Hyponormal Toeplitz operators with polynomial symbols on weighted Bergman spaces

Abstract

In this note we consider the hyponormality of Toeplitz operators $T_{\phi }$ on weighted Bergman space $A_{\alpha }^2(\mathbb{D})$ with symbol in the class of functions $f+\overline{g}$ with polynomials f and g.

MSC:47B20, 47B35.

1 Introduction

Let $\mathbb{D}$ be the open unit disk in the complex plane. For $-1<\alpha <\mathrm{\infty }$, the weighted Bergman space $A_{\alpha }^2(\mathbb{D})$ of the unit disk $\mathbb{D}$ is the space of analytic functions in $L^2(\mathbb{D},d{A}_{\alpha })$, where

$$d{A}_{\alpha }(z)=(\alpha +1){(1-{|z|}^2)}^{\alpha }dA(z).$$

The space $L^2(\mathbb{D},d{A}_{\alpha })$ is a Hilbert space with the inner product

$${〈f,g〉}_{\alpha }={\int }_{\mathbb{D}}f(z)\overline{g(z)}d{A}_{\alpha }(z)(f,g\in L^2(\mathbb{D},d{A}_{\alpha })).$$

If $\alpha =0$ then $A_0^2(\mathbb{D})$ is the Bergman space $A^2(\mathbb{D})$. For any nonnegative integer n, let

$$e_n(z)=\sqrt{\frac{\mathrm{\Gamma }(n+\alpha +2)}{\mathrm{\Gamma }(n+1)\mathrm{\Gamma }(\alpha +2)}}{z}^n(z\in \mathbb{D}),$$

where $\mathrm{\Gamma }(s)$ stands for the usual Gamma function. It is easy to check that $\{e_n\}$ is an orthonormal basis for $A_{\alpha }^2(\mathbb{D})$ [1]. For $\phi \in L^{\mathrm{\infty }}(\mathbb{D})$, the Toeplitz operator $T_{\phi }$, and the Hankel operator $H_{\phi }$ on $A_{\alpha }^2(\mathbb{D})$ are defined by

$$T_{\phi }f:=P_{\alpha }(\phi \cdot f)\text{and}{H}_{\phi }f:=(I-P_{\alpha })(\phi \cdot f)(f\in A_{\alpha }^2(\mathbb{D})),$$

where $P_{\alpha }$ denotes the orthogonal projection that maps from $L^2(\mathbb{D},d{A}_{\alpha })$ onto $A_{\alpha }^2(\mathbb{D})$. The reproducing kernel in $A_{\alpha }^2(\mathbb{D})$ is given by

$$K_z^{(\alpha )}(\omega )=\frac{1}{{(1-z\overline{\omega })}^{2+\alpha }},$$

for $z,\omega \in \mathbb{D}$. We thus have

$$(T_{\phi }f)(z)={\int }_{\mathbb{D}}\frac{\phi (\omega )f(\omega )}{{(1-z\overline{\omega })}^{2+\alpha }}d{A}_{\alpha }(\omega ),$$

for $f\in A_{\alpha }^2(\mathbb{D})$ and $\omega \in \mathbb{D}$.

A bounded linear operator A on a Hilbert space is said to be hyponormal if its selfcommutator $[A^{\ast },A]:=A^{\ast }A-A{A}^{\ast }$ is positive (semidefinite). The hyponormality of Toeplitz operators on the Hardy space $H^2(\mathbb{T})$ of the unit circle $\mathbb{T}=\partial \mathbb{D}$ has been studied by Cowen [2], Curto, Hwang and Lee [3–5] and others [6]. Recently, in [7] and [8], the hyponormality of $T_{\phi }$ on the weighted Bergman space $A_{\alpha }^2(\mathbb{D})$ was studied. In [2], Cowen characterized the hyponormality of Toeplitz operator $T_{\phi }$ on $H^2(\mathbb{T})$ by properties of the symbol $\phi \in L^{\mathrm{\infty }}(\mathbb{T})$. Here we shall employ an equivalent variant of Cowen’s theorem that was first proposed by Nakazi and Takahashi [9].

