Complex symmetric Toeplitz operators on the generalized derivative Hardy space

Abstract

The generalized derivative Hardy space $S^{2}_{\alpha ,\beta}(\mathbb{D})$ consists of all functions whose derivatives are in the Hardy and Bergman spaces as follows:

for positive integers α, β,

$$ S^{2}_{\alpha ,\beta}(\mathbb{D})= \biggl\{ f\in H(\mathbb{D}) : \Vert {f} \Vert ^{2}_{S^{2}_{ \alpha ,\beta}}= \Vert {f} \Vert ^{2}_{H^{2}}+{ \frac{{\alpha +\beta}}{\alpha \beta}} \bigl\Vert {f'} \bigr\Vert ^{2}_{A^{2}}+ \frac{1}{\alpha \beta} \bigl\Vert {f'} \bigr\Vert ^{2}_{H^{2}}< \infty \biggr\} , $$

where $H({\mathbb{D}})$ denotes the space of all functions analytic on the open unit disk ${\mathbb{D}}$. In this paper, we study characterizations for Toeplitz operators to be complex symmetric on the generalized derivative Hardy space $S^{2}_{\alpha ,\beta}(\mathbb{D})$ with respect to some conjugations $C_{\xi}$, $C_{\mu , \lambda}$. Moreover, for any conjugation C, we consider the necessary and sufficient conditions for complex symmetric Toeplitz operators with the symbol φ of the form $\varphi (z)=\sum_{n=1}^{\infty}\overline{\hat{\varphi}(-n)} \overline{z}^{n}+\sum_{n=0}^{\infty}\hat{\varphi}(n)z^{n}$. Next, we also study complex symmetric Toeplitz operators with non-harmonic symbols on the generalized derivative Hardy space $S^{2}_{\alpha ,\beta}(\mathbb{D})$.

1 Introduction

Let ${\mathcal {H}}$ be a separable complex Hilbert space and let ${\mathcal{{L}}}({\mathcal{{H}}})$ be the algebra of bounded linear operators on ${\mathcal{{H}}}$. We say that an anti-linear operator C on $\mathcal{H}$ is a conjugation if ${C}^{2}=I$ and $\langle {C}x, {C}y \rangle =\langle y, x\rangle$ for all $x,y\in {\mathcal{H}}$. If C is a conjugation on ${\mathcal {H}}$, there exists an orthonormal basis $\{e_{n}\}_{n=0}^{\infty}$ for ${\mathcal{H}}$ such that ${C}e_{n}=e_{n}$ for all n (see [5]). We say that an operator $T\in{\mathcal{L(H)}}$ is complex symmetric if $T= {C}T^{\ast}{C}$ for a conjugation operator C on ${\mathcal{H}}$. The topic of complex symmetric operators, which includes all truncated Toeplitz operators, Hankel operators, normal operators, and some Volterra integration operators, has been studied by many authors (see [4, 5], and [8] for more details).

For the open unit disk $\mathbb{D}$ in ${\mathbb{C}}$, let $H({\mathbb{D}})$ be the space of all analytic functions on ${\mathbb{D}}$. Let $L^{2}(\mathbb{D}, dA)$ be a Hilbert space with the inner product

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

where $f, g \in L^{2}(\mathbb{D}, dA)$ and dA is the area measure of ${\mathbb{D}}$. The Hilbert Hardy space $H^{2}({\mathbb{D}})$ contains all functions f analytic on $\mathbb{D}$ with

$$ f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}, \quad \text{where } \sum_{n=0}^{ \infty} \vert a_{n} \vert ^{2}< \infty .$$

The Bergman space $A^{2}(\mathbb{D})$ consists of the space of analytic functions f in $L^{2}(\mathbb{D}, dA)$ with

$$ f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}, \quad \text{where } \sum_{n=0}^{ \infty} \frac{1}{n+1} \vert a_{n} \vert ^{2}< \infty .$$

The Dirichlet space $D^{2}(\mathbb{D})$ is given by

$$ D^{2}(\mathbb{D})=\Biggl\{ f\in H(\mathbb{D}): \Vert f \Vert _{D^{2}}^{2}= \Vert f \Vert _{H^{2}}^{2}+ \bigl\Vert f^{\prime } \bigr\Vert _{A^{2}}^{2}= \sum_{n=0}^{\infty}(n+1) \vert f_{n} \vert ^{2}< \infty \Biggr\} .$$

The reproducing kernels of the spaces $H^{2}(\mathbb{D})$, $A^{2}(\mathbb{D})$, and $D^{2}(\mathbb{D})$ have the following forms:

$$ K^{1}_{w}(z)=\frac{1}{1-\overline{w}z}, \qquad K^{2}_{w}(z)= \frac{1}{(1-\overline{w}z)^{2}}, \quad \text{and} \quad K^{3}_{w}(z)= \frac{1}{\overline{w}z} \ln \frac{1}{1-\overline{w}z}, $$

respectively. Many authors in [1–3] and [11] studied intensively multiplication and Toeplitz operators on the Hardy space, Bergman space, and Dirichlet space.

In 2019, Gu and Luo [6] introduced the derivative Hardy space $S^{2}_{1}(\mathbb{D})$ as follows:

$$ \begin{aligned} S^{2}_{1}(\mathbb{D})&= \biggl\{ f\in H(\mathbb{D}) : \Vert f \Vert ^{2}_{S^{2}_{1}}= \Vert f \Vert ^{2}_{H^{2}}+{\frac{3}{2}} \bigl\Vert f' \bigr\Vert ^{2}_{A^{2}}+ \frac{1}{2} \bigl\Vert f' \bigr\Vert ^{2}_{H^{2}}< \infty \biggr\} \\ &= \Biggl\{ f\in H(\mathbb{D}) : \Vert f \Vert ^{2}_{S^{2}_{1}}= \sum_{n=0}^{\infty} \frac{(n+1)(n+2)}{2} \vert a_{n} \vert ^{2}< \infty \Biggr\} , \end{aligned} $$

where $f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}$. The reproducing kernel of the derivative Hardy space $S^{2}_{1}(\mathbb{D})$ is given by

$$ K_{w}(z)=\frac{2}{(\overline{w}z)^{2}}\biggl[\overline{w}z+( \overline{w}z-1) \ln \frac{1}{1-\overline{w}z}\biggr].$$

Recently, the authors in [9] defined the generalized derivative Hardy space $S^{2}_{\alpha ,\beta}(\mathbb{D})$ for $\alpha ,\beta \in{\mathbb{N}}$ as

$$\begin{aligned} S^{2}_{\alpha ,\beta}(\mathbb{D}) =& \biggl\{ f\in H(\mathbb{D}) : \Vert f \Vert ^{2}_{S^{2}_{ \alpha ,\beta}}= \Vert f \Vert ^{2}_{H^{2}}+{ \frac{{\alpha +\beta}}{\alpha \beta}} \bigl\Vert f' \bigr\Vert ^{2}_{A^{2}}+ \frac{1}{\alpha \beta} \bigl\Vert f' \bigr\Vert ^{2}_{H^{2}}< \infty \biggr\} \\ =& \Biggl\{ f\in H(\mathbb{D}) : \Vert f \Vert ^{2}_{S^{2}_{\alpha ,\beta}}= \sum_{n=0}^{ \infty}\frac{(n+\alpha )(n+\beta )}{\alpha \beta} \vert f_{n} \vert ^{2}< \infty \Biggr\} , \end{aligned}$$

where $f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}$. Since $S^{2}_{\alpha ,\beta}(\mathbb{D})=S^{2}_{\beta ,\alpha}(\mathbb{D})$ clearly holds, we focus on the space $S^{2}_{\alpha ,\beta}(\mathbb{D})$ for $\alpha <\beta$. Especially, if $\alpha =1$ and $\beta =2$, then $S^{2}_{\alpha ,\beta}(\mathbb{D})$ becomes $S^{2}_{1}(\mathbb{D})$.

