Hyponormal Toeplitz operators with non-harmonic symbols on the weighted Bergman spaces

Abstract

In this paper, we consider the hyponormality of Toeplitz operators acting on the weighted Bergman space $A_{\alpha }^2({\mathbb{D}}).$ We establish necessary or sufficient conditions for the hyponormality of Toeplitz operators with non-harmonic symbols $\varphi (z)$ that are bivariate polynomials in z and ${\overline{z}}$ (i.e., $\varphi (z)=az^m{\overline{z}}^n+bz^s{\overline{z}}^t$). In particular, we present some characteristics according to the coefficients and degree of $\varphi (z).$

1 Introduction

For $-1<\alpha < \infty ,$ the weighted Bergman space $A_{\alpha }^2({\mathbb{D}})$ is defined by the space of all analytic functions in $L^2({\mathbb{D}}, {\mathrm{d}}A_{\alpha }),$ where

$$\begin{aligned} {\mathrm{d}}A_{\alpha }(z)=(\alpha +1)(1-|z|^2)^{\alpha }{\mathrm{d}}A(z). \end{aligned}$$

If $f, \ g \in L^2({\mathbb{D}}, {\mathrm{d}}A_{\alpha }),$ we write

$$\begin{aligned} \langle f, \ g \rangle =\int _{{\mathbb{D}}}f(z) \overline{g(z)}{\mathrm{d}}A_{\alpha }(z) \quad \text{and} \quad \Vert f\Vert =\left( \int _{{\mathbb{D}}}|f(z)|^2{\mathrm{d}}A_{\alpha }(z)\right) ^{\frac{1}{2}}. \end{aligned}$$

The space $L^2({\mathbb{D}}, {\mathrm{d}}A_{\alpha })$ is a Hilbert space with the above inner product. For any nonnegative integer n and $z \in {\mathbb{D}},$ let

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

where $\Gamma (\cdot )$ is the usual Gamma function. Then $\{e_n\}$ is an orthonormal basis for the weighted Bergman spaces (cf. [4]). If $f,g\in A_{\alpha }^2({\mathbb{D}})$ of the form

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

then, the inner product of f and g is

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

(1.1)

Given a bounded measurable function $\varphi \in L^{\infty }(\mathbb D),$ the Toeplitz operator $T_{\varphi }$ with symbol $\varphi$ on $A_{\alpha }^2({\mathbb{D}})$ is defined by

$$\begin{aligned} T_{\varphi }g := P({\varphi }\cdot g)\quad (g\in A_{\alpha }^2({\mathbb{D}})) \end{aligned}$$

where P denote the orthogonal projection from $L^2 ({\mathbb{D}}, {\mathrm{d}}A_{\alpha })$ onto $A_{\alpha }^2({\mathbb{D}}).$

For bounded linear operators A, B on a Hilbert space, we let $[A,B]:=AB-BA.$ A bounded linear operator T is said to be hyponormal if its selfcommutator $[T^*,T]$ is positive (semidefinite). Hyponormal operators theory is a broad and highly advanced field, that has contributed a lot of problems in operator theory, functional analysis, and mathematical physics. On the Hardy space $H^2({\mathbb{T}}),$ the hyponormal Toeplitz operators has been researched in [1, 2, 5]. In [1], the author characterized the hyponormality of $T_{\varphi }$ on $H^2({{\mathbb{T}}})$ by using the properties of the symbol $\varphi \in L^\infty ({\mathbb{T}})$ as follows.

Cowen’sTheorem

[1] For $\varphi \in L^{\infty }({\mathbb{T}}),$ let

$$\begin{aligned} {{\mathcal {E}}}(\varphi ):=\{k\in H^{\infty }:\Vert k\Vert _{\infty }\le 1 \quad \text{and}\quad \varphi -k{{\overline{\varphi }}}\in H^{\infty }({\mathbb{T}})\}. \end{aligned}$$

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

The main idea of the proof of Cowen’s Theorem is a dilation theorem as in [14]. However, the Cowen’s theorem cannot be utilized to $A_{\alpha }^2({\mathbb{D}}).$ So, for the weighted Bergman space, determining the hyponormality of Toeplitz operators is very difficult. In fact, there seems to be very little study of hyponormal Toeplitz operators on $A_{\alpha }^2(\mathbb D)$ in the literature. In [7, 8, 10, 11, 13], the authors characterized the hyponormality of Toeplitz operators $T_{\varphi },$ in terms of the coefficient of the symbol $\varphi$ under certain assumptions on $A_{\alpha }^2({\mathbb{D}}).$ Moreover, since the hyponormality of Toeplitz operators on $A_{\alpha }^2({\mathbb{D}})$ is translation invariant, we can assume that the constant term is zero. By the definition and properties of Toeplitz operators, we have that for $g, h\in L^{\infty }(\mathbb D)$ and $\alpha ,\beta \in {\mathbb C},$ we have that $T_{\alpha g+\beta h}=\alpha T_g+\beta T_h,$ $T_g^{*}=T_{{\overline{g}}},$ and $T_{{\overline{g}}}T_h=T_{{\overline{g}}h}$ if g or h is analytic.

