduction

Let $\mathbb{C}$ denote the complex plane, ${\mathbb{C}}^N$ the N-dimensional complex Euclidean space with the inner product $\langle z,w\rangle ={\sum }_{j=1}^N{z}_j{\overline{w}}_j$, ${\left|z\right|}^2=\langle z,z\rangle$, and ${\mathbb{B}}^N=\{z\in {\mathbb{C}}^N:\left|z\right|<1\}$ the unit ball in ${\mathbb{C}}^N$. Let $\mathcal{B}\left(H\right)$ be the set of all bounded linear operators on a separable complex Hilbert space H. For an operator $T\in \mathcal{B}\left(H\right)$, let $T^{\ast }$ denote the adjoint operator of T.

A conjugation on H is an antilinear operator $C:H\to H$ which satisfies $\langle Cx,Cy\rangle =\langle y,x\rangle$ for all $x,y\in H$, and $C^2=I$. For any conjugation C, there is an orthonormal basis ${\left\{e_n\right\}}_{n=1}^{\infty }$ for H such that $C{e}_n=e_n$ for each $n\in \mathbb{N}$ (see [1]). An operator $T\in \mathcal{B}\left(H\right)$ is said to be complex symmetric if there exists a conjugation C on H such that $CTC=T^{\ast }$. At this point, we say that T is complex symmetric with the conjugation C. This concept is due to the fact that T is a complex symmetric operator if and only if it is unitarily equivalent to a symmetric matrix with complex entries, regarded as an operator acting on an $l^2$-space of the appropriate dimension (see [2]).

The class of complex symmetric operators includes all normal operators, binormal operators, Hankel operators, compressed Toeplitz operators and Volterra integration operators. The general study of complex symmetric operators originated from the works of Garcia, Putinar, and Wogen in [1,2,3,4]. Subsequent research, including contributions by Gao et al. [5] and Han [6], extended these ideas to composition operators on classical function spaces. More studies about complex symmetric operators on function spaces can be found in [7,8,9,10,11,12,13]. However, as noted in the aforementioned literature, the majority of results remain confined to Toeplitz operators and composition operators, leaving weighted composition operators (a natural generalization combining composition and multiplication), relatively unexplored. Therefore, we will focus on these operators as a primary research objective.

Helton, in [14], studied the operators $T\in \mathcal{B}\left(H\right)$, which satisfy an identity of the form

$$\sum _{j=0}^m{(-1)}^{m-j}{C}_m^j{T}^{\ast j}{T}^{m-j}=0,$$

(1)

where $C_m^j={\displaystyle \frac{m(m-1)\cdots (m-j+1)}{j!}}$. According to complex symmetric operators, Chō et al., in [15], introduced the following definition by using the identity (1).

Definition 1. 

Let m be a positive integerand C be a conjugation on H. An operator $T\in \mathcal{B}\left(H\right)$ is said to be m-complex symmetric with C if

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

By the definition, we see that if $m=2$, then the operator $T\in \mathcal{B}\left(H\right)$ is 2-complex symmetric with the conjugation C if and only if

$$C{T}^2-2T^{\ast }CT+T^{\ast 2}C=0,$$

and 1-complex symmetric operator is just the complex symmetric operator. All complex symmetric operators are 2-complex symmetric (see [16]). It is not difficult to prove that all complex symmetric operators are m-complex symmetric. In this paper, we will prove that complex symmetry and 2-complex symmetry are equivalent for the weighted composition operators discussed. However, whether complex symmetry is equivalent to m-complex symmetry for such operators remains an open problem. At the moment, we have not solved this problem.

Over the past few decades, weighted composition operators and some related operators have been extensively studied on holomorphic function spaces (see [17,18,19,20,21,22,23,24,25,26,27]). From the aforementioned studies, one can find that there is a significant amount of research on boundedness, compactness, essential norm, closed range, and so on. However, little progress has been made in the study of 2-complex symmetry. The literature we were able to find includes [28,29]. Among others, in [28], Hu obtained the necessary and sufficient conditions for the composition operator $C_{\phi }$ to be 2-complex symmetric with the conjugation $\mathcal{J}$ when $\phi$ is an automorphism of the unit disk. They also characterized 2-complex symmetric operator $C_{\phi }$ with $\mathcal{J}$ when $\phi$ is a linear fractional self-map of the unit disk. Therefore, we focus on the study of 2-complex symmetric weighted composition operators induced by the linear fractional self-mappings.

It is well known that linear fractional mapping $\phi \left(z\right)={\displaystyle \frac{Az+B}{\langle z,C\rangle +D}}$ plays an important role in the study of holomorphic function spaces on ${\mathbb{B}}^N$ and has been studied (see [30,31,32]), which also motivates us to study 2-complex symmetric weighted composition operators induced by the linear fractional self-mappings of ${\mathbb{B}}^N$. On the other hand, in order to characterize 2-complex symmetric weighted composition operators on the weighted Bergman space of the unit ball, we need to determine the adjoint of the weighted composition operators. However, we find that it is impossible to study this problem for weighted composition operators induced by general symbols. Therefore, the following problem arises: which symbols are appropriate? At this very moment, we find that the author in [33] gives a sufficient condition for the boundedness of weighted composition operators on the weighted Bergman space of the unit ball. This result allows us to identify specific bounded weighted composition operators on the space. This is one of the keys to this paper. In short, we characterize 2-complex symmetric weighted composition operators induced by the symbols $\phi \left(z\right)={\displaystyle \frac{Az+b}{\langle z,c\rangle +1}}$ and $u\left(z\right)={\displaystyle \frac{1}{{(1+\langle z,c\rangle )}^{N+1+\alpha }}}$.

2. Preliminaries

Let $\overline{z}$ denote the conjugation of the complex number z. For $z=(z_1,z_2,\dots ,z_N)\in {\mathbb{B}}^N$, we define

$$\kappa \left(z\right)=({\overline{z}}_1,{\overline{z}}_2,\dots ,{\overline{z}}_N).$$

A calculation shows that $\mathcal{J}f\left(z\right)=\overline{f\left(\kappa \right(z\left)\right)}$ is a conjugation on weighted Bergman space. As a generalization of $\mathcal{J}$, Cao, in [34], proved that the operator ${\mathcal{J}}_{\tau ,\zeta }f\left(z\right)=\tau \overline{f\left({\tau }_{\zeta }\kappa \left(z\right)\right)}$ is a conjugation on weighted Bergman space, where $\tau \in \mathbb{C}$ with $\left|\tau \right|=1$, $\zeta =({\zeta }_1,\dots ,{\zeta }_N)\in {\mathbb{R}}^N$ and ${\tau }_{\zeta }=\mathrm{diag}\{e^{i{\zeta }_1},\dots ,e^{i{\zeta }_N}\}$. In this paper, we characterize complex symmetric and 2-complex symmetric weighted composition operators on Bergman space with $\mathcal{J}$ and ${\mathcal{J}}_{\tau ,\zeta }$, respectively.

Let A be a linear operator on ${\mathbb{C}}^N$. We know that there exists an $N\times N$ complex matrix ${\left(a_{jk}\right)}_{N\times N}$ such that $Az={\left(a_{jk}\right)}_{N\times N}z$ for all $z\in {\mathbb{C}}^N$. This allows us to regard A as the complex matrix ${\left(a_{jk}\right)}_{N\times N}$, and then the norm of A is defined by

$${\parallel A\parallel }^2=\sum _{k=1}^N\sum _{j=1}^N{\left|a_{jk}\right|}^2.$$

Let $A^{\ast }$ be the adjoint operator of A. Then, $A^{\ast }=\kappa \left(A^T\right)$, where $A^T$ denotes the transpose of A.

Let $\mathcal{O}\left({\mathbb{B}}^N\right)$ be the set of all holomorphic functions on ${\mathbb{B}}^N$ and $dv$ be the normalized volume measure on ${\mathbb{B}}^N$. For $\alpha >-1$, let

$$d{v}_{\alpha }\left(z\right)=c_{\alpha }{(1-|z|}^2{)}^{\alpha }dv\left(z\right),$$

where $c_{\alpha }$ is a constant such that $v_{\alpha }\left({\mathbb{B}}^N\right)=1$. For $\alpha \in \mathbb{N}$, the weighted Bergman space $A_{\alpha }^2\left({\mathbb{B}}^N\right)$ consists of all $f\in H\left({\mathbb{B}}^N\right)$ such that

$$\begin{array}{c}\hfill {\parallel f\parallel }^2={\int }_{{\mathbb{B}}^N}{\left|f\left(z\right)\right|}^{2}d{v}_{\alpha }\left(z\right)<\infty .\end{array}$$

The norm is induced by the inner product

$$\begin{array}{c}\hfill \langle f,g\rangle ={\int }_{{\mathbb{B}}^N}f\left(z\right)\overline{g\left(z\right)}d{v}_{\alpha }\left(z\right).\end{array}$$

The space $A_{\alpha }^2\left({\mathbb{B}}^N\right)$ is a Hilbert space where the reproducing kernel function is given by

$$\begin{array}{c}\hfill K_w\left(z\right)={\displaystyle \frac{1}{{(1-\langle z,w\rangle )}^{N+1+\alpha }}}\end{array}$$

for each w, $z\in {\mathbb{B}}^N$. See [35,36,37] for more information on the space.

Let $u\in \mathcal{O}\left({\mathbb{B}}^N\right)$ and $\phi$ be a holomorphic self-mapping of ${\mathbb{B}}^N$. Then, the weighted composition operator on (or between) some subspaces of $\mathcal{O}\left({\mathbb{B}}^N\right)$ is defined by

$$\begin{array}{c}\hfill W_{u,\phi }f\left(z\right)=u\left(z\right)f\left(\phi \left(z\right)\right).\end{array}$$

If $u\equiv 1$, $W_{u,\phi }$ is the composition operator $C_{\phi }$. While $\phi \left(z\right)=z$, $W_{u,\phi }$ is the multiplication operator $M_u$. Since $W_{u,\phi }=M_u·C_{\phi }$, $W_{u,\phi }$ can be regarded as the product of $M_u$ and $C_{\phi }$. An elementary problem is to provide function-theoretic characterizations of when the symbols u and $\phi$ induce bounded or compact weighted composition operators on holomorphic function spaces. Weighted composition operators also have many applications in partial differential equations (see [38,39]). Therefore, the research on weighted composition operators has attracted the interest of many scholars and yielded numerous results (see [40,41,42,43,44]).

With significant progress in the study of a single weighted composition operator, the study of differences of composition operators or weighted composition operators acting on various Banach spaces of holomorphic functions has been receiving more attention. Among these, the topological properties of such differences have been studied, and some results have been obtained (see [45,46,47,48,49,50]). Although the topological properties have been extensively studied, the algebraic properties of such operators, such as normality and complex symmetry, remain relatively unexplored. Motivated by this, a natural problem is to consider 2-complex symmetric difference of composition operators or weighted composition operators on holomorphic function spaces, including Hardy spaces, weighted Bergman spaces, weighted Bergman–Orlicz spaces, and so on. Here, we try to consider such problems on $A_{\alpha }^2\left({\mathbb{B}}^N\right)$. Our consideration is also influenced by Yang’s work [48]. Yang in, [48], proved that when $0<p\le q<\infty$, the difference of two weighted composition operators from Fock spaces ${\mathcal{F}}^p$ to ${\mathcal{F}}^q$ is bounded if and only if each weighted composition operator is bounded from ${\mathcal{F}}^p$ to ${\mathcal{F}}^q$. Inspired by this result, we study the relationship between a single complex symmetric weighted composition operator and the difference of two complex symmetric weighted composition operators and find a similar relationship.

Let $\mathbb{N}$ denote the set of all nonzero integers. Throughout the paper, we always assume that $\alpha \in \mathbb{N}$, since the equality ${\langle w_1,z_1\rangle }^{\alpha }{\langle w_2,z_2\rangle }^{\alpha }={\left(\langle w_1,z_1\rangle \langle w_2,z_2\rangle \right)}^{\alpha }$ holds when $\alpha \in \mathbb{N}$.

Lemma 1. 

Let b, $c\in {\mathbb{B}}^N$, A be an $N\times N$ matrix, and $\parallel A\parallel +\left|b\right|+\left|c\right|<1$. If

$$\begin{array}{c}\hfill \phi \left(z\right)={\displaystyle \frac{Az+b}{\langle z,c\rangle +1}},\end{array}$$

then φ is a holomorphic self-mapping of ${\mathbb{B}}^N$.

Proof. 

For $c\in {\mathbb{B}}^N$, we have $\langle z,c\rangle +1\ne 0$ for all $z\in {\mathbb{B}}^N$, which means that $\phi$ is holomorphic on ${\mathbb{B}}^N$. Since $\parallel A\parallel +\left|b\right|+\left|c\right|<1$ implies that

$$\parallel A\parallel +\left|b\right|<1-\left|c\right|,$$

we obtain

$$\begin{array}{c}\hfill |Az+b|\le \left|Az\right|+\left|b\right|\le \parallel A\parallel \left|z\right|+\left|b\right|\le \parallel A\parallel +\left|b\right|\end{array}$$

(2)

for $z\in {\mathbb{B}}^N$. From the elementary inequality $|\langle z,w\rangle |\le \left|z\right|\left|w\right|$, we have

$$\begin{array}{c}\hfill 1-\left|c\right|<1-\left|c\right|\left|z\right|\le 1-|\sum _{j=1}^N{z}_j{\overline{c}}_j|\le |\langle z,c\rangle +1|\end{array}$$

(3)

for $z\in {\mathbb{B}}^N$. From (2) and (3), it follows that

$$\begin{array}{c}\hfill \left|Az\right|+\left|b\right|<1-|\sum _{j=1}^N{z}_j{\overline{c}}_j|.\end{array}$$

(4)

Then, it follows from (2)–(4) that $|Az+b|<|\langle z,c\rangle +1|$, which shows

$$\begin{array}{c}\hfill \left|\phi \left(z\right)\right|={\displaystyle \frac{|Az+b|}{|\langle z,c\rangle +1|}}<1.\end{array}$$

From this, the desired result follows. □

The following result was obtained in [33].

Lemma 2. 

Let $u\in \mathcal{O}\left({\mathbb{B}}^N\right)$ and φ be a holomorphic self-mapping of ${\mathbb{B}}^N$. If

$$\begin{array}{c}\hfill \underset{z\in {\mathbb{B}}^N}{sup}{\displaystyle \frac{\left|u(z)\right|(1-{\left|z\right|}^2{)}^{\frac{\alpha }{2}}}{{(1-|\phi \left(z\right)|}^2{)}^{\frac{N+1+\alpha }{2}}}}<\infty ,\end{array}$$

then, the operator $W_{u,\phi }$ is bounded on $A_{\alpha }^2\left({\mathbb{B}}^N\right)$.

From Lemma 2, we obtain the following result.

Corollary 1.

Let $u\left(z\right)={\displaystyle \frac{1}{{(1+\langle z,c\rangle )}^{N+1+\alpha }}}$, $c\in {\mathbb{B}}^N$, and φ be defined in Lemma 1. Then, the operator $W_{u,\phi }$ is bounded on $A_{\alpha }^2\left({\mathbb{B}}^N\right)$.

Proof. 

For each $z\in {\mathbb{B}}^N$, we have

$$\begin{array}{cc}\hfill {\displaystyle \frac{\left|u\left(z\right)\right|(1-{\left|z\right|}^2{)}^{\frac{\alpha }{2}}}{{(1-|\phi \left(z\right)|}^2{)}^{\frac{N+1+\alpha }{2}}}}& ={\displaystyle \frac{|\frac{1}{{(1+\langle z,c\rangle )}^{N+1+\alpha }}{|(1-{\left|z\right|}^2)}^{\frac{\alpha }{2}}}{(1-|\frac{Az+b}{\langle z,c\rangle +1}{{|}^2)}^{\frac{N+1+\alpha }{2}}}}\hfill \\ \hfill & ={\displaystyle \frac{{(1-|z|}^2{)}^{\frac{\alpha }{2}}}{{\left(\right|1+\langle z,c\rangle |}^2{-|Az+b|}^2{)}^{\frac{N+1+\alpha }{2}}}}\hfill \\ \hfill & \le {\displaystyle \frac{1}{\left[\right(1-{\left|\langle z,c\rangle \right|)}^2-\left(\right|Az|+{\left|b\right|)}^2{]}^{\frac{N+1+\alpha }{2}}}}\hfill \\ \hfill & \le {\displaystyle \frac{1}{\left[\right(1-{\left|c\right|)}^2-\left(\right|\left|A\right||+{\left|b\right|)}^2{]}^{\frac{N+1+\alpha }{2}}}}.\hfill \end{array}$$

By Lemma 2, the operator $W_{u,\phi }$ is bounded on $A_{\alpha }^2\left({\mathbb{B}}^N\right)$. □

Since the linear span of the set $\{K_w:w\in {\mathbb{B}}^N\}$ is dense in $A_{\alpha }^2\left({\mathbb{B}}^N\right)$, we have the following two direct results.

Lemma 3. 

