2-Complex Symmetric Composition Operators on H²

Abstract

In this paper, we study 2-complex symmetric composition operators with the conjugation J, defined by $Jf\left(z\right)=\overline{\left(f\left(\overline{z}\right)\right)}$, on the Hardy space $H^2$. More precisely, we obtain the necessary and sufficient condition for the composition operator $C_{\varphi }$ to be 2-complex symmetric with J when $\varphi$ is an automorphism of $\mathbb{D}$. We also characterize 2-complex symmetric with J when $\varphi$ is a linear fractional self-map of $\mathbb{D}$.

1. Introduction

Throughout this paper, H and $\mathcal{B}\left(H\right)$ will always denote a separable complex Hilbert space and the set of all continuous linear operators on H, respectively.

Definition 1.

An operator $C:H\to H$ is said to be a conjugation on H if it is

(a)

anti-linear or conjugate-linear: $C(\alpha x+\beta y)=\overline{\alpha }C\left(x\right)+\overline{\beta }C\left(y\right)$, for all $\alpha ,\beta \in \mathbb{C}$ and $x,y\in H$,

(b)

isometric: $\parallel Cx\parallel =\parallel x\parallel$, for all $x\in H$,

(c)

involutive: $C^2=I_d$, where $I_d$ is an identity operator.

It is easy to check that $\left(Jf\right)\left(z\right)=\overline{f\left(\overline{z}\right)}$ is a conjugation on the Hardy space $H^2$.

Definition 2.

For a conjugation C on H, an operator $T\in \mathcal{B}\left(H\right)$ is said to be complex symmetric (complex symmetric with C or a C-symmetric operator) if $T=C{T}^{*}C$.

The class of complex symmetric operators includes all normal operators, binormal operators, Hankel operators, compressed Toeplitz operators and Volterra integration operators. The study of complex symmetric operators was initiated by Garcia, Putinar, and Wogen in [1,2,3,4]. See [5,6,7,8,9,10,11,12,13,14] for more results on complex symmetric operators.

Definition 3.

Let m be a positive integer and $T\in \mathcal{B}\left(H\right)$. T is said to be a m-complex symmetric operator (m-complex symmetric with C) if there exists a conjugation C such that

$$\sum _{j=0}^m{(-1)}^{m-j}\left(\begin{array}{c}m\\ j\end{array}\right){T*}^{j}C{T}^{m-j}C=0.$$

The above definition was introduced by Chō, Ko, and Lee in [15]. When $m=2$, we obtain

$$C{T}^{2}C-2T^{*}CTC+{T^{*}}^2=0,$$

which is equivalent to

$$C{T}^2-2T^{*}CT+{T^{*}}^{2}C=0.$$

It is clear that 1-complex symmetric operator is just the complex symmetric operator. From [15], we see that all complex symmetric operators are 2-complex symmetric operators. Thus, the set of all 2-complex symmetric operators is larger than the set of all complex symmetric operators. In [16], the authors studied m-complex symmetric weighted shifts on ${\mathbb{C}}^n$. We refer the reader to [15,16,17,18] for more results about m-complex symmetric operators.

Inspired by these papers, in this paper, we study 2-complex symmetric composition operators, induced by linear fractional self-maps of $\mathbb{D}$, with J on the Hardy space $H^2$. When the symbol $\varphi$ is an automorphism of $\mathbb{D}$, we show that the composition operator $C_{\varphi }$ is 2-complex symmetric with J if and only if $C_{\varphi }$ is normal. Furthermore, we also characterize 2-complex symmetric composition operators with J on $H^2$ when the induced maps are linear fractional self-maps of $\mathbb{D}$.

2. Preliminaries

Let $\mathbb{D}$ and $\partial \mathbb{D}$ be the open unit disk $\mathbb{C}$ and the unit circle in the complex plane, respectively. Let $H\left(\mathbb{D}\right)$ be the set of all analytic functions on $\mathbb{D}$. The Hardy space $H^2\left(\mathbb{D}\right)$ is the space of all $f\in H\left(\mathbb{D}\right)$ such that

$${\parallel f\parallel }^2=\underset{0\le r<1}{sup}\frac{1}{2\pi }{\int }_0^{2\pi }{\left|f\left(r{e}^{i\theta }\right)\right|}^{2}d\theta <\infty .$$

The space $H^2\left(\mathbb{D}\right)$ is a reproducing kernel Hilbert space, that is, for each $w\in \mathbb{D}$ and $f\in H^2\left(\mathbb{D}\right)$, there is a unique function $K_w$ such that

$$\langle f,K_w\rangle =f\left(w\right),$$

where $K_w\left(z\right)=\frac{1}{1-\overline{w}z}$ is said to be the reproducing kernel at w. For $g\in L^{\infty }(\partial \mathbb{D})$, the Toeplitz operator $T_g$ is defined as $T_{g}f=P\left(fg\right)$ for $f\in H^2\left(\mathbb{D}\right)$, where P is the orthogonal projection of $L^2$ onto $H^2\left(\mathbb{D}\right)$. Recall that

$$T_g^{*}{K}_w=\overline{f\left(w\right)}{K}_w$$

for each $w\in \mathbb{D}$ and $f\in H^{\infty }$.

Let $\varphi$ be an analytic self-map of $\mathbb{D}$. Recall that the composition operator $C_{\varphi }$ is defined by

$$C_{\varphi }f\left(z\right)=f\left(\varphi \left(z\right)\right),f\in H\left(\mathbb{D}\right),z\in \mathbb{D}.$$

It is easy to see that $C_{\varphi }^{*}{K}_w=K_{\varphi \left(w\right)}$ for each $w\in \mathbb{D}$.

3. Main Results

We begin this section with Cowen’s formula for the adjoint of a linear fractional self-map. For a linear fractional self-map $\varphi \left(z\right)=\frac{az+b}{cz+d}$, Cowen in [19] obtained the following important formula:

$$C_{\varphi }^{*}=T_g{C}_{\sigma }{T}_h^{*},$$

where $a,b,c,d\in \mathbb{C}$, $ad-bc\ne 0$ and

$$\sigma \left(z\right)=\frac{\overline{a}z-\overline{c}}{-\overline{b}z+\overline{d}},g\left(z\right)=\frac{1}{-\overline{b}z+\overline{d}},h\left(z\right)=cz+d.$$

Next, we state some lemmas which will be used in our mian results.

Lemma 1

([20]). Let ϕ be an analytic self-map of $\mathbb{D}$. Then, $C_{\varphi }$ is normal if and only if $\varphi \left(z\right)=az$ with $\left|a\right|\le 1$.

Lemma 2.