We briefly summarize the results relating to the hyponormality of Toeplitz operator with non-harmonic symbols on the Bergman space, which have been recently obtained in [3, 9, 15]. In [15], the author provide a necessary condition on the complex constant C for the operator $T_{z^n+C|z|^s}$ to be hyponormal on the Bergman space, and after that, in [12], hyponormality of Toeplitz operator $T_{z^n+C|z|^s}$ is characterized on the weighted bergman spaces. In [3], the authors consider the sufficient conditions of hyponormality of $T_{f+g}$ on the Bergman space where $f(z)=a_{m,n}z^m{\overline{z}}^n \ (m>n)$ and $g(z)=a_{i,j}z^i{\overline{z}}^j \ (i>j)$ with $m-n>i-j.$ Recently, the authors as in [9] characterized the necessary conditions for the hyponormality of Toeplitz operators $T_{\varphi }$ on the Bergman space where $\varphi (z)=az^m{\overline{z}}^n+bz^s{\overline{z}}^t \ (m\ge n,\ t\ge s)$ with $m\ne t$ and $m-n=t-s.$

In this paper, we consider the hyponormality of Toeplitz operators when $\varphi$ is bivariate polynomials in z and ${\overline{z}}$ acting on the weighted Bergman space $A_{\alpha }^2({\mathbb{D}}).$ First, we briefly recall some basic consequences of the hyponormality of Toeplitz operators on $A^2_{\alpha }({\mathbb{D}}).$ Next, we study either necessary or sufficient condition for the hyponormality of Toeplitz operators $T_{\varphi }$ with non-harmonic symbols $\varphi (z)=az^m{\overline{z}}^n+bz^s{\overline{z}}^t$ on the weighted Bergman space.

2 Main results

Firstly, we need the following lemmas for our program. We recall the properties of projection and norm on $A^2_{\alpha }({\mathbb{D}}).$

Lemma 2.1

[7] For any nonnegative integers s, t, 

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

Notation 2.2

For our convenience, we shall use the following notations.

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

Set $k_i(z):=\sum _{i=0}^{\infty }c_iz^i \in L^2({\mathbb{D}}, {\mathrm{d}}A_{\alpha }).$ By using the Lemma 2.1 and the relation (1.1), we have the results.

Lemma 2.3

[7] For any nonnegative m,  we deduce that

• (i) ${\Vert {\overline{z}}^m k_i(z)\Vert ^2 =\sum _{i=0}^{\infty }\Lambda _{\alpha }(i+m)|c_{i}|^2},$

• (ii) ${\Vert P({\overline{z}}^m k_i(z) )\Vert ^2}=\left\{ \begin{array}{ll} \sum _{i=0}^{\infty }\Lambda _{\alpha }(i,m)|c_{i}|^2 &{}\quad \text{if}\ m \le i \\ \sum _{i=1}^{\infty }\Lambda _{\alpha }(i,m)|c_{i}|^2 &{}\quad \text{if}\ m > i. \end{array}\right.$

Next, we deal with the hyponormality of Toeplitz operators with non-harmonic symbols $\varphi (z)$ on $A^2_{\alpha }({\mathbb{D}}).$ In particular, $\varphi (z)$ be a bivariate polynomials in z and ${\overline{z}}$ of the form $\varphi (z)=az^m{\overline{z}}^n+bz^s{\overline{z}}^t.$ Many researchers have used the Proposition 1.1 as in [6] to consider the hyponormality of $T_{\varphi }.$ However, in the case of Toeplitz operators $T_{\varphi }$ with non-harmonic symbols ${\varphi },$ we cannot apply the well-known consequences of Proposition 1.1 as in [6], since the symbol $\varphi$ cannot separated into analytic and coanalytic parts. So, we have to find the self-commutator of $T_{\varphi }$ directly. First, we will consider some of the relation introduced in Notation 2.2.

Lemma 2.4

Let m, n, s, t are any nonnegative integers.

• (i) If $i\ge m-n,$ then $\Lambda _{\alpha }(m+i,n)\ge \Lambda _{\alpha }(n+i,m).$

• (ii) If $i\ge m-n=s-t>0,$ then

  $$\begin{aligned} \frac{\Lambda _{\alpha }(m+i)\Lambda _{\alpha }(s+i)}{\Lambda _{\alpha }(m-n+i)}\ge \frac{\Lambda _{\alpha }(n+i)\Lambda _{\alpha }(t+i)}{\Lambda _{\alpha }(n-m+i)}. \end{aligned}$$

  (2.1)

• (iii) For $0 \le i<m-n=t-s,$ if $t>m,$ then $\frac{\Lambda _{\alpha }(t+i,s)}{\Lambda _{\alpha }(m+i,n)}$ is increasing in i,  and if $t<m,$ then $\frac{\Lambda _{\alpha }(t+i,s)}{\Lambda _{\alpha }(m+i,n)}$ is decreasing in i.

Proof

(i) Let $m\ge n.$ For $\ell =1,2,\ldots , m-n,$ by calculating the inequality, we have that

$$\begin{aligned} (i+n+\ell )(i+n-m+\ell +\alpha +1)\ge (i+n-m+\ell )(i+n+\ell +\alpha +1) \end{aligned}$$

for $i\ge m-n$ and so

$$\begin{aligned} \prod _{\ell =1}^{m-n}(i+n+\ell )(i+n-m+\ell +\alpha +1)\ge \prod _{\ell =1}^{m-n}(i+n-m+\ell )(i+n+\ell +\alpha +1) \end{aligned}$$

(2.2)

for $i\ge m-n.$ Similarly, for $\ell =1,2,\ldots , m-n,$ we get that

$$\begin{aligned} (i+n+\ell )(i+\ell +\alpha +1)\ge (i+\ell )(i+n+\ell +\alpha +1) \end{aligned}$$

for $i\ge m-n$ and so

$$\begin{aligned} \prod _{\ell =1}^{m-n}(i+n+\ell )(i+\ell +\alpha +1)\ge \prod _{\ell =1}^{m-n}(i+\ell )(i+n+\ell +\alpha +1) \end{aligned}$$

(2.3)

for $i\ge m-n.$ From the relations (2.2) and (2.3), we deduced that

$$\begin{aligned}&\prod _{\ell =1}^{m-n}(i+n+\ell )^2(i+n-m+\ell +\alpha +1)(i+\ell +\alpha +1) \\ &\quad \ge \prod _{\ell =1}^{m-n}(i+\ell )(i+n-m+\ell )(i+n+\ell +\alpha +1)^2. \end{aligned}$$

