Normal Toeplitz Operators on the Bergman Space

Abstract

In this paper, we give a characterization of normality of Toeplitz operator $T_{\phi }$ on the Bergman space $A^2\left(\mathbb{D}\right)$. First, we state basic properties for Toeplitz operator $T_{\phi }$ on $A^2\left(\mathbb{D}\right)$. Next, we consider the normal Toeplitz operator $T_{\phi }$ on $A^2\left(\mathbb{D}\right)$ in terms of harmonic symbols $\phi$. Finally, we characterize the normal Toeplitz operators $T_{\phi }$ with non-harmonic symbols acting on $A^2\left(\mathbb{D}\right)$.

1. Introduction

The purpose of this paper is to study the normality of Toeplitz operators acting on the Bergman space. Our interest is focused on Toeplitz operators with harmonic and non-harmonic symbols. Let $\mathcal{L}\left(\mathcal{H}\right)$ be the algebra of bounded linear operators on a separable complex Hilbert space $\mathcal{H}$. An operator $T\in \mathcal{L}\left(\mathcal{H}\right)$ is normal if its self-commutator $[T^{*},T]:=T^{*}T-T{T}^{*}=0$, where $T^{*}$ denotes the adjoint of T. Let $\mathbb{D}$ denote the open unit disk in $\mathbb{C}$, $dA$ be the normalized area measure on $\mathbb{D}$. The space $L^2\left(\mathbb{D}\right)$ is a Hilbert space with the inner product

$$\langle f,g\rangle =\frac{1}{\pi }{\int }_{\mathbb{D}}f\left(z\right)\overline{g\left(z\right)}dA\left(z\right).$$

The Bergman space $A^2\left(\mathbb{D}\right)$ is the subspace of $L^2\left(\mathbb{D}\right)$ consisting of functions analytic on $\mathbb{D}$ and $L^{\infty }\left(\mathbb{D}\right)$ be the space of bounded area measurable functions on $\mathbb{D}$. The multiplication operator $M_{\psi }$ with symbol $\psi \in L^{\infty }\left(\mathbb{D}\right)$ is defined by $M_{\psi }f=\psi f$ for $f\in A^2\left(\mathbb{D}\right).$ For any $\phi \in L^{\infty }\left(\mathbb{D}\right)$, the Toeplitz operator $T_{\phi }$ on the Bergman space is defined by the formula

$$T_{\phi }f=P\left(\phi f\right)$$

for $f\in A^2\left(\mathbb{D}\right)$ and P denotes the orthogonal projection of $L^2\left(\mathbb{D}\right)$ onto $A^2\left(\mathbb{D}\right)$. In this case, the function $\phi$ is called the symbol of $T_{\phi }$. It is clear that those operators are bounded if $\phi$ is in $L^{\infty }\left(\mathbb{D}\right)$. Recall that the Bergman space $A^2\left(\mathbb{D}\right)$ is the space of analytic functions in $L^2(\mathbb{D},dA)$ with the power series representation

$$f\left(z\right)={\displaystyle \sum _{n=0}^{\infty }}{a}_n{z}^n\mathrm{where}{\displaystyle \sum _{n=0}^{\infty }}\frac{1}{n+1}{\left|a_n\right|}^2<\infty .$$

We will now consider the normality of Toeplitz operators on the Bergman space with various symbols. We shall list the well-known properties of Toeplitz operators $T_{\phi }$ on the Bergman space.

Let $f,g$ are in $L^{\infty }\left(\mathbb{D}\right)$ and $\alpha ,\beta \in \mathbb{C}$, then we can easily check that

$$\begin{array}{c}\left(\mathrm{i}\right)T_{\alpha f+\beta g}=\alpha T_f+\beta T_g\hfill \\ \left(\mathrm{ii}\right)T_f^{*}=T_{\overline{f}}\hfill \\ \left(\mathrm{iii}\right)T_{\overline{f}}{T}_g=T_{\overline{f}g},\mathrm{if}f\mathrm{or}g\mathrm{is}\mathrm{analytic}.\hfill \end{array}$$

Many authors in [1,2,3,4,5,6] studied intensively Normal operators and Toeplitz operators. It is a natural ask of when Toeplitz operator becomes normal. In 1963, A. Brown and P. Halmos [7] characterized normal Toeplitz operators on the Hardy space. This contains many of the fundamental results on the algebraic properties of Toeplitz operators. It has been central significance in operator theory. Thus we will focus primarily on normality of Toeplitz operators on the Bergman space.

Recently, the authors in [3] gave a properties of $T_{\phi }$ with non-harmonic symbols on the Bergman space. In view of this, first, we characterize the normal Toeplitz operator $T_{\phi }$ on $A^2\left(\mathbb{D}\right)$ in terms of harmonic symbols $\phi$. Next, we characterize the normal Toeplitz operators $T_{\phi }$ with non-harmonic symbols acting on $A^2\left(\mathbb{D}\right)$.

The remainder of the paper is organized as follows. In Section 2, we study the normal Toeplitz operators $T_{\phi }$ on $A^2\left(\mathbb{D}\right)$ with harmonic symbols $\phi$. In Section 3, we consider the normal Toeplitz operators $T_{\phi }$ with non-harmonic symbols acting on $A^2\left(\mathbb{D}\right)$ and their applications.