Let $\varphi \left(z\right)=\frac{az+b}{cz+d}$ be a linear fractional self-map of $\mathbb{D}$. If $w\in \mathbb{D}$ satisfies $\overline{a}w\ne \overline{c}$, then the following statements hold:

(a)

$J{C}_{\varphi }^2{K}_w(z)=K_{\overline{w}}(\overline{{\varphi }_2(\overline{z})})$,

(b)

$C_{\varphi }^{*}J{C}_{\varphi }{K}_w\left(z\right)=-\overline{c}\frac{g\left(w\right)}{\sigma \left(w\right)}{K}_{\varphi \left(0\right)}\left(z\right)+\overline{h\left(\frac{1}{\overline{\sigma \left(w\right)}}\right)}g\left(w\right)K_{\varphi (\overline{\sigma \left(w\right)})}\left(z\right),$

(c)

${C_{\varphi }^{*}}^{2}J{K}_w\left(z\right)=K_{{\varphi }_2\left(\overline{w}\right)}\left(z\right),$

(d)

$J{C}_{\varphi }J{C}_{\varphi }^{*}J{K}_w\left(z\right)=K_{\varphi \left(\overline{w}\right)}(\overline{\varphi \left(\overline{z}\right)}),$

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

Proof. 

Since $Jf\left(z\right)=\overline{f\left(\overline{z}\right)}$ and $f\left(w\right)=\langle f,K_w\rangle$ for any $w\in \mathbb{D}$ and $f\in H^2$, we see that $\left(a\right)$ and $\left(c\right)$ hold obviously. Now, we only verify $\left(b\right)$ and $\left(d\right)$. For any $w\in \mathbb{D}$ with $\overline{a}w\ne \overline{c}$, we obtain

$$\begin{array}{cc}\hfill C_{\varphi }^{*}J{C}_{\varphi }{K}_w\left(z\right)=& C_{\varphi }^{*}{K}_{\overline{w}}(\overline{\varphi \left(\overline{z}\right)})\hfill \\ \hfill =& C_{\varphi }^{*}\frac{\overline{c}z+\overline{d}}{(\overline{c}-\overline{a}w)z+(\overline{d}-\overline{b}w)}\hfill \\ \hfill =& C_{\varphi }^{*}\left(\frac{\overline{c}}{\overline{c}-\overline{a}w}+\left(\overline{d}-\frac{\overline{c}(\overline{d}-\overline{b}w)}{\overline{c}-\overline{a}w}\right)\frac{1}{\overline{d}-\overline{b}w}\frac{1}{1-\frac{\overline{a}w-\overline{c}}{\overline{d}-\overline{b}w}z}\right)\hfill \\ \hfill =& \frac{\overline{c}}{\overline{c}-\overline{a}w}{C}_{\varphi }^{*}{K}_0\left(z\right)+\left(\overline{d}-\frac{\overline{c}(\overline{d}-\overline{b}w)}{\overline{c}-\overline{a}w}\right)\frac{1}{\overline{d}-\overline{b}w}{C}_{\varphi }^{*}{K}_{\overline{\sigma \left(w\right)}}\left(z\right)\hfill \\ \hfill =& -\overline{c}\frac{g\left(w\right)}{\sigma \left(w\right)}{K}_{\varphi \left(0\right)}\left(z\right)+\overline{h\left(\frac{1}{\overline{\sigma \left(w\right)}}\right)}g\left(w\right)K_{\varphi (\overline{\sigma \left(w\right)})}\left(z\right).\hfill \end{array}$$

Clearly,

$$\begin{array}{cc}\hfill J{C}_{\varphi }J{C}_{\varphi }^{*}J{K}_w\left(z\right)& =J{C}_{\varphi }J{C}_{\varphi }^{*}{K}_{\overline{w}}\left(z\right)=J{C}_{\varphi }J{K}_{\varphi \left(\overline{w}\right)}\left(z\right)\hfill \\ \hfill & =J{C}_{\varphi }{K}_{\overline{\varphi \left(\overline{w}\right)}}\left(z\right)=J{K}_{\overline{\varphi \left(\overline{w}\right)}}\left(\varphi \left(z\right)\right)\hfill \\ & =K_{\varphi \left(\overline{w}\right)}(\overline{\varphi \left(\overline{z}\right)}).\hfill \end{array}$$

The proof is complete. □

Lemma 3.

Let ϕ be an analytic self-map of $\mathbb{D}$. If $C_{\varphi }$ is normal, then $C_{\varphi }^{*}$ and $C_{\varphi }$ is 2-complex symmetric with J.

Proof. 

Assume that $C_{\varphi }$ is normal. From Corollary 3.10 in [21], we see that $C_{\varphi }$ is complex symmetric with J. Hence, $C_{\varphi }$ is 2-complex symmetric with J. Write $\varphi \left(z\right)=az$ with $\left|a\right|\le 1$ from Lemma 1. We notice that $T^{*}$ is 2-complex symmetric with J if and only if

$${T^{*}}^{2}J-2JTJ{T}^{*}J+J{T}^2=0.$$

In view of Lemma 2, set $\sigma \left(z\right)=\overline{a}z$, $g\left(z\right)=h\left(z\right)=1$, we have

$${C_{\varphi }^{*}}^{2}J{K}_w\left(z\right)=K_{{\varphi }_2\left(\overline{w}\right)}\left(z\right)=\frac{1}{1-{\overline{a}}^{2}wz},$$

$$\begin{array}{cc}\hfill J{C}_{\varphi }J{C}_{\varphi }^{*}J{K}_w\left(z\right)& =K_{\varphi \left(\overline{w}\right)}(\overline{\varphi \left(\overline{z}\right)})=\frac{1}{1-{\overline{a}}^{2}wz}\hfill \end{array}$$

and

$$J{C}_{\varphi }^2{K}_w\left(z\right)=K_{\overline{w}}(\overline{{\varphi }_2\left(\overline{z}\right)})=\frac{1}{1-w\overline{{\varphi }_2\left(\overline{z}\right)}}=\frac{1}{1-{\overline{a}}^{2}wz}.$$

Thus, we obtain that ${C_{\varphi }^{*}}^{2}J-2J{C}_{\varphi }J{C}_{\varphi }^{*}J+J{C}_{\varphi }^2=0$, which means that $C_{\varphi }^{*}$ is 2-complex symmetric with J. □

Lemma 4.

Let ϕ be an analytic self-map of $\mathbb{D}$. If $C_{\varphi }$ is 2-complex symmetric with J, then

$$2{\varphi }_2\left(0\right)-4\varphi \left(0\right)+\varphi \left(0\right){\varphi }_2\left(0\right)=0.$$

Proof. 

