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Hyponormality of Toeplitz operators with non-harmonic symbols on the Bergman spaces

Abstract

In this paper, we present some necessary and sufficient conditions for the hyponormality of Toeplitz operator \(T_{\varphi }\) on the Bergman space \(A^{2}(\mathbb{D})\) with non-harmonic symbols under certain assumptions.

Introduction

Let \({\mathcal{{H}}}\) be a separable complex Hilbert space and \({\mathcal{{L}}}({\mathcal{{H}}})\) be the set of bounded linear operators on \({\mathcal{{H}}}\). An operator \(T\in {\mathcal{{L}}}({\mathcal{H}})\) is hyponormal if its self-commutator \([T^{\ast }, T]:=T^{\ast }T-TT^{\ast }\) is positive semidefinite. Let dA be the normalized area measure on the open unit disk \({\mathbb{D}}\) in \({\mathbb{C}}\) and \(L^{2}(\mathbb{D})\) be a Hilbert space of square-integrable measurable functions on \(\mathbb{D}\) with the inner product

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

The Bergman space \(A^{2}(\mathbb{D})\) is the space of analytic functions in \(L^{2}(\mathbb{D})\). The multiplication operator \(M_{\psi }\) with symbol \(\psi \in L^{\infty }(\mathbb{D})\) is defined by \(M_{\psi }f=\psi f\) for \(f\in A^{2}(\mathbb{D})\). For any \(\varphi \in L^{\infty }(\mathbb{D})\), the Toeplitz operator \(T_{\varphi }\) on the Bergman space is defined by \(T_{\varphi }{f}=P(\varphi f)\) for \(f\in A^{2}(\mathbb{D})\) and P is the orthogonal projection that maps \(L^{2}(\mathbb{D})\) onto \(A^{2}(\mathbb{D})\). Recall that the power series representation of \(f\in A^{2}(\mathbb{D})\) is

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

In [1, 4, 5], and [7], the basic properties of the Bergman space and the Hardy space are well known. The hyponormality of Toeplitz operators on the Hardy space has been developed in [2, 3, 10], and [12]. In [2], Cowen characterized the hyponormality of Toeplitz operator \(T_{\varphi }\) on \(H^{2}(\mathbb{T})\) by the properties of the symbol \(\varphi \in L^{\infty }(\mathbb{T})\). Cowen’s method is to reconstruct the operator-theoretic problem of hyponormal Toeplitz operator into the problem of finding a solution of equations of functionals. Recently, in [8, 9], the authors characterized the hyponormality of Toeplitz operators on the Bergman space with harmonic symbols.

Proposition 1.1

([8])

Let \(\varphi (z)=\overline{g(z)}+f(z)\), where \(f(z)=a_{m}z^{m}+a_{N}z^{N}\) and \(g(z)=a_{-m}z^{m}+a_{-N}z^{N}\) (\(0< m< N\)). If \(a_{m}\overline{a_{N}}=a_{-m}\overline{a_{-N}}\), then \(T_{\varphi }\) is hyponormal

$$ \Longleftrightarrow \quad \textstyle\begin{cases} \frac{1}{N+1}( \vert a_{N} \vert ^{2}- \vert a_{-N} \vert ^{2})\geq \frac{1}{m+1}( \vert a_{-m} \vert ^{2}- \vert a_{m} \vert ^{2}) &\textit{if } \vert a_{-N} \vert \leq \vert a_{N} \vert , \\ N^{2}( \vert a_{-N} \vert ^{2}- \vert a_{N} \vert ^{2})\leq m^{2}( \vert a_{m} \vert ^{2}- \vert a_{-m} \vert ^{2}) & \textit{if } \vert a_{N} \vert \leq \vert a_{-N} \vert . \end{cases} $$

Proposition 1.2

([9])

Let \(\varphi (z)=\overline{g(z)}+f(z)\), where \(f(z)=a_{m}z^{m}+a_{N}z^{N}\) and \(g(z)=a_{-m}z^{m}+a_{-N}z^{N}\) (\(0< m< N\)). If \(T_{\varphi }\) is hyponormal and \(|a_{N}|\leq |a_{-N}|\), then we have

$$ N^{2}\bigl( \vert a_{-N} \vert ^{2}- \vert a_{N} \vert ^{2}\bigr)\leq m^{2} \bigl( \vert a_{m} \vert ^{2}- \vert a_{-m} \vert ^{2}\bigr). $$

Since the hyponormality of operators is translation invariant, we may assume that constant term is zero. We shall list the well-known properties of Toeplitz operators \(T_{\varphi }\) on the Bergman space. Let f, g be in \(L^{\infty }(\mathbb{D})\) and \(\alpha ,\beta \in {\mathbb{C}}\), then we can easily check that \(T_{\alpha f+\beta g}=\alpha T_{f}+\beta T_{g}\), \(T_{f}^{\ast }=T_{\overline{f}}\), and \(T_{\overline{f}}T_{g}=T_{\overline{f}g}\) if f or g is analytic.

We briefly summarize a number of partial results relating to the hyponormality of Toeplitz operator with non-harmonic symbols, which have been recently developed in [6] and [14].

Proposition 1.3

([6])

  1. (i)

    Suppose \(f=a_{m,n}z^{m}\overline{z}^{n}\) and \(g=a_{i,j}z^{i}\overline{z}^{j}\) with \(m>n\), \(i>j\) and \(m-n>i-j\). Then \(T_{f+g}\) is hyponormal if, for each \(k\geq 0\), the term

    $$ \biggl\vert \frac{a_{m,n}}{a_{i,j}} \biggr\vert \frac{m-n+k+1}{(m+k+1)^{2}}+ \biggl\vert \frac{a_{i,j}}{a_{m,n}} \biggr\vert \frac{i-j+k+1}{(i+k+1)^{2}} $$

    is sufficiently large.

  2. (ii)

    Suppose \(f=a_{m,n}z^{m}\overline{z}^{n}\) and \(g=a_{i,j}z^{i}\overline{z}^{j}\) with \(m>n\) and \(i>j\). Then \(T_{f+g}\) is hyponormal if, for each \(k\geq 0\),

    $$ \biggl\vert \frac{a_{m,n}}{a_{i,j}} \biggr\vert \frac{m-n+k+1}{(m+k+1)^{2}}- \biggl\vert \frac{a_{i,j}}{a_{m,n}} \biggr\vert \frac{i-j+k+1}{(i+k+1)^{2}} $$

    is sufficiently large.

