H-Toeplitz operators on the function spaces

Abstract

In this paper, we study several classes of H-Toeplitz operators (defined below) on the Hardy space $H^2$. In particular, we prove that, for $\varphi \in L^{\infty }$, the adjoint of H-Toeplitz operators is hyponormal. Next, we investigate several properties of H-Toeplitz operators on the weighted Bergman spaces. Finally, we give necessary and sufficient conditions for H-Toeplitz operators to be contractive and expansive on the weighted Bergman spaces.

1 Introduction

Let $\mathcal{{L}}(\mathcal{{H}})$ be the algebra of all bounded linear operators on a separable complex Hilbert space $\mathcal{{H}}$. For $T\in {\mathcal {L}}(\mathcal {H})$, $T^{*}$ denotes the adjoint of T. An operator $T\in {\mathcal {L}}(\mathcal {H})$ is said to be self-adjoint if $T=T^{*}$, isometric if $T^{*}T=I$, normal if $[T^{*},T]=0$, hyponormal if $[T^{*},T]\ge 0$, quasinormal if $[T^{*}T, T]=0$, and binormal if $[T^{*}T, TT^{*}]=0$, respectively, where $[R,S]:=RS-SR.$

H-Toeplitz operators have been studied in various spaces. Recently, the authors in [13] studied the essential conditions for H-Toeplitz operators to become a co-isometry and a partial isometry, explored their invariant subspaces and kernels, and investigated their compactness and Fredholmness. In particular, they showed a nonzero H-Toeplitz operator cannot be a Fredholm operator on the Bergman space. Moreover, they considered the necessary and sufficient conditions for the commutativity of H-Toeplitz operators. In [25], the authors provided a characterization of the commutativity of H-Toeplitz operators with quasihomogeneous symbols on the Bergman space. In [22], the authors explored the characteristics of H-Toeplitz operators on the Bergman space and offered essential criteria for identifying both contractive and expansive operators. Additionally, the authors in [14] studied the slant Toeplitz operators on the Hardy space.

Basic properties of Toeplitz operators on the Hardy space and (weighted) Begman space can be found in [2, 7, 8, 18, 20, 28]. Recently, many authors have characterized the hyponormality of Toeplitz operators on the Bergman spaces and the weighted Bergman spaces (cf. [16, 17, 19, 21, 26, 27, 29]). The theory of Toeplitz operators is a vast and significant field that has made fundamental contributions to several problems in functional analysis and mathematical physics.

Several decades ago, researchers extensively studied contractive and expansive operators (cf. [3, 5, 6]). In particular, in [9], the authors investigated the problem of invariant subspaces for contractive operators. In [22], the authors studied the contractivity and expansivity of H-Toeplitz operators with analytic, co-analytic and harmonic symbols on the Bergman spaces.

In this paper, we study several classes of H-Toeplitz operators on the function spaces. In Sect. 2, we focus on the self-adjointness of H-Toeplitz operators on the Hardy space $H^2$. Moreover, we consider complex symmetric H-Toeplitz operator on $H^2.$ Furthermore, we investigate hyponormality, quasinormality, and binormality of H-Toeplitz operators. In particular, we show that for $\varphi \in L^{\infty }$ the adjoint of H-Toeplitz operators is hyponormal. As an application of this, such an operator has a nontrivial invariant subspace. In Sect. 3, we will investigate the the algebraic properties of H-Toeplitz operators on the weighted Bergman spaces $A^2_{\alpha }(\mathbb D)$. More concretely, we introduce the notion of H-Toeplitz operators on the weighted Bergman spaces, which combine the properties of both Toeplitz and Hankel operators. The importance of this notion is that it provides a unifying framework for a class of operators on the weighted Bergman spaces, which includes both Toeplitz and Hankel operators. Furthermore, we establish a convenient and explicit criterion for determining the contractivity and expansivity of H-Toeplitz operators.

2 H-Toeplitz operators on the Hardy spaces

Let $\mathbb D$ be the open unit disk in the complex plane and let $\mathbb T (\equiv \partial \mathbb D)$ be the unit circle. Let $L^{\infty }(\mathbb T)$ denote the set of all essentially bounded measurable functions on ${\mathbb T}$. The Hilbert Hardy space $H^2({\mathbb T})$ consists of all analytic functions f with the power series representation

$$\begin{aligned} f(z)=\sum _{n=0}^{\infty }a_nz^n\ \text{ where }\ \sum _{n=0}^{\infty }|a_n|^2<\infty . \end{aligned}$$

For a convenience, we denote $L^{\infty }(\mathbb T)$ and $H^{2}(\mathbb T)$ by $L^{\infty }$ and $H^2$, respectively. For any $\varphi \in L^{\infty }$, the multiplication operator $M_{\varphi }$ is defined by $M_{\varphi }(f)=\varphi f$ for $f\in H^2$, the Toeplitz operator $T_{\varphi }:{H^2}\rightarrow {H^2}$ is defined by

$$\begin{aligned} T_{\varphi }{f}=P(\varphi f) \end{aligned}$$

for $f\in {H^2}$ where P denotes the orthogonal projection of $L^2$ onto $H^2$, and the Hankel operator $H_{\varphi }:{H^2}\rightarrow {H^2}$ is defined by

$$\begin{aligned} H_{\varphi }{f}=PM_{\varphi }Jf \end{aligned}$$

where $J:{H^2}\rightarrow ({H^2})^{\perp }$ denotes the flip operator given by $J(e_n)=e_{-n-1}$ for all $n\ge 0$ where $\{e_n\}_{n=-\infty }^{\infty }$ is an orthonormal basis for $L^2$. Note that $T_{\varphi }$ is bounded if and only if $\varphi \in L^{\infty }$ and, in which case, $\Vert T_{\varphi } \Vert =\Vert {\varphi }\Vert _{\infty }$.

Notation 2.1

Throughout this paper, a dilation operator K from $H^2$ to $L^2$ is denoted as $K(e_{2n})=e_n$ and $K(e_{2n+1})=e_{-n-1}$ for all $n=0,1,2,\ldots$ where $\{e_n\}_{n=-\infty }^{\infty }$ is an orthonormal basis for $L^2$.

Let ${\mathbb N}$, ${\mathbb N}_0$, ${\mathbb Z}$, ${\mathbb R}$, and ${\mathbb C}$ be the set of positive integers, $\text {non-negative integers}$, integers, real numbers, and complex numbers, respectively. A dilation operator K is bounded from $H^2$ to $L^2$ with $\Vert K\Vert =1$ and its adjoint $K^{*}$ from $L^2$ to $H^2$ is defined as

$$\begin{aligned} K^{*}(e_n)=e_{2n}~\text{ and }~K^{*}(e_{-n-1})=e_{2n+1} \end{aligned}$$

for all $n=0,1,2,\ldots$. Thus $K^{*}K=I$ on $H^2$ and $K^{*}K=I$ on $L^2$. Indeed, since $KK^{*}e_n=Ke_{2n}=e_n$ for each $n\ge 0$, it follows that $KK^{*}=I$ on $H^2$. Moreover, since $KK^{*}e_{-n-1}=Ke_{2n+1}=e_{-n-1}$ for each $n\in {\mathbb N}$, we know that $KK^{*}=I$ on $(H^2)^{\bot }$. Thus $KK^{*}=I$ on $L^2$. Hence K is unitary from $H^2$ to $L^2$.

The authors in [1] have introduced “H-Toeplitz operators" motivated by the Toeplitz, Hankel, and Slant Toeplitz operators.

Definition 2.2

For $\varphi \in L^{\infty }$, an H-Toeplitz operator $S_{\varphi }$ with the symbol $\varphi$ on $H^2$ is defined by

$$\begin{aligned} S_{\varphi }f=PM_{\varphi }Kf \end{aligned}$$

for each $f\in H^2$ where P denotes the orthogonal projection of $L^2$ onto $H^2$.

In this case, $\Vert S_{\varphi }\Vert =\Vert PM_{\varphi }K\Vert \le \Vert M_{\varphi }\Vert =\Vert \varphi \Vert _{\infty }.$ Note that if $\{e_n\}_{n=0}^{\infty }$ denotes the orthonormal basis for $H^2$, then

$$\begin{aligned} S_{\varphi }e_{2n}=PM_{\varphi }K{e_{2n}}=PM_{\varphi }e_n=T_{\varphi }e_n \end{aligned}$$

and

$$\begin{aligned} S_{\varphi }e_{2n+1}=PM_{\varphi }K{e_{2n+1}}=PM_{\varphi }e_{-n-1}=PM_{\varphi }Je_{n}=H_{\varphi }e_n \end{aligned}$$

for each $n=0,1,2\ldots .$ Note that for $\varphi \in L^{\infty }$, the adjoint of $S_{\varphi }$ on $H^2$ is given by

$$\begin{aligned} S_{\varphi }^{*}=K^{*}M_{\overline{\varphi }}. \end{aligned}$$

2.1 Basic properties of an H-Toeplitz operator

In this section, we consider the basic properties of an H-Toeplitz operator. We first study the self-adjointness of H-Toeplitz operators on $H^2$.

Theorem 2.3

If $\varphi (z)=\sum _{j=-\infty }^{\infty }a_je_j$ with respect to the orthonormal basis $\mathcal {B}=\{e_n\}_{n=0}^{\infty }$ in $L^{\infty }$ and $S_{\varphi }$ is an H-Toeplitz operator on $H^2$, then the matrices of $S_{\varphi }$ and $S_{\varphi }^{*}$ are represented as

$$\begin{aligned}{}[S_{\varphi }]_\mathcal {B}=\begin{pmatrix} a_0 &{}\quad a_1 &{}\quad a_{-1} &{}\quad a_2 &{}\quad a_{-2} &{}\quad a_3 &{}\quad a_{-3} &{}\quad \cdots \\ a_1 &{}\quad a_2 &{}\quad a_{0} &{}\quad a_{3} &{}\quad a_{-1} &{}\quad a_4 &{}\quad a_{-2} &{}\quad \cdots \\ a_2 &{}\quad a_3 &{}\quad a_{1} &{}\quad a_{4} &{}\quad a_{0} &{}\quad a_5 &{}\quad a_{-1} &{}\quad \cdots \\ a_3 &{}\quad a_4 &{}\quad a_{2} &{}\quad a_{5} &{}\quad a_{1} &{}\quad a_6 &{}\quad a_{0} &{}\quad \cdots \\ a_4 &{}\quad a_5 &{}\quad a_{3} &{}\quad a_{6} &{}\quad a_{2} &{}\quad a_7 &{}\quad a_{1} &{}\quad \cdots \\ a_5 &{}\quad a_6 &{}\quad a_{4} &{}\quad a_{7} &{}\quad a_{3} &{}\quad a_8 &{}\quad a_{2} &{}\quad \cdots \\ a_6 &{}\quad a_7 &{}\quad a_{5} &{}\quad a_{8} &{}\quad a_{4} &{}\quad a_9 &{}\quad a_{3} &{}\quad \cdots \\ \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \ddots \end{pmatrix} \end{aligned}$$

(1)

and

$$\begin{aligned}{}[S_{\varphi }^{*}]_\mathcal {B}=\begin{pmatrix} \overline{a_0} &{}\quad \overline{a_1} &{}\quad \overline{a_{2}} &{}\quad \overline{a_3} &{}\quad \overline{a_{4}} &{}\quad \overline{a_5} &{}\quad \overline{a_{6}} &{}\quad \cdots \\ \overline{a_1} &{}\quad \overline{a_2} &{}\quad \overline{a_{3}} &{}\quad \overline{a_4} &{}\quad \overline{a_{5}} &{}\quad \overline{a_6} &{}\quad \overline{a_{7}} &{}\quad \cdots \\ \overline{a_{-1}} &{}\quad \overline{a_0} &{}\quad \overline{a_{1}} &{}\quad \overline{a_2} &{}\quad \overline{a_{3}} &{}\quad \overline{a_4} &{}\quad \overline{a_{5}} &{}\quad \cdots \\ \overline{a_{2}} &{}\quad \overline{a_3} &{}\quad \overline{a_{4}} &{}\quad \overline{a_5} &{}\quad \overline{a_{6}} &{}\quad \overline{a_7} &{}\quad \overline{a_{8}} &{}\quad \cdots \\ \overline{a_{-2}} &{}\quad \overline{a_{-1}} &{}\quad \overline{a_{0}} &{}\quad \overline{a_1} &{}\quad \overline{a_{2}} &{}\quad \overline{a_3} &{}\quad \overline{a_{4}} &{}\quad \cdots \\ \overline{a_{3}} &{}\quad \overline{a_{4}} &{}\quad \overline{a_{5}} &{}\quad \overline{a_6} &{}\quad \overline{a_{7}} &{}\quad \overline{a_8} &{}\quad \overline{a_{9}} &{}\quad \cdots \\ \overline{a_{-3}} &{}\quad \overline{a_{-2}} &{}\quad \overline{a_{-1}} &{}\quad \overline{a_0} &{}\quad \overline{a_{1}} &{}\quad \overline{a_2} &{}\quad \overline{a_{3}} &{}\quad \cdots \\ \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \ddots \end{pmatrix}. \end{aligned}$$

Furthermore, $[S_{\varphi }]_\mathcal {B}$ is self-adjoint if and only if $[S_{\varphi }]_\mathcal {B}=0.$

Proof

We know that $S_{\varphi }$ is self-adjoint if and only if $a_0,a_1,a_2,a_3,a_5,a_8,\ldots$ are real and $a_2=\overline{a_{-1}}$, $a_3=\overline{a_0}$, $a_2=\overline{a_3}$, $a_{-2}=\overline{a_4}$, $a_3=\overline{a_5},$ $a_{-3}=\overline{a_6}$, and so on (cf. [1, Page 151]).

On the other hand, the (i, j) entry of the matrix $[S_{\varphi }]_\mathcal {B}$ is given by

$$\begin{aligned} a_{i,j}={\left\{ \begin{array}{ll} a_i~~~~~~~&{}\text{ if }~j=0\\ a_{i-n}~~~~&{}\text{ if }~j=2n\\ a_{i+n+1}~&{}\text{ if }~j=2n+1 \end{array}\right. } \end{aligned}$$

(see [1]). And the (i, j) entry of the matrix $[S_{\varphi }^{*}]_\mathcal {B}$ is given by

$$\begin{aligned} \overline{a_{j,i}}={\left\{ \begin{array}{ll} \overline{a_j}~~~~~~~&{}\text{ if }~i=0\\ \overline{a_{j-n}}~~~~&{}\text{ if }~i=2n\\ \overline{a_{j+n+1}}~&{}\text{ if }~i=2n+1. \end{array}\right. } \end{aligned}$$

(2)

Thus $[S_{\varphi }]_\mathcal {B}=[S_{\varphi }^{*}]_\mathcal {B}$ if and only if $a_{i,j}=\overline{a_{j,i}}$ for all i, j. Hence $[S_{\varphi }]_\mathcal {B}$ is self-adjoint if and only if $a_j=a_0\in {\mathbb R}$ for all $j\in {\mathbb Z}$ if and only if $a_j=0$ for all j since $[S_{\varphi }]_\mathcal {B}$ is bounded. $\square$

Proposition 2.4

Let $\varphi \in L^{\infty }$ and $S_{\varphi }$ be an H-Toeplitz operator on $H^2$. Then $S_{\varphi }$ is an isometry on $H^2$ if and only if $M_{\overline{\varphi }}PM_{\varphi }=I$ on $L^2$. In particular, $\varphi$ is not inner.

