加权Bergman空间上以调和多项式为符号函数的Toeplitz算子的亚正规性

Hyponormality of the Toeplitz Operators with the Harmonic Polynomial Functions on Weighted Bergman Spaces

摘要: 本文刻画了在复平面上开单位圆盘中一般加权Bergman空间上以调和多项式为符号的Toeplitz算子的亚正规性。

Abstract: In this paper, we describe the hyponormality of the Toeplitz operators with harmonic polynomialsas symbols on the weighted Bergman space in the open unit disk on the complex plane.

1. 引言

Toeplitz算子的研究一直是算子理论研究的重点内容之一。Toeplitz算子理论将Toeplitz矩阵和函数空间建立纽带，使能够对无限维矩阵进行变换运算，为一般算子的研究提供模板和技术方法。Toeplitz算子也广泛应用于数值计算、信号检测与处理、和量子物理学等学科中。

设H表示无穷维复可分Hilbert空间， $B\left(H\right)$ 表示其上所有有界线性算子构成的Banach代数。 $T\in B\left(H\right)$ 称为正规算子，若 $T^{\ast }T=T{T}^{\ast }$ ； $T\in B\left(H\right)$ 称为亚正规算子若 $T^{\ast }T-T{T}^{\ast }\ge 0$ 。亚正规算子是正规算子，正规算子作为亚正规算子的一个特殊情况。由于正规算子理论完备性，非正规算子的研究更能激发学者们的兴趣。亚正规算子是重要的非正规算子。此外，亚正规算子的研究与量子力学紧密联系，如海森堡对易关系、波算子、散射矩阵和扰动等。Bergman空间是指区域上平方可积的解析空间，Bergman空间是重要的解析函数空间理论，与算子理论一直以来的公开问题——不变子空间问题相关联。在20世纪80年代，与Bergman空间相关的算子理论研究蓬勃发展，这一时期的成果斐然，体现在1990年出版的《函数空间中的算子理论》一书中。在20世纪90年代，学者们对Bergman空间的研究取得了函数论和算子论两方面的突破。随着研究的进步，人们不再满足把算子理论局限于经典Bergman空间，进一步将算子理论上升到加权Bergman空间。诸多学者在加权Bergman空间上的研究，尽管面临着许多挑战，但是收获了丰富的理论成果，极大地推动了算子理论的发展。本文主要研究复平面上开单位圆盘上一般加权Bergman空间上Toeplitz算子的亚正规性，Bergman空间和加权Bergman空间上的关于亚正规的Toeplitz算子的研究，见参考文献 [1] [2] [3] [4] [5] 。

记 $\mathbb{D}$ 为复平面上的开单位圆盘。将 $\mathbb{D}$ 上的测度v定义为 $\text{d}v\left(r{\text{e}}^{i\theta }\right)=\text{d}\eta \left(r\right)\times \frac{\text{d}\theta }{2\pi }$ ，这里 $\text{d}\eta$ 代表 $\left[0,1\right)$ 上的概率测度。Bergman空间是 $\mathbb{D}$ 上关于 $\text{d}v$ 测度的平方可积空间 $L^2\left(\mathbb{D},\text{d}v\right)$ 的解析闭子空间，记为 $A_v^2\left(\mathbb{D}\right)$ 。序列 ${\left\{{\tau }_t\right\}}_{t=0}^{\infty }$ 定义为

${\tau }_t:={\displaystyle {\int }_{\mathbb{D}}{\left|z\right|}^t\text{d}v\left(z\right)}={\displaystyle {\int }_{\left[0,1\right)}{r}^t\text{d}\eta }\left(r\right).$

这样Bergman空间 $A_v^2\left(\mathbb{D}\right)$ 可表示成

$A_v^2\left(\mathbb{D}\right)=\left\{f\left(z\right)={\displaystyle \underset{n=0}{\overset{\infty }{\sum }}{a}_n}{z}^n:{\displaystyle \underset{n=0}{\overset{\infty }{\sum }}{\left|a_n\right|}^2{\tau }_{2n}}<\infty \right\},$

其上内积有对应的表示

$\langle {\displaystyle \underset{n=0}{\overset{\infty }{\sum }}{a}_n{z}^n},{\displaystyle \underset{n=0}{\overset{\infty }{\sum }}{b}_n}{z}^n\rangle ={\displaystyle \underset{n=0}{\overset{\infty }{\sum }}{a}_n\overline{b_n}{\tau }_{2n}}.$

