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CFI算子與(ω)性質和(ω1)性質的判定

殷樂 曹小紅

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CFI算子與(ω)性質和(ω1)性質的判定

    作者簡介: 殷 樂(1994?),女,陜西人,碩士生,主要從事算子理論方面的研究. E-mail:yinle@snnu.edu.cn;
    通訊作者: 曹小紅, xiaohongcao@snnu.edu.cn
  • 中圖分類號: O177.2

CFI operators and the judgements of property (ω) and property (ω1)

    Corresponding author: CAO Xiao-hong, xiaohongcao@snnu.edu.cn
  • CLC number: O177.2

  • 摘要: 利用一致Fredholm指標性質定義了一個新的譜集,根據這個譜集給岀了算子T及其共軛算子T*滿足$\left( {{\omega _1}} \right)$性質和$\left( \omega \right)$性質的判定條件.并且,對Hilbert空間上有界線性算子的$\left( \omega \right)$性質的與T交換的有限秩攝動F進行了討論.
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    [8] Berberian S K. The Weyl spectrum of an operator[J]. Indiana University Mathematics Journal, 1970, 20(6): 529-544. DOI:  10.1512/iumj.1971.20.20044.
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    [13] Sun C H, Cao X H, Dai L. Property $\left( {{\omega _1}} \right)$ and Weyl type theorem[J]. Journal of Mathematical Analysis and Applications, 2010, 363(1): 1-6. DOI:  10.1016/j.jmaa.2009.07.045.
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出版歷程
  • 收稿日期:  2019-07-01
  • 錄用日期:  2020-01-19
  • 網絡出版日期:  2020-04-11
  • 刊出日期:  2020-07-01

CFI算子與(ω)性質和(ω1)性質的判定

    作者簡介:殷 樂(1994?),女,陜西人,碩士生,主要從事算子理論方面的研究. E-mail:yinle@snnu.edu.cn
    通訊作者: 曹小紅, xiaohongcao@snnu.edu.cn
  • 陜西師范大學 數學與信息科學學院,陜西 西安 710119

摘要: 利用一致Fredholm指標性質定義了一個新的譜集,根據這個譜集給岀了算子T及其共軛算子T*滿足$\left( {{\omega _1}} \right)$性質和$\left( \omega \right)$性質的判定條件.并且,對Hilbert空間上有界線性算子的$\left( \omega \right)$性質的與T交換的有限秩攝動F進行了討論.

English Abstract

  • 1909年,Weyl檢查Hilbert空間上正規算子的譜時發現了Weyl定理[1]. 自此譜理論逐漸發展成為算子理論和算子代數中的一個重要分支,與解算子方程聯系密切,且應用廣泛[2-3]. 之后許多文獻都討論了Weyl定理的變式[4-6]. Rako?evic[5]研究了其中一個變式,即 $\left( \omega \right)$ 性質. 在本文中,H表示復的無限維可分的Hilbert空間,$B\left( H \right)$ 表示 $H$ 上有界線性算子的全體. 對于 $T \in B\left( H \right)$,$n\left( T \right)$$d\left( T \right)$ 分別表示算子 $T$ 的零空間 $N\left( T \right)$ 的維數和值域 $R\left( T \right)$ 的余維數. 由文獻[7]中推論1.15可知,若 $d\left( T \right) < 0$,則 $R\left( T \right)$ 閉. 并且,我們定義:

    $\qquad {\sigma _d}(T) = \left\{ {\lambda \in {\bf{C}}:R\left( {T - \lambda I} \right){\text{不閉}}} \right\}. $

    回顧可知,算子 $T \in B\left( H \right)$ 稱為上半Fredholm算子,若 $T$ 的值域 $R\left( T \right)$ 閉且其零空間 $N\left( T \right)$ 是有限維的[8-10];算子 $T \in B\left( H \right)$ 稱為下半Fredholm算子,若 $T$ 的值域 $R\left( T \right)$ 的余維數是有限維的. 特別地,若 $R\left( T \right)$ 閉且 $n\left( T \right) = 0$,則稱算子 $T$ 為下有界算子;若 $d\left( T \right) = 0$,則稱算子 $T$ 為滿算子. 稱算子 $T$ 是可逆的,當且僅當 $N\left( T \right) = d\left( T \right) = 0$. 設算子 $T$ 的譜 $\sigma \left( T \right)$,逼近點譜 ${\sigma _a}\left( T \right)$ 和滿譜 ${\sigma _s}\left( T \right)$ 定義如下:

    $\qquad \sigma (T) = \left\{ {\lambda \in {\bf{C}}:T - \lambda I{\text{不為可逆算子}}} \right\}; $

    $\qquad {\sigma _a}(T) = \left\{ {\lambda \in {\bf{C}}:T - \lambda I{\text{不為下有界算子}}} \right\}; $

    $\qquad {\sigma _s}(T) = \{ \lambda \in {\bf{C}}:T - \lambda I{\text{不為滿算子}}\}. $