Proposition 1.4

([14])

Suppose \(c\in \mathbb{C}\), \(s\in (0, \infty )\) and \(n\in \mathbb{N}\). If \(T_{z^{n}+C|z|^{s}}\) is hyponormal, then \(|C|\leq \frac{n}{s}\). If \(s\geq 2n\), then the converse is also true (i.e., \(T_{z+C|z|^{2}}\) is hyponormal \(\Longleftrightarrow |C|\leq \frac{1}{2}\)).

Furthermore, in [11], the authors extended Proposition 1.4 to the weighted Bergman spaces. The purpose of this paper is to characterize the hyponormal Toeplitz operators \(T_{\varphi }\) with non-harmonic symbols acting on \(A^{2}(\mathbb{D})\).

Toeplitz operators with non-harmonic symbols

We need several auxiliary lemmas to prove the main theorem in this section. We begin with the following.

Lemma 2.1

([8])

For any \(s, t\in \mathbb{N}\),

$$ P\bigl(\overline{z}^{t} z^{s}\bigr)= \textstyle\begin{cases} \frac{s-t+1}{s+1} z^{s-t} & \textit{if } s \geq t, \\ 0 & \textit{if } s < t. \end{cases} $$

The proof for Lemma 2.2 follows the proof of Lemma 2.1 in [8].

Lemma 2.2

For \(0\leq m \leq N\), we deduce that

  1. (i)

    \(\Vert \overline{z}^{m} \sum_{i=0}^{ \infty }c_{i}z^{i} \Vert ^{2} =\sum_{i=0}^{\infty } \frac{1}{i+m+1}|c_{i}|^{2}\),

  2. (ii)

    \(\Vert P(\overline{z}^{m} \sum_{i=0}^{ \infty }c_{i}z^{i}) \Vert ^{2}= \sum_{i=m}^{\infty } \frac{i-m+1}{(i+1)^{2}}|c_{i}|^{2}\).

In [13], the author characterized the hyponormality of Toeplitz operators \(T_{\overline{g}+f}\) with bounded and analytic functions f and g by \(\Vert (I-P)(\overline{g}k) \Vert \leq \Vert (I-P)( \overline{f}k) \Vert \) for every k in \(A^{2}(\mathbb{D})\). Furthermore, many authors have used the inequality to study the hyponormal Toeplitz operators. However, we consider the hyponormality of \(T_{\varphi }\) on \(A^{2}(\mathbb{D})\) with the non-analytic symbol φ. So, in our case, we cannot apply that inequality to φ since we cannot separate φ to analytic and coanalytic parts. Therefore we directly calculate the self-commutator of \(T_{\varphi }\). First, we consider the symbol φ of the form \(\varphi (z)=a_{m,n}z^{m}\overline{z}^{n}\) with \(a_{m,n}\in \mathbb{C}\).

Theorem 2.3

Let \(\varphi (z)= a_{m,n}z^{m}\overline{z}^{n}\) with \(a_{m,n}\in \mathbb{C}\). Then \(T_{\varphi }\) on \(A^{2}(\mathbb{D})\) is hyponormal if and only if \(m\geq n\).

Proof

If \(m\ge n\), then the authors as in [6] proved that \(T_{\varphi }\) is hyponormal. Suppose that \(T_{\varphi }\) is hyponormal. By the definition of hyponormal Toeplitz operators, \(T_{\varphi }\) is hyponormal if and only if

$$ \Biggl\langle \bigl(T_{ \varphi }^{*} T_{\varphi } -T_{\varphi } T_{\varphi }^{*}\bigr) \sum_{i=0}^{\infty }c_{i}z^{i} , \sum_{i=0}^{\infty }c_{i}z^{i} \Biggr\rangle \geq 0 $$

for all \(c_{i} \in \mathbb{C}\). Using Lemmas 2.1 and 2.2, we have that

$$ \begin{aligned} & \Biggl\Vert T_{\varphi } \sum_{i=0}^{\infty }c_{i}z^{i} \Biggr\Vert ^{2}- \Biggl\Vert T_{\varphi }^{*} \sum_{i=0}^{\infty }c_{i}z^{i} \Biggr\Vert ^{2} \\ &\quad = \Biggl\Vert T_{a_{m,n}z^{m}\overline{z}^{n}} \sum_{i=0}^{\infty }c_{i}z^{i} \Biggr\Vert ^{2}- \Biggl\Vert T_{\overline{a}_{m,n}\overline{z}^{m}z^{n}} \sum _{i=0}^{\infty }c_{i}z^{i} \Biggr\Vert ^{2} \\ &\quad = \Biggl\Vert P \Biggl({a_{m,n}z^{m} \overline{z}^{n}} \sum_{i=0}^{ \infty }c_{i}z^{i} \Biggr) \Biggr\Vert ^{2}- \Biggl\Vert P \Biggl( \overline{a}_{m,n} \overline{z}^{m}z^{n} \sum _{i=0}^{\infty }c_{i}z^{i} \Biggr) \Biggr\Vert ^{2} \\ &\quad = \vert a_{m,n} \vert ^{2}\sum _{i=\max \{n-m,0\}}^{\infty } \frac{m+i-n+1}{(m+i+1)^{2}} \vert c_{i} \vert ^{2} - \vert a_{m,n} \vert ^{2}\sum_{i=\max \{m-n,0\}}^{\infty } \frac{n+i-m+1}{(n+i+1)^{2}} \vert c_{i} \vert ^{2}\geq 0. \end{aligned} $$

Hence \(T_{\varphi }\) is hyponormal if and only if

$$ \sum_{i=\max \{n-m,0\}}^{\infty }\frac{m+i-n+1}{(m+i+1)^{2}} \vert c_{i} \vert ^{2} \geq \sum _{i=\max \{m-n,0\}}^{\infty } \frac{n+i-m+1}{(n+i+1)^{2}} \vert c_{i} \vert ^{2} $$

for all \(c_{i} \in \mathbb{C}\). Since \(c_{i}\)s are arbitrary, we have that \(T_{\varphi }\) is hyponormal if and only if \(m\geq n\). This completes the proof. □

We now consider the hyponormality of Toeplitz operators with two terms non-harmonic symbols.