对于自然数n，

$e_n\left(z\right)=\frac{z^n}{\sqrt{{\tau }_{2n}}}\left(z\in \mathbb{D}\right)$

是 $A_v^2\left(\mathbb{D}\right)$ 上的正规正交基。进而 $A_v^2\left(\mathbb{D}\right)$ 上再生核为

$K_z^{\left(v\right)}\left(w\right)={\displaystyle \underset{n=0}{\overset{\infty }{\sum }}\overline{e_n\left(z\right)}{e}_n\left(w\right)}={\displaystyle \underset{n=0}{\overset{\infty }{\sum }}\frac{1}{{\tau }_{2n}}{\bar{z}}^n{w}^n}\left(z,w\in \mathbb{D}\right)$

本文总记 $k_i\left(z\right)={\displaystyle \underset{n=0}{\overset{\infty }{\sum }}{c}_{2n+i}}{z}^{2n+i}\in A_v^2\left(\mathbb{D}\right),i=0,1$ 。这样 $A_v^2\left(\mathbb{D}\right)$ 中的每个函数都有形如 $k_0\left(z\right)$ 和 $k_1\left(z\right)$ 这样的分解表达式。

记 $L^{\infty }\left(\mathbb{D},\text{d}v\right)$ 是 $\mathbb{D}$ 上关于测度 $\text{d}v$ 的本性可测函数空间。若 $\phi \in L^{\infty }\left(\mathbb{D},\text{d}v\right)$ ，称 $T_{\phi }$ 为在 $A_v^2\left(\mathbb{D}\right)$ 上以 $\phi$ 为符号的Toeplitz算子， $H_{\phi }$ 为以 $\phi$ 为符号的Hankel算子

$T_{\phi }f=P\left(\phi f\right),H_{\phi }f=\left(I-P\right)\left(\phi f\right),$

其中 $f\in A_v^2\left(\mathbb{D}\right)$ ，P是从 $L^2\left(\mathbb{D},\text{d}v\right)$ 到 $A_v^2\left(\mathbb{D}\right)$ 的正交投影。

2. 一般加权Bergman空间的结果

令 $$ ，Toeplitz算子 $T_{\phi }$ 是亚正规算子，那么就有 $T_{\phi }^{\ast }{T}_{\phi }-T_{\phi }{T}_{\phi }^{\ast }\ge 0$ 。

为了简化本文定理的计算过程，引入下面两个引理。

引理2.1 若 $k,m$ 为自然数，则

$P\left(z^k{\bar{z}}^m\right)\left(w\right)=\{\begin{array}{l}0,k<m,\\ \frac{{\tau }_{2k}}{{\tau }_{2\left(k-m\right)}}{w}^{k-m},k\ge m.\end{array}$ (2.1)

证明：对于自然数 $k,m$ ，

$\begin{array}{c}P\left(z^k{\bar{z}}^m\right)\left(w\right)=\langle z^k{\bar{z}}^m,K_z^{\left(v\right)}\left(w\right)\rangle ={\displaystyle \underset{j=0}{\overset{\infty }{\sum }}\langle z^k{\bar{z}}^m,e_j\left(z\right)\rangle e_j\left(w\right)}\\ ={\displaystyle \underset{j=0}{\overset{\infty }{\sum }}\frac{1}{{\tau }_{2j}}\langle z^k,z^{j+m}\rangle w^j}=\{\begin{array}{l}\frac{{\tau }_{2k}}{{\tau }_{2\left(k-m\right)}}{w}^{k-m},k\ge m,\\ 0,k<m.\end{array}\end{array}$

引理2.2 令 $i\in \left\{0,1\right\}$ ， $m\ge 1$ ， $k<m$ 。记 $\left[\frac{m-i}{2}\right]$ 表示 $\frac{m-i}{2}$ 的整数部分。则