Since $f\left(w\right)=\langle f,K_w\rangle$ for any $w\in \mathbb{D}$ and $f\in H^2$, we have

$$\langle J{C}_{\varphi }^2{K}_0,K_{\frac{1}{2}}\rangle =\langle J{K}_0,K_{\frac{1}{2}}\rangle =1,$$

$$\langle C_{\varphi }^{*}J{C}_{\varphi }{K}_0,K_{\frac{1}{2}}\rangle =\langle C_{\varphi }^{*}{K}_0,K_{\frac{1}{2}}\rangle =\langle K_{\varphi \left(0\right)},K_{\frac{1}{2}}\rangle =\frac{2}{2-\overline{\varphi \left(0\right)}}$$

and

$$\langle {C_{\varphi }^{*}}^{2}J{K}_0,K_{\frac{1}{2}}\rangle =\langle {C_{\varphi }^{*}}^2{K}_0,K_{\frac{1}{2}}\rangle =\langle K_{{\varphi }_2\left(0\right)},K_{\frac{1}{2}}\rangle =\frac{2}{2-\overline{{\varphi }_2\left(0\right)}}.$$

By the assumption that $C_{\varphi }$ is 2-complex symmetric with J, we obtain that

$$1-\frac{4}{2-\overline{\varphi \left(0\right)}}+\frac{2}{2-\overline{{\varphi }_2\left(0\right)}}=0.$$

By a simple calculation, we obtain the desired result. □

The following result gives a necessary and sufficient condition for $C_{\varphi }$ to be 2-complex symmetric with J when $\varphi$ is an automorphism of $\mathbb{D}$.

Theorem 1.

Let ϕ be an automorphism of $\mathbb{D}$. Then, $C_{\varphi }$ is 2-complex symmetric with J if and only if $C_{\varphi }$ is normal.

Proof. 

Assume that $C_{\varphi }$ is normal. From Lemma 3, we see that $C_{\varphi }$ is 2-complex symmetric with J.

Now, suppose that $\varphi \left(z\right)=\lambda \frac{a-z}{1-\overline{a}z}$ with $\left|\lambda \right|=1$, $a\in \mathbb{D}$ and $C_{\varphi }$ is 2-complex symmetric with J. Then, $\sigma \left(z\right)=\frac{-\overline{\lambda }z+a}{-\overline{\lambda a}z+1}$, $g\left(z\right)=\frac{1}{-\overline{\lambda a}z+1}$ and $h\left(z\right)=-\overline{a}z+1$. Noting that

$${\varphi }_2\left(z\right)=\frac{\lambda (\lambda z-\lambda a)-\lambda a(\overline{a}z-1)}{\overline{a}(\lambda z-\lambda a)-(\overline{a}z-1)},$$

then, for $\overline{\lambda }w\ne a$, Lemma 2 gives that

$$\begin{array}{c}\hfill J{C}_{\varphi }^2{K}_w\left(z\right)=K_{\overline{w}}(\overline{{\varphi }_2\left(\overline{z}\right)})=\frac{a(\overline{\lambda }z-\overline{\lambda a})-(az-1)}{a(\overline{\lambda }z-\overline{\lambda a})-(az-1)-w[\overline{\lambda }(\overline{\lambda }z-\overline{\lambda a})-\overline{\lambda a}(az-1)]},\end{array}$$

(1)

$$\begin{array}{cc}\hfill C_{\varphi }^{*}J{C}_{\varphi }{K}_w\left(z\right)=& a\frac{g\left(w\right)}{\sigma \left(w\right)}{K}_{\varphi \left(0\right)}\left(z\right)+\overline{h\left(\frac{1}{\overline{\sigma \left(w\right)}}\right)}g\left(w\right)K_{\varphi (\overline{\sigma \left(w\right)})}(z)\hfill \\ \hfill =& \frac{a}{(-\overline{\lambda }w+a)(1-\overline{\lambda a}z)}+\frac{\overline{\lambda }{w(|a|}^2-1)}{(-\overline{\lambda }w+a)(-\overline{\lambda a}w+1)}\hfill \\ \hfill & ·\frac{-\overline{\lambda a}w+1+a\overline{\lambda }w-a^2}{-\overline{\lambda a}w+1+a\overline{\lambda }w-a^2-({\overline{\lambda }}^{2}w-{\overline{\lambda }}^2{\overline{a}}^{2}w-\overline{\lambda }a+\overline{\lambda a})z}\hfill \end{array}$$

(2)

and

$$\begin{array}{c}\hfill {C_{\varphi }^{*}}^{2}J{K}_w(z)=K_{{\varphi }_2(\overline{w})}(z)=\frac{a(\overline{\lambda }w-\overline{\lambda a})-(aw-1)}{a(\overline{\lambda }w-\overline{\lambda a})-(aw-1)-[\overline{\lambda }(\overline{\lambda }w-\overline{\lambda a})-\overline{\lambda a}(aw-1)]z}\end{array}$$

(3)

for any $w,z\in \mathbb{D}$. Taking $w=0$ in (1)–(3), we have

$$J{C}_{\varphi }^2{K}_0\left(z\right)=1,C_{\varphi }^{*}J{C}_{\varphi }{K}_0\left(z\right)=\frac{1}{1-\overline{\lambda a}z},$$

and

$${C_{\varphi }^{*}}^{2}J{K}_0\left(z\right)=\frac{-\overline{\lambda }{\left|a\right|}^2+1}{-\overline{\lambda }{\left|a\right|}^2+1+({\overline{\lambda }}^2\overline{a}-\overline{\lambda a})z}$$

for any $z\in \mathbb{D}$. Since $C_{\varphi }$ is 2-complex symmetric with J, we obtain

$$\begin{array}{c}\hfill \frac{1+\overline{\lambda a}z}{1-\overline{\lambda a}z}=\frac{1-\overline{\lambda }{\left|a\right|}^2}{-\overline{\lambda }{\left|a\right|}^2+1+({\overline{\lambda }}^2\overline{a}-\overline{\lambda a})z}.\end{array}$$

(4)

Calculating and noting that the coefficient of $z^2$ must be 0, we obtain that ${\lambda }^2{a}^2(\lambda -1)=0$. Since $\lambda$ and a are non-zero, then $\lambda =1$. Hence, (4) becomes

$$\frac{1+\overline{a}z}{1-\overline{a}z}=1,$$

which implies that $\overline{a}z=0$ for all $z\in \mathbb{D}$. Thus, $a=0$. Hence, $\varphi \left(z\right)=-\lambda z$ with $\left|\lambda \right|=1$. Lemma 1 gives that $C_{\varphi }$ is normal. The proof is complete. □

4. Linear Fractional Self-Maps

In this section, we first consider 2-complex symmetric composition operators with J which are induced linear fractional self-maps with $\varphi \left(0\right)=0$. In the following, we obtain a sufficient and necessary condition for this case.

Theorem 2.

Let $\varphi \left(z\right)=\frac{az+b}{cz+d}$ be a linear fractional self-map of $\mathbb{D}$ and $\varphi \left(0\right)=0$. Then, $C_{\varphi }$ ($C_{\varphi }^{*}$) is 2-complex symmetric with J if and only if $C_{\varphi }$ is normal.

Proof. 

Assume first that $C_{\varphi }$ is normal. Lemma 3 gives that $C_{\varphi }$ ($C_{\varphi }^{*}$) is 2-complex symmetric with J.