Let T be a bounded operator on $A_{\alpha }^2\left({\mathbb{B}}^N\right)$ and C be a conjugation on $A_{\alpha }^2\left({\mathbb{B}}^N\right)$. Then, the operator T is complex symmetric with C on $A_{\alpha }^2\left({\mathbb{B}}^N\right)$ if and only if

$$\begin{array}{c}\hfill (CT-T^{\ast }C)K_w\left(z\right)=0\end{array}$$

for all $w,z\in {\mathbb{B}}^N$.

Proof. 

We denote M by the linear span of the reproducing kernel functions $\{K_w:w\in {\mathbb{B}}^N\}$. Then, M is dense in $A_{\alpha }^2\left({\mathbb{B}}^N\right)$. For each $f\in A_{\alpha }^2\left({\mathbb{B}}^N\right)$, there exists a sequence $\left\{f_j\right\}$ in M such that $\parallel f_j-f\parallel \to 0$ as $j\to \infty$. Since each $f_j$ is a finite linear combination of the functions $K_w$, we have that if $(CT-T^{\ast }C)K_w\left(z\right)=0$ for all $w,z\in {\mathbb{B}}^N$, then

$$\begin{array}{c}\hfill (CT-T^{\ast }C)f_j\left(z\right)=0\end{array}$$

(5)

for all $z\in {\mathbb{B}}^N$.

Now, assume that $(CT-T^{\ast }C)K_w\left(z\right)=0$ for all $w,z\in {\mathbb{B}}^N$. Then, it follows from (5) that

$$\begin{array}{cc}\hfill \parallel (CT-T^{\ast }C)f\parallel & =\parallel (CT-T^{\ast }C)f-(CT-T^{\ast }C)f_j+(CT-T^{\ast }C)f_j\parallel \hfill \\ \hfill & \le \parallel (CT-T^{\ast }C)f-(CT-T^{\ast }C)f_j\parallel +\parallel (CT-T^{\ast }C)f_j\parallel \hfill \\ \hfill & \le \parallel CT-T^{\ast }C\parallel \parallel f-f_j\parallel ,\hfill \end{array}$$

which implies that $\parallel (CT-T^{\ast }C)f\parallel =0$; that is, $(CT-T^{\ast }C)f=0$. This shows that the operator T is complex symmetric with C on $A_{\alpha }^2\left({\mathbb{B}}^N\right)$.

Conversely, if the operator T is complex symmetric with C on $A_{\alpha }^2\left({\mathbb{B}}^N\right)$, then $(CT-T^{\ast }C)K_w\left(z\right)=0$ for all $w,z\in {\mathbb{B}}^N$. □

Lemma 4. 

Let T be a bounded operator on $A_{\alpha }^2\left({\mathbb{B}}^N\right)$ and C be a conjugation on $A_{\alpha }^2\left({\mathbb{B}}^N\right)$. Then, the operator T is 2-complex symmetric with C on $A_{\alpha }^2\left({\mathbb{B}}^N\right)$ if and only if

$$\begin{array}{c}\hfill (C{T}^2-2T^{\ast }CT+T^{\ast 2}C)K_w\left(z\right)=0\end{array}$$

for all $w,z\in {\mathbb{B}}^N$.

Proof. 

The proof is similar to Lemma 3, so the details are omitted. □

We have the following formula of weighted composition operators on $A_{\alpha }^2\left({\mathbb{B}}^N\right)$.

Lemma 5. 

Let φ and u be the symbols defined in Lemma 1 and Corollary 1, respectively. Then, on $A_{\alpha }^2\left({\mathbb{B}}^N\right)$, it holds

$$\begin{array}{c}\hfill W_{u,\phi }^{\ast }=W_{{\displaystyle \frac{1}{{(1-\langle z,b\rangle )}^{N+1+\alpha }}},{\displaystyle \frac{A^{\ast }z-c}{1-\langle z,b\rangle }}}.\end{array}$$

Proof. 

From Corollary 1, it follows that $W_{u,\phi }$ is bounded on $A_{\alpha }^2\left({\mathbb{B}}^N\right)$. Then, for each $f\in A_{\alpha }^2\left({\mathbb{B}}^N\right)$ and $z\in {\mathbb{B}}^N$, we have

$$\begin{array}{cc}\hfill W_{u,\phi }^{\ast }f\left(z\right)& =\langle W_{u,\phi }^{\ast }f,K_z\rangle =\langle f,W_{u,\phi }{K}_z\rangle ={\int }_{{\mathbb{B}}^N}f\left(w\right)\overline{W_{u,\phi }{K}_z\left(w\right)}d{v}_{\alpha }\left(w\right)\hfill \\ \hfill & ={\int }_{{\mathbb{B}}^N}f\left(w\right){\displaystyle \frac{1}{{(1+\langle c,w\rangle )}^{N+1+\alpha }}}\overline{K_z\left(\phi \left(w\right)\right)}d{v}_{\alpha }\left(w\right)\hfill \\ \hfill & ={\int }_{{\mathbb{B}}^N}f\left(w\right){\displaystyle \frac{1}{{(1+\langle c,w\rangle )}^{N+1+\alpha }}}{\displaystyle \frac{1}{{\left(1-\langle z,\frac{Aw+b}{\langle w,c\rangle +1}\rangle \right)}^{N+1+\alpha }}}d{v}_{\alpha }\left(w\right)\hfill \\ \hfill & ={\int }_{{\mathbb{B}}^N}f\left(w\right){\displaystyle \frac{1}{{(1+\langle c,w\rangle -\langle z,Aw+b\rangle )}^{N+1+\alpha }}}d{v}_{\alpha }\left(w\right)\hfill \\ \hfill & ={\int }_{{\mathbb{B}}^N}f\left(w\right){\displaystyle \frac{1}{{(1+\langle c-A^{\ast }z,w\rangle -\langle z,b\rangle )}^{N+1+\alpha }}}d{v}_{\alpha }\left(w\right)\hfill \\ \hfill & ={\int }_{{\mathbb{B}}^N}f\left(w\right){\displaystyle \frac{1}{{(1-\langle z,b\rangle -\langle A^{\ast }z-c,w\rangle )}^{N+1+\alpha }}}d{v}_{\alpha }\left(w\right)\hfill \\ \hfill & ={\int }_{{\mathbb{B}}^N}f\left(w\right){\displaystyle \frac{1}{{(1-\langle z,b\rangle )}^{N+1+\alpha }}}{\displaystyle \frac{1}{{(1-\frac{\langle A^{\ast }z-c,w\rangle }{1-\langle z,b\rangle })}^{N+1+\alpha }}}d{v}_{\alpha }\left(w\right)\hfill \\ \hfill & ={\displaystyle \frac{1}{{(1-\langle z,b\rangle )}^{N+1+\alpha }}}{\int }_{{\mathbb{B}}^N}f\left(w\right){\displaystyle \frac{1}{{\left(1-\langle \frac{A^{\ast }z-c}{1-\langle z,b\rangle },w\rangle \right)}^{N+1+\alpha }}}d{v}_{\alpha }\left(w\right)\hfill \\ \hfill & ={\displaystyle \frac{1}{{(1-\langle z,b\rangle )}^{N+1+\alpha }}}\langle f,K_{{\displaystyle \frac{A^{\ast }z-c}{1-\langle z,b\rangle }}}\rangle ={\displaystyle \frac{1}{{(1-\langle z,b\rangle )}^{N+1+\alpha }}}f\left({\displaystyle \frac{A^{\ast }z-c}{1-\langle z,b\rangle }}\right)\hfill \\ \hfill & =W_{{\displaystyle \frac{1}{{(1-\langle z,b\rangle )}^{N+1+\alpha }}},{\displaystyle \frac{A^{\ast }z-c}{1-\langle z,b\rangle }}}f\left(z\right),\hfill \end{array}$$

from which the desired result follows. □

3. Complex Symmetric and 2-Complex Symmetric Weighted Composition Operators

First, we characterize complex symmetric weighted composition operators on $A_{\alpha }^2\left({\mathbb{B}}^N\right)$ when the conjugation is of the form $\mathcal{J}f\left(z\right)=\overline{f\left(\kappa \right(z\left)\right)}$.

Theorem 1.

Let φ and u be the symbols in Lemma 1 and Corollary 1, respectively, and A be symmetric. Then, the operator $W_{u,\phi }$ is complex symmetric on $A_{\alpha }^2\left({\mathbb{B}}^N\right)$ with the conjugation $\mathcal{J}$ if and only if $b=-\kappa \left(c\right)$.

Proof. 

For all $w,z\in {\mathbb{B}}^N$, it follows from Lemma 5 that

$$\begin{array}{cc}\hfill \mathcal{J}{W}_{u,\phi }{K}_w\left(z\right)& =\mathcal{J}\left({\displaystyle \frac{1}{{(1+\langle z,c\rangle )}^{N+1+\alpha }}}{\displaystyle \frac{1}{{(1-\langle \frac{Az+b}{\langle z,c\rangle +1},w\rangle )}^{N+1+\alpha }}}\right)\hfill \\ \hfill & =\mathcal{J}\left({\displaystyle \frac{1}{{(1+\langle z,c\rangle -\langle Az+b,w\rangle )}^{N+1+\alpha }}}\right)\hfill \\ \hfill & =\mathcal{J}\left({\displaystyle \frac{1}{{(1+\langle z,c\rangle -\langle z,A^{\ast }w\rangle -\langle b,w\rangle )}^{N+1+\alpha }}}\right)\hfill \\ \hfill & ={\displaystyle \frac{1}{{(1+\langle z,\kappa \left(c\right)-\kappa \left(A^{\ast }w\right)\rangle -\langle w,b\rangle )}^{N+1+\alpha }}}\hfill \end{array}$$

(6)

and

$$\begin{array}{cc}\hfill W_{u,\phi }^{\ast }\mathcal{J}{K}_w\left(z\right)& =W_{u,\phi }^{\ast }\left({\displaystyle \frac{1}{{(1-\langle z,\kappa \left(w\right)\rangle )}^{N+1+\alpha }}}\right)\hfill \\ \hfill & =W_{{\displaystyle \frac{1}{{(1-\langle z,b\rangle )}^{N+1+\alpha }}},{\displaystyle \frac{A^{\ast }z-c}{1-\langle z,b\rangle }}}\left({\displaystyle \frac{1}{{(1-\langle z,\kappa \left(w\right)\rangle )}^{N+1+\alpha }}}\right)\hfill \\ \hfill & ={\displaystyle \frac{1}{{(1-\langle z,b\rangle -\langle A^{\ast }z-c,\kappa \left(w\right)\rangle )}^{N+1+\alpha }}}\hfill \\ \hfill & ={\displaystyle \frac{1}{{(1-\langle z,b+A\kappa \left(w\right)\rangle +\langle c,\kappa \left(w\right)\rangle )}^{N+1+\alpha }}}.\hfill \end{array}$$

(7)

Then, from (6), (7) and Lemma 3, it follows that the operator $W_{u,\phi }$ is complex symmetric on $A_{\alpha }^2\left({\mathbb{B}}^N\right)$ with the conjugation $\mathcal{J}$ if and only if

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{{(1+\langle z,\kappa \left(c\right)-\kappa \left(A^{\ast }w\right)\rangle -\langle w,b\rangle )}^{N+1+\alpha }}}={\displaystyle \frac{1}{{(1-\langle z,b+A\kappa \left(w\right)\rangle +\langle c,\kappa \left(w\right)\rangle )}^{N+1+\alpha }}}\end{array}$$

(8)

for all w, $z\in {\mathbb{B}}^N$.

Assume that the operator $W_{u,\phi }$ is complex symmetric on $A_{\alpha }^2\left({\mathbb{B}}^N\right)$ with the conjugation $\mathcal{J}$. Then, letting $w=0$ in (8), we obtain

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{{(1+\langle z,\kappa \left(c\right)\rangle )}^{N+1+\alpha }}}={\displaystyle \frac{1}{{(1-\langle z,b\rangle )}^{N+1+\alpha }}}\end{array}$$

for all $z\in {\mathbb{B}}^N$. That is,

$$\begin{array}{c}\hfill {(1+\langle z,\kappa \left(c\right)\rangle )}^{N+1+\alpha }={(1-\langle z,b\rangle )}^{N+1+\alpha },\end{array}$$

which implies

$$\begin{array}{c}\hfill 1+\langle z,\kappa \left(c\right)\rangle =\sigma (0,\dots ,z_i,\dots ,0)(1-\langle z,b\rangle ),\end{array}$$

(9)

where $\sigma {(0,\dots ,z_i,\dots ,0)}^{N+1+\alpha }=1$. Letting $z=(0,\dots ,z_i,\dots ,0)$ in (9), we obtain

$$\begin{array}{c}\hfill 1+z_i{c}_i=\sigma (0,\dots ,z_i,\dots ,0)(1-z_i\overline{b_i}),\end{array}$$

where $b_i,c_i$ are the $ith$ components of b and c, respectively. From the definition of $\sigma (0,\dots ,z_i,\dots ,0)$, we obtain that $\sigma (0,\dots ,z_i,\dots ,0)$ has at most $N+1+\alpha$ different values. Since $z_i$ exists infinite values, there are infinite $z_i$ such as $1+z_i{c}_i={\sigma }_k(1-z_i\overline{b_i})$, where ${\sigma }_k$ is a fix constant. From the constant term and the first term coefficient of $z_i$, we have ${\sigma }_k=1$ and $c_i={\sigma }_k(-\overline{b_i})$, which implies $c_i=-\overline{b_i}$, $i=\overline{1,N}$, and then $b=-\kappa \left(c\right)$.

Conversely, if $b=-\kappa \left(c\right)$, then the left of (8) equals

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{{(1+\langle z,-b-\kappa \left(A^{\ast }w\right)\rangle -\langle w,b\rangle )}^{N+1+\alpha }}},\end{array}$$

and the right of (8) also does. From this, it follows that (8) holds. This shows that the operator $W_{u,\phi }$ is complex symmetric on $A_{\alpha }^2\left({\mathbb{B}}^N\right)$ with the conjugation $\mathcal{J}$. □

Next, we characterize the complex symmetric weighted composition operators on $A_{\alpha }^2\left({\mathbb{B}}^N\right)$ when the conjugation is of the form ${\mathcal{J}}_{\tau ,\zeta }f\left(z\right)=\tau \overline{f\left({\tau }_{\zeta }\kappa \left(z\right)\right)}$, where $\tau \in \mathbb{C}$ with $\left|\tau \right|=1$, $\zeta =({\zeta }_1,\dots ,{\zeta }_N)\in {\mathbb{R}}^N$ and ${\tau }_{\zeta }=diag\{e^{i{\zeta }_1},\dots ,e^{i{\zeta }_N}\}$.

Theorem 2.

Let φ and u be the symbols in Lemma 1 and Corollary 1, respectively, and A be symmetric. Then, the operator $W_{u,\phi }$ is complex symmetric on $A_{\alpha }^2\left({\mathbb{B}}^N\right)$ with the conjugation ${\mathcal{J}}_{\tau ,\zeta }$ if and only if $b=-{\tau }_{\zeta }\kappa \left(c\right)$.

Proof. 

For all $w,z\in {\mathbb{B}}^N$, it follows from Lemma 5 that

$$\begin{array}{cc}\hfill {\mathcal{J}}_{\tau ,\zeta }{W}_{u,\phi }{K}_w\left(z\right)& ={\mathcal{J}}_{\tau ,\zeta }\left({\displaystyle \frac{1}{{(1+\langle z,c\rangle )}^{N+1+\alpha }}}{\displaystyle \frac{1}{{(1-\langle \frac{Az+b}{\langle z,c\rangle +1},w\rangle )}^{N+1+\alpha }}}\right)\hfill \\ \hfill & ={\mathcal{J}}_{\tau ,\zeta }\left({\displaystyle \frac{1}{{(1+\langle z,c\rangle -\langle Az+b,w\rangle )}^{N+1+\alpha }}}\right)\hfill \\ \hfill & ={\mathcal{J}}_{\tau ,\zeta }\left({\displaystyle \frac{1}{{(1+\langle z,c\rangle -\langle z,A^{\ast }w\rangle -\langle b,w\rangle )}^{N+1+\alpha }}}\right)\hfill \\ \hfill & ={\displaystyle \frac{\tau }{{(1+\langle z,\kappa \left({\tau }_{\zeta }^{\ast }c\right)-\kappa \left({\tau }_{\zeta }^{\ast }{A}^{\ast }w\right)\rangle -\langle w,b\rangle )}^{N+1+\alpha }}}\hfill \end{array}$$

(10)

and

$$\begin{array}{cc}\hfill W_{u,\phi }^{\ast }{\mathcal{J}}_{\tau ,\zeta }{K}_w\left(z\right)& =W_{u,\phi }^{\ast }\left({\displaystyle \frac{\tau }{{(1-\langle z,\kappa \left({\tau }_{\zeta }^{\ast }w\right)\rangle )}^{N+1+\alpha }}}\right)\hfill \\ \hfill & =W_{{\displaystyle \frac{1}{{(1-\langle z,b\rangle )}^{N+1+\alpha }}},{\displaystyle \frac{A^{\ast }z-c}{1-\langle z,b\rangle }}}\left({\displaystyle \frac{\tau }{{(1-\langle z,\kappa \left({\tau }_{\zeta }^{\ast }w\right)\rangle )}^{N+1+\alpha }}}\right)\hfill \\ \hfill & ={\displaystyle \frac{\tau }{{(1-\langle z,b\rangle -\langle A^{\ast }z-c,\kappa \left({\tau }_{\zeta }^{\ast }w\right)\rangle )}^{N+1+\alpha }}}\hfill \\ \hfill & ={\displaystyle \frac{\tau }{{(1-\langle z,b+A\kappa \left({\tau }_{\zeta }^{\ast }w\right)\rangle +\langle c,\kappa \left({\tau }_{\zeta }^{\ast }w\right)\rangle )}^{N+1+\alpha }}}.\hfill \end{array}$$

(11)

Then, it follows from (10), (11) and Lemma 3 that the operator $W_{u,\phi }$ is complex symmetric on $A_{\alpha }^2\left({\mathbb{B}}^N\right)$ with the conjugation ${\mathcal{J}}_{\tau ,\zeta }$ if and only if

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{{(1+\langle z,\kappa \left({\tau }_{\zeta }^{\ast }c\right)-\kappa \left({\tau }_{\zeta }^{\ast }{A}^{\ast }w\right)\rangle -\langle w,b\rangle )}^{N+1+\alpha }}}={\displaystyle \frac{1}{{(1-\langle z,b+A\kappa \left({\tau }_{\zeta }^{\ast }w\right)\rangle +\langle c,\kappa \left({\tau }_{\zeta }^{\ast }w\right)\rangle )}^{N+1+\alpha }}}\end{array}$$

(12)

for all w, $z\in {\mathbb{B}}^N$.