Proof

Since $S_{\varphi }^{*}=K^{*}M_{\overline{\varphi }}$, we have $S_{\varphi }^{*}S_{\varphi }=K^{*}M_{\overline{\varphi }}PM_{\varphi }K$. Then $K^{*}M_{\overline{\varphi }}PM_{\varphi }K=I$ on $H^2$. Hence $M_{\overline{\varphi }}PM_{\varphi }=I$ on $L^2$ since K is unitary from $H^2$ to $L^2.$ Thus $S_{\varphi }$ is an isometry on $H^2$ if and only if $M_{\overline{\varphi }}PM_{\varphi }=I$ on $L^2$.

If $\varphi$ is inner, then $M_{\overline{\varphi }}PM_{\varphi }-I=M_{\overline{\varphi }}M_{\varphi }-I=M_{|{\varphi }|^2}-I=0$. Thus $S_{\varphi }$ is an isometry on $H^2$. But, if $\varphi$ is inner, then $S_{\varphi }^{*}$ is an isometry on $H^2$ (cf. [1]), and so $S_{\varphi }^{*}$ is normal. Therefore, $\varphi =0$ from [1], which is a contradiction. $\square$

Next, we study complex symmetric H-Toeplitz operator on $H^2$. A conjugation on $\mathcal H$ is an antilinear operator ${C}: \mathcal{H}\rightarrow \mathcal{H}$ which satisfies ${C}^{2}=I$ and $\langle {C}x, {C}y \rangle =\langle y, x\rangle$ for all $x,y\in \mathcal{H}$. If C is a conjugation on $\mathcal {H}$, then there exists an orthonormal basis $\{e_n\}_{n=0}^{\infty }$ for $\mathcal{H}$ such that ${C}e_n=e_n$ for all n (see [10]). An operator $T\in \mathcal{L(H)}$ is complex symmetric if there exists a conjugation C on $\mathcal{H}$ such that $T= {C}T^{*}{C}$. Complex symmetric operators have been widely studied by several mathematicians (see [10,11,12, 23, 24] for more details).

Proposition 2.5

For $\varphi \in L^{\infty }$, let $S_{\varphi }$ be an H-Toeplitz operator on $H^2$ and C be a conjugation on $L^2$ given by $Cf(z)=\overline{f(\overline{z})}$ for $f\in H^2$. Then $S_{\varphi }$ is complex symmetric with the conjugation C if and only if $T_{\overline{\varphi (\overline{z})}}e_{n}=K^{*}M_{\overline{\varphi (z)}}e_{2n}$ and $H_{\overline{\varphi (\overline{z})}}e_n=K^{*}M_{\overline{\varphi (z)}}e_{2n+1}$ for $n\in {\mathbb N}_0$.

Proof

Let C be a conjugation on $L^2$ given by $Cf(z)=\overline{f(\overline{z})}$ for $f\in H^2$. Then $Ce_n=e_n$ for $n\ge 0$ and so $CP=PC$ on $L^2$ from [24]. Thus for $n\ge 0,$

$$\begin{aligned} CS_{\varphi }Ce_{2n}=CS_{\varphi }e_{2n}=CT_{\varphi }(e_n)=CP(\varphi e_n)=PC(\varphi e_n)=T_{\overline{\varphi (\overline{z})}}e_n \end{aligned}$$

and

$$\begin{aligned} CS_{\varphi }Ce_{2n+1}=CH_{\varphi }(e_n)=CPM_{\varphi }(Je_{n})=CP{\varphi }(Je_{n})=H_{\overline{\varphi (\overline{z})}}e_n \end{aligned}$$

hold. Since $S_{\varphi }^{*}=K^{*}M_{\overline{\varphi }}$ for $\varphi \in L^{\infty }$, we obtain that $S_{\varphi }$ is complex symmetric with the conjugation C if and only if $T_{\overline{\varphi (\overline{z})}}e_{n}=K^{*}M_{\overline{\varphi (z)}}e_{2n}$ and $H_{\overline{\varphi (\overline{z})}}e_n=K^{*}M_{\overline{\varphi (z)}}e_{2n+1}$ for $n\in {\mathbb N}_0$. $\square$

Theorem 2.6

For $\varphi \in L^{\infty }$, let $S_{\varphi }$ be an H-Toeplitz operator on $H^2$. Assume that C is a conjugation on $L^2$ given by $Cf(z)=\overline{f(\overline{z})}$ for $f\in H^2$ and $C_{\mu ,\lambda }$ is a conjugation on $L^2$ given by $C_{\mu , \lambda }f(z)=\mu \overline{f(\lambda \overline{z})}$ for $f\in H^2$ with $|\lambda |=|\mu |=1$. Then the following statements are equivalent:

(i) $S_{\varphi }$ is complex symmetric with the conjugation C.

(ii) $S_{\varphi }$ is complex symmetric with the conjugation $C_{\mu , \lambda }$.

(iii) $\varphi =0.$

Proof

(i) $\Leftrightarrow$ (iii) Let $\varphi (z)=\sum _{j=-\infty }^{\infty }a_je_j$ be with respect to the basis $\mathcal {B}=\{e_n\}_{n=0}^{\infty }$. Since the matrix of $S_{\varphi }$ is of the form (1), it follows that the matrix of $CS_{\varphi }C$ is the followings:

$$\begin{aligned}{}[CS_{\varphi }C]_\mathcal {B}=\begin{pmatrix} \overline{a_0} &{}\quad \overline{a_1} &{}\quad \overline{a_{-1}} &{}\quad \overline{a_2} &{}\quad \overline{a_{-2}} &{}\quad \overline{a_3} &{}\quad \overline{a_{-3}} &{}\quad \cdots \\ \overline{a_1} &{}\quad \overline{a_2} &{}\quad \overline{a_{0}} &{}\quad \overline{a_{3}} &{}\quad \overline{a_{-1}} &{}\quad \overline{a_4} &{}\quad \overline{a_{-2}} &{}\quad \cdots \\ \overline{a_2} &{}\quad \overline{a_3} &{}\quad \overline{a_{1}} &{}\quad \overline{a_{4}} &{}\quad \overline{a_{0}} &{}\quad \overline{a_5} &{}\quad \overline{a_{-1}} &{}\quad \cdots \\ \overline{a_3} &{}\quad \overline{a_4} &{}\quad \overline{a_{2}} &{}\quad \overline{a_{5}} &{}\quad \overline{a_{1}} &{}\quad \overline{a_6} &{}\quad \overline{a_{0}} &{}\quad \cdots \\ \overline{a_4} &{}\quad \overline{a_5} &{}\quad \overline{a_{3}} &{}\quad \overline{a_{6}} &{}\quad \overline{a_{2}} &{}\quad \overline{a_7} &{}\quad \overline{a_{1}} &{}\quad \cdots \\ \overline{a_5} &{}\quad \overline{a_6} &{}\quad \overline{a_{4}} &{}\quad \overline{a_{7}} &{}\quad \overline{a_{3}} &{}\quad \overline{a_8} &{}\quad \overline{a_{2}} &{}\quad \cdots \\ \overline{a_6} &{}\quad \overline{a_7} &{}\quad \overline{a_{5}} &{}\quad \overline{a_{8}} &{}\quad \overline{a_{4}} &{}\quad \overline{a_9} &{}\quad \overline{a_{3}} &{}\quad \cdots \\ \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \ddots \end{pmatrix}. \end{aligned}$$

Then $[S_{\varphi }]_\mathcal {B}$ is complex symmetric with the conjugation C if and only if $a_j=a_0\in {\mathbb C}$ for all $j\in {\mathbb Z}$. Hence $\varphi$ is of the form $\varphi =\sum _{j=-\infty }^{\infty }\hat{\varphi }(0)e_j$ and so $\varphi =0$ since $\varphi \in L^{\infty }$.

(ii) $\Leftrightarrow$ (iii) Let $\varphi (z)=\sum _{j=-\infty }^{\infty }a_je_j$ be with respect to the basis $\mathcal {B}=\{e_n\}_{n=0}^{\infty }$. It is known from [24] that $C_{\mu ,\lambda }$ is unitarily equivalent to $C_{1,\lambda }$. Since the matrix of $S_{\varphi }$ is the form of (1), it follows that the matrix of $C_{1,\lambda }S_{\varphi }C_{1,\lambda }$ is the followings:

$$\begin{aligned}{}[C_{1,\lambda }S_{\varphi }C_{1,\lambda }]_\mathcal {B}=\lambda I\begin{pmatrix} \overline{a_0} &{}\quad \overline{a_1} &{}\quad \overline{a_{-1}} &{}\quad \overline{a_2} &{}\quad \overline{a_{-2}} &{}\quad \overline{a_3} &{}\quad \overline{a_{-3}} &{}\quad \cdots \\ \overline{a_1} &{}\quad \overline{a_2} &{}\quad \overline{a_{0}} &{}\quad \overline{a_{3}} &{}\quad \overline{a_{-1}} &{}\quad \overline{a_4} &{}\quad \overline{a_{-2}} &{}\quad \cdots \\ \overline{a_2} &{}\quad \overline{a_3} &{}\quad \overline{a_{1}} &{}\quad \overline{a_{4}} &{}\quad \overline{a_{0}} &{}\quad \overline{a_5} &{}\quad \overline{a_{-1}} &{}\quad \cdots \\ \overline{a_3} &{}\quad \overline{a_4} &{}\quad \overline{a_{2}} &{}\quad \overline{a_{5}} &{}\quad \overline{a_{1}} &{}\quad \overline{a_6} &{}\quad \overline{a_{0}} &{}\quad \cdots \\ \overline{a_4} &{}\quad \overline{a_5} &{}\quad \overline{a_{3}} &{}\quad \overline{a_{6}} &{}\quad \overline{a_{2}} &{}\quad \overline{a_7} &{}\quad \overline{a_{1}} &{}\quad \cdots \\ \overline{a_5} &{}\quad \overline{a_6} &{}\quad \overline{a_{4}} &{}\quad \overline{a_{7}} &{}\quad \overline{a_{3}} &{}\quad \overline{a_8} &{}\quad \overline{a_{2}} &{}\quad \cdots \\ \overline{a_6} &{}\quad \overline{a_7} &{}\quad \overline{a_{5}} &{}\quad \overline{a_{8}} &{}\quad \overline{a_{4}} &{}\quad \overline{a_9} &{}\quad \overline{a_{3}} &{}\quad \cdots \\ \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \ddots \end{pmatrix}=\lambda [CS_{\varphi }C]_\mathcal {B}. \end{aligned}$$

Then $[S_{\varphi }]_\mathcal {B}$ is complex symmetric with the conjugation $C_{1,\lambda }$ if and only if $a_j=a_0\in {\mathbb C}$ for all $j\in {\mathbb Z}$ and $\lambda =1$. Hence, in this case, $\varphi$ is of the form $\varphi =\sum _{j=-\infty }^{\infty }\hat{\varphi }(0)e_j$ and so $\varphi =0$ since $\varphi \in L^{\infty }$. $\square$

Remark that if $\varphi \in L^2$, then an unbounded H-Toeplitz operator is complex symmetric with the conjugation C if and only if $\hat{\varphi }(j)=\hat{\varphi }(0)\in {\mathbb C}$ for all $j\in {\mathbb Z}$. In previous theorem, if $\hat{\varphi }(0)\not =0$, then $\varphi =\sum _{j=-\infty }^{\infty }\hat{\varphi }(0)e_j$ does not belong to $L^{\infty }$.

2.2 Hyponormal, quasinormal, and binormal H-Toeplitz operators

In this section, we study hyponormal, quasinormal, and binormal H-Toeplitz operators.

Lemma 2.7

For $\varphi \in L^{\infty }$, let $S_{\varphi }$ be an H-Toeplitz operator on $H^2$. Then the following statements hold.

(i) $S_{\varphi }S_{\varphi }^{*}=T_{|\varphi |^2}$, $S_{\varphi }^{*}S_{\varphi }e_{2n}=K^{*}M_{\overline{\varphi }}T_{\varphi }e_n$, and $S_{\varphi }^{*}S_{\varphi }e_{2n+1}=K^{*}M_{\overline{\varphi }}T_{\varphi }e_n$ hold for each $n\in {\mathbb N}_0$.

(ii) $S_{\varphi }^{*}$ is hyponormal if and only if the following equations hold.

$$\begin{aligned} {\left\{ \begin{array}{ll} T_{|\varphi |^2}e_{2n}\ge K^{*}M_{\overline{\varphi }}T_{\varphi }e_n\\ T_{|\varphi |^2}e_{2n+1}\ge K^{*}M_{\overline{\varphi }}H_{\varphi }e_n\end{array}\right. } \end{aligned}$$

(3)

for each $n\in {\mathbb N}_0$. In particular, the equalities in (3) hold if and only if $S_{\varphi }$ is normal.

Proof

(i) For $\varphi \in L^{\infty }$, let $S_{\varphi }$ be an H-Toeplitz operator on $H^2$. Since $S_{\varphi }^{*}=K^{*}M_{\overline{\varphi }}$, it follows that $S_{\varphi }^{*}S_{\varphi }=K^{*}M_{\overline{\varphi }}PM_{\varphi }K$ and

$$\begin{aligned} S_{\varphi }S_{\varphi }^{*}=(PM_{\varphi }K)(K^{*}M_{\overline{\varphi }})=PM_{\varphi }M_{\overline{\varphi }}=PM_{|\varphi |^2}=T_{|\varphi |^2}. \end{aligned}$$

On the other hand, since $S_{\varphi }^{*}S_{\varphi }=K^{*}M_{\overline{\varphi }}PM_{\varphi }K$, it follows that

$$\begin{aligned} S_{\varphi }^{*}S_{\varphi }e_{2n}= & {} K^{*}M_{\overline{\varphi }}PM_{\varphi }Ke_{2n}\\= & {} K^{*}M_{\overline{\varphi }}PM_{\varphi }e_{n}\\= & {} K^{*}M_{\overline{\varphi }}T_{\varphi }e_n \end{aligned}$$

and

$$\begin{aligned} S_{\varphi }^{*}S_{\varphi }e_{2n+1}= & {} K^{*}M_{\overline{\varphi }}PM_{\varphi }Ke_{2n+1}\\= & {} K^{*}M_{\overline{\varphi }}PM_{\varphi }e_{-n-1}\\= & {} K^{*}M_{\overline{\varphi }}H_{\varphi }e_n \end{aligned}$$

for each $n\in {\mathbb N}_0$.