2. Toeplitz Operators with Harmonic Symbols

First, we study the normality of Toeplitz operators on $A^2\left(\mathbb{D}\right)$ with the general symbol. In [2], the authors study the normality of Toeplitz operators on the weighted Bergman spaces as follows.

Lemma 1

(Theorem 3.2 in [2]). Let $\phi =\alpha z^n+\beta z^m+\gamma {\overline{z}}^p+\delta {\overline{z}}^q$ with $n<m,p<q,n-m=q-p$ and for nonzero $\alpha ,\beta ,\gamma ,\delta$ and $\phi +\overline{\phi }$ is a constant. Then $T_{\phi }$ is normal if and only if $\lambda \in \mathbb{T}$ and $\phi$ is one of exactly three types;

(i) 

$\phi =\alpha z^n-\lambda {\overline{\alpha z}}^n$,

(ii) 

$\phi =\alpha z^n+\beta z^m-\lambda (\overline{\alpha }{\overline{z}}^n+\overline{\beta }{\overline{z}}^m)$,

(iii) 

$\phi =\beta z^m-\lambda \overline{\beta }{\overline{z}}^m$.

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

Lemma 2

([8]). For any $s,t\in \mathbb{N}$,

$$P\left({\overline{z}}^t{z}^s\right)=\left\{\begin{array}{c}\frac{s-t+1}{s+1}{z}^{s-t}ifs\ge t\hfill \\ 0ifs<t.\hfill \end{array}\right.$$

The proof for the Lemma 3 follows the proof of Theorem 2.1 in [8].

Lemma 3.

For 0$\le m\le N$, we deduce that

(i)

${\displaystyle \left|\right|{\overline{z}}^m{\displaystyle \sum _{i=0}^{\infty }}{c}_i{z}^i{\left|\right|}^2={\displaystyle \sum _{i=0}^{\infty }}\frac{1}{i+m+1}{\left|c_i\right|}^2}$,

(ii)

${\displaystyle \left|\right|P({\overline{z}}^m{\displaystyle \sum _{i=0}^{\infty }}{c}_i{z}^i){\left|\right|}^2={\displaystyle \sum _{i=m}^{\infty }}\frac{i-m+1}{{(i+1)}^2}{\left|c_i\right|}^2.}$

Next, we characterize the normality of Toeplitz operators with harmonic symbols on $A^2\left(\mathbb{D}\right)$.

Theorem 1.

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

$$f\left(z\right)={\displaystyle \sum _{n=0}^N}{a}_n{z}^{n}andg\left(z\right)={\displaystyle \sum _{n=1}^m}{a}_{-n}{z}^n$$

with$a_{-m}{a}_N\ne 0$. Then$T_{\phi }$is normal on$A^2\left(\mathbb{D}\right)$if and only if$g\left(z\right)=e^{i\theta }f\left(z\right)$.

Proof. 

Observe that $T_{\phi }$ is normal if and only if $T_{\overline{f}f}+T_g{T}_{\overline{g}}=T_f{T}_{\overline{f}}+T_{\overline{g}g}$. First we show $N=m$. Without loss of generality, we assume that $N>m$. Then by acting $z^m$ to the both side in the above relation, we have $T_{\overline{f}f}{z}^m=T_f{T}_{\overline{f}}{z}^m$, and so

$$P\left({\displaystyle \sum _{n=0}^N}\overline{a_n}{\overline{z}}^n{\displaystyle \sum _{n=0}^N}{a}_n{z}^n{z}^m\right)={\displaystyle \sum _{n=0}^N}{a}_n{z}^{n}P\left({\displaystyle \sum _{n=0}^N}\overline{a_n}{\overline{z}}^n{z}^m\right).$$

Since

$$P\left({\displaystyle \sum _{n=0}^N}\overline{a_n}{\overline{z}}^n{\displaystyle \sum _{n=0}^N}{a}_n{z}^n{z}^m\right)=P\left({\displaystyle \sum _{i=0}^N}{\displaystyle \sum _{j=max\{0,i-m\}}^N}\overline{a_i}{\overline{z}}^i{a}_j{z}^{j+m}\right)$$

and

$${\displaystyle \sum _{n=0}^N}{a}_n{z}^{n}P\left({\displaystyle \sum _{n=0}^N}\overline{a_n}{\overline{z}}^n{z}^m\right)={\displaystyle \sum _{n=0}^N}{a}_n{z}^{n}P\left({\displaystyle \sum _{n=0}^m}\overline{a_n}{\overline{z}}^n{z}^m\right),$$

for the coefficient of $z^m$, we deduce that

$${\displaystyle \sum _{j=0}^N}\frac{m+1}{m+j+1}|a_j{|}^2={\displaystyle \sum _{j=0}^m}\frac{m-j+1}{m+1}{\left|a_j\right|}^2.$$

Moreover, since

$$\frac{m+1}{m+j+1}\ge \frac{m-j+1}{m+1}$$

for $1\le j\le N$, we have $a_j=0$ for all $1\le j\le N$. This contradicts the assumption $a_N\ne 0$.

