辽宁师范大学数学学院，辽宁 大连

收稿日期：2023年8月13日；录用日期：2023年9月14日；发布日期：2023年9月22日

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。

关键词

加权Bergman空间，Toeplitz算子，亚正规，调和多项式

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

Yi Zheng, Jilong Yang

School of Mathematics, Liaoning Normal University, Dalian Liaoning

Received: Aug. 13^th, 2023; accepted: Sep. 14^th, 2023; published: Sep. 22^nd, 2023

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.

Keywords:Weighted Bergman Space, Toeplitz Operator, Hyponormal, Harmonic Polynomial

Copyright © 2023 by author(s) and Hans Publishers Inc.

This work is licensed under the Creative Commons Attribution International License (CC BY 4.0).

http://creativecommons.org/licenses/by/4.0/

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。

文章引用

郑 益,杨纪龙. 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[J]. 理论数学, 2023, 13(09): 2683-2689. https://doi.org/10.12677/PM.2023.139275

参考文献