本文中,设$X$是无限维复Banach空间, $L(X)$表示从$X$到$X$的有界线性算子的全体.对于$T\in L(X)$, $T^*$表示$T$的共轭算子, $\alpha(T)$表示零空间$N(T)$的维数, $\beta(T)$表示值域$R(T)$的亏维. $p(T)$和$q(T)$分别表示$T$的升指数和降指数,即

(若下确界不存在,记$p(T)=\infty$, $q(T)=\infty$).若$T$的升指数和降指数都是有限的,那么它们相等(见文献[1,定理1.19]).称$T$是Fredholm算子(上半Fredholm算子),若$\alpha(T)$和$\beta(T)$都有限($R(T)$闭且$\alpha(T) < \infty$). $T$的Fredholm指标定义为$ind(T)=\alpha(T)-\beta(T)$.指标为0的Fredholm算子称为Weyl算子.若$T$是Fredholm算子且$p(T)=q(T) < \infty$,则称$T$是Browder算子.若$p(T)=q(T) < \infty$,则称$T$为Drazin可逆.若$\lambda I-T$不可逆但Drazin可逆,则称$\lambda\in{\Bbb C}$为$T$的极点,记$\Pi(T)$为$T$的所有极点组成的集合.同时,用$\Pi^0(T)$表示$T$的所有有限秩的极点,即

此外,定义

显然$\Pi^0(T)\subseteq E^0(T)$.定义$\sigma(T)$表示$T$的谱集, $T$的本性谱$\sigma_f(T)$, Weyl谱$\sigma_w(T)$, Browder谱$\sigma_b(T)$分别定义为:

记$\rho_{f}(T)={\Bbb C}\setminus\sigma_{f}(T), \rho_{w}(T)={\Bbb C}\setminus\sigma_{w}(T), \rho_{b}(T)={\Bbb C}\setminus\sigma_{b}(T)$为这些谱子集相对应的预解集.

Dunford在文献[2]和[3]中引入了单值扩张性的概念(简称SVEP),它作为研究算子的谱的有力工具极大地丰富了算子谱结构的经典研究.称$T$在$\lambda_0\in{\Bbb C}$处有SVEP,如果对$\lambda_0$的任何开邻域$U$,满足式子$(\lambda I-T)f(\lambda)=0 (\forall\lambda\in U)$的唯一解析函数$f:U\longrightarrow X$是$U$上的零函数.称$T$有SVEP,若$T$在任意的$\lambda\in{\Bbb C}$处都有SVEP.显然, $T$和$T^*$在$\lambda\in iso\sigma(T)$处都有SVEP.

近年来,许多学者对Weyl型定理从不同角度进行了研究, Coburn在文献[4]中称$T$满足Weyl定理,若$\sigma(T)\setminus\sigma_{w}(T)= E^{0}(T),$ Harte和Lee在文献[5]中称$T$满足Browder定理,若$\sigma(T)\setminus\sigma_{w}(T)=\Pi^{0}(T)$.本文利用算子的广义Kato分解特征,从广义Kato谱的角度继续探讨了有界线性算子满足Browder定理和Weyl定理的充要条件.

设$T\in L(X)$,称$T$为半正则算子,如果对任意的$n\in{\Bbb N}$, $R(T^n)$闭且$N(T^n)\subseteq R(T)$.称$T$有广义Kato分解(简记为GKD),如果存在$T$的不变子空间$(M, N)$使得$T=M\oplus N$,使得$T|_M$半正则且$T|_N$拟幂零.称$T$是Kato型算子,如果$T$有GKD(M, N)使得$T|_N$是幂零的.称$T$是广义Drazin可逆的,如果存在$S\in L(X)$使得$ST=TS, STS=S, TST=T+U$,其中$U$是拟幂零的. Drazin可逆的算子必定是广义Drazin可逆的算子.记$T$的广义Kato谱和广义Drazin谱分别为:

记$\rho_{gk}(T)={\Bbb C}\setminus\sigma_{gk}(T)$.由文献[6,定理4.2]知$\lambda\notin\sigma_{gd}(T)$当且仅当$\lambda\notin acc\sigma(T)$, $\rho_{gd}(T)\subseteq\rho_{gk}(T)$.下面先引入有关$\sigma_{gk}(T)$与$\sigma_{gd}(T)$的一个重要引理:

引理2.1  设$T\in L(X)$,则$\sigma_{gd}(T)=\sigma_{gk}(T)\cup int\sigma(T)$.