(ii) By (i), we obtain that $S_{\varphi }^{*}$ is hyponormal if and only if for each n, it holds that

$$\begin{aligned} {\left\{ \begin{array}{ll} T_{|\varphi |^2}e_{2n}\ge K^{*}M_{\overline{\varphi }}T_{\varphi }e_n\\ T_{|\varphi |^2}e_{2n+1}\ge K^{*}M_{\overline{\varphi }}H_{\varphi }e_{n}.\end{array}\right. } \end{aligned}$$

In particular, we get that $S_{\varphi }$ is normal if and only if

$$\begin{aligned} {\left\{ \begin{array}{ll} T_{|\varphi |^2}e_{2n}=K^{*}M_{\overline{\varphi }}T_{\varphi }e_n\\ T_{|\varphi |^2}e_{2n+1}=K^{*}M_{\overline{\varphi }}H_{\varphi }e_{n}\end{array}\right. } \end{aligned}$$

for each $n\in {\mathbb N}_0$. $\square$

Using Lemma 2.7, we show that every H-Toeplitz operator on $H^2$ is cohyponormal.

Theorem 2.8

For $\varphi \in L^{\infty }$, let $S_{\varphi }$ be an H-Toeplitz operator on $H^2$. Then $S_{\varphi }^{*}$ is hyponormal.

Proof. Set $\varphi =\sum _{j=-\infty }^{\infty }\hat{\varphi }(j)e_j\in L^{\infty }$. Let $S_{\varphi }$ be an H-Toeplitz operator on $H^2$. Then $S_{\varphi }^{*}$ is hyponormal if and only if

$$\begin{aligned} \Vert S_{\varphi }f\Vert ^2\le \Vert S_{\varphi }^{*}f\Vert ^2 \end{aligned}$$

for each $f\in H^2$. Taking $f=e_{2n}$ for each n, Lemma 2.7 implies that

$$\begin{aligned} \Vert S_{\varphi }^{*}e_{2n}\Vert ^2-\Vert S_{\varphi }e_{2n}\Vert ^2= & {} \Vert K^{*}M_{\varphi }^{*}e_{2n}\Vert ^2-\Vert PT_{\varphi }e_n\Vert ^2\\= & {} \Vert M_{\varphi }^{*}e_{2n}\Vert ^2-\Vert PT_{\varphi }e_n\Vert ^2\\= & {} \bigg \Vert \sum _{j=-\infty }^{\infty }\overline{\hat{\varphi }(j)}e_{2n-j}\bigg \Vert ^2-\bigg \Vert \sum _{j=-n}^{\infty }\hat{\varphi }(j)e_{n+j} \bigg \Vert ^2\\= & {} \sum _{j=-\infty }^{\infty }|{\hat{\varphi }(j)}|^2-\sum _{j=-n}^{\infty }|\hat{\varphi }(j)|^2\\= & {} \sum _{j=n-1}^{\infty }|\hat{\varphi }(-j)|^2\ge 0 \end{aligned}$$

since $K^{*}$ is unitary. Put $f(z)=e_{2n+1}$ for each n. Then Lemma 2.7 ensures that

$$\begin{aligned} \Vert S_{\varphi }^{*}e_{2n+1}\Vert ^2-\Vert S_{\varphi }e_{2n+1}\Vert ^2= & {} \Vert S_{\varphi }^{*}e_{2n+1}\Vert ^2-\Vert H_{\varphi }e_{n}\Vert ^2\\= & {} \Vert K^{*}M_{\varphi }^{*}e_{2n+1}\Vert ^2-\Vert PM_{\varphi }Je_{n}\Vert ^2\\= & {} \Vert M_{\varphi }^{*}e_{2n+1}\Vert ^2-\Vert PM_{\varphi }Je_{n}\Vert ^2\\= & {} \bigg \Vert \sum _{j=-\infty }^{\infty }\overline{\hat{\varphi }(j)}e_{2n+1-j}\bigg \Vert ^2-\bigg \Vert P(\sum _{j=-\infty }^{\infty }\hat{\varphi }(j)e_{j-n-1})\bigg \Vert ^2\\= & {} \sum _{j=-\infty }^{\infty }|{\hat{\varphi }(j)}|^2-\sum _{j=n+1}^{\infty }|\hat{\varphi }(j)|^2\\= & {} \sum _{j=-\infty }^{n}|\hat{\varphi }(j)|^2\ge 0. \end{aligned}$$

Hence we conclude that $S_{\varphi }^{*}$ is hyponormal. $\square$

Theorem 2.9

Let $\varphi \in L^{\infty }$ and $S_{\varphi }$ be an H-Toeplitz operator on $H^2$. Then the following statement hold.

(i) If $\varphi$ is a nonzero constant function, then $S_{\varphi }$ is not quasinormal, but its adjoint $S_{\varphi }^{*}$ is quasinormal.

(ii) If $\varphi =\lambda u$ for an inner function u and $\lambda \in {\mathbb C}$, then $S_{\varphi }^{*}$ is quasinormal.

Proof

(i) Let $\varphi =\varphi _1+\overline{\varphi _2}\in L^{\infty }$ where $\varphi _1,\varphi _2\in H^{\infty }$. Then $S_{\varphi }$ is quasinormal if and only if, for each $n\in {\mathbb N}_0$,

$$\begin{aligned} 0= & {} (S_{\varphi }^{*}S_{\varphi }S_{\varphi }-S_{\varphi }S_{\varphi }^{*}S_{\varphi })e_{2n}\\= & {} K^{*}M_{\overline{\varphi }}PM_{\varphi }KPM_{\varphi }Ke_{2n}-PM_{\varphi }KK^{*}M_{\overline{\varphi }}PM_{\varphi }Ke_{2n} \\= & {} K^{*}M_{\overline{\varphi }}PM_{\varphi }KPM_{\varphi }e_{n}-PM_{\varphi }M_{\overline{\varphi }}PM_{\varphi }e_{n} \\= & {} K^{*}M_{\overline{\varphi }}PM_{\varphi }KPM_{\varphi }e_{n}-PM_{|\varphi |^2}PM_{\varphi }e_{n} \end{aligned}$$

(4)

and

$$\begin{aligned} 0= & {} (S_{\varphi }^{*}S_{\varphi }S_{\varphi }-S_{\varphi }S_{\varphi }^{*}S_{\varphi })e_{2n+1}\\= & {} K^{*}M_{\overline{\varphi }}PM_{\varphi }KPM_{\varphi }Ke_{2n+1}-PM_{\varphi }KK^{*}M_{\overline{\varphi }}PM_{\varphi }Ke_{2n+1} \\= & {} K^{*}M_{\overline{\varphi }}PM_{\varphi }KPM_{\varphi }e_{-n-1}-PM_{\varphi }M_{\overline{\varphi }}PM_{\varphi }e_{-n-1} \\= & {} K^{*}M_{\overline{\varphi }}PM_{\varphi }KPM_{\varphi }e_{-n-1}-PM_{|\varphi |^2}PM_{\varphi }e_{-n-1}. \end{aligned}$$

If $\varphi =c$ is nonzero constant and n is odd, then (4) becomes

$$\begin{aligned} K^{*}M_{\overline{\varphi }}PM_{\varphi }KPM_{\varphi }e_{n}-PM_{|\varphi |^2}PM_{\varphi }e_{n}= & {} K^{*}M_{\overline{\varphi }}PM_{\varphi }Kce_n-PM_{|c|^2}ce_n\\= & {} K^{*}M_{\overline{\varphi }}P(c^2e_{\frac{-n-1}{2}})-{|c|^2}ce_n=-{|c|^2}ce_n\\\ne & {} 0. \end{aligned}$$

Hence $S_{\varphi }$ is not quasinormal.

On the other hand, $S_{\varphi }^{*}$ is quasinormal if and only if $S_{\varphi }S_{\varphi }^{*}S_{\varphi }^{*}-S_{\varphi }^{*}S_{\varphi }S_{\varphi }^{*}=0.$ Since $S_{\varphi }S_{\varphi }^{*}=T_{|\varphi |^2}$, it follows that $S_{\varphi }^*$ is quasinormal if and only if

$$\begin{aligned} T_{|\varphi |^2}S_{\varphi }^{*}=S_{\varphi }^{*}T_{|\varphi |^2}. \end{aligned}$$

(5)

If $\varphi$ is a constant function, i.e. $\varphi =c$, then

$$\begin{aligned} (T_{|\varphi |^2}S_{\varphi }^{*}-S_{\varphi }^{*}T_{|\varphi |^2})e_{2n}= & {} (T_{|c|^2}S_{c}^{*}-S_{c}^{*}T_{|c|^2})e_{2n}\\= & {} T_{|c|^2}K^{*}M_{\overline{c}}e_{2n}-K^{*}M_{\overline{c}}T_{|c|^2}e_{2n}\\= & {} P(\overline{c}|c|^2 K^{*}e_{2n})-\overline{c}|c|^2 K^{*}e_{2n}=0 \end{aligned}$$

and

$$\begin{aligned} (T_{|\varphi |^2}S_{\varphi }^{*}-S_{\varphi }^{*}T_{|\varphi |^2})e_{2n+1}= & {} (T_{|c|^2}S_{c}^{*}-S_{c}^{*}T_{|c|^2})e_{2n+1}\\= & {} T_{|c|^2}K^{*}M_{\overline{c}}e_{2n+1}-K^{*}M_{\overline{c}}T_{|c|^2}e_{2n+1}\\= & {} P(\overline{c}|c|^2 K^{*}e_{2n+1})-\overline{c}|c|^2 K^{*}e_{2n+1}=0 \end{aligned}$$

for each $n\in {\mathbb N}_0$. Therefore, $S_{\varphi }^{*}$ is quasinormal.

(ii) Since $\varphi =\lambda u$ for an inner function u and $\lambda \in {\mathbb C}$, it follows that

$$\begin{aligned} S_{\varphi }S_{\varphi }^{*}=(PM_{\varphi }K)(K^{*}M_{\overline{\varphi }})=PM_{\lambda u}M_{\overline{\lambda u}}=PM_{|\lambda |^2|u|^2}=T_{|\lambda |^2}=|\lambda |^2 I. \end{aligned}$$

Thus (5) holds. Hence $S_{\varphi }^{*}$ is quasinormal. $\square$

We next consider the hyponormality and the binormality of $S_{\varphi }$.

Proposition 2.10

For $\varphi \in L^{\infty }$, let $S_{\varphi }$ be an H-Toeplitz operator on $H^2$. Then the following statements are equivalent.

(i) $S_{\varphi }$ is normal.

(ii) $S_{\varphi }$ is hyponormal.

(iii) $\varphi =0$.

Proof

If $\varphi =0$, then $S_{\varphi }$ is normal, and hence hyponormal. If $S_{\varphi }$ is hyponormal, the proof follows from [1]. $\square$

Theorem 2.11

Let $\varphi \in L^{\infty }$and $S_{\varphi }$ be an H-Toeplitz operator on $H^2$. Assume that one of the following statements hold.

(i) $\varphi$ is a constant function.

(ii) $\varphi =\lambda u$ for an inner function u and $\lambda \in {\mathbb C}$.

(iii) $\varphi =\lambda \overline{u}$ for an inner function u and $\lambda \in {\mathbb C}$. Then $S_{\varphi }$ is binormal.

Proof

Let $\varphi \in L^{\infty }$. Then $S_{\varphi }$ is binormal if and only if $S_{\varphi }^{*}S_{\varphi }$ and $S_{\varphi }S_{\varphi }^{*}$ commute. This is equivalent to $S_{\varphi }^{*}S_{\varphi }$ and $T_{|\varphi |^2}$ commute. Thus $S_{\varphi }$ is binormal if and only if

$$\begin{aligned}{}[S_{\varphi }^{*}S_{\varphi },S_{\varphi }S_{\varphi }^{*}]=[S_{\varphi }^{*}S_{\varphi }, T_{|\varphi |^2}]=[K^{*}M_{\overline{\varphi }}PM_{\varphi }K,T_{|\varphi |^2}]=0. \end{aligned}$$

(6)

(i) If $\varphi$ is a constant function, then (6) clearly holds.

(ii) If $\varphi =\lambda u$ for an inner function u and $\lambda \in {\mathbb C}$, then $S_{\varphi }^{*}$ is quasinormal and so $S_{\varphi }^{*}$ is binormal. Hence $S_{\varphi }$ is binormal.

(iii) If $\varphi =\lambda \overline{u}$ for an inner function u and $\lambda \in {\mathbb C}$, then

$$\begin{aligned} S_{\varphi }S_{\varphi }^{*}=(PM_{\varphi }K)(K^{*}M_{\overline{\varphi }})=PM_{\lambda \overline{u}}M_{\overline{\lambda } u}=PM_{|\lambda |^2|u|^2}=T_{|\lambda |^2}=|\lambda |^2 I. \end{aligned}$$

Thus (6) clearly holds. Hence $S_{\varphi }$ is binormal. $\square$

Example 2.12

If $\varphi (z)=z^m$ for some m, then by Theorem 2.11, $S_{z^m}$ is binormal and by Theorem 2.9, $S_{z^m}$ is not quasinormal and $S_{{z}^m}^{*}$ is quasinormal.

Example 2.13

Let $\varphi (z)=\lambda \Big (\frac{z-\mu }{1-\overline{\mu }z}\Big )$ for $\mu \in {\mathbb D}$ and $\lambda \in {\mathbb C}$. Then $S_{\varphi }$ is binormal from Theorem 2.11.

Corollary 2.14

For $\varphi \in L^{\infty }$, let $S_{\varphi }$ be an H-Toeplitz operator on $H^2$. Assume that one of the following statements hold.

(i) $\varphi$ is a constant function.

(ii) $\varphi =\lambda u$ for an inner function u and $\lambda \in {\mathbb C}$.

(iii) $\varphi =\lambda \overline{u}$ for an inner function u and $\lambda \in {\mathbb C}$.

Then $S_{\varphi }^{*}$ has a nontrivial invariant subspace.