1) $H_{{\bar{z}}^m}{k}_i\left(z\right)=\{\begin{array}{l}{\bar{z}}^m{k}_i\left(z\right)-{\displaystyle \underset{k=\left[\frac{m-i}{2}\right]+1}{\overset{\infty }{\sum }}{c}_{2k+i}\frac{{\tau }_{2\left(2k+i\right)}{z}^{2k+i-m}}{{\tau }_{2\left(2k+i-m\right)}},m>i,}\\ {\bar{z}}^m{k}_i\left(z\right)-{\displaystyle \underset{k=0}{\overset{\infty }{\sum }}{c}_{2k+i}\frac{{\tau }_{2\left(2k+i\right)}{z}^{2k+i-m}}{{\tau }_{2\left(2k+i-m\right)}},m\le i;}\end{array}$ (2.2)

2) ${\Vert H_{{\bar{z}}^m}{k}_i\Vert }^2=\{\begin{array}{l}{\displaystyle \underset{k=0}{\overset{\left[\frac{m-i}{2}\right]}{\sum }}{\left|c_{2k+i}\right|}^2{\tau }_{2\left(2k+i+m\right)}+{\displaystyle \underset{k=\left[\frac{m-i}{2}\right]+1}{\overset{\infty }{\sum }}\left({\tau }_{2\left(2k+i+m\right)}-\frac{{\tau }_{2\left(2k+i\right)}^2}{{\tau }_{2\left(2k+i-m\right)}}\right)}}{\left|c_{2k+i}\right|}^2,m>i,\\ {\displaystyle \underset{k=0}{\overset{\infty }{\sum }}\left({\tau }_{2\left(2k+i+m\right)}-\frac{{\tau }_{2\left(2k+i\right)}^2}{{\tau }_{2\left(2k+i-m\right)}}\right)}{\left|c_{2k+i}\right|}^2,m\le i;\end{array}$ (2.3)

3) $\langle H_{\bar{z}}{k}_0,H_{\bar{z}}{k}_1\rangle =0$ ;

4) $\langle H_{{\bar{z}}^2}{k}_0,H_{{\bar{z}}^2}{k}_1\rangle =0$ .

证明：(1) 当 $i=0,1$ 时，由Hankel算子的概念，

$H_{{\bar{z}}^m}{k}_i\left(z\right)=\left(I-P\right)\left({\bar{z}}^m{k}_i\left(z\right)\right)={\bar{z}}^m{k}_i\left(z\right)-P\left({\bar{z}}^m{k}_i\left(z\right)\right).$

应用 $k_i\left(z\right)$ 的展开式和引理2.1得

$P\left({\bar{z}}^m{k}_i\right)\left(z\right)=\{\begin{array}{l}{\displaystyle \underset{k=\left[\frac{m-i}{2}\right]+1}{\overset{\infty }{\sum }}{c}_{2k+i}\frac{{\tau }_{2\left(2k+i\right)}{z}^{2k+i-m}}{{\tau }_{2\left(2k+i-m\right)}},m>i,}\\ {\displaystyle \underset{k=0}{\overset{\infty }{\sum }}{c}_{2k+i}\frac{{\tau }_{2\left(2k+i\right)}{z}^{2k+i-m}}{{\tau }_{2\left(2k+i-m\right)}},m\le i,}\end{array}$

将 $P\left({\bar{z}}^m{k}_i\right)$ 带入，则结果得证。

(2) 对于 $i=0$ 或 $i=1$ ， ${\Vert H_{{\bar{z}}^m}{k}_i\Vert }^2=\langle H_{{\bar{z}}^m}{k}_i,H_{{\bar{z}}^m}{k}_i\rangle$ 。当 $m\le i$ 时，带入式(2.2)得