证  对任意的$\lambda\notin\sigma_{gd}(T)$,则$\lambda\notin acc\sigma(T)$.若$\lambda\notin\sigma(T)$,则$\lambda\notin int\sigma(T)$且$\lambda I-T$有广义Kato分解,即$\lambda\notin\sigma_{gk}(T)\cup int\sigma(T)$.若$\lambda\in\sigma(T)$,显然$\lambda\in iso\sigma(T)$,则$\lambda\notin int\sigma(T)$且$T$和$T^*$都在$\lambda$处有SVEP,由文献[7,定理3.1.11]可知, $\lambda I-T$有广义Kato分解,即$\lambda\notin\sigma_{gk}(T)\cup int\sigma(T)$.所以$\sigma_{gk}(T)\cup int\sigma(T)\subseteq\sigma_{gd}(T)$.

反之,对任意的$\lambda\notin\sigma_{gk}(T)\cup int\sigma(T)$,则$\lambda\notin\sigma_{gk}(T)$且$\lambda\in\rho(T)\cup\partial\sigma(T)$.若$\lambda\notin\sigma_{gk}(T)$且$\lambda\in\rho(T)$,显然$\lambda\notin\sigma_{gd}(T)$.若$\lambda\notin\sigma_{gk}(T)$且$\lambda\in\partial\sigma(T)$,由文献[7,推论3.1.12]可知, $\lambda\in iso\sigma(T)$,即$\lambda\notin acc\sigma(T)$,则$\lambda\notin\sigma_{gd}(T)$.所以, $\sigma_{gd}(T)\subseteq\sigma_{gk}(T)\cup int\sigma(T)$.

下面利用$\sigma_{gk}(T)$和$\sigma_{gd}(T)$刻画Browder定理和Weyl定理.

定理2.1  $T$满足Browder定理当且仅当$\sigma_{gd}(T)\subseteq\sigma_{gk}(T)\cup int\sigma_w(T)$.

证  $\Rightarrow)$设$T$满足Browder定理,即有$\sigma_b(T)=\sigma_w(T)$.设$\lambda_0\notin\sigma_{gk}(T)\cup int\sigma_w(T)$,有$\lambda_0\notin\sigma_{gk}(T)$且$\lambda_0\in\rho_w(T)\cup\partial\sigma_w(T)$.

若$\lambda_0\notin\sigma_{gk}(T)$且$\lambda_0\in\rho_w(T)$,则$\lambda_0\notin\sigma_w(T)=\sigma_{b}(T)$,即$\lambda_0 I-T$是Browder算子,从而$\lambda_0 I-T$是Drazin可逆算子,所以$\lambda_0\notin\sigma_{gd}(T)$.

若$\lambda_0\notin\sigma_{gk}(T)$且$\lambda_0\in\partial\sigma_w(T)$,因$\sigma_b(T)=\sigma_w(T)$,有$\lambda_0\in\partial\sigma_b(T)$,则对$\forall\varepsilon>0$,都存在$\lambda_1$满足$0 < |\lambda_1-\lambda_0| < \varepsilon$,使得$\lambda_1 I-T$是Browder算子,即$\lambda_1 I-T$是Fredholm算子且$p(\lambda_1 I-T)=q(\lambda_1 I-T) < \infty$.下面分情况讨论.若$p(\lambda_1 I-T)=q(\lambda_1 I-T)=0$,则$\lambda_1 I-T$可逆,所以$\lambda_0\in\partial\sigma(T)$.若$0 < p(\lambda_1 I-T)=q(\lambda_1 I-T) < \infty$,即$\lambda_1$是$T$的极点.因$\lambda_1 I-T$是Fredholm算子,由文献[1,定义1.76]知,存在$d\in{\Bbb N}$使得$\lambda_1 I-T$是对$n\geq d$有拓扑一致降指数的算子.于是,由文献[8,推论4.8]知,对任意满足$(\varepsilon-|\lambda_1-\lambda_0|)>\varepsilon_1>0$的$\varepsilon_1$,都存在$\lambda_2$满足$0 < |\lambda_2-\lambda_1| < \varepsilon_1$,使得$\lambda_2 I-T$可逆.因