Proof

By Theorem 2.11, $S_{\varphi }$ is binormal. Hence $S_{\varphi }^{*}$ is binormal. Since $S_{\varphi }^{*}$ is hyponormal by Theorem 2.8, we conclude that $S_{\varphi }^{*}$ has a nontrivial invariant subspace from [4]. $\square$

3 H-Toeplitz operators on the weighted Bergman spaces

3.1 Preliminaries and auxiliary lemmas

For $-1<\alpha <\infty$, the weighted Bergman spaces $A_{\alpha }^2(\mathbb D)$ is the space of analytic functions in $L^2(\mathbb D)\equiv L^2(\mathbb D, dA_{\alpha })$, where

$$\begin{aligned} dA_{\alpha }(z)=(\alpha +1)(1-|z|^2)^{\alpha }dA(z). \end{aligned}$$

The inner product on $L^2(\mathbb D)$ is given by

$$\begin{aligned} \langle f, \ g \rangle _{\alpha }=\int _{\mathbb D}f(z)\overline{g(z)}dA_{\alpha }(z) \qquad (f, \ g \in L^2(\mathbb D, dA_{\alpha })). \end{aligned}$$

If $\alpha =0$, then, $A_{0}^2(\mathbb D)$ is the Bergman spaces. For $n\in {\mathbb N}_0,$ let

$$\begin{aligned} e_n(z)=\sqrt{\frac{\Gamma (n+\alpha +2)}{\Gamma (n+1)\Gamma (\alpha +2)}}~z^n \ \ (z \in \mathbb D). \end{aligned}$$

Here, $\Gamma (s)$ stands for the usual Gamma functions. It is easy to check that $\{e_n\}_{n=0}^\infty$ be an orthonormal set in $A_{\alpha }^2(\mathbb D)$ ([15]). Because the set of polynomials is dense in $A_{\alpha }^2(\mathbb D)$, we conclude that ${e_n}$ forms an orthonormal basis for $A_{\alpha }^2(\mathbb D)$. If $f, g\in A_{\alpha }^2(\mathbb D)$ are functions of the form

$$\begin{aligned} f(z)=\sum _{n=0}^{\infty }a_nz^n \quad \hbox {and} \quad g(z)=\sum _{n=0}^{\infty }b_nz^n, \end{aligned}$$

then

$$\begin{aligned} \langle f, \ g \rangle _{\alpha }=\sum _{n=0}^{\infty }\frac{\Gamma (n+1)\Gamma (\alpha +2)}{\Gamma (n+\alpha +2)}a_n\overline{b}_n. \end{aligned}$$

The weighted harmonic Bergman spaces $L^2_{\alpha }(\mathbb D)$ denote the space of all harmonic functions f on $\mathbb D$ such that

$$\begin{aligned} |\!|f|\!|:=\Biggl ( \int _{\mathbb D}|f(z)|^2dA_{\alpha }(z)\Biggr )^{1/2}<\infty . \end{aligned}$$

The space $L^2_{\alpha }(\mathbb D)$ is a closed subspace of $L^2(\mathbb D)$ and therefore inherits the structure of a Hilbert space from $L^2(\mathbb D)$. Let $P_{ {harm}}$ denote the orthogonal projection from $L^2(\mathbb D)$ onto $L^2_{\alpha }(\mathbb D)$.

For $\varphi \in L^{\infty }(\mathbb D),$ the multiplication operators $M_{\varphi }$ on $A_{\alpha }^2(\mathbb D)$ is defined by $M_{\varphi }(f)=\varphi f$, and the Toeplitz operators $T_{\varphi }$ on $A_{\alpha }^2(\mathbb D)$ is defined by

$$\begin{aligned} T_{\varphi } (f) = P_{\alpha }({\varphi } f), \end{aligned}$$

where $P_{\alpha }$ denotes the orthogonal projection of $L^2 (\mathbb D)$ onto $A_{\alpha }^2(\mathbb D)$ and $f\in A_{\alpha }^2(\mathbb D)$. It is evident that those operators are bounded when $\varphi \in L^{\infty }(\mathbb D)$. The Hankel operators $H_{\varphi }$ on the $A_{\alpha }^2(\mathbb D)$ is defined by

$$\begin{aligned} H_{\varphi } (f) = P_{\alpha }M_{\varphi }J(f), \end{aligned}$$

where the operators $J:A_{\alpha }^2(\mathbb D) \rightarrow \overline{A_{\alpha }^2(\mathbb D)}$ is given by $J(e_n(z))=\overline{e_{n+1}(z)}$ for all $n\in {\mathbb N}_0$.

Now, we introduce the notion of H-Toeplitz operators on the weighted Bergman spaces and discuss their various familiar properties. First of all, we recall the well-known facts.

Lemma 3.1

[19] For any $s, t\in {\mathbb N}_0$,

$$\begin{aligned} P_{\alpha }(\overline{z}^t z^s)={\left\{ \begin{array}{ll} \frac{\Gamma (s+1)\Gamma (s-t+\alpha +2)}{\Gamma (s+\alpha +2)\Gamma (s-t+1)} z^{s-t} \, \, \, \, \qquad &{}\text {if} \ s \ge t \\ 0 \qquad \qquad \qquad \qquad \qquad \, \, \, &{}\text {if} \ s < t. \end{array}\right. } \end{aligned}$$

In [13], the orthogonal projection from the space $L^2(\mathbb D)$ onto the harmonic Bergman space is given. Using a similar method, the following results can be induced.

Lemma 3.2

In the weighted harmonic Bergman spaces $L^2_{\alpha }(\mathbb D)$, for $s, t\in {\mathbb N}_0$,

$$\begin{aligned} P_{ {harm}}(\overline{z}^t z^s)={\left\{ \begin{array}{ll} \frac{\Gamma (s-t+\alpha +2)\Gamma (s+1)}{\Gamma (s+\alpha +2)\Gamma (s-t+1)} z^{s-t} \qquad &{}\text { if} \ s \ge t \\ \frac{\Gamma (t-s+\alpha +2)\Gamma (t+1)}{\Gamma (t+\alpha +2)\Gamma (t-s+1)} \overline{z}^{t-s} \qquad \, &{}\text { if} \ s < t. \end{array}\right. } \end{aligned}$$

Proof

If $s\ge t$, then

$$\begin{aligned} \begin{aligned} \langle P_{ {harm}}(\overline{z}^tz^s), z^k\rangle&=\langle \overline{z}^tz^s, z^k\rangle \\&={\left\{ \begin{array}{ll} \frac{\Gamma (\alpha +2)\Gamma (s+1)}{\Gamma (s+\alpha +2)} &{} \qquad \qquad \quad \text {if} \ k=s-t \\ 0\ \ {} &{}\qquad \qquad \quad \text {otherwise} \end{array}\right. }\\&={\left\{ \begin{array}{ll} \frac{\Gamma (k+\alpha +2)\Gamma (s+1)}{\Gamma (s+\alpha +2)\Gamma (k+1)}\langle z^k, z^k\rangle &{}\text {if} \ k=s-t \\ 0 &{}\text {otherwise} \end{array}\right. }\\&=\frac{\Gamma (s-t+\alpha +2)\Gamma (s+1)}{\Gamma (s+\alpha +2)\Gamma (s-t+1)}\langle z^{s-t}, z^k\rangle . \end{aligned} \end{aligned}$$

On the other hands, if $s< t$, then

$$\begin{aligned} \langle P_{ {harm}}(\overline{z}^tz^s), \overline{z}^k\rangle= & {} \langle \overline{z}^tz^s, \overline{z}^k\rangle \\= & {} {\left\{ \begin{array}{ll} \frac{\Gamma (\alpha +2)\Gamma (t+1)}{\Gamma (t+\alpha +2)} &{}\qquad \qquad \quad \text {if} \ k=t-s \\ 0\ \ &{}\qquad \qquad \quad \text {otherwise} \end{array}\right. }\\= & {} {\left\{ \begin{array}{ll} \frac{\Gamma (k+\alpha +2)\Gamma (t+1)}{\Gamma (t+\alpha +2)\Gamma (k+1)}\langle \overline{z}^k, \overline{z}^k\rangle &{}\text {if} \ k=t-s \\ 0 &{}\text {otherwise} \end{array}\right. }\\= & {} \frac{\Gamma (t-s+\alpha +2)\Gamma (t+1)}{\Gamma (t+\alpha +2)\Gamma (t-s+1)}\langle \overline{z}^{t-s},\overline{z}^k\rangle . \end{aligned}$$

$\square$

Next, we find the matrix representations of Toeplitz operators $T_{\varphi }$ and of Hankel operators $H_{\varphi }$ with harmonic symbols $\varphi$ on the weighted Bergman spaces. For the harmonic symbol $\varphi (z)=\sum _{i=0}^{\infty }a_iz^i+\sum _{j=1}^{\infty }b_j\overline{z}^j\in L^{\infty }(\mathbb D)$, the $(m, n)^{th}$ entry of the matrix of $T_{\varphi }$ with respect to orthonormal basis ${\mathcal B}=\{e_n\}_{n=0}^\infty$ of $A_{\alpha }^2(\mathbb D)$ is given by

$$\begin{aligned} \begin{aligned} \langle T_{\varphi }e_n, e_m\rangle&= \langle P_{\alpha }({\varphi }e_n), e_m\rangle \\&=\sqrt{\frac{\Gamma (n+\alpha +2)}{\Gamma (n+1)\Gamma (\alpha +2)}}\sqrt{\frac{\Gamma (m+\alpha +2)}{\Gamma (m+1)\Gamma (\alpha +2)}}\Biggl \langle \Biggl (\sum _{i=0}^{\infty }a_iz^i+\sum _{j=1}^{\infty }b_j\overline{z}^j\Biggr )z^n,z^m\Biggr \rangle \\&=\sqrt{\frac{\Gamma (n+\alpha +2)}{\Gamma (n+1)\Gamma (\alpha +2)}}\sqrt{\frac{\Gamma (m+\alpha +2)}{\Gamma (m+1)\Gamma (\alpha +2)}} \Biggl (\sum _{i=0}^{\infty }a_i\langle z^{i+n},z^m\rangle +\sum _{j=1}^{\infty } b_j\langle {z}^n, z^{m+j}\rangle \Biggr ). \end{aligned} \end{aligned}$$

There are two cases to consider. If $m\ge n$, then we have

$$\begin{aligned} \langle T_{\varphi }e_n, e_m\rangle= & {} \sqrt{\frac{\Gamma (n+\alpha +2)}{\Gamma (n+1)\Gamma (\alpha +2)}}\sqrt{\frac{\Gamma (m+\alpha +2)}{\Gamma (m+1)\Gamma (\alpha +2)}} \sum _{i=0}^{\infty }a_i\langle z^{i+n},z^m\rangle \\= & {} \sqrt{\frac{\Gamma (n+\alpha +2)}{\Gamma (n+1)\Gamma (\alpha +2)}}\sqrt{\frac{\Gamma (m+\alpha +2)}{\Gamma (m+1)\Gamma (\alpha +2)}} \frac{\Gamma (m+1)\Gamma (\alpha +2)}{\Gamma (m+\alpha +2)}a_{m-n}\\= & {} \sqrt{\frac{\Gamma (n+\alpha +2)\Gamma (m+1)}{\Gamma (n+1)\Gamma (m+\alpha +2)}}a_{m-n}. \end{aligned}$$

If $m<n$, then we have

$$\begin{aligned} \begin{aligned} \langle T_{\varphi }e_n, e_m\rangle&=\sqrt{\frac{\Gamma (n+\alpha +2)}{\Gamma (n+1)\Gamma (\alpha +2)}}\sqrt{\frac{\Gamma (m+\alpha +2)}{\Gamma (m+1)\Gamma (\alpha +2)}} \sum _{j=1}^{\infty }b_j\langle z^{n},z^{m+j}\rangle \\&=\sqrt{\frac{\Gamma (n+\alpha +2)}{\Gamma (n+1)\Gamma (\alpha +2)}}\sqrt{\frac{\Gamma (m+\alpha +2)}{\Gamma (m+1)\Gamma (\alpha +2)}} \frac{\Gamma (n+1)\Gamma (\alpha +2)}{\Gamma (n+\alpha +2)}b_{n-m}\\&=\sqrt{\frac{\Gamma (m+\alpha +2)\Gamma (n+1)}{\Gamma (m+1)\Gamma (n+\alpha +2)}}b_{n-m}. \end{aligned} \end{aligned}$$

Thus, we have

$$\begin{aligned} \langle T_{\varphi }e_n, e_m\rangle ={\left\{ \begin{array}{ll}\sqrt{\frac{\Gamma (n+\alpha +2)\Gamma (m+1)}{\Gamma (n+1)\Gamma (m+\alpha +2)}}a_{m-n} \ \ \text {for} \ \ m\ge n\\ \sqrt{\frac{\Gamma (m+\alpha +2)\Gamma (n+1)}{\Gamma (m+1)\Gamma (n+\alpha +2)}}b_{n-m} \ \, \ \text {for} \ \ m<n, \end{array}\right. } \end{aligned}$$

where $m, n\in {\mathbb N}_0$. Therefore, the matrix representation of $T_{\varphi }$ is given by

$$\begin{aligned}{}[T_{\varphi }]_\mathcal {B}=\begin{pmatrix} a_0 &{}\quad \sqrt{\frac{1}{\alpha +2}}b_1 &{}\quad \sqrt{\frac{1\cdot 2}{(\alpha +2)(\alpha +3)}}b_2 &{}\quad \sqrt{\frac{1\cdot 2\cdot 3}{(\alpha +2)(\alpha +3)(\alpha +4)}}b_3 &{}\quad \cdots \\ \sqrt{\frac{1}{\alpha +2}}a_1 &{}\quad a_0 &{}\quad \sqrt{\frac{1\cdot 2}{\alpha +3}}b_1 &{}\quad \sqrt{\frac{1\cdot 2\cdot 3}{(\alpha +3)(\alpha +4)}}b_2 &{}\quad \cdots \\ \sqrt{\frac{1\cdot 2}{(\alpha +2)(\alpha +3)}}a_2 &{}\quad \sqrt{\frac{1\cdot 2}{\alpha +3}}a_1 &{}\quad a_0 &{}\quad \sqrt{\frac{3}{\alpha +4}}b_1 &{}\quad \cdots \\ \sqrt{\frac{1\cdot 2\cdot 3}{(\alpha +2)(\alpha +3)(\alpha +4)}}a_3 &{}\quad \sqrt{\frac{1\cdot 2\cdot 3}{(\alpha +3)(\alpha +4)}}a_2 &{}\quad \sqrt{\frac{3}{\alpha +4}}a_1 &{}\quad a_0 &{}\quad \cdots \\ \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \ddots \end{pmatrix} \end{aligned}$$

and the adjoint of the matrix representation of $T_{\varphi }$ is given by

$$\begin{aligned}{}[T_{\varphi }^*]_\mathcal {B}=\begin{pmatrix} \overline{a_0} &{}\quad \sqrt{\frac{1}{\alpha +2}}\overline{a_1} &{}\quad \sqrt{\frac{1\cdot 2}{(\alpha +2)(\alpha +3)}}\overline{a_2} &{}\quad \sqrt{\frac{1\cdot 2\cdot 3}{(\alpha +2)(\alpha +3)(\alpha +4)}}\overline{a_3} &{}\quad \cdots \\ \sqrt{\frac{1}{\alpha +2}}\overline{b_1} &{}\quad \overline{a_0} &{}\quad \sqrt{\frac{1\cdot 2}{\alpha +3}}\overline{a_1} &{}\quad \sqrt{\frac{1\cdot 2\cdot 3}{(\alpha +3)(\alpha +4)}}\overline{a_2} &{}\quad \cdots \\ \sqrt{\frac{1\cdot 2}{(\alpha +2)(\alpha +3)}}\overline{b_2} &{}\quad \sqrt{\frac{1\cdot 2}{\alpha +3}}\overline{b_1} &{}\quad \overline{a_0} &{}\quad \sqrt{\frac{3}{\alpha +4}}\overline{a_1} &{}\quad \cdots \\ \sqrt{\frac{1\cdot 2\cdot 3}{(\alpha +2)(\alpha +3)(\alpha +4)}}\overline{b_3} &{}\quad \sqrt{\frac{1\cdot 2\cdot 3}{(\alpha +3)(\alpha +4)}}\overline{b_2} &{}\quad \sqrt{\frac{3}{\alpha +4}}\overline{b_1} &{}\quad \overline{a_0} &{}\quad \cdots \\ \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \ddots \end{pmatrix} \end{aligned}$$

and hence, we check that $T_{\varphi }^*=T_{\overline{\varphi }}$.