Next, we find the necessary and sufficient condition of normality. Since $T_{\phi }$ is normal if and only if $T_{\overline{f}f}-T_f{T}_{\overline{f}}=T_{\overline{g}g}-T_g{T}_{\overline{g}}$, for any $k\in \mathbb{N}$,

$$\begin{array}{cc}\hfill (T_{\overline{f}f}-T_f{T}_{\overline{f}})z^k& =P\left({\displaystyle \sum _{n=0}^N}\overline{a_n}{\overline{z}}^n{\displaystyle \sum _{n=0}^N}{a}_n{z}^n{z}^k\right)-{\displaystyle \sum _{n=0}^N}{a}_n{z}^{n}P\left({\displaystyle \sum _{n=0}^N}\overline{a_n}{\overline{z}}^n{z}^k\right)\hfill \\ & ={\displaystyle \sum _{i=0}^{min\{j+k,N\}}}{\displaystyle \sum _{j=0}^N}\frac{k+j-i+1}{k+j+1}\overline{a_i}{a}_j{z}^{j+k-i}-{\displaystyle \sum _{j=0}^N}{a}_j{z}^j{\displaystyle \sum _{i=0}^k}\frac{k-i+1}{k+1}\overline{a_i}{z}^{k-i}\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill (T_{\overline{g}g}-T_g{T}_{\overline{g}})z^k& =P\left({\displaystyle \sum _{n=1}^N}\overline{a_{-n}}{\overline{z}}^n{\displaystyle \sum _{n=1}^N}{a}_{-n}{z}^n{z}^k\right)-{\displaystyle \sum _{n=1}^N}{a}_{-n}{z}^{n}P\left({\displaystyle \sum _{n=1}^N}\overline{a_{-n}}{\overline{z}}^n{z}^k\right)\hfill \\ & ={\displaystyle \sum _{i=1}^{min\{j+k,N\}}}{\displaystyle \sum _{j=1}^N}\frac{k+j-i+1}{k+j+1}\overline{a_{-i}}{a}_{-j}{z}^{j+k-i}-{\displaystyle \sum _{j=1}^N}{a}_{-j}{z}^j{\displaystyle \sum _{i=1}^k}\frac{k-i+1}{k+1}\overline{a_{-i}}{z}^{k-i}.\hfill \end{array}$$

By direct calculations of coefficients of $z^{k+N-2}$, we have

$$\begin{array}{c}\overline{a_0}{a}_{N-2}+\overline{a_1}{a}_{N-1}\frac{N+k-1}{N+k}+\overline{a_2}{a}_N\frac{N+k-1}{N+k+1}-\left(\overline{a_0}{a}_{N-2}+\overline{a_1}{a}_{N-1}\frac{k}{k+1}+\overline{a_2}{a}_N\frac{k-1}{k+1}\right)\hfill \\ =\overline{a_{-1}}{a}_{-(N-1)}\frac{N+k-1}{N+k}+\overline{a_{-2}}{a}_{-N}\frac{N+k-1}{N+k+1}-\left(\overline{a_{-1}}{a}_{-(N-1)}\frac{k}{k+1}+\overline{a_{-2}}{a}_{-N}\frac{k-1}{k+1}\right)\hfill \end{array}$$

for any $k\in \mathbb{N}$. Therefore

$$\left(\overline{a_1}{a}_{N-1}-\overline{a_{-1}}{a}_{-(N-1)}\right)\left(\frac{N+k-1}{N+k}-\frac{k}{k+1}\right)+\left(\overline{a_2}{a}_N-\overline{a_{-2}}{a}_{-N}\right)\left(\frac{N+k-1}{N+k+1}-\frac{k-1}{k+1}\right)=0$$

for any $k\in \mathbb{N}$. Since k is arbitrary, so that $\overline{a_1}{a}_{N-1}=\overline{a_{-1}}{a}_{-(N-1)}$ and $\overline{a_2}{a}_N=\overline{a_{-2}}{a}_{-N}$. By the similar arguments, we have

$$\overline{a_i}{a}_j=\overline{a_{-i}}{a}_{-j}$$

for all $i,j$ with $1\le i,j\le N$. Therefore,

$$a_{-j}=e^{i\theta }{a}_j$$

for all $1\le j\le N$ and so $g\left(z\right)=e^{i\theta }f\left(z\right)$.

If $g\left(z\right)=e^{i\theta }f\left(z\right)$, then

$$T_{\phi }^{*}{T}_{\phi }=T_{\overline{f}+g}{T}_{\overline{g}+f}=T_{\overline{f}+e^{i\theta }f}{T}_{e^{-i\theta }\overline{f}+f}=T_{e^{-i\theta }\overline{f}+f}{T}_{\overline{f}+e^{i\theta }f}=T_{\phi }{T}_{\phi }^{*}.$$

Thus, $T_{\phi }$ is normal on $A^2\left(\mathbb{D}\right)$. This completes the proof. □

3. Toeplitz Operators with Non-Harmonic Symbols

In this section, we study the normality of $T_{\phi }$ on $A^2\left(\mathbb{D}\right)$ with the non-harmonic symbol $\phi$. In this case, we cannot apply the tool as in Theorem 1 to $\phi$, since it cannot be devided into analytic and co-analytic parts. Hence we should calculate the self-commutator of $T_{\phi }$ for non-harmonic symbol $\phi$. For this, we discuss about the symbol $\phi$ of the form $\phi \left(z\right)=a_{m,n}{z}^m{\overline{z}}^n$ with $a_{m,n}\in \mathbb{C}.$

Lemma 4.