$\begin{array}{c}{\Vert H_{{\bar{z}}^m}{k}_i\Vert }^2=\langle {\bar{z}}^m{k}_i\left(z\right)-{\displaystyle \underset{k=0}{\overset{\infty }{\sum }}{c}_{2k+i}\frac{{\tau }_{2\left(2k+i\right)}{z}^{2k+i-m}}{{\tau }_{2\left(2k+i-m\right)}},{\bar{z}}^m{k}_i\left(z\right)-{\displaystyle \underset{k=0}{\overset{\infty }{\sum }}{c}_{2k+i}\frac{{\tau }_{2\left(2k+i\right)}{z}^{2k+i-m}}{{\tau }_{2\left(2k+i-m\right)}}}}\rangle \\ ={\Vert {\bar{z}}^m{k}_i\Vert }^2+{\displaystyle \underset{k=0}{\overset{\infty }{\sum }}{\left|c_{2k+i}\right|}^2\frac{{\tau }_{2\left(2k+i\right)}^2}{{\tau }^2{}_{2\left(2k+i-m\right)}}{\tau }_{2\left(2k+i-m\right)}}-\langle {\bar{z}}^m{k}_i\left(z\right),{\displaystyle \underset{k=\left[\frac{m-i}{2}\right]+1}{\overset{\infty }{\sum }}{c}_{2k+i}\frac{{\tau }_{2\left(2k+i\right)}{z}^{2k+i-m}}{{\tau }_{2\left(2k+i-m\right)}}}\rangle \\ -\langle {\displaystyle \underset{k=\left[\frac{m-i}{2}\right]+1}{\overset{\infty }{\sum }}{c}_{2k+i}\frac{{\tau }_{2\left(2k+i\right)}{z}^{2k+i-m}}{{\tau }_{2\left(2k+i-m\right)}}},{\bar{z}}^m{k}_i\left(z\right)\rangle \\ ={\displaystyle \underset{k=0}{\overset{\infty }{\sum }}{\left|c_{2k+i}\right|}^2{\tau }_{2\left(2k+i-m\right)}-{\displaystyle \underset{k=0}{\overset{\infty }{\sum }}{\left|c_{2k+i}\right|}^2}}\frac{{\tau }_{2\left(2k+i\right)}^2}{{\tau }_{2\left(2k+i-m\right)}}\\ ={\displaystyle \underset{k=0}{\overset{\infty }{\sum }}\left({\tau }_{2\left(2k+i+m\right)}-\frac{{\tau }_{2\left(2k+i\right)}^2}{{\tau }_{2\left(2k+i-m\right)}}\right)}{\left|c_{2k+i}\right|}^2.\end{array}$

相似地，当 $m\le i$ 时

${\Vert H_{{\bar{z}}^m}{k}_i\Vert }^2={\displaystyle \underset{k=0}{\overset{\left[\frac{m-i}{2}\right]}{\sum }}{\left|c_{2k+i}\right|}^2{\tau }_{2\left(2k+i+m\right)}+{\displaystyle \underset{k=\left[\frac{m-i}{2}\right]+1}{\overset{\infty }{\sum }}\left({\tau }_{2\left(2k+i+m\right)}-\frac{{\tau }_{2\left(2k+i\right)}^2}{{\tau }_{2\left(2k+i+m\right)}}\right)}}{\left|c_{2k+i}\right|}^2.$

(3)与(4)由式(2.2)直接可得。

定理2.3 设 $\phi \left(z\right)=f\left(z\right)+\overline{g\left(z\right)}$ ，其中 $f\left(z\right)=a_{1}z+a_2{z}^2$ ， $g\left(z\right)=a_{-1}z+a_{-2}{z}^2$ 且 $a_1{\bar{a}}_2=a_{-1}{\bar{a}}_{-2}$ 。则 $T_{\phi }$ 在 $A_v^2\left(\mathbb{D}\right)$ 上是亚正规算子当且仅当

$\begin{array}{l}\left({\left|a_1\right|}^2-{\left|a_{-1}\right|}^2\right){\tau }_2+\left({\left|a_2\right|}^2-{\left|a_{-2}\right|}^2\right){\tau }_4\ge 0,\\ \left({\left|a_1\right|}^2-{\left|a_{-1}\right|}^2\right)\left({\tau }_4-\frac{{\tau }_2^2}{{\tau }_0}\right)+\left({\left|a_2\right|}^2-{\left|a_{-2}\right|}^2\right){\tau }_6\ge 0,\\ \left({\left|a_1\right|}^2-{\left|a_{-1}\right|}^2\right)\left({\tau }_6-\frac{{\tau }_4^2}{{\tau }_2}\right)+\left({\left|a_2\right|}^2-{\left|a_{-2}\right|}^2\right){\tau }_8\ge 0,\end{array}$