    令相應的預解集為 $\rho (T) ={\bf{C}}\backslash \sigma (T)$,${\rho _a}(T) = {\bf{C}}\backslash {\sigma _a}(T)$${\rho _s}(T) = {\bf{C}}\backslash {\sigma _s}(T)$. 若算子 $T$ 是半(上半或下半)Fredholm算子,則可定義算子 $T$ 的指標為 ${\rm{ind}}\left( T \right) = n\left( T \right) - d\left( T \right)$. 若 ${\rm{ind}}\left( T \right) < \infty $,則稱算子 $T$ 為Fredholm算子. 特別地,當 ${\rm{ind}}\left( T \right) = 0$ 時,稱 $T$ 為Weyl算子. 設算子 $T$ 的本質譜 ${\sigma _e}\left( T \right)$ 和Weyl譜 ${\sigma _w}\left( T \right)$ 定義如下:

    $\qquad {\sigma _e}(T) = \{ \lambda \in {\bf{C}}:T - \lambda I{\text{不為}}{\rm{ Fredholm }}{\text{算子}}\}; $

    $\qquad {\sigma _w}(T) = \{ \lambda \in {\bf{C}}:T - \lambda I{\text{不為}}{\rm{Weyl }}{\text{算子}}\}. $

    $T$ 是Fredholm算子且有有限的升標 ${\rm{asc}}\left( T \right)$ 和降標 ${\rm{des}}\left( T \right)$ 時,稱 $T$ 是Browder算子,其中 ${\rm{asc}}\left( T \right) = \inf \left\{ {n \in {\bf{N}}:N\left( {{T^n}} \right) = N\left( {{T^{n + 1}}} \right)} \right\}$${\rm{des}}\left( T \right) = \inf \left\{ {n \in {\bf{N}}:R\left( {{T^n}} \right) = R\left( {{T^{n + 1}}} \right)} \right\}$. 由文獻[11]中推論4.4可知Browder算子就是有有限升降標的Weyl算子. 我們定義Browder譜為

      ${\sigma _b}(T) = \{ \lambda \in {\bf{C}}:T - \lambda I{\text{不為}}{\rm{Browder}}{\text{算子}}\} $,

    ${\;\rho _b}(T) = {\bf{C}}\backslash {\sigma _b}(T)$.給定一個子集 $G \subseteq {\bf{C}}$,令 $\operatorname{int} G$$G$ 的所有內點之集,${\rm{iso}}G$$G$ 的所有孤立點之集. 令 ${\rm{acc}}G = \bar G\backslash {\rm{iso}}G$,其中 $\bar G$$G$ 的閉包. 并且,稱 $\partial G$$G$的邊界. 稱 $T \in B\left( H \right)$ 為isoloid算子,若 ${\rm{iso}}\sigma (T) \subseteq {\sigma _p}(T)$,其中 ${\sigma _p}(T) = $$ \{ \lambda \in {\bf{C}}:n\left( {T - \lambda I} \right) > 0\} $.

    在文獻[12~13]中,作者介紹了 $\left( \omega \right)$ 性質和 $\left( {{\omega _1}} \right)$ 性質的相關概念. 稱 $T$ 滿足 $\left( \omega \right)$ 性質,若

    $\qquad{\sigma _a}(T)\backslash {\sigma _{ea}}(T) = {{\rm{\pi }}_{00}}(T),$

    其中 ${{\rm{\pi }}_{00}}(T) = \left\{ {\lambda \in {\rm{iso}}\sigma (T):0 < n(T - \lambda I) < \infty } \right\}$,${\sigma _{ea}}(T) = {\bf{C}}\backslash {\rho _{ea}}(T)$${\rho _{ea}}(T) = \{ \lambda \in {\bf{C}}:T - \lambda I{\text{是上半}}{\rm{Fredholm}}$算子且 ${{\rm{ind}}( {T - \lambda I{\rm{ }}} ) \leqslant 0}\} $. 稱 $T$ 滿足 $\left( {{\omega _1}} \right)$ 性質,若 ${\sigma _a}(T)\backslash {\sigma _{ea}}(T) \subseteq {{\rm{\pi }}_{00}}(T). $

    在本文中,我們用CFI算子[14]定義了一個新的譜集,利用這個譜集討論了算子 $T$ 及其共軛算子 ${T^ * }$ 滿足 $\left( \omega \right)$ 性質的判定條件,并且討論了算子 $T$ 關于 $\left( \omega \right)$ 性質的有限秩攝動.

    在文獻[14]中,作者定義了一致Fredholm指標算子(簡寫為CFI算子),即算子 $T \in B\left( H \right)$ 稱為CFI算子,若對任意的 $S \in B\left( H \right)$,下列情況之一成立:

    (1)$TS$$ST$ 均為Fredholm算子且 ${\rm{ind}}\left( {TS} \right) = {\rm{ind}}\left( {ST} \right) = {\rm{ind}}\left( S \right)$;

    (2)$TS$$ST$ 都不為Fredholm算子.