Since

$$\begin{aligned}&\prod _{\ell =1}^{m-n}(i+n+\ell )^2(i+n-m+\ell +\alpha +1)(i+\ell +\alpha +1)\\ &\quad =\frac{\Gamma (m+i+1)^2\Gamma (i+m-n+\alpha +2)}{\Gamma (n+i+1)^2\Gamma (i+n-m+\alpha +2)} \end{aligned}$$

and

$$\begin{aligned}&\prod _{\ell =1}^{m-n}(i+\ell )(i+n-m+\ell )(i+n+\ell +\alpha +1)^2\\ &\quad =\frac{\Gamma (m-n+i+1)\Gamma (i+m+\alpha +2)^2}{\Gamma (i+n-m+1)\Gamma (i+n+\alpha +2)^2}, \end{aligned}$$

we have

$$\begin{aligned} \Lambda _{\alpha }(m+i,n)\ge \Lambda _{\alpha }(n+i,m) \end{aligned}$$

for $i\ge m-n.$

(ii) By Notation 2.2, the desired inequality is equivalent to the inequality

$$\begin{aligned}&\frac{\Gamma {(m+i+1)}\Gamma {(s+i+1)}\Gamma {(m-n+i+\alpha +2)}}{\Gamma {(m+i+\alpha +2)}\Gamma {(s+i+\alpha +2)}\Gamma {(m-n+i+1)}}\\ &\quad \ge \frac{\Gamma {(n+i+1)}\Gamma {(t+i+1)}\Gamma {(n-m+i+\alpha +2)}}{\Gamma {(n+i+\alpha +2)}\Gamma {(t+i+\alpha +2)}\Gamma {(n-m+i+1)}}, \end{aligned}$$

for $i\ge m-n.$ This is equivalent to

$$\begin{aligned}&\frac{\Gamma {(m+i+1)}\Gamma {(s+i+1)}\Gamma {(m-n+i+\alpha +2)}}{\Gamma {(n+i+1)}\Gamma {(t+i+1)}\Gamma {(n-m+i+\alpha +2)}}\\ &\quad \ge \frac{\Gamma {(m+i+\alpha +2)}\Gamma {(s+i+\alpha +2)} \Gamma {(m-n+i+1)}}{\Gamma {(n+i+\alpha +2)}\Gamma {(t+i+\alpha +2)}\Gamma {(n-m+i+1)}}. \end{aligned}$$

Using properties of the Gamma function and the fact $m-n=s-t,$ this last inequality is equivalent to

$$\begin{aligned}&\prod _{\ell =1}^{m-n}(n+i+\ell )(t+i+\ell )(n-m+i+\ell +\alpha +1)(i+\ell +\alpha +1)\\ &\quad \ge \prod _{\ell =1}^{m-n}(n+i+\ell +\alpha +1)(t+i+\ell +\alpha +1)(n-m+i+\ell )(i+\ell ). \end{aligned}$$

For each $i\ge m-n,$ we claim that

$$\begin{aligned} {}&(n+i+\ell )(t+i+\ell )(n-m+i+\ell +\alpha +1)(i+\ell +\alpha +1) \\ &\quad \ge (n+i+\ell +\alpha +1)(t+i+\ell +\alpha +1)(n-m+i+\ell )(i+\ell ). \end{aligned}$$

(2.4)

By simple calculations, we have

$$\begin{aligned}&(n+i+\ell )(t+i+\ell )(n-m+i+\ell +\alpha +1)(i+\ell +\alpha +1)\\ &\qquad - (n+i+\ell +\alpha +1)(t+i+\ell +\alpha +1)(n-m+i+\ell )(i+\ell )\\ &\quad =(\alpha +1)^2\{(n+i+\ell )(t+i+\ell )-(i+\ell )(n-m+i+\ell )\}\\ &\qquad +(\alpha +1)\{(n+i+\ell )(t+i+\ell )(n-m+i+\ell )+(i+\ell )(n+i+\ell )(t+i+\ell )\}\\ &\qquad -(\alpha +1)\{(i+\ell )(t+i+\ell )(n-m+i+\ell )+(n+i+\ell )(n-m+i+\ell )(i+\ell )\}\\ &\quad =(\alpha +1)^2\{(m+t)(i+\ell )+nt\}\\ &\qquad +(\alpha +1)\{(m+t)(i+\ell )^2+2nt(i+\ell )+nt(n-m)\}. \end{aligned}$$

Since $\alpha +1>0,$ inequality (2.4) equivalent to

$$\begin{aligned} (\alpha +1)\{(m+t)(i+\ell )+nt\}+\{(m+t)(i+\ell )^2+2nt(i+\ell )+nt(n-m)\}\ge 0, \end{aligned}$$

(2.5)

or equivalently

$$\begin{aligned} (\alpha +1)\ge -\frac{(m+t)(i+\ell )^2+(2nt)(i+\ell )+nt(n-m)}{(m+t)(i+\ell )+nt}. \end{aligned}$$

Set $i+\ell =x$ and

$$\begin{aligned} f(x)=-\frac{(m+t)x^2+2ntx+nt(n-m)}{(m+t)x+nt} \quad \quad (x\ge m-n). \end{aligned}$$

So that

$$\begin{aligned} f'(x)=-\frac{(m+t)^2x^2+2nt(m+t)x+nt(m-n)(m+t)+2n^2t^2}{\{(m+t)x+nt\}^2} \end{aligned}$$

for $x\ge m-n.$ Thus f(x) is decreasing for all $x\ge m-n.$ Furthermore, f(x) has a maximum at $x=m-n,$ so

$$\begin{aligned} f(m-n)=-\frac{(m+t)(m-n)^2+nt(m-n)}{(m+t)(m-n)+nt}<0, \end{aligned}$$

since $(\alpha +1)>0\ge f(x)$ for all $x\ge m-n,$ and hence (2.5) holds.