Let$\phi \left(z\right)=a_{m,n}{z}^m{\overline{z}}^n$with$a_{m,n}\in \mathbb{C}$. Then$T_{\phi }$on$A^2\left(\mathbb{D}\right)$is normal if and only if$m=n$.

Proof. 

For $\phi \left(z\right)=a_{m,n}{z}^m{\overline{z}}^n$, $T_{\phi }$ is normal if and only if

$$〈(T_{\phi }^{*}{T}_{\phi }-T_{\phi }{T}_{\phi }^{*}){\displaystyle \sum _{i=0}^{\infty }}{c}_i{z}^i,{\displaystyle \sum _{i=0}^{\infty }}{c}_i{z}^i〉=0$$

for all $c_i\in \mathbb{C}.$ Using Lemmas 2 and 3, we have that

$$\begin{array}{l}\left|\right|T_{\phi }{\displaystyle \sum _{i=0}^{\infty }}{c}_i{z}^i{\left|\right|}^2-\left|\right|T_{\phi }^{*}{\displaystyle \sum _{i=0}^{\infty }}{c}_i{z}^i{\left|\right|}^2\hfill \\ =\left|\right|T_{a_{m,n}{z}^m{\overline{z}}^n}{\displaystyle \sum _{i=0}^{\infty }}{c}_i{z}^i{\left|\right|}^2-\left|\right|T_{{\overline{a}}_{m,n}{\overline{z}}^m{z}^n}{\displaystyle \sum _{i=0}^{\infty }}{c}_i{z}^i{\left|\right|}^2\hfill \\ ={\left|\right|P\left(a_{m,n}{z}^m{\overline{z}}^n{\displaystyle \sum _{i=0}^{\infty }}{c}_i{z}^i\right)\left|\right|}^2-{\left|\right|P\left({\overline{a}}_{m,n}{\overline{z}}^m{z}^n{\displaystyle \sum _{i=0}^{\infty }}{c}_i{z}^i\right)\left|\right|}^2\hfill \\ =|a_{m,n}{|}^2{\displaystyle \sum _{i=max\{n-m,0\}}^{\infty }}\frac{m+i-n+1}{{(m+i+1)}^2}|c_i{|}^2-|a_{m,n}{|}^2{\displaystyle \sum _{i=max\{m-n,0\}}^{\infty }}\frac{n+i-m+1}{{(n+i+1)}^2}{|c_i|}^2=0.\hfill \end{array}$$

Hence $T_{\phi }$ is normal if and only if

$${\displaystyle \sum _{i=max\{n-m,0\}}^{\infty }}\frac{m+i-n+1}{{(m+i+1)}^2}|c_i{|}^2={\displaystyle \sum _{i=max\{m-n,0\}}^{\infty }}\frac{n+i-m+1}{{(n+i+1)}^2}{\left|c_i\right|}^2$$

for all $c_i\in \mathbb{C}$. Since $c_i$’s are arbitrary, we have that $T_{\phi }$ is normal if and only if $m=n$. This completes the proof. □

We now consider the normality of Toeplitz operators with two terms of non-harmonic symbols. The following theorem gives a general characterization of normal Toeplitz operators with the symbols is of the form $\phi \left(z\right)=a{z}^m{\overline{z}}^n+b{z}^s{\overline{z}}^t$ $(m\ge n\ge 0,t\ge s\ge 0)$ with some assumptions.

Theorem 2.

Let$\phi \left(z\right)=a{z}^m{\overline{z}}^n+b{z}^s{\overline{z}}^t$with$m\ge n\ge 0,t\ge s\ge 0$and nonzeros$a,b\in \mathbb{C}$. Then$T_{\phi }$is normal if and only if$\phi \left(z\right)$is of the form

$$\phi \left(z\right)={(a+b)\left|z\right|}^{2m}or\phi \left(z\right)=a{z}^m{\overline{z}}^n+b{z}^n{\overline{z}}^m\left(\right|a|=\left|b\right|).$$

Proof. 