    并且,作者證明了 $T$ 是CFI算子當且僅當以下情況之一成立:

    (1)$T$ 是Weyl算子;

    (2)$R\left( T \right)$不閉;

    (3)$R\left( T \right)$ 閉且 $n\left( T \right) = d\left( T \right) = \infty $.

    顯然,$T$ 不是CFI算子當且僅當T是半Fredholm算子且 ${\rm{ind}}\left( T \right) \ne 0$.

    利用CFI算子,我們定義了一個新的譜集:${\sigma _1}\left( T \right) = \left\{ {\lambda \in {\bf{C}}:T - \lambda I{\text{不為}}{\rm{CFI}}{\text{算子}}} \right\}$. 顯然 ${\sigma _1}\left( T \right)$ 是一個開集. 令 ${\rho _1}(T) = {\bf{C}}\backslash {\sigma _1}(T)$. 接下來,我們利用 ${\sigma _1}\left( T \right)$ 分別研究算子 $T$ 及其共軛算子 ${T^ * }$$\left( {{\omega _1}} \right)$ 性質和 $\left( \omega \right)$ 性質.

    定理1 設 $T \in B\left( H \right)$. $T$ 滿足 $\left( {{\omega _1}} \right)$ 性質 $ \Leftrightarrow {\sigma _b}\left( T \right) = $    $\partial {\sigma _1}\left( T \right) \cup \left\{ {\lambda \in {\bf{C}}:n\left( {T - \lambda I} \right)} \right. > d\left( T \right. - \left. {\left. {\lambda I} \right)} \right\} \cup \{ \lambda \in $${\mathop{\rm int}} {\rho _1}\left( T \right):n\left( {T - \lambda I} \right) = \infty \} \cup \left[ {{\rho _a}\left( T \right) \cap \sigma \left( T \right)} \right] \cup {\sigma _d}\left( T \right) $.

    證明 顯然,對任意 $T \in B\left( H \right)$,有    ${\sigma _b}\left( T \right) \supseteq \partial {\sigma _1}\left( T \right) \cup \left\{ {\lambda \in{\bf{C}}:n\left( {T \!-\! \lambda I} \right)} \right. \!>\! \left. {d\left( {T \!-\! \lambda I} \right)} \right\} \cup \left\{ \lambda \in \operatorname{int} {\rho _1}\left( T \right)\!:\right.$ $\left.n\left( {T \!-\! \lambda I} \right) = { \infty } \right\} \cup \left[ {{\rho _a}\left( T \right) \cap \sigma \left( T \right)} \right] \cup {\sigma _d}\left( T \right)$.

    對于反包含,我們假設     ${\lambda _0} \notin \partial {\sigma _1}\left( T \right) \cup \left\{ {\lambda \in {\bf{C}}:n\left( {T - \lambda I} \right)} \right. > \left. {d\left( {T - \lambda I} \right)} \right\} \cup \left\{ {\lambda \in \operatorname{int} {\rho _1}} \right.\left( T \right):n\left( T \right. \left. { - \lambda I} \right)\left. { = \infty } \right\} \cup $$\left[ {{\rho _a}\left( T \right) \cap \sigma \left( T \right)} \right] \cup{\sigma _d}\left( T \right) $,則有 $n\left( {T - {\lambda _0}I} \right) \leqslant d\left( {T - {\lambda _0}I} \right)$$R\left( {T - {\lambda _0}I} \right)$ 閉. 不妨設 ${\lambda _0} \in {\sigma _a}\left( T \right)$. 由于 ${\lambda _0} \notin \partial {\sigma _1}\left( T \right)$,故 ${\lambda _0} \in {\sigma _1}\left( T \right) \cup \operatorname{int} {\rho _1}\left( T \right)$. 若 ${\lambda _0} \in {\sigma _1}\left( T \right)$,則 ${\lambda _0} \in {\sigma _a}(T)\backslash {\sigma _{ea}}(T)$,由 $T$ 滿足 $\left( {{\omega _1}} \right)$ 性質可知 ${\lambda _0} \in {\rho _b}\left( T \right)$,這與 ${\lambda _0} \in {\sigma _1}\left( T \right)$ 矛盾. 若 ${\lambda _0} \in \operatorname{int} {\rho _1}\left( T \right)$,則 $n\left( {T - {\lambda _0}I} \right) < \infty $. 因此 ${\lambda _0} \in {\rho _w}\left( T \right)$,即 ${\lambda _0} \in {\sigma _a}(T)\backslash {\sigma _{ea}}(T)$. 同理,由 $T$ 滿足 $\left( {{\omega _1}} \right)$ 性質可知 ${\lambda _0} \in {\rho _b}\left( T \right)$.