Theorem 2.4

Let \(\varphi (z)=az^{m}\overline{z}^{n}+bz^{n}\overline{z}^{m}\) with nonnegative integers m, n with \(m\ge n\) and nonzeros \(a, b\in \mathbb{C}\). Then \(T_{\varphi }\) on \(A^{2}(\mathbb{D})\) is hyponormal if and only if \(|a|\geq |b|\).

Proof

In a similar way to the proof of Theorem 2.3, \(T_{\varphi }\) is hyponormal if and only if

$$ \begin{aligned} & \Biggl\Vert T_{\varphi } \sum_{i=0}^{\infty }c_{i}z^{i} \Biggr\Vert ^{2}- \Biggl\Vert T_{\varphi }^{*} \sum_{i=0}^{\infty }c_{i}z^{i} \Biggr\Vert ^{2} \\ &\quad = \Biggl\Vert P \Biggl({a\overline{z}^{n}} \sum _{i=0}^{\infty }c_{i}z^{i+m} \Biggr)+P \Biggl({b\overline{z}^{m}} \sum _{i=0}^{\infty }c_{i}z^{i+n} \Biggr) \Biggr\Vert ^{2} \\ &\qquad {} - \Biggl\Vert P \Biggl(\overline{a}\overline{z}^{m} \sum _{i=0}^{ \infty }c_{i}z^{i+n} \Biggr)+P \Biggl(\overline{b}\overline{z}^{n} \sum _{i=0}^{\infty }c_{i}z^{i+m} \Biggr) \Biggr\Vert ^{2} \\ &\quad = \vert a \vert ^{2}\sum_{i=0}^{\infty } \frac{m+i-n+1}{(m+i+1)^{2}} \vert c_{i} \vert ^{2}+ \vert b \vert ^{2} \sum_{i=m-n}^{\infty } \frac{n+i-m+1}{(n+i+1)^{2}} \vert c_{i} \vert ^{2} \\ &\qquad {} - \vert a \vert ^{2}\sum_{i=m-n}^{\infty } \frac{n+i-m+1}{(n+i+1)^{2}} \vert c_{i} \vert ^{2}- \vert b \vert ^{2} \sum_{i=0}^{\infty } \frac{m+i-n+1}{(m+i+1)^{2}} \vert c_{i} \vert ^{2} \\ &\quad =\bigl( \vert a \vert ^{2}- \vert b \vert ^{2} \bigr) \Biggl[\sum_{i=0}^{m-n-1} \frac{m+i-n+1}{(m+i+1)^{2}} \vert c_{i} \vert ^{2} \\ &\qquad {} +\sum_{i=m-n}^{\infty } \biggl( \frac{m+i-n+1}{(m+i+1)^{2}}-\frac{n+i-m+1}{(n+i+1)^{2}} \biggr) \vert c_{i} \vert ^{2} \Biggr]\geq 0. \end{aligned} $$

Since \(\frac{m+i-n+1}{(m+i+1)^{2}}\) and \(\frac{m+i-n+1}{(m+i+1)^{2}}-\frac{n+i-m+1}{(n+i+1)^{2}}\) are positive for all \(i\ge 0\) and \(i\ge m-n\), respectively, \(T_{\varphi }\) is hyponormal if and only if \(|a|\geq |b|\). □

The following theorem gives a general characterization of hyponormal Toeplitz operators with the symbols of the form \(\varphi (z)= az^{m}\overline{z}^{n}+bz^{s}\overline{z}^{t}\) (\(m\ge n\ge 0\), \(t\ge s\ge 0\)) with some conditions.

Theorem 2.5

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

$$ \textstyle\begin{cases} \vert a \vert ^{2}\geq \max \{ \frac{(2m-n)^{2}}{(t+m-n)^{2}}, \Lambda (m,n,t,s) \} \vert b \vert ^{2} &\textit{if } t> m, \\ \vert a \vert ^{2}\geq \max \{ \frac{(m+1)^{2}}{(t+1)^{2}}, \Lambda (m,n,t,s) \} \vert b \vert ^{2} &\textit{if } t< m, \end{cases} $$

where \(\Lambda (m,n,t,s)=\max_{i\in [m-n,\infty )} \frac{\frac{(t+i-s+1)}{(t+i+1)^{2}}-\frac{(s+i-t+1)}{(s+i+1)^{2}}}{\frac{(m+i-n+1)}{(m+i+1)^{2}}-\frac{(n+i-m+1)}{(n+i+1)^{2}}}\).

Proof

In a similar way to the proof of Theorem 2.4, \(T_{\varphi }\) is hyponormal if and only if

$$ \begin{aligned} & \Biggl\Vert T_{\varphi } \sum_{i=0}^{\infty }c_{i}z^{i} \Biggr\Vert ^{2}- \Biggl\Vert T_{\varphi }^{*} \sum_{i=0}^{\infty }c_{i}z^{i} \Biggr\Vert ^{2} \\ &\quad = \vert a \vert ^{2}\sum_{i=0}^{\infty } \frac{m+i-n+1}{(m+i+1)^{2}} \vert c_{i} \vert ^{2}+ \vert b \vert ^{2} \sum_{i=t-s}^{\infty } \frac{s+i-t+1}{(s+i+1)^{2}} \vert c_{i} \vert ^{2} \\ &\qquad {} - \vert a \vert ^{2}\sum_{i=m-n}^{\infty } \frac{n+i-m+1}{(n+i+1)^{2}} \vert c_{i} \vert ^{2}- \vert b \vert ^{2} \sum_{i=0}^{\infty } \frac{t+i-s+1}{(t+i+1)^{2}} \vert c_{i} \vert ^{2} \\ &\qquad {} +2\operatorname{Re} \Biggl(a\overline{b}\sum_{i=m-n}^{\infty } \frac{i+1}{(n+i+1)(t+i+1)}c_{i-m+n}\overline{c}_{t-s+i} \Biggr) \\ &\qquad {} -2\operatorname{Re} \Biggl(a\overline{b}\sum_{i=t-s}^{\infty } \frac{i+1}{(m+i+1)(s+i+1)}\overline{c}_{i+m-n}{c}_{i-t+s} \Biggr) \geq 0 \end{aligned} $$
(2.1)

for any \(c_{i}\in \mathbb{C}\) (\(i=0,1,2,\ldots \)).