$\begin{array}{l}\left({\left|a_1\right|}^2-{\left|a_{-1}\right|}^2\right)\left({\tau }_{2\left(2k+1\right)}-\frac{{\tau }_{2\left(2k\right)}^2}{{\tau }_{2\left(2k-1\right)}}\right)+\left({\left|a_2\right|}^2-{\left|a_{-2}\right|}^2\right)\left({\tau }_{2\left(2k+2\right)}-\frac{{\tau }_{2\left(2k\right)}^2}{{\tau }_{2\left(2k-2\right)}}\right)\left(k\ge 2\right)\\ \left({\left|a_1\right|}^2-{\left|a_{-1}\right|}^2\right)\left({\tau }_{2\left(2k+2\right)}-\frac{{\tau }_{2\left(2k+1\right)}^2}{{\tau }_{2\left(2k\right)}}\right)+\left({\left|a_2\right|}^2-{\left|a_{-2}\right|}^2\right)\left({\tau }_{2\left(2k+3\right)}-\frac{{\tau }_{2\left(2k+1\right)}^2}{{\tau }_{2\left(2k-1\right)}}\right)\left(k\ge 1\right)\end{array}$ (2.4)

同时成立。

证明：由Toeplitz算子与Hankel算子的基本性质，得 $T_{\phi }^{*}{T}_{\phi }-T_{\phi }{T}_{\phi }^{*}=H_{\bar{f}}^{*}{H}_{\bar{f}}-H_{\bar{g}}^{*}{H}_{\bar{g}}.$

带入 $f\left(z\right)$ 有， $H_{\bar{f}}={\bar{a}}_1{H}_{\bar{z}}+{\bar{a}}_2{H}_{{\bar{z}}^2}$ 。进一步，

$\begin{array}{c}{H}_{\bar{f}}^{*}{H}_{\bar{f}}=\left(a_1{H}_{\bar{z}}^{*}+a_2{H}_{{\bar{z}}^2}\right)\left({\bar{a}}_1{H}_{\bar{z}}+{\bar{a}}_2{H}_{{\bar{z}}^2}\right)\\ ={\left|a_1\right|}^2{H}_{\bar{z}}^{*}{H}_{\bar{z}}+{\left|a_2\right|}^2{H}_{\bar{z}}^{*}{H}_{{\bar{z}}^2}+a_1{\bar{a}}_2{H}_{\bar{z}}^{*}{H}_{{\bar{z}}^2}+{\bar{a}}_1{a}_2{H}_{{\bar{z}}^2}^{*}{H}_{\bar{z}}.\end{array}$ (2.5)

上式用 $g\left(z\right)$ 代替 $f\left(z\right)$ ，结合式(2.6)得到

$\begin{array}{c}{H}_{\bar{f}}^{*}{H}_{\bar{f}}-H_{\bar{g}}^{*}{H}_{\bar{g}}=\left({\left|a_1\right|}^2-{\left|a_{-1}\right|}^2\right)H_{\bar{z}}^{*}{H}_{\bar{z}}+\left({\left|a_2\right|}^2-{\left|a_{-2}\right|}^2\right)H_{{\bar{z}}^2}^{*}{H}_{{\bar{z}}^2}\\ +\left(a_1{\bar{a}}_2-a_{-1}{\bar{a}}_{-2}\right)H_{\bar{z}}^{*}{H}_{{\bar{z}}^2}+\left({\bar{a}}_1{a}_2-{\bar{a}}_{-1}{a}_{-2}\right)H_{{\bar{z}}^2}{H}_{\bar{z}}\\ =\left({\left|a_1\right|}^2-{\left|a_{-1}\right|}^2\right)H_{\bar{z}}^{*}{H}_{\bar{z}}+\left({\left|a_2\right|}^2-{\left|a_{-2}\right|}^2\right)H_{{\bar{z}}^2}^{*}{H}_{{\bar{z}}^2}.\end{array}$ (2.6)

$T_{\phi }$ 在 $A_v^2\left(\mathbb{D}\right)$ 上是亚正规的当且仅当 $\langle H_{\bar{f}}^{*}{H}_{\bar{f}}-H_{\bar{g}}^{*}{H}_{\bar{g}}\left(k_{1k_0+}\right),\left(k_0+k_1\right)\rangle \ge 0$ 。而