Next, for the harmonic symbol $\varphi (z)=\sum _{i=0}^{\infty }a_iz^i+\sum _{j=1}^{\infty }b_j\overline{z}^j\in L^{\infty }(\mathbb D)$, the $(m, n)^{th}$ entry of the matrix of $H_{\varphi }$ with respect to orthonormal basis ${\mathcal B}=\{e_n\}_{n=0}^\infty$ of $A_{\alpha }^2(\mathbb D)$ is given by

$$\begin{aligned} \begin{aligned} \langle H_{\varphi }e_n, e_m\rangle&= \langle P_{\alpha }M_{\varphi }Je_n, e_m\rangle \\&=\sqrt{\frac{\Gamma (n+\alpha +3)}{\Gamma (n+2)\Gamma (\alpha +2)}}\sqrt{\frac{\Gamma (m+\alpha +2)}{\Gamma (m+1)\Gamma (\alpha +2)}}\Biggl \langle \Biggl (\sum _{i=0}^{\infty }a_iz^i+\sum _{j=1}^{\infty }b_j\overline{z}^j\Biggr )\overline{z}^{n+1},z^m\Biggr \rangle \\&=\sqrt{\frac{\Gamma (n+\alpha +3)}{\Gamma (n+2)\Gamma (\alpha +2)}}\sqrt{\frac{\Gamma (m+\alpha +2)}{\Gamma (m+1)\Gamma (\alpha +2)}}\\&\qquad \bigg (\sum _{i=0}^{\infty }a_i\langle z^{i},z^{m+n+1}\rangle +\sum _{j=1}^{\infty } b_j\langle \overline{z}^j, z^{m+n+1}\rangle \bigg )\\&=\sqrt{\frac{\Gamma (n+\alpha +3)\Gamma (m+\alpha +2)}{\Gamma (n+2)\Gamma (m+1)}}\frac{\Gamma (m+n+2)}{\Gamma (m+n+\alpha +3)}a_{m+n+1} \end{aligned} \end{aligned}$$

for $m, n\in {\mathbb N}_0$. Therefore, the matrix representation of $H_{\varphi }$ is given by

$$\begin{aligned} {[}H_{\varphi }]_\mathcal {B}=\begin{pmatrix} \sqrt{\frac{1}{\alpha +2}}a_1 &{}\quad \sqrt{\frac{1\cdot 2}{(\alpha +2)(\alpha +3)}}a_2 &{}\quad \sqrt{\frac{1\cdot 2\cdot 3}{(\alpha +2)(\alpha +3)(\alpha +4)}}a_3 &{}\quad \cdots \\ {\frac{1\cdot 2}{\alpha +3}}a_2 &{}\quad \frac{3\sqrt{2}}{(\alpha +4)\sqrt{\alpha +3}}a_3 &{}\quad \frac{4\sqrt{6}}{(\alpha +5)\sqrt{(\alpha +3)(\alpha +4)}}a_4 &{}\quad \cdots \\ \frac{3\sqrt{2}}{(\alpha +4)\sqrt{\alpha +3}}a_3 &{}\quad \frac{12}{(\alpha +4)(\alpha +5)}a_4 &{}\quad \frac{20\sqrt{3}}{(\alpha +5)(\alpha +6)\sqrt{\alpha +4}}a_5 &{}\quad \cdots \\ \frac{4\sqrt{6}}{(\alpha +5)\sqrt{(\alpha +3)(\alpha +4)}}a_4 &{}\quad \frac{20\sqrt{3}}{(\alpha +5)(\alpha +6)\sqrt{\alpha +4}}a_5 &{}\quad \frac{4\cdot 5\cdot 6}{(\alpha +5)(\alpha +6)(\alpha +7)}a_6&{}\quad \cdots \\ \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \ddots \end{pmatrix}. \end{aligned}$$

Notation 3.3

For our convenience, we introduce the following notations:

$$\begin{aligned} \Lambda _{\alpha }(s)=\frac{\Gamma (s+1)\Gamma (\alpha +2)}{\Gamma (s+\alpha +2)} \quad \text {and} \quad \Lambda _{\alpha }(s,t)=\frac{\Gamma (s+1)^2\Gamma (s-t+\alpha +2)\Gamma (\alpha +2)}{\Gamma (s+\alpha +2)^2 \Gamma (s-t+1)}. \end{aligned}$$

Lemma 3.4

[19] For $m\ge 0$, we have that

(i) $\displaystyle {\mathinner {\Big |\!\Big |\overline{z}^m \sum _ {j=0}^{\infty } c_{j}z^{j}\Big |\!\Big |}^2 =\sum _{j=0}^{\infty } \Lambda _{\alpha }(j+m)|c_{j}|^2}$, and

(ii) $\displaystyle {\mathinner {\Big |\!\Big |P_{\alpha }\bigg (\overline{z}^m \sum _ {j=0}^{\infty } c_{j}z^{j} \bigg )\Big |\!\Big |}^2}={\left\{ \begin{array}{ll} \displaystyle {\sum _{j=0}^{\infty } \Lambda _{\alpha }(j,m)|c_{j}|^2 \quad \text {if} \quad m \le j} \\ \displaystyle {\sum _{j=1}^{\infty } \Lambda _{\alpha }(j,m)|c_{j}|^2 \quad \text {if} \quad m > j}. \end{array}\right. }$

Applying Lemmas 3.2 and 3.4, we obtain the following Remarks.

Remark 3.5

For $m \ge 0$, we have

$$\begin{aligned} \displaystyle {|\!|P_{ {harm}}\bigg (\overline{z}^m \sum _ {j=0}^{\infty } c_{j}z^{j}\bigg )|\!|^2} =\displaystyle {\sum _{j=0}^{m} \Lambda _{\alpha }(m,j)|c_{j}|^2} + \sum _{j=m+1}^{\infty } \Lambda _{\alpha }(j,m)|c_{j}|^2. \end{aligned}$$

To define the notion of H-Toeplitz operators on $A_{\alpha }^2(\mathbb D)$, we start by considering the operators $K: A_{\alpha }^2(\mathbb D) \rightarrow L^2_{\alpha }(\mathbb D)$ defined by

$$\begin{aligned} K(e_{2n}(z)) = e_n(z) ~\text{ and } \,\,K(e_{2n+1}(z)) = \overline{e_{n+1}(z)} \end{aligned}$$

(7)

for all $n\ge 0$ and $z\in \mathbb D$. The operator K can be shown to be a bounded linear operator on $A_{\alpha }^2(\mathbb D)$ with $|\!|K|\!| = 1$. Furthermore, the adjoint operator $K^*$ is given by

$$\begin{aligned} K^*(e_{n}(z)) = e_{2n}(z) \ \text {and} \ K^*(\overline{e_{n+1}(z)}) = e_{2n+1}(z) \quad \end{aligned}$$

for all $n\ge 0$. From the definitions of the operators K and $K^*$, we have that $KK^*=I_{L^2_{\alpha }(\mathbb D)}$ and $K^*K=I_{A_{\alpha }^2(\mathbb D)}$.

Remark 3.6

It follows from the definition of operator K, we have

$$\begin{aligned} K(z^{2n})=\frac{\sqrt{\Gamma (2n+1)\Gamma (n+\alpha +2)}}{\sqrt{\Gamma (n+1)\Gamma (2n+\alpha +2)}}z^n, K(z^{2n+1})=\frac{\sqrt{\Gamma (2n+2)\Gamma (n+\alpha +3)}}{\sqrt{\Gamma (n+2)\Gamma (2n+\alpha +3)}}\overline{z}^{n+1}, \end{aligned}$$

$K^*(z^{n})=\frac{\sqrt{\Gamma (n+1)\Gamma (2n+\alpha +2)}}{\sqrt{\Gamma (2n+1)\Gamma (n+\alpha +2)}}z^{2n},$ and $K^*(\overline{z}^{n+1})=\frac{\sqrt{\Gamma (n+2)\Gamma (2n+\alpha +3)}}{\sqrt{\Gamma (2n+2)\Gamma (n+\alpha +3)}}{z}^{2n+1}.$

We next define H-Toeplitz operators on the weighted Bergman spaces $A_{\alpha }^2(\mathbb D)$.

Definition 3.7

For $\varphi \in L^{\infty }(\mathbb D)$, the H-Toeplitz operator $B_{\varphi }$ on the weighted Bergman space is defined by $B_{\varphi }(f) = P_{\alpha }M_{\varphi }K(f)$ for all $f \in A_{\alpha }^2(\mathbb D)$ where K is defined as in (7).

We find the matrix representation of H-Toeplitz operators $B_{\varphi }$ with harmonic symbol $\varphi$ on the weighted Bergman spaces. If the harmonic symbol of the form $\varphi (z)=\sum _{i=0}^{\infty }a_iz^i+\sum _{j=1}^{\infty }b_j\overline{z}^j\in L^{\infty }(\mathbb D)$, then

$$\begin{aligned} B_{\varphi }(e_{2n})=P_{\alpha }M_{\varphi }K(e_{2n})=P_{\alpha }M_{\varphi }(e_{n})=T_{\varphi }(e_n) \end{aligned}$$

and

$$\begin{aligned} B_{\varphi }(e_{2n+1})=P_{\alpha }M_{\varphi }K(e_{2n+1})=P_{\alpha }M_{\varphi }(\overline{e_{n+1}})=P_{\alpha }M_{\varphi }J(e_n)=H_{\varphi }(e_n) \end{aligned}$$

where $\{e_n\}_{n=0}^\infty$ is an orthonormal set in $A_{\alpha }^2(\mathbb D)$. Thus

$$\begin{aligned} \begin{aligned} \langle B_{\varphi }e_{2n}, e_m\rangle&=\langle T_{\varphi }e_{n}, e_m\rangle \\&={\left\{ \begin{array}{ll}\sqrt{\frac{\Gamma (n+\alpha +2)\Gamma (m+1)}{\Gamma (n+1)\Gamma (m+\alpha +2)}}a_{m-n} \quad \text {for} \ m\ge n\\ \sqrt{\frac{\Gamma (m+\alpha +2)\Gamma (n+1)}{\Gamma (m+1)\Gamma (n+\alpha +2)}}b_{n-m} \quad \, \text {for} \ m<n, \end{array}\right. } \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} \langle B_{\varphi }e_{2n+1}, e_m\rangle&=\langle H_{\varphi }e_{n}, e_m\rangle \\&=\sqrt{\frac{\Gamma (n+\alpha +3)\Gamma (m+\alpha +2)}{\Gamma (n+2)\Gamma (m+1)}}\frac{\Gamma (m+n+2)}{\Gamma (m+n+\alpha +3)}a_{m+n+1} \end{aligned} \end{aligned}$$

where $m, n\in {\mathbb N}_0$. Thus $(m, n)^{th}$ entry of the matrix representation of $B_{\varphi }$ with respect to orthonormal basis $\mathcal {B}=\{e_n\}_{n=0}^\infty$ of $A_{\alpha }^2(\mathbb D)$ is given by

$$\begin{aligned} {[}B_{\varphi }]_{\mathcal B}=\begin{pmatrix} a_0 &{}\quad \sqrt{\frac{1}{\alpha +2}}a_1 &{}\quad \sqrt{\frac{1}{\alpha +2}}b_1 &{}\quad \sqrt{\frac{1\cdot 2}{(\alpha +2)(\alpha +3)}}a_2 &{}\quad \cdots \\ \sqrt{\frac{1}{\alpha +2}}a_1 &{}\quad {\frac{1\cdot 2}{\alpha +3}}a_2 &{}\quad a_0 &{}\quad \frac{3\sqrt{2}}{(\alpha +4)\sqrt{\alpha +3}}a_3 &{}\quad \cdots \\ \sqrt{\frac{1\cdot 2}{(\alpha +2)(\alpha +3)}}a_2 &{}\quad \frac{3\sqrt{2}}{(\alpha +4)\sqrt{\alpha +3}}a_3 &{}\quad \sqrt{\frac{1\cdot 2}{\alpha +3}}a_1 &{}\quad \frac{12}{(\alpha +4)(\alpha +5)}a_4&{}\quad \cdots \\ \sqrt{\frac{1\cdot 2\cdot 3}{(\alpha +2)(\alpha +3)(\alpha +4)}}a_3 &{}\quad \frac{4\sqrt{6}}{(\alpha +5)\sqrt{(\alpha +3)(\alpha +4)}}a_4 &{}\quad \sqrt{\frac{1\cdot 2\cdot 3}{(\alpha +3)(\alpha +4)}}a_2 &{}\quad \frac{20\sqrt{3}}{(\alpha +5)(\alpha +6)\sqrt{\alpha +4}}a_5 &{}\quad \cdots \\ \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \ddots \end{pmatrix}. \end{aligned}$$

The following proposition presents some basic properties of H-Toeplitz operators on the weighted Bergman spaces (cf. [13]).