Let $L^{\infty}(\mathbb{D})$ be the set of all essentially bounded measurable functions in $\mathbb{D}$, and let P be the orthogonal projection from $L^{2} (\mathbb{D}, dA)$ onto $S^{2}_{\alpha ,\beta}(\mathbb{D})$. For $\varphi \in L^{\infty}(\mathbb{D})$, the Toeplitz operator $T_{\varphi}$ on $S^{2}_{\alpha ,\beta}(\mathbb{D})$ is defined by

$$ T_{\varphi}f := P({\varphi}\cdot f)\quad \text{for } f\in S^{2}_{\alpha , \beta}(\mathbb{D}). $$

Note that, for $\varphi , \psi \in L^{\infty}(\mathbb{D})$, from the definition of the Toeplitz operator, $T_{\varphi +\psi}=T_{\varphi}+T_{\psi} \ \text{ and } \ T^{*}_{ \varphi}=T_{\overline{\varphi}}$ where φ̅ is a complex conjugation of φ. The reproducing kernel $K_{w}(z)$ of the space $S^{2}_{\alpha ,\beta}(\mathbb{D})$ is

$$\begin{aligned} K_{w}(z)={}&\frac{\alpha \beta}{\beta -\alpha} \ln \frac{1}{1-\overline{w}z} \biggl( \frac{1}{(\overline{w}z)^{\alpha}}- \frac{1}{(\overline{w}z)^{\beta}} \biggr) \\ &{}+\frac{\alpha \beta}{\beta -\alpha} \biggl\{ \frac{1}{(\overline{w}z)^{\alpha}}\delta _{\alpha -1}( \overline{w}z)- \frac{1}{(\overline{w}z)^{\beta}}\delta _{\beta -1}(\overline{w}z) \biggr\} , \end{aligned}$$

where $\delta _{m}(z)=\sum_{r=1}^{m}\frac{_{m}C_{r}(-1)^{r}}{r}{[(1-z)^{r}-1]}$ (see [9, Lemma 2.1]). Thus we have

$$ (T_{\varphi}f) (z)= \int _{\mathbb{D}}{\varphi (\omega )f(\omega )}K_{z}(w)\,dA( \omega ) $$

for $f \in S^{2}_{\alpha ,\beta}(\mathbb{D})$ and $\omega \in \mathbb{D}$.

This paper is organized as follows. First, we study characterizations for Toeplitz operators to be complex symmetric on the generalized derivative Hardy space $S^{2}_{\alpha ,\beta}(\mathbb{D})$ with respect to some conjugations. Moreover, we also focus on complex symmetric Toeplitz operators with non-harmonic symbols on the generalized derivative Hardy space $S^{2}_{\alpha ,\beta}(\mathbb{D})$.

2 Complex symmetric Toeplitz operators

In this section, we study complex symmetry of Toeplitz operators on the space $S^{2}_{\alpha ,\beta}(\mathbb{D})$. For the convenience of readers, we begin with the following lemma which comes from [9]. Let ${\mathbb{N}}$ be the natural numbers and let ${\mathbb{N}}_{0}={\mathbb{N}}\cup \{0\}$.

Lemma 2.1

([9])

For $s, t\in{{\mathbb{N}}_{0}}$, the following statements hold:

$$\begin{aligned}& \mathrm{(i)}\quad \bigl\langle {z}^{t}, z^{s}\bigr\rangle = \textstyle\begin{cases} \frac{(s+{\alpha })(s+{\beta })}{{\alpha }{\beta }} & \textit{if } s = t, \\ 0 & \textit{if } s \neq t. \end{cases}\displaystyle \\& \mathrm{(ii)}\quad P\bigl(\overline{z}^{t} z^{s} \bigr)= \textstyle\begin{cases} \frac{(s+{\alpha })(s+{\beta })}{(s-t+{\alpha })(s-t+{\beta })} z^{s-t} &\textit{if } s \geq t, \\ 0 &\textit{if } s < t. \end{cases}\displaystyle \end{aligned}$$

Remark 2.2

We mentioned in [9] that there is the difference between the Hardy space $H^{2}({\mathbb{D}})$ (or Bergman space $A^{2}({\mathbb{D}})$) and the generalized derivative Hardy space $S^{2}_{\alpha ,\beta}(\mathbb{D})$. Indeed, for $s, t\in{{\mathbb{N}}_{0}}$, the inequality $\|\overline{z}^{t}z^{s}\|\ge \|P(\overline{z}^{t}z^{s})\|$ holds on $H^{2}({\mathbb{D}})$ and $A^{2}({\mathbb{D}})$. However, it holds that

$$ \bigl\Vert \overline{z}^{t}z^{s} \bigr\Vert \le \bigl\Vert P\bigl(\overline{z}^{t}z^{s}\bigr) \bigr\Vert \quad \text{on } S^{2}_{ \alpha ,\beta}(\mathbb{D})$$

because of $\frac{(s+{\alpha })(s+{\beta })}{(s-t+{\alpha })(s-t+{\beta })}>1$.

Theorem 2.3

For $n\in{{\mathbb{N}}_{0}}$, let $\{e_{n}\}$ on $S^{2}_{\alpha ,\beta}(\mathbb{D})$ be given by

$$ e_{n}(z)=\sqrt{\frac{\alpha \beta}{(n+\alpha )(n+\beta )}} z^{n}\quad \textit{for }z\in{\mathbb{D}}. $$

If C is anti-linear on $S^{2}_{\alpha ,\beta}(\mathbb{D})$ such that $Ce_{n}=\overline{\delta _{n}}e_{n}$ with $|\delta _{n}|=1$, then the following statements hold:

• (i) Parseval’s identity $\sum_{n=0}^{\infty}|\langle f, Ce_{n}\rangle |^{2}=\sum_{n=0}^{ \infty}|\langle f, e_{n}\rangle |^{2}=\|f\|_{2}^{2}$ holds for every $f\in S^{2}_{\alpha ,\beta}(\mathbb{D})$.