所以$\lambda_0\in\partial\sigma(T)$.又因$\lambda_0\notin\sigma_{gk}(T)$,由引理2.1可知, $\lambda_0\notin\sigma_{gd}(T)$.因此

$\Leftarrow)$显然$\sigma_w(T)\subseteq\sigma_b(T)$,下证: $\sigma_b(T)\subseteq\sigma_w(T)$.设$\lambda_0\notin\sigma_w(T)$,即$\lambda_0 I-T$是Weyl算子,则$\lambda_0\notin\sigma_{gk}(T)\cup int\sigma_w(T)$,从而$\lambda_0\notin\sigma_{gd}(T)$,即有$\lambda_0\notin acc\sigma(T)$.若$\lambda_0\notin\sigma(T)$,显然$\lambda_0\notin\sigma_b(T)$.若$\lambda_0\in\sigma(T)$,则$\lambda_0\in iso\sigma(T)$.由文献[1,定理2.66]可知, $\lambda_0 I-T$是Browder算子,即$\lambda\notin\sigma_b(T)$.所以$\sigma_b(T)\subseteq\sigma_w(T)$.因此, $T$满足Browder定理.

定理2.2   $T$满足Weyl定理当且仅当$\sigma_{gd}(T)\subseteq\sigma_{gk}(T)\cup int\sigma_w(T)$且$\sigma_f(T)\cap E^0(T)=\emptyset$.

证   $\Rightarrow)$设$T$满足Weyl定理,则$T$满足Browder定理,由定理2.1知, $\sigma_{gd}(T)\subseteq\sigma_{gk}(T)\cup int\sigma_w(T)$.又由于$E^0(T)=\sigma(T)\setminus\sigma_w(T)\subseteq\rho_f(T)$,所以, $\sigma_f(T)\cap E^0(T)=\emptyset$.

$\Leftarrow)$由定理2.1可知,当$\sigma_{gd}(T)\subseteq\sigma_{gk}(T)\cup int\sigma_w(T)$时, $T$满足Browder定理,即

下面证明: $E^0(T)\subseteq\sigma(T)\setminus\sigma_w(T)$.对任意的$\lambda\in E^0(T)$,由于$\sigma_f(T)\cap E^0(T)=\emptyset$,则$\lambda I-T$是Fredholm算子.因$\lambda\in iso\sigma(T)$,由文献[1,定理2.66]可知, $\lambda I-T$是Browder算子,所以$\lambda\in\sigma(T)\setminus\sigma_b(T)\subseteq\sigma(T)\setminus\sigma_w(T)$.因此, $\sigma(T)\setminus\sigma_w(T)=E^0(T)$,即$T$满足Weyl定理.

在定理2.1和定理2.2中,若$\sigma_{gk}(T)=\emptyset$,能否得到$T$满足Browder定理或者Weyl定理?

定理2.3  设$T\in L(X)$, $T$满足Browder定理,如果$T$满足下列条件之一:

$(1)$$\sigma_{gk}(T)=\emptyset$;

$(2)$任意的$\lambda\in\partial\sigma(T)$都是孤立的;

$(3)$任意的$\lambda\in\sigma(T)$都是孤立的;

$(4)$$\sigma(T)$是一个有限集;

$(5)$$\sigma_{gd}(T)=\emptyset$.

证  由文献[7,定理3.2.17]知,条件$(1)(2)(3)(4)(5)$等价.若$\sigma_{gd}(T)=\emptyset$,显然$\sigma_{gd}(T)\subseteq\sigma_{gk}(T)\cup int\sigma_w(T)$,所以由定理2.1知, $T$满足Browder定理.