Let $\phi \left(z\right)=a{z}^m{\overline{z}}^n+b{z}^s{\overline{z}}^t$ with $m\ge n,t\ge s$. By similar arguments as in the proof of Lemma 4, $T_{\phi }$ is normal if and only if

$$\begin{array}{l}{\left|\right|T_{\phi }{\displaystyle \sum _{i=0}^{\infty }}{c}_i{z}^i\left|\right|}^2-{\left|\right|T_{\phi }^{*}{\displaystyle \sum _{i=0}^{\infty }}{c}_i{z}^i\left|\right|}^2\hfill \\ ={\left|\right|T_{a{z}^m{\overline{z}}^n}{\displaystyle \sum _{i=0}^{\infty }}{c}_i{z}^i+T_{b{z}^s{\overline{z}}^t}{\displaystyle \sum _{i=0}^{\infty }}{c}_i{z}^i\left|\right|}^2-{\left|\right|T_{\overline{a}{\overline{z}}^m{z}^n}{\displaystyle \sum _{i=0}^{\infty }}{c}_i{z}^i+T_{\overline{b}{\overline{z}}^s{z}^t}{\displaystyle \sum _{i=0}^{\infty }}{c}_i{z}^i\left|\right|}^2\hfill \\ ={\left|\right|P\left(a{\overline{z}}^n{\displaystyle \sum _{i=0}^{\infty }}{c}_i{z}^{i+m}\right)+P\left(b{\overline{z}}^t{\displaystyle \sum _{i=0}^{\infty }}{c}_i{z}^{i+s}\right)\left|\right|}^2\hfill \\ \hspace{1em}-{\left|\right|P\left(\overline{a}{\overline{z}}^m{\displaystyle \sum _{i=0}^{\infty }}{c}_i{z}^{i+n}\right)+P\left(\overline{b}{\overline{z}}^s{\displaystyle \sum _{i=0}^{\infty }}{c}_i{z}^{i+t}\right)\left|\right|}^2\hfill \\ ={\left|a\right|}^2{\displaystyle \sum _{i=0}^{\infty }}\frac{m+i-n+1}{{(m+i+1)}^2}|c_i{|}^2+{\left|b\right|}^2{\displaystyle \sum _{i=t-s}^{\infty }}\frac{s+i-t+1}{{(s+i+1)}^2}|c_i{|}^2\hfill \\ \hspace{1em}-{\left|a\right|}^2{\displaystyle \sum _{i=m-n}^{\infty }}\frac{n+i-m+1}{{(n+i+1)}^2}|c_i{|}^2-{\left|b\right|}^2{\displaystyle \sum _{i=0}^{\infty }}\frac{t+i-s+1}{{(t+i+1)}^2}{\left|c_i\right|}^2=0\hfill \end{array}$$

(1)

for any $c_i\in \mathbb{C}(i=0,1,2,\dots )$.

Case (1) If $m=n,t=s$, then the equality (1) holds, and so $T_{\phi }$ is normal.

Case (2) If $m>n,t>s$, set $c_0=1$ and $c_i=0$ for $i\ge 1$, then

$$\frac{m-n+1}{{(m+1)}^2}{\left|a\right|}^2=\frac{t-s+1}{{(t+1)}^2}{\left|b\right|}^2.$$

(2)

(i) If $m=n+1,t=s+1$, then we have

$$\frac{i+2}{{(m+i+1)}^2}{\left|a\right|}^2+\frac{i}{{(t+i)}^2}{\left|b\right|}^2=\frac{i}{{(m+i)}^2}{\left|a\right|}^2+\frac{i+2}{{(t+i+1)}^2}{\left|b\right|}^2$$

for all $i\ge 1$. From the equality (2), we have

$$\frac{i+2}{{(m+i+1)}^2}\frac{{(m+1)}^2}{{(t+1)}^2}+\frac{i}{{(t+i)}^2}=\frac{i}{{(m+i)}^2}\frac{{(m+1)}^2}{{(t+1)}^2}+\frac{i+2}{{(t+i+1)}^2}$$

for all $i\ge 1$. Hence, for $i=1$,

$$\frac{3}{{(m+2)}^2}\frac{{(m+1)}^2}{{(t+1)}^2}+\frac{1}{{(t+1)}^2}=\frac{1}{{(m+1)}^2}\frac{{(m+1)}^2}{{(t+1)}^2}+\frac{3}{{(t+2)}^2}$$

or equivalently,

$$\frac{{(m+1)}^2}{{(t+1)}^2}\left\{\frac{3}{{(m+2)}^2}-\frac{1}{{(m+1)}^2}\right\}=\frac{3}{{(t+2)}^2}-\frac{1}{{(t+1)}^2}.$$

By a direct calculation, we have

$$m=t\mathrm{or}m=-\frac{3t+4}{2t+3}.$$

Since m and t are nonnegative, we have $m=t$ and so $n=s$.

(ii) Let $m>n+1,t=s+1$. Suppose that $T_{\phi }$ is normal. For a sufficiently large k,

$$\begin{array}{ccc}\hfill T_{\phi }{T}_{\overline{\phi }}{z}^k& =& T_{\phi }P(\overline{a}{\overline{z}}^m{z}^n{z}^k+\overline{b}{\overline{z}}^s{z}^{s+1}{z}^k)\hfill \\ & =& T_{\phi }\left(\overline{a}\frac{n+k-m+1}{n+k+1}{z}^{n+k-m}+\overline{b}\frac{k+2}{s+k+2}{z}^{k+1}\right)\hfill \\ & =& {\left|a\right|}^2\frac{(n+k-m+1)(k+1)}{{(n+k+1)}^2}{z}^k+\overline{a}b\frac{(n+k-m+1)(n+k-m)}{(n+k+1)(n+k-m+s+1)}{z}^{n+k-m-1}\hfill \\ & & +a\overline{b}\frac{(k+2)(m+k-n+2)}{(s+k+2)(m+k+2)}{z}^{m+k-n+1}+{\left|b\right|}^2\frac{(k+1)(k+2)}{{(s+k+2)}^2}{z}^k,\hfill \end{array}$$