• (ii) The set of functions $\{ Ce_{n}(z):=\sqrt{\frac{\alpha \beta}{(n+\alpha )(n+\beta )}}Cz^{n} \}$ forms an orthonormal basis for $S^{2}_{\alpha ,\beta}(\mathbb{D})$.

Proof

(i) From Lemma 2.1, we have

$$ \langle e_{n}, e_{m} \rangle =\sqrt{ \frac{\alpha \beta}{(n+\alpha )(n+\beta )}}\sqrt{ \frac{\alpha \beta}{(m+\alpha )(m+\beta )}} \bigl\langle z^{n}, z^{m} \bigr\rangle =\delta _{nm}, $$

(1)

where $\delta _{nm}=1$ if $n=m$, and $\delta _{nm}=0$ if $n\neq m$. Thus $\{e_{n}\}$ is an orthonormal sequence for $S^{2}_{\alpha ,\beta}(\mathbb{D})$. Since C is anti-linear on $S^{2}_{\alpha ,\beta}(\mathbb{D})$ such that $Ce_{n}={\overline{\delta _{n}}}e_{n}$ with $|\delta _{n}|=1$, it follows that C is a conjugation on $S^{2}_{\alpha ,\beta}(\mathbb{D})$. Hence

$$\begin{aligned} \langle Ce_{n}, Ce_{m} \rangle =&\langle e_{m}, e_{n}\rangle \\ =&\sqrt{\frac{\alpha \beta}{(m+\alpha )(m+\beta )}} \sqrt{ \frac{\alpha \beta}{(n+\alpha )(n+\beta )}}\bigl\langle z^{m}, z^{n} \bigr\rangle =\delta _{mn}. \end{aligned}$$

First, we will show that Parseval’s identity $\sum_{n=0}^{\infty}|\langle f, Ce_{n}\rangle |^{2}=\|f\|^{2}$ holds for every $f\in S^{2}_{\alpha ,\beta}(\mathbb{D})$. Let $Cf(z)=\sum_{k=0}^{\infty}\widetilde{a_{k}} z^{k}$. Then

$$ \Vert f \Vert ^{2}= \Vert Cf \Vert ^{2}=\sum _{k=0}^{\infty} \frac{(k+\alpha )(k+\beta )}{\alpha \beta} \vert \widetilde{a_{k}} \vert ^{2}$$

and

$$\begin{aligned} \bigl\langle f(z),Ce_{n}(z)\bigr\rangle =&\bigl\langle e_{n}(z),Cf(z)\bigr\rangle \\ =& \Biggl\langle \sqrt{\frac{\alpha \beta}{(n+\alpha )(n+\beta )}}z^{n}, \sum _{k=0}^{\infty}\widetilde{a_{k}} z^{k} \Biggr\rangle \\ =&\sqrt{\frac{\alpha \beta}{(n+\alpha )(n+\beta )}}\sum_{k=0}^{ \infty} \overline{\widetilde{a_{k}}}\bigl\langle z^{n}, z^{k}\bigr\rangle \\ =&\sqrt{\frac{\alpha \beta}{(n+\alpha )(n+\beta )}} \overline{\widetilde{a_{n}}}\bigl\langle z^{n}, z^{n}\bigr\rangle \\ =&\sqrt{\frac{(n+\alpha )(n+\beta )}{\alpha \beta}} \overline{\widetilde{a_{n}}}. \end{aligned}$$

Therefore

$$ \sum_{n=0}^{\infty} \bigl\vert \langle f, Ce_{n}\rangle \bigr\vert ^{2}=\sum _{n=0}^{ \infty}\frac{(n+\alpha )(n+\beta )}{\alpha \beta} \vert \widetilde{a_{n}} \vert ^{2}= \Vert f \Vert ^{2}, $$

and by a similar calculation, we get $\|f\|^{2}=\sum_{n=0}^{\infty}|\langle f, e_{n}\rangle |^{2}$. Hence Parseval’s identity holds.

(ii) Since Parseval’s identity holds by (i), $f=\sum_{n=0}^{\infty}\langle f,Ce_{n}\rangle Ce_{n}$ for every $f\in S^{2}_{\alpha ,\beta}(\mathbb{D})$. Hence $\{Ce_{n}\}$ forms an orthonormal basis for $S^{2}_{\alpha ,\beta}(\mathbb{D})$. □

Especially, if $\alpha =1$ and $\beta =2$ in Theorem 2.3, then we get the following result.

Corollary 2.4

For $n\in{{\mathbb{N}}_{0}}$, let $\{e_{n}\}$ on $S^{2}_{1}(\mathbb{D})$ be given by

$$ e_{n}(z)=\sqrt{\frac{2}{(n+1)(n+2)}} z^{n}\quad \textit{for } z\in{\mathbb{D}}. $$

Let C be anti-linear on $S^{2}_{1}(\mathbb{D})$ such that $Ce_{n}=\overline{\delta _{n}}e_{n}$ with $|\delta _{n}|=1$. Then Parseval’s identity $\sum_{n=0}^{\infty}|\langle f, Ce_{n}\rangle |^{2}=\sum_{n=0}^{ \infty}|\langle f, e_{n}\rangle |^{2}=\|f\|_{2}^{2}$ holds for every $f\in S^{2}_{1}(\mathbb{D})$. Moreover, the set of functions $\{ Ce_{n}(z):=\sqrt{\frac{2}{(n+1)(n+2)}}Cz^{n}\}$ forms an orthonormal basis for $S^{2}_{1}(\mathbb{D})$.

In 2016, the authors in [8] introduced the conjugation $C_{\mu ,\lambda}$ on the Hardy space $H^{2}$ as in (3). Remark that the space $H^{2}(\mathbb{D})$ has the reproducing kernel $K^{1}_{w}(z)$ and the normalized reproducing kernel $k_{w}(z)$ given by

$$ K^{1}_{w}(z)=\frac{1}{1-\overline{w}z}\quad \text{and}\quad k_{w}(z)= \frac{\sqrt{1- \vert w \vert ^{2}}}{1-\overline{w}z}\quad \text{for } w\in{\mathbb{D}},$$

respectively. Recently, the authors in [10] gave the conjugation $C_{\xi}$ which has the form as in (2) on the Hardy space $H^{2}(\mathbb{D})$. We can easily show that the following operator as in (2) is the conjugation on $S^{2}_{\alpha ,\beta}(\mathbb{D})$.

Lemma 2.5

(i) Let $\xi \in{\mathbb{R}}$ be with $|\xi |<1$. Assume that the operator $C_{\xi}$ is defined by

$$ C_{\xi}f(z)=-k_{\xi}(z)\overline{f\bigl( \psi _{\xi}(\overline{z})\bigr)} $$

(2)

for some $f\in S^{2}_{\alpha ,\beta}(\mathbb{D})$ where $\psi _{\xi}(z)=\frac{\xi -z}{1-\overline{\xi}z}$. Then $C_{\xi}$ is a conjugation on $S^{2}_{\alpha ,\beta}(\mathbb{D})$.