Assume that the operator $W_{u,\phi }$ is complex symmetric on $A_{\alpha }^2\left({\mathbb{B}}^N\right)$ with the conjugation ${\mathcal{J}}_{\tau ,\zeta }$. Letting $w=0$ in (12), we obtain

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{{(1+\langle z,\kappa \left({\tau }_{\zeta }^{\ast }c\right)\rangle )}^{N+1+\alpha }}}={\displaystyle \frac{1}{{(1-\langle z,b\rangle )}^{N+1+\alpha }}}\end{array}$$

for all $z\in {\mathbb{B}}^N$, which shows

$${(1+\langle z,\kappa \left({\tau }_{\zeta }^{\ast }c\right)\rangle )}^{N+1+\alpha }={(1-\langle z,b\rangle )}^{N+1+\alpha }$$

for all $z\in {\mathbb{B}}^N$. This implies

$$\begin{array}{c}\hfill 1+\langle z,\kappa \left({\tau }_{\zeta }^{\ast }c\right)\rangle =\varrho (0,\dots ,z_l,\dots ,0)(1-\langle z,b\rangle ),\end{array}$$

(13)

where $\varrho {(0,\dots ,z_l,\dots ,0)}^{N+1+\alpha }=1$. Letting $z=(0,\dots ,z_l,\dots ,0)$ in (13), we obtain

$$\begin{array}{c}\hfill 1+z_l{c}_l{e}^{-i{\zeta }_l}=\varrho (0,\dots ,z_l,\dots ,0)(1-z_l\overline{b_l}),\end{array}$$

(14)

where $b_l,c_l$ are the $lth$ components of b and c, respectively. From the definition of $\varrho (0,\dots ,z_l,\dots ,0)$, we obtain that $\varrho (0,\dots ,z_l,\dots ,0)$ has at most $N+1+\alpha$ different values. Since $z_l$ exists infinite values, there are infinite $z_l$ such as $1+z_l{c}_l{e}^{-i{\zeta }_l}={\varrho }_k(1-z_l\overline{b_l})$, where ${\varrho }_k$ is a fix constant. From the constant term and the first term coefficient of $z_l$, we have ${\varrho }_k=1$ and $c_l{e}^{-i{\zeta }_l}={\varrho }_k(-\overline{b_l})$, which implies $c_l{e}^{-i{\zeta }_l}=-\overline{b_l}$, $l=\overline{1,N}$, and then $b=-{\tau }_{\zeta }\kappa \left(c\right)$.

Conversely, if $b=-{\tau }_{\zeta }\kappa \left(c\right)$ , we see that the left of (12) equals

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{{(1+\langle z,{\tau }_{\zeta }\kappa \left(c\right)-{\tau }_{\zeta }A\kappa \left(w\right)\rangle +\langle w,{\tau }_{\zeta }\kappa \left(c\right)\rangle )}^{N+1+\alpha }}},\end{array}$$

and the right of (12) also does. From this, it follows that (12) holds. This shows that the operator $W_{u,\phi }$ is complex symmetric on $A_{\alpha }^2\left({\mathbb{B}}^N\right)$ with the conjugation ${\mathcal{J}}_{\tau ,\zeta }$. □

Now, we characterize 2-complex symmetric weighted composition operators on $A_{\alpha }^2\left({\mathbb{B}}^N\right)$ with the conjugations $\mathcal{J}f\left(z\right)=\overline{f\left(\kappa \right(z\left)\right)}$ and ${\mathcal{J}}_{\tau ,\zeta }f\left(z\right)=\tau \overline{f\left({\tau }_{\zeta }\kappa \left(z\right)\right)}$ on $A_{\alpha }^2\left({\mathbb{B}}^N\right)$.

Theorem 3.

Let φ and u be the symbols in Lemma 1 and Corollary 1, respectively, and A be symmetric. Then, the operator $W_{u,\phi }$ is 2-complex symmetric on $A_{\alpha }^2\left({\mathbb{B}}^N\right)$ with the conjugation $\mathcal{J}$ if and only if $b=-\kappa \left(c\right)$.

Proof. 

For all $w,z\in {\mathbb{B}}^N$, it follows from Lemma 5 that

$$\begin{array}{cc}\hfill & \mathcal{J}{W}_{u,\phi }^2{K}_w\left(z\right)=\mathcal{J}{W}_{u,\phi }\left({\displaystyle \frac{1}{{(1+\langle z,c\rangle )}^{N+1+\alpha }}}{K}_w\left({\displaystyle \frac{Az+b}{\langle z,c\rangle +1}}\right)\right)\hfill \\ \hfill & =\mathcal{J}{W}_{u,\phi }\left({\displaystyle \frac{1}{{(1+\langle z,c\rangle )}^{N+1+\alpha }}}{\displaystyle \frac{1}{{(1-\langle \frac{Az+b}{\langle z,c\rangle +1},w\rangle )}^{N+1+\alpha }}}\right)\hfill \\ \hfill & =\mathcal{J}{W}_{u,\phi }\left({\displaystyle \frac{1}{{(1+\langle z,c\rangle -\langle Az+b,w\rangle )}^{N+1+\alpha }}}\right)\hfill \\ \hfill & =\mathcal{J}{W}_{u,\phi }\left({\displaystyle \frac{1}{{(1+\langle z,c-A^{\ast }w\rangle -\langle b,w\rangle )}^{N+1+\alpha }}}\right)\hfill \\ \hfill & =\mathcal{J}\left({\displaystyle \frac{1}{{(1+\langle z,c\rangle )}^{N+1+\alpha }}}{\displaystyle \frac{1}{{(1+\langle \frac{Az+b}{\langle z,c\rangle +1},c-A^{\ast }w\rangle -\langle b,w\rangle )}^{N+1+\alpha }}}\right)\hfill \\ \hfill & =\mathcal{J}\left({\displaystyle \frac{1}{{(1+\langle z,c\rangle +\langle Az+b,c-A^{\ast }w\rangle -(1+\langle z,c\rangle )\langle b,w\rangle )}^{N+1+\alpha }}}\right)\hfill \\ \hfill & =\mathcal{J}\left({\displaystyle \frac{1}{{(1+\langle z,c\rangle +\langle z,A^{\ast }c-A^{{\ast }^2}w\rangle +\langle b,c-A^{\ast }w\rangle -\langle b,w\rangle -\langle z,c\rangle \langle b,w\rangle )}^{N+1+\alpha }}}\right)\hfill \\ \hfill & ={\displaystyle \frac{1}{{(1+\langle z,\kappa (c+A^{\ast }c-A^{{\ast }^2}w)\rangle +\langle c-A^{\ast }w,b\rangle -\langle w,b\rangle -\langle z,\kappa \left(c\right)\rangle \langle w,b\rangle )}^{N+1+\alpha }}}\hfill \\ \hfill & ={\displaystyle \frac{1}{{(1+\langle z,\kappa (c+A^{\ast }c)\rangle -\langle w,A^2\kappa \left(z\right)+b+Ab\rangle +\langle c,b\rangle -\langle z,\kappa \left(c\right)\rangle \langle w,b\rangle )}^{N+1+\alpha }}},\hfill \end{array}$$

(15)

$$\begin{array}{cc}\hfill & W_{u,\phi }^{\ast }\mathcal{J}{W}_{u,\phi }{K}_w\left(z\right)=W_{u,\phi }^{\ast }\mathcal{J}\left({\displaystyle \frac{1}{{(1+\langle z,c\rangle )}^{N+1+\alpha }}}{\displaystyle \frac{1}{{(1-\langle \frac{Az+b}{\langle z,c\rangle +1},w\rangle )}^{N+1+\alpha }}}\right)\hfill \\ \hfill & =W_{u,\phi }^{\ast }\mathcal{J}\left({\displaystyle \frac{1}{{(1+\langle z,c\rangle -\langle Az+b,w\rangle )}^{N+1+\alpha }}}\right)\hfill \\ \hfill & =W_{u,\phi }^{\ast }\mathcal{J}\left({\displaystyle \frac{1}{{(1+\langle z,c-A^{\ast }w\rangle -\langle b,w\rangle )}^{N+1+\alpha }}}\right)\hfill \\ \hfill & =W_{u,\phi }^{\ast }\left({\displaystyle \frac{1}{{(1+\langle z,\kappa (c-A^{\ast }w)\rangle -\langle w,b\rangle )}^{N+1+\alpha }}}\right)\hfill \\ \hfill & ={\displaystyle \frac{1}{{(1-\langle z,b\rangle )}^{N+1+\alpha }}}{\displaystyle \frac{1}{{(1+\langle \frac{A^{\ast }z-c}{1-\langle z,b\rangle },\kappa (c-A^{\ast }w)\rangle -\langle w,b\rangle )}^{N+1+\alpha }}}\hfill \\ \hfill & ={\displaystyle \frac{1}{{(1-\langle z,b\rangle +\langle A^{\ast }z-c,\kappa (c-A^{\ast }w)\rangle -(1-\langle z,b\rangle )\langle w,b\rangle )}^{N+1+\alpha }}}\hfill \\ \hfill & ={\displaystyle \frac{1}{{(1-\langle z,b-A\kappa \left(c\right)\rangle -\langle A^{\ast }z,\kappa \left(A^{\ast }w\right)\rangle -\langle c,\kappa \left(c\right)\rangle +\langle w,A\kappa \left(c\right)-b\rangle +\langle w,b\rangle \langle z,b\rangle )}^{N+1+\alpha }}},\hfill \end{array}$$

(16)

and

$$\begin{array}{cc}\hfill & W_{u,\phi }^{\ast 2}\mathcal{J}{K}_w\left(z\right)=W_{u,\phi }^{\ast }\left({\displaystyle \frac{1}{{(1-\langle z,b\rangle )}^{N+1+\alpha }}}{\displaystyle \frac{1}{{(1-\langle \frac{A^{\ast }z-c}{1-\langle z,b\rangle },\kappa \left(w\right)\rangle )}^{N+1+\alpha }}}\right)\hfill \\ \hfill & =W_{u,\phi }^{\ast }\left({\displaystyle \frac{1}{{(1-\langle z,b\rangle -\langle A^{\ast }z-c,\kappa \left(w\right)\rangle )}^{N+1+\alpha }}}\right)\hfill \\ \hfill & =W_{u,\phi }^{\ast }\left({\displaystyle \frac{1}{{(1-\langle z,b+A\kappa \left(w\right)\rangle +\langle c,\kappa \left(w\right)\rangle )}^{N+1+\alpha }}}\right)\hfill \\ \hfill & =W_{{\displaystyle \frac{1}{{(1-\langle z,b\rangle )}^{N+1+\alpha }}},{\displaystyle \frac{A^{\ast }z-c}{1-\langle z,b\rangle }}}\left({\displaystyle \frac{1}{{(1-\langle z,b+A\kappa \left(w\right)\rangle +\langle c,\kappa \left(w\right)\rangle )}^{N+1+\alpha }}}\right)\hfill \\ \hfill & ={\displaystyle \frac{1}{{(1-\langle z,b\rangle )}^{N+1+\alpha }}}{\displaystyle \frac{1}{{(1-\langle \frac{A^{\ast }z-c}{1-\langle z,b\rangle },b+A\kappa \left(w\right)\rangle +\langle c,\kappa \left(w\right)\rangle )}^{N+1+\alpha }}}\hfill \\ \hfill & ={\displaystyle \frac{1}{{(1-\langle z,b+Ab\rangle -\langle z,A^2\kappa \left(w\right)\rangle +\langle c,b\rangle +\langle w,\kappa (A^{\ast }c+c)\rangle -\langle z,b\rangle \langle w,\kappa \left(c\right)\rangle )}^{N+1+\alpha }}}.\hfill \end{array}$$

(17)

Then, from (15)–(17) and Lemma 4, it follows that the operator $W_{u,\phi }$ is 2-complex symmetric on $A_{\alpha }^2\left({\mathbb{B}}^N\right)$ with the conjugation $\mathcal{J}$ if and only if

$$\begin{array}{cc}\hfill & {\displaystyle \frac{1}{{(1+\langle z,\kappa (c+A^{\ast }c)\rangle -\langle w,A^2\kappa \left(z\right)+b+Ab\rangle +\langle c,b\rangle -\langle z,\kappa \left(c\right)\rangle \langle w,b\rangle )}^{N+1+\alpha }}}\hfill \\ \hfill & +{\displaystyle \frac{1}{{(1-\langle z,b+Ab\rangle -\langle z,A^2\kappa \left(w\right)\rangle +\langle c,b\rangle +\langle w,\kappa (A^{\ast }c+c)\rangle -\langle z,b\rangle \langle w,\kappa \left(c\right)\rangle )}^{N+1+\alpha }}}\hfill \\ \hfill & ={\displaystyle \frac{2}{{(1-\langle z,b-A\kappa \left(c\right)\rangle -\langle A^{\ast }z,\kappa \left(A^{\ast }w\right)\rangle -\langle c,\kappa \left(c\right)\rangle +\langle w,A\kappa \left(c\right)-b\rangle +\langle w,b\rangle \langle z,b\rangle )}^{N+1+\alpha }}}\hfill \end{array}$$

(18)

for all w, $z\in {\mathbb{B}}^N$.