    反之,假設 ${\lambda _0} \in {\sigma _a}(T)\backslash {\sigma _{ea}}(T)$. 由半Fredholm算子的攝動定理可知 ${\lambda _0} \in {\sigma _1}\left( T \right) \cup \operatorname{int} {\rho _1}\left( T \right)$,故 ${\lambda _0} \notin \partial {\sigma _1}\left( T \right)$. 因此     ${\lambda _0} \notin \partial {\sigma _1}\left( T \right) \cup \left\{ {\lambda \in {\bf{C}}:n\left( {T - \lambda I} \right)} \right. > \left. {d\left( {T - \lambda I} \right)} \right\} \cup \left\{ {\lambda \in \operatorname{int} {\rho _1}} \right.\left( T \right):n\left( T \right.\left. { - \lambda I} \right) = \left. \infty \right\} \cup \left[ {{\rho _a}\left( T \right) \cap \sigma \left( T \right)} \right] \cup$$ {\sigma _d}\left( T \right)$. 故 ${\lambda _0} \notin {\sigma _b}\left( T \right)$,即 ${\lambda _0} \in {{\rm{\pi }}_{00}}(T)$. 證畢.

    由定理1可得,下列敘述等價:

    (1) $T$ 滿足 $\left( {{\omega _1}} \right)$ 性質;

    (2) $\sigma \left( T \right) \subseteq {\rho _1}\left( T \right) \cup \left\{ {\lambda \in {\bf{C}}:n\left( {T - \lambda I} \right)} \right. > \left. {d\left( {T - \lambda I} \right)} \right\} \cup \left[ {{\rho _a}\left( T \right) \cap \sigma \left( T \right)} \right]$${\sigma _w}\left( T \right) = {\sigma _b}\left( T \right)$;

    (3) $\sigma \left( T \right) = \partial {\sigma _1}\left( T \right) \cup \left\{ {\lambda \in {\bf{C}}:n\left( {T - \lambda I} \right)} \right. > \left. {d\left( {T - \lambda I} \right)} \right\} \cup \left\{ {\lambda \in \operatorname{int} {\rho _1}\left( T \right):n\left( T \right.\left. { - \lambda I} \right) = \infty } \right\} \cup \left. {\left[ {{\rho _a}\left( T \right)} \right. \cap \sigma \left( T \right)} \right] \cup $       $ {\sigma _d}\left( T \right) \cup{\sigma _0}\left( T \right) $,

    其中 ${\sigma _0}\left( T \right) = \sigma \left( T \right)\backslash {\sigma _b}\left( T \right)$.

    類似定理1的證明,我們可以得到 ${T^ * }$ 滿足 $\left( {{\omega _1}} \right)$ 性質的充要條件.

    推論1 設 $T \in B\left( H \right)$. ${T^ * }$ 滿足 $\left( {{\omega _1}} \right)$ 性質 $ \Leftrightarrow {\sigma _b}\left( T \right) =$    $ \partial {\sigma _1}\left( T \right) \cup \left\{ {\lambda \in {\bf{C}}:n\left( {T - \lambda I} \right)} \right. < d\left( T \right. - \left. {\left. {\lambda I} \right)} \right\} \cup \{ \lambda \in {\mathop{\rm int}} {\rho _1}\left( T \right):d\left( {T - \lambda I} \right) = \infty \} \cup \left[ {{\rho _s}\left( T \right) \cap \sigma \left( T \right)} \right]$.

    根據之前的結果,我們很自然地考慮了 $T$${T^ * }$ 均滿足 $\left( {{\omega _1}} \right)$ 性質時 ${\sigma _b}\left( T \right)$ 的結構.

    推論2 設 $T \in B\left( H \right)$. 則下列命題成立:

    (1)$T$${T^ * }$ 均滿足 $\left( {{\omega _1}} \right)$ 性質 $ \Leftrightarrow {\sigma _b}\left( T \right) = $    $ \partial {\sigma _1}\left( T \right) \cup \left\{ {\lambda \in \operatorname{int} {\rho _1}\left( T \right):d\left( {T - \lambda I} \right) = \infty } \right\} \cup \left[ {{\rho _a}} \right.\left( T \right)\left. { \cap \sigma \left( T \right)} \right] \cup\left[ {{\rho _s}\left( T \right) \cap \sigma \left( T \right)} \right]$;

    (2)若 $T$ 滿足 $\left( {{\omega _1}} \right)$ 性質. 則 ${T^ * }$ 滿足 $\left( {{\omega _1}} \right)$ 性質 $ \Leftrightarrow {\sigma _b}\left( T \right) = $    $ \partial {\sigma _1}\left( T \right) \cup \left\{ {\lambda \in {\bf{C}}:d\left( {T - \lambda I} \right) = \infty } \right\} \cup \left[ {{\rho _a}\left( T \right) \cap \sigma \left( T \right)} \right] \cup\left[ {{\rho _s}\left( T \right) \cap \sigma \left( T \right)} \right]$;