(ii) For every μ and λ in $\mathbb{C}$ with $|\mu |=|\lambda |=1$, let $C_{\mu ,\lambda}: {S^{2}_{\alpha ,\beta}(\mathbb{D})}\rightarrow S^{2}_{ \alpha ,\beta}(\mathbb{D})$ be given by

$$ C_{\mu ,\lambda}f(z)=\mu \overline{f(\lambda \overline{z})}. $$

(3)

Then $C_{\mu ,\lambda}$ is a conjugation on $S^{2}_{\alpha ,\beta}(\mathbb{D})$.

Now, we establish a necessary and sufficient condition for a Toeplitz operator $T_{\varphi}$ on $S^{2}_{\alpha ,\beta}(\mathbb{D})$ to be complex symmetric with respect to the above conjugations.

Theorem 2.6

Let $\varphi \in L^{\infty}(\mathbb{D})$. Then the following statements hold:

(i) If $\xi \in{\mathbb{R}}$ with $|\xi |<1$, then $T_{\varphi}$ on $S^{2}_{\alpha ,\beta}(\mathbb{D})$ is complex symmetric with the conjugation $C_{\xi}$ if and only if $\varphi (z)=\varphi (\overline{\psi _{\xi}(z)})$, where $\psi _{\xi}(z)=\frac{\xi -z}{1-\overline{\xi}z}$.

(ii) $T_{\varphi}$ on $S^{2}_{\alpha ,\beta}(\mathbb{D})$ is complex symmetric with the conjugation $C_{\mu ,\lambda}$ if and only if $\varphi (z)=\varphi (\lambda \overline{z})$.

Proof

(i) Let $T_{\varphi}$ on $S^{2}_{\alpha ,\beta}(\mathbb{D})$ be complex symmetric with the conjugation $C_{\xi}$. Since ξ is real, it follows that $|1-\xi \overline{z}|=|1-\xi{z}|$, and so $|k_{\xi}(\overline{\psi _{\xi}({z})})|^{2} |k_{\xi}(z)|^{2}=1$. Thus, by Lemma 2.5, we have

$$\begin{aligned} \bigl\langle T_{{\varphi}}^{\ast}f, g \bigr\rangle =&\langle C_{\xi}T_{\varphi}C_{ \xi}f, g \rangle \\ =&\langle C_{\xi}g, T_{\varphi}C_{\xi}f \rangle = \bigl\langle C_{\xi}g, P({ \varphi}C_{\xi}f) \bigr\rangle = \langle C_{\xi}g, {\varphi}C_{\xi}f \rangle \\ =& \int _{\mathbb{D}}k_{\xi}(z)\overline{g\bigl(\psi _{\xi}(\overline{z})\bigr)} \cdot \overline{\varphi (z)} \overline{k_{\xi}(z)} {f\bigl(\psi _{\xi}( \overline{z}) \bigr)} \,dA(z) \\ =& \int _{\mathbb{D}} \bigl\vert k_{\xi}(\overline{z}) \bigr\vert ^{2} \overline{g\bigl(\psi _{\xi}({z})\bigr)} \overline{\varphi (\overline{z})} {f\bigl( \psi _{\xi}({z})\bigr)} \,dA(z) \\ =& \int _{\mathbb{D}} \bigl\vert k_{\xi}\bigl(\overline{ \psi _{\xi}({z})}\bigr) \bigr\vert ^{2} \overline{\varphi \bigl(\overline{\psi _{\xi}(z)}\bigr)} {f(z)}\overline{g(z)} \bigl\vert k_{ \xi}(z) \bigr\vert ^{2}\,dA(z) \\ =& \int _{\mathbb{D}}\overline{\varphi \bigl(\overline{\psi _{\xi}(z)}\bigr)} {f(z)} \overline{g(z)} \,dA(z) \\ =&\bigl\langle \overline{\varphi \bigl(\overline{\psi _{\xi}(z)}\bigr)}f,g \bigr\rangle \\ =&\bigl\langle f,P\bigl(\varphi \bigl(\overline{\psi _{\xi}(z)} \bigr)\bigr)g \bigr\rangle \end{aligned}$$

for $f,g\in S^{2}_{\alpha ,\beta}(\mathbb{D})$. Hence we get that $T_{\varphi}g=T_{\varphi (\overline{\psi _{\xi}(z)})}g$ for all $g\in S^{2}_{\alpha ,\beta}(\mathbb{D})$ and then $\varphi (z)=\varphi (\overline{\psi _{\xi}(z)})$. The converse implications clearly hold by a similar method.

(ii) We claim that if P denotes the orthogonal projection of $L^{2}$ onto $S^{2}_{\alpha ,\beta}(\mathbb{D})$, then the operators $C_{\mu ,\lambda}$ and P commute.

Since $e_{n}(z)=\sqrt{\frac{\alpha \beta}{(n+\alpha )(n+\beta )}}z^{n}$, it follows that, for $n\geq 0$,

$$\begin{aligned} PC_{\mu ,\lambda}e_{n} =&P\mu \overline{ \lambda}^{n}e_{n} \\ =&{ \sqrt{\frac{\alpha \beta}{(n+\alpha )(n+\beta )}}}\mu \overline{\lambda}^{n} z^{n} \\ =&{ \sqrt{\frac{\alpha \beta}{(n+\alpha )(n+\beta )}}}C_{\mu , \lambda}z^{n} =C_{\mu ,\lambda}Pe_{n}, \end{aligned}$$

and for $n<0$, we have that $PC_{\mu ,\lambda}e_{n}= P\mu \overline{\lambda}^{n}e_{n}=0=C_{\mu , \lambda}Pe_{n}$. Hence

$$ C_{\mu ,\lambda}P=PC_{\mu ,\lambda}. $$

(4)

Thus we complete the proof for the claim. By a similar way of the proof of [7, Theorem 2.2], we know from (4) that $T_{\varphi}$ on $S^{2}_{\alpha ,\beta}(\mathbb{D})$ is complex symmetric with the conjugation $C_{\mu ,\lambda}$ if and only if $T_{\varphi}$ on $S^{2}_{\alpha ,\beta}(\mathbb{D})$ is complex symmetric with the conjugation $C_{1,\lambda}$ if and only if $\varphi (z)=\varphi (\lambda \overline{z})$. □

As some applications of Theorem 2.6, we get the following corollaries.

Corollary 2.7

Let $\varphi \in L^{\infty}(\mathbb{D})$ and let the operator $C_{\frac{1}{2}}$ be defined by

$$ C_{\frac{1}{2}}f(z)=- \biggl(\frac{\sqrt{3}}{2-z} \biggr)\cdot \overline{f \biggl(\frac{1-2\overline{z}}{2-\overline{z}} \biggr)}$$

for some $f\in S^{2}_{\alpha ,\beta}(\mathbb{D})$. Then $T_{\varphi}$ on $S^{2}_{\alpha ,\beta}(\mathbb{D})$ is complex symmetric with the conjugation $C_{\frac{1}{2}}$ if and only if $\varphi (z)=\varphi (\frac{1-2\overline{z}}{2-\overline{z}})$.