Assume that the operator $W_{u,\phi }$ is 2-complex symmetric on $A_{\alpha }^2\left({\mathbb{B}}^N\right)$ with the conjugation $\mathcal{J}$. Since $A=A^T$, we have $A=\kappa \left(A^{\ast }\right)$. From this and letting $w=z$ in (18), we obtain

$$\begin{array}{cc}\hfill & {\displaystyle \frac{1}{{(1+\langle z,\kappa \left(c\right)+A\kappa \left(c\right)-b-Ab\rangle -\langle z,A^2\kappa \left(z\right)\rangle +\langle c,b\rangle -\langle z,\kappa \left(c\right)\rangle \langle z,b\rangle )}^{N+1+\alpha }}}\hfill \\ \hfill & ={\displaystyle \frac{1}{{(1+\langle z,-2b+2A\kappa \left(c\right)\rangle -\langle z,A^2\kappa \left(z\right)\rangle -\langle c,\kappa \left(c\right)\rangle +\langle z,b\rangle \langle z,b\rangle )}^{N+1+\alpha }}}.\hfill \end{array}$$

(19)

for all $z\in {\mathbb{B}}^N$, which implies

$$\begin{array}{cc}\hfill & 1+\langle z,\kappa \left(c\right)+A\kappa \left(c\right)-b-Ab\rangle -\langle z,A^2\kappa \left(z\right)\rangle +\langle c,b\rangle -\langle z,\kappa \left(c\right)\rangle \langle z,b\rangle \chi (0,\dots ,z_i,\dots ,0)\hfill \\ \hfill & =(1+\langle z,-2b+2A\kappa \left(c\right)\rangle -\langle z,A^2\kappa \left(z\right)\rangle -\langle c,\kappa \left(c\right)\rangle +\langle z,b\rangle \langle z,b\rangle ),\hfill \end{array}$$

(20)

where $\chi {(0,\dots ,z_i,\dots ,0)}^{N+1+\alpha }=1$. Write $A={\left\{a_{ij}\right\}}_{N\times N}$. Letting $z=(0,\dots ,z_i,\dots ,0)$, $b=(0,\dots ,b_i,\dots ,0)$ and $c=(0,\dots ,c_i,\dots ,0)$ in (20), we obtain

$$\begin{array}{cc}\hfill [& 1+z_i\left(c_i-\overline{b_i}+\overline{a_{ii}}{c}_i-\overline{a_{ii}{b}_i}\right)-z_i^2\sum _{j=1}^N\overline{a_{ij}{a}_{ji}}+\langle c,b\rangle -z_i^2\overline{b_i}{c}_i]\chi (0,\dots ,z_i,\dots ,0)\hfill \\ \hfill & =1+z_i(-2\overline{b_i}+2\overline{a_{ii}}{c}_i)-z_i^2\sum _{j=1}^N\overline{a_{ij}{a}_{ji}}-\langle c,\kappa \left(c\right)\rangle -z_i^2\overline{b_i^2},\hfill \end{array}$$

(21)

From the definition of $\chi (0,\dots ,z_i,\dots ,0)$, we obtain that $\chi (0,\dots ,z_i,\dots ,0)$ has at most $N+1+\alpha$ different values. Since $z_i$ exists infinite values, there are infinite $z_i$ such as

$$\begin{array}{cc}\hfill & \left[1+z_i\left(c_i-\overline{b_i}+\overline{a_{ii}}{c}_i-\overline{a_{ii}{b}_i}\right)-z_i^2\sum _{j=1}^N\overline{a_{ij}{a}_{ji}}+\langle c,b\rangle -z_i^2\overline{b_i}{c}_i\right]{\chi }_k\hfill \\ \hfill & =1+z_i(-2\overline{b_i}+2\overline{a_{ii}}{c}_i)-z_i^2\sum _{j=1}^N\overline{a_{ij}{a}_{ji}}-\langle c,\kappa \left(c\right)\rangle -z_i^2\overline{b_i^2},\hfill \end{array}$$

(22)

where ${\chi }_k$ is a fix constant. From the coefficients of $z_i$, $z_i^2$ and the constant terms, we have

$$\left\{\begin{array}{c}(c_i-\overline{b_i}+\overline{a_{ii}}{c}_i-\overline{a_{ii}{b}_i}){\chi }_k=-2\overline{b_i}+2\overline{a_{ii}}{c}_i\\ ({\displaystyle \sum _{j=1}^N}\overline{a_{ij}{a}_{ji}}+\overline{b_i}{c}_i){\chi }_k={\displaystyle \sum _{j=1}^N}\overline{a_{ij}{a}_{ji}}+\overline{b_i^2}\\ (1+\langle c,b\rangle ){\chi }_k=1-\langle c,\kappa \left(c\right)\rangle .\end{array}\right.$$

(23)

Let $G_1=c_i+\overline{a_{ii}}{c}_i$, $G_2=-\overline{b_i}-\overline{a_{ii}{b}_i}$ and $H_1=1+\langle c,b\rangle$. Then, from (23), we obtain

$$\begin{array}{c}\hfill \overline{a_{ii}}{c}_i-\overline{b_i}={\displaystyle \frac{G_1+G_2}{2}}{\chi }_k\mathrm{and}1-\langle c,\kappa \left(c\right)\rangle =H_1{\chi }_k.\end{array}$$

(24)

Also, letting $w=0$ in (18), we obtain

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{{(1+\langle z,\kappa (c+A^{\ast }c)\rangle +\langle c,b\rangle )}^{N+1+\alpha }}}+{\displaystyle \frac{1}{{(1-\langle z,b+Ab\rangle +\langle c,b\rangle )}^{N+1+\alpha }}}={\displaystyle \frac{2}{{(1-\langle z,b-A\kappa \left(c\right)\rangle -\langle c,\kappa \left(c\right)\rangle )}^{N+1+\alpha }}}\end{array}$$

(25)

for all $z\in {\mathbb{B}}^N$. Combining (24) with (25), from ${\chi }_k^{N+1+\alpha }=1$ we obtain

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{{(G_{1}z+H_1)}^{N+1+\alpha }}}+{\displaystyle \frac{1}{{(G_{2}z+H_1)}^{N+1+\alpha }}}-{\displaystyle \frac{2}{{(\frac{G_1+G_2}{2}z+H_1)}^{N+1+\alpha }}}=0\end{array}$$

(26)

for all $z\in {\mathbb{B}}^N$.

Taking two consecutive derivatives of the variable z on the both sides of (26), we obtain

$$\begin{array}{c}\hfill {\displaystyle \frac{G_1^2}{{(G_{1}z+H_1)}^{N+3+\alpha }}}+{\displaystyle \frac{G_2^2}{{(G_{2}z+H_1)}^{N+3+\alpha }}}-{\displaystyle \frac{{(G_1+G_2)}^2}{2{(\frac{G_1+G_2}{2}z+H_1)}^{N+3+\alpha }}}=0.\end{array}$$

(27)

for all $z\in {\mathbb{B}}^N$. Choosing $z=0$ in (27), we have $G_1-G_2=0$, which implies $(\overline{a_{ii}}+1)(\overline{b_i}+c_i)=0$. Since $\parallel A\parallel <1$ implies that $(\overline{a_{ii}}+1)\ne 0$. Since $i=\overline{1,N}$, we have $b=-\kappa \left(c\right)$.

Conversely, if $b=-\kappa \left(c\right)$ , we see that the left of (18) equals

$$\begin{array}{c}\hfill {\displaystyle \frac{2}{{(1+\langle z,\kappa \left(c\right)+A\kappa \left(c\right)\rangle -\langle w,A^2\kappa \left(z\right)\rangle +\langle w,\kappa \left(c\right)+A\kappa \left(c\right)\rangle -\langle c,\kappa \left(c\right)\rangle +\langle z,\kappa \left(c\right)\rangle \langle w,\kappa \left(c\right)\rangle )}^{N+1+\alpha }}},\end{array}$$

and the right of (18) also does. From this, it follows that (18) holds. This shows that the operator $W_{u,\phi }$ is 2-complex symmetric on $A_{\alpha }^2\left({\mathbb{B}}^N\right)$ with the conjugation $\mathcal{J}$. □

From Theorems 1 and 3, we have the following

Corollary 2.

Let φ and u be the symbols in Lemma 1 and Corollary 1, respectively. The following are equivalent:

(a) 

$W_{u,\phi }$ is 2-complex symmetric on $A_{\alpha }^2\left({\mathbb{B}}^N\right)$ with the conjugation $\mathcal{J}$;

(b) 

$W_{u,\phi }$ is complex symmetric on $A_{\alpha }^2\left({\mathbb{B}}^N\right)$ with the conjugation $\mathcal{J}$;

(c) 

$b=-\kappa \left(c\right)$.

By Theorem 3, we can give some examples. To this end, we need to construct $A,b,c$ such that

$$\begin{array}{c}\hfill A=A^T,\parallel A\parallel +\left|b\right|+\left|c\right|<1,b=-\kappa \left(c\right).\end{array}$$

To simplify the proof, we only give an example for $N=2$.

Example 1.

Let $u\left(z\right)={\displaystyle \frac{1}{{(1+\langle z,c\rangle )}^{3+\alpha }}}$ and $\phi \left(z\right)={\displaystyle \frac{Az+b}{\langle z,c\rangle +1}}$, where

$$\begin{array}{c}\hfill A=\left(\begin{array}{cc}{\displaystyle \frac{1}{13}}+i{\displaystyle \frac{1}{13}}& {\displaystyle \frac{\sqrt{2}}{13}}\\ {\displaystyle \frac{\sqrt{2}}{13}}& 0\end{array}\right),b={\left(\begin{array}{cc}{\displaystyle \frac{1}{9}},& 0\end{array}\right)}^T,c={\left(\begin{array}{cc}{\displaystyle \frac{-1}{9}},& 0\end{array}\right)}^T.\end{array}$$

Then, the operator $W_{u,\phi }$ is 2-complex symmetric on $A_{\alpha }^2\left({\mathbb{B}}^2\right)$ with the conjugation $\mathcal{J}$.

Proof. 

From the direct calculations, it follows that $A=A^T$ and

$$\begin{array}{c}\hfill \parallel A\parallel ={\displaystyle \frac{\sqrt{6}}{13}},\left|b\right|={\displaystyle \frac{1}{9}},\left|c\right|={\displaystyle \frac{1}{9}},b=-\kappa \left(c\right).\end{array}$$

Thus, we obtain that $\parallel A\parallel +\left|b\right|+\left|c\right|<1$. By Theorem 3, the operator is 2-complex symmetric on $A_{\alpha }^2\left({\mathbb{B}}^2\right)$ with the conjugation $\mathcal{J}$. □

We begin to characterize 2-complex symmetric weighted composition operators on $A_{\alpha }^2\left({\mathbb{B}}^N\right)$ with the conjugation ${\mathcal{J}}_{\tau ,\zeta }$.

Theorem 4.

Let φ and u be the symbols in Lemma 1 and Corollary 1, respectively, and A be symmetric. Then, the operator $W_{u,\phi }$ is 2-complex symmetric on $A_{\alpha }^2\left({\mathbb{B}}^N\right)$ with the conjugation ${\mathcal{J}}_{\tau ,\zeta }$ if and only if $b=-{\tau }_{\zeta }\kappa \left(c\right)$.

Proof. 

For all $w,z\in {\mathbb{B}}^N$, from Lemma 5 we have

$$\begin{array}{cc}\hfill & {\mathcal{J}}_{\tau ,\zeta }{W}_{u,\phi }^2{K}_w\left(z\right)={\mathcal{J}}_{\tau ,\zeta }\left({\displaystyle \frac{1}{{(1+\langle z,c\rangle +\langle z,A^{\ast }c-A^{{\ast }^2}w\rangle +\langle b,c-A^{\ast }w\rangle -\langle b,w\rangle -\langle z,c\rangle \langle b,w\rangle )}^{N+1+\alpha }}}\right)\hfill \\ \hfill & ={\displaystyle \frac{\tau }{{(1+\langle z,\kappa ({\tau }_{\zeta }^{\ast }c+{\tau }_{\zeta }^{\ast }{A}^{\ast }c-{\tau }_{\zeta }^{\ast }{A}^{{\ast }^2}w)\rangle +\langle c-A^{\ast }w,b\rangle -\langle w,b\rangle -\langle z,\kappa \left({\tau }_{\zeta }^{\ast }c\right)\rangle \langle w,b\rangle )}^{N+1+\alpha }}}\hfill \\ \hfill & ={\displaystyle \frac{\tau }{{(1+\langle z,\kappa ({\tau }_{\zeta }^{\ast }c+{\tau }_{\zeta }^{\ast }{A}^{\ast }c)\rangle -\langle w,A^2{\tau }_{\zeta }\kappa \left(z\right)+b+Ab\rangle +\langle c,b\rangle -\langle z,\kappa \left({\tau }_{\zeta }^{\ast }c\right)\rangle \langle w,b\rangle )}^{N+1+\alpha }}},\hfill \end{array}$$

(28)

$$\begin{array}{cc}\hfill & W_{u,\phi }^{\ast }{\mathcal{J}}_{\tau ,\zeta }{W}_{u,\phi }{K}_w\left(z\right)=W_{u,\phi }^{\ast }{\mathcal{J}}_{\tau ,\zeta }\left({\displaystyle \frac{1}{{(1+\langle z,c-A^{\ast }w\rangle -\langle b,w\rangle )}^{N+1+\alpha }}}\right)\hfill \\ \hfill & =W_{u,\phi }^{\ast }\left({\displaystyle \frac{\tau }{{(1+\langle z,\kappa ({\tau }_{\zeta }^{\ast }c-{\tau }_{\zeta }^{\ast }{A}^{\ast }w)\rangle -\langle w,b\rangle )}^{N+1+\alpha }}}\right)\hfill \\ \hfill & ={\displaystyle \frac{1}{{(1-\langle z,b\rangle )}^{N+1+\alpha }}}{\displaystyle \frac{\tau }{{(1+\langle \frac{A^{\ast }z-c}{1-\langle z,b\rangle },\kappa ({\tau }_{\zeta }^{\ast }c-{\tau }_{\zeta }^{\ast }{A}^{\ast }w)\rangle -\langle w,b\rangle )}^{N+1+\alpha }}}\hfill \\ \hfill & ={\displaystyle \frac{\tau }{{(1-\langle z,b\rangle +\langle A^{\ast }z-c,\kappa ({\tau }_{\zeta }^{\ast }c-{\tau }_{\zeta }^{\ast }{A}^{\ast }w)\rangle -(1-\langle z,b\rangle )\langle w,b\rangle )}^{N+1+\alpha }}},\hfill \end{array}$$

(29)

and

$$\begin{array}{cc}\hfill & W_{u,\phi }^{\ast 2}{\mathcal{J}}_{\tau ,\zeta }{K}_w\left(z\right)=W_{u,\phi }^{\ast }\left({\displaystyle \frac{1}{{(1-\langle z,b\rangle )}^{N+1+\alpha }}}{\displaystyle \frac{\tau }{{(1-\langle \frac{A^{\ast }z-c}{1-\langle z,b\rangle },\kappa \left({\tau }_{\zeta }^{\ast }w\right)\rangle )}^{N+1+\alpha }}}\right)\hfill \\ \hfill & =W_{u,\phi }^{\ast }\left({\displaystyle \frac{\tau }{{(1-\langle z,b\rangle -\langle A^{\ast }z-c,\kappa \left({\tau }_{\zeta }^{\ast }w\right)\rangle )}^{N+1+\alpha }}}\right)\hfill \\ \hfill & =W_{u,\phi }^{\ast }\left({\displaystyle \frac{\tau }{{(1-\langle z,b+A\kappa \left({\tau }_{\zeta }^{\ast }w\right)\rangle +\langle c,\kappa \left({\tau }_{\zeta }^{\ast }w\right)\rangle )}^{N+1+\alpha }}}\right)\hfill \\ \hfill & =W_{{\displaystyle \frac{1}{{(1-\langle z,b\rangle )}^{N+1+\alpha }}},{\displaystyle \frac{A^{\ast }z-c}{1-\langle z,b\rangle }}}\left({\displaystyle \frac{\tau }{{(1-\langle z,b+A\kappa \left({\tau }_{\zeta }^{\ast }w\right)\rangle +\langle c,\kappa \left({\tau }_{\zeta }^{\ast }w\right)\rangle )}^{N+1+\alpha }}}\right)\hfill \\ \hfill & ={\displaystyle \frac{1}{{(1-\langle z,b\rangle )}^{N+1+\alpha }}}{\displaystyle \frac{\tau }{{(1-\langle \frac{A^{\ast }z-c}{1-\langle z,b\rangle },b+A\kappa \left({\tau }_{\zeta }^{\ast }w\right)\rangle +\langle c,\kappa \left({\tau }_{\zeta }^{\ast }w\right)\rangle )}^{N+1+\alpha }}}\hfill \\ \hfill & ={\displaystyle \frac{\tau }{{(1-\langle z,b+Ab\rangle -\langle z,A^2\kappa \left({\tau }_{\zeta }^{\ast }w\right)\rangle +\langle c,b\rangle +\langle w,{\tau }_{\zeta }\kappa (A^{\ast }c+c)\rangle -\langle z,b\rangle \langle w,{\tau }_{\zeta }\kappa \left(c\right)\rangle )}^{N+1+\alpha }}}.\hfill \end{array}$$

(30)

From a direct calculation, (29) is equivalent to

$$\begin{array}{c}\hfill {\displaystyle \frac{\tau }{{(1-\langle z,b-A\kappa \left({\tau }_{\zeta }^{\ast }c\right)\rangle -\langle A^{\ast }z,\kappa \left({\tau }_{\zeta }^{\ast }{A}^{\ast }w\right)\rangle -\langle c,\kappa \left({\tau }_{\zeta }^{\ast }c\right)\rangle +\langle c,\kappa \left({\tau }_{\zeta }^{\ast }{A}^{\ast }w\right)\rangle -\langle w,b\rangle +\langle w,b\rangle \langle z,b\rangle )}^{N+1+\alpha }}}.\end{array}$$

(31)

Then, from (28), (30), (31) and Lemma 4, it follows that the operator $W_{u,\phi }$ is 2-complex symmetric on $A_{\alpha }^2\left({\mathbb{B}}^N\right)$ with the conjugation ${\mathcal{J}}_{\tau ,\zeta }$ if and only if

$$\begin{array}{cc}\hfill & {\displaystyle \frac{1}{{(1+\langle z,\kappa ({\tau }_{\zeta }^{\ast }c+{\tau }_{\zeta }^{\ast }{A}^{\ast }c)\rangle -\langle w,A^2{\tau }_{\zeta }\kappa \left(z\right)+b+Ab\rangle +\langle c,b\rangle -\langle z,\kappa \left({\tau }_{\zeta }^{\ast }c\right)\rangle \langle w,b\rangle )}^{N+1+\alpha }}}\hfill \\ \hfill & +{\displaystyle \frac{1}{{(1-\langle z,b+Ab\rangle -\langle z,A^2\kappa \left({\tau }_{\zeta }^{\ast }w\right)\rangle +\langle c,b\rangle +\langle w,{\tau }_{\zeta }\kappa (A^{\ast }c+c)\rangle -\langle z,b\rangle \langle w,{\tau }_{\zeta }\kappa \left(c\right)\rangle )}^{N+1+\alpha }}}\hfill \\ \hfill & ={\displaystyle \frac{2}{{(1-\langle z,b-A\kappa \left({\tau }_{\zeta }^{\ast }c\right)\rangle -\langle A^{\ast }z,\kappa \left({\tau }_{\zeta }^{\ast }{A}^{\ast }w\right)\rangle -\langle c,\kappa \left({\tau }_{\zeta }^{\ast }c\right)\rangle +\langle c,\kappa \left({\tau }_{\zeta }^{\ast }{A}^{\ast }w\right)\rangle -\langle w,b\rangle +\langle w,b\rangle \langle z,b\rangle )}^{N+1+\alpha }}}\hfill \end{array}$$

(32)

for all w, $z\in {\mathbb{B}}^N$.

