## Essential and Weyl Spectra of 2×2 Bounded Block Operator Matrices

Li Lin,1,2, Alatancang ,3

 基金资助: 国家自然科学基金.  11761029国家自然科学基金.  11861048内蒙古自然科学基金.  2016MS0105河套学院自然科学基金.  HYZY201702

 Fund supported: the NSFC.  11761029the NSFC.  11861048the Natural Science Foundation of Inner Mongolia.  2016MS0105the Natural Science Foundation of Hetao College.  HYZY201702

Abstract

This paper is concerned with the necessary and sufficient conditions that a class of bounded 2×2 block operator matrices are Fredholm operators or Weyl operators. Some necessary and sufficient conditions are given under which the essential spectrum and the Weyl spectrum of the block operator matrix coincide with the essential spectrum and the Weyl spectrum of its entries.

Keywords： Essential spectrum ; Weyl spectrum ; Block operator matrices

Li Lin, Alatancang . Essential and Weyl Spectra of 2×2 Bounded Block Operator Matrices. Acta Mathematica Scientia[J], 2019, 39(3): 431-440 doi:

## 1 引言

(ⅰ)若$T$的值域是闭的且$\alpha(T) < \infty$,则称$T$为左Fredholm算子;

(ⅱ)若$T$$\beta(T) < \infty,则称T为右Fredholm算子; (ⅲ)若T既是左Fredholm算子又是右Fredholm算子,则称T是Fredholm算子. 定义2.2 T\in B(X), (ⅰ)如果T是左Fredholm算子且ind(T)\leq 0,则称T为左Weyl算子; (ⅱ)如果T是右Fredholm算子且ind(T)\geq 0,则称T为右Weyl算子; (ⅲ)如果T既是左Weyl算子又是右Weyl算子,则称T为Weyl算子. 定义2.3 T\in B(X),则T的左本质谱、右本质谱、本质谱和Weyl谱定义为 定义2.4 设A\in B(X, Y), X=N(A)\oplus X_{0}, Y=R(A)\oplus Y_{0},其中X_{0}$$Y_{0}$分别是$X$$Y的闭子空间.定义映射\widetilde{A}:X_{0}\times Y_{0}\rightarrow Y $$\widetilde{A}(x_{0}, y_{0})=Ax_{0}+y_{0},$$ \widetilde{A}$$A$生成的双射.

(ⅰ) ${\cal A}-\lambda$是左Fredholm算子当且仅当$C-(D-\lambda)B^{-1}(A-\lambda)$是左Fredholm算子;

(ⅱ) ${\cal A}-\lambda$是右Fredholm算子当且仅当$C-(D-\lambda)B^{-1}(A-\lambda)$是右Fredholm算子.

因为$B$有有界逆算子,所以我们可以得到

$$$\label{eq:0} {\cal A}-\lambda=\left( \begin{array}{cc} I~~& 0 \\ (D-\lambda)B^{-1} ~~& I \\ \end{array} \right) \left( \begin{array}{cc} 0~~&B \\ T_{1}(\lambda) ~ & 0 \\ \end{array} \right) \left( \begin{array}{cc} I~~& 0 \\ B^{-1}(A-\lambda) ~~& I \\ \end{array} \right),$$$

(ⅰ) ${\cal A}-\lambda$是左Fredholm算子当且仅当$B-(A-\lambda)C^{-1}(D-\lambda)$是左Fredholm算子;

(ⅱ) ${\cal A}-\lambda$是右Fredholm算子当且仅当$B-(A-\lambda)C^{-1}(D-\lambda)$是右Fredholm算子.

(ⅰ) ${\cal A}-\lambda$是Fredholm算子;

(ⅱ) $ind({\cal A}-\lambda)=ind C$;

(ⅲ)若$C$是Weyl算子,则${\cal A}-\lambda$是Weyl算子.