Corollary 2.8

Let $\varphi (z)=\sum_{n=-\infty}^{\infty}b_{n}z^{n}\in L^{\infty}( \mathbb{D})$. If $T_{\varphi}$ is a Toeplitz operator on $S^{2}_{\alpha ,\beta}(\mathbb{D})$, then $T_{\varphi}$ is complex symmetric with the conjugation $C_{\mu ,\lambda}$ if and only if $\varphi (z)=b_{0}+\sum_{n=1}^{\infty}b_{n}(z^{n}+\lambda ^{n} \overline{z}^{n})\ \textit{with}\ |\lambda |=1$. Moreover, if $T_{\varphi}$ is complex symmetric with the conjugation $C_{\mu ,\lambda}$, then $T_{\varphi}$ is normal if and only if $\lambda ^{n} b_{n}=\overline{b_{n}}=\lambda ^{n}b_{-n}$ for all $n\in{{\mathbb{N}}_{0}}$ with $|\lambda |=1$.

Proof

The proof follows from Theorem 2.6 and [8]. □

Theorem 2.9

Let φ be in $L^{\infty}(\mathbb{D})$ such that $\varphi (z)=\sum_{n=1}^{\infty}\overline{\hat{\varphi}(-n)} \overline{z}^{n}+\sum_{n=0}^{\infty}\hat{\varphi}(n)z^{n}$, and let C be a conjugation on $S^{2}_{\alpha ,\beta}(\mathbb{D})$. Then $T_{\varphi}$ on $S^{2}_{\alpha ,\beta}(\mathbb{D})$ is a complex symmetric operator with the conjugation C if and only if $\hat{\varphi}(-k)=C{{\hat{\varphi}(k)}}$ for all $k\in{{\mathbb{N}}_{0}}$.

Proof

Let $f(z)=\sum_{j=0}^{\infty}a_{j}z^{j}$ and $Cf(z)=\sum_{j=0}^{\infty}\widetilde{a_{j}}z^{j}$. Denote

$$ \varphi _{+}(z)=\sum_{n=0}^{\infty} \hat{\varphi}(n)z^{n}\quad \text{and}\quad \varphi _{-}(z)= \sum_{n=1}^{\infty}\overline{\hat{ \varphi}(-n)} \overline{z}^{n}.$$

By the proof of Theorem 2.3, we obtain that

$$ \textstyle\begin{cases} \varphi _{+}f=\sum_{k=0}^{\infty}\sum_{n=0}^{\infty}\hat{\varphi}(k)a_{n}z^{k+n}, \\ P({\varphi _{-}}f)=\sum_{k=1}^{\infty}\sum_{n=k}^{\infty} \frac{(n+{\alpha })(n+{\beta })}{(n-k+{\alpha })(n-k+{\beta })} \overline{\hat{\varphi}(-k)}a_{n}z^{n-k}, \\ P(\overline{\varphi _{+}}Cf)=\sum_{k=0}^{\infty}\sum_{n=k}^{\infty} \frac{(n+{\alpha })(n+{\beta })}{(n-k+{\alpha })(n-k+{\beta })} { \overline{\hat{\varphi}(k)}}\widetilde{a_{n}}z^{n-k}, \\ \overline{\varphi _{-}}Cf=\sum_{k=1}^{\infty}\sum_{n=0}^{\infty}{ \hat{\varphi}(-k)}\widetilde{a_{n}}z^{k+n}. \end{cases} $$

(5)

Since $T_{\varphi}$ is complex symmetric with the conjugation C if and only if

$$ \varphi _{+}f+P({\varphi _{-}}f)=CP( \overline{\varphi _{+}}Cf)+C( \overline{\varphi _{-}}Cf), $$

(6)

thus equation (6) gives that

$$\begin{aligned} &\sum_{k=0}^{\infty} \sum_{n=0}^{\infty}\hat{\varphi}(k)a_{n}z^{k+n}+ \sum_{k=1}^{\infty}\sum _{n=k}^{\infty} \frac{(n+{\alpha })(n+{\beta })}{(n-k+{\alpha })(n-k+{\beta })} \overline{\hat{ \varphi}(-k)}a_{n}z^{n-k}, \\ &\quad =\sum_{k=0}^{\infty}\sum _{n=k}^{\infty} \frac{(n+{\alpha })(n+{\beta })}{(n-k+{\alpha })(n-k+{\beta })} { \widetilde{ \overline{\hat{\varphi}(k)}}} {a_{n}}z^{n-k} +\sum _{k=1}^{ \infty}\sum_{n=0}^{\infty} \widetilde{\hat{\varphi}(-k)} {a_{n}}z^{k+n}. \end{aligned}$$

(7)

From the constant term in (7), we have

$$ \sum_{k=0}^{\infty} \frac{(n+{\alpha })(n+{\beta })}{(n-k+{\alpha })(n-k+{\beta })} \overline{\hat{\varphi}(-k)}a_{k} =\sum _{k=0}^{\infty} \frac{(n+{\alpha })(n+{\beta })}{(n-k+{\alpha })(n-k+{\beta })} { \widetilde{\overline{\hat{\varphi}(k)}}} {a_{k}}. $$

Since $a_{k}$ is arbitrary, we have $\hat{\varphi}(-k)=C{{\hat{\varphi}(k)}}$ for all $k\in{{\mathbb{N}}_{0}}$. Conversely, if $\hat{\varphi}(-k)=C{{\hat{\varphi}(k)}}$ for all $k\in{{\mathbb{N}}_{0}}$, then $T_{\varphi}$ is complex symmetric with the conjugation C. □

Corollary 2.10

Let φ be in $L^{\infty}(\mathbb{D})$ such that $\varphi (z)=\sum_{n=1}^{\infty}\overline{\hat{\varphi}(-n)} \overline{z}^{n}+\sum_{n=0}^{\infty}\hat{\varphi}(n)z^{n}$. If C is anti-linear on $S^{2}_{\alpha ,\beta}(\mathbb{D})$ such that $Ce_{n}=\overline{\delta _{n}}e_{n}$ with $|\delta _{n}|=1$, then $T_{\varphi}$ on $S^{2}_{\alpha ,\beta}(\mathbb{D})$ is a complex symmetric operator with the conjugation C if and only if $\hat{\varphi}(-k)=\overline{\delta _{k}{{\hat{\varphi}(k)}}}$ for all $k\in{\mathbb{N}}_{0}$.

3 Complex symmetry with non-harmonic symbols

In this section, we study the complex symmetry with non-harmonic symbols. In the Hardy space $H^{2}(\mathbb{T})$, $\overline{z}^{n}z^{m}$ is equal to $z^{m-n}$, but in the generalized derivative Hardy $S^{2}_{\alpha ,\beta}(\mathbb{D})$, $\overline{z}^{n}z^{m}\neq z^{m-n}$ since $z\in \mathbb{D}$. The following result gives a necessary and sufficient condition for complex symmetric Toeplitz operators with non-harmonic symbols.