$\begin{array}{l}\langle \left(H_{\bar{f}}^{*}{H}_{\bar{f}}-H_{\bar{g}}^{*}{H}_{\bar{g}}\right)\left(k_0+k_1\right),\left(k_0+k_1\right)\rangle \\ =\left({\left|a_1\right|}^2-{\left|a_{-1}\right|}^2\right)\left({\Vert H_{\bar{z}}{k}_0\Vert }^2+{\Vert H_{\bar{z}}{k}_1\Vert }^2\right)+\left({\left|a_2\right|}^2-{\left|a_{-2}\right|}^2\right)\left({\Vert H_{{\bar{z}}^2}{k}_0\Vert }^2+{\Vert H_{{\bar{z}}^2}{k}_1\Vert }^2\right).\end{array}$ (2.7)

应用引理2.2(2)得到

${\Vert H_{\bar{z}}{k}_0\Vert }^2+{\Vert H_{\bar{z}}{k}_1\Vert }^2={\left|c_0\right|}^2{\tau }_2+{\displaystyle \underset{k=1}{\overset{\infty }{\sum }}\left({\tau }_{2\left(2k+1\right)}-\frac{{\tau }_{2\left(2k\right)}^2}{{\tau }_{2\left(2k-1\right)}}\right)}{\left|c_{2k}\right|}^2+{\displaystyle \underset{k=0}{\overset{\infty }{\sum }}\left({\tau }_{2\left(2k+2\right)}-\frac{{\tau }_{2\left(2k+1\right)}^2}{{\tau }_{2\left(2k\right)}}\right)}{\left|c_{2k+1}\right|}^2$ (2.8)

和

$\begin{array}{c}{\Vert H_{{\bar{z}}^2}{k}_0\Vert }^2+{\Vert H_{{\bar{z}}^2}{k}_1\Vert }^2={\left|c_0\right|}^2{\tau }_4+{\left|c_2\right|}^2{\tau }_8+{\displaystyle \underset{k=2}{\overset{\infty }{\sum }}\left({\tau }_{2\left(2k+2\right)}-\frac{{\tau }_{2\left(2k\right)}^2}{{\tau }_{2\left(2k-2\right)}}\right)}{\left|c_{2k}\right|}^2\\ +{\left|c_1\right|}^2{\tau }_6+{\displaystyle \underset{k=1}{\overset{\infty }{\sum }}\left({\tau }_{2\left(2k+3\right)}-\frac{{\tau }_{2\left(2k+1\right)}^2}{{\tau }_{2\left(2k-1\right)}}\right)}\left|c_{2k+1}\right|.\end{array}$ (2.9)

所以

$\begin{array}{l}\langle \left(H_{\bar{f}}^{*}{H}_{\bar{f}}-H_{\bar{g}}^{*}{H}_{\bar{g}}\right)\left(k_0+k_1\right),\left(k_0+k_1\right)\rangle \\ =\left({\left|a_1\right|}^2-{\left|a_{-1}\right|}^2\right)\left\{{\left|c_0\right|}^2{\tau }_2+{\displaystyle \underset{k=1}{\overset{\infty }{\sum }}\left({\tau }_{2\left(2k+1\right)}-\frac{{\tau }_{2\left(2k\right)}^2}{{\tau }_{2\left(2k-1\right)}}\right)}{\left|c_{2k}\right|}^2+{\displaystyle \underset{k=0}{\overset{\infty }{\sum }}\left({\tau }_{2\left(2k+2\right)}-\frac{{\tau }_{2\left(2k+1\right)}^2}{{\tau }_{2\left(2k\right)}}\right)}{\left|c_{2k+1}\right|}^2\right\}\\ +\left({\left|a_2\right|}^2-{\left|a_{-2}\right|}^2\right)\left\{{\left|c_0\right|}^2{\tau }_4+{\left|c_2\right|}^2{\tau }_8+{\displaystyle \underset{k=2}{\overset{\infty }{\sum }}\left({\tau }_{2\left(2k+2\right)}-\frac{{\tau }_{2\left(2k\right)}^2}{{\tau }_{2\left(2k-2\right)}}\right)}{\left|c_{2k}\right|}^2\right\}\\ +\left({\left|a_2\right|}^2-{\left|a_{-2}\right|}^2\right)\left\{{\left|c_1\right|}^2{\tau }_6+{\displaystyle \underset{k=1}{\overset{\infty }{\sum }}\left({\tau }_{2\left(2k+3\right)}-\frac{{\tau }_{2\left(2k+1\right)}^2}{{\tau }_{2\left(2k-1\right)}}\right)}\left|c_{2k+1}\right|\right\}\end{array}$