Assume that the operator $W_{u,\phi }$ is 2-complex symmetric on $A_{\alpha }^2\left({\mathbb{B}}^N\right)$ with the conjugation ${\mathcal{J}}_{\tau ,\zeta }$. Since $A=A^T$ and ${\tau }_{\zeta }$ is a diagonal matrix, we have $A=\kappa \left(A^{\ast }\right)$, ${\tau }_{\zeta }=\kappa \left({\tau }_{\zeta }^{\ast }\right)$ and $A{\tau }_{\zeta }={\tau }_{\zeta }A$. Letting $w=z$ in (32), we obtain

$$\begin{array}{cc}\hfill & {\displaystyle \frac{1}{{(1+\langle z,{\tau }_{\zeta }\kappa \left(c\right)+A{\tau }_{\zeta }\kappa \left(c\right)-b-Ab\rangle -\langle z,A^2{\tau }_{\zeta }\kappa \left(z\right)\rangle +\langle c,b\rangle -\langle z,{\tau }_{\zeta }\kappa \left(c\right)\rangle \langle z,b\rangle )}^{N+1+\alpha }}}\hfill \\ \hfill & ={\displaystyle \frac{1}{{(1+\langle z,2A{\tau }_{\zeta }\kappa \left(c\right)-2b\rangle -\langle z,A^2{\tau }_{\zeta }\kappa \left(z\right)\rangle -\langle c,{\tau }_{\zeta }\kappa \left(c\right)\rangle +\langle z,b\rangle \langle z,b\rangle )}^{N+1+\alpha }}}\hfill \end{array}$$

(33)

for all $z\in {\mathbb{B}}^N$, which implies

$$\begin{array}{cc}\hfill & (1+\langle z,{\tau }_{\zeta }\kappa \left(c\right)+A{\tau }_{\zeta }\kappa \left(c\right)-b-Ab\rangle -\langle z,A^2{\tau }_{\zeta }\kappa \left(z\right)\rangle +\langle c,b\rangle -\langle z,{\tau }_{\zeta }\kappa \left(c\right)\rangle \langle z,b\rangle )\varsigma (0,\dots ,z_l,\dots ,0)\hfill \\ \hfill & =1+\langle z,2A{\tau }_{\zeta }\kappa \left(c\right)-2b\rangle -\langle z,A^2{\tau }_{\zeta }\kappa \left(z\right)\rangle -\langle c,{\tau }_{\zeta }\kappa \left(c\right)\rangle +\langle z,b\rangle \langle z,b\rangle ,\hfill \end{array}$$

(34)

where $\varsigma {(0,\dots ,z_l,\dots ,0)}^{N+1+\alpha }=1$. Write $A={\left\{a_{ij}\right\}}_{N\times N}$ . Letting $z=(0,\dots ,z_l,\dots ,0)$ , $b=(0,\dots ,b_l,\dots ,0)$ and $c=(0,\dots ,c_l,\dots ,0)$ in (34), we obtain

$$\begin{array}{cc}\hfill & \left[1+z_l\left(c_l{e}^{-i{\zeta }_l}-\overline{b_l}+\overline{a_{ll}}{c}_l{e}^{-i{\zeta }_l}-\overline{a_{ll}{b}_l}\right)-z_l^2{e}^{-i{\zeta }_l}\sum _{j=1}^N\overline{a_{lj}{a}_{jl}}-z_l^2\overline{b_l}{c}_l{e}^{-i{\zeta }_l}+\langle c,b\rangle \right]\varsigma (0,\dots ,z_l,\dots ,0)\hfill \\ \hfill & =1+z_l\left(2\overline{a_{ll}}{c}_l{e}^{-i{\zeta }_l}-2\overline{b_l}\right)-z_l^2{e}^{-i{\zeta }_l}\sum _{j=1}^N\overline{a_{lj}{a}_{jl}}+z_l^2\overline{b_l^2}-\langle c,{\tau }_{\zeta }\kappa \left(c\right)\rangle ,\hfill \end{array}$$

(35)

From the definition of $\varsigma (0,\dots ,z_l,\dots ,0)$, we obtain that $\varsigma (0,\dots ,z_l,\dots ,0)$ has at most $N+1+\alpha$ different values. Since $z_l$ exists infinite values, there are infinite $z_l$ such as

$$\begin{array}{cc}\hfill & \left[1+z_l\left(c_l{e}^{-i{\zeta }_l}-\overline{b_l}+\overline{a_{ll}}{c}_l{e}^{-i{\zeta }_l}-\overline{a_{ll}{b}_l}\right)-z_l^2{e}^{-i{\zeta }_l}\sum _{j=1}^N\overline{a_{lj}{a}_{jl}}-z_l^2\overline{b_l}{c}_l{e}^{-i{\zeta }_l}+\langle c,b\rangle \right]{\varsigma }_k\hfill \\ \hfill & =1+z_l\left(2\overline{a_{ll}}{c}_l{e}^{-i{\zeta }_l}-2\overline{b_l}\right)-z_l^2{e}^{-i{\zeta }_l}\sum _{j=1}^N\overline{a_{lj}{a}_{jl}}+z_l^2\overline{b_l^2}-\langle c,{\tau }_{\zeta }\kappa \left(c\right)\rangle ,\hfill \end{array}$$

(36)

where ${\varsigma }_k$ is a fixed constant. From the coefficients of $z_l$, $z_l^2$ and the constant terms, we have

$$\left\{\begin{array}{c}(c_l{e}^{-i{\zeta }_l}-\overline{b_l}+\overline{a_{ll}}{c}_l{e}^{-i{\zeta }_l}-\overline{a_{ll}{b}_l}){\varsigma }_k=2\overline{a_{ll}}{c}_l{e}^{-i{\zeta }_l}-2\overline{b_l}\\ (-e^{-i{\zeta }_l}{\displaystyle \sum _{j=1}^N}\overline{a_{lj}{a}_{jl}}-\overline{b_l}{c}_l{e}^{-i{\zeta }_l}){\varsigma }_k=-e^{-i{\zeta }_l}{\displaystyle \sum _{j=1}^N}\overline{a_{lj}{a}_{jl}}+\overline{b_l^2}\\ (1+\langle c,b\rangle ){\varsigma }_k=1-\langle c,{\tau }_{\zeta }\kappa \left(c\right)\rangle .\end{array}\right.$$

(37)

Let $G_3=c_l{e}^{-i{\zeta }_l}+\overline{a_{ll}}{c}_l{e}^{-i{\zeta }_l}$, $G_4=-\overline{b_l}-\overline{a_{ll}{b}_l}$ and $H_1=1+\langle c,b\rangle$. Then, from (37), we obtain

$$\begin{array}{c}\hfill \overline{a_{ll}}{c}_l{e}^{-i{\zeta }_l}-\overline{b_l}={\displaystyle \frac{G_3+G_4}{2}}{\varsigma }_k\mathrm{and}1-\langle c,{\tau }_{\zeta }\kappa \left(c\right)\rangle =H_1{\varsigma }_k.\end{array}$$

(38)

Also, letting $w=0$ in (32), we obtain

$$\begin{array}{cc}\hfill & {\displaystyle \frac{1}{{(1+\langle z,\kappa ({\tau }_{\zeta }^{\ast }c+{\tau }_{\zeta }^{\ast }{A}^{\ast }c)\rangle +\langle c,b\rangle )}^{N+1+\alpha }}}+{\displaystyle \frac{1}{{(1-\langle z,b+Ab\rangle +\langle c,b\rangle )}^{N+1+\alpha }}}\hfill \\ \hfill & ={\displaystyle \frac{2}{{(1-\langle z,b-A\kappa \left({\tau }_{\zeta }^{\ast }c\right)\rangle -\langle c,{\tau }_{\zeta }\kappa \left(c\right)\rangle )}^{N+1+\alpha }}}\hfill \end{array}$$

(39)

for all $z\in {\mathbb{B}}^N$. Combining (38) with (39), from ${\varsigma }_k^{N+1+\alpha }=1$ we obtain

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{{(G_{3}z+H_1)}^{N+1+\alpha }}}+{\displaystyle \frac{1}{{(G_{4}z+H_1)}^{N+1+\alpha }}}-{\displaystyle \frac{2}{{(\frac{G_3+G_4}{2}z+H_1)}^{N+1+\alpha }}}=0\end{array}$$

(40)

for all $z\in {\mathbb{B}}^N$.

Taking two consecutive derivatives of the variable z on the both sides of (40), we obtain

$$\begin{array}{c}\hfill {\displaystyle \frac{G_3^2}{{(G_{3}z+H_1)}^{N+3+\alpha }}}+{\displaystyle \frac{G_4^2}{{(G_{4}z+H_1)}^{N+3+\alpha }}}-{\displaystyle \frac{{(G_3+G_4)}^2}{2{(\frac{G_3+G_4}{2}z+H_1)}^{N+3+\alpha }}}=0.\end{array}$$

(41)

for all $z\in {\mathbb{B}}^N$. Choosing $z=0$ in (41), we have $G_3-G_4=0$, which implies $(\overline{a_{ll}}+1)(\overline{b_l}+c_l{e}^{-i{\zeta }_l})=0$. Since $\parallel A\parallel <1$ implies that $(\overline{a_{ll}}+1)\ne 0$. Since $i=\overline{1,N}$, we have $b=-{\tau }_{\zeta }\kappa \left(c\right)$.

Conversely, if $b=-{\tau }_{\zeta }\kappa \left(c\right)$, we see that the right of (32) equals

$$\begin{array}{c}\hfill {\displaystyle \frac{2}{{(1+\langle z,{\tau }_{\zeta }\kappa \left(c\right)+A{\tau }_{\zeta }\kappa \left(c\right)\rangle -\langle z,A^2\kappa \left({\tau }_{\zeta }^{\ast }w\right)\rangle -\langle c,{\tau }_{\zeta }\kappa \left(c\right)\rangle +\langle w,{\tau }_{\zeta }\kappa (A^{\ast }c+c)\rangle +\langle z,{\tau }_{\zeta }\kappa \left(c\right)\rangle \langle w,{\tau }_{\zeta }\kappa \left(c\right)\rangle )}^{N+1+\alpha }}},\end{array}$$

and the right of (32) also does. From this, it follows that (32) holds. This shows that the operator $W_{u,\phi }$ is 2-complex symmetric on $A_{\alpha }^2\left({\mathbb{B}}^N\right)$ with the conjugation ${\mathcal{J}}_{\tau ,\zeta }$. □

We have the following Corollary 3 from Theorems 2 and 4.

Corollary 3.

Let φ and u be the symbols in Lemma 1 and Corollary 1, respectively. The following are equivalent:

(a) 

$W_{u,\phi }$ is 2-complex symmetric on $A_{\alpha }^2\left({\mathbb{B}}^N\right)$ with the conjugation ${\mathcal{J}}_{\tau ,\zeta }$;

(b) 

$W_{u,\phi }$ is complex symmetric on $A_{\alpha }^2\left({\mathbb{B}}^N\right)$ with the conjugation ${\mathcal{J}}_{\tau ,\zeta }$;

(c) 

$b=-{\tau }_{\zeta }\kappa \left(c\right)$.

By Theorem 4, we also can give some examples. From Theorem 4, first, we need to construct $A,b,c$ such that

$$\begin{array}{c}\hfill A=A^T,\parallel A\parallel +\left|b\right|+\left|c\right|<1,b=-{\tau }_{\zeta }\kappa \left(c\right).\end{array}$$

At the same time, in order to guarantee ${\mathcal{J}}_{\tau ,\zeta }$ is a conjugation on $A_{\alpha }^2\left({\mathbb{B}}^N\right)$, it also must satisfy

$$\begin{array}{c}\hfill \left|\tau \right|=1,\zeta =({\zeta }_1,\dots ,{\zeta }_n)\in {\mathbb{R}}^n,{\tau }_{\zeta }=diag\{e^{i{\zeta }_1},\dots ,e^{i{\zeta }_n}\}.\end{array}$$

We give some additional examples for the case $N=2$.

Example 2.

(a) Let $u\left(z\right)={\displaystyle \frac{1}{{(1+\langle z,c\rangle )}^{3+\alpha }}}$ and $\phi \left(z\right)={\displaystyle \frac{Az+b}{\langle z,c\rangle +1}}$, where

$$\begin{array}{c}\hfill A=\left(\begin{array}{cc}{\displaystyle \frac{-i}{15}}& {\displaystyle \frac{\sqrt{2}i}{20}}\\ {\displaystyle \frac{\sqrt{2}i}{20}}& {\displaystyle \frac{1}{15}}\end{array}\right),c={\left(\begin{array}{cc}{\displaystyle \frac{i}{8}},& {\displaystyle \frac{i}{15}}\end{array}\right)}^T,{\tau }_{\zeta }=\left(\begin{array}{cc}{e}^{{\displaystyle \frac{i}{3}}}& 0\\ 0& e^{{\displaystyle \frac{i}{4}}}\end{array}\right)\end{array}$$

and

$$\begin{array}{c}\hfill b=-{\tau }_{\zeta }\kappa \left(c\right)={\left(\begin{array}{cc}{\displaystyle \frac{i}{8}}{e}^{{\displaystyle \frac{i}{3}}},& {\displaystyle \frac{i}{15}}{e}^{{\displaystyle \frac{i}{4}}}\end{array}\right)}^T.\end{array}$$

Then, the operator $W_{u,\phi }$ is 2-complex symmetric on $A_{\alpha }^2\left({\mathbb{B}}^2\right)$ with the conjugation ${\mathcal{J}}_{\tau ,\zeta }$.

(b) Let $u\left(z\right)={\displaystyle \frac{1}{{(1+\langle z,c\rangle )}^{3+\alpha }}}$ and $\phi \left(z\right)={\displaystyle \frac{Az+b}{\langle z,c\rangle +1}}$, where

$$\begin{array}{c}\hfill A=\left(\begin{array}{cc}{\displaystyle \frac{1}{13}}+{\displaystyle \frac{1}{13}}i& {\displaystyle \frac{\sqrt{2}}{13}}\\ {\displaystyle \frac{\sqrt{2}}{13}}& 0\end{array}\right),c={\left(\begin{array}{cc}{\displaystyle \frac{i}{12}},& 0\end{array}\right)}^T,{\tau }_{\zeta }=\left(\begin{array}{cc}{e}^{{\displaystyle \frac{i}{3}}}& 0\\ 0& e^{{\displaystyle \frac{i}{4}}}\end{array}\right)\end{array}$$

and

$$\begin{array}{c}\hfill b=-{\tau }_{\zeta }\kappa \left(c\right)={\left(\begin{array}{cc}{\displaystyle \frac{i}{12}}{e}^{{\displaystyle \frac{i}{3}}},& 0\end{array}\right)}^T.\end{array}$$

Then, the operator $W_{u,\phi }$ is 2-complex symmetric on $A_{\alpha }^2\left({\mathbb{B}}^2\right)$ with the conjugation ${\mathcal{J}}_{\tau ,\zeta }$.

Proof. 

(a) Calculations show that $A=A^T$ and

$$\begin{array}{c}\hfill \parallel A\parallel ={\displaystyle \frac{\sqrt{6}}{25}},\left|b\right|={\displaystyle \frac{\sqrt{46}}{23}},\left|c\right|={\displaystyle \frac{\sqrt{2}}{17}}.\end{array}$$

Then, $\parallel A\parallel +\left|b\right|+\left|c\right|<1$. By Theorem 4, the operator $W_{u,\phi }$ is 2-complex symmetric on $A_{\alpha }^2\left({\mathbb{B}}^2\right)$ with the conjugation ${\mathcal{J}}_{\tau ,\zeta }$.

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

4. Complex Symmetric Difference of Weighted Composition Operators

In this section, we study the complex symmetry of the operator $W_{u_1,{\phi }_1}-W_{u_2,{\phi }_2}$ on $A_{\alpha }^2\left({\mathbb{B}}^N\right)$ with the conjugations $\mathcal{J}$ and ${\mathcal{J}}_{\tau ,\zeta }$, respectively.

To avoid triviality, we assume that $W_{u_1,{\phi }_1}-W_{u_2,{\phi }_2}$ is not the zero operator.

Theorem 5.