Since \(m-n=t-s\) and \(m\neq t\), from (2.1), \(T_{\varphi }\) is hyponormal if and only if

$$ \begin{aligned} & \vert a \vert ^{2} \Biggl\{ \sum_{i=0}^{\infty } \frac{m+i-n+1}{(m+i+1)^{2}} \vert c_{i} \vert ^{2}- \sum _{i=m-n}^{\infty }\frac{n+i-m+1}{(n+i+1)^{2}} \vert c_{i} \vert ^{2} \Biggr\} \\ &\quad \geq \vert b \vert ^{2} \Biggl\{ \sum _{i=0}^{\infty }\frac{m+i-n+1}{(t+i+1)^{2}} \vert c_{i} \vert ^{2}- \sum_{i=m-n}^{\infty } \frac{n+i-m+1}{(s+i+1)^{2}} \vert c_{i} \vert ^{2} \Biggr\} \\ &\qquad {} +2\operatorname{Re} \Biggl(a\overline{b}\sum_{i=m-n}^{\infty } \biggl( \frac{i+1}{(m+i+1)(s+i+1)}-\frac{i+1}{(n+i+1)(t+i+1)} \biggr) \overline{c}_{i+m-n}{c}_{i-m+n} \Biggr) \end{aligned} $$
(2.2)

for any \(c_{i}\in \mathbb{C}\) (\(i=0,1,2,\ldots \)). Since \(c_{i}\)s are arbitrary, set \(\operatorname{Re}(a\overline{b}\overline{c}_{i+m-n}{c}_{i-m+n})=0\) for any i, \(i\ge m-n\). If \(0 \leq i < m-n\), then (2.2) implies

$$ \vert a \vert ^{2}\geq \frac{(m+i+1)^{2}}{(t+i+1)^{2}} \vert b \vert ^{2}. $$

There are two cases to consider. If \(t> m\), then \(\frac{(m+i+1)^{2}}{(t+i+1)^{2}}\) is increasing in i, and hence \(|a|^{2}\geq \frac{(2m-n)^{2}}{(t+m-n)^{2}}|b|^{2}\). If \(t< m\), then \(\frac{(m+i+1)^{2}}{(t+i+1)^{2}}\) is decreasing in i and hence

$$ \vert a \vert ^{2}\geq \frac{(m+1)^{2}}{(t+1)^{2}} \vert b \vert ^{2}. $$

For \(i\geq m-n=t-s\),

$$ \vert a \vert ^{2}\geq \max_{i\in [m-n,\infty )} \frac{\frac{t+i-s+1}{(t+i+1)^{2}}-\frac{s+i-t+1}{(s+i+1)^{2}}}{\frac{m+i-n+1}{(m+i+1)^{2}}-\frac{n+i-m+1}{(n+i+1)^{2}}} \vert b \vert ^{2}. $$

Hence, if \(T_{\varphi }\) is hyponormal, then

$$ \textstyle\begin{cases} \vert a \vert ^{2}\geq \max \{ \frac{(2m-n)^{2}}{(t+m-n)^{2}}, \Lambda (m,n,t,s) \} \vert b \vert ^{2} &\text{if } t > m, \\ \vert a \vert ^{2}\geq \max \{ \frac{(m+1)^{2}}{(t+1)^{2}}, \Lambda (m,n,t,s) \} \vert b \vert ^{2} &\text{if } t< m, \end{cases} $$

where \(\Lambda (m,n,t,s)=\max_{i\in [m-n,\infty )} \frac{\frac{(t+i-s+1)}{(t+i+1)^{2}}-\frac{(s+i-t+1)}{(s+i+1)^{2}}}{\frac{(m+i-n+1)}{(m+i+1)^{2}}-\frac{(n+i-m+1)}{(n+i+1)^{2}}}\). □

Corollary 2.6

Let \(\varphi (z)=az^{m}\overline{z}^{n}+bz^{s}\overline{z}^{t}\) with nonnegative integers m, n, s, t with \(m\ge n\), \(t\ge s\), \(m> t\), \(m-n=t-s\) and nonzeros \(a, b\in \mathbb{C}\). If

$$ \vert a \vert ^{2}< \max \biggl\{ \frac{(m+1)^{2}}{(t+1)^{2}}, \frac{(m+1)^{2}(s+t)(2m-n+1)^{2}}{(t+1)^{2}(m+n)(2t-s+1)^{2}} \biggr\} \vert b \vert ^{2}, $$

then \(T_{\varphi }\) on \(A^{2}(\mathbb{D})\) is never hyponormal.

Proof

By a direct calculation,

$$ \frac{\frac{(t+i-s+1)}{(t+i+1)^{2}}-\frac{(s+i-t+1)}{(s+i+1)^{2}}}{\frac{(m+i-n+1)}{(m+i+1)^{2}} -\frac{(n+i-m+1)}{(n+i+1)^{2}}} = \frac{(m+i+1)^{2}(n+i+1)^{2}\{(s+t)i+(t^{2}+s^{2}+t+s)\}}{(t+i+1)^{2}(s+i+1)^{2}\{(m+n)i+(n^{2}+m^{2}+n+m)\}}. $$

For convenience, we set

$$ G(i)=\frac{(m+i+1)(n+i+1)}{(t+i+1)(s+i+1)} \quad \text{and}\quad H(i)=\frac{(s+t)i+(t^{2}+s^{2}+t+s)}{(m+n)i+(n^{2}+m^{2}+n+m)}, $$

then

$$ \Lambda (m,n,t,s)=\max_{i\in [m-n,\infty )}G^{2}(i)H(i). $$

By direct calculations,

$$\begin{aligned} G'(i)&= \frac{(s+t-m-n)i^{2}+2\{(t+1)(s+1)-(m+1)(n+1)\}i}{(t+i+1)^{2}(s+i+1)^{2}} \\ &\quad {}+\frac{(m+n+2)(t+1)(s+1)-(s+t+2)(m+1)(n+1)}{(t+i+1)^{2}(s+i+1)^{2}}. \end{aligned}$$