(iii) For $0 \le i<m-n=t-s,$

$$\begin{aligned} \frac{\Lambda _{\alpha }(t+i+1,s)}{\Lambda _{\alpha }(m+i+1,n)}&=\frac{\Gamma (m+i+\alpha +3)^2\Gamma (t+i+2)^2}{\Gamma (t+i+\alpha +3)^2\Gamma (m+i+2)^2}\\ &=\frac{\Gamma (m+i+\alpha +2)^2\Gamma (t+i+1)^2}{\Gamma (t+i+\alpha +2)^2\Gamma (m+i+1)^2}\cdot \frac{(m+i+\alpha +2)^2(t+i+1)^2}{(t+i+\alpha +2)^2(m+i+1)^2}_. \end{aligned}$$

If $t>m,$ then by the simple calculations, we have that $\frac{m+i+\alpha +2}{m+i+1}>\frac{t+i+\alpha +2}{t+i+1}$ and hence

$$\begin{aligned} \frac{\Lambda _{\alpha }(t+i+1,s)}{\Lambda _{\alpha }(m+i+1,n)}>\frac{\Gamma (m+i+\alpha +2)^2 \Gamma (t+i+1)^2}{\Gamma (t+i+\alpha +2)^2\Gamma (m+i+1)^2}= \frac{\Lambda _{\alpha }(t+i,s)}{\Lambda _{\alpha }(m+i,n)}. \end{aligned}$$

Therefore, $\frac{\Lambda _{\alpha }(t+i,s)}{\Lambda _{\alpha }(m+i,n)}$ is increasing in i. By the similar arguments, if $t<m,$ $\frac{\Lambda _{\alpha }(t+i,s)}{\Lambda _{\alpha }(m+i,n)}$ is decreasing in i. $\square$

In the next theorem, we study the case where the symbol $\varphi$ is of the form $\varphi (z)=a_{m,n}z^m{\overline{z}}^n$ with $a_{m,n}\in \mathbb C.$

Theorem 2.5

If $\varphi (z)= a_{m,n}z^m{\overline{z}}^n$ with $a_{m,n}\in \mathbb C,$ then, $T_{\varphi }$ on $A^2_{\alpha }({\mathbb{D}})$ is hyponormal if and only if $m\ge n.$

Proof

For $\varphi (z)= a_{m,n}z^m{\overline{z}}^n,$ then $T_{\varphi }$ is hyponormal if and only if

$$\begin{aligned} \left\langle (T_{ \varphi }^* T_{\varphi } -T_{\varphi } T_{\varphi }^*) \sum _{i=0}^{\infty }c_iz^i , \sum _{i=0}^{\infty }c_iz^i\right\rangle \ge 0 \end{aligned}$$

(2.6)

for all $c_i \in \mathbb C.$ Without loss of generality, we assume that $a_{m,n}=1.$ By using Lemma 2.3, we have that

$$\begin{aligned}&\left\| T_{\varphi } \sum _{i=0}^{\infty }c_iz^i\right\| ^2-\left\| T_{\varphi }^* \sum _{i=0}^{\infty }c_iz^i\right\| ^2 \\ &\quad =\left\| T_{a_{m,n}z^m{\overline{z}}^n} \sum _{i=0}^{\infty }c_iz^i\right\| ^2-\left\| T_{{\overline{a}}_{m,n}{\overline{z}}^mz^n} \sum _{i=0}^{\infty }c_iz^i\right\| ^2 \\ &\quad =\left\| P\left( {a_{m,n}z^m{\overline{z}}^n} \sum _{i=0}^{\infty }c_iz^i\right) \right\| ^2-\left\| P\left( {\overline{a}}_{m,n}{\overline{z}}^mz^n \sum _{i=0}^{\infty }c_iz^i\right) \right\| ^2 \\ &\quad =\sum _{i=0}^{\infty }\Lambda _{\alpha }(m+i,n)|c_{i}|^2 -\sum _{i=m-n}^{\infty }\Lambda _{\alpha }(n+i,m)|c_{i}|^2\ge 0. \end{aligned}$$

(2.7)

If $m\ge n,$ by Lemma 2.4(i), we deduced that

$$\begin{aligned} \Lambda _{\alpha }(m+i,n)\ge \Lambda _{\alpha }(n+i,m) \end{aligned}$$

for $i\ge m-n,$ and so the relation (2.7) holds. Hence $T_{\varphi }$ is hyponormal. If $m<n,$ then $T_{\varphi }$ is the adjoint of a hyponormal operator and thus the self-commutators $[T_{\varphi }^*,T_{\varphi }]$ is negative (semidefinite), so $T_{\varphi }$ is never hyponormal. This completes the proof. $\square$

Combined with Theorem 2.5, we can generalize to the following corollary:

Corollary 2.6

If $\varphi (z)= az^m{\overline{z}}^n+bz^n{\overline{z}}^m$ with $m>n,$ then, $T_{\varphi }$ on $A^2_{\alpha }({\mathbb{D}})$ is hyponormal if and only if $|a|\ge |b|.$

Proof

By the properties of Toeplitz operators, for $h\in A^2_{\alpha }(\mathbb D),$ the inequality

$$\begin{aligned} \Vert {T_{\varphi }h}\Vert =\Vert aT_{z^m{\overline{z}}^n}h + bT_{z^m{\overline{z}}^n}^*h\Vert ^2 \ge \Vert {\overline{a}}T_{z^m{\overline{z}}^n}^*h +{\overline{b}}T_{z^m{\overline{z}}^n}h\Vert ^2=\Vert T_{\varphi }^*h\Vert \end{aligned}$$

is equivalent to

$$\begin{aligned} (|a|^2 - |b|^2)(\Vert {T_{z^m{\overline{z}}^n}h}\Vert ^2 - \Vert {T_{z^m{\overline{z}}^n}^*h}\Vert ^2) \ge 0. \end{aligned}$$