然而,当$\sigma_{gk}(T)=\emptyset$时, $T$不一定满足Weyl定理.例如:设$T(x_1, x_2, x_3, \cdots)=(\frac{1}{3}x_2, \frac{1}{4}x_3, \cdots), (x_n)\in l^2({\Bbb N})$,则$\sigma(T)=\sigma_{w}(T)=\{0\}$且$E^0(T)=\{0\}$,由文献[7,定理3.2.17]可知$\sigma_{gk}(T)=\emptyset$,但是$\sigma(T)\setminus\sigma_{w}(T)=\emptyset\neq E^0(T)$,即$T$不满足Weyl定理.

定理2.3的逆命题也不一定成立.例如:设$T\in L(l^2({\Bbb N}))$是单边右移位算子,则$\sigma(T)=\sigma_w(T)={\Bbb D}(0, 1)$,其中${\Bbb D}(0, 1)$是${\Bbb C}$中的单位闭圆盘,且$\Pi^0(T)=\emptyset$.所以$\sigma(T)\setminus\sigma_w(T)=\emptyset=\Pi^0(T)$,即$T$满足Browder定理.但是, $\sigma(T)={\Bbb D}(0, 1)$不是有限集,即$(4)$不成立,又由文献[7,定理3.2.17]可知, $(1)(2)(3)(5)$也不成立.

下面给出$T$满足Browder定理与$\sigma_{gk}(T)=\emptyset$的等价条件.

定理2.4  $\sigma_{gk}(T)=\emptyset$当且仅当$T$满足Browder定理且$\sigma_b(T)\cap\sigma_{gd}(T)=\emptyset$.

证  $\Rightarrow)$若$\sigma_{gk}(T)=\emptyset$,由定理2.3可知, $T$满足Browder定理.又由文献[7,定理3.2.17]可知, $\sigma_{gd}(T)=\emptyset$,即有$\sigma_b(T)\cap\sigma_{gd}(T)=\emptyset$.

$\Leftarrow)$由于$T$满足Browder定理,则$\Pi^0(T)=\sigma(T)\setminus\sigma_w(T)\subseteq\rho_{gk}(T)$.因$\sigma_b(T)\cap\sigma_{gd}(T)=\emptyset$,则$\sigma_b(T)\subseteq\rho_{gd}(T)$.又因$\rho_{gd}(T)\subseteq\rho_{gk}(T)$,所以$\sigma_b(T)\subseteq\rho_{gk}(T)$.显然$\rho(T)\subseteq\rho_{gk}(T)$.又由$\Pi^0(T)=\sigma(T)\setminus\sigma_b(T)$,所以

因此, $\rho_{gk}(T)={\Bbb C}$,即$\sigma_{gk}(T)=\emptyset$.

下面进一步利用$\sigma_{gk}(T)$来得到$T$满足Browder定理和Weyl定理的充要条件.

定理2.5  设$T\in L(X)$,则下列叙述等价:

$(1)$$T$满足Browder定理;

$(2)$$\sigma_{b}(T)=[\sigma_{gk}(T)\cap acc\sigma(T)]\cup int\sigma_w(T)\cup\{\lambda\in{\Bbb C}: \alpha(\lambda I-T)=\infty\}\cup\{\lambda\in{\Bbb C}: R(\lambda I-T)$不闭$\}$;

$(3)$$\sigma_{b}(T)\subseteq[\sigma_{gk}(T)\cap acc\sigma(T)]\cup int\sigma_w(T)\cup\{\lambda\in{\Bbb C}: \alpha(\lambda I-T)=\infty\}\cup\{\lambda\in{\Bbb C}: R(\lambda I-T)$不闭$\}$.

证  $(1)\Rightarrow(2)$若$T$满足Browder定理,即有$\sigma_b(T)=\sigma_w(T)$.显然$[\sigma_{gk}(T)\cap acc\sigma(T)]\cup int\sigma_w(T)\cup\{\lambda\in{\Bbb C}: \alpha(\lambda I-T)=\infty\}\cup\{\lambda\in{\Bbb C}: R(\lambda I-T)$不闭$\}\subseteq\sigma_w(T)=\sigma_b(T)$.