Let $u_j\left(z\right)={\displaystyle \frac{1}{{(1+\langle z,c_j\rangle )}^{N+1+\alpha }}}$, ${\phi }_j\left(z\right)={\displaystyle \frac{A_{j}z+b_j}{\langle z,c_j\rangle +1}}$, $A_j=A_j^T$, and $A_j,b_j,c_j$ satisfy $\parallel A_j\parallel +\left|b_j\right|+\left|c_j\right|<1$, where $j=1,2$. Then, the operator $W_{u_1,{\phi }_1}-W_{u_2,{\phi }_2}$ is complex symmetric on $A_{\alpha }^2\left({\mathbb{B}}^N\right)$ with the conjugation $\mathcal{J}$ if and only if $b_1=-\kappa \left(c_1\right)$ and $b_2=-\kappa \left(c_2\right)$.

Proof. 

For each $w,z\in {\mathbb{B}}^N$, it follows from Lemma 5 that

$$\begin{array}{cc}\hfill & \mathcal{J}(W_{u_1,{\phi }_1}-W_{u_2,{\phi }_2})K_w\left(z\right)=\mathcal{J}{W}_{u_1,{\phi }_1}{K}_w\left(z\right)-\mathcal{J}{W}_{u_2,{\phi }_2}{K}_w\left(z\right)\hfill \\ \hfill & =\mathcal{J}\left({\displaystyle \frac{1}{{(1+\langle z,c_1\rangle )}^{N+1+\alpha }}}{\displaystyle \frac{1}{{(1-\langle \frac{A_{1}z+b_1}{\langle z,c_1\rangle +1},w\rangle )}^{N+1+\alpha }}}\right)-\mathcal{J}\left({\displaystyle \frac{1}{{(1+\langle z,c_2\rangle )}^{N+1+\alpha }}}{\displaystyle \frac{1}{{(1-\langle \frac{A_{2}z+b_2}{\langle z,c_2\rangle +1},w\rangle )}^{N+1+\alpha }}}\right)\hfill \\ \hfill & =\mathcal{J}\left({\displaystyle \frac{1}{{(1+\langle z,c_1\rangle -\langle A_{1}z+b_1,w\rangle )}^{N+1+\alpha }}}\right)-\mathcal{J}\left({\displaystyle \frac{1}{{(1+\langle z,c_2\rangle -\langle A_{2}z+b_2,w\rangle )}^{N+1+\alpha }}}\right)\hfill \\ \hfill & =\mathcal{J}\left({\displaystyle \frac{1}{{(1+\langle z,c_1\rangle -\langle z,A_1^{\ast }w\rangle -\langle b_1,w\rangle )}^{N+1+\alpha }}}\right)-\mathcal{J}\left({\displaystyle \frac{1}{{(1+\langle z,c_2\rangle -\langle z,A_2^{\ast }w\rangle -\langle b_2,w\rangle )}^{N+1+\alpha }}}\right)\hfill \\ \hfill & ={\displaystyle \frac{1}{{(1+\langle z,\kappa \left(c_1\right)-\kappa \left(A_1^{\ast }w\right)\rangle -\langle w,b_1\rangle )}^{N+1+\alpha }}}-{\displaystyle \frac{1}{{(1+\langle z,\kappa \left(c_2\right)-\kappa \left(A_2^{\ast }w\right)\rangle -\langle w,b_2\rangle )}^{N+1+\alpha }}}\hfill \end{array}$$

(42)

and

$$\begin{array}{cc}\hfill & (W_{u_1,{\phi }_1}^{\ast }-W_{u_2,{\phi }_2}^{\ast })\mathcal{J}{K}_w\left(z\right)=(W_{u_1,{\phi }_1}^{\ast }-W_{u_2,{\phi }_2}^{\ast })\left({\displaystyle \frac{1}{{(1-\langle z,\kappa \left(w\right)\rangle )}^{N+1+\alpha }}}\right)\hfill \\ \hfill & =W_{u_1,{\phi }_1}^{\ast }\left({\displaystyle \frac{1}{{(1-\langle z,\kappa \left(w\right)\rangle )}^{N+1+\alpha }}}\right)-W_{u_2,{\phi }_2}^{\ast }\left({\displaystyle \frac{1}{{(1-\langle z,\kappa \left(w\right)\rangle )}^{N+1+\alpha }}}\right)\hfill \\ \hfill & =W_{{\displaystyle \frac{1}{{(1-\langle z,b_1\rangle )}^{N+1+\alpha }}},{\displaystyle \frac{A_1^{\ast }z-c_1}{1-\langle z,b_1\rangle }}}\left({\displaystyle \frac{1}{{(1-\langle z,\kappa \left(w\right)\rangle )}^{N+1+\alpha }}}\right)-W_{{\displaystyle \frac{1}{{(1-\langle z,b_2\rangle )}^{N+1+\alpha }}},{\displaystyle \frac{A_2^{\ast }z-c_2}{1-\langle z,b_2\rangle }}}\left({\displaystyle \frac{1}{{(1-\langle z,\kappa \left(w\right)\rangle )}^{N+1+\alpha }}}\right)\hfill \\ \hfill & ={\displaystyle \frac{1}{{(1-\langle z,b_1\rangle -\langle A_1^{\ast }z-c_1,\kappa \left(w\right)\rangle )}^{N+1+\alpha }}}-{\displaystyle \frac{1}{{(1-\langle z,b_2\rangle -\langle A_2^{\ast }z-c_2,\kappa \left(w\right)\rangle )}^{N+1+\alpha }}}\hfill \\ \hfill & ={\displaystyle \frac{1}{{(1-\langle z,b_1+A_1\kappa \left(w\right)\rangle +\langle c_1,\kappa \left(w\right)\rangle )}^{N+1+\alpha }}}-{\displaystyle \frac{1}{{(1-\langle z,b_2+A_2\kappa \left(w\right)\rangle +\langle c_2,\kappa \left(w\right)\rangle )}^{N+1+\alpha }}}.\hfill \end{array}$$

(43)

Then, by Lemma 3, the operator $W_{u_1,{\phi }_1}-W_{u_2,{\phi }_2}$ is complex symmetric on $A_{\alpha }^2\left({\mathbb{B}}^N\right)$ with the conjugation $\mathcal{J}$ if and only if

$$\begin{array}{cc}\hfill & {\displaystyle \frac{1}{{(1+\langle z,\kappa \left(c_1\right)-\kappa \left(A_1^{\ast }w\right)\rangle -\langle w,b_1\rangle )}^{N+1+\alpha }}}-{\displaystyle \frac{1}{{(1+\langle z,\kappa \left(c_2\right)-\kappa \left(A_2^{\ast }w\right)\rangle -\langle w,b_2\rangle )}^{N+1+\alpha }}}\hfill \\ \hfill & ={\displaystyle \frac{1}{{(1-\langle z,b_1+A_1\kappa \left(w\right)\rangle +\langle c_1,\kappa \left(w\right)\rangle )}^{N+1+\alpha }}}-{\displaystyle \frac{1}{{(1-\langle z,b_2+A_2\kappa \left(w\right)\rangle +\langle c_2,\kappa \left(w\right)\rangle )}^{N+1+\alpha }}}\hfill \end{array}$$

(44)

for all $w,z\in {\mathbb{B}}^N$.

Assume that the operator $W_{u_1,{\phi }_1}-W_{u_2,{\phi }_2}$ is complex symmetric on $A_{\alpha }^2\left({\mathbb{B}}^N\right)$ with the conjugation $\mathcal{J}$. Letting $w=0$ in (44), we see that (44) becomes

$$\begin{array}{cc}\hfill & {\displaystyle \frac{1}{{(1+\langle z,\kappa \left(c_1\right)\rangle )}^{N+1+\alpha }}}-{\displaystyle \frac{1}{{(1+\langle z,\kappa \left(c_2\right)\rangle )}^{N+1+\alpha }}}={\displaystyle \frac{1}{{(1-\langle z,b_1\rangle )}^{N+1+\alpha }}}-{\displaystyle \frac{1}{{(1-\langle z,b_2\rangle )}^{N+1+\alpha }}}\hfill \end{array}$$

(45)

for all $z\in {\mathbb{B}}^N$. From a direct calculation, (45) is equivalent to

$$\begin{array}{cc}\hfill & {\displaystyle \frac{\langle z,b_1+\kappa \left(c_1\right)\rangle {\displaystyle \sum _{j=0}^{N+\alpha }}{(1-\langle z,b_1\rangle )}^{N+\alpha -j}{(1+\langle z,\kappa \left(c_1\right)\rangle )}^j}{{(1+\langle z,\kappa \left(c_1\right)\rangle )}^{N+1+\alpha }{(1-\langle z,b_1\rangle )}^{N+1+\alpha }}}\hfill \\ \hfill & ={\displaystyle \frac{\langle z,b_2+\kappa \left(c_2\right)\rangle {\displaystyle \sum _{j=0}^{N+\alpha }}{(1-\langle z,b_2\rangle )}^{N+\alpha -j}{(1+\langle z,\kappa \left(c_2\right)\rangle )}^j}{{(1+\langle z,\kappa \left(c_2\right)\rangle )}^{N+1+\alpha }{(1-\langle z,b_2\rangle )}^{N+1+\alpha }}}.\hfill \end{array}$$

(46)

Letting $z=(0,\dots ,z_i,\dots ,0)$ in (46), we obtain

$$\begin{array}{c}\hfill {\displaystyle \frac{z_i({\overline{b}}_{1,i}+c_{1,i}){\displaystyle \sum _{j=0}^{N+\alpha }}{(1-z_i{\overline{b}}_{1,i})}^{N+\alpha -j}{(1+z_i{c}_{1,i})}^j}{{(1+z_i{c}_{1,i})}^{N+1+\alpha }{(1-z_i{\overline{b}}_{1,i})}^{N+1+\alpha }}}={\displaystyle \frac{z_i({\overline{b}}_{2,i}+c_{2,i}){\displaystyle \sum _{j=0}^{N+\alpha }}(1-z_i{\overline{b}}_{2,i}\rangle {)}^{N+\alpha -j}{(1+z_i{c}_{2,i})}^j}{{(1+z_i{c}_{2,i})}^{N+1+\alpha }{(1-z_i{\overline{b}}_{2,i})}^{N+1+\alpha }}},\end{array}$$

(47)

where $b_{1,i},b_{2,i},c_{1,i},c_{2,i}$ are the $ith$ components of $b_1,b_2,c_1$ and $c_2$, respectively.

Comparing the coefficients of $z_i$, we obtain ${\overline{b}}_{1,i}+c_{1,i}={\overline{b}}_{2,i}+c_{2,i}$, which implies $b_1+\kappa \left(c_1\right)=b_2+\kappa \left(c_2\right)$. This allows us to divide into the following two cases for consideration.

Case 1. Assume that $b_1+\kappa \left(c_1\right)=b_2+\kappa \left(c_2\right)=0$. From a calculation, we obtain that (44) holds for this case.

Case 2. Assume that $b_1+\kappa \left(c_1\right)=b_2+\kappa \left(c_2\right)\ne 0$. Then, (47) is reduced to

$$\begin{array}{c}\hfill {\displaystyle \frac{{\displaystyle \sum _{j=0}^{N+\alpha }}{(1-z_i{\overline{b}}_{1,i})}^{N+\alpha -j}{(1+z_i{c}_{1,i})}^j}{{(1+z_i{c}_{1,i})}^{N+1+\alpha }{(1-z_i{\overline{b}}_{1,i})}^{N+1+\alpha }}}={\displaystyle \frac{{\displaystyle \sum _{j=0}^{N+\alpha }}(1-z_i{\overline{b}}_{2,i}\rangle {)}^{N+\alpha -j}{(1+z_i{c}_{2,i})}^j}{{(1+z_i{c}_{2,i})}^{N+1+\alpha }{(1-z_i,{\overline{b}}_{2,i})}^{N+1+\alpha }}}.\end{array}$$

Comparing the coefficients of $z_i$, we obtain

$$\begin{array}{cc}\hfill & \sum _{j=0}^{N+\alpha }\left[(N+\alpha -j)(-b_{1,i})+j{\overline{c}}_{1,i}\right]+{(N+1+\alpha )}^2{\overline{c}}_{2,i}-{(N+1+\alpha )}^2{b}_{2,i}\hfill \\ \hfill & =\sum _{j=0}^{N+\alpha }\left[(N+\alpha -j)(-b_{2,i})+j{\overline{c}}_{2,i}\right]+{(N+1+\alpha )}^2{\overline{c}}_{1,i}-{(N+1+\alpha )}^2{b}_{1,i},\hfill \end{array}$$

which implies ${\overline{c}}_{1,i}-b_{1,i}={\overline{c}}_{2,i}-b_{2,i}$; that is, $\kappa \left(c_1\right)-b_1=\kappa \left(c_2\right)-b_2$. Since $b_1+\kappa \left(c_1\right)=b_2+\kappa \left(c_2\right)\ne 0$, we obtain $\kappa \left(c_1\right)=\kappa \left(c_2\right)$ and $b_1=b_2$. Bringing these two equalities into (44), we have

$$\begin{array}{cc}\hfill & {\displaystyle \frac{\langle z,(A_1-A_2)\kappa \left(w\right)\rangle {\displaystyle \sum _{j=0}^{N+\alpha }}{(1+\langle z,\kappa \left(c_1\right)-A_2\kappa \left(w\right)\rangle -\langle w,b_1\rangle )}^{N+\alpha -j}{(1+\langle z,\kappa \left(c_1\right)-A_1\kappa \left(w\right)\rangle -\langle w,b_1\rangle )}^j}{{(1+\langle z,\kappa \left(c_1\right)-A_1\kappa \left(w\right)\rangle -\langle w,b_1\rangle )}^{N+1+\alpha }{(1+\langle z,\kappa \left(c_1\right)-A_2\kappa \left(w\right)\rangle -\langle w,b_1\rangle )}^{N+1+\alpha }}}\hfill \\ \hfill & ={\displaystyle \frac{\langle z,(A_1-A_2)\kappa \left(w\right)\rangle {\displaystyle \sum _{j=0}^{N+\alpha }}{(1-\langle z,b_1+A_2\kappa \left(w\right)\rangle +\langle w,\kappa \left(c_1\right)\rangle )}^{N+\alpha -j}{(1-\langle z,b_1+A_1\kappa \left(w\right)\rangle +\langle w,\kappa \left(c_1\right)\rangle )}^j}{{(1-\langle z,b_1+A_1\kappa \left(w\right)\rangle +\langle w,\kappa \left(c_1\right)\rangle )}^{N+1+\alpha }{(1-\langle z,b_1+A_2\kappa \left(w\right)\rangle +\langle w,\kappa \left(c_1\right)\rangle )}^{N+1+\alpha }}}\hfill \end{array}$$

(48)

for all w, $z\in {\mathbb{B}}^N$. Since $W_{u_1,{\phi }_1}-W_{u_2,{\phi }_2}$ is nonzero, we have $A_1\ne A_2$. By a calculation, (48) is reduced to

$$\begin{array}{cc}\hfill & {\displaystyle \frac{{\displaystyle \sum _{j=0}^{N+\alpha }}{(1+\langle z,\kappa \left(c_1\right)-A_2\kappa \left(w\right)\rangle -\langle w,b_1\rangle )}^{N+\alpha -j}{(1+\langle z,\kappa \left(c_1\right)-A_1\kappa \left(w\right)\rangle -\langle w,b_1\rangle )}^j}{{(1+\langle z,\kappa \left(c_1\right)-A_1\kappa \left(w\right)\rangle -\langle w,b_1\rangle )}^{N+1+\alpha }{(1+\langle z,\kappa \left(c_1\right)-A_2\kappa \left(w\right)\rangle -\langle w,b_1\rangle )}^{N+1+\alpha }}}\hfill \\ \hfill & ={\displaystyle \frac{{\displaystyle \sum _{j=0}^{N+\alpha }}{(1-\langle z,b_1+A_2\kappa \left(w\right)\rangle +\langle w,\kappa \left(c_1\right)\rangle )}^{N+\alpha -j}{(1-\langle z,b_1+A_1\kappa \left(w\right)\rangle +\langle w,\kappa \left(c_1\right)\rangle )}^j}{{(1-\langle z,b_1+A_1\kappa \left(w\right)\rangle +\langle w,\kappa \left(c_1\right)\rangle )}^{N+1+\alpha }{(1-\langle z,b_1+A_2\kappa \left(w\right)\rangle +\langle w,\kappa \left(c_1\right)\rangle )}^{N+1+\alpha }}}\hfill \end{array}$$

(49)

for all w, $z\in {\mathbb{B}}^N$.