Theorem 3.1

Let $\varphi (z)=\sum_{i=0}^{\infty}(a_{i}\overline{z}^{n_{i}}z^{m_{i}}+b_{i} \overline{z}^{s_{i}}z^{t_{i}})$ for $a_{i},b_{i}\in{\mathbb{C}}$, and let $n_{i}-m_{i}=t_{i}-s_{i}$ hold. Then $T_{\varphi}$ on $S^{2}_{\alpha ,\beta}(\mathbb{D})$ is complex symmetric with the conjugation $C_{\mu ,\lambda}$ if and only if φ is either

$$ \varphi (z)=\sum_{i=0}^{\infty} \bigl(a_{i} \vert z \vert ^{2m_{i}}+b_{i} \vert z \vert ^{2t_{i}}\bigr) $$

or

$$ \varphi (z)=\sum_{i=0}^{\infty}a_{i} \bigl(\overline{z}^{n_{i}}z^{m_{i}}+ \lambda ^{m_{i}-n_{i}} \overline{z}^{m_{i}}z^{n_{i}}\bigr) $$

for $a_{i},b_{i}\in{\mathbb{C}}$.

Proof

Assume that $n_{i}>m_{i}$ for $i\in \mathbb{N}$ and $T_{\varphi}$ is complex symmetric with the conjugation $C_{\mu ,\lambda}$. If $k\ge \max_{i\in \mathbb{N}}\{n_{i}-m_{i}\}$, then

$$\begin{aligned} C_{\mu ,\lambda}T_{\varphi}z^{k} =&C_{\mu ,\lambda} P\Biggl(\sum_{i=0}^{ \infty} \bigl(a_{i}\overline{z}^{n_{i}}z^{m_{i}+k}+b_{i} \overline{z}^{s_{i}}z^{t_{i}+k}\bigr)\Biggr) \\ =&C_{\mu ,\lambda} \Biggl(\sum_{i=0}^{\infty} \biggl[ \frac{(m_{i}+k+\alpha )(m_{i}+k+\beta )}{(m_{i}+k-n_{i}+\alpha )(m_{i}+k-n_{i}+\beta )}a_{i}z^{m_{i}+k-n_{i}} \\ &{} + \frac{(t_{i}+k+\alpha )(t_{i}+k+\beta )}{(t_{i}+k-s_{i}+\alpha )(t_{i}+k-s_{i}+\beta )}b_{i}z^{t_{i}+k-s_{i}} \biggr] \Biggr) \\ =&\sum_{i=0}^{\infty} \biggl[\mu \frac{(m_{i}+k+\alpha )(m_{i}+k+\beta )}{(m_{i}+k-n_{i}+\alpha )(m_{i}+k-n_{i}+\beta )} \overline{a_{i}}\overline{\lambda}^{m_{i}+k-n_{i}}z^{m_{i}+k-n_{i}} \\ &{} + \mu \frac{(t_{i}+k+\alpha )(t_{i}+k+\beta )}{(t_{i}+k-s_{i}+\alpha )(t_{i}+k-s_{i}+\beta )} \overline{b_{i}}\overline{ \lambda}^{t_{i}+k-s_{i}}z^{t_{i}+k-s_{i}} \biggr] \end{aligned}$$

and

$$\begin{aligned} T_{\varphi}^{\ast}C_{\mu ,\lambda}z^{k} =& T_{\overline{\varphi}}\mu \overline{\lambda}^{k}z^{k} \\ =&\mu \overline{\lambda}^{k} P\Biggl(\sum _{i=0}^{\infty} \bigl[ \overline{a_{i}} \overline{z}^{m_{i}}z^{n_{i}+k}+\overline{b_{i}} \overline{z}^{t_{i}}z^{s_{i}+k} \bigr]\Biggr) \\ =&\sum_{i=0}^{\infty} \biggl[\mu \overline{ \lambda}^{k} \frac{(n_{i}+k+\alpha )(n_{i}+k+\beta )}{(n_{i}+k-m_{i}+\alpha )(n_{i}+k-m_{i}+\beta )} \overline{a_{i}}z^{n_{i}+k-m_{i}} \\ &{} + \mu \overline{\lambda}^{k} \frac{(s_{i}+k+\alpha )(s_{i}+k+\beta )}{(s_{i}+k-t_{i}+\alpha )(s_{i}+k-t_{i}+\beta )} \overline{b_{i}}z^{s_{i}+k-t_{i}} \biggr]. \end{aligned}$$

Since $T_{\varphi}$ is a complex symmetric operator with the conjugation $C_{\mu ,\lambda}$, we have that

$$ n_{i}=m_{i} \quad \text{and} \quad s_{i}=t_{i} $$

for any $i\ge 0$, i.e., φ is of the form $\varphi (z)=\sum_{i=0}^{\infty}(a_{i}|z|^{2m_{i}}+b_{i}|z|^{2t_{i}})$ or

$$ \frac{(m_{i}+k+\alpha )(m_{i}+k+\beta )}{(m_{i}+k-n_{i}+\alpha )(m_{i}+k-n_{i}+\beta )} \overline{a_{i}}\overline{ \lambda}^{m_{i}-n_{i}}= \frac{(s_{i}+k+\alpha )(s_{i}+k+\beta )}{(s_{i}+k-t_{i}+\alpha )(s_{i}+k-t_{i}+\beta )} \overline{b_{i}} $$

(8)

and

$$ \frac{(t_{i}+k+\alpha )(t_{i}+k+\beta )}{(t_{i}+k-s_{i}+\alpha )(t_{i}+k-s_{i}+\beta )} \overline{b_{i}}\overline{ \lambda}^{t_{i}-s_{i}}= \frac{(n_{i}+k+\alpha )(n_{i}+k+\beta )}{(n_{i}+k-m_{i}+\alpha )(n_{i}+k-m_{i}+\beta )} \overline{a_{i}} $$

(9)

for all $i\in \mathbb{N}$. By equations (8) and (9), we obtain $s_{i}=m_{i}$, $t_{i}=n_{i}$, and ${a_{i}}={b_{i}}{\lambda}^{n_{i}-m_{i}}$ for all $i\ge 0$, and so φ is of the form

$$ \varphi (z)=\sum_{i=0}^{\infty}a_{i} \bigl(\overline{z}^{n_{i}}z^{m_{i}}+ \lambda ^{m_{i}-n_{i}} \overline{z}^{m_{i}}z^{n_{i}}\bigr).$$

On the one hand, suppose that φ is of the form

$$ \varphi (z)=\sum_{i=0}^{\infty}a_{i} \bigl(\overline{z}^{n_{i}}z^{m_{i}}+ \lambda ^{m_{i}-n_{i}} \overline{z}^{m_{i}}z^{n_{i}}\bigr).$$