and

$$\begin{array}{ccc}\hfill T_{\overline{\phi }}{T}_{\phi }{z}^k& =& T_{\overline{\phi }}P(a{z}^m{\overline{z}}^n{z}^k+b{z}^s{\overline{z}}^{s+1}{z}^k)\hfill \\ & =& T_{\overline{\phi }}\left(a\frac{m+k-n+1}{m+k+1}{z}^{m+k-n}+b\frac{k}{s+k+1}{z}^{k-1}\right)\hfill \\ & =& {\left|a\right|}^2\frac{(m+k-n+1)(k+1)}{{(m+k+1)}^2}{z}^k+\overline{a}b\frac{k(n+k-m)}{(s+k+1)(n+k)}{z}^{n+k-m-1}\hfill \\ & & +a\overline{b}\frac{(m+k-n+1)(m+k-n+2)}{(m+k+1)(m+k-n+s+2)}{z}^{m+k-n+1}+{\left|b\right|}^2\frac{k(k+1)}{{(s+k+1)}^2}{z}^k.\hfill \end{array}$$

Since $T_{\phi }$ is normal, we have that

$$\frac{(k+2)(m+k-n+2)}{(s+k+2)(m+k+2)}=\frac{(m+k-n+1)(m+k-n+2)}{(m+k+1)(m+k-n+s+2)}$$

for all sufficiently large k. By a direct calculation,

$$\begin{array}{c}(m^2-mn+ms+7m-3n+3s+8)k+2(m+1)(m-n+s+2)\hfill \\ =(m^2-mn+2ms-ns+3s+7m-4n+8)k+(m-n+1)(s+2)(m+2).\hfill \end{array}$$

Since k is arbitrary,

$$m^2-mn+ms+7m-3n+3s+8=m^2-mn+2ms-ns+3s+7m-4n+8$$

and

$$2(m+1)(m-n+s+2)=(m-n+1)(s+2)(m+2),$$

and so

$$ms-ns-n=0\mathrm{and}{m}^{2}s-mns-2n-2ns+ms=0.$$

By the first equality, $s=\frac{n}{m-n}$, and so $mn(m-n-1)=0$. Therefore, $m=0$ or $n=0$ or $m=n+1$, a contradiction. Hence $T_{\phi }$ is not normal.

(iii) If $m>n+1,t>s+1$, set $c_1=1$ and $c_i=0$ for $i\ne 1$, then by a similar argument as in (i), $m=t$.

By (i)–(iii), $m=t$ and so $T_{\phi }$ is normal if and only if

$$\begin{array}{c}{\left|a\right|}^2{\displaystyle \sum _{i=0}^{\infty }}\frac{m+i-n+1}{{(m+i+1)}^2}|c_i{|}^2+{\left|b\right|}^2{\displaystyle \sum _{i=m-s}^{\infty }}\frac{s+i-m+1}{{(s+i+1)}^2}|c_i{|}^2\hfill \\ -{\left|a\right|}^2{\displaystyle \sum _{i=m-n}^{\infty }}\frac{n+i-m+1}{{(n+i+1)}^2}|c_i{|}^2-{\left|b\right|}^2{\displaystyle \sum _{i=0}^{\infty }}\frac{m+i-s+1}{{(m+i+1)}^2}{|c_i|}^2=0\hfill \end{array}$$

(3)

for any $c_i\in \mathbb{C}(i=0,1,2,\dots )$. If $n\ne s$, let $n>s$ and from (3) with $c_1=1$ and $c_i=0$ with $i\ne 1$. If $m-n=1$, then

$${\left|a\right|}^2\frac{m-n+2}{{(m+2)}^2}-{\left|a\right|}^2\frac{n-m+2}{{(n+2)}^2}-{\left|b\right|}^2\frac{m-s+2}{{(m+2)}^2}=0,$$

or equivalently,

$${\left|a\right|}^2\frac{3}{{(m+2)}^2}-{\left|a\right|}^2\frac{1}{{(m+1)}^2}-{\left|b\right|}^2\frac{m-s+2}{{(m+2)}^2}=0.$$

By direct calculations with (2), we have

$$s=\frac{(m+1)(4m+5)}{2m+3}=2m+1+\frac{m+2}{2m+3}.$$

Therefore s is not a nonnegative integer. If $m-n>1$, then

$${\left|a\right|}^2\frac{m-n+2}{{(m+2)}^2}-{\left|b\right|}^2\frac{m-s+2}{{(m+2)}^2}=0.$$