对$\lambda\in {\Bbb C}$,根据定理$3.1$${\cal A}-\lambda是Fredholm算子当且仅当C-(D-\lambda)B^{-1}(A-\lambda)是Fredholm算子.因为 $$\|(D-\lambda)B^{-1}(A-\lambda)\|\leq\|\widetilde{C}^{-1}\|^{-1},$$ 所以根据引理2.2$$C-(D-\lambda)B^{-1}(A-\lambda)$是Fredholm算子,并且

$$$\label{eq:t2}ind\left(C-(D-\lambda)B^{-1}(A-\lambda)\right)=ind(C).$$$

$$$\label{eq:t1} {\cal A}-\lambda=\left( \begin{array}{cc} I~~& 0 \\ (D-\lambda)B^{-1} ~~& I \\ \end{array} \right) \left( \begin{array}{cc} 0~~&B \\ T_{1}(\lambda) ~ & 0 \\ \end{array} \right) \left( \begin{array}{cc} I~~& 0 \\ B^{-1}(A-\lambda) ~~& I \\ \end{array} \right),$$$

$$$ind\left({\cal A}-\lambda\right)= ind\left(C-(D-\lambda)B^{-1}(A-\lambda)\right).$$$

(ⅲ)因为$C$是Weyl算子,所以$B$$C都是Fredholm算子,并且 $$ind(B)=ind(C)=0.$$ 由(ⅰ)和(ⅱ)知{\cal A}-\lambda是Fredholm算子,并且 $$ind({\cal A}-\lambda)=0.$$ 因此(ⅲ)成立.结论证毕. 推论3.2 设{\cal A}=\left(\begin{array}{cc} A~~&B \\ C~~&D \\ \end{array} \right)$${\cal H}_{1}\times {\cal H}_{2}$上的有界线性算子, $B$有有界逆算子,则

(ⅰ) ${\cal A}-\lambda$是左Weyl算子当且仅当$C-(D-\lambda)B^{-1}(A-\lambda)$是左Weyl算子;

(ⅱ) ${\cal A}-\lambda$是右Weyl算子当且仅当$C-(D-\lambda)B^{-1}(A-\lambda)$是右Weyl算子.

令$T_{1}(\lambda)=C-(D-\lambda)B^{-1}(A-\lambda)$.

$$$ind({\cal A}-\lambda)=ind\left( \begin{array}{cc} 0~~&B \\ T_{1}(\lambda) ~ & 0 \\ \end{array} \right)=ind(T_{1}(\lambda)).$$$

$$$\label{eq:01}\sigma_{e}({\cal A})=\{\lambda\in{\Bbb C} : C-(D-\lambda)B^{-1}(A-\lambda) \hbox{不是Fredholm算子}\};$$$

$$$\label{eq:02}\sigma_{w}({\cal A})=\{\lambda\in{\Bbb C} : C-(D-\lambda)B^{-1}(A-\lambda) \hbox{不是Weyl算子}\}.$$$

因为$B$有有界逆算子,所以我们可以得到

$$$\label{eq:3} {\cal A}-\lambda=\left( \begin{array}{cc} I~~& 0 \\ (D-\lambda)B^{-1} ~~& I \\ \end{array} \right) \left( \begin{array}{cc} 0~~&B \\ T_{1}(\lambda) ~ & 0 \\ \end{array} \right) \left( \begin{array}{cc} I~~& 0 \\ B^{-1}(A-\lambda) ~~& I \\ \end{array} \right),$$$

$$$\alpha({\cal A}-\lambda)=\alpha(\left( \begin{array}{cc} 0~~&B \\ T_{1}(\lambda) ~ & 0 \\ \end{array} \right))=\alpha\left(T_{1}(\lambda)\right),$$$

$$$\beta({\cal A}-\lambda)=\beta(\left( \begin{array}{cc} 0~~&B \\ T_{1}(\lambda) ~ & 0 \\ \end{array} \right))=\beta(T_{1}(\lambda)).$$$