Then, by similar calculations, we have that

$$\begin{aligned}& C_{\mu ,\lambda}T_{\varphi}\sum _{k=0}^{\infty}c_{k}z^{k} \\& \quad =C_{\mu ,\lambda} P\Biggl(\sum_{i=0}^{\infty} \sum_{k=0}^{\infty}\bigl(a_{i}c_{k} \overline{z}^{n_{i}}z^{m_{i}+k}+a_{i}c_{k}{ \lambda}^{m_{i}-n_{i}} \overline{z}^{m_{i}}z^{n_{i}+k}\bigr) \Biggr) \\& \quad =C_{\mu ,\lambda} \Biggl(\sum_{i=0 }^{\infty} \Biggl[\sum_{k=n_{i}-m_{i}}^{ \infty} \frac{(m_{i}+k+\alpha )(m_{i}+k+\beta )}{(m_{i}+k-n_{i}+\alpha )(m_{i}+k-n_{i}+\beta )}a_{i}c_{k}z^{m_{i}+k-n_{i}} \\& \qquad {} +\sum_{k=0}^{\infty} \frac{(n_{i}+k+\alpha )(n_{i}+k+\beta )}{(n_{i}+k-m_{i}+\alpha )(n_{i}+k-m_{i}+\beta )}a_{i}{ \lambda}^{m_{i}-n_{i}}c_{k}z^{n_{i}+k-m_{i}} \Biggr] \Biggr) \\& \quad =\sum_{i=0}^{\infty} \Biggl[\sum _{k=n_{i}-m_{i}}^{\infty}\mu \frac{(m_{i}+k+\alpha )(m_{i}+k+\beta )}{(m_{i}+k-n_{i}+\alpha )(m_{i}+k-n_{i}+\beta )} \overline{a_{i}c_{k}}\overline{\lambda}^{m_{i}+k-n_{i}}z^{m_{i}+k-n_{i}} \\& \qquad {} +\sum_{k=0}^{\infty}\mu \frac{(n_{i}+k+\alpha )(n_{i}+k+\beta )}{(n_{i}+k-m_{i}+\alpha )(n_{i}+k-m_{i}+\beta )} \overline{a_{i}c_{k}}\overline{ \lambda}^{k}z^{n_{i}+k-m_{i}} \Biggr] \end{aligned}$$

and

$$\begin{aligned}& T_{\varphi}^{\ast}C_{\mu ,\lambda}\sum _{k=0}^{\infty}c_{k}z^{k} \\& \quad = T_{\overline{\varphi}}\sum_{k=0}^{\infty} \overline{c_{k}}\mu \overline{\lambda}^{k}z^{k} \\& \quad = P \Biggl(\sum_{i=0}^{\infty}\sum _{k=0}^{\infty}\overline{c_{k}} \mu \overline{\lambda}^{k} \bigl[\overline{a_{i}} \overline{z}^{m_{i}}z^{n_{i}+k}+ \overline{a_{i}} \overline{{\lambda}}^{m_{i}-n_{i}}\overline{z}^{n_{i}}z^{m_{i}+k} \bigr] \Biggr) \\& \quad =\sum_{i=0}^{\infty} \Biggl[\sum _{k=0}^{\infty}\overline{c_{k}}\mu \overline{\lambda}^{k} \frac{(n_{i}+k+\alpha )(n_{i}+k+\beta )}{(n_{i}+k-m_{i}+\alpha )(n_{i}+k-m_{i}+\beta )} \overline{a_{i}}z^{n_{i}+k-m_{i}} \\& \qquad {}+\sum_{k=n_{i}-m_{i}}^{\infty}\overline{c_{k}} \mu \overline{\lambda}^{k} \frac{(m_{i}+k+\alpha )(m_{i}+k+\beta )}{(m_{i}+k-n_{i}+\alpha )(m_{i}+k-n_{i}+\beta )} \overline{a_{i}}\overline{{\lambda}}^{m_{i}-n_{i}}z^{m_{i}+k-n_{i}} \Biggr]. \end{aligned}$$

Therefore, we know that

$$ C_{\mu ,\lambda}T_{\varphi}\sum_{k=0}^{\infty}c_{k}z^{k}=T_{\varphi}^{ \ast}C_{\mu ,\lambda} \sum_{k=0}^{\infty}c_{k}z^{k}, $$

and hence $T_{\varphi}$ is complex symmetric with the conjugation $C_{\mu ,\lambda}$.

Similarly, if φ is of the form $\varphi (z)=\sum_{i=0}^{\infty}(a_{i}|z|^{2m_{i}}+b_{i}|z|^{2t_{i}})$, then we can show that $T_{\varphi}$ is complex symmetric with the conjugation $C_{\mu ,\lambda}$. This completes the proof. □

Remark 3.2

Let

$$ \varphi (z)=\sum_{i=0}^{\infty} \bigl(a_{i}\overline{z}^{n_{i}}z^{m_{i}}+b_{i} \overline{z}^{s_{i}}z^{t_{i}}\bigr)$$

for $a_{i},b_{i}\in{\mathbb{C}}$ and let $n_{i}-m_{i}=t_{i}-s_{i}$ hold. By Theorem 2.6, $T_{\varphi}$ on $S^{2}_{\alpha ,\beta}(\mathbb{D})$ is complex symmetric with the conjugation $C_{\mu ,\lambda}$ if and only if $\varphi (\lambda \overline{z})=\varphi (z)$. Indeed, since

$$\begin{aligned} \varphi (\lambda \overline{z})-\varphi (z) =&\sum_{i=0}^{\infty} \bigl(a_{i} \overline{(\lambda \overline{z})}^{n_{i}}( \lambda \overline{z})^{m_{i}}+b_{i} \overline{(\lambda \overline{z})}^{s_{i}}(\lambda \overline{z})^{t_{i}}\bigr)- \sum _{i=0}^{\infty}\bigl(a_{i} \overline{z}^{n_{i}}z^{m_{i}}+b_{i} \overline{z}^{s_{i}}z^{t_{i}}\bigr) \\ =&\sum_{i=0}^{\infty}\bigl(a_{i} \overline{\lambda}^{n_{i}}{z}^{n_{i}}( \lambda \overline{z})^{m_{i}}+b_{i}\overline{\lambda}^{s_{i}}{z}^{s_{i}} \bigl( \lambda ^{t_{i}}\overline{z}^{t_{i}}\bigr)\bigr)-\sum _{i=0}^{\infty}\bigl(a_{i} \overline{z}^{n_{i}}z^{m_{i}}+b_{i} \overline{z}^{s_{i}}z^{t_{i}}\bigr) \\ =&\sum_{i=0}^{\infty} \bigl[{a_{i}\lambda ^{m_{i}-n_{i}}{z}^{n_{i}} \overline{z}^{m_{i}}+b_{i}\lambda ^{t_{i}-s_{i}}{z}^{s_{i}} \overline{z}^{t_{i}}}-a_{i}\overline{z}^{n_{i}}z^{m_{i}}-b_{i} \overline{z}^{s_{i}}z^{t_{i}}\bigr]=0 \end{aligned}$$

if and only if $s_{i}=m_{i}$, $t_{i}=n_{i}$, and $a_{i}=b_{i}{\lambda}^{n_{i}-m_{i}}$.