Conversely, suppose that $C_{\varphi }$ is 2-complex symmetric with J. Since $a\ne 0$ and $\varphi \left(0\right)=0$, set $\varphi \left(z\right)=\frac{z}{sz+t}$, where $s=\frac{c}{a}$ and $t=\frac{d}{a}$ satisfy $\left|t\right|\ge \left|s\right|+1$. Then, $\sigma \left(z\right)=\frac{z-\overline{s}}{\overline{t}}$, $g\left(z\right)=\frac{1}{\overline{t}}$ and $h\left(z\right)=sz+t$. If $s=0$, then $\varphi \left(z\right)=\frac{1}{t}z$. Therefore, Lemma 1 gives that $C_{\varphi }$ is normal. Now, we suppose that $s\ne 0$. After a calculation, we have that

$${\varphi }_2\left(z\right)=\frac{z}{(s+st)z+t^2}$$

and

$$\varphi (\overline{\sigma \left(z\right)})=\frac{\overline{z}-s}{s\overline{z}-s^2+t^2},h\left(\frac{1}{\overline{\sigma \left(z\right)}}\right)=\frac{st}{\overline{z}-s}+t.$$

For any $w,z\in \mathbb{D}$ with $w\ne \overline{s}$, employing Lemma 2, we obtain that

$$\begin{array}{c}\hfill J{C}_{\varphi }^2{K}_w\left(z\right)=K_{\overline{w}}(\overline{{\varphi }_2\left(\overline{z}\right)})=\frac{1}{1-\frac{zw}{\overline{s}z+\overline{st}z+{\overline{t}}^2}}=\frac{(\overline{s}+\overline{st})z+{\overline{t}}^2}{(\overline{s}+\overline{st})z+{\overline{t}}^2-wz},\end{array}$$

(5)

$$\begin{array}{cc}\hfill C_{\varphi }^{*}J{C}_{\varphi }{K}_w\left(z\right)=& -\overline{s}\frac{g\left(w\right)}{\sigma \left(w\right)}{K}_{\varphi \left(0\right)}\left(z\right)+\overline{h\left(\frac{1}{\overline{\sigma \left(w\right)}}\right)}g\left(w\right)K_{\varphi (\overline{\sigma \left(w\right)})}\left(z\right)\hfill \\ \hfill =& \frac{-\overline{s}}{w-\overline{s}}+\left(\frac{\overline{st}}{w-\overline{s}}+\overline{t}\right)·\frac{1}{\overline{t}}\frac{1}{1-\frac{w-\overline{s}}{\overline{s}w-{\overline{s}}^2+{\overline{t}}^2}z}\hfill \\ \hfill =& \frac{\overline{s}}{\overline{s}-w}+\frac{w\overline{t}}{(w-\overline{s})\overline{t}}\frac{\overline{s}w-{\overline{s}}^2+{\overline{t}}^2}{\overline{s}w-{\overline{s}}^2+{\overline{t}}^2-(w-\overline{s})z},\hfill \end{array}$$

(6)

and

$$\begin{array}{c}\hfill {C_{\varphi }^{*}}^{2}J{K}_w\left(z\right)=K_{{\varphi }_2\left(\overline{w}\right)}\left(z\right)=\frac{1}{1-\frac{zw}{\overline{s}w+\overline{st}w+{\overline{t}}^2}}=\frac{(\overline{s}+\overline{st})w+{\overline{t}}^2}{(\overline{s}+\overline{st})w+{\overline{t}}^2-wz}.\end{array}$$

(7)

Since $C_{\varphi }$ is 2-complex symmetric with J, we obtain from (5)–(7) that

$$\frac{(\overline{s}+\overline{st})z+{\overline{t}}^2}{(\overline{s}+\overline{st}-w)z+{\overline{t}}^2}+\frac{(\overline{s}+\overline{st})w+{\overline{t}}^2}{(\overline{s}+\overline{st})w+{\overline{t}}^2-wz}$$

$$=\frac{2\overline{s}}{\overline{s}-w}+\frac{2w\overline{t}}{(w-\overline{s})\overline{t}}\frac{\overline{s}w-{\overline{s}}^2+{\overline{t}}^2}{\overline{s}w-{\overline{s}}^2+{\overline{t}}^2-(w-\overline{s})z}$$

for any $w,z\in \mathbb{D}$ with $w\ne \overline{s}$, which gives that

$$\begin{array}{cc}\hfill & \frac{[(\overline{s}+\overline{st})z+{\overline{t}}^2][(\overline{s}+\overline{st})w+{\overline{t}}^2-wz]+[(\overline{s}+\overline{st})w+{\overline{t}}^2][(\overline{s}+\overline{st}-w)z+{\overline{t}}^2]}{[(\overline{s}+\overline{st}-w)z+{\overline{t}}^2][(\overline{s}+\overline{st})w+{\overline{t}}^2-wz]}\hfill \\ \hfill =& \frac{2\overline{st}(w-\overline{s})[\overline{s}w-{\overline{s}}^2+{\overline{t}}^2-(w-\overline{s})z]+2w\overline{t}(\overline{s}-w)(\overline{s}w-{\overline{s}}^2+{\overline{t}}^2)}{-{(w-\overline{s})}^2\overline{t}[\overline{s}w-{\overline{s}}^2+{\overline{t}}^2-(w-\overline{s})z]}\hfill \end{array}$$

(8)

for any $w,z\in \mathbb{D}$ with $w\ne \overline{s}$. Taking $w=0$ in (8), then

$$\frac{{\overline{t}}^2[(\overline{s}-\overline{st})z+{\overline{t}}^2]+{\overline{t}}^2[(\overline{s}+\overline{st})z+{\overline{t}}^2]}{{\overline{t}}^2[(\overline{s}+\overline{st})z+{\overline{t}}^2]}=\frac{2{\overline{s}}^2\overline{t}(-{\overline{s}}^2+{\overline{t}}^2+\overline{s}z)}{{\overline{s}}^2\overline{t}(-{\overline{s}}^2+{\overline{t}}^2+\overline{s}z)}=2.$$

Therefore, $(\overline{s}+\overline{st})z+{\overline{t}}^2=(\overline{s}-\overline{st})z+{\overline{t}}^2$ for all $z\in \mathbb{D}$. Since $s\ne 0$, then $1+t=1-t$. Hence, $t=0$, a contradiction. Thus, the hypothesis is not true that is $s=0$.