By the Theorem 2.5, $T_{\varphi }$ on $A^2_{\alpha }({\mathbb{D}})$ is hyponormal if and only if $|a|\ge |b|.$ $\square$

2.1 Sufficient condition for hyponormal Toeplitz operators

Now, we consider the sufficient condition for hyponormal Toeplitz operators $T_{\varphi }$ on $A^2_{\alpha }({\mathbb{D}})$ where the symbols is of the form $\varphi (z)=az^m{\overline{z}}^n+bz^s{\overline{z}}^t$ under certain assumptions about the coefficients and degree of $\varphi (z).$

Theorem 2.7

For any $a,b\in \mathbb C$ and nonnegative integers m, n, s, t,  if $\varphi (z)=az^m{\overline{z}}^n+bz^s{\overline{z}}^t$ with $m-n=s-t \ge 0$ and Re$(a{\overline{b}})>0,$ then $T_{\varphi }$ on $A^2_{\alpha }({\mathbb{D}})$ is hyponormal.

Proof

For $\varphi (z)= az^m{\overline{z}}^n+bz^s{\overline{z}}^t,$ $T_{\varphi }$ is hyponormal if and only if

$$\begin{aligned} \left\langle (T_{ \varphi }^* T_{\varphi } -T_{\varphi } T_{\varphi }^*) \sum _{i=0}^{\infty }c_iz^i , \sum _{i=0}^{\infty }c_iz^i\right\rangle \ge 0 \end{aligned}$$

for any $c_i \in \mathbb C.$ We get that

$$\begin{aligned}&\left\| T_{\varphi } \sum _{i=0}^{\infty }c_iz^i\right\| ^2-\left\| T_{\varphi }^* \sum _{i=0}^{\infty }c_iz^i\right\| ^2\\ &\quad =\left\| P\left( a{\overline{z}}^n\sum _{i=0}^{\infty }c_iz^{i+m}\right) +P\left( b{\overline{z}}^t\sum _{i=0}^{\infty }c_iz^{i+s}\right) \right\| ^2\\ &\qquad -\left\| P\left( {\overline{a}}{\overline{z}}^m\sum _{i=0}^{\infty } c_iz^{i+n}\right) +P\left( {\overline{b}}{\overline{z}}^s\sum _{i=0} ^{\infty }c_iz^{i+t}\right) \right\| ^2. \end{aligned}$$

We let a notation $P(a,n,m):=P(a{\bar{z}}^n\sum _{i=0}^{\infty }c_{i}z^{m+i})$ and assume that $c_k=0$ when $k\le 0.$ Then

$$\begin{aligned} P(a,n,m)= & {} P\left( a{\overline{z}}^n\sum _{i=m}^{\infty }c_{-m+i}z^{i}\right) = \sum _{i=m}^{\infty }\frac{\Lambda _{\alpha }(i)}{\Lambda _{\alpha }(i-n)}ac_{-m+i}z^{i-n}\\= & {} \sum _{i=m-n}^{\infty }\frac{\Lambda _{\alpha }(n+i)}{\Lambda _{\alpha }(i)}ac_{n-m+i}z^{i}. \end{aligned}$$

Thus

$$\begin{aligned} \langle P(a,n,m),P(b,t,s)\rangle =\sum _{i=\max \{m-n,s-t\}}^{\infty }\frac{\Lambda _{\alpha }(n+i)\Lambda _{\alpha }(t+i)}{\Lambda _{\alpha }(i)}a{\overline{b}}c_{n-m+i}{\overline{c}}_{t-s+i}. \end{aligned}$$

Similarly, we get

$$\begin{aligned} \langle P({\overline{a}},m,n),P({\overline{b}},s,t) \rangle =\sum _{i=0}^{\infty } \frac{\Lambda _{\alpha }(m+i)\Lambda _{\alpha }(s+i)}{\Lambda _{\alpha }(i)}{\overline{a}}bc_{m-n+i}{\overline{c}}_{s-t+i}. \end{aligned}$$

Thus

$$\begin{aligned} {}&\left\| T_{\varphi } \sum _{i=0}^{\infty }c_iz^i\right\| ^2-\left\| T_{\varphi }^* \sum _{i=0}^{\infty }c_iz^i\right\| ^2 \\ &\quad =\Vert P(a,n,m)+P(b,t,s)\Vert ^2 -\Vert P({\overline{a}},m,n)+P({\overline{b}},s,t)\Vert ^2 \\ &\quad =\Vert P(a,n,m)\Vert ^2+\Vert P(b,t,s)\Vert ^2+2\text{Re}\langle P(a,n,m),P(b,t,s)\rangle \\ &\qquad -\Vert P({\overline{a}},m,n)\Vert ^2-\Vert P({\overline{b}},s,t)\Vert ^2-2\text{Re}\langle P({\overline{a}},m,n),P({\overline{b}},s,t)\rangle \\ &\quad =|a|^2\sum _{i=0}^{\infty }\Lambda _{\alpha }(m+i,n)|c_i|^2+|b|^2\sum _{i=0}^{\infty }\Lambda _{\alpha }(s+i,t)|c_i|^2 \\ &\qquad +\sum _{i=\max \{m-n, s-t\}}^{\infty }\frac{2\Lambda _{\alpha }(n+i)\Lambda _{\alpha }(t+i)}{\Lambda _{\alpha }(i)} \text{Re}(a{\overline{b}}c_{n-m+i}{\overline{c}}_{t-s+i}) \\ &\qquad -|a|^2\sum _{i=m-n}^{\infty }\Lambda _{\alpha }(n+i,m) |c_i|^2-|b|^2\sum _{i=s-t}^{\infty }\Lambda _{\alpha }(t+i,s)|c_i|^2 \\ &\qquad -\sum _{i=0}^{\infty }\frac{2\Lambda _{\alpha }(m+i)\Lambda _{\alpha }(s+i)}{\Lambda _{\alpha }(i)} \text{Re}({\overline{a}}b{\overline{c}}_{m-n+i}c_{s-t+i}). \end{aligned}$$