Proposition 3.8

For $\varphi , \psi \in L^{\infty }(\mathbb D)$, the operator $B_{\varphi }$ satisfies the following:

(i) $B_{\varphi }$ is a bounded linear operators on $A_{\alpha }^2(\mathbb D)$ with $|\!|B_{\varphi }|\!|\le |\!|\varphi |\!|_{\infty }$.

(ii) For any scalars $\alpha$ and $\beta$, it holds $B_{\alpha \varphi +\beta \psi }=\alpha B_{\varphi }+\beta B_{\psi }$.

(iii) The adjoint of the H-Toeplitz operators $B_{\varphi }$ is given by $B^*_{\varphi }=K^*P_{ {harm}}M_{\overline{\varphi }}$.

The following remark provides an important information regarding adjoint operators, showing the difference between the adjoint of Toeplitz operators and the adjoint of H-Toeplitz operators.

Remark 3.9

If f, g are in $L^{\infty }(\mathbb D)$, then, by the definition of Toeplitz operators, we have that

$$\begin{aligned} T^*_f = T_{\overline{f}} \quad \text {and} \quad T_{\overline{f}}T_g = T_{\overline{f}g} \ \text {if} \ f \ \text {or} \ g \ \text {is analytic.} \end{aligned}$$

However, for the case of the H-Toeplitz operators,

$$\begin{aligned} B^*_z(az)=K^*P_{ {harm}}M_{\overline{z}}(az)=K^*P_{ {harm}}({a\overline{z}}z)=K^*\left( \frac{\Gamma (2)\Gamma (\alpha +2)}{\Gamma (\alpha +3)}a\right) =\frac{a}{\alpha +2} \end{aligned}$$

and

$$\begin{aligned} B_{\overline{z}}(az)=P_{\alpha }M_{\overline{z}}K(az)=P_{\alpha }M_{\overline{z}}a\overline{z}=P_{\alpha }(a\overline{z}^2)=0. \end{aligned}$$

Therefore, $B^*_z(az)\ne B_{\overline{z}}(az)$. It can be easily verified by computation that $B_zB_z\ne B_{z^2}$.

Recall that a bounded linear operator T on a Hilbert space is called expansive if $T^{*}T\ge I$, contractive if $T^{*}T\le I$, and isometric if $T^{*}T=I$, respectively. For $k\in A_{\alpha }^2(\mathbb D)$, let $k(z)=k_e(z)+k_o(z)$, where

$$\begin{aligned} k_e(z):= \sum _ {n=0}^{\infty } c_{2n}z^{2n} \quad \text {and} \quad k_o(z):= \sum _ {n=0}^{\infty } c_{2n+1}z^{2n+1}. \end{aligned}$$

3.2 H-Toeplitz operators with analytic symbols

In this subsection, we examine the characteristics of H-Toeplitz operators $B_{\varphi }$ with analytic symbol functions $\varphi$. First, we study the necessary condition for contractivity and expansivity of $B_{\varphi }$ where $\varphi (z)= \sum _{j=0}^{\infty }a_jz^j$ with $a_j\in \mathbb C$ under a certain additional assumptions concerning the symbol $\varphi$.

Theorem 3.10

Let $\varphi (z)= \sum _{j=0}^{\infty }a_jz^j$ and $a_j\in \mathbb C$.

(i) If $B_{\varphi }$ is contractive, then

$$\begin{aligned} \sum _{j=0}^{\infty }|a_j|^2\le 1 \quad \text { and }\quad \sum _{j=s+1}^{\infty }\frac{\Lambda _{\alpha }(j,s+1)}{\Lambda _{\alpha }(s+1)}|a_{j}|^2 \le 1 \end{aligned}$$

for any $s\in {\mathbb N}_0$.

(ii) If $B_{\varphi }$ is expansive, then

$$\begin{aligned} \sum _{j=0}^{\infty }{\Lambda _{\alpha }(j)}|a_j|^2\ge 1 \quad \text { and }\quad \sum _{j=s+1}^{\infty }\frac{\Lambda _{\alpha }(j,s+1)}{\Lambda _{\alpha }(s+1)}|a_{j}|^2 \ge 1 \end{aligned}$$

for any $s\in {\mathbb N}_0$.

Proof

For any $k\in A_{\alpha }^2(\mathbb D)$, we have

$$\begin{aligned} \begin{aligned} B_{\varphi }k(z)&=P_{\alpha }M_{\varphi }K(k_e(z)+k_o(z))\\&=P_{\alpha }M_{\varphi }\sum _{n=0}^{\infty }\left( \frac{\sqrt{\Gamma (2n+1)\Gamma (n+\alpha +2)}}{\sqrt{\Gamma (n+1)\Gamma (2n+\alpha +2)}}c_{2n}z^n+\frac{\sqrt{\Gamma (2n+2)\Gamma (n+\alpha +3)}}{\sqrt{\Gamma (n+2)\Gamma (2n+\alpha +3)}}c_{2n+1}\overline{z}^{n+1}\right) \\&=\sum _{j=0}^{\infty }\sum _{n=0}^{\infty }\frac{\sqrt{\Gamma (2n+1)\Gamma (n+\alpha +2)}}{\sqrt{\Gamma (n+1)\Gamma (2n+\alpha +2)}}a_jc_{2n}z^{n+j}\\&\quad +\sum _{j=1}^{\infty }\sum _{n=0}^{j-1}\frac{\sqrt{\Gamma (2n+2)\Gamma (n+\alpha +3)}}{\sqrt{\Gamma (n+2)\Gamma (2n+\alpha +3)}}\cdot \frac{\Gamma (j+1)\Gamma (j-n+\alpha +1)}{\Gamma (j+\alpha +2)\Gamma (j-n)}a_jc_{2n+1}{z}^{j-n-1} \end{aligned} \nonumber \\ \end{aligned}$$

(8)

for any $c_k\in \mathbb C \ (k=0,1,2,\ldots ).$ Then, from (8), the coefficient of $z^m$ is

$$\begin{aligned} \begin{aligned}&\quad \sum _{n=0}^m\frac{\sqrt{\Gamma (2n+1)\Gamma (n+\alpha +2)}}{\sqrt{\Gamma (n+1)\Gamma (2n+\alpha +2)}}a_{m-n}c_{2n}\\&\quad +\sum _{n=0}^{\infty }\frac{\sqrt{\Gamma (2n+2)\Gamma (n+\alpha +3)}}{\sqrt{\Gamma (n+2)\Gamma (2n+\alpha +3)}}\cdot \frac{\Gamma (m+n+2)\Gamma (m+\alpha +2)}{\Gamma (m+n+\alpha +3)\Gamma (m+1)}a_{n+m+1}c_{2n+1}. \end{aligned} \end{aligned}$$

For a fixed $\ell \in {\mathbb N}_0$, set $c_{\ell }\ne 0$ and $c_k=0$ for any $k\ne \ell$. We consider the following two cases:

Case 1: If $\ell =2s$ for any $s\in {\mathbb N}_0$, then

$$\begin{aligned} B_{\varphi }k(z)=\sum _{j=0}^{\infty }\frac{\sqrt{\Gamma (2s+1)\Gamma (s+\alpha +2)}}{\sqrt{\Gamma (s+1)\Gamma (2s+\alpha +2)}}a_jc_{2s}z^{s+j}. \end{aligned}$$

If $B_{\varphi }$ on $A_{\alpha }^2(\mathbb D)$ is contractive, then

$$\begin{aligned} \sum _{j=0}^{\infty }\frac{{\Gamma (2s+1)\Gamma (s+\alpha +2)}}{{\Gamma (s+1)\Gamma (2s+\alpha +2)}}\Lambda _{\alpha }(s+j)|a_j|^2|c_{2s}|^2\le \Lambda _{\alpha }(2s)|c_{2s}|^2. \end{aligned}$$

Thus

$$\begin{aligned} {} \sum _{j=0}^{\infty }\Lambda _{\alpha }(s+j)|a_j|^2\le \frac{{\Gamma (s+1)\Gamma (2s+\alpha +2)}}{{\Gamma (2s+1)\Gamma (s+\alpha +2)}}\Lambda _{\alpha }(2s)=\Lambda _{\alpha }(s) \end{aligned}$$

(9)

for any $s\in {\mathbb N}_0$. By a direct calculation, $\frac{\Lambda _{\alpha }(s+j)}{\Lambda _{\alpha }(s)}$ is increasing for $s\in {\mathbb N}_0$ and

$$\begin{aligned} \lim _{s\rightarrow \infty }\frac{\Lambda _{\alpha }(s+j)}{\Lambda _{\alpha }(s)}=1, \end{aligned}$$

and so (9) implies that

$$\begin{aligned} \sum _{j=0}^{\infty }|a_j|^2\le 1. \end{aligned}$$

Similarly, if $B_{\varphi }$ on $A_{\alpha }^2(\mathbb D)$ is expansive, then

$$\begin{aligned} {} \sum _{j=0}^{\infty }\Lambda _{\alpha }(s+j)|a_j|^2\ge \Lambda _{\alpha }(s) \end{aligned}$$

(10)

for any $s\in {\mathbb N}_0$. By setting $s=0$ in (10), we have the results.

Case 2: If $\ell =2s+1$ for any $s\in {\mathbb N}_0$, then

$$\begin{aligned} B_{\varphi }k(z)=\sum _{j=s+1}^{\infty }\frac{\sqrt{\Gamma (2s+2)\Gamma (s+\alpha +3)}}{\sqrt{\Gamma (s+2)\Gamma (2s+\alpha +3)}}\cdot \frac{\Gamma (j+1)\Gamma (j-s+\alpha +1)}{\Gamma (j+\alpha +2)\Gamma (j-s)}a_{j}c_{2s+1}z^{j-s-1}. \end{aligned}$$

If $B_{\varphi }$ on $A_{\alpha }^2(\mathbb D)$ is contractive, then

$$\begin{aligned} \sum _{j=s+1}^{\infty }\frac{{\Gamma (2s+2)\Gamma (s+\alpha +3)}}{{\Gamma (s+2)\Gamma (2s+\alpha +3)}}\cdot \frac{\Gamma (j+1)^2\Gamma (j-s+\alpha +1)^2}{\Gamma (j+\alpha +2)^2\Gamma (j-s)^2}\cdot \frac{\Lambda _{\alpha }(j-s-1)}{\Lambda _{\alpha }(2s+1)}|a_{j}|^2\le 1. \end{aligned}$$

Thus

$$\begin{aligned} \sum _{j=s+1}^{\infty }\frac{\Lambda _{\alpha }(j,s+1)}{\Lambda _{\alpha }(s+1)}|a_{j}|^2 \le 1 \end{aligned}$$

for any $s\in {\mathbb N}_0$. Similarly, if $B_{\varphi }$ on $A_{\alpha }^2(\mathbb D)$ is expansive, then

$$\begin{aligned} \sum _{j=s+1}^{\infty }\frac{\Lambda _{\alpha }(j,s+1)}{\Lambda _{\alpha }(s+1)}|a_{j}|^2 \ge 1 \end{aligned}$$

for any $s\in {\mathbb N}_0$. This completes the proof. $\square$

Example 3.11

Let $\varphi (z)= \sum _{j=1}^{\infty }\frac{1}{j^{n/2}}z^j$ for any $n\in \mathbb N$. Then,

$$\begin{aligned} \sum _{j=1}^{\infty }\frac{1}{j^n}=\zeta (n)>1, \end{aligned}$$

where $\zeta (n)$ is the Riemann-zeta function for $n\in \mathbb N$. Thus $B_{\varphi }$ is not contractive from Theorem 3.10.

Example 3.12

Let $\varphi (z)= \sum _{j=0}^{\infty }c^jz^j$ with any $|c|< 1$. Then

$$\begin{aligned} \sum _{j=0}^{\infty }{|c|^{2j}}=\frac{1}{1-|c|^2}>1. \end{aligned}$$

Hence $B_{\varphi }$ is not contractive from Theorem 3.10.

We give a description on the contractivity and the expansivity of H-Toeplitz operators in terms of the coefficients for the polynomial symbol $\varphi$ of degree n on the Bergman spaces $A_{0}^2(\mathbb D)$.

Corollary 3.13

Let $\varphi (z)= \sum _{j=1}^{n}a_jz^j$ with any $a_j\in \mathbb C$ and $n\ge 1$. If $B_{\varphi }$ is contractive on $A_{0}^2(\mathbb D)$, then $\sum _{j=1}^{n}|a_j|^2\le 1.$

Proof

From the case 1 in the proof of Theorem 3.10, if $B_{\varphi }$ is contractive, then

$$\begin{aligned} \sum _{j=1}^{n}|a_j|^2\le 1. \end{aligned}$$

(11)

Since $\frac{\Lambda _0(j,s+1)}{\Lambda _0(s+1)}$ is increasing for $j\le 2s$ and decreasing for $j\ge 2s+1$, we have

$$\begin{aligned} \max _{j\ge s+1}\frac{\Lambda _0(j,s+1)}{\Lambda _0(s+1)}=\max \Bigl \{\frac{\Lambda _0(2s,s+1)}{\Lambda _0(s+1)}, \frac{\Lambda _0(2s+1,s+1)}{\Lambda _0(s+1)}\Bigr \}=\frac{s+2}{4(s+1)} \end{aligned}$$

for any $s\in {\mathbb N}_0$, and

$$\begin{aligned} \max _{s\ge 0}\Bigl \{\frac{s+2}{4(s+1)}\Bigl \}=\frac{1}{2}. \end{aligned}$$

Thus, for any $s\in {\mathbb N}_0$, the inequality given by

$$\begin{aligned} \sum _{j=s+1}^{n}\frac{\Lambda _0(j,s+1)}{\Lambda _0(s+1)}|a_{j}|^2 \le 1 \end{aligned}$$

implies that

$$\begin{aligned} \sum _{j=1}^{n}\frac{1}{2}|a_{j}|^2 \le 1. \end{aligned}$$

(12)

From (11) and (12), we have complete the proof. $\square$

Next, we consider the necessary and sufficient condition for the contractivity and the expansivity of $B_{\varphi }$ with $\varphi (z)= az^N$ for $N\in \mathbb N$ and $a\in \mathbb C$.