Write \(G'(i)=\frac{P(i)}{Q(i)}\). Since \(s+t-m-n<0\), \(P(i)\) has a maximum at \(i=-\frac{(t+1)(s+1)-(m+1)(n+1)}{s+t-m-n}<0\), and since

$$ \begin{aligned} P(0)&=(m+n+2) (t+1) (s+1)-(s+t+2) (m+1) (n+1) \\ & =(m+1) (t+1) (s-n)+(n+1) (s+1) (t-m)< 0, \end{aligned} $$

\(P(i)<0\) in \(i\ge m-n\) and \(Q(i)>0\) in \(i\ge m-n\). Hence \(G(i)\) is decreasing in \(i\ge m-n\). Similarly,

$$ H'(i)= \frac{ms(m-s)+nt(n-t)+mt(m-t)+ns(n-s)}{\{(m+n)i+(n^{2}+m^{2}+n+m)\}^{2}}, $$

and since \(m>s\), \(m>t\), \(n>s\), and

$$ nt(n-t)+mt(m-t)>mt\bigl(m-t- \vert n-t \vert \bigr)>0, $$

\(H'(i)>0\) and so \(H(i)\) is increasing in \(i, i\ge m-n\). Furthermore, \(\lim_{i\to \infty }H(i)=\frac{s+t}{m+n}\), we have that

$$ \max_{i\in [m-n,\infty )}G^{2}(i)H(i)\le \frac{s+t}{m+n} \max_{i\in [m-n, \infty )}G^{2}(i) \le \frac{(m+1)^{2}(s+t)(2m-n+1)^{2}}{(t+1)^{2}(m+n)(2t-s+1)^{2}}. $$

Hence, by Theorem 2.5, we have the results. □

Theorem 2.7

Let \(\varphi (z)=az^{m}\overline{z}^{m-1}+b\overline{z}^{m-1}z^{m-2}\) with \(m> 0\) and nonzeros \(a, b\in \mathbb{C}\). If \(T_{\varphi }\) on \(A^{2}(\mathbb{D})\) is hyponormal, then

$$ \vert a \vert ^{2}\geq \frac{(m+1)^{2}}{m^{2}} \vert b \vert ^{2}. $$

Proof

Let \(\varphi (z)=az^{m}\overline{z}^{m-1}+b\overline{z}^{m-1}z^{m-2}\). From (2.1), if \(T_{\varphi }\) is hyponormal, then

$$ \begin{aligned} & \vert a \vert ^{2} \Biggl\{ \sum _{i=0}^{\infty }\frac{i+2}{(m+i+1)^{2}} \vert c_{i} \vert ^{2}- \sum_{i=1}^{\infty } \frac{i}{(m+i)^{2}} \vert c_{i} \vert ^{2} \Biggr\} \\ &\quad \geq \vert b \vert ^{2} \Biggl\{ \sum _{i=0}^{\infty }\frac{i+2}{(m+i)^{2}} \vert c_{i} \vert ^{2}- \sum_{i=1}^{\infty } \frac{i}{(m+i-1)^{2}} \vert c_{i} \vert ^{2} \Biggr\} \end{aligned} $$

for any \(c_{i}\in \mathbb{C} \) with \(\operatorname{Re}(a\overline{b}\overline{c}_{i+2}{c}_{i})=0 \) (\(i=0,1,2, \ldots \)). If \(c_{0}\neq 0\) and \(c_{i}=0\) for \(i\ge 0\), then \(|a|^{2}\ge \frac{(m+1)^{2}}{m^{2}}|b|^{2}\) and if \(c_{0}=0\) and \(c_{i}\neq 0\) for \(i\geq 1\),

$$ \vert a \vert ^{2}\geq \max_{i\in [1, \infty )} \frac{\frac{i+2}{(m+i)^{2}}-\frac{i}{(m+i-1)^{2}}}{\frac{i+2}{(m+i+1)^{2}} -\frac{i}{(m+i)^{2}}} \vert b \vert ^{2}. $$

If \(i\geq 1\), then we can easily check that \(\frac{\frac{i+2}{(m+i)^{2}}-\frac{i}{(m+i-1)^{2}}}{\frac{i+2}{(m+i+1)^{2}}-\frac{i}{(m+i)^{2}}} \) is decreasing in i. Hence, if \(T_{\varphi }\) is hyponormal, then

$$ \vert a \vert ^{2}\geq \max \biggl\{ \frac{(m+1)^{2}}{m^{2}}, \frac{(m+2)^{2}(2m^{2}-2m-1)}{m^{2}(2m^{2}+2m-1)} \biggr\} \vert b \vert ^{2}. $$

Since for every nonnegative integer m,

$$ \frac{(m+1)^{2}}{m^{2}}> \frac{(m+2)^{2}(2m^{2}-2m-1)}{m^{2}(2m^{2}+2m-1)}, $$

this completes the proof. □

Now we give the example mentioned above.

Example 2.8

Let \(\varphi (z)=az^{3}\overline{z}^{2}+bz\overline{z}^{2}\) with nonzeros \(a, b\in \mathbb{C}\). Then, by Theorem 2.7,

$$ \frac{(m+1)^{2}}{m^{2}}=\frac{16}{9}, $$

and so if \(T_{\varphi }\) is hyponormal, then

$$ \vert a \vert ^{2}\geq \frac{16}{9} \vert b \vert ^{2}. $$

Theorem 2.9

Let \(\varphi (z)=az^{m}\overline{z}^{n}+bz^{s}\overline{z}^{m}\) with nonnegative integers m, n, s with \(m\ge s> n\) and nonzeros \(a, b\in \mathbb{C}\). If \(T_{\varphi }\) on \(A^{2}(\mathbb{D})\) is hyponormal, then

$$ \vert a \vert ^{2}\geq \max \biggl\{ \frac{2m-2s}{2m-n-s}, \frac{\frac{(2m-n-s)}{(2m-n)^{2}}-\frac{(s-n)}{(s+m-n)^{2}}}{\frac{2(m-n)}{(2m-n)^{2}}}, \Lambda (m,n,m,s) \biggr\} \vert b \vert ^{2}, $$

where \(\Lambda (m,n,m,s)\) is given in Theorem 2.5.