(2.8)

By Lemma 2.4(i),

$$\begin{aligned} \Lambda _{\alpha }(n+i,m)\le \Lambda _{\alpha }(m+i,n) \end{aligned}$$

for $i\ge m-n$ and

$$\begin{aligned} \Lambda _{\alpha }(t+i,s)\le \Lambda _{\alpha }(s+i,t) \end{aligned}$$

for $i\ge s-t.$ So we have

$$\begin{aligned}&|a|^2\sum _{i=0}^{\infty }\Lambda _{\alpha }(m+i,n)|c_i|^2+|b|^2 \sum _{i=0}^{\infty }\Lambda _{\alpha }(s+i,t)|c_i|^2\\ &\quad -|a|^2\sum _{i=m-n}^{\infty }\Lambda _{\alpha }(n+i,m)|c_i|^2 -|b|^2\sum _{i=s-t}^{\infty }\Lambda _{\alpha }(t+i,s)|c_i|^2 \ge 0. \end{aligned}$$

Now we consider the term of

$$\begin{aligned}&\sum _{i=\max \{m-n, s-t\}}^{\infty }\frac{2\Lambda _{\alpha }(n+i)\Lambda _{\alpha }(t+i)}{\Lambda _{\alpha }(i)}\text{Re}(a{\overline{b}}c_{n-m+i}{\overline{c}}_{t-s+i})\\ &\qquad -\sum _{i=0}^{\infty }\frac{2\Lambda _{\alpha }(m+i)\Lambda _{\alpha }(s+i)}{\Lambda _{\alpha }(i)} \text{Re}({\overline{a}}b{\overline{c}}_{m-n+i}c_{s-t+i}). \end{aligned}$$

Using the condition $m-n=s-t$ then we have

$$\begin{aligned}&\sum _{i=m-n}^{\infty }\frac{2\Lambda _{\alpha }(n+i)\Lambda _{\alpha }(t+i)}{\Lambda _{\alpha }(i)}\text{Re}(a{\overline{b}} |c_{n-m+i}|^2)-\sum _{i=0}^{\infty }\frac{2\Lambda _{\alpha }(m+i)\Lambda _{\alpha }(s+i)}{\Lambda _{\alpha }(i)} \text{Re}({\overline{a}}b|c_{m-n+i}|^2)\\ &\quad =\sum _{i=0}^{\infty }\frac{2\Lambda _{\alpha }(m+i)\Lambda _{\alpha }(s+i)}{\Lambda _{\alpha }(m-n+i)}\text{Re}(a{\overline{b}} |c_{i}|^2) -\sum _{i=m-n}^{\infty }\frac{2\Lambda _{\alpha }(n+i)\Lambda _{\alpha }(t+i)}{\Lambda _{\alpha }(n-m+i)} \text{Re}({\overline{a}}b|c_{i}|^2). \end{aligned}$$

By Lemma 2.4(ii), we have that for $i\ge m-n,$

$$\begin{aligned} \frac{\Lambda _{\alpha }(m+i)\Lambda _{\alpha }(s+i)}{\Lambda _{\alpha }(m-n+i)}\ge \frac{\Lambda _{\alpha }(n+i)\Lambda _{\alpha }(t+i)}{\Lambda _{\alpha }(n-m+i)}. \end{aligned}$$

Therefore $T_{\varphi }$ on $A^2_{\alpha }({\mathbb{D}})$ is hyponormal. This completes the proof. $\square$

Example

If $\varphi (z)=iz^2{\overline{z}}+(2+i)z^3{\overline{z}}^2,$ then, by Theorem 2.7, $T_{\varphi }$ on $A^2_{\alpha }({\mathbb{D}})$ is hyponormal.

In [3], showed that the sufficient conditions for the hyponormality of Toeplitz operators with nonharmonic symbols of fixed related degrees on the Bergman space. The following Example generalizes the result of [3, Theorem 12] to the space $A_{\alpha }^2(\mathbb D).$

Example

Let $\varphi (z)=r_1e^{i \theta _1}z^m{\overline{z}}^n+r_2e^{i \theta _2}z^s{\overline{z}}^t$ with $m-n=s-t \ge 0.$ If $|\theta _1-\theta _2|<\frac{\pi }{2}$ then $T_{\varphi }$ on $A^2_{\alpha }({\mathbb{D}})$ is hyponormal.

2.2 Necessary condition for hyponormal Toeplitz operators

Now, we consider some necessary conditions for hyponormality of $T_{\varphi }$ with more general symbol of the form $\varphi (z)=az^m{\overline{z}}^n+bz^s{\overline{z}}^t$ with $m\ge n$ and $t\ge s.$

Theorem 2.8

Let $\varphi (z)=az^m{\overline{z}}^n+bz^s{\overline{z}}^t$ with $m\ge n, t\ge s$ and $a, b\in \mathbb C.$ If $m-n=t-s$ and $m\ne t,$ $T_{\varphi }$ on $A^2_{\alpha }({\mathbb{D}})$ is hyponormal then

$$\begin{aligned} \left\{ \begin{array}{ll} |a|^2\ge \max \left\{ \frac{\Lambda _{\alpha }(2t-s-1,s)}{\Lambda _{\alpha }(2m-n-1,n)}, W(m,n,t,s)\right\} |b|^2&{}\quad \text{if}\ t> m\\ |a|^2\ge \max \left\{ \frac{\Lambda _{\alpha }(t,s)}{\Lambda _{\alpha }(m,n)}, W(m,n,t,s)\right\} |b|^2&{}\quad \text{if}\ t<m, \end{array}\right. \end{aligned}$$

where $W(m,n,t,s)=\max _{i\in [m-n,\infty )}\frac{\Lambda _{\alpha }(t+i,s)- \Lambda _{\alpha }(s+i,t)}{\Lambda _{\alpha }(m+i,n)-\Lambda _{\alpha }(n+i,m)}.$