Now, assume that $C_{\varphi }^{*}$ is 2-complex symmetric with J. We also assume that $s\ne 0$. By Lemma 2, we have

$$\begin{array}{cc}\hfill J{C}_{\varphi }J{C}_{\varphi }^{*}J{K}_w\left(z\right)& =K_{\varphi \left(\overline{w}\right)}(\overline{\varphi \left(\overline{z}\right)})=\frac{1}{1-\frac{w}{\overline{s}w+\overline{t}}·\frac{z}{\overline{s}z+\overline{t}}}\hfill \\ \hfill & =\frac{(\overline{s}w+\overline{t})(\overline{s}z+\overline{t})}{(\overline{s}w+\overline{t})(\overline{s}z+\overline{t})-wz}\hfill \end{array}$$

(9)

for any $w,z\in \mathbb{D}$. Since $C_{\varphi }^{*}$ is 2-complex symmetric with J, we obtain from (5), (7), and (9) that

$$\begin{array}{cc}\hfill & \frac{[(\overline{s}+\overline{st})z+{\overline{t}}^2][(\overline{s}+\overline{st})w+{\overline{t}}^2-wz]+[(\overline{s}+\overline{st})w+{\overline{t}}^2][(\overline{s}+\overline{st}-w)z+{\overline{t}}^2]}{[(\overline{s}+\overline{st}-w)z+{\overline{t}}^2][(\overline{s}+\overline{st})w+{\overline{t}}^2-wz]}\hfill \\ \hfill =& \frac{2(\overline{s}w+\overline{t})(\overline{s}z+\overline{t})}{(\overline{s}w+\overline{t})(\overline{s}z+\overline{t})-wz}\hfill \end{array}$$

(10)

for any $w,z\in \mathbb{D}$ with $w\ne \overline{s}$. Taking $w=0$ in (10), we also have that

$$\frac{{\overline{t}}^2[(\overline{s}-\overline{st})z+{\overline{t}}^2]+{\overline{t}}^2[(\overline{s}+\overline{st})z+{\overline{t}}^2]}{{\overline{t}}^2[(\overline{s}+\overline{st})z+{\overline{t}}^2]}=2.$$

The other arguments are similar to the case of $C_{\varphi }$. Then, we obtain the desired result. The proof is complete. □

We now consider 2-complex symmetric composition operators with J which are induced linear fractional self-maps with $c=0$.

Lemma 5

([22]). Let $\varphi \left(z\right)=\frac{az+b}{cz+d}$ be a linear fractional self-mapof $\mathbb{D}$ such that $c=0$. Then, $C_{\varphi }=C_{\xi }^{*}{T}_{\tau }^{*}$, where $\xi \left(z\right)=\frac{\overline{a}z}{-\overline{b}z+\overline{d}}$ and $\tau \left(z\right)=\frac{\overline{d}}{-\overline{b}z+\overline{d}}$.

In the next two results, we give the necessary and sufficient conditions for $C_{\varphi }$ and its adjoint operator $C_{\varphi }^{*}$ to be 2-complex symmetric with J when $c=0$.

Theorem 3.

Let $\varphi \left(z\right)=\frac{az+b}{cz+d}$ be a linear fractional self-map of $\mathbb{D}$ such that $c=0$. Then, $C_{\varphi }$ is 2-complex symmetric with J if and only if $C_{\varphi }$ is normal.

Proof. 

Assume first that $C_{\varphi }$ is normal. Lemma 3 gives that $C_{\varphi }$ is 2-complex symmetric with J.

Conversely, suppose that $C_{\varphi }$ is 2-complex symmetric with J. Since $c=0$, then $\varphi \left(z\right)=mz+n$, where $m=\frac{a}{d}$ and $n=\frac{b}{d}$ satisfy $\left|m\right|+\left|n\right|\le 1$. Let $\xi \left(z\right)=\frac{\overline{m}z}{-\overline{n}z+1}$ and $\tau \left(z\right)=\frac{1}{-\overline{n}z+1}$. Note that

$${\varphi }_2\left(z\right)=m^{2}z+mn+n$$

and

$${\xi }_2\left(z\right)=\frac{{\overline{m}}^{2}z}{-\overline{mn}z-\overline{n}z+1},\varphi (\overline{\xi \left(z\right)})=\frac{m^2\overline{z}-n^2\overline{z}+n}{-n\overline{z}+1},\tau \left(\xi \left(z\right)\right)=\frac{-\overline{n}z+1}{-\overline{mn}z-\overline{n}z+1}.$$

Lemmas 2 and 5 give that

$$\begin{array}{cc}\hfill J{C}_{\varphi }^2{K}_w\left(z\right)& =J{C}_{\xi }^{*}{T}_{\tau }^{*}{C}_{\xi }^{*}{T}_{\tau }^{*}{K}_w\left(z\right)=J\overline{\tau \left(w\right)}{C}_{\xi }^{*}{T}_{\tau }^{*}{K}_{\xi \left(w\right)}\left(z\right)\hfill \\ \hfill & =J\overline{\tau \left(w\right)\tau \left(\xi \right(w\left)\right)}{K}_{{\xi }_2\left(w\right)}\left(z\right)=\tau \left(w\right)\tau \left(\xi \left(w\right)\right)K_{\overline{{\xi }_2\left(w\right)}}\left(z\right)\hfill \\ \hfill & =\frac{1}{-\overline{n}w+1}\frac{-\overline{n}w+1}{-\overline{mn}w-\overline{n}w+1}·\frac{1}{1-\frac{{\overline{m}}^{2}w}{-\overline{mn}w-\overline{n}w+1}z}\hfill \\ \hfill & =\frac{1}{-\overline{mn}w-\overline{n}w+1-{\overline{m}}^{2}wz},\hfill \end{array}$$

(11)

$$\begin{array}{cc}\hfill C_{\varphi }^{*}J{C}_{\varphi }{K}_w\left(z\right)& =C_{\varphi }^{*}J{C}_{\xi }^{*}{T}_{\tau }^{*}{K}_w\left(z\right)=C_{\varphi }^{*}J\overline{\tau \left(w\right)}{K}_{\xi \left(w\right)}\left(z\right)=C_{\varphi }^{*}\tau \left(w\right)\overline{K_{\xi \left(w\right)}\left(\overline{z}\right)}\hfill \\ & =C_{\varphi }^{*}\tau \left(w\right)K_{\overline{\xi \left(w\right)}}\left(z\right)=\tau \left(w\right)K_{\varphi (\overline{\xi \left(w\right)})}\left(z\right)\hfill \\ \hfill & =\frac{1}{-\overline{n}w+1}·\frac{1}{1-\frac{{\overline{m}}^{2}w-{\overline{n}}^{2}w+\overline{n}}{-\overline{n}w+1}z}\hfill \\ \hfill & =\frac{1}{-\overline{n}w+1-({\overline{m}}^{2}w-{\overline{n}}^{2}w+\overline{n})z}\hfill \end{array}$$

(12)

and

$$\begin{array}{c}\hfill {C_{\varphi }^{*}}^{2}J{K}_w\left(z\right)=K_{{\varphi }_2\left(\overline{w}\right)}\left(z\right)=\frac{1}{1-({\overline{m}}^{2}w+\overline{mn}+\overline{n})z}\end{array}$$

(13)

for any $w,z\in \mathbb{D}$. Since $C_{\varphi }$ is 2-complex symmetric with J, then we obtain from (11)–(13) that

$$\begin{array}{cc}\hfill & \frac{1}{-\overline{mn}w-\overline{n}w+1-{\overline{m}}^{2}wz}+\frac{1}{1-({\overline{m}}^{2}w+\overline{mn}+\overline{n})z}\hfill \\ \hfill =& \frac{2}{-\overline{n}w+1-({\overline{m}}^{2}w-{\overline{n}}^{2}w+\overline{n})z}\hfill \end{array}$$