Cowen’s theorem ([2, 9])

For $\phi \in L^{\mathrm{\infty }}(\mathbb{T})$, write

$$\mathcal{E}(\phi ):=\{k\in H^{\mathrm{\infty }}:{\parallel k\parallel }_{\mathrm{\infty }}\le 1\mathit{\text{ and }}\phi -k\overline{\phi }\in H^{\mathrm{\infty }}(\mathbb{T})\}.$$

Then $T_{\phi }$ is hyponormal if and only if $\mathcal{E}(\phi )$ is nonempty.

The solution is based on a dilation theorem of Sarason [10]. For the weighted Bergman space, no dilation theorem (similar to Sarason’s theorem) is available. In [11], the first named author characterized the hyponormality of $T_{\phi }$ on $A_{\alpha }^2(\mathbb{D})$ in terms of the coefficients of the trigonometric polynomial φ under certain assumptions as regards the coefficients of φ on the weighted Bergman space when $\alpha \ge 0$ and in [12], extended for all $-1<\alpha <\mathrm{\infty }$.

Theorem A ([12])

Let $\phi (z)=\overline{g(z)}+f(z)$, where $f(z)=a_{1}z+a_2{z}^2$ $g(z)=a_{-1}z+a_{-2}{z}^2$. If $a_1\overline{a_2}=a_{-1}\overline{a_{-2}}$ and $-1<\alpha <\mathrm{\infty }$, then

$$\begin{array}{c}{T}_{\phi }\mathit{\text{ on }}{A}_{\alpha }^2(\mathbb{D})\mathit{\text{ is hyponormal }}\hfill \\ ⟺\{\begin{array}{cc}\frac{1}{\alpha +3}({|a_2|}^2-{|a_{-2}|}^2)\ge \frac{1}{2}({|a_{-1}|}^2-{|a_1|}^2)\hfill & \mathit{\text{if }}|a_{-2}|\le |a_2|,\hfill \\ 4({|a_{-2}|}^2-{|a_2|}^2)\le {|a_1|}^2-{|a_{-1}|}^2\hfill & \mathit{\text{if }}|a_2|\le |a_{-2}|.\hfill \end{array}\hfill \end{array}$$

In this note we consider the hyponormality of Toeplitz operators $T_{\phi }$ on $A_{\alpha }^2(\mathbb{D})$ with symbol in the class of functions $f+\overline{g}$ with polynomials f and g. Since the hyponormality of operators is translation invariant we may assume that $f(0)=g(0)=0$. The following relations can easily be proved:

$$T_{\phi +\psi }=T_{\phi }+T_{\psi }(\phi ,\psi \in L^{\mathrm{\infty }});$$

(1.1)

$$T_{\phi }^{\ast }=T_{\overline{\phi }}(\phi \in L^{\mathrm{\infty }});$$

(1.2)

$$T_{\overline{\phi }}{T}_{\psi }=T_{\overline{\phi }\psi }\text{if }\phi \text{ or }\psi \text{ is analytic}.$$

(1.3)

The purpose of this paper is to prove Theorem A for the Toeplitz operators on $A_{\alpha }^2(\mathbb{D})$ when f and g of degree N.

2 Main result

In this section we establish a necessary and sufficient condition for the hyponormality of the Toeplitz operator $T_{\phi }$ on the weighted Bergman space under a certain additional assumption concerning the symbol φ. The assumption is related on the symmetry, so it is reasonable in view point of the Hardy space [13]. We expect that this approach would provide some clue for the future study of the symmetry case.

Lemma 1 ([11]) For any s, t nonnegative integers,

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

For $0\le i\le N-1$, write

$$k_i(z):=\sum _{n=0}^{\mathrm{\infty }}{c}_{Nn+i}{z}^{Nn+i}.$$

The following two lemmas will be used for proving the main result of this section.