$\begin{array}{l}=\left\{\left({\left|a_1\right|}^2-{\left|a_{-1}\right|}^2\right){\tau }_2+\left({\left|a_2\right|}^2-{\left|a_{-2}\right|}^2\right){\tau }_4\right\}{\left|c_0\right|}^2+\left\{\left({\left|a_1\right|}^2-{\left|a_{-1}\right|}^2\right)\left({\tau }_4-\frac{{\tau }_2^2}{{\tau }_0}\right)+\left({\left|a_2\right|}^2-{\left|a_{-2}\right|}^2\right){\tau }_6\right\}{\left|c_1\right|}^2\\ +\left\{\left({\left|a_1\right|}^2-{\left|a_{-1}\right|}^2\right)\left({\tau }_6-\frac{{\tau }^2{}_4}{{\tau }_2}\right)+\left({\left|a_2\right|}^2-{\left|a_{-2}\right|}^2\right){\tau }_8\right\}{\left|c_2\right|}^2\\ +{\displaystyle \underset{k=2}{\overset{\infty }{\sum }}\left\{\left({\left|a_1\right|}^2-{\left|a_{-1}\right|}^2\right)\left({\tau }_{2\left(2k+1\right)}-\frac{{\tau }_{2\left(2k\right)}^2}{{\tau }_{2\left(2k-1\right)}}\right)+\left({\left|a_2\right|}^2-{\left|a_{-2}\right|}^2\right)\left({\tau }_{2\left(2k+2\right)}-\frac{{\tau }_{2\left(2k\right)}^2}{{\tau }_{2\left(2k-2\right)}}\right)\right\}}{\left|c_{2k}\right|}^2\\ +{\displaystyle \underset{k=1}{\overset{\infty }{\sum }}\left\{\left({\left|a_1\right|}^2-{\left|a_{-1}\right|}^2\right)\left({\tau }_{2\left(2k+2\right)}-\frac{{\tau }_{2\left(2k+1\right)}^2}{{\tau }_{2(2k)}}\right)+\left({\left|a_2\right|}^2-{\left|a_{-2}\right|}^2\right)\left({\tau }_{2\left(2k+3\right)}-\frac{{\tau }_{2\left(2k+1\right)}^2}{{\tau }_{2\left(2k-1\right)}}\right)\right\}}\left|c_{2k+1}\right|.\end{array}$

分别取 $k_0+k_1$ 为 $e_n\left(z\right)\left(n=0,1,\cdots \right)$ 可得定理，证毕。

若 $\eta \left(r\right)=r^2$ ，则

$\text{d}v\left(r{\text{e}}^{i\theta }\right)=\text{d}A\left(z\right)=\frac{r}{\pi }\text{d}r\text{d}\theta .$

$\phi \left(z\right)$ 同定理2.3， $T_{\phi }$ 为 $A_v^2\left(\mathbb{D}\right)$ 上亚正规算子见参考文献 [1] ；若 $\eta \left(r\right)=-{\left(1-r^2\right)}^{\alpha +1}$ ，则

$\text{d}v\left(r{\text{e}}^{i\theta }\right)=\text{d}{A}_{\alpha }\left(z\right)=\left(\alpha +1\right){\left(1-{\left|z\right|}^2\right)}^{\alpha }\text{d}A\left(z\right)=\left(\alpha +1\right){\left(1-r^2\right)}^{\alpha }\frac{r}{\pi }\text{d}r\text{d}\theta .$

$\phi \left(z\right)$ 同定理2.3， $T_{\phi }$ 为 $A_v^2\left(\mathbb{D}\right)$ 上亚正规算子见参考文献 [2] 。

3. 特殊加权Bergman空间的结果

以下研究由 $\eta \left(r\right)=r^{\beta +1}\left(\beta >-1\right)$ 诱导的 $\mathbb{D}$ 上测度

$\text{d}v\left(r{\text{e}}^{i\theta }\right)=\left(\beta +1\right)r^{\beta }\frac{1}{2\pi }\text{d}r\text{d}\theta ,$

对应的加权Bergman空间上的亚正规Toeplitz算子。此时，

${\tau }_t={\displaystyle {\int }_{\mathbb{D}}{\left|z\right|}^t\text{d}v\left(z\right)}={\displaystyle {\int }_{\left[0,1\right)}{r}^t\text{d}{r}^{\beta +1}}=\frac{\beta +1}{t+\beta +1}.$