(14)

for any $w,z\in \mathbb{D}$. Taking $w=0$ in (14), we have that

$$1+\frac{1}{1-(\overline{mn}+\overline{n})z}=\frac{2}{1-\overline{n}z}$$

for any $z\in \mathbb{D}$. Then,

$$\frac{2-(\overline{mn}+\overline{n})z}{1-(\overline{mn}+\overline{n})z}=\frac{2}{1-\overline{n}z}.$$

Noting that the coefficient of $z^2$ must be 0, then $n^2(m+1)=0$, which means that $n=0$ or $m=-1$. Similarly, noting that the coefficient of z must be also 0, then $n=mn$, which means that $n=0$ or $m=1$. Therefore, $n=0$. This implies that $\varphi \left(z\right)=mz$ with $\left|m\right|\le 1$. Lemma 1 gives that $C_{\varphi }$ is normal. □

Theorem 4.

Let $\varphi \left(z\right)=\frac{az+b}{cz+d}$ be a linear fractional self-map of $\mathbb{D}$ such that $c=0$. Then, $C_{\varphi }^{*}$ is 2-complex symmetric with J if and only if $C_{\varphi }$ is normal.

Proof. 

Since $c=0$, we can set $\varphi \left(z\right)=mz+n$ where $m=\frac{a}{d}$ and $n=\frac{b}{d}$ satisfy $\left|m\right|+\left|n\right|\le 1$. Let $\xi \left(z\right)=\frac{\overline{m}z}{-\overline{n}z+1}$ and $\tau \left(z\right)=\frac{1}{-\overline{n}z+1}$. Noting that

$${\varphi }_2\left(z\right)=m^{2}z+mn+n,$$

and

$$\tau (\overline{\varphi \left(\overline{z}\right)})=\frac{1}{-\overline{n}(\overline{m}z+\overline{n})+1},\xi (\overline{\varphi \left(\overline{z}\right)})=\frac{\overline{m}(\overline{m}z+\overline{n})}{-\overline{n}(\overline{m}z+\overline{n})+1}.$$

Using Lemma 5 and Theorem 3, we obtain that

$$\begin{array}{cc}\hfill J{C}_{\varphi }^2{K}_w\left(z\right)=J{C}_{\xi }^{*}{T}_{\tau }^{*}{C}_{\xi }^{*}{T}_{\tau }^{*}{K}_w\left(z\right)& =\frac{1}{-\overline{mn}w-\overline{n}w+1-{\overline{m}}^{2}wz},\hfill \end{array}$$

(15)

$$\begin{array}{cc}\hfill J{C}_{\varphi }J{C}_{\varphi }^{*}J{K}_w\left(z\right)& =J{C}_{\xi }^{*}{T}_{\tau }^{*}{K}_{\overline{\varphi (\overline{w})}}\left(z\right)\hfill \\ \hfill & =J\overline{\tau (\overline{\varphi \left(\overline{w}\right)})}{K}_{\xi (\overline{\varphi \left(\overline{w}\right)})}\left(z\right)=\tau (\overline{\varphi \left(\overline{w}\right)})K_{\overline{\xi (\overline{\varphi \left(\overline{w}\right)})}}\left(z\right)\hfill \\ \hfill & =\frac{1}{-\overline{n}(\overline{m}w+\overline{n})+1}·\frac{1}{1-\frac{\overline{m}(\overline{m}w+\overline{n})z}{-\overline{n}(\overline{m}w+\overline{n})+1}}\hfill \\ & =\frac{1}{1-(\overline{m}w+\overline{n})(\overline{m}z+\overline{n})}\hfill \end{array}$$

(16)

and

$$\begin{array}{c}\hfill {C_{\varphi }^{*}}^{2}J{K}_w\left(z\right)=K_{{\varphi }_2\left(\overline{w}\right)}\left(z\right)=\frac{1}{1-({\overline{m}}^{2}w+\overline{mn}+\overline{n})z}\end{array}$$

(17)

for any $w,z\in \mathbb{D}$. Since $C_{\varphi }^{*}$ is 2-complex symmetric with J, then we obtain from (15)–(17) that

$$\frac{1}{-\overline{mn}w-\overline{n}w+1-{\overline{m}}^{2}wz}+\frac{1}{1-({\overline{m}}^{2}w+\overline{mn}+\overline{n})z}=\frac{2}{1-(\overline{m}w+\overline{n})(\overline{m}z+\overline{n})}$$

for any $w,z\in \mathbb{D}$. Taking $w=0$, we get

$$\frac{2-(\overline{mn}+\overline{n})z}{1-(\overline{mn}+\overline{n})z}=\frac{2}{1-\overline{n}(\overline{m}z+\overline{n})}.$$

Noting that the coefficient of constant term must be 0, then ${\overline{n}}^2=0$, which means that $n=0$. Thus, $\varphi \left(z\right)=mz$ with $\left|m\right|\le 1$. Lemma 1 assures that $C_{\varphi }$ is normal.

Conversely, assume that $C_{\varphi }$ is normal. Lemma 3 gives that $C_{\varphi }^{*}$ is 2-complex symmetric with J. □

Furthermore, we prove that there is no 2-complex symmetric composition operators with J which are induced linear fractional self-maps with $a=0$ and $\varphi \left(0\right)\ne 0$.

Lemma 6.

Let $\varphi :\mathbb{D}\to \mathbb{D}$ be a constant function. Then, $C_{\varphi }$$(C_{\varphi }^{*})$ is 2-complex symmetric with J if and only if $\varphi \left(z\right)\equiv 0$.

Proof. 

The sufficiency is obvious.

Now, we assume that $\varphi \left(z\right)\equiv c$ for some $c\in \mathbb{D}$ and $C_{\varphi }$ is 2-complex symmetric with J. Then, we obtain that

$$J{C}_{\varphi }^2{K}_w\left(z\right)=J{K}_w\left({\varphi }_2\left(z\right)\right)=J\frac{1}{1-\overline{w}c}=\frac{1}{1-w\overline{c}},$$

$$C_{\varphi }^{*}J{C}_{\varphi }{K}_w\left(z\right)=C_{\varphi }^{*}J{K}_w\left(\varphi \left(z\right)\right)=\frac{1}{1-w\overline{c}}{C}_{\varphi }^{*}{K}_0\left(z\right)=\frac{1}{(1-w\overline{c})(1-\overline{c}z)},$$

$${C_{\varphi }^{*}}^{2}J{K}_w\left(z\right)=K_{{\varphi }_2\left(\overline{w}\right)}\left(z\right)=\frac{1}{1-\overline{c}z}$$

and

$$J{C}_{\varphi }J{C}_{\varphi }^{*}J{K}_w\left(z\right)=K_{\varphi (\overline{w})}(\overline{\varphi \left(\overline{z}\right)})=\frac{1}{1-{\overline{c}}^2}$$

for any $w,z\in \mathbb{D}$. Since $C_{\varphi }$ is 2-complex symmetric with J, we obtain

$$\frac{1}{1-w\overline{c}}-\frac{2}{(1-w\overline{c})(1-\overline{c}z)}+\frac{1}{1-\overline{c}z}=0$$

for any $w,z\in \mathbb{D}$. By a simple calculation, we see that $c=0$.