    (3)若 ${T^ * }$ 滿足 $\left( {{\omega _1}} \right)$ 性質. 則 $T$ 滿足 $\left( {{\omega _1}} \right)$ 性質 $ \Leftrightarrow {\sigma _b}\left( T \right) = $    $\partial {\sigma _1}\left( T \right) \cup \left\{ {\lambda \in {\bf{C}}:d\left( {{T^ * } - \bar \lambda I} \right) = \infty } \right\}\cup \left[ {{\rho _a}\left( T \right) \cap \sigma \left( T \right)} \right] \cup \left[ {{\rho _s}\left( T \right) \cap \sigma \left( T \right)} \right]$;

    (4)若 ${T^ * }$ 滿足 $\left( {{\omega _1}} \right)$ 性質. 則 $T$ 滿足 $\left( {{\omega _1}} \right)$ 性質 $ \Leftrightarrow {\sigma _b}\left( T \right) = $    $\partial {\sigma _1}\left( T \right) \cup \left\{ {\lambda \in {\bf{C}}:d\left( {{T^ * } - \bar \lambda I} \right) = \infty } \right\} \cup \left[ {{\rho _a}\left( T \right) \cap \sigma \left( T \right)} \right] \cup\left[ {{\rho _s}\left( T \right) \cap \sigma \left( T \right)} \right] \cup {\sigma _d}\left( T \right)$.

    證明?。?)根據定理1和推論1,可得    ${\sigma _b}( T ) = \{ \partial {\sigma _1}( T ) \cup \{ \lambda \in {\bf{C}}:n( {T - \lambda I} ) > d( {T - \lambda I} ) \} \cup \{ \lambda \in \operatorname{int} {\rho _1}( T ):$ $n( {T - \lambda I} ) = \infty \} \cup [ {\rho _a}( T ) \cap \sigma ( T ) ] \cup $    ${\sigma _d}( T )\}\cap \{ \partial {\sigma _1}( T ) \cup\{ \lambda \in {\bf{C}}:n( {T - \lambda I} ) < d {( T { - \lambda I} )} \} \cup \{ \lambda \in \operatorname{int} {\rho _1}( T ):d( {T - \lambda I} ) = \infty \} \cup [ {\rho _s}( T ) \cap\sigma ( T ) ] \} = $    $ \partial {\sigma _1}( T ) \cup $$ \{ \lambda \in \partial {\sigma _1}( T ):n( {T - \lambda I} ) < d( {T - \lambda I} ) \} \cup[ {{\rho _s}( T ) \cap \sigma ( T )} ] \cup $    $ \{ \lambda \in {\mathop{\rm int}} {\rho _1}( T ):R( {T - \lambda I} ){\text{不閉且}}n( {T - \lambda I} ) =d( {T - \lambda I} ) = \infty \} \cup $    $\{ {\lambda \in {\mathop{\rm int}} {\rho _1}( T ):R( {T - \lambda I} ){\text{閉且}}n( {T \!-\! \lambda I} ) \!=\! d( {T \!-\! \lambda I} ) \!=\! \infty } \} \cup \{ {\lambda \in {\mathop{\rm int}} {\rho _1}( T ):R( {T \!-\! \lambda I} ){\text{不閉且}}} n({T \!-\! \lambda I} ){\rm{ = }}\infty \} \cup$    $ [ {{\rho _a}( T ) \cap \sigma ( T )} ] \cup [ {\partial {\sigma _1}( T ) \cap {\sigma _d}( T )} ] \cup \{ {\lambda \in {\sigma _d}( T ):n( {T - \lambda I} ) < d( {T - \lambda I} )} \} \cup \left\{ \lambda \in {\mathop{\rm int}} {\rho _1}( T ):R( {T - \lambda I} )\right. $不閉且    $\left.d( {T - \lambda I} ) = \infty \right\} \subseteq $    $ \partial {\sigma _1}( T ) \cup\{ {\lambda \in {\mathop{\rm int}} {\rho _1}( T ):d( {T - \lambda I} ) = \infty } \} \cup [ {{\rho _a}} ( T ) { \cap \sigma ( T )} ] \cup [ {{\rho _s}( T ) \cap \sigma ( T )} ] \cup {\sigma _d}( T ){\rm{ = }}$    $ \partial {\sigma _1}( T ) \cup \{ {\lambda \in {\mathop{\rm int}} {\rho _1}( T ):d( {T - \lambda I} ) = \infty } \} \cup [ {{\rho _a}} ( T ) \cap\sigma ( T ) ] \cup [ {\rho _s}( T ) \cap \sigma ( T ) ] \cup $    $[ \partial {\sigma _1}( T ) \cap {\sigma _d}( T ) ] \cup [ \operatorname{int} {\rho _1}( T ) \cap {\sigma _d}( T ) ] = $    $ \partial {\sigma _1}( T ) \cup \{ \lambda \in \operatorname{int} {\rho _1}( T ): d( T - {\lambda I} ) = \infty \} \cup [ {\rho _a}( T ) \cap \sigma ( T ) ] \cup [ {{\rho _s}( T ) \cap \sigma ( T )} ] \subseteq {\sigma _b}( T )$. 因此可得