Corollary 3.3

(i) Let $\varphi (z)=\sum_{i=0}^{\infty}(a_{i}\overline{z}^{n_{i}}z^{m_{i}}+b_{i} \overline{z}^{m_{i}}z^{n_{i}})$ for some $a_{i},b_{i}\in{\mathbb{C}}$. Then $T_{\varphi}$ on $S^{2}_{\alpha ,\beta}(\mathbb{D})$ is complex symmetric with the conjugation $C_{\mu ,\lambda}$ if and only if $a_{i}=b_{i}{\lambda}^{n_{i}-m_{i}}$.

(ii) If $\varphi (z)=\sum_{i=0}^{\infty}(a_{i}\overline{z}^{m_{i}+1}z^{m_{i}}+b_{i} \overline{z}^{m_{i}}z^{m_{i}+1})$ for some $a_{i},b_{i}\in{\mathbb{C}}$, then $T_{\varphi}$ on $S^{2}_{\alpha ,\beta}(\mathbb{D})$ is complex symmetric with the conjugation $C_{\mu ,\lambda}$ if and only if $a_{i}=b_{i}{\lambda}$ for all i.

(iii) If $\varphi (z)=\sum_{i=0}^{\infty}2a_{i}|z|^{2i}$ for $a_{i}\in{\mathbb{C}}$, then $T_{\varphi}$ on $S^{2}_{\alpha ,\beta}(\mathbb{D})$ is complex symmetric with the conjugation $C_{\mu ,\lambda}$.

Proof

(i) If $s_{i}=m_{i}$ and $t_{i}=n_{i}$ in Theorem 3.1, then we obtain statement (i).

(ii) If we put $n_{i}=m_{i}+1$ in (i), then we get statement (ii).

(iii) If $s_{i}=m_{i}$, $t_{i}=n_{i}$, and $a_{i}=b_{i}$, then we have this result. □

Example 3.4

Let $\varphi (z)=\sum_{j=0}^{\infty}(a_{j}\overline{z}^{n_{j}}z^{m_{j}}+a_{j} e^{i\theta}\overline{z}^{m_{j}}z^{n_{j}})$ for some $a_{i}\in{\mathbb{C}}$ and for some real θ. Then $T_{\varphi}$ on $S^{2}_{\alpha ,\beta}(\mathbb{D})$ is complex symmetric with the conjugation $C_{\mu ,\lambda}$.

Corollary 3.5

Let $\varphi (z)=\sum_{i=0}^{\infty}(a_{i}\overline{z}^{n_{i}}z^{m_{i}}+b_{i} \overline{z}^{s_{i}}z^{t_{i}})$ for $a_{i},b_{i}\in{\mathbb{C}}$ and let $n_{i}-m_{i}=t_{i}-s_{i}$ hold. If one of $s_{i}=m_{i}$, $t_{i}=n_{i}$, and $a_{i}=b_{i}{\lambda}^{n_{i}-m_{i}}$ does not hold, then $T_{\varphi}$ on $S^{2}_{\alpha ,\beta}(\mathbb{D})$ is not complex symmetric with the conjugation $C_{\mu ,\lambda}$.

Proof

The proof follows from Theorem 3.1. □

Remark 3.6

In Theorem 3.1, the condition “$n_{i}-m_{i}=t_{i}-s_{i}$” is a necessary condition. If not, we may not consider complex symmetric Toeplitz operators with such non-harmonic symbols. For example, let $\varphi (z)=a\overline{z}z^{3}+b\overline{z}^{2}z$ for $a,b\in{\mathbb{C}}$. Then, for $k\ge 2$, we have

$$ C_{\mu ,\lambda}T_{\varphi}z^{k}=\mu \frac{(k+\alpha +3)(k+\beta +3)}{(k+\alpha +2)(k+\beta +2)} \overline{a}\overline{\lambda}^{k+2}z^{k+2}+\mu \frac{(k+\alpha +1)(k+\beta +1)}{(k+\alpha -1)(k+\beta -1)} \overline{b}\overline{\lambda}^{k-1}z^{k-1} $$

and

$$ T_{\varphi}^{\ast}C_{\mu ,\lambda}z^{k}=\mu \overline{\lambda}^{k} \frac{(k+\alpha +1)(k+\beta +1)}{(k+\alpha -2)(k+\beta -2)} \overline{a}z^{k-2}+ \mu \overline{\lambda}^{k} \frac{(k+\alpha +2)(k+\beta +2)}{(k+\alpha +1)(k+\beta +1)} \overline{b}z^{k+1}.$$

Thus $C_{\mu ,\lambda}T_{\varphi}z^{k}\neq T_{\varphi}^{\ast}C_{\mu , \lambda}z^{k}$ for any $k\ge 2$, and so $T_{\varphi}$ is not complex symmetric Toeplitz operators.

Corollary 3.7

Let $\varphi (z)=a\overline{z}^{n}z^{m}+bz^{k}$, where $n,m,k \in{\mathbb{N}}$ with $n> m$ and $a,b\in{\mathbb{C}}$ with $|a|\neq|b|$. Then $T_{\varphi}$ on $S^{2}_{\alpha ,\beta}(\mathbb{D})$ is never complex symmetric with the conjugation $C_{\mu ,\lambda}$.

Remark 3.8

If $\varphi (z)=a\overline{z}^{n}z^{m}$ for $a\in{\mathbb{C}}$ and $m,n\in{\mathbb{N}}$ with $m\neq n$ or $\varphi (z)=\overline{z}^{2}z+bz$ for $b\in{\mathbb{C}}$ with $b\neq1$. By Theorem 3.1 and Corollary 3.7, $T_{\varphi}$ is never complex symmetric with the conjugation $C_{\mu ,\lambda}$ in $S^{2}_{\alpha ,\beta}(\mathbb{D})$ and the Hardy space $H^{2}(\mathbb{T})$.

4 Conclusion

In this paper, we make characterizations for Toeplitz operators to be complex symmetric on the generalized derivative Hardy space $S^{2}_{\alpha ,\beta}(\mathbb{D})$ with respect to the conjugations $C_{\xi}$, $C_{\mu , \lambda}$ as in Theorem 2.6. Moreover, in Theorem 2.9, we deduce the necessary and sufficient conditions for complex symmetric Toeplitz operators with any conjugation C. Next, for the conjugation $C_{\mu , \lambda}$, we also obtain complex symmetric Toeplitz operators with non-harmonic symbols of the form $\varphi (z)=\sum_{i=0}^{\infty}(a_{i}\overline{z}^{n_{i}}z^{m_{i}}+b_{i} \overline{z}^{s_{i}}z^{t_{i}})$ in Theorem 3.1. The results of this paper provide an answer in the generalized derivative Hardy space $S^{2}_{\alpha ,\beta}(\mathbb{D})$ as in the question raised in [8].