定理3.1设 $\phi \left(z\right)=f\left(z\right)+\overline{g\left(z\right)}$ ，其中 $f\left(z\right)=a_{1}z+a_2{z}^2$ ， $g\left(z\right)=a_{-1}z+a_{-2}{z}^2$ 且 $a_1{\bar{a}}_2=a_{-1}{\bar{a}}_{-2}$ 。则 $T_{\phi }$ 在 $A_v^2\left(\mathbb{D}\right)$ 上是亚正规算子当且仅当

1) 若 $\left|a_2\right|>\left|a_{-2}\right|$ ， $\left({\left|a_1\right|}^2-{\left|a_{-1}\right|}^2\right)+\left({\left|a_2\right|}^2-{\left|a_{-2}\right|}^2\right){\delta }_{\beta }\ge 0$ ，

2) 若 $\left|a_2\right|\le \left|a_{-2}\right|$ ， $\left({\left|a_1\right|}^2-{\left|a_{-1}\right|}^2\right)+\left({\left|a_2\right|}^2-{\left|a_{-2}\right|}^2\right){\lambda }_{\beta }\ge 0$ ，

其中

${\delta }_{\beta }=\frac{4\left(\beta +1\right)}{\left(4+\beta +1\right){\left(2+\beta +1\right)}^2},$

${\lambda }_{\beta }=\max \left\{\frac{\left(6+\beta +1\right){\left(4+\beta +1\right)}^2}{4\left(8+\beta +1\right)},4\right\}.$

证明：运用定理2.3，直接计算得

$\frac{{\tau }_4}{{\tau }_2}=\frac{2+\beta +1}{4+\beta +1}<1,$

${\tau }_4-\frac{{\tau }^2{}_2}{{\tau }_0}=\frac{\beta +1}{4+\beta +1}-\frac{{\left(\beta +1\right)}^2}{{\left(2+\beta +1\right)}^2}=\frac{4\left(\beta +1\right)}{\left(4+\beta +1\right){\left(2+\beta +1\right)}^2}<\frac{2+\beta +1}{4+\beta +1},$

${\tau }_6/\left({\tau }_4-\frac{{\tau }_2^2}{{\tau }_0}\right)=\frac{\beta +1}{6+\beta +1}/\frac{4\left(\beta +1\right)}{\left(4+\beta +1\right){\left(2+\beta +1\right)}^2}=\frac{\left(4+\beta +1\right){\left(2+\beta +1\right)}^2}{4\left(6+\beta +1\right)},$

${\tau }_8/\left({\tau }_6-\frac{{\tau }^2{}_4}{{\tau }_2}\right)=\frac{\beta +1}{8+\beta +1}/\frac{4\left(\beta +1\right)}{\left(6+\beta +1\right){\left(4+\beta +1\right)}^2}=\frac{\left(6+\beta +1\right){\left(4+\beta +1\right)}^2}{4\left(8+\beta +1\right)},$

$1<\left({\tau }_{2\left(2k+2\right)}-\frac{{\tau }_{2\left(2k\right)}^2}{{\tau }_{2\left(2k-2\right)}}\right)/\left({\tau }_{2\left(2k+1\right)}-\frac{{\tau }_{2\left(2k\right)}^2}{{\tau }_{2\left(2k-1\right)}}\right)=\frac{4\left(4k+2+\beta +1\right)}{\left(4k+4+\beta +1\right)}\le 4,\left(k\ge 2\right)$

$1<\left({\tau }_{2\left(2k+3\right)}-\frac{{\tau }_{2\left(2k+1\right)}^2}{{\tau }_{2\left(2k-1\right)}}\right)/\left({\tau }_{2\left(2k+2\right)}-\frac{{\tau }_{2\left(2k+1\right)}^2}{{\tau }_{2\left(2k\right)}}\right)=\frac{4\left(4k+4+\beta +1\right)}{\left(4k+6+\beta +1\right)}\le 4,\left(k\ge 1\right)$

定理得证。

本文研究了复平面上开单位圆盘中一般和特殊加权Bergman空间上以调和多项式为符号的Toeplitz算子的亚正规性的充分必要条件。