    $\qquad{\sigma _b}\left( T \right) = \partial {\sigma _1}\left( T \right) \cup \left\{ {\lambda \in \operatorname{int} {\rho _1}\left( T \right):d\left( {T - \lambda I} \right) = \infty } \right\} \cup \left[ {{\rho _a}\left( T \right) \cap \sigma \left( T \right)} \right] \cup \left[ {{\rho _s}\left( T \right) \cap \sigma \left( T \right)} \right].$

    反之,由所給條件可得     ${\sigma _b}\left( T \right) = \partial {\sigma _1}\left( T \right) \cup \left\{ {\lambda \in \operatorname{int} {\rho _1}\left( T \right):d\left( {T - \lambda I} \right) = \infty } \right\} \cup \left[ {{\rho _a}\left( T \right) \cap \sigma \left( T \right)} \right]\cup \left[ {{\rho _s}\left( T \right) \cap \sigma \left( T \right)} \right] \subseteq \partial {\sigma _1}\left( T \right) \cup$        $ \left\{ {\lambda \in {\mathop{\rm int}} {\rho _1}\left( T \right):R\left( {T - \lambda I} \right){\text{閉且}}n\left( {T - \lambda I} \right) = d\left( {T - \lambda I} \right) = \infty } \right\} \cup$        $ \left\{ {\lambda \in } {\mathop{\rm int}} {\rho _1}\left( T \right):R\left( {T - \lambda I} \right){\text{不閉且}}n\left( {T - \lambda I} \right) = d\left( {T - \lambda I} \right) =\right.\left. \infty \right\} \cup $        $\left\{ {\lambda \in {\mathop{\rm int}} {\rho _1}\left( T \right):R\left( {T - \lambda I} \right){\text{不閉且}}} \right.n ( T { - \lambda I}) < d\left( {T - \lambda I} \right) = \infty \} \cup $        $\left[ {{\rho _a}\left( T \right) \cap \sigma \left( T \right)} \right] \cup \{ \lambda \in {{\bf{C}}}:n\left( {T - \lambda I} \right) > d\left( {T - \lambda I} \right)\} \subseteq $        $\partial {\sigma _1}\left( T \right) \cup\{ \lambda \in{{\bf{C}}}:$$n\left( {T \!-\! \lambda I} \right)\left. { > {d\left( {T\! -\! \lambda I} \right)} } \right\} \cup \left\{ {\lambda \in {\mathop{\rm int}} {\rho _1}\left( T \right):n\left( {T \!-\! \lambda I} \right) \!=\! \infty } \right\} \cup $        $\left[ {{\rho _a}\left( T \right) \cap \sigma \left( T \right)} \right] \cup {\sigma _d}\left( T \right) \subseteq {\sigma _b}\left( T \right) $. 因此,由定理1可得 $T$ 滿足 $\left( {{\omega _1}} \right)$ 性質. 同樣地,由推論1我們可以證明 ${T^ * }$ 滿足 $\left( {{\omega _1}} \right)$ 性質.

    (2)根據推論2(1)的必要性,可得     ${\sigma _b}\left( T \right) = \partial {\sigma _1}\left( T \right) \cup \left\{ {\lambda \in \operatorname{int} {\rho _1}\left( T \right):d\left( {T - \lambda I} \right) = \infty } \right\} \cup \left[ {{\rho _a}} \left( T \right)\cap \sigma \left( T \right) \right] \cup \left[ {{\rho _s}\left( T \right) \cap \sigma \left( T \right)} \right] \subseteq$    $ \partial {\sigma _1}\left( T \right) \cup \left\{ {\lambda \in {\bf{C}}:d\left( {T - \lambda I} \right) = \infty } \right\} \cup \left[ {{\rho _a}} \right.\left( T \right)\left. { \cap \sigma \left( T \right)} \right] \cup \left[ {{\rho _s}\left( T \right) \cap } \right.$$\left. {\sigma \left( T \right)} \right]$,且其反包含顯然成立.