Lemma 2 For $0\le m\le N$, we have

$$\begin{array}{rl}(\mathrm{i})& {\parallel {\overline{z}}^m{k}_i(z)\parallel }_{\alpha }^2=\sum _{n=0}^{\mathrm{\infty }}\frac{\mathrm{\Gamma }(Nn+i+m+1)\mathrm{\Gamma }(\alpha +2)}{\mathrm{\Gamma }(Nn+i+m+\alpha +2)}{|c_{Nn+i}|}^2,\\ (\mathrm{ii})& {\parallel P_{\alpha }({\overline{z}}^m{k}_i(z))\parallel }_{\alpha }^2=\{\begin{array}{cc}{\sum }_{n=0}^{\mathrm{\infty }}\frac{\mathrm{\Gamma }{(Nn+i+1)}^2\mathrm{\Gamma }(Nn+i-m+\alpha +2)\mathrm{\Gamma }(\alpha +2)}{\mathrm{\Gamma }{(Nn+i+\alpha +2)}^2\mathrm{\Gamma }(Nn+i-m+1)}{|c_{Nn+i}|}^2\hfill & \mathit{\text{if }}m\le i,\hfill \\ {\sum }_{n=1}^{\mathrm{\infty }}\frac{\mathrm{\Gamma }{(Nn+i+1)}^2\mathrm{\Gamma }(Nn+i-m+\alpha +2)\mathrm{\Gamma }(\alpha +2)}{\mathrm{\Gamma }{(Nn+i+\alpha +2)}^2\mathrm{\Gamma }(Nn+i-m+1)}{|c_{Nn+i}|}^2\hfill & \mathit{\text{if }}m>i.\hfill \end{array}\end{array}$$

Proof Let $0\le m\le N$. Then we have

$$\begin{array}{rl}{\parallel {\overline{z}}^m{k}_i(z)\parallel }_{\alpha }^2& ={\parallel \sum _{n=0}^{\mathrm{\infty }}{c}_{Nn+i}{z}^{Nn+i+m}\parallel }_{\alpha }^2\\ =\sum _{n=0}^{\mathrm{\infty }}{|c_{Nn+i}|}^2{\parallel z^{Nn+i+m}\parallel }_{\alpha }^2\\ =\sum _{n=0}^{\mathrm{\infty }}\frac{\mathrm{\Gamma }(Nn+i+m+1)\mathrm{\Gamma }(\alpha +2)}{\mathrm{\Gamma }(Nn+i+m+\alpha +2)}{|c_{Nn+i}|}^2.\end{array}$$

This proves (i). For (ii), if $m\le i$ then by Lemma 1 we have

$$\begin{array}{rl}{\parallel P_{\alpha }({\overline{z}}^m{k}_i(z))\parallel }_{\alpha }^2& ={\parallel \sum _{n=0}^{\mathrm{\infty }}\frac{\mathrm{\Gamma }(Nn+i+1)\mathrm{\Gamma }(Nn+i-m+\alpha +2)}{\mathrm{\Gamma }(Nn+i+\alpha +2)\mathrm{\Gamma }(Nn+i-m+1)}{c}_{Nn+i}{z}^{Nn+i-m}\parallel }_{\alpha }^2\\ =\sum _{n=0}^{\mathrm{\infty }}\frac{\mathrm{\Gamma }{(Nn+i+1)}^2\mathrm{\Gamma }(Nn+i-m+\alpha +2)\mathrm{\Gamma }(\alpha +2)}{\mathrm{\Gamma }{(Nn+i+\alpha +2)}^2\mathrm{\Gamma }(Nn+i-m+1)}{|c_{Nn+i}|}^2.\end{array}$$

If instead $m>i$, a similar argument gives the result. □

Lemma 3 ([14])

Let $f(z)=a_{N-1}{z}^{N-1}+a_N{z}^N$ and $g(z)=a_{-(N-1)}{z}^{N-1}+a_{-N}{z}^N$. If $a_{N-1}\overline{a_N}=a_{-(N-1)}\overline{a_{-N}}$, then for $i\ne j$, we have

$${〈H_{\overline{f}}{k}_i(z),H_{\overline{f}}{k}_j(z)〉}_{\alpha }={〈H_{\overline{g}}{k}_i(z),H_{\overline{g}}{k}_j(z)〉}_{\alpha }.$$

Our main result now follows.