反之,设$\lambda_0\notin[\sigma_{gk}(T)\cap acc\sigma(T)]\cup int\sigma_w(T)\cup\{\lambda\in{\Bbb C}: \alpha(\lambda I-T)=\infty\}\cup\{\lambda\in{\Bbb C}: R(\lambda I-T)$不闭$\}$,则$\lambda_0\in\rho_w(T)\cup\partial\sigma_w(T)$, $\alpha(\lambda_0 I-T) < \infty$且$R(\lambda_0 I-T)$闭,从而$\lambda_0 I-T$是上半Fredholm算子.下面对$\lambda_0\notin \sigma_{gk}(T)\cap acc\sigma(T)$分两种情形讨论:

情形1    $\lambda_0\notin\sigma_{gk}(T)$, $\lambda_0\in\rho_w(T)\cup\partial\sigma_w(T)$且$\lambda_0 I-T$是上半Fredholm算子.

若$\lambda_0\in\rho_w(T)$,因$\sigma_w(T)=\sigma_b(T)$,则$\lambda_0\notin\sigma_b(T)$.

若$\lambda_0\in\partial\sigma_w(T)$,因$\sigma_b(T)=\sigma_w(T)$,由定理2.1的证明过程可知, $\lambda_0\in\partial\sigma(T)$.

因$\lambda_0\notin\sigma_{gk}(T)$,由文献[7,推论3.1.12]知, $\lambda_0\in iso\sigma(T)$.又因$\lambda_0 I-T$是上半Fredholm算子,由文献[1,定理2.66]可知, $\lambda_0 I-T$是Browder算子,即$\lambda_0\notin\sigma_b(T)$.

情形2   $\lambda_0\notin acc\sigma(T)$, $\lambda_0\in\rho_w(T)\cup\partial\sigma_w(T)$且$\lambda_0 I-T$是上半Fredholm算子.

若$\lambda_0\notin\sigma(T)$,显然$\lambda_0\notin\sigma_b(T)$.若$\lambda_0\in\sigma(T)$,则$\lambda_0\in iso\sigma(T)$,由情形1的证明过程可知, $\lambda_0\notin\sigma_b(T)$.

由情形1和情形2知, $\sigma_{b}(T)\subseteq[\sigma_{gk}(T)\cap acc\sigma(T)]\cup int\sigma_w(T)\cup\{\lambda\in{\Bbb C}: \alpha(\lambda I-T)=\infty\}\cup\{\lambda\in{\Bbb C}: R(\lambda I-T)$不闭$\}$.

$(2)\Rightarrow(3)$显然成立.

$(3)\Rightarrow(1)$显然$\sigma_w(T)\subseteq\sigma_b(T)$,又因$\sigma_b(T)\subseteq[\sigma_{gk}(T)\cap acc\sigma(T)]\cup int\sigma_w(T)\cup\{\lambda\in{\Bbb C}: \alpha(\lambda I-T)=\infty\}\cup\{\lambda\in{\Bbb C}: R(\lambda I-T)$不闭$\}\subseteq\sigma_w(T)$,即有$\sigma_b(T)\subseteq\sigma_w(T)$.因此, $\sigma_w(T)=\sigma_b(T)$,即$T$满足Browder定理.

定理2.6  设$T\in L(X)$,则下列叙述等价:

$(1)$$T$满足Weyl定理;

$(2)$$\sigma_{b}(T)=[\sigma_{gk}(T)\cap acc\sigma(T)]\cup int\sigma_w(T)\cup\{\lambda\in\sigma(T): \alpha(\lambda I-T)=\infty$或$\alpha(\lambda I-T)=0\}$;

$(3)$$\sigma_{b}(T)\subseteq[\sigma_{gk}(T)\cap acc\sigma(T)]\cup int\sigma_w(T)\cup\{\lambda\in{\Bbb C}: \alpha(\lambda I-T)=\infty$或$\alpha(\lambda I-T)=0\}$.