    反之,因為 $T$ 滿足 $\left( {{\omega _1}} \right)$ 性質,則由定理1可得     ${\sigma _b}\left( T \right) \!=\! \partial {\sigma _1}\left( T \right) \cup \left\{ {\lambda \in {\bf{C}}:n\left( {T \!-\! \lambda I} \right)} \right. > d\left( T \!-\! \left. {\lambda I}\right) \right\} \cup \{ \lambda \in$$ \operatorname{int} {\rho _1}\left( T \right):n\left( {T - \lambda I} \right) = \infty \} \cup \left[ {{\rho _a}\left( T \right) \cap \sigma \left( T \right)} \right] \cup {\sigma _d}\left( T \right)$. 由條件得     ${\sigma _b}\left( T \right) = \left\{ {\partial {\sigma _1}\left( T \right)} \right. \cup \left\{ {\lambda \in {\bf{C}}:n\left( {T - \lambda I} \right)} \right. > d\left( T \right. - \left. {\left. {\lambda I} \right)} \right\} \cup\{ \lambda \in \operatorname{int} {\rho _1}( T ):n( {T - \lambda I}) = \infty \} \cup $        $ [ {{\rho _a}( T ) \cap \sigma ( T )} ] \cup {\sigma _d}( T ) \} \cap\{ {\partial {\sigma _1}( T ) \cup \{ {\lambda \in {\bf{C}}:d( {T - \lambda I}) = \infty } \}} \cup $        $ [ {{\rho _a}( T ) \cap \sigma ( T )} ] \cup[ {\rho _s}( T ) \cap\sigma ( T ) ] \} = $        $\partial {\sigma _1}( T ) \cup $        $\{ {\lambda \in } \partial {\sigma _1}( T ):d\left( T \right. - \left. {\left. {\lambda I} \right) = \infty } \right\} \cup $        $\left[ {{\rho _s}\left( T \right) \cap \sigma \left( T \right)} \right] \cup \left[ {\partial {\sigma _1}\left( T \right) \cap {\sigma _d}\left( T \right)} \right] \cup $        $\left\{ {\lambda \in \operatorname{int} {\rho _1}\left( T \right):n\left( {T\!\! -\! \!\lambda I} \right) \!=\! d\left( {T \!\!-\!\! \lambda I} \right)} \right.\!=\! \infty \} \cup \{ \lambda \in {\sigma _d}( T ):d( T \!-\! {\lambda I}) \!=\! \infty \} \cup $        $ \left\{ {\lambda \in {\rho _a}\left( T \right) \cap \sigma \left( T \right):d\left( T \right. \!\!-\!\! \left. {\left. {\lambda I} \right) \!=\! \infty } \right\}} \right. \cup\left[ {{\rho _a}\left( T \right) \cap \sigma \left( T \right)} \right]\subseteq $        $ \partial {\sigma _1}\left( T \right) \cup \{ \lambda \in {\bf{C}}:n( {T - \lambda I} ) < d( T - {\lambda I} ) \} \cup \left\{ {\lambda \in \operatorname{int} {\rho _1}\left( T \right):d\left( {T - \lambda I} \right) = \infty } \right\} \cup $        $\left[ {{\rho _s}\left( T \right) \cap \sigma \left( T \right)} \right] \subseteq $${\sigma _b}\left( T \right)$. 因此,由推論1可得 ${T^ * }$ 滿足 $\left( {{\omega _1}} \right)$ 性質.

    類似于推論2(2)的證明,易得推論2(3)(4)成立. 證畢.

    接下來,我們討論 $T$ 滿足 $\left( \omega \right)$ 性質時的相關結論.

    定理2 若 ${T^ * }$ 滿足 $\left( {{\omega _1}} \right)$ 性質. 則 $T$ 滿足 $\left( \omega \right)$ 性質且為isoloid算子 $ \Leftrightarrow {\sigma _b}\left( T \right) = $    ${\left[ \rho \right._1}\left( T \right) \cap \left. {{\rm{acc}}\sigma \left( T \right)} \right] \cup\left\{ {\lambda \in {\bf{C}}:n\left( {T - \lambda I} \right)} \right. > \left. {d\left( {T - \lambda I} \right)} \right\} \cup \left. {\left\{ {\lambda \in {\bf{C}}:n\left( {T - \lambda I} \right)} \right. = \infty } \right\} \cup \left[ {{\rho _a}\left( T \right) \cap \sigma \left( T \right)} \right]$.