Assume that $C_{\varphi }^{*}$ is 2-complex symmetric with J. Similarly, we have

$$\frac{1}{1-w\overline{c}}-\frac{2}{1-{\overline{c}}^2}+\frac{1}{1-\overline{c}z}=0$$

for any $w,z\in \mathbb{D}$. Therefore, $c=0$. The proof is complete. □

Theorem 5.

Let $\varphi \left(z\right)=\frac{az+b}{cz+d}$ be a linear fractional self-map of $\mathbb{D}$ such that $a=0$ and $\varphi \left(0\right)\ne 0$. Then, $C_{\varphi }$ is not 2-complex symmetric with J.

Proof. 

Since $a=0$ and $\varphi \left(0\right)\ne 0$, we set $\varphi \left(z\right)=\frac{1}{mz+n}$, where $m=\frac{c}{b}$ and $n=\frac{d}{b}$. Then, $\sigma \left(z\right)=\frac{-\overline{m}}{-z+\overline{n}}$, $g\left(z\right)=\frac{1}{-z+\overline{n}}$ and $h\left(z\right)=mz+n$. When $m=0$, $\varphi \left(z\right)=\frac{1}{n}\ne 0$, then Lemma 6 gives that $C_{\varphi }$ is not 2-complex symmetric with J. Now, we assume that $m\ne 0$, $n\ne 0$ and $C_{\varphi }$ is 2-complex symmetric with J. Note that

$${\varphi }_2\left(z\right)=\frac{mz+n}{m+mnz+n^2},h\left(\frac{1}{\overline{\sigma \left(z\right)}}\right)=\overline{z},\varphi (\overline{\sigma \left(z\right)})=\frac{-\overline{z}+n}{-m^2-n\overline{z}+n^2}.$$

Lemma 2 gives that

$$\begin{array}{c}\hfill J{C}_{\varphi }^2{K}_w\left(z\right)=K_{\overline{w}}(\overline{{\varphi }_2\left(\overline{z}\right)})=\frac{1}{1-w\frac{\overline{m}z+\overline{n}}{\overline{m}+\overline{mn}z+{\overline{n}}^2}}=\frac{\overline{m}+\overline{mn}z+{\overline{n}}^2}{\overline{m}+\overline{mn}z+{\overline{n}}^2-(\overline{m}z+\overline{n})w},\end{array}$$

(18)

$$\begin{array}{cc}\hfill C_{\varphi }^{*}J{C}_{\varphi }{K}_w\left(z\right)=& -\overline{m}\frac{g\left(w\right)}{\sigma \left(w\right)}{K}_{\varphi \left(0\right)}\left(z\right)+\overline{h\left(\frac{1}{\overline{\sigma \left(w\right)}}\right)}g\left(w\right)K_{\varphi (\overline{\sigma \left(w\right)})}\left(z\right)\hfill \\ \hfill =& K_{\varphi \left(0\right)}\left(z\right)+\frac{w}{-w+\overline{n}}{K}_{\varphi (\overline{\sigma \left(w\right)})}\left(z\right)\hfill \\ \hfill =& \frac{1}{1-\frac{z}{\overline{n}}}+\frac{w}{-w+\overline{n}}·\frac{1}{1-\frac{-w+\overline{n}}{-{\overline{m}}^2-\overline{n}w+{\overline{n}}^2}z}\hfill \\ \hfill =& \frac{\overline{n}}{\overline{n}-z}+\frac{w}{-w+\overline{n}}·\frac{-{\overline{m}}^2-\overline{n}w+{\overline{n}}^2}{-{\overline{m}}^2-\overline{n}w+{\overline{n}}^2-(-w+\overline{n})z}\hfill \end{array}$$

(19)

and

$$\begin{array}{c}\hfill {C_{\varphi }^{*}}^{2}J{K}_w\left(z\right)=K_{{\varphi }_2\left(\overline{w}\right)}\left(z\right)=\frac{1}{1-z\frac{\overline{m}w+n}{\overline{m}+\overline{mn}w+{\overline{n}}^2}}=\frac{\overline{m}+\overline{mn}w+{\overline{n}}^2}{\overline{m}+\overline{mn}w+{\overline{n}}^2-(\overline{m}w+\overline{n})z}\end{array}$$

(20)

for any $w,z\in \mathbb{D}$. Since $C_{\varphi }$ is 2-complex symmetric with J, we obtain from (18)–(20) that

$$\begin{array}{cc}\hfill & \frac{\overline{m}+\overline{mn}z+{\overline{n}}^2}{\overline{m}+\overline{mn}z+{\overline{n}}^2-(\overline{m}z+\overline{n})w}+\frac{\overline{m}+\overline{mn}w+{\overline{n}}^2}{\overline{m}+\overline{mn}w+{\overline{n}}^2-(\overline{m}w+n)z}\hfill \\ \hfill =& 2\left(\frac{\overline{n}}{\overline{n}-z}+\frac{w}{-w+\overline{n}}·\frac{-{\overline{m}}^2-\overline{n}w+{\overline{n}}^2}{-{\overline{m}}^2-\overline{n}w+{\overline{n}}^2-(-w+\overline{n})z}\right)\hfill \end{array}$$

(21)

for any $w,z\in \mathbb{D}$. Taking $w=0$ in (21), we have that

$$\begin{array}{c}\hfill \frac{(-\overline{n}z+\overline{m}+{\overline{n}}^2)+(\overline{m}+{\overline{n}}^2)}{-\overline{n}z+\overline{m}+{\overline{n}}^2}=\frac{2\overline{n}}{\overline{n}-kz}\end{array}$$

for any $z\in \mathbb{D}$. Thus, the coefficient of $z^2$ must be 0. This implies that $n=0$, which is a contradiction. The proof is complete. □

In the remainder of this paper, we consider 2-complex symmetric composition operators with J which are induced linear fractional self-maps with $a\ne 0$, $c\ne 0$ and $\varphi \left(0\right)\ne 0$ and obtain that all composition operators are not 2-complex symmetric with J.

Theorem 6.

Let $\varphi \left(z\right)=\frac{az+b}{cz+d}$ be a linear fractional self-map of $\mathbb{D}$ such that $a\ne 0$, $c\ne 0$ and $\varphi \left(0\right)\ne 0$. Then, $C_{\varphi }$ is not 2-complex symmetric with J.

Proof. 