Theorem 4 Let $\phi (z)=\overline{g(z)}+f(z)$, where

$$f(z)=a_{N-1}{z}^{N-1}+a_N{z}^N\mathit{\text{and}}g(z)=a_{-(N-1)}{z}^{N-1}+a_{-N}{z}^N.$$

If $a_{N-1}\overline{a_N}=a_{-(N-1)}\overline{a_{-N}}$ and $|a_{-N}|\le |a_N|$, then $T_{\phi }$ on $A_{\alpha }^2(\mathbb{D})$ is hyponormal if and only if

$$\frac{1}{N+\alpha +1}({|a_N|}^2-{|a_{-N}|}^2)\ge \frac{1}{N}({|a_{-(N-1)}|}^2-{|a_{N-1}|}^2).$$

Proof For $0\le i<N$, put

$$K_i:=\{k_i(z)\in A_{\alpha }^2(\mathbb{D}):k_i(z)=\sum _{n=0}^{\mathrm{\infty }}{c}_{Nn+i}{z}^{Nn+i}\}.$$

Then a straightforward calculation shows that $T_{\phi }$ is hyponormal if and only if

$${〈(H_{\overline{f}}^{\ast }{H}_{\overline{f}}-H_{\overline{g}}^{\ast }{H}_{\overline{g}})\sum _{i=0}^{N-1}{k}_i(z),\sum _{i=0}^{N-1}{k}_i(z)〉}_{\alpha }\ge 0\text{for all }{k}_i\in K_i(i=0,1,\dots ,N-1).$$

(2.1)

Also we have

$$\begin{array}{c}{〈H_{\overline{f}}^{\ast }{H}_{\overline{f}}\sum _{i=0}^{N-1}{k}_i(z),\sum _{i=0}^{N-1}{k}_i(z)〉}_{\alpha }\hfill \\ =\sum _{i=0}^{N-1}{〈H_{\overline{f}}{k}_i(z),H_{\overline{f}}{k}_i(z)〉}_{\alpha }+\sum _{i\ne j,i,j\ge 0}^{N-1}{〈H_{\overline{f}}{k}_i(z),H_{\overline{f}}{k}_k(z)〉}_{\alpha }\hfill \end{array}$$

(2.2)

and

$$\begin{array}{c}{〈H_{\overline{g}}^{\ast }{H}_{\overline{g}}\sum _{i=0}^{N-1}{k}_i(z),\sum _{i=0}^{N-1}{k}_i(z)〉}_{\alpha }\hfill \\ =\sum _{i=0}^{N-1}{〈H_{\overline{g}}{k}_i(z),H_{\overline{g}}{k}_i(z)〉}_{\alpha }+\sum _{i\ne j,i,j\ge 0}^{N-1}{〈H_{\overline{g}}{k}_i(z),H_{\overline{g}}{k}_k(z)〉}_{\alpha }.\hfill \end{array}$$

(2.3)

Substituting (2.2) and (2.3) into (2.1), it follows from Lemma 3 that

$$\begin{array}{rl}{T}_{\phi }\text{: hyponormal}& ⟺\sum _{i=0}^{N-1}{〈(H_{\overline{f}}^{\ast }{H}_{\overline{f}}-H_{\overline{g}}^{\ast }{H}_{\overline{g}})k_i(z),k_i(z)〉}_{\alpha }\ge 0\\ ⟺\sum _{i=0}^{N-1}({\parallel \overline{f}{k}_i\parallel }_{\alpha }^2-{\parallel \overline{g}{k}_i\parallel }_{\alpha }^2+{\parallel P_{\alpha }(\overline{g}{k}_i)\parallel }_{\alpha }^2-{\parallel P_{\alpha }(\overline{f}{k}_i)\parallel }_{\alpha }^2)\ge 0.\end{array}$$