By (2), we have $n=s$, a contradiction. Therefore $n=s$ and so

$$\begin{array}{c}{\left|a\right|}^2{\displaystyle \sum _{i=0}^{\infty }}\frac{m+i-n+1}{{(m+i+1)}^2}|c_i{|}^2+{\left|b\right|}^2{\displaystyle \sum _{i=m-n}^{\infty }}\frac{n+i-m+1}{{(n+i+1)}^2}|c_i{|}^2\hfill \\ -{\left|a\right|}^2{\displaystyle \sum _{i=m-n}^{\infty }}\frac{n+i-m+1}{{(n+i+1)}^2}|c_i{|}^2-{\left|b\right|}^2{\displaystyle \sum _{i=0}^{\infty }}\frac{m+i-n+1}{{(m+i+1)}^2}{\left|c_i\right|}^2=0,\hfill \end{array}$$

or equivalently,

$${\left(\right|a|}^2-{\left|b\right|}^2)\left({\displaystyle \sum _{i=0}^{\infty }}\frac{m+i-n+1}{{(m+i+1)}^2}|c_i{|}^2-{\displaystyle \sum _{i=m-n}^{\infty }}\frac{n+i-m+1}{{(n+i+1)}^2}{\left|c_i\right|}^2\right)=0,$$

and hence we have the results. □

Corollary 1.

Consider the polynomial$\phi \left(z\right)=z^m+{\overline{z}}^m$with$m>0$. Then by Theorem 2,$T_{\phi }$is normal.

The matrix representation of the Toeplitz operator $T_{\phi }$ with non-harmonic symbol $\phi$ has the following form:

Lemma 5.

Suppose that$\phi \left(z\right)=a_{m,n}{z}^m{\overline{z}}^n+e^{i\theta }{a}_{n,m}{z}^n{\overline{z}}^m$with$0\le n\le m$and$\theta \in [0,2\pi ).$Then the Toeplitz matrix with symbol$\phi$given by

$$T_{\phi }=\left(\begin{array}{cccccc}0& \cdots & \frac{1}{m+1}{e}^{i\theta }{a}_{n,m}& 0& 0& \cdots \\ ⋮& 0& \cdots & \frac{2}{m+2}{a}_{n.m}& 0& \ddots \\ \frac{m-n+1}{m+1}{a}_{m,n}& ⋮& 0& \ddots & \ddots & \ddots \\ 0& \frac{m-n+2}{m+2}{a}_{m,n}& ⋮& \ddots & \ddots & \ddots \\ 0& 0& \ddots & \ddots & \ddots & \ddots \\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots \end{array}\right)$$

where$\frac{1}{m+1}{e}^{i\theta }{a}_{n,m}$is a$(1,m-n)$-th entry and$\frac{m-n+1}{m+1}{a}_{m,n}$is a$(m-n,1)$-th entry.

We will give a necessary and sufficient condition that the sum of two normal Toeplitz operator with non-harmonic symbol is again a normal Toeplitz operator.

Theorem 3.

For$a,b\in \mathbb{C}$, let$\phi \left(z\right)=a{\phi }_1\left(z\right)+b{\phi }_2\left(z\right)$be of the form

$${\phi }_1\left(z\right)=z^m{\overline{z}}^n+e^{i{\theta }_1}{z}^n{\overline{z}}^m,{\phi }_2\left(z\right)=z^s{\overline{z}}^t+e^{i{\theta }_2}{z}^t{\overline{z}}^s$$

where${\theta }_1,{\theta }_2\in [0,2\pi ),m\ge n\ge 0$and$t\ge s\ge 0$. Then$T_{\phi }$is normal if and only if either$\overline{a}b{e}^{i{\theta }_1}=a\overline{b}{e}^{i{\theta }_2}$or$\phi \left(z\right)=a(1+e^{i{\theta }_1}){\left|z\right|}^{2m}+b(1+e^{i{\theta }_2}){\left|z\right|}^{2s}.$

Proof. 

Let $\phi \left(z\right)=a{\phi }_1\left(z\right)+b{\phi }_2\left(z\right)$ then $T_{\phi }$ is normal if and only if

$$T_{a{\phi }_1}{T}_{\overline{b{\phi }_2}}+T_{b{\phi }_2}{T}_{\overline{a{\phi }_1}}=T_{\overline{a{\phi }_1}}{T}_{b{\phi }_2}+T_{\overline{b{\phi }_2}}{T}_{a{\phi }_1},$$

or equivalently,

$$a\overline{b}{e}^{i{\theta }_2}{T}_{{\phi }_1}{T}_{{\phi }_2}+\overline{a}b{e}^{i{\theta }_1}{T}_{{\phi }_2}{T}_{{\phi }_1}=\overline{a}b{e}^{i{\theta }_1}{T}_{{\phi }_1}{T}_{{\phi }_2}+a\overline{b}{e}^{i{\theta }_2}{T}_{{\phi }_2}{T}_{{\phi }_1},$$

we have

$$(\overline{a}b{e}^{i{\theta }_1}-a\overline{b}{e}^{i{\theta }_2})(T_{{\phi }_1}{T}_{{\phi }_2}-T_{{\phi }_2}{T}_{{\phi }_1})=0.$$