Proof

In a similar way to the proof of Theorem 2.5, if \(T_{\varphi }\) is hyponormal, then

$$ \begin{aligned} & \Biggl\Vert T_{\varphi } \sum_{i=0}^{\infty }c_{i}z^{i} \Biggr\Vert ^{2}- \Biggl\Vert T_{\varphi }^{*} \sum_{i=0}^{\infty }c_{i}z^{i} \Biggr\Vert ^{2} \\ &\quad = \vert a \vert ^{2}\sum_{i=0}^{\infty } \frac{m+i-n+1}{(m+i+1)^{2}} \vert c_{i} \vert ^{2}+ \vert b \vert ^{2} \sum_{i=m-s}^{\infty } \frac{s+i-m+1}{(s+i+1)^{2}} \vert c_{i} \vert ^{2} \\ &\qquad {} - \vert a \vert ^{2}\sum_{i=m-n}^{\infty } \frac{n+i-m+1}{(n+i+1)^{2}} \vert c_{i} \vert ^{2}- \vert b \vert ^{2} \sum_{i=0}^{\infty } \frac{m+i-s+1}{(m+i+1)^{2}} \vert c_{i} \vert ^{2}\geq 0 \end{aligned} $$

or, equivalently,

$$ \begin{aligned} & \vert a \vert ^{2} \Biggl(\sum_{i=0}^{\infty } \frac{m+i-n+1}{(m+i+1)^{2}} \vert c_{i} \vert ^{2}- \sum _{i=m-n}^{\infty }\frac{n+i-m+1}{(n+i+1)^{2}} \vert c_{i} \vert ^{2} \Biggr) \\ &\quad \geq \vert b \vert ^{2} \Biggl(\sum _{i=0}^{\infty } \frac{m+i-s+1}{(m+i+1)^{2}} \vert c_{i} \vert ^{2}-\sum_{i=m-s}^{\infty } \frac{s+i-m+1}{(s+i+1)^{2}} \vert c_{i} \vert ^{2} \Biggr) \end{aligned} $$
(2.3)

for any \(c_{i}\in \mathbb{C}\) with \(\operatorname{Re}(a\overline{b}{c}_{i-m+n}\overline{c}_{i-m-s})=0\) and \(\operatorname{Re}(a\overline{b}\overline{c}_{i+m-n}{c}_{i-m+s})=0\) (\(i=0,1,2, \ldots \)). If \(c_{i}\neq 0\) for \(0\le i< m-s\) and \(c_{i}=0\) for \(i\ge m-s\), then (2.3) implies

$$ \vert a \vert ^{2}\geq \frac{m+i-s+1}{m+i-n+1} \vert b \vert ^{2}, $$

and since \(\frac{m+i-s+1}{m+i-n+1}\) is increasing in i, we have that

$$ \vert a \vert ^{2}\geq \frac{2m-2s}{2m-n-s} \vert b \vert ^{2}. $$

If \(c_{i}\neq 0\) for \(m-s\le i< m-n\) and \(c_{i}=0\) for \(i< m-s\) or \(i\ge m-n\), then

$$ \vert a \vert ^{2}\ge \frac{\frac{m+i-s+1}{(m+i+1)^{2}}-\frac{s+i-m+1}{(s+i+1)^{2}}}{\frac{m+i-n+1}{(m+i+1)^{2}}} \vert b \vert ^{2}. $$

By direct calculations, \(\frac{\frac{m+i-s+1}{(m+i+1)^{2}}-\frac{s+i-m+1}{(s+i+1)^{2}}}{\frac{m+i-n+1}{(m+i+1)^{2}}}\) is increasing and hence

$$ \vert a \vert ^{2}\ge \frac{\frac{(2m-n-s)}{(2m-n)^{2}}-\frac{(s-n)}{(s+m-n)^{2}}}{\frac{2(m-n)}{(2m-n)^{2}}} \vert b \vert ^{2}. $$

If \(c_{i}\neq 0\) for \(i\ge m-n\) and \(c_{i}=0\) for \(i< m-n\), then

$$ \vert a \vert ^{2}\geq \Lambda (m,n,m,s) \vert b \vert ^{2}, $$

where \(\Lambda (m,n,m,s)\) is given in Theorem 2.5. □

Corollary 2.10

Let \(\varphi (z)=az^{m}\overline{z}^{n}+bz^{s}\overline{z}^{m}\) with nonnegative integers m, n, s with \(m\ge s> n\) and nonzeros \(a, b\in \mathbb{C}\). If \(|a|^{2}< C|b|^{2}\), where

$$\begin{aligned} C =&\max \biggl\{ \frac{2m-2s}{2m-n-s}, \frac{\frac{(2m-n-s)}{(2m-n)^{2}}- \frac{(s-n)}{(s+m-n)^{2}}}{\frac{2(m-n)}{(2m-n)^{2}}}, \\ & \frac{(m-s)\{2m^{2}+(s-n+1)m+s^{2}+s-sn\}}{(m-n)(2m^{2}+m+n)}, \frac{m^{2}-s^{2}}{m^{2}-n^{2}} \biggr\} , \end{aligned}$$

then \(T_{\varphi }\) on \(A^{2}(\mathbb{D})\) is never hyponormal.