Therefore it follows from Lemma 2 that $T_{\phi }$ is hyponormal if and only if

$$\begin{array}{c}({|a_{N-1}|}^2-{|a_{-(N-1)}|}^2)\left[\sum _{i=0}^{N-2}\right\{\frac{\mathrm{\Gamma }(i+N)\mathrm{\Gamma }(\alpha +2)}{\mathrm{\Gamma }(i+N+\alpha +1)}{|c_i|}^2+\sum _{n=1}^{\mathrm{\infty }}(\frac{\mathrm{\Gamma }(Nn+i+N)\mathrm{\Gamma }(\alpha +2)}{\mathrm{\Gamma }(Nn+i+N+\alpha +1)}\hfill \\ -\frac{\mathrm{\Gamma }{(Nn+i+1)}^2\mathrm{\Gamma }(Nn+i-N+\alpha +3)\mathrm{\Gamma }(\alpha +2)}{\mathrm{\Gamma }{(Nn+i+\alpha +2)}^2\mathrm{\Gamma }(Nn+i-N+2)}\left){|c_{Nn+i}|}^2\right\}\hfill \\ +\sum _{n=0}^{\mathrm{\infty }}(\frac{\mathrm{\Gamma }(Nn+2N-1)\mathrm{\Gamma }(\alpha +2)}{\mathrm{\Gamma }(Nn+2N+\alpha )}-\frac{\mathrm{\Gamma }{(Nn+N)}^2\mathrm{\Gamma }(Nn+\alpha +2)\mathrm{\Gamma }(\alpha +2)}{\mathrm{\Gamma }{(Nn+N+\alpha +1)}^2\mathrm{\Gamma }(Nn+1)}){|c_{Nn+N-1}|}^2]\hfill \\ +({|a_N|}^2-{|a_{-N}|}^2)\left[\sum _{i=0}^{N-1}\right\{\frac{\mathrm{\Gamma }(N+i+1)\mathrm{\Gamma }(\alpha +2)}{\mathrm{\Gamma }(i+n+\alpha +2)}{|c_i|}^2\hfill \\ +\sum _{n=1}^{\mathrm{\infty }}(\frac{\mathrm{\Gamma }(Nn+i+N+1)\mathrm{\Gamma }(\alpha +2)}{\mathrm{\Gamma }(Nn+i+N+\alpha +2)}\hfill \\ -\frac{\mathrm{\Gamma }{(Nn+i+1)}^2\mathrm{\Gamma }(Nn+i-N+\alpha +2)\mathrm{\Gamma }(\alpha +2)}{\mathrm{\Gamma }{(Nn+i+\alpha +2)}^2\mathrm{\Gamma }(Nn+i-N+1)}\left){|c_{Nn+i}|}^2\right\}]\ge 0,\hfill \end{array}$$

or equivalently

$$\begin{array}{c}({|a_{N-1}|}^2-{|a_{-(N-1)}|}^2)\{\sum _{n=0}^{N-2}\frac{\mathrm{\Gamma }(N+n)\mathrm{\Gamma }(\alpha +2)}{\mathrm{\Gamma }(N+n+\alpha +1)}{|c_n|}^2+\sum _{n=N-1}^{\mathrm{\infty }}(\frac{\mathrm{\Gamma }(N+n)\mathrm{\Gamma }(\alpha +2)}{\mathrm{\Gamma }(N+n+\alpha +1)}\hfill \\ -\frac{\mathrm{\Gamma }{(n+1)}^2\mathrm{\Gamma }(n-N+\alpha +3)\mathrm{\Gamma }(\alpha +2)}{\mathrm{\Gamma }{(n+\alpha +2)}^2\mathrm{\Gamma }(n-N+2)}\left){|c_n|}^2\right\}\hfill \\ +({|a_N|}^2-{|a_{-N}|}^2)\{\sum _{n=0}^{N-1}\frac{\mathrm{\Gamma }(n+N+1)\mathrm{\Gamma }(\alpha +2)}{\mathrm{\Gamma }(N+n+\alpha +2)}{|c_n|}^2+\sum _{n=N}^{\mathrm{\infty }}(\frac{\mathrm{\Gamma }(N+n+1)\mathrm{\Gamma }(\alpha +2)}{\mathrm{\Gamma }(N+n+\alpha +2)}\hfill \\ -\frac{\mathrm{\Gamma }{(n+1)}^2\mathrm{\Gamma }(n-N+\alpha +2)\mathrm{\Gamma }(\alpha +2)}{\mathrm{\Gamma }{(n+\alpha +2)}^2\mathrm{\Gamma }(n-N+1)}\left){|c_n|}^2\right\}\ge 0.\hfill \end{array}$$

(2.4)