若$\lambda\notin \sigma_{e}(A)\cup\sigma_{e}(D)$,则$(D-\lambda)$$(A-\lambda)都是Fredholm算子.因为B有有界逆算子,所以B是Fredholm算子.根据引理2.1知, (D-\lambda)B^{-1}(A-\lambda)是Fredholm算子.因为C是紧算子,所以根据引理2.2知T_{1}(\lambda)是Fredholm算子.由定理3.2知\lambda\notin \sigma_{e}({\cal A}). 根据引理2.3,我们得到 $$ind({\cal A}-\lambda)=ind(\left( \begin{array}{cc} 0~~&B \\ T_{1}(\lambda) ~ & 0 \\ \end{array} \right))=ind(T_{1}(\lambda))=ind(A-\lambda)+ind(D-\lambda).$$ \lambda\notin \sigma_{w}(A)\cup\sigma_{w}(D),则\lambda\notin \sigma_{e}({\cal A})$$ind(A-\lambda)=ind(D-\lambda)=0.$因此$\lambda\notin \sigma_{w}({\cal A}).$结论证毕.

(ⅰ) ${\cal A}-\lambda$是左Fredholm算子当且仅当$A-\lambda-B(D-\lambda)^{-1}C$是左Fredholm算子;

(ⅱ) ${\cal A}-\lambda$是右Fredholm算子当且仅当$A-\lambda-B(D-\lambda)^{-1}C$是右Fredholm算子.

对任意$\lambda\in\rho(D)$,我们有

$$$\label{eq:4.01} {\cal A}-\lambda =\left( \begin{array}{cc} I~~ & B(D-\lambda)^{-1} \\ 0 ~~& I \\ \end{array} \right) \left( \begin{array}{cc} S_{1}(\lambda) ~~& 0 \\ 0 ~~& D-\lambda \\ \end{array} \right) \left( \begin{array}{cc} I~~& 0 \\ (D-\lambda)^{-1}C~~& I \\ \end{array} \right),$$$

(ⅰ)若$A-\lambda$是Fredholm算子, $\widetilde{A-\lambda}$$A-\lambda生成的双射, \lambda\in\rho(D),且\|B(D-\lambda)^{-1}C\|\leq\|(\widetilde{A-\lambda})^{-1}\|^{-1},则{\cal A}-\lambda是Fredholm算子; (ⅱ)若\lambda\in(\rho(A)\cap\rho(D)),且\|B(D-\lambda)^{-1}C\|\leq\|(\widetilde{A-\lambda})^{-1}\|^{-1},则{\cal A}-\lambda是Fredholm算子. (ⅰ)对\lambda\in\rho(D),根据定理4.1知{\cal A}-\lambda是Fredholm算子当且仅当A-\lambda-B(D-\lambda)^{-1}C是Fredholm算子.因为A-\lambda是Fredholm算子,并且\|B(D-\lambda)^{-1}C\|\leq\|(\widetilde{A-\lambda})^{-1}\|^{-1},所以由引理2.2知A-\lambda-B(D-\lambda)^{-1}C是Fredholm算子.因此{\cal A}-\lambda是Fredholm算子. (ⅱ)对\lambda\in(\rho(A)\cap\rho(D)),则A-\lambda是双射.因此 $$(\widetilde{A-\lambda})(x_{0}, y_{0})=Ax_{0},$$ $$\|(\widetilde{A-\lambda})^{-1}\|=\|(A-\lambda)^{-1}\|.$$ 从而 $$\|B(D-\lambda)^{-1}C\|\leq\|(A-\lambda)^{-1}\|^{-1} =\|(\widetilde{A-\lambda})^{-1}\|^{-1}.$$ 由(ⅰ)知{\cal A}-\lambda是Fredholm算子.结论证毕. 注4.2 设{\cal A}=\left(\begin{array}{cc} A~~&B \\ C~~&D \\ \end{array} \right)$${\cal H}_{1}\times{\cal H}_{2}$上的有界线性算子.