Theorem 3.14

For $\varphi (z)= az^N$ with $N\in \mathbb N$ and $a\in \mathbb C$, $B_{\varphi }$ is contractive if and only if $|a|\le 1.$

Proof

From the proof of Theorem 3.10, for any $k\in A_{\alpha }^2(\mathbb D)$, we get that

$$\begin{aligned} \begin{aligned} B_{\varphi }k(z)&=P_{\alpha }M_{\varphi }K(k(z))\\&=\sum _{n=0}^{\infty }\frac{\sqrt{\Gamma (2n+1)\Gamma (n+\alpha +2)}}{\sqrt{\Gamma (n+1)\Gamma (2n+\alpha +2)}}ac_{2n}z^{n+N}\\&\quad +\sum _{n=0}^{N-1}\frac{\sqrt{\Gamma (2n+2)\Gamma (n+\alpha +3)}}{\sqrt{\Gamma (n+2)\Gamma (2n+\alpha +3)}}\cdot \frac{\Gamma (N+1)\Gamma (N-n+\alpha +1)}{\Gamma (N+\alpha +2)\Gamma (N-n)}ac_{2n+1}{z}^{N-n-1} \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} |\!|B_{\varphi }k(z)|\!|^2&=|a|^2\sum _{n=0}^{\infty }\frac{{\Gamma (2n+1)\Gamma (n+\alpha +2)}}{{\Gamma (n+1)\Gamma (2n+\alpha +2)}}\Lambda _{\alpha }(n+N)|c_{2n}|^2\\&\quad +|a|^2\sum _{n=0}^{N-1}\frac{{\Gamma (2n+2)\Gamma (n+\alpha +3)}}{{\Gamma (n+2)\Gamma (2n+\alpha +3)}}\Lambda _{\alpha }(N,n+1)|c_{2n+1}|^2. \end{aligned} \end{aligned}$$

Thus the contractivity of $B_{\varphi }$ on $A_{\alpha }^2(\mathbb D)$ is equivalent to

$$\begin{aligned} \begin{aligned}&|a|^2\sum _{n=0}^{\infty }\frac{{\Gamma (2n+1)\Gamma (n+\alpha +2)}}{{\Gamma (n+1)\Gamma (2n+\alpha +2)}}\Lambda _{\alpha }(n+N)|c_{2n}|^2\\&\quad +|a|^2\sum _{n=0}^{N-1}\frac{{\Gamma (2n+2)\Gamma (n+\alpha +3)}}{{\Gamma (n+2)\Gamma (2n+\alpha +3)}}\Lambda _{\alpha }(N,n+1)|c_{2n+1}|^2\\&\quad \le \sum _{j=0}^{\infty }\Lambda _{\alpha }(j)|c_j|^2. \end{aligned} \nonumber \\ \end{aligned}$$

(13)

There are two possibilities to consider. The first case is when $c_{\ell }\ne 0$ for $\ell$ is even, and $c_{\ell }=0$ for $\ell$ is odd, then by (13), we have

$$\begin{aligned} |a|^2\frac{{\Gamma (2n+1)\Gamma (n+\alpha +2)}}{{\Gamma (n+1)\Gamma (2n+\alpha +2)}}\Lambda _{\alpha }(n+N)|c_{2n}|^2\le \Lambda _{\alpha }(2n)|c_{2n}|^2, \end{aligned}$$

or equivalently,

$$\begin{aligned} |a|^2\le \frac{\Lambda _{\alpha }(n)}{\Lambda _{\alpha }(n+N)} \end{aligned}$$

for any $n\in {\mathbb N}_0$. By a direct calculation, $\frac{\Lambda _{\alpha }(n)}{\Lambda _{\alpha }(n+N)}$ is decreasing for n, and

$$\begin{aligned} |a|^2\le \min _{n\ge 0}\frac{\Lambda _{\alpha }(n)}{\Lambda _{\alpha }(n+N)}=\lim _{n\rightarrow \infty }\frac{\Lambda _{\alpha }(n)}{\Lambda _{\alpha }(n+N)}=1. \end{aligned}$$

(14)

The second case is when $c_{\ell }\ne 0$ for $\ell$ is odd, and $c_{\ell }=0$ for $\ell$ is even, then from (13), we have

$$\begin{aligned} |a|^2\frac{{\Gamma (2n+2)\Gamma (n+\alpha +3)}}{{\Gamma (n+2)\Gamma (2n+\alpha +3)}}\Lambda _{\alpha }(N,n+1)|c_{2n+1}|^2 \le \Lambda _{\alpha }(2n+1)|c_{2n+1}|^2, \end{aligned}$$

or equivalently,

$$\begin{aligned} |a|^2\le \frac{\Lambda _{\alpha }(n+1)}{\Lambda _{\alpha }(N,n+1)} \end{aligned}$$

for any $0\le n\le N-1$. By a simple calculation,

$$\begin{aligned} \frac{\Lambda _{\alpha }(n+1)}{\Lambda _{\alpha }(N,n+1)}=\frac{\Gamma (N+\alpha +2)^2}{\Gamma (N+1)^2}\cdot \frac{\Gamma (N-n)}{\Gamma (N-n+\alpha +1)}\cdot \frac{\Gamma (n+2)}{\Gamma (n+\alpha +3)}>1, \nonumber \\ \end{aligned}$$

(15)

since $(N+j+1)^2>(N-n+j)(n+j+2)$ for any $j\in \mathbb R$ and for all $0\le n\le N-1$. From (14) and (15), $B_{\varphi }$ is contractive if and only if $|a|\le 1$. This completes the proof. $\square$

Corollary 3.15

If $\varphi (z)= az^N$ with $N\in \mathbb N$ and $a\in \mathbb C$, then $B_{\varphi }$ is a neither expansive nor isometric operator.

Proof

It follows from the proof of Theorem 3.14 that if $B_{\varphi }$ is expansive, then

$$\begin{aligned} \begin{aligned}&|a|^2\sum _{n=0}^{\infty }\frac{{\Gamma (2n+1)\Gamma (n+\alpha +2)}}{{\Gamma (n+1)\Gamma (2n+\alpha +2)}}\Lambda _{\alpha }(n+N)|c_{2n}|^2\\&\qquad +|a|^2\sum _{n=0}^{N-1}\frac{{\Gamma (2n+2)\Gamma (n+\alpha +3)}}{{\Gamma (n+2)\Gamma (2n+\alpha +3)}}\Lambda _{\alpha }(N,n+1)|c_{2n+1}|^2\\&\quad \ge \sum _{j=0}^{\infty }\Lambda _{\alpha }(j)|c_j|^2. \end{aligned} \end{aligned}$$

(16)

If we substitute $c_j=0$ for $j\ne 2N+1$ in (16), then we obtain that

$$\begin{aligned} \Lambda _{\alpha }(2N+1)|c_{2N+1}|^2\le 0, \end{aligned}$$

which is a contradiction. $\square$

Corollary 3.16

Let $\varphi (z)= az^N$ with $N\in \mathbb N$ and $a\in \mathbb C$. Then $B_{\varphi }$ is not self-adjoint.

Proof

By the definition of the adjoint of $B_{\varphi }$, we deduce that

$$\begin{aligned} \begin{aligned} B^*_{\varphi }k(z)&=K^*P_{ {harm}}M_{\overline{\varphi }}k(z)\\&=\overline{a}K^*\Biggl ( \sum _ {n=0}^{N-1} \frac{{\Gamma (N-n+\alpha +2)\Gamma (N+1)}}{{\Gamma (N-n+1)\Gamma (N+\alpha +2)}}c_{n}\overline{z}^{N-n}\\&\quad + \sum _ {n=N}^{\infty } \frac{{\Gamma (n-N+\alpha +2)\Gamma (n+1)}}{{\Gamma (n-N+1)\Gamma (n+\alpha +2)}}c_{n}{z}^{n-N}\Biggr )\\&= \overline{a}\sum _ {n=0}^{N-1} \frac{\Gamma (N+1)\sqrt{\Gamma (N-n+\alpha +2)\Gamma (2N-2n+\alpha +1)}}{\Gamma (N+\alpha +2)\sqrt{\Gamma (N-n+1)\Gamma (2N-2n)}}c_{n}{z}^{2N-2n-1}\\&\quad + \overline{a}\sum _ {n=N}^{\infty } \frac{\Gamma (n+1)\sqrt{\Gamma (n-N+\alpha +2)\Gamma (2n-2N+\alpha +2)}}{\Gamma (n+\alpha +2)\sqrt{\Gamma (n-N+1)\Gamma (2n-2N+1)}}c_{n}{z}^{2n-2N}. \end{aligned} \end{aligned}$$

Comparing constant terms in $B_{\varphi }k(z)$ and $B^*_{\varphi }k(z)$, they are

$$\begin{aligned} \frac{\sqrt{\Gamma (2N)\Gamma (N+1)}\Gamma (\alpha +2)}{\sqrt{\Gamma (2N+\alpha +1)\Gamma (N+\alpha +2)}}ac_{2N-1} \quad \text {and} \quad \frac{{\Gamma (N+1)\Gamma (\alpha +2)}}{{\Gamma (N+\alpha +2)}}\overline{a}c_{N}, \end{aligned}$$

respectively. As $c_{2N-1}$ and $c_{N}$ can be chosen arbitrarily, it follows that the constant terms in $B_{\varphi }k(z)$ and $B^*_{\varphi }k(z)$ are different, and hence $B_{\varphi }$ is not self-adjoint. $\square$

Corollary 3.17

For $\varphi (z)= az^N$ with $N\in \mathbb N$ and $a\in \mathbb C$, $B_{\varphi }$ is not normal.

Proof

For any $k \in A_{\alpha }^2(\mathbb D)$, the normality of $B_{\varphi }$ is equivalent to $B^*_{\varphi }B_{\varphi }k(z)=B_{\varphi }B_{\varphi }^*k(z)$ or $|\!|B_{\varphi }k(z)|\!|^2=|\!|B^*_{\varphi }k(z)|\!|^2$. Using the proof of Theorem 3.14 and Corollary 3.16, we get

$$\begin{aligned} \begin{aligned} |\!|B_{\varphi }k(z)|\!|^2&=|a|^2\sum _{n=0}^{\infty }\frac{{\Gamma (2n+1)\Gamma (n+\alpha +2)}}{{\Gamma (n+1)\Gamma (2n+\alpha +2)}}\Lambda _{\alpha }(n+N)|c_{2n}|^2\\&\quad +|a|^2\sum _{n=0}^{N-1}\frac{{\Gamma (2n+2)\Gamma (n+\alpha +3)}}{{\Gamma (n+2)\Gamma (2n+\alpha +3)}}\Lambda _{\alpha }(N,n+1)|c_{2n+1}|^2 \end{aligned} \end{aligned}$$

(17)

and

$$\begin{aligned} |\!|B^*_{\varphi }k(z)|\!|^2=|a|^2 \sum _ {n=0}^{N-1} \frac{\Lambda _{\alpha }^2(N)}{\Lambda _{\alpha }(N-n)}|c_{n}|^2+ \sum _ {n=N}^{\infty } \frac{\Lambda _{\alpha }^2(n)}{\Lambda _{\alpha }(n-N)}|c_{n}|^2. \end{aligned}$$

(18)

If we substitute $c_i=0$ for $i\ne 2N+1$ in (17) and (18), then we obtain that $|\!|B_{\varphi }k(z)|\!|^2=0$ and $|\!|B^*_{\varphi }k(z)|\!|^2= \frac{\Lambda _{\alpha }^2(2N+1)}{\Lambda _{\alpha }(N+1)}|c_{2N+1}|^2\ne 0,$ which gives the results. $\square$

3.3 H-Toeplitz operators with coanalytic symbols

In this subsection, we examine the characteristics of H-Toeplitz operators $B_{\varphi }$ with coanalytic symbol $\varphi$. First, we examine the contractivity and the expansivity of $B_{\varphi }$ where $\varphi$ is of the form $\varphi (z)= \sum _{j=1}^{\infty }b_j\overline{z}^j$ with $b_j\in \mathbb C$.

Theorem 3.18

Let $\varphi (z)= \sum _{j=1}^{\infty }b_j\overline{z}^j$ with $b_j\in \mathbb C$. If $B_{\varphi }$ is contractive, then

$$\begin{aligned} \sum _{j=1}^{s}\frac{1}{\Lambda _{\alpha }(s-j)}|b_j|^2\le \frac{1}{\Lambda _{\alpha }(s)} \end{aligned}$$

for any $s\in \mathbb N$.

Proof

For any $k\in A_{\alpha }^2(\mathbb D)$,

$$\begin{aligned} \begin{aligned} B_{\varphi }k(z)&=P_{\alpha }M_{\varphi }\left[ \sum _{n=0}^{\infty }\left( \frac{\sqrt{\Gamma (2n+1)\Gamma (n+\alpha +2)}}{\sqrt{\Gamma (n+1)\Gamma (2n+\alpha +2)}}c_{2n}z^n+\frac{\sqrt{\Gamma (2n+2)\Gamma (n+\alpha +3)}}{\sqrt{\Gamma (n+2)\Gamma (2n+\alpha +3)}}c_{2n+1}\overline{z}^{n+1}\right) \right] \\&=P_{\alpha }\left( \sum _{j=1}^{\infty }\sum _{n=0}^{\infty }\frac{\sqrt{\Gamma (2n+1)\Gamma (n+\alpha +2)}}{\sqrt{\Gamma (n+1)\Gamma (2n+\alpha +2)}}b_jc_{2n}z^n\overline{z}^j\right) \\&=\sum _{n=1}^{\infty }\sum _{j=1}^{n}\frac{\sqrt{\Gamma (2n+1)\Gamma (n+1)}}{\sqrt{\Gamma (n+\alpha +2)\Gamma (2n+\alpha +2)}}\cdot \frac{{\Gamma (n-j+\alpha +2)}}{{\Gamma (n-j+1)}}b_jc_{2n}z^{n-j}. \end{aligned} \nonumber \\ \end{aligned}$$

(19)

It follows from (19) that, the coefficient of $z^m$ is

$$\begin{aligned} \sum _{n=m+1}^{\infty }\frac{\sqrt{\Gamma (2n+1)\Gamma (n+1)}}{\sqrt{\Gamma (n+\alpha +2)\Gamma (2n+\alpha +2)}}\cdot \frac{{\Gamma (m+\alpha +2)}}{{\Gamma (m+1)}}b_{n-m}c_{2n}. \end{aligned}$$

For some $s\in \mathbb N$, we set $c_{\ell }\ne 0$ if $\ell =2s$ and $c_{\ell }=0$ if $\ell \ne 2s$. If $B_{\varphi }$ on $A_{\alpha }^2(\mathbb D)$ is contractive, then

$$\begin{aligned} \sum _{j=1}^{s}\frac{\Lambda _{\alpha }(2s)\Lambda _{\alpha }(s)}{\Lambda _{\alpha }(s-j)}|b_j|^2|c_{2s}|^2\le \Lambda _{\alpha }(2s)|c_{2s}|^2. \end{aligned}$$

Therefore,

$$\begin{aligned} \sum _{j=1}^{s}\frac{1}{\Lambda _{\alpha }(s-j)}|b_j|^2\le \frac{1}{\Lambda _{\alpha }(s)}. \end{aligned}$$

This completes the proof. $\square$

Corollary 3.19

For $\varphi (z)= \sum _{j=1}^{\infty }b_j\overline{z}^i$ with $b_j\in \mathbb C$, $B_{\varphi }$ is not an expansive operator.