Proof

By a similar argument to the proof of Theorem 2.7, $T_{\varphi }$ is hyponormal if and only if

$$\begin{aligned}&\left\| T_{\varphi } \sum _{i=0}^{\infty }c_iz^i\right\| ^2-\left\| T_{\varphi }^* \sum _{i=0}^{\infty }c_iz^i\right\| ^2 \\ &\quad =|a|^2\sum _{i=0}^{\infty }\Lambda _{\alpha }(m+i,n)|c_i|^2 +|b|^2\sum _{i=m-n}^{\infty }\Lambda _{\alpha }(s+i,t)|c_i|^2 \\ &\qquad -|a|^2\sum _{i=m-n}^{\infty }\Lambda _{\alpha }(n+i,m) |c_i|^2-|b|^2\sum _{i=0}^{\infty }\Lambda _{\alpha }(t+i,s)|c_i|^2 \\ &\qquad +\sum _{i=m-n}^{\infty }\frac{2\Lambda _{\alpha }(n+i)\Lambda _{\alpha }(t+i)}{\Lambda _{\alpha }(i)} \text{Re}(a{\overline{b}}c_{n-m+i}{\overline{c}}_{t-s+i}) \\ &\qquad -\sum _{i=t-s}^{\infty }\frac{2\Lambda _{\alpha }(m+i)\Lambda _{\alpha }(s+i)}{\Lambda _{\alpha }(i)} \text{Re}({\overline{a}}b{\overline{c}}_{m-n+i}c_{s-t+i})\ge 0, \end{aligned}$$

(2.9)

for any $c_i\in \mathbb C.$ Since $m-n=t-s$ and $m\ne t,$ from (2.9), $T_{\varphi }$ is hyponormal if and only if

$$\begin{aligned}&|a|^2\left\{ \sum _{i=0}^{\infty }\Lambda _{\alpha }(m+i,n)|c_i|^2- \sum _{i=m-n}^{\infty }\Lambda _{\alpha }(n+i,m)|c_i|^2\right\} \\ &\quad \ge |b|^2\left\{ \sum _{i=0}^{\infty }\Lambda _{\alpha } (t+i,s)|c_i|^2-\sum _{i=m-n}^{\infty }\Lambda _{\alpha }(s+i,t)|c_i|^2\right\} \\ &\qquad +\sum _{i=m-n}^{\infty }\frac{2\Lambda _{\alpha }(m+i)\Lambda _{\alpha }(s+i)}{\Lambda _{\alpha }(i)} \text{Re}({\overline{a}}bc_{n-m+i}{\overline{c}}_{m-n+i}) \\ &\qquad -\sum _{i=m-n}^{\infty }\frac{2\Lambda _{\alpha }(n+i)\Lambda _{\alpha }(t+i)}{\Lambda _{\alpha }(i)} \text{Re}(a{\overline{b}}{\overline{c}}_{m-n+i}c_{n-m+i}) \end{aligned}$$

(2.10)

for all $c_i\in \mathbb C \ (i=0,1,2,\ldots ).$ Since $c_i$ are arbitrary, set $\text{Re}(a{\overline{b}}{\overline{c}}_{m-n+i}c_{n-m+i})=\text{Re}({\overline{a}}bc_{n-m+i}{\overline{c}}_{m-n+i})=0$ for any $i \ge m-n.$ If $0 \le i<m-n$ then (2.10) implies

$$\begin{aligned} |a|^2\ge \max _{0\le i<m-n}\frac{\Lambda _{\alpha }(t+i,s)}{\Lambda _{\alpha }(m+i,n)}|b|^2. \end{aligned}$$

We can think of it in two cases. If $t>m,$ then by Lemma 2.4(iii), $\frac{\Lambda _{\alpha }(t+i,s)}{\Lambda _{\alpha }(m+i,n)}$ is increasing in i,  and hence

$$\begin{aligned} |a|^2 \ge \frac{\Lambda _{\alpha }(2t-s-1,s)}{\Lambda _{\alpha }(2m-n-1,n)}|b|^2. \end{aligned}$$

If $t<m,$ then by Lemma 2.4(iii), $\frac{\Lambda _{\alpha }(t+i,s)}{\Lambda _{\alpha }(m+i,n)}$ is decreasing in i,  and so

$$\begin{aligned} |a|^2 \ge \frac{\Lambda _{\alpha }(t,s)}{\Lambda _{\alpha }(m,n)}|b|^2. \end{aligned}$$

For $i\ge m-n$ with ${\overline{c}}_{m-n+i}c_{n-m+i}=0,$

$$\begin{aligned} |a|^2\ge \max _{ i\in [m-n,\infty )}\frac{\Lambda _{\alpha }(t+i,s)- \Lambda _{\alpha }(s+i,t)}{\Lambda _{\alpha }(m+i,n)-\Lambda _{\alpha }(n+i,m)}|b|^2. \end{aligned}$$