证  $(1)\Rightarrow(2)$若$T$满足Weyl定理,则$T$满足Browder定理,即有$\sigma_b(T)=\sigma_w(T)$.于是,显然$[\sigma_{gk}(T)\cap acc\sigma(T)]\cup int\sigma_w(T)\cup\{\lambda\in\sigma(T): \alpha(\lambda I-T)=\infty$或$\alpha(\lambda I-T)=0\}\subseteq\sigma_w(T)=\sigma_b(T)$.

设$\lambda_0\notin[\sigma_{gk}(T)\cap acc\sigma(T)]\cup int\sigma_w(T)\cup\{\lambda\in\sigma(T): \alpha(\lambda I-T)=\infty$或$\alpha(\lambda I-T)=0\}$,若$\lambda_0\notin\sigma(T)$,则$\lambda_0\notin\sigma_b(T)$.不妨设$\lambda_0\in\{\lambda\in{\Bbb C}: 0 < \alpha(\lambda I-T) < \infty\}$.分两种情形讨论:

情形1  $\lambda_0\notin\sigma_{gk}(T)$, $\lambda_0\in\rho_w(T)\cup\partial\sigma_w(T)$且$0 < \alpha(\lambda_0 I-T) < \infty$.

若$\lambda_0\in\rho_w(T)$,因$\sigma_w(T)=\sigma_b(T)$,则$\lambda_0\notin\sigma_b(T)$.

若$\lambda_0\in\partial\sigma_w(T)$,因$\sigma_b(T)=\sigma_w(T)$,由定理2.1的证明过程可知, $\lambda_0\in\partial\sigma(T)$.因$\lambda_0\notin\sigma_{gk}(T)$,由文献[7,推论3.1.12]可知, $\lambda_0\in iso\sigma(T)$.又因$0 < \alpha(\lambda_0 I-T) < \infty$,则$\lambda_0\in E^0(T)$.由于$T$满足Weyl定理,所以$\lambda_0\notin\sigma_w(T)=\sigma_b(T)$.

情形2  $\lambda_0\notin acc\sigma(T)$, $\lambda_0\in\rho_w(T)\cup\partial\sigma_w(T)$且$0 < \alpha(\lambda_0 I-T) < \infty$.

若$\lambda_0\notin\sigma(T)$,显然$\lambda_0\notin\sigma_b(T)$.若$\lambda_0\in\sigma(T)$,则$\lambda_0\in iso\sigma(T)$,由情形1的证明过程可知, $\lambda_0\notin\sigma_b(T)$.

由情形1和情形2知, $\sigma_{b}(T)\subseteq[\sigma_{gk}(T)\cap acc\sigma(T)]\cup int\sigma_w(T)\cup\{\lambda\in\sigma(T): \alpha(\lambda I-T)=\infty$或$\alpha(\lambda I-T)=0\}.$

$(2)\Rightarrow(3)$显然成立.

$(3)\Rightarrow(1)$由于$\{[\sigma(T)\setminus\sigma_w(T)]\cup E^0(T)\}\cap\{[\sigma_{gk}(T)\cap acc\sigma(T)]\cup int\sigma_w(T)\cup\{\lambda\in{\Bbb C}: \alpha(\lambda I-T)=\infty$或$\alpha(\lambda I-T)=0\}\}=\emptyset$,则$\{[\sigma(T)\setminus\sigma_w(T)]\cup E^0(T)\}\cap\sigma_b(T)=\emptyset$,从而$[\sigma(T)\setminus\sigma_w(T)]\cap\sigma_b(T)=\emptyset$且$E^0(T)\cap\sigma_b(T)=\emptyset$,所以$\sigma_w(T)=\sigma_b(T)$且$E^0(T)\subseteq\sigma(T)/\sigma_w(T)$.又因$\sigma(T)/\sigma_w(T)=\sigma(T)\setminus\sigma_b(T)=\Pi^0(T)\subseteq E^0(T)$,所以$\sigma(T)/\sigma_w(T)=E^0(T)$,即$T$满足Weyl定理.

设$T\in L(X)$,称$T$是isoloid算子,若$iso\sigma(T)\subseteq E(T)$.