Write $A_1={{\{}_1{a}_{ij}\}}_{N\times N}$ and $A_2={{\{}_2{a}_{ij}\}}_{N\times N}$. Letting $z=(0,\dots ,z_i,\dots ,0)$ and $w=(0,\dots ,w_i,···,0)$ in (49), we obtain

$$\begin{array}{cc}\hfill & {\displaystyle \frac{{\displaystyle \sum _{j=0}^{N+\alpha }}{[1+z_i(c_{1,i}-{\overline{2a}}_{ii}{w}_i)-w_i{\overline{b}}_{1,i}]}^{N+\alpha -j}{[1+z_i(c_{1,i}-{\overline{1a}}_{ii}{w}_i)-w_i{\overline{b}}_{1,i}]}^j}{{[1+z_i(c_{1,i}-{\overline{1a}}_{ii}{w}_i)-w_i{\overline{b}}_{1,i}]}^{N+1+\alpha }{[1+z_i(c_{1,i}-{\overline{2a}}_{ii}{w}_i)-w_i{\overline{b}}_{1,i}]}^{N+1+\alpha }}}\hfill \\ \hfill & ={\displaystyle \frac{{\displaystyle \sum _{j=0}^{N+\alpha }}{[1-z_i({\overline{b}}_{1,i}+{\overline{2a}}_{ii}{w}_i)+w_i{c}_{1,i}]}^{N+\alpha -j}{[1-z_i({\overline{b}}_{1,i}+{\overline{1a}}_{ii}{w}_i)+w_i{c}_{1,i}]}^j}{{[1-z_i({\overline{b}}_{1,i}+{\overline{1a}}_{ii}{w}_i)+w_i{c}_{1,i}]}^{N+1+\alpha }{[1-z_i({\overline{b}}_{1,i}+{\overline{2a}}_{ii}{w}_i)+w_i{c}_{1,i}]}^{N+1+\alpha }}},\hfill \end{array}$$

where $b_{1,i},c_{1,i}$ are the $ith$ components of $b_1,c_1$. Comparing the coefficients of $w_i$, we have

$$\begin{array}{c}\hfill \sum _{j=0}^{N+\alpha }(N+\alpha )(-{\overline{b}}_{1,i})+2{(N+1+\alpha )}^2{c}_{1,i}=\sum _{j=0}^{N+\alpha }(N+\alpha )c_{1,i}+2{(N+1+\alpha )}^2(-{\overline{b}}_{1,i}),\end{array}$$

which shows $b_{1,i}+{\overline{c}}_{1,i}=0$; that is, $b_1+\kappa \left(c_1\right)=0$. This contradicts Case 2. Therefore, we deduce that $b_1=-\kappa \left(c_1\right)$ and $b_2=-\kappa \left(c_2\right)$.

Conversely, if $b_1=-\kappa \left(c_1\right)$ and $b_2=-\kappa \left(c_2\right)$, we see that the left of (44) equals

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{{(1+\langle z,\kappa \left(c_1\right)-\kappa \left(A_1^{\ast }w\right)\rangle +\langle w,\kappa \left(c_1\right)\rangle )}^{N+1+\alpha }}}-{\displaystyle \frac{1}{{(1+\langle z,\kappa \left(c_2\right)-\kappa \left(A_2^{\ast }w\right)\rangle +\langle w,\kappa \left(c_2\right)\rangle )}^{N+1+\alpha }}},\end{array}$$

and the right of (44) also does. From this, it follows that (44) holds. This shows that $W_{u_1,{\phi }_1}-W_{u_2,{\phi }_2}$ is complex symmetric on $A_{\alpha }^2\left({\mathbb{B}}^N\right)$ with the conjugation $\mathcal{J}$. □

From Theorems 1 and 5, we obtain the following interesting result.

Corollary 4.

Let $u_j\left(z\right)={\displaystyle \frac{1}{{(1+\langle z,c_j\rangle )}^{N+1+\alpha }}}$ and ${\phi }_j\left(z\right)={\displaystyle \frac{A_{j}z+b_j}{\langle z,c_j\rangle +1}}$, $A_j=A_j^T$ and $A_j,b_j,c_j$ satisfy $\parallel A_j\parallel +\left|b_j\right|+\left|c_j\right|<1$, where $j=1,2$. Then, the operator $W_{u_1,{\phi }_1}-W_{u_2,{\phi }_2}$ is complex symmetric on $A_{\alpha }^2\left({\mathbb{B}}^N\right)$ with the conjugation $\mathcal{J}$ if and only if $W_{u_1,{\phi }_1}$ and $W_{u_2,{\phi }_2}$ are complex symmetric on $A_{\alpha }^2\left({\mathbb{B}}^N\right)$ with the conjugation $\mathcal{J}$.

By Corollary 2 and Corollary 4, we give the following example for $N=2$.

Example 3.

Let $u_j\left(z\right)={\displaystyle \frac{1}{{(1+\langle z,c_j\rangle )}^{3+\alpha }}}$, ${\phi }_j\left(z\right)={\displaystyle \frac{A_{j}z+b_j}{\langle z,c_j\rangle +1}}$, $j=1,2$, where

$$\begin{array}{c}\hfill A_1=\left(\begin{array}{cc}{\displaystyle \frac{1}{13}}+i{\displaystyle \frac{1}{13}}& {\displaystyle \frac{\sqrt{2}}{13}}\\ {\displaystyle \frac{\sqrt{2}}{13}}& 0\end{array}\right),b_1={\left(\begin{array}{cc}{\displaystyle \frac{1}{9}},& 0\end{array}\right)}^T,c_1={\left(\begin{array}{cc}{\displaystyle \frac{-1}{9}},& 0\end{array}\right)}^T,\end{array}$$

$$\begin{array}{c}\hfill A_2=\left(\begin{array}{cc}{\displaystyle \frac{-i}{15}}& {\displaystyle \frac{\sqrt{2}i}{20}}\\ {\displaystyle \frac{\sqrt{2}i}{20}}& {\displaystyle \frac{1}{15}}\end{array}\right),b_2={\left(\begin{array}{cc}{\displaystyle \frac{i}{25}},& 0\end{array}\right)}^T,c_2={\left(\begin{array}{cc}{\displaystyle \frac{-i}{25}},& 0\end{array}\right)}^T,\end{array}$$

Then, the operator $W_{u_1,{\phi }_1}-W_{u_2,{\phi }_2}$ is complex symmetric on $A_{\alpha }^2\left({\mathbb{B}}^2\right)$ with the conjugation $\mathcal{J}$.

Proof. 

We have proved that $W_{u_1,{\phi }_1}$ is complex symmetric with the conjugation $\mathcal{J}$ on $A_{\alpha }^2\left({\mathbb{B}}^2\right)$ in Example 1. Now, we prove that $W_{u_2,{\phi }_2}$ is also complex symmetric with the conjugation $\mathcal{J}$ on $A_{\alpha }^2\left({\mathbb{B}}^2\right)$. This can be realized by a calculation following Theorem 1, so the details are omitted. □

Theorem 6.

Let $u_j\left(z\right)={\displaystyle \frac{1}{{(1+\langle z,c_j\rangle )}^{N+1+\alpha }}}$ and ${\phi }_j\left(z\right)={\displaystyle \frac{A_{j}z+b_j}{\langle z,c_j\rangle +1}}$, $A_j=A_j^T$ and $A_j,b_j,c_j$ satisfy $\parallel A_j\parallel +\left|b_j\right|+\left|c_j\right|<1$, where $j=1,2$. Then, the operator $W_{u_1,{\phi }_1}-W_{u_2,{\phi }_2}$ is complex symmetric on $A_{\alpha }^2\left({\mathbb{B}}^N\right)$ with the conjugation ${\mathcal{J}}_{\tau ,\zeta }$ if and only if $b_1=-{\tau }_{\zeta }\kappa \left(c_1\right)$ and $b_2=-{\tau }_{\zeta }\kappa \left(c_2\right)$.

Proof. 

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

$$\begin{array}{cc}\hfill & {\mathcal{J}}_{\tau ,\zeta }(W_{u_1,{\phi }_1}-W_{u_2,{\phi }_2})K_w\left(z\right)={\mathcal{J}}_{\tau ,\zeta }{W}_{u_1,{\phi }_1}{K}_w\left(z\right)-{\mathcal{J}}_{\tau ,\zeta }{W}_{u_2,{\phi }_2}{K}_w\left(z\right)\hfill \\ \hfill & ={\mathcal{J}}_{\tau ,\zeta }\left({\displaystyle \frac{1}{{(1+\langle z,c_1\rangle )}^{N+1+\alpha }}}{\displaystyle \frac{1}{{(1-\langle \frac{A_{1}z+b_1}{\langle z,c_1\rangle +1},w\rangle )}^{N+1+\alpha }}}\right)-{\mathcal{J}}_{\tau ,\zeta }\left({\displaystyle \frac{1}{{(1+\langle z,c_2\rangle )}^{N+1+\alpha }}}{\displaystyle \frac{1}{{(1-\langle \frac{A_{2}z+b_2}{\langle z,c_2\rangle +1},w\rangle )}^{N+1+\alpha }}}\right)\hfill \\ \hfill & ={\mathcal{J}}_{\tau ,\zeta }\left({\displaystyle \frac{1}{{(1+\langle z,c_1\rangle -\langle A_{1}z+b_1,w\rangle )}^{N+1+\alpha }}}\right)-{\mathcal{J}}_{\tau ,\zeta }\left({\displaystyle \frac{1}{{(1+\langle z,c_2\rangle -\langle A_{2}z+b_2,w\rangle )}^{N+1+\alpha }}}\right)\hfill \\ \hfill & ={\mathcal{J}}_{\tau ,\zeta }\left({\displaystyle \frac{1}{{(1+\langle z,c_1\rangle -\langle z,A_1^{\ast }w\rangle -\langle b_1,w\rangle )}^{N+1+\alpha }}}\right)-{\mathcal{J}}_{\tau ,\zeta }\left({\displaystyle \frac{1}{{(1+\langle z,c_2\rangle -\langle z,A_2^{\ast }w\rangle -\langle b_2,w\rangle )}^{N+1+\alpha }}}\right)\hfill \\ \hfill & ={\displaystyle \frac{\tau }{{(1+\langle z,\kappa \left({\tau }_{\zeta }^{\ast }{c}_1\right)-\kappa \left({\tau }_{\zeta }^{\ast }{A}_1^{\ast }w\right)\rangle -\langle w,b_1\rangle )}^{N+1+\alpha }}}-{\displaystyle \frac{\tau }{{(1+\langle z,\kappa \left({\tau }_{\zeta }^{\ast }{c}_2\right)-\kappa \left({\tau }_{\zeta }^{\ast }{A}_2^{\ast }w\right)\rangle -\langle w,b_2\rangle )}^{N+1+\alpha }}}\hfill \end{array}$$

(50)

and

$$\begin{array}{cc}\hfill & (W_{u_1,{\phi }_1}^{\ast }-W_{u_2,{\phi }_2}^{\ast }){\mathcal{J}}_{\tau ,\zeta }{K}_w\left(z\right)=(W_{u_1,{\phi }_1}^{\ast }-W_{u_2,{\phi }_2}^{\ast })\left({\displaystyle \frac{\tau }{{(1-\langle z,\kappa \left({\tau }_{\zeta }^{\ast }w\right)\rangle )}^{N+1+\alpha }}}\right)\hfill \\ \hfill & =W_{u_1,{\phi }_1}^{\ast }\left({\displaystyle \frac{\tau }{{(1-\langle z,\kappa \left({\tau }_{\zeta }^{\ast }w\right)\rangle )}^{N+1+\alpha }}}\right)-W_{u_2,{\phi }_2}^{\ast }\left({\displaystyle \frac{\tau }{{(1-\langle z,\kappa \left({\tau }_{\zeta }^{\ast }w\right)\rangle )}^{N+1+\alpha }}}\right)\hfill \\ \hfill & =W_{{\displaystyle \frac{1}{{(1-\langle z,b_1\rangle )}^{N+1+\alpha }}},{\displaystyle \frac{A_1^{\ast }z-c_1}{1-\langle z,b_1\rangle }}}\left({\displaystyle \frac{\tau }{{(1-\langle z,\kappa \left({\tau }_{\zeta }^{\ast }w\right)\rangle )}^{N+1+\alpha }}}\right)-W_{{\displaystyle \frac{1}{{(1-\langle z,b_2\rangle )}^{N+1+\alpha }}},{\displaystyle \frac{A_2^{\ast }z-c_2}{1-\langle z,b_2\rangle }}}\left({\displaystyle \frac{\tau }{{(1-\langle z,\kappa \left({\tau }_{\zeta }^{\ast }w\right)\rangle )}^{N+1+\alpha }}}\right)\hfill \\ \hfill & ={\displaystyle \frac{\tau }{{(1-\langle z,b_1\rangle -\langle A_1^{\ast }z-c_1,\kappa \left({\tau }_{\zeta }^{\ast }w\right)\rangle )}^{N+1+\alpha }}}-{\displaystyle \frac{\tau }{{(1-\langle z,b_2\rangle -\langle A_2^{\ast }z-c_2,\kappa \left({\tau }_{\zeta }^{\ast }w\right)\rangle )}^{N+1+\alpha }}}\hfill \\ \hfill & ={\displaystyle \frac{\tau }{{(1-\langle z,b_1+A_1\kappa \left({\tau }_{\zeta }^{\ast }w\right)\rangle +\langle c_1,\kappa \left({\tau }_{\zeta }^{\ast }w\right)\rangle )}^{N+1+\alpha }}}-{\displaystyle \frac{\tau }{{(1-\langle z,b_2+A_2\kappa \left({\tau }_{\zeta }^{\ast }w\right)\rangle +\langle c_2,\kappa \left({\tau }_{\zeta }^{\ast }w\right)\rangle )}^{N+1+\alpha }}}.\hfill \end{array}$$

(51)

Then, by Lemma 3, the operator $W_{u_1,{\phi }_1}-W_{u_2,{\phi }_2}$ is complex symmetric on $A_{\alpha }^2\left({\mathbb{B}}^N\right)$ with the conjugation ${\mathcal{J}}_{\tau ,\zeta }$ if and only if

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{{(1+\langle z,\kappa \left({\tau }_{\zeta }^{\ast }{c}_1\right)-\kappa \left({\tau }_{\zeta }^{\ast }{A}_1^{\ast }w\right)\rangle -\langle w,b_1\rangle )}^{N+1+\alpha }}}-{\displaystyle \frac{1}{{(1+\langle z,\kappa \left({\tau }_{\zeta }^{\ast }{c}_2\right)-\kappa \left({\tau }_{\zeta }^{\ast }{A}_2^{\ast }w\right)\rangle -\langle w,b_2\rangle )}^{N+1+\alpha }}}\end{array}$$

(52)

$$\begin{array}{c}\hfill ={\displaystyle \frac{1}{{(1-\langle z,b_1+A_1\kappa \left({\tau }_{\zeta }^{\ast }w\right)\rangle +\langle c_1,\kappa \left({\tau }_{\zeta }^{\ast }w\right)\rangle )}^{N+1+\alpha }}}-{\displaystyle \frac{1}{{(1-\langle z,b_2+A_2\kappa \left({\tau }_{\zeta }^{\ast }w\right)\rangle +\langle c_2,\kappa \left({\tau }_{\zeta }^{\ast }w\right)\rangle )}^{N+1+\alpha }}}\end{array}$$

(53)

for all w, $z\in {\mathbb{B}}^N$. Letting $w=0$ in (53), we obtain

$$\begin{array}{cc}\hfill & {\displaystyle \frac{1}{{(1+\langle z,{\tau }_{\zeta }\kappa \left(c_1\right)\rangle )}^{N+1+\alpha }}}-{\displaystyle \frac{1}{{(1+\langle z,{\tau }_{\zeta }\kappa \left(c_2\right)\rangle )}^{N+1+\alpha }}}={\displaystyle \frac{1}{{(1-\langle z,b_1\rangle )}^{N+1+\alpha }}}-{\displaystyle \frac{1}{{(1-\langle z,b_2\rangle )}^{N+1+\alpha }}}\hfill \end{array}$$

(54)

for all $z\in {\mathbb{B}}^N$. Since $\tau \ne 0$, from a direct calculation, (54) is equivalent to

$$\begin{array}{cc}\hfill & {\displaystyle \frac{\langle z,b_1+{\tau }_{\zeta }\kappa \left(c_1\right)\rangle {\displaystyle \sum _{j=0}^{N+\alpha }}{(1-\langle z,b_1\rangle )}^{N+\alpha -j}{(1+\langle z,{\tau }_{\zeta }\kappa \left(c_1\right)\rangle )}^j}{{(1+\langle z,{\tau }_{\zeta }\kappa \left(c_1\right)\rangle )}^{N+1+\alpha }{(1-\langle z,b_1\rangle )}^{N+1+\alpha }}}\hfill \\ \hfill & ={\displaystyle \frac{\langle z,b_2+{\tau }_{\zeta }\kappa \left(c_2\right)\rangle {\displaystyle \sum _{j=0}^{N+\alpha }}{(1-\langle z,b_2\rangle )}^{N+\alpha -j}{(1+\langle z,{\tau }_{\zeta }\kappa \left(c_2\right)\rangle )}^j}{{(1+\langle z,{\tau }_{\zeta }\kappa \left(c_2\right)\rangle )}^{N+1+\alpha }{(1-\langle z,b_2\rangle )}^{N+1+\alpha }}}.\hfill \end{array}$$

(55)

Replacing z by $z=(0,\dots ,z_l,\dots ,0)$ in (55), we obtain

$$\begin{array}{cc}\hfill & {\displaystyle \frac{z_l({\overline{b}}_{1,l}+e^{-i{\zeta }_l}{c}_{1,l}){\displaystyle \sum _{j=0}^{N+\alpha }}{(1-z_l{\overline{b}}_{1,l})}^{N+\alpha -j}{(1+z_l{e}^{-i{\zeta }_l}{c}_{1,l})}^j}{{(1+z_l{e}^{-i{\zeta }_l}{c}_{1,l})}^{N+1+\alpha }{(1-z_l{\overline{b}}_{1,l})}^{N+1+\alpha }}}\hfill \\ \hfill & ={\displaystyle \frac{z_2({\overline{b}}_{2,l}+e^{-i{\zeta }_l}{c}_{2,l}){\displaystyle \sum _{j=0}^{N+\alpha }}(1-z_l{\overline{b}}_{2,l}\rangle {)}^{N+\alpha -j}{(1+z_l{e}^{-i{\zeta }_l}{c}_{2,l})}^j}{{(1+z_l{e}^{-i{\zeta }_l}{c}_{2,l})}^{N+1+\alpha }{(1-z_l{\overline{b}}_{2,l})}^{N+1+\alpha }}},\hfill \end{array}$$

(56)

where $b_{1,l}$, $b_{2,l}$, $c_{1,l}$, $c_{2,l}$ are the $lth$ components of $b_1$, $b_2$, $c_1$ and $c_2$, respectively. By comparing the coefficients of $z_l$, we have ${\overline{b}}_{1,l}+e^{-i{\zeta }_l}{c}_{1,l}={\overline{b}}_{1,l}+e^{-i{\zeta }_l}{c}_{1,l}$, which implies $b_1+{\tau }_{\zeta }\kappa \left(c_1\right)=b_2+{\tau }_{\zeta }\kappa \left(c_2\right)$. This allows us to divide into the following two cases for consideration.