Proof

By a direct calculation,

$$ \frac{\frac{(m+i-s+1)}{(m+i+1)^{2}}-\frac{(s+i-m+1)}{(s+i+1)^{2}}}{\frac{(m+i-n+1)}{(m+i+1)^{2}}-\frac{(n+i-m+1)}{(n+i+1)^{2}}} = \frac{(n+i+1)^{2}\{(m+s)(m-s)i+(m-s)(m^{2}+s^{2}+m+s)\}}{(s+i+1)^{2}\{(m+n)(m-n)i+(m-n)(n^{2}+m^{2}+n+m)\}}. $$

For convenience, we set

$$ G(i)=\frac{n+i+1}{s+i+1} \quad \text{and}\quad H(i)= \frac{(m+s)(m-s)i+(m-s)(m^{2}+s^{2}+m+s)}{(m+n)(m-n)i+(m-n)(n^{2}+m^{2}+n+m)}, $$

then

$$ \Lambda (m,n,m,s)=\max_{i\in [m-n,\infty )}G^{2} (i)H(i). $$

Since

$$ G'(i)=\frac{s-n}{(s+i+1)^{2}}, $$

\(G(i)\) is increasing and \(\lim_{i\to \infty }G(i)=1\). Similarly,

$$ H'(i)= \frac{(m-s)\{(m+s)(m^{2}+n^{2}+m+n)-(m+n)(m^{2}+s^{2}+m+s)\}}{(m-n)\{(m+n)i+(m^{2}+n^{2}+m+n)\}^{2}}, $$

and thus \(H(i)\) is monotone for \(i\ge m-n\). Therefore

$$ \max_{i\in [m-n,\infty )}H(i)\le \max \biggl\{ \frac{(m-s)\{2m^{2}+(s-n+1)m+s^{2}+s-sn\}}{(m-n)(2m^{2}+m+n)}, \frac{m^{2}-s^{2}}{m^{2}-n^{2}} \biggr\} . $$

Furthermore, we have that

$$ \begin{aligned} \max_{i\in [m-n,\infty )}G^{2}(i)H(i)& \le \max_{i\in [m-n,\infty )}H(i) \\ & \le \max \biggl\{ \frac{(m-s)\{2m^{2}+(s-n+1)m+s^{2}+s-sn\}}{(m-n)(2m^{2}+m+n)}, \frac{m^{2}-s^{2}}{m^{2}-n^{2}} \biggr\} . \end{aligned} $$

Hence, by Theorem 2.9, we have the results. □

Example 2.11

Let \(\varphi (z)=az^{3}\overline{z}+bz^{2}\overline{z}^{3}\) with nonzeros \(a, b\in \mathbb{C}\). Then, by Theorem 2.9,

$$\begin{aligned}& \frac{2m-2s}{2m-n-s}=\frac{2}{3},\qquad \frac{\frac{(2m-n-s)}{(2m-n)^{2}}-\frac{(s-n)}{(s+m-n)^{2}}}{\frac{2(m-n)}{(2m-n)^{2}}}= \frac{23}{64}, \\& \Lambda (m,n,t,s)=\max_{i\in [2,\infty )} \frac{(i+2)^{2}(5i+18)}{4(i+3)^{2}(2i+7)}. \end{aligned}$$

Since \(\frac{(i+2)^{2}(5i+18)}{4(i+3)^{2}(2i+7)}\) is increasing for \(i\ge 2\), we have that

$$ \Lambda (m,n,t,s)=\lim_{i\to \infty } \frac{(i+2)^{2}(5i+18)}{4(i+3)^{2}(2i+7)}= \frac{5}{8}. $$

Therefore, if \(T_{\varphi }\) is hyponormal, then

$$ \vert a \vert ^{2}\geq \frac{2}{3} \vert b \vert ^{2}. $$

Theorem 2.12

Let \(\varphi (z)=az^{m}\overline{z}^{n}+bz^{s}\overline{z}^{m}\) with nonnegative integers m, n, s with \(m\ge n > s\) and nonzeros \(a, b\in \mathbb{C}\). If \(T_{\varphi }\) on \(A^{2}(\mathbb{D})\) is hyponormal, then

$$ \vert a \vert ^{2}\geq \max \biggl\{ \frac{m-s+1}{m-n+1}, \frac{\frac{2m-2s}{(2m-s)^{2}}}{\frac{2m-s-n}{(2m-s)^{2}}-\frac{n-s}{(n+m-s)^{2}}}, \Lambda (m,n,m,s) \biggr\} \vert b \vert ^{2}, $$

where \(\Lambda (m,n,m,s)\) is given in Theorem 2.5.

Proof

In a similar way to the proof of Theorem 2.4, \(T_{\varphi }\) is hyponormal if and only if

$$ \begin{aligned} & \vert a \vert ^{2} \Biggl(\sum_{i=0}^{\infty } \frac{m+i-n+1}{(m+i+1)^{2}} \vert c_{i} \vert ^{2}- \sum _{i=m-n}^{\infty }\frac{n+i-m+1}{(n+i+1)^{2}} \vert c_{i} \vert ^{2} \Biggr) \\ &\quad \geq \vert b \vert ^{2} \Biggl(\sum _{i=0}^{\infty } \frac{m+i-s+1}{(m+i+1)^{2}} \vert c_{i} \vert ^{2}-\sum_{i=m-s}^{\infty } \frac{s+i-m+1}{(s+i+1)^{2}} \vert c_{i} \vert ^{2} \Biggr) \end{aligned} $$
(2.4)

for any \(c_{i}\in \mathbb{C}\) with \(\operatorname{Re}(a\overline{b}{c}_{i-m+n}\overline{c}_{i-m-s})=0\) and \(\operatorname{Re}(a\overline{b}\overline{c}_{i+m-n}{c}_{i-m+s})=0\) (\(i=0,1,2,\ldots \)). If \(c_{i}\neq 0\) for \(0\le i< m-n\) and \(c_{i}=0\) for \(i\ge m-n\), then (2.4) implies

$$ \vert a \vert ^{2}\geq \frac{m+i-s+1}{m+i-n+1} \vert b \vert ^{2}, $$

and since \(\frac{m+i-s+1}{m+i-n+1}\) is decreasing in i, we have that

$$ \vert a \vert ^{2}\geq \frac{m-s+1}{m-n+1} \vert b \vert ^{2}. $$

If \(c_{i}\neq 0\) for \(m-n\le i< m-s\) and \(c_{i}=0\) for \(i< m-n\) or \(i\ge m-s\), then

$$ \vert a \vert ^{2}\ge \frac{\frac{m+i-s+1}{(m+i+1)^{2}}}{\frac{m+i-n+1}{(m+i+1)^{2}}-\frac{n+i-m+1}{(n+i+1)^{2}}} \vert b \vert ^{2}. $$