We prove it by contradiction. Assume that $C_{\varphi }$ is 2-complex symmetric with J. Since $a\ne 0$, $c\ne 0$ and $\varphi \left(0\right)\ne 0$, set $\varphi \left(z\right)=\frac{mz+n}{sz+1}$, where $m=\frac{a}{d}$, $n=\frac{b}{d}$ and $s=\frac{c}{d}$. Then, $\sigma \left(z\right)=\frac{\overline{m}z-\overline{s}}{-\overline{n}z+1}$, $g\left(z\right)=\frac{1}{-\overline{n}z+1}$ and $h\left(z\right)=sz+1$. After a calculation, we obtain

$${\varphi }_2\left(z\right)=\frac{(m^2+ns)z+mn+n}{(ms+s)z+ns+1},h\left(\frac{1}{\overline{\sigma \left(z\right)}}\right)=\frac{(m-ns)\overline{z}}{m\overline{z}-s}$$

and

$$\varphi (\overline{\sigma \left(z\right)})=\frac{(m^2-n^2)\overline{z}+n-ms}{(ms-n)\overline{z}+1-s^2}.$$

Lemma 2 gives that

$$\begin{array}{cc}\hfill J{C}_{\varphi }^2{K}_w\left(z\right)& =K_{\overline{w}}(\overline{{\varphi }_2\left(\overline{z}\right)})=\frac{1}{1-w\frac{({\overline{m}}^2+\overline{ns})z+\overline{mn}+\overline{n}}{(\overline{ms}+\overline{s})z+\overline{ns}+1}}\hfill \\ & =\frac{(\overline{ms}+\overline{s})z+\overline{ns}+1}{(\overline{ms}+\overline{s})z+\overline{ns}+1-[({\overline{m}}^2+\overline{ns})z+\overline{mn}+\overline{n}]w},\hfill \end{array}$$

(22)

$$\begin{array}{cc}\hfill C_{\varphi }^{*}J{C}_{\varphi }{K}_w\left(z\right)=& -\overline{s}\frac{g\left(w\right)}{\sigma \left(w\right)}{K}_{\varphi \left(0\right)}\left(z\right)+\overline{h\left(\frac{1}{\overline{\sigma \left(w\right)}}\right)}g\left(w\right)K_{\varphi (\overline{\sigma \left(w\right)})}\left(z\right)\hfill \\ \hfill =& \frac{-\overline{s}}{\overline{m}w-\overline{s}}\frac{1}{1-\overline{n}z}+\frac{(\overline{m}-\overline{ns})w}{\overline{m}w-\overline{s}}\frac{1}{-\overline{n}w+1}·\frac{1}{1-\frac{({\overline{m}}^2-{\overline{n}}^2)w-\overline{ms}+\overline{n}}{(\overline{ms}-\overline{n})w+1-{\overline{s}}^2}z}\hfill \\ \hfill =& \frac{-\overline{s}}{(\overline{m}w-\overline{s})(1-\overline{n}z)}+\frac{(\overline{m}-\overline{ns})w}{(\overline{m}w-\overline{s})(-\overline{n}w+1)}\hfill \\ \hfill & ·\frac{(\overline{ms}-\overline{n})w+1-{\overline{s}}^2}{(\overline{ms}-\overline{n})w+1-{\overline{s}}^2-[({\overline{m}}^2-{\overline{n}}^2)w-\overline{ms}+\overline{n}]z}\hfill \end{array}$$

(23)

and

$$\begin{array}{cc}\hfill {C_{\varphi }^{*}}^{2}J{K}_w\left(z\right)& =K_{{\varphi }_2\left(\overline{w}\right)}\left(z\right)=\frac{1}{1-z\frac{({\overline{m}}^2+\overline{ns})w+\overline{mn}+\overline{n}}{(\overline{ms}+\overline{s})w+\overline{ns}+1}}\hfill \\ & =\frac{(\overline{ms}+\overline{s})w+\overline{ns}+1}{(\overline{ms}+\overline{s})w+\overline{ns}+1-[({\overline{m}}^2+\overline{ns})w+\overline{mn}+\overline{n}]z}\hfill \end{array}$$

(24)

for any $w,z\in \mathbb{D}$. Taking $w=0$, then we obtain from (22)–(24) that

$$J{C}_{\varphi }^2{K}_0\left(z\right)=1,C_{\varphi }^{*}J{C}_{\varphi }{K}_0\left(z\right)=\frac{1}{1-\overline{n}z}$$

and

$${C_{\varphi }^{*}}^{2}J{K}_0\left(z\right)=\frac{\overline{ns}+1}{\overline{ns}+1-(\overline{mn}+\overline{n})z}$$

for any $z\in \mathbb{D}$. Since $C_{\varphi }$ is 2-complex symmetric with J, we obtain that

$$\begin{array}{c}\hfill 1+\frac{\overline{ns}+1}{\overline{ns}+1-(\overline{mn}+\overline{n})z}=\frac{2}{1-\overline{n}z}\end{array}$$

for any $z\in \mathbb{D}$, which implies that

$$\begin{array}{c}\hfill \frac{2\overline{ns}+2-(\overline{mn}+\overline{n})z}{\overline{ns}+1-(\overline{mn}+\overline{n})z}=\frac{2}{1-\overline{n}z}\end{array}$$

for any $z\in \mathbb{D}$. Noting that the coefficient of $z^2$ must be 0, then we have that ${\overline{n}}^2(m+1)=0$. Since $\varphi \left(0\right)\ne 0$, then $m=-1$.

Noting that ${\varphi }_2\left(z\right)=\frac{(m^2+ns)z+mn+n}{(ms+s)z+ns+1}$, then Lemma 4 gives that

$$2{\varphi }_2\left(0\right)-4\varphi \left(0\right)+\varphi \left(0\right){\varphi }_2\left(0\right)=\frac{2(mn+n)}{ns+1}-4n+\frac{n(mn+n)}{ns+1}=0.$$

By a simple calculation, we see that $2m-4ns+mn+n=2$. Since $m=-1$, we obtain that $ns=-1$. Therefore, $m-ns=0$, which means that $\varphi$ is not a linear fractional self-map of $\mathbb{D}$, a contradiction. The proof is complete. □

5. Conclusions

In this paper, we obtained some characterizations for composition operators $C_{\varphi }$ to be 2-complex symmetric with respect to the conjugation J on the Hardy space $H^2$. To be specific, in Theorem 1, we obtain the necessary and sufficient condition for $C_{\varphi }$ to be 2-complex symmetric with J when $\varphi$ is an automorphism of $\mathbb{D}$. Next, we discuss 2-complex symmetric composition operators $C_{\varphi }$ which were induced by linear fractional self-maps of $\mathbb{D}$ with four different forms. The necessary and sufficient conditions for $C_{\varphi }$ to be 2-complex symmetric with J in the first two cases are given in Theorems 2 and 3, respectively. In Theorems 5 and 6, we obtain that there is no 2-complex symmetric composition operators with J in the latter two cases.