Define ${\zeta }_{\alpha }$ by

$${\zeta }_{\alpha }(n):=\frac{\frac{\mathrm{\Gamma }(N+n)\mathrm{\Gamma }(\alpha +2)}{\mathrm{\Gamma }(N+n+\alpha +1)}-\frac{\mathrm{\Gamma }{(n+1)}^2\mathrm{\Gamma }(n-N+\alpha +3)\mathrm{\Gamma }(\alpha +2)}{\mathrm{\Gamma }{(n+\alpha +2)}^2\mathrm{\Gamma }(n-N+2)}}{\frac{\mathrm{\Gamma }(N+n+1)\mathrm{\Gamma }(\alpha +2)}{\mathrm{\Gamma }(N+n+\alpha +2)}-\frac{\mathrm{\Gamma }{(n+1)}^2\mathrm{\Gamma }(n-N+\alpha +2)\mathrm{\Gamma }(\alpha +2)}{\mathrm{\Gamma }{(n+\alpha +2)}^2\mathrm{\Gamma }(n-N+1)}}(n\ge 1).$$

Then a direct calculation gives

$${\zeta }_{\alpha }(n)<\frac{\frac{\mathrm{\Gamma }(N+n)\mathrm{\Gamma }(\alpha +2)}{\mathrm{\Gamma }(N+n+\alpha +1)}}{\frac{\mathrm{\Gamma }(N+n+1)\mathrm{\Gamma }(\alpha +2)}{\mathrm{\Gamma }(N+n+\alpha +2)}}.$$

Observe that

$$\begin{array}{rl}\frac{N+\alpha +1}{N}& \ge \frac{N+n+\alpha +1}{N+n}\ge \frac{N+N_i+\alpha +1}{N+N_i}\\ \ge {\zeta }_{\alpha }(N_i)\text{for all }{N}_i\ge N\text{ and }n=1,2,\dots ,N-1;\end{array}$$

(2.5)

and

$$\frac{N+\alpha +1}{N}\ge \frac{\frac{\mathrm{\Gamma }(2N-1)\mathrm{\Gamma }(\alpha +2)}{\mathrm{\Gamma }(2N+\alpha )}-\frac{\mathrm{\Gamma }{(N)}^2\mathrm{\Gamma }{(\alpha +2)}^2}{\mathrm{\Gamma }{(N+\alpha +1)}^2}}{\frac{\mathrm{\Gamma }(2N)\mathrm{\Gamma }(\alpha +2)}{\mathrm{\Gamma }(2N+\alpha +1)}}.$$

Therefore (2.4) and (2.5) show that $T_{\phi }$ is hyponormal if and only if

$$\frac{1}{N+\alpha +1}({|a_N|}^2-{|a_{-N}|}^2)\ge \frac{1}{N}({|a_{-(N-1)}|}^2-{|a_{N-1}|}^2).$$

This completes the proof. □

Remark 5 Let $\phi (z)=\overline{g(z)}+f(z)$, where

$$f(z)=a_{N-1}{z}^{N-1}+a_N{z}^N\text{and}g(z)=a_{-(N-1)}{z}^{N-1}+a_{-N}{z}^N.$$

If $a_{N-1}\overline{a_N}=a_{-(N-1)}\overline{a_{-N}}$, $|a_N|\le |a_{-N}|$, and $T_{\phi }$ on $A_{\alpha }^2(\mathbb{D})$ is hyponormal. Then

$${|a_{-N}|}^2-{|a_N|}^2\le \{\frac{2N+\alpha }{2N-1}-\frac{\mathrm{\Gamma }{(N)}^2\mathrm{\Gamma }(2N+\alpha +1)\mathrm{\Gamma }(\alpha +2)}{\mathrm{\Gamma }(2N)\mathrm{\Gamma }{(N+\alpha +1)}^2}\}({|a_{N-1}|}^2-{|a_{-(N-1)}|}^2).$$