推论2.1  $T$是$isoloid$算子且满足Weyl定理当且仅当$\sigma_{b}(T)=[\sigma_{gk}(T)\cap acc\sigma(T)]\cup int\sigma_w(T)\cup\{\lambda\in{\Bbb C}: \alpha(\lambda I-T)=\infty\}$.

证   $\Rightarrow)$设$T$是$isoloid$算子且满足Weyl定理,由定理2.6知, $\sigma_{b}(T)=[\sigma_{gk}(T)\cap acc\sigma(T)]\cup int\sigma_w(T)\cup\{\lambda\in\sigma(T): \alpha(\lambda I-T)=\infty$或$\alpha(\lambda I-T)=0\}=[\sigma_{gk}(T)\cap acc\sigma(T)]\cup int\sigma_w(T)\cup\{\lambda\in{\Bbb C}: \alpha(\lambda I-T)=\infty\}\cup\{\lambda\in\sigma(T): \alpha(\lambda I-T)=0\}$.下面证明: $\{\lambda\in\sigma(T): \alpha(\lambda I-T)=0\}\subseteq[\sigma_{gk}(T)\cap acc\sigma(T)]\cup int\sigma_w(T)$.

设$\lambda_0\notin[\sigma_{gk}(T)\cap acc\sigma(T)]\cup int\sigma_w(T)$,由$\lambda_0\notin int\sigma_w(T)$,有$\lambda_0\in\rho_w(T)\cup\partial\sigma_w(T)$.若$\lambda_0\in\rho_w(T)$,显然$\lambda_0\notin\{\lambda\in\sigma(T): \alpha(\lambda I-T)=0\}$.若$\lambda_0\in\partial\sigma_w(T)$,因$T$满足Weyl定理,则$T$满足Browder定理,即有$\sigma_b(T)=\sigma_w(T)$.由定理2.1的证明过程可知, $\lambda_0\in\partial\sigma(T)$.又由$\lambda_0\notin\sigma_{gk}(T)\cap acc\sigma(T)$,下面分两种情形讨论:

情形1  当$\lambda_0\notin acc\sigma(T)$时,若$\lambda_0\notin\sigma(T)$,则$\lambda_0\notin\{\lambda\in\sigma(T): \alpha(\lambda I-T)=0\}$.若$\lambda_0\in\sigma(T)$,有$\lambda_0\in iso\sigma(T)$,因$T$是isoloid算子,则$\alpha(\lambda_0 I-T)>0$,所以$\lambda_0\notin\{\lambda\in\sigma(T): \alpha(\lambda I-T)=0\}$.

情形2  若$\lambda_0\notin\sigma_{gk}(T)$,因$\lambda_0\in\partial\sigma(T)$,由文献[7,推论3.1.12]可知, $\lambda_0\in iso\sigma(T)$.同情形1的证明过程可知, $\lambda_0\notin\{\lambda\in\sigma(T): \alpha(\lambda I-T)=0\}$.

因此

所以

$\Leftarrow)$设$\sigma_{b}(T)=[\sigma_{gk}(T)\cap acc\sigma(T)]\cup int\sigma_w(T)\cup\{\lambda\in{\Bbb C}: \alpha(\lambda I-T)=\infty\}$,由定理2.6可知, $T$满足Weyl定理,下面证明$T$是isoloid算子.

对任意的$\lambda\in iso\sigma(T)$,若$\alpha(\lambda I-T)=0$,则$\lambda\notin[\sigma_{gk}(T)\cap acc\sigma(T)]\cup int\sigma_w(T)\cup\{\lambda\in{\Bbb C}: \alpha(\lambda I-T)=\infty\}=\sigma_b(T)$.又因$T$满足Weyl定理,则$T$满足Browder定理,即有$\lambda\notin\sigma_b(T)=\sigma_w(T)$,所以$\lambda I-T$是Weyl算子.又由于$\alpha(\lambda I-T)=0$,则$\lambda I-T$可逆,这与$\lambda\in iso\sigma(T)$矛盾.因此, $\alpha(\lambda I-T)>0$,即$T$是isoloid算子.