Proof

From the Eq. (19), if we substitute $c_j=0$ for j is even, then we obtain that $B_{\varphi }k(z)=0.$ Thus $B_{\varphi }$ on $A_{\alpha }^2(\mathbb D)$ is not an expansive operator. $\square$

Example 3.20

For $\varphi (z)= \sqrt{\alpha +3}\overline{z}+ \sqrt{\alpha +2}\overline{z}^2$, we have

$$\begin{aligned} \sum _{j=1}^{2}\frac{1}{\Lambda _{\alpha }(2-j)}|b_j|^2=(\alpha +2)(\alpha +4) > \frac{(\alpha +2)(\alpha +3)}{2}=\frac{1}{\Lambda _{\alpha }(2)}. \end{aligned}$$

Hence by Theorem 3.18, $B_{\varphi }$ is not contractive.

Next, we study the necessary and sufficient condition for the contractivity and the expansivity of $B_{\varphi }$ with $\varphi =b\overline{z}^N$ for $N\in \mathbb N$ and $b\in \mathbb C$.

Theorem 3.21

Let $\varphi (z)= b\overline{z}^N$ with $N\in \mathbb N$ and $b\in \mathbb C$. Then $B_{\varphi }$ is contractive if and only if $|b|\le 1.$

Proof

From the proof of Theorem 3.18, for any $k\in A_{\alpha }^2(\mathbb D)$,

$$\begin{aligned} B_{\varphi }k(z)=\sum _{n=N}^{\infty }\frac{\sqrt{\Gamma (2n+1)\Gamma (n+1)}}{\sqrt{\Gamma (n+\alpha +2)\Gamma (2n+\alpha +2)}}\cdot \frac{{\Gamma (n-N+\alpha +2)}}{{\Gamma (n-N+1)}}bc_{2n}z^{n-N}. \end{aligned}$$

Thus

$$\begin{aligned} |\!|B_{\varphi }k(z)|\!|^2=|b|^2\sum _{n=N}^{\infty }\frac{\Lambda _{\alpha }(2n)\Lambda _{\alpha }(n)}{\Lambda _{\alpha }(n-N)}|c_{2n}|^2. \end{aligned}$$

Hence the contractivity of $B_{\varphi }$ is equivalent to

$$\begin{aligned} |b|^2\sum _{n=N}^{\infty }\frac{\Lambda _{\alpha }(2n)\Lambda _{\alpha }(n)}{\Lambda _{\alpha }(n-N)}|c_{2n}|^2\le \sum _{n=0}^{\infty }\Lambda _{\alpha }(n)|c_n|^2. \end{aligned}$$

If we compare the terms involving $|c_{2n}|^2$, then we have

$$\begin{aligned} \frac{\Lambda _{\alpha }(2n)\Lambda _{\alpha }(n)}{\Lambda _{\alpha }(n-N)}|b|^2|c_{2n}|^2\le \Lambda _{\alpha }(2n)|c_{2n}|^2, \end{aligned}$$

and so

$$\begin{aligned} |b|^2\le \frac{\Lambda _{\alpha }(n-N)}{\Lambda _{\alpha }(n)} \end{aligned}$$

for any $n\ge N$. Since $\frac{\Lambda _{\alpha }(n-N)}{\Lambda _{\alpha }(n)}$ is decreasing for $n\ge N$, $B_{\varphi }$ is contractive if and only if

$$\begin{aligned} |b|^2\le \min _{n\ge 0}\frac{\Lambda _{\alpha }(n-N)}{\Lambda _{\alpha }(n)}=\lim _{n\rightarrow \infty }\frac{\Lambda _{\alpha }(n-N)}{\Lambda _{\alpha }(n)}=1. \end{aligned}$$

This completes the proof. $\square$

Corollary 3.22

For $\varphi (z)= b\overline{z}^N$ with $N\in \mathbb N$ and $b\in \mathbb C$, $B_{\varphi }$ is neither expansive nor isometric.

Proof

Using the result as in the proof of Theorem 3.21, the expansivity of $B_{\varphi }$ is equivalent to

$$\begin{aligned} |b|^2\sum _{n=N}^{\infty }\frac{\Lambda _{\alpha }(2n)\Lambda _{\alpha }(n)}{\Lambda _{\alpha }(n-N)}|c_{2n}|^2\ge \sum _{n=0}^{\infty }\Lambda _{\alpha }(n)|c_n|^2. \end{aligned}$$

If we substitute $c_n= 0$ for $n\ge 2N$, then we deduce that $\sum _{n=0}^{2N-1}\Lambda _{\alpha }(n)|c_n|^2\le 0,$ which is a contradiction. $\square$

3.4 H-Toeplitz operators with harmonic symbols

Finally, we analyze the properties of H-Toeplitz operators $B_{\varphi }$ that have harmonic symbols of the form $\varphi (z)= \sum _{j=0}^{\infty }a_j{z}^j+\sum _{j=1}^{\infty }b_j{\overline{z}}^j$ with $a_j, b_j \in \mathbb C$. Our focus is on determining the necessary and sufficient conditions for the contractivity and the expansivity of $B_{\varphi }$.

Theorem 3.23

Let $\varphi (z)= \sum _{j=0}^{\infty }a_j{z}^j+\sum _{j=1}^{\infty }b_j{\overline{z}}^j$ and $a_j, b_j \in \mathbb C$.

(i) If $B_{\varphi }$ is contractive, then

$$\begin{aligned} \sum _{j=0}^{\infty }\Lambda _{\alpha }(j)|a_{j}|^2\le 1, \quad \sum _{j=0}^{\infty }\frac{\Lambda _{\alpha }(s+j)}{\Lambda _{\alpha }(s)}|a_{j}|^2+\sum _{j=1}^{s}\frac{\Lambda _{\alpha }(s)}{\Lambda _{\alpha }(s-j)}|b_{j}|^2 \le 1 \end{aligned}$$

and

$$\begin{aligned} \sum _{j=s}^{\infty }\Lambda _{\alpha }(j,s)|a_{j}|^2 \le \Lambda _{\alpha }(s) \end{aligned}$$

for any $s\in \mathbb N$.

(ii) If $B_{\varphi }$ is expansive, then

$$\begin{aligned} \sum _{j=0}^{\infty }\Lambda _{\alpha }(j)|a_{j}|^2\ge 1, \quad \sum _{j=0}^{\infty }\frac{\Lambda _{\alpha }(s+j)}{\Lambda _{\alpha }(s)}|a_{j}|^2+\sum _{j=1}^{s}\frac{\Lambda _{\alpha }(s)}{\Lambda _{\alpha }(s-j)}|b_{j}|^2 \ge 1 \end{aligned}$$

and

$$\begin{aligned} \sum _{j=s}^{\infty }\Lambda _{\alpha }(j,s)|a_{j}|^2 \ge \Lambda _{\alpha }(s) \end{aligned}$$

for any $s\in \mathbb N$.

Proof

By the similar arguments as in the proof of Theorems 3.10 and 3.18, for any $k\in A_{\alpha }^2(\mathbb D)$,

$$\begin{aligned} \begin{aligned} B_{\varphi }k(z)&=\sum _{j=0}^{\infty }\sum _{n=0}^{\infty }\frac{\sqrt{\Gamma (2n+1)\Gamma (n+\alpha +2)}}{\sqrt{\Gamma (n+1)\Gamma (2n+\alpha +2)}}a_jc_{2n}z^{n+j}\\&\quad +\sum _{j=1}^{\infty }\sum _{n=0}^{j-1}\frac{\sqrt{\Gamma (2n+2)\Gamma (n+\alpha +3)}}{\sqrt{\Gamma (n+2)\Gamma (2n+\alpha +3)}}\cdot \frac{\Gamma (j+1)\Gamma (j-n+\alpha +1)}{\Gamma (j+\alpha +2)\Gamma (j-n)}a_jc_{2n+1}{z}^{j-n-1}\\&\quad +\sum _{n=1}^{\infty }\sum _{j=1}^{n}\frac{\sqrt{\Gamma (2n+1)\Gamma (n+1)}}{\sqrt{\Gamma (n+\alpha +2)\Gamma (2n+\alpha +2)}}\cdot \frac{{\Gamma (n-j+\alpha +2)}}{{\Gamma (n-j+1)}}b_jc_{2n}z^{n-j} \end{aligned} \end{aligned}$$

for any $c_j\in \mathbb C \ (j=0,1,2,\ldots ).$ For some $\ell \in {\mathbb N}_0$, set $c_{\ell }\ne 0$ and $c_j=0$ for any $j\ne \ell$. Next, we examine the two cases below:

Case 1:  If $\ell =0$, then $B_{\varphi }k(z)=\sum _{j=0}^{\infty }a_jc_{0}z^{j}.$ Thus if $B_{\varphi }$ on $A_{\alpha }^2(\mathbb D)$ is contractive, then

$$\begin{aligned} \sum _{j=0}^{\infty }\Lambda _{\alpha }(j)|a_j|^2|c_{0}|^2 \le \Lambda _{\alpha }(0)|c_0|^2, \end{aligned}$$

or equivalently $\sum _{j=0}^{\infty }\Lambda _{\alpha }(j)|a_j|^2\le 1$. Similarly, if $B_{\varphi }$ on $A_{\alpha }^2(\mathbb D)$ is expansive, then $\sum _{j=0}^{\infty }\Lambda _{\alpha }(j)|a_j|^2\ge 1$.

Case 2:  If $\ell =2s$ for any $s\in \mathbb N$ and $c_{2\,s}\ne 0$, then

$$\begin{aligned} B_{\varphi }k(z)=&\sum _{j=0}^{\infty }\frac{\sqrt{\Gamma (2s+1)\Gamma (s+\alpha +2)}}{\sqrt{\Gamma (s+1)\Gamma (2s+\alpha +2)}}a_jc_{2s}z^{s+j}\\&\quad +\sum _{j=1}^{s}\frac{\sqrt{\Gamma (2s+1)\Gamma (s+1)}}{\sqrt{\Gamma (s+\alpha +2)\Gamma (2s+\alpha +2)}}\cdot \frac{{\Gamma (s-j+\alpha +2)}}{{\Gamma (s-j+1)}}b_jc_{2s}z^{s-j}. \end{aligned}$$

If $B_{\varphi }$ on $A_{\alpha }^2(\mathbb D)$ is contractive, then

$$\begin{aligned}{} & {} \sum _{j=0}^{\infty }\frac{{\Gamma (2s+1)\Gamma (s+\alpha +2)}}{{\Gamma (s+1)\Gamma (2s+\alpha +2)}}\Lambda _{\alpha }(s+j)|a_j|^2|c_{2s}|^2\\{} & {} \quad +\sum _{j=1}^{s}\frac{\Lambda _{\alpha }(2s)\Lambda _{\alpha }(s)}{\Lambda _{\alpha }(s-j)}|b_j|^2|c_{2s}|^2\le \Lambda _{\alpha }(2s)|c_{2s}|^2 \end{aligned}$$

or equivalently

$$\begin{aligned} \sum _{j=0}^{\infty }\frac{\Lambda _{\alpha }(s+j)}{\Lambda _{\alpha }(s)}|a_{j}|^2+\sum _{j=1}^{s}\frac{\Lambda _{\alpha }(s)}{\Lambda _{\alpha }(s-j)}|b_{j}|^2 \le 1. \end{aligned}$$

Similarly, if $B_{\varphi }$ on $A_{\alpha }^2(\mathbb D)$ is expansive, then

$$\begin{aligned} \sum _{j=0}^{\infty }\frac{\Lambda _{\alpha }(s+j)}{\Lambda _{\alpha }(s)}|a_{j}|^2+\sum _{j=1}^{s}\frac{\Lambda _{\alpha }(s)}{\Lambda _{\alpha }(s-j)}|b_{j}|^2 \ge 1. \end{aligned}$$

Case 3:  If $\ell =2s-1$ for any $s\in \mathbb N$ and $c_{2s-1}\ne 0$, then by the case 2 of Theorem 3.10, we have the results. This completes the proof. $\square$

The following results can be easily derived from Theorem 3.23.

Corollary 3.24

Let $\varphi (z)= a_1z+b_1\overline{z}$ and $a_1, b_1 \in \mathbb C$. If $B_{\varphi }$ is contractive, then

$$\begin{aligned} |a_1|^2 \le \alpha +2, \quad \frac{2}{\alpha +3}|a_1|^2+\frac{1}{\alpha +2}|b_1|^2 \le 1 \end{aligned}$$

and

$$\begin{aligned} \frac{s+1}{s+\alpha +2}|a_1|^2+\frac{s}{s+\alpha +1}|b_1|^2 \le 1 \end{aligned}$$

for any $s\ge 2$.

Example 3.25

For $\varphi (z)= \frac{\sqrt{\alpha +2}}{\sqrt{2}}z+\frac{\sqrt{(\alpha +3)(3\alpha +7)}}{4}z^2-\frac{3\sqrt{(\alpha +1)(\alpha +2)}}{2\sqrt{2(\alpha +3)}}\overline{z}$, we have

$$\begin{aligned} \sum _{j=0}^{\infty }\Lambda _{\alpha }(j)|a_j|^2=\frac{1}{2}+\frac{3\alpha +7}{8(\alpha +2)}<1=\Lambda _{\alpha }(0), \\ \sum _{j=1}^{\infty }\Lambda _{\alpha }(j,1)|a_j|^2=\frac{1}{2(\alpha +2)}+\frac{3\alpha +7}{4(\alpha +2)(\alpha +3)} >\frac{1}{\alpha +2}=\Lambda _{\alpha }(1), \end{aligned}$$

and

$$\begin{aligned} \sum _{j=0}^{\infty }\frac{\Lambda _{\alpha }(1+j)}{\Lambda _{\alpha }(1)}|a_{j}|^2+\frac{\Lambda _{\alpha }(1)}{\Lambda _{\alpha }(0)}|b_{1}|^2=\frac{\alpha +2}{\alpha +3}+\frac{3(3\alpha +7)}{8(\alpha +4)}-\frac{9(\alpha +1)}{8(\alpha +3)} < 1. \end{aligned}$$

Hence, by the Theorem 3.23, $B_{\varphi }$ is neither contractive nor expansive.