(ⅰ)若$\lambda\in\rho(A)$, $D-\lambda$是Fredholm算子, $\widetilde{D-\lambda}$$D-\lambda生成的双射,且\|C(A-\lambda)^{-1}B\|\leq\|(\widetilde{D-\lambda})^{-1}\|^{-1},则{\cal A}-\lambda是Fredholm算子; (ⅱ)若\lambda\in(\rho(A)\cap\rho(D)),且\|C(A-\lambda)^{-1}B\|\leq\|(\widetilde{D-\lambda})^{-1}\|^{-1},则{\cal A}-\lambda是Fredholm算子. 推论4.2 设{\cal A}=\left(\begin{array}{cc} A~~&B \\ C~~&D \\ \end{array} \right)$${\cal H}_{1}\times{\cal H}_{2}$上的有界线性算子,则对任意$\lambda\in \rho(D)$,

(ⅰ) ${\cal A}-\lambda$是左Weyl算子当且仅当$A-\lambda-B(D-\lambda)^{-1}C$是左Weyl算子;

(ⅱ) ${\cal A}-\lambda$是右Weyl算子当且仅当$A-\lambda-B(D-\lambda)^{-1}C$是右Weyl算子.

对任意$\lambda\in\rho(D)$,有

$$$\label{eq:4.02} {\cal A}-\lambda =\left( \begin{array}{cc} I~~ & B(D-\lambda)^{-1} \\ 0~~ & I \\ \end{array} \right) \left( \begin{array}{cc} S_{1}(\lambda) ~ & 0 \\ 0~~ & D-\lambda \\ \end{array} \right) \left( \begin{array}{cc} I~~& 0 \\ (D-\lambda)^{-1}C~~& I \\ \end{array} \right),$$$

$$$ind({\cal A}-\lambda)=ind \left( \begin{array}{cc} S_{1}(\lambda) ~~& 0 \\ 0 ~~& D-\lambda \\ \end{array} \right)=ind(S_{1}(\lambda))+ind(D-\lambda)).$$$

$$$ind({\cal A}-\lambda)=ind(S_{1}(\lambda)).$$$

(ⅰ) $\rho(D)\cap\sigma_{e}({\cal A})=\rho(D)\cap\{\lambda\in{\Bbb C} : A-\lambda-B(D-\lambda)^{-1}C$不是Fredholm算子};

(ⅱ) $\rho(D)\cap\sigma_{w}({\cal A})=\rho(D)\cap\{\lambda\in{\Bbb C} : A-\lambda-B(D-\lambda)^{-1}C$不是Weyl算子}.

对$\lambda\in(\rho(D)\cap\sigma_{e}({\cal A}))$,得到

$$$\label{eq:4.4} {\cal A}-\lambda =\left( \begin{array}{cc} I ~~& B(D-\lambda)^{-1} \\ 0 ~~& I \\ \end{array} \right) \left( \begin{array}{cc} S_{1}(\lambda) ~~& 0 \\ 0 ~~& D-\lambda \\ \end{array} \right) \left( \begin{array}{cc} I~~& 0 \\ (D-\lambda)^{-1}C~~& I \\ \end{array} \right),$$$

$$$ind({\cal A}-\lambda)=ind(S_{1}(\lambda)).$$$

(ⅰ) $\rho(A)\cap\sigma_{e}({\cal A})=\rho(A)\cap\{\lambda\in{\Bbb C} : D-\lambda-C(A-\lambda)^{-1}B$不是Fredholm算子};

(ⅱ) $\rho(A)\cap\sigma_{w}({\cal A})=\rho(A)\cap\{\lambda\in{\Bbb C} : D-\lambda-C(A-\lambda)^{-1}B$不是Weyl算子}.

$$$\lim\alpha_{n}=0,$$$

$$$\{\lambda\in{\Bbb C} : D-\lambda-C(A-\lambda)^{-1}B不是{\rm{Fredholm}}算子\}=\{1\}.$$$

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