Proof If we let $c_j=1$ for $0\le j\le N-1$ and the other $c_j$’s be 0 into (2.4), then we have

$$\begin{array}{c}({|a_{N-1}|}^2-{|a_{-(N-1)}|}^2)\{\sum _{n=0}^{N-2}\frac{\mathrm{\Gamma }(N+n)\mathrm{\Gamma }(\alpha +2)}{\mathrm{\Gamma }(N+n+\alpha +1)}\hfill \\ +(\frac{\mathrm{\Gamma }(2N-1)\mathrm{\Gamma }(\alpha +2)}{\mathrm{\Gamma }(2N+\alpha )}-\frac{\mathrm{\Gamma }{(N)}^2\mathrm{\Gamma }{(\alpha +2)}^2}{\mathrm{\Gamma }{(N+\alpha +1)}^2\mathrm{\Gamma }(1)})\}\hfill \\ +({|a_N|}^2-{|a_{-N}|}^2)\sum _{n=0}^{N-1}\frac{\mathrm{\Gamma }(n+N+1)\mathrm{\Gamma }(\alpha +2)}{\mathrm{\Gamma }(N+n+\alpha +2)}\ge 0.\hfill \end{array}$$

(2.6)

Define ${\xi }_{\alpha }$ by

$${\xi }_{\alpha }(n):=\frac{\frac{\mathrm{\Gamma }(N+n)\mathrm{\Gamma }(\alpha +2)}{\mathrm{\Gamma }(N+n+\alpha +1)}}{\frac{\mathrm{\Gamma }(N+n+1)\mathrm{\Gamma }(\alpha +2)}{\mathrm{\Gamma }(N+n+\alpha +2)}}(0\le n\le N-1).$$

Then ${\xi }_{\alpha }(n)$ is a strictly decreasing function and

$$\begin{array}{c}\frac{N+n+\alpha +1}{N+n}\ge \frac{2N+\alpha }{2N-1}\ge \frac{2N+\alpha }{2N-1}-\frac{\mathrm{\Gamma }{(N)}^2\mathrm{\Gamma }(2N+\alpha +1)\mathrm{\Gamma }(\alpha +2)}{\mathrm{\Gamma }(2N)\mathrm{\Gamma }{(N+\alpha +1)}^2}\hfill \\ \text{for all }n=0,1,\dots ,N-1.\hfill \end{array}$$

(2.7)

Therefore (2.6) and (2.7) give that if $T_{\phi }$ is hyponormal then

$$\{\frac{2N+\alpha }{2N-1}-\frac{\mathrm{\Gamma }{(N)}^2\mathrm{\Gamma }(2N+\alpha +1)\mathrm{\Gamma }(\alpha +2)}{\mathrm{\Gamma }(2N)\mathrm{\Gamma }{(N+\alpha +1)}^2}\}({|a_{N-1}|}^2-{|a_{-(N-1)}|}^2)\ge {|a_{-N}|}^2-{|a_N|}^2.$$

This completes the proof. □

Example 6 Let $\phi (z)=2{\overline{z}}^2+\frac{3}{2}\overline{z}+\frac{7}{2}z+\frac{6}{7}{z}^2$ and $\alpha =0$. Then by Theorem A, $T_{\phi }$ is not hyponormal. But φ satisfies the inequality in Remark 5, hence the inverse of Remark 5 is not satisfied.

Remark 7 Let $\phi (z)={\sum }_{n=-m}^N{a}_n{z}^n$, where $a_{-m}$ and $a_N$ are nonzero. Suppose $T_{\phi }$ on $H^2(\mathbb{T})$ is hyponormal. It is well known [15] that

$$N-m\le rank[T_{\phi }^{\ast },T_{\phi }]\le N.$$

However, the result cannot be extended to the case of $A^2(\mathbb{D})$; for example, if $\phi (z)=a_{-1}\overline{z}+a_{1}z$ then a straightforward calculation shows that the selfcommutator of Toeplitz operator $T_{\phi }$ on $A^2(\mathbb{D})$ is given by

$$[T_{\phi }^{\ast },T_{\phi }]=({|a_1|}^2-{|a_{-1}|}^2)\left[\begin{array}{cccc}{\alpha }_1& 0& 0& \cdots \\ 0& {\alpha }_2& 0& \cdots \\ 0& 0& {\alpha }_3& \cdots \\ ⋮& ⋮& ⋮& \ddots \end{array}\right],$$

where ${\alpha }_n=\frac{1}{n(n+1)}$. Thus $rank[T_{\phi }^{\ast },T_{\phi }]=\mathrm{\infty }$ and the trace of the selfcommutator $tr[T_{\phi }^{\ast },T_{\phi }]=1$.