Therefore, if $T_{\varphi }$ is hyponormal, then

$$\begin{aligned} |a|^2\ge \max \left\{ \frac{\Lambda _{\alpha }(2t-s-1,s)}{\Lambda _{\alpha }(2m-n-1,n)}, W(m,n,t,s)\right\} |b|^2 \end{aligned}$$

if $t> m,$ and

$$\begin{aligned} |a|^2\ge \max \left\{ \frac{\Lambda _{\alpha }(t,s)}{\Lambda _{\alpha }(m,n)}, W(m,n,t,s)\right\} |b|^2 \end{aligned}$$

if $t<m,$ where $W(m,n,t,s)=\max _{i\in [m-n,\infty )}\frac{\Lambda _{\alpha }(t+i,s)-\Lambda _{\alpha }(s+i,t)}{\Lambda _{\alpha }(m+i,n)-\Lambda _{\alpha }(n+i,m)}.$ This completes the proof. $\square$

Corollary 2.9

Let $\varphi (z)=a|z|^{2m}+b|z|^{2s}$ with $a, b\in \mathbb C.$ Then $T_{\varphi }$ on $A^2_{\alpha }({\mathbb{D}})$ is normal and hence hyponormal.

Proof

From (2.9) as in Theorem 2.8, if $m=n$ and $s=t,$ then

$$\begin{aligned} \left\| T_{\varphi } \sum _{i=0}^{\infty }c_iz^i\right\| ^2=\left\| T_{\varphi }^* \sum _{i=0}^{\infty }c_iz^i\right\| ^2 \end{aligned}$$

for any $c_i\in \mathbb C.$ $\square$

Example

Let $\varphi (z)=az^3{\overline{z}}^2+b{\overline{z}}^2z$ with nonzeros $a, b\in \mathbb C.$ Then,

$$\begin{aligned} \frac{\Lambda _{\alpha }(2,1)}{\Lambda _{\alpha }(3,2)}=\left( \frac{\alpha +4}{3}\right) ^2, \end{aligned}$$

and

$$\begin{aligned} \frac{\Lambda _{\alpha }(2+i,1)-\Lambda _{\alpha }(1+i,2)}{\Lambda _{\alpha }(3+i,2)- \Lambda _{\alpha }(2+i,3)}= & {} \frac{\frac{\Gamma (i+3)^2\Gamma (i+\alpha +3)}{\Gamma (i+\alpha +4)^2\Gamma (i+2)} -\frac{\Gamma (i+2)^2\Gamma (i+\alpha +1)}{\Gamma (i+\alpha +3)^2\Gamma (i)}}{\frac{\Gamma (i+4)^2\Gamma (i+\alpha +3)}{\Gamma (i+\alpha +5)^2\Gamma (i+2)}-\frac{\Gamma (i+3)^2\Gamma (i+\alpha +1)}{\Gamma (i+\alpha +4)^2\Gamma (i)}}\\= & {} \frac{(i+\alpha +4)^2(3i^2+(3\alpha +11)i+4(\alpha +2))}{(i+2)^2(5i^2+(5\alpha +23)i+9(\alpha +2))}. \end{aligned}$$

Set

$$\begin{aligned} f(i)=\frac{(i+\alpha +4)^2(3i^2+(3\alpha +11)i+4(\alpha +2))}{(i+2)^2(5i^2+(5\alpha +23)i+9(\alpha +2))}. \end{aligned}$$

Then the denominator of $f'(i)$ is

$$\begin{aligned} (i+2)^4(5i^2+(5\alpha +23)i+9(\alpha +2))^2, \end{aligned}$$

which is always positive. Note that the numerator of $f'(i)$ is

$$\begin{aligned}&(i+2)(i+\alpha +4)[\{14-30(\alpha +2)\}i^4+2\{(14\alpha +56)-(\alpha +2)(30\alpha +124)\}i^3\\ &\quad +\{(21\alpha ^2+161\alpha +294)-(\alpha +2)(30\alpha ^2+342\alpha +694)\}i^2\\ &\quad +(\alpha +2)\{7\alpha ^2+77\alpha +154-(\alpha +2) (94\alpha +382)\}i-2(\alpha +2)(29\alpha ^2+109\alpha +116)]. \end{aligned}$$

Since $\alpha >-1,$ then

$$\begin{aligned}&14-30(\alpha +2)=-30\alpha -46<0,\\ &(14\alpha +56)-(\alpha +2)(30\alpha +124)=-30{\alpha }^2-170\alpha -192<0,\\ &(21\alpha ^2+161\alpha +294)-(\alpha +2)(30\alpha ^2+342\alpha +694)\\ &\quad =-(30{\alpha }^3+381{\alpha }^2+1217{\alpha }+1094)<0 \end{aligned}$$

and

$$\begin{aligned} 7\alpha ^2+77\alpha +154-(\alpha +2)(94\alpha +382)=-(87{\alpha }^2+493{\alpha }+610)<0. \end{aligned}$$

The numerator of $f'(i)$ is negative and hence f(i) is decreasing. Thus

$$\begin{aligned} W(3,2,2,1)= & {} \max _{i\in [1,\infty )}\frac{\Lambda _{\alpha }(2+i,1)-\Lambda _{\alpha }(1+i,2)}{\Lambda _{\alpha }(3+i,2)-\Lambda _{\alpha }(2+i,3)}\\= & {} \frac{\Lambda _{\alpha }(3,1)-\Lambda _{\alpha }(2,2)}{\Lambda _{\alpha }(4,2)-\Lambda _{\alpha }(3,3)}\\= & {} \frac{(\alpha +5)^2(7\alpha +22)}{9(14\alpha +46)}. \end{aligned}$$

Hence by Theorem 2.8, if $T_{\varphi }$ is hyponormal, then

$$\begin{aligned} |a|^2\ge \max \Bigg \{\left( \frac{\alpha +4}{3}\right) ^2, \frac{(\alpha +5)^2(7\alpha +22)}{9(14\alpha +46)}\Bigg \} |b|^2=\left( \frac{\alpha +4}{3}\right) ^2|b|^2. \end{aligned}$$