By direct calculations, \(\frac{\frac{m+i-s+1}{(m+i+1)^{2}}}{\frac{m+i-n+1}{(m+i+1)^{2}}-\frac{n+i-m+1}{(n+i+1)^{2}}}\) is increasing and hence

$$ \vert a \vert ^{2}\ge \frac{\frac{2m-2s}{(2m-s)^{2}}}{\frac{2m-s-n}{(2m-s)^{2}}-\frac{n-s}{(n+m-s)^{2}}} \vert b \vert ^{2}. $$

If \(c_{i}\neq 0\) for \(i\ge m-s\) and \(c_{i}=0\) for \(i< m-s\), then

$$ \vert a \vert ^{2}\geq \Lambda (m,n,m,s) \vert b \vert ^{2}, $$

where \(\Lambda (m,n,m,s)\) is given in Theorem 2.5. □

Corollary 2.13

Let \(\varphi (z)=az^{m}\overline{z}^{n}+bz^{s}\overline{z}^{m}\) with nonnegative integers m, n, s with \(m\ge n > s\) and nonzeros \(a, b\in \mathbb{C}\). If

$$ \vert a \vert ^{2}< \max \biggl\{ \frac{m-s+1}{m-n+1}, \frac{\frac{2m-2s}{(2m-s)^{2}}}{\frac{2m-s-n}{(2m-s)^{2}}-\frac{n-s}{(n+m-s)^{2}}}, \frac{C_{1}(m+1)^{2}}{(s+m-n+1)^{2}} \biggr\} \vert b \vert ^{2}, $$

where \(C_{1}=\max \{ \frac{(m-s)\{2m^{2}+(s-n+1)m+s^{2}+s-sn\}}{(m-n)(2m^{2}+m+n)}, \frac{m^{2}-s^{2}}{m^{2}-n^{2}} \} \), then \(T_{\varphi }\) on \(A^{2}(\mathbb{D})\) is never hyponormal.

Proof

$$ \frac{\frac{(m+i-s+1)}{(m+i+1)^{2}}-\frac{(s+i-m+1)}{(s+i+1)^{2}}}{\frac{(m+i-n+1)}{(m+i+1)^{2}}-\frac{(n+i-m+1)}{(n+i+1)^{2}}} = \frac{(n+i+1)^{2}\{(m+s)(m-s)i+(m-s)(m^{2}+s^{2}+m+s)\}}{(s+i+1)^{2}\{(m+n)(m-n)i+(m-n)(n^{2}+m^{2}+n+m)\}}. $$

For convenience, we set

$$ G(i)=\frac{n+i+1}{s+i+1} \quad \text{and}\quad H(i)= \frac{(m+s)(m-s)i+(m-s)(m^{2}+s^{2}+m+s)}{(m+n)(m-n)i+(m-n)(n^{2}+m^{2}+n+m)}, $$

then

$$ \Lambda (m,n,m,s)=\max_{i\in [m-n,\infty )}G^{2} (i)H(i). $$

Since

$$ G'(i)=\frac{s-n}{(s+i+1)^{2}}, $$

\(G(i)\) is decreasing. Similarly,

$$ H'(i)= \frac{(m-s)\{(m+s)(m^{2}+n^{2}+m+n)-(m+n)(m^{2}+s^{2}+m+s)\}}{(m-n)\{(m+n)i+(m^{2}+n^{2}+m+n)\}^{2}}, $$

and thus \(H(i)\) is monotone for \(i\ge m-n\). Therefore

$$ \max_{i\in [m-n,\infty )}H(i)\le \max \biggl\{ \frac{(m-s)\{2m^{2}+(s-n+1)m+s^{2}+s-sn\}}{(m-n)(2m^{2}+m+n)}, \frac{m^{2}-s^{2}}{m^{2}-n^{2}} \biggr\} . $$

Hence

$$ \max_{i\in [m-n,\infty )}G^{2}(i)H(i)\le C_{1}\max _{i\in [m-n, \infty )}G^{2}(i) \le \frac{C_{1}(m+1)^{2}}{(s+m-n+1)^{2}}, $$

where \(C_{1}=\max \{ \frac{(m-s)\{2m^{2}+(s-n+1)m+s^{2}+s-sn\}}{(m-n)(2m^{2}+m+n)}, \frac{m^{2}-s^{2}}{m^{2}-n^{2}} \} \). Hence, by Theorem 2.12, we have the results. □

Example 2.14

Let \(\varphi (z)=az^{3}\overline{z}^{2}+bz\overline{z}^{3}\) with nonzeros \(a, b\in \mathbb{C}\). Then, by Theorem 2.12,

$$\begin{aligned}& \frac{m-s+1}{m-n+1}=\frac{3}{2},\qquad \frac{\frac{(2m-n-s+1)}{(2m-n+1)^{2}}}{\frac{(2m-2n+1)}{(2m-n+1)^{2}}-\frac{1}{(m+1)^{2}}}= \frac{64}{23}, \\& \Lambda (m,n,t,s)=\max_{i\in [1,\infty )} \frac{4(i+3)^{2}(2i+7)}{(i+2)^{2}(5i+18)}. \end{aligned}$$

Since \(\frac{4(i+3)^{2}(2i+7)}{(i+2)^{2}(5i+18)}\) is decreasing for \(i\ge 1\), we have that

$$ \Lambda (m,n,t,s)=\frac{64}{23}. $$

Therefore, if \(T_{\varphi }\) is hyponormal, then

$$ \vert a \vert ^{2}\geq \frac{64}{23} \vert b \vert ^{2}. $$

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Acknowledgements

The authors would like to thank the reviewers for their suggestions that helped improve the original manuscript in its present form.

Funding

The first author was supported by the Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education (2020R1I1A1A01053085 and 2019R1A6A1A10073079). The second author was supported by the Basic Science Research Program to Research Institute for Basic Sciences (RIBS) of Jeju National University through the National Research Foundation of Korea (NRF) funded by the Ministry of Education (2019R1A6A1A10072987) and the Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education (No. 2018R1D1A1B07048620).

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Kim, S., Lee, J. Hyponormality of Toeplitz operators with non-harmonic symbols on the Bergman spaces. J Inequal Appl 2021, 67 (2021). https://doi.org/10.1186/s13660-021-02602-1

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MSC

  • 47B35
  • 47B20
  • 30H20

Keywords

  • Toeplitz operator
  • Hyponormal operator
  • Bergman space