Case 1. Assume that $b_1+{\tau }_{\zeta }\kappa \left(c_1\right)=b_2+{\tau }_{\zeta }\kappa \left(c_2\right)=0$. For this case, it is easy to see that (53) holds if $b_1=-{\tau }_{\zeta }\kappa \left(c_1\right)$ and $b_2=-{\tau }_{\zeta }\kappa \left(c_2\right)$.

Case 2. Assume that $b_1+{\tau }_{\zeta }\kappa \left(c_1\right)=b_2+{\tau }_{\zeta }\kappa \left(c_2\right)\ne 0$. Then, (56) is reduced to

$$\begin{array}{c}\hfill {\displaystyle \frac{\sum _{j=0}^{N+\alpha }{(1-z_l{\overline{b}}_{1,l})}^{N+\alpha -j}{(1+z_l{e}^{-i{\zeta }_l}{c}_{1,l})}^j}{{(1+z_l{e}^{-i{\zeta }_l}{c}_{1,l})}^{N+1+\alpha }{(1-z_l{\overline{b}}_{1,l})}^{N+1+\alpha }}}={\displaystyle \frac{\sum _{j=0}^{N+\alpha }(1-z_l{\overline{b}}_{2,l}\rangle {)}^{N+\alpha -j}{(1+z_l{e}^{-i{\zeta }_l}{c}_{2,l})}^j}{{(1+z_l{e}^{-i{\zeta }_l}{c}_{2,l})}^{N+1+\alpha }{(1-z_l{\overline{b}}_{2,l})}^{N+1+\alpha }}},\end{array}$$

comparing the coefficients of $z_l$, we have

$$\begin{array}{cc}\hfill & \sum _{j=0}^{N+\alpha }[(N+\alpha -j)(-{\overline{b}}_{1,l})+j{e}^{-i{\zeta }_l}{c}_{1,l}]+{(N+1+\alpha )}^2{e}^{-i{\zeta }_l}{c}_{2,l}-{(N+1+\alpha )}^2{\overline{b}}_{2,l}\hfill \\ \hfill & =\sum _{j=0}^{N+\alpha }[(N+\alpha -j)(-{\overline{b}}_{2,l})+j{e}^{-i{\zeta }_l}{c}_{2,l}]+{(N+1+\alpha )}^2{e}^{-i{\zeta }_l}{c}_{1,l}-{(N+1+\alpha )}^2{\overline{b}}_{1,l},\hfill \end{array}$$

which implies $e^{i{\zeta }_l}{\overline{c}}_{1,l}-b_{1,l}=e^{i{\zeta }_l}{\overline{c}}_{2,l}-b_{2,l}$. So, we obtain ${\tau }_{\zeta }\kappa \left(c_1\right)-b_1={\tau }_{\zeta }\kappa \left(c_2\right)-b_2$. Since $b_1+{\tau }_{\zeta }\kappa \left(c_1\right)=b_2+{\tau }_{\zeta }\kappa \left(c_2\right)\ne 0$, we therefore have ${\tau }_{\zeta }\kappa \left(c_1\right)={\tau }_{\zeta }\kappa \left(c_2\right)$ and $b_1=b_2$. From this and $W_{u_1,{\phi }_1}\ne W_{u_1,{\phi }_2}$, we have $A_1\ne A_2$. Then, (53) becomes

$$\begin{array}{cc}\hfill & {\displaystyle \frac{{\displaystyle \sum _{j=0}^{N+\alpha }}{(1+\langle z,\kappa ({\tau }_{\zeta }^{\ast }{c}_1-{\tau }_{\zeta }^{\ast }{A}_2^{\ast }w)\rangle -\langle w,b_1\rangle )}^{N+\alpha -j}{(1+\langle z,\kappa ({\tau }_{\zeta }^{\ast }{c}_1-{\tau }_{\zeta }^{\ast }{A}_1^{\ast }w)\rangle -\langle w,b_1\rangle )}^j}{{(1+\langle z,\kappa ({\tau }_{\zeta }^{\ast }{c}_1-{\tau }_{\zeta }^{\ast }{A}_1^{\ast }w)\rangle -\langle w,b_1\rangle )}^{N+\alpha +1}{(1+\langle z,\kappa ({\tau }_{\zeta }^{\ast }{c}_1-{\tau }_{\zeta }^{\ast }{A}_2^{\ast }w)\rangle -\langle w,b_1\rangle )}^{N+\alpha +1}}}\hfill \\ \hfill & ={\displaystyle \frac{{\displaystyle \sum _{j=0}^{N+\alpha }}(1-{\langle z,b_1+A_2\kappa ({\tau }_{\zeta }^{\ast }w)\rangle +\langle w,{\tau }_{\zeta }\kappa \left(c_1\right)\rangle )}^{N+\alpha -j}(1-{\langle z,b_1+A_1\kappa ({\tau }_{\zeta }^{\ast }w)\rangle +\langle w,{\tau }_{\zeta }\kappa \left(c_1\right)\rangle )}^j}{(1-{\langle z,b_1+A_1\kappa ({\tau }_{\zeta }^{\ast }w)\rangle +\langle w,{\tau }_{\zeta }\kappa \left(c_1\right)\rangle )}^{N+\alpha +1}(1-{\langle z,b_1+A_2\kappa ({\tau }_{\zeta }^{\ast }w)\rangle +\langle w,{\tau }_{\zeta }\kappa \left(c_1\right)\rangle )}^{N+\alpha +1}}}.\hfill \end{array}$$

(57)

Also, write $A_1={{\{}_1{a}_{ij}\}}_{N\times N}$ and $A_2={{\{}_2{a}_{ij}\}}_{N\times N}$. Replacing z and w by $z=(0,\dots ,z_l,\dots ,0)$ and $w=(0,\dots ,w_l,\dots ,0)$, respectively, in (57), (57) becomes

$$\begin{array}{cc}\hfill & {\displaystyle \frac{{\displaystyle \sum _{j=0}^{N+\alpha }}{[1+z_l(e^{-i{\zeta }_l}{c}_{1,l}-e^{-i{\zeta }_l}{\overline{2a}}_{ll}{w}_l)-w_l{\overline{b}}_{1,l}]}^{N+\alpha -j}{[1+z_l(e^{-i{\zeta }_l}{c}_{1,l}-e^{-i{\zeta }_l}{\overline{1a}}_{ll}{w}_l)-w_l{\overline{b}}_{1,l}]}^j}{{[1+z_l(e^{-i{\zeta }_l}{c}_{1,l}-e^{-i{\zeta }_l}{\overline{1a}}_{ll}{w}_l)-w_l{\overline{b}}_{1,l}]}^{N+1+\alpha }{[1+z_l(e^{-i{\zeta }_l}{c}_{1,l}-e^{-i{\zeta }_l}{\overline{2a}}_{ll}{w}_l)-w_l{\overline{b}}_{1,l}]}^{N+1+\alpha }}}\hfill \\ \hfill & ={\displaystyle \frac{{\displaystyle \sum _{j=0}^{N+\alpha }}{[1-z_l({\overline{b}}_{1,l}+{\overline{2a}}_{ll}{e}^{-i{\zeta }_l}{w}_l)+w_l{e}^{-i{\zeta }_l}{c}_{1,l}]}^{N+\alpha -j}{[1-z_l({\overline{b}}_{1,l}+{\overline{1a}}_{ll}{e}^{-i{\zeta }_l}{w}_l)+w_l{e}^{-i{\zeta }_l}{c}_{1,l}]}^j}{{[1-z_l({\overline{b}}_{1,l}+{\overline{1a}}_{ll}{e}^{-i{\zeta }_l}{w}_l)+w_l{e}^{-i{\zeta }_l}{c}_{1,l}]}^{N+1+\alpha }{[1-z_l({\overline{b}}_{1,l}+{\overline{2a}}_{ll}{e}^{-i{\zeta }_l}{w}_l)+w_l{e}^{-i{\zeta }_l}{c}_{1,l}]}^{N+1+\alpha }}},\hfill \end{array}$$

where $b_{1,l},c_{1,l}$ are the $lth$ components of $b_1,c_1$. From the coefficients of $w_l$, we have

$$\begin{array}{c}\hfill \sum _{j=0}^{N+\alpha }(N+\alpha )(-b_{1,l})+2{(N+1+\alpha )}^2{e}^{i{\zeta }_l}{\overline{c}}_{1,l}=\sum _{j=0}^{N+\alpha }(N+\alpha )e^{i{\zeta }_l}{\overline{c}}_{1,l}+2{(N+1+\alpha )}^2(-b_{1,l}),\end{array}$$

which shows that $b_{1,l}+e^{i{\zeta }_l}{\overline{c}}_{1,l}=0$; that is, $b_1+{\tau }_{\zeta }\kappa \left(c_1\right)=0$, which contradicts Case 2. Combing Cases 1 and 2, we see that the (53) holds if $b_1=-{\tau }_{\zeta }\kappa \left(c_1\right)$ and $b_2=-{\tau }_{\zeta }\kappa \left(c_2\right)$.

Conversely, if $b_1=-{\tau }_{\zeta }\kappa \left(c_1\right)$ and $b_2=-{\tau }_{\zeta }\kappa \left(c_2\right)$, then the left of (53) equals

$$\begin{array}{c}\hfill {\displaystyle \frac{\tau }{{(1+\langle z,\kappa \left({\tau }_{\zeta }^{\ast }{c}_1\right)-\kappa \left({\tau }_{\zeta }^{\ast }{A}_1^{\ast }w\right)\rangle +\langle w,{\tau }_{\zeta }\kappa \left(c_1\right)\rangle )}^{N+1+\alpha }}}-{\displaystyle \frac{\tau }{{(1+\langle z,\kappa \left({\tau }_{\zeta }^{\ast }{c}_2\right)-\kappa \left({\tau }_{\zeta }^{\ast }{A}_2^{\ast }w\right)\rangle +\langle w,{\tau }_{\zeta }\kappa \left(c_2\right)\rangle )}^{N+1+\alpha }}},\end{array}$$

and the right of (53) also does. From this, it follows that (53) holds. This shows that the operator $W_{u_1,{\phi }_1}-W_{u_2,{\phi }_2}$ is complex symmetric on $A_{\alpha }^2\left({\mathbb{B}}^N\right)$ with the conjugation ${\mathcal{J}}_{\tau ,\zeta }$. □

From Theorems 2 and 6, we also have the following result.

Corollary 5.

Let $u_j\left(z\right)={\displaystyle \frac{1}{{(1+\langle z,c_j\rangle )}^{N+1+\alpha }}}$ and ${\phi }_j\left(z\right)={\displaystyle \frac{A_{j}z+b_j}{\langle z,c_j\rangle +1}}$, $A_j=A_j^T$ and $A_j,b_j,c_j$ satisfy $\parallel A_j\parallel +\left|b_j\right|+\left|c_j\right|<1$, where $j=1,2$. Then, the operator $W_{u_1,{\phi }_1}-W_{u_2,{\phi }_2}$ is complex symmetric on $A_{\alpha }^2\left({\mathbb{B}}^N\right)$ with the conjugation ${\mathcal{J}}_{\tau ,\zeta }$ if and only if $W_{u_1,{\phi }_1}$ and $W_{u_2,{\phi }_2}$ are complex symmetric on $A_{\alpha }^2\left({\mathbb{B}}^N\right)$ with the conjugation ${\mathcal{J}}_{\tau ,\zeta }$.

By Corollary 3 and Corollary 5, we give an example as follows.

Example 4.

Let $u_j\left(z\right)={\displaystyle \frac{1}{{(1+\langle z,c_j\rangle )}^{3+\alpha }}}$, ${\phi }_j\left(z\right)={\displaystyle \frac{A_{j}z+b_j}{\langle z,c_j\rangle +1}}$, $j=1,2$, where

$$\begin{array}{c}\hfill A_1=\left(\begin{array}{cc}{\displaystyle \frac{-i}{15}}& {\displaystyle \frac{\sqrt{2}i}{20}}\\ {\displaystyle \frac{\sqrt{2}i}{20}}& {\displaystyle \frac{1}{15}}\end{array}\right),c_1={\left(\begin{array}{cc}{\displaystyle \frac{i}{8}},& {\displaystyle \frac{i}{15}}\end{array}\right)}^T,{\tau }_{\zeta }=\left(\begin{array}{cc}{e}^{{\displaystyle \frac{i}{3}}}& 0\\ 0& e^{{\displaystyle \frac{i}{4}}}\end{array}\right),\\ \hfill A_2=\left(\begin{array}{cc}{\displaystyle \frac{1}{13}}+i{\displaystyle \frac{1}{13}}& {\displaystyle \frac{\sqrt{2}}{13}}\\ {\displaystyle \frac{\sqrt{2}}{13}}& 0\end{array}\right),c_2={\left(\begin{array}{cc}{\displaystyle \frac{i}{12}},& 0\end{array}\right)}^T,\end{array}$$

and

$$\begin{array}{c}\hfill b_1=-{\tau }_{\zeta }\kappa \left(c_1\right)={\left(\begin{array}{cc}{\displaystyle \frac{i}{8}}{e}^{{\displaystyle \frac{i}{3}}},& {\displaystyle \frac{i}{15}}{e}^{{\displaystyle \frac{i}{4}}}\end{array}\right)}^T,b_2=-{\tau }_{\zeta }\kappa \left(c_2\right)={\left(\begin{array}{cc}{\displaystyle \frac{i}{12}}{e}^{{\displaystyle \frac{i}{3}}},& 0\end{array}\right)}^T.\end{array}$$

Then, $W_{u_1,{\phi }_1}-W_{u_2,{\phi }_2}$ is complex symmetric on $A_{\alpha }^2\left({\mathbb{B}}^2\right)$ with the conjugation ${\mathcal{J}}_{\tau ,\zeta }$.

Proof. 

From Example 2 and Corollary 5, the desired result of this example follows. □

5. Conclusions

In this paper, we obtain some characterizations for weighted composition operators to be 2-complex symmetric on $A_{\alpha }^2\left({\mathbb{B}}^N\right)$ with respect to the conjugations $\mathcal{J}$ and ${\mathcal{J}}_{\tau ,\zeta }$. To be specific, in Theorems 1 and 2, we characterize the complex symmetric weighted composition operators with the conjugations $\mathcal{J}$ and ${\mathcal{J}}_{\tau ,\zeta }$ on $A_{\alpha }^2\left({\mathbb{B}}^N\right)$, respectively. In Theorems 3 and 4, we characterize 2-complex symmetric weighted composition operators with the same conjugations. We also characterize the complex symmetric differences of weighted composition operators in Theorems 5 and 6. It is worth noting that we only characterize complex symmetric and 2-complex symmetric weighted composition operators induced by specific symbols with the conjugations $\mathcal{J}$ and ${\mathcal{J}}_{\tau ,\zeta }$ on $A_{\alpha }^2\left({\mathbb{B}}^N\right)$. The current results naturally lead to several research extensions: characterizing 2-complex symmetric weighted composition operators induced by general symbols on $A_{\alpha }^2\left({\mathbb{B}}^N\right)$ or on some other spaces, and applying these findings to distributed parameter systems in Hilbert spaces. We regard these as the next research projects.