Therefore $T_{{\phi }_1}{T}_{{\phi }_2}=T_{{\phi }_2}{T}_{{\phi }_1}$ if and only if

$$\begin{array}{c}\frac{m-n+1}{m+1}\frac{s+m-n-t+1}{s+m-n+1}{z}^{s+m-n-t}+e^{i{\theta }_1}\frac{m-n+1}{m+1}\frac{t+m-n-s+1}{t+m-n+1}{z}^{m-n+t-s}\hfill \\ =\frac{s-t+1}{s+1}\frac{s+m-t-n+1}{s+m-t+1}{z}^{s+m-n-t}+e^{i{\theta }_2}\frac{s-t+1}{s+1}\frac{s+n-t-m+1}{s+n-t+1}{z}^{s+n-m-t}.\hfill \end{array}$$

If $m-n=s-t$ then

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

or equivalently,

$$(m+1)(s+m-n+1)=(s+1)(s+m-t+1).$$

By a direct calculation, $m=n$ and $s=t$, and so

$$\phi \left(z\right)=a(1+e^{i{\theta }_1}){\left|z\right|}^{2m}+b(1+e^{i{\theta }_2}){\left|z\right|}^{2s}.$$

This completes the proof. □

Corollary 2.

Let$\phi \left(z\right)=\overline{a}{\overline{z}}^m+\overline{b}{\overline{z}}^N+c{z}^m+d{z}^N.$Then$T_{\phi }$is normal on$A^2\left(\mathbb{D}\right)$if and only if$\left|b\right|=\left|d\right|$and$\overline{c}d=\overline{a}b$.

Example 1.

Let $\phi \left(z\right)=a(z{\overline{z}}^2+e^{i{\theta }_1}{z}^2\overline{z})+b(z{\overline{z}}^3+e^{i{\theta }_2}{z}^3\overline{z})$. It follows from Theorem 3, $T_{\phi }$ is normal if and only if $\phi \left(z\right)$ is of the form

$$\phi \left(z\right)=1(z{\overline{z}}^2+\overline{z}{z}^2)+(1+i)(z{\overline{z}}^3+i\overline{z}{z}^3).$$

We thus have:

Corollary 3.

Let$\phi \left(z\right)=a(z^m{\overline{z}}^n+e^{i\theta }{z}^n{\overline{z}}^m)$with$m>n$and nonzero$a\in \mathbb{C}$. Then$T_{{\phi }^2}$is normal if and only if$\theta =\frac{k}{2}\pi (k=Z)$.

Proof. 

Let $\phi \left(z\right)=a(z^m{\overline{z}}^n+e^{i\theta }{z}^n{\overline{z}}^m)$ then

$${\phi }^2\left(z\right)=a^2(z^{2m}{\overline{z}}^{2n}+e^{2i\theta }{\overline{z}}^{2m}{z}^{2n})+2a^2{e}^{i\theta }{\left|z\right|}^{2(m+n)}.$$

Put

$${\phi }_1\left(z\right)=a^2(z^{2m}{\overline{z}}^{2n}+e^{2i\theta }{\overline{z}}^{2m}{z}^{2n}),$$

$${\phi }_2\left(z\right)=2a^2{e}^{i\theta }{\left|z\right|}^{2(m+n)}=a^2{e}^{i\theta }(z^{m+n}{\overline{z}}^{m+n}+z^{m+n}{\overline{z}}^{m+n}).$$

Then by Theorem 3, we have $e^{4i\theta }=1$. □

As some applications of Theorems 2 and 3, we get the following result.

Corollary 4.

Let

$$\phi \left(z\right)={\displaystyle \sum _{i=1}^n}{a}_i{\phi }_i\left(z\right)$$

where${\phi }_i\left(z\right)=z^{m_i}{\overline{z}}^{n_i}+e^{i{\theta }_i}{z}^{n_i}{\overline{z}}^{m_i}$with nonzero$m_i$and$n_i$, and${\theta }_i\in \mathbb{R}$. If$a_i\overline{a_j}{e}^{i{\theta }_j}-\overline{a_i}{a}_j{e}^{i{\theta }_i}$for all$i,j$, then$T_{\phi }$is normal.

Proof. 

Let $\phi \left(z\right)={\sum }_{i=1}^n{a}_i{\phi }_i\left(z\right)$ where ${\phi }_i\left(z\right)=z^{m_i}{\overline{z}}^{n_i}+e^{i{\theta }_i}{z}^{n_i}{\overline{z}}^{m_i}$ with nonzero $m_i$ and $n_i$, and ${\theta }_i\in \mathbb{R}$. Then by the similar arguments as in Theorem 3, $T_{\phi }$ is normal if and only if

$${\displaystyle \sum _{i,j=1,i\le j}^n}(a_i\overline{a_j}{e}^{i{\theta }_j}-\overline{a_i}{a}_j{e}^{i{\theta }_i})(T_{{\phi }_i}{T}_{{\phi }_j}-T_{{\phi }_j}{T}_{{\phi }_i})=0.$$

(4)

For any $i,j$, if $a_i\overline{a_j}{e}^{i{\theta }_j}-\overline{a_i}{a}_j{e}^{i{\theta }_i}$, then the relation (4) hold and so $T_{\phi }$ is normal. This completes the proof. □