    證明 由推論2(3)可知,當 $T$${T^ * }$ 滿足 $\left( {{\omega _1}} \right)$ 性質時,有    ${\sigma _b}\left( T \right) = \partial {\sigma _1}\left( T \right) \cup \left\{ {\lambda \in {{\bf{C}}}:} \right.d\left( {{T^ * } - \bar \lambda I} \right)\left. { = \infty } \right\} \cup$$ \left[ {{\rho _a}\left( T \right) \cap \sigma \left( T \right)} \right] \cup \left[ {{\rho _s}\left( T \right) \cap \sigma \left( T \right)} \right]$. 因為     $\partial {\sigma _1}\left( T \right) \!=\!\! \left\{ {\lambda \in\! \partial {\rho _1}\left( T \right) \cap {\rm{iso}}\sigma \left( T \right):\!} \right.\!n\left. {\left( {T \!\!-\!\! \lambda I} \right) \!=\! \infty } \right\}\cup \{ \lambda \in\! \partial {\rho _1}\left( T \right) \cap {\rm{iso}}\sigma ( T ): $ $ n ( {T \!\!-\! \lambda I} ){\rm{<}}\infty \} \!\cup\! \left[ {\partial {\rho _1}\left( T \right) \!\cap \!{\rm{acc}}\sigma \left( T \right)} \right] \!\subseteq $        $\left\{ {\lambda \!\in\! {\bf{C}}:} \right.n\left. {\left( {T \!-\! \lambda I} \right) \!= \!\infty } \right\} \!\cup\! \left\{ {\lambda \!\in } \right.\partial {\rho _1}\left( T \right) \cap {\rm{iso}}\sigma \left( T \right):n\left. {\left( {T - \lambda I} \right){\rm{ < }}\infty } \right\} \cup \left[ {{\rho _1}\left( T \right) \cap {\rm{acc}}\sigma \left( T \right)} \right]$,并且有     $\left\{ {\lambda \!\in\! {\bf{C}}:d\left( {{T^ * } \!-\! \bar \lambda I} \right) \!=\! \infty } \right\} \!=\! \left\{ {\lambda \notin } \right.\left. {{\sigma _d}\left( T \right):n\left( {T - \lambda I} \right) \!=\! \infty } \right\} \cup \left\{ {\lambda \in {\sigma _d}\left( T \right):d\left( {{T^ * } - \bar \lambda I} \right) = \infty } \right\} \subseteq$    $ \left\{ {\lambda \notin } {\sigma _d}\left( T \right):n\left( {T - \lambda I} \right) =\right.\left. { \infty } \right\} \cup \left[ {{\rho _1}\left( T \right)} \right.\left. { \cap \sigma \left( T \right)} \right] \subseteq \left\{ {\lambda \in {\bf{C}}:n\left( {T - \lambda I} \right) = \infty } \right\} \cup$    $ \left[ {{\rho _1}\left( T \right) \cap } \right.{\rm{acc}}\left. {\sigma \left( T \right)} \right] \cup \left[ {{\rho _1}\left( T \right) \cap } \right.{\rm{iso}}\left. {\sigma \left( T \right)} \right] \subseteq $    $\left\{ {\lambda \in {\bf{C}}:n} \right.\left( T \right. -\left. {\left. {\lambda I} \right) = \infty } \right\} \cup \left[ {{\rho _1}\left( T \right) \cap } \right.{\rm{acc}}\left. {\sigma \left( T \right)} \right] \cup \left\{ {\lambda \in {\rho _1}\left( T \right) \cap {\rm{iso}}\sigma \left( T \right):} \right.n\left. {\left( {T - \lambda I} \right) < \infty } \right\} \cup $    $\left\{ {\lambda \in {\rho _1}\left( T \right) \cap {\rm{iso}}\sigma \left( T \right):} \right.n\left. {\left( {T - \lambda I} \right) = \infty } \right\} \subseteq $    $\{ \lambda \in {\bf{C}}:n( {T - \lambda I} ) = \infty \} \cup \left[ {{\rho _1}\left( T \right) \cap {\rm{acc}}\sigma \left( T \right)} \right] \cup \left\{ {\lambda \in {\rho _1}\left( T \right) \cap {\rm{iso}}} \right.\sigma \left( T \right):n\left( T \right. - $$\left. {\left. {\lambda I} \right)< \infty } \right\}.$

    $T$ 滿足 $\left( \omega \right)$ 性質且為isoloid算子,則有 $\left\{ {\lambda \in \partial {\rho _1}\left( T \right) \cap {\rm{iso}}\sigma \left( T \right):} \right.n\left. {\left( {T - \lambda I} \right){\rm{ < }}\infty } \right\} \cap {\sigma _b}\left( T \right) = \emptyset$,且 $ \left\{ {\lambda \in {\rho _1}\left( T \right) \cap} \right.$$\left. {\rm{iso}}\sigma \left( T \right):n {\left( {T \!-\! \lambda I} \right){\rm{ < }}\infty } \right\} \!\cap \!{\sigma _b}\left( T \right) \!=\! \emptyset $. 故有     ${\sigma _b}\left( T \right) \!=\! \left\{ {\partial {\sigma _1}\left( T \right)\! \cup\! } \right.\left\{ {\lambda \!\in\! {\bf{C}}:} \right.d\left( {{T^ * }} \right. \!-\! \bar \lambda \left. {\left. I \right)\left. { \!=\! \infty } \right\} \!\cup \!\left[ {{\rho _a}\left( T \right) \cap \sigma \left( T \right)} \right] \!\cup\! \left[ {{\rho _s}\left( T \right) \!\cap\! \sigma \left( T \right)} \right]} \right\} \!\cap {\sigma _b}\left( T \right) = $        $\left[ {{\rho _1}\left( T \right) \cap {\rm{acc}}\sigma \left( T \right)} \right] \cup \left\{ {\lambda \in {\bf{C}}:n} \right.\left( T \right. -\left. {\lambda I} \right) = \left. \infty \right\} \cup \left[ {{\rho _a}\left( T \right) \cap \sigma \left( T \right)} \right] \cup \left[ {{\rho _s}\left( T \right) \cap \sigma \left( T \right)} \right] \subseteq $        $\left[ {{\rho _1}\left( T \right) \cap {\rm{acc}}\sigma \left( T \right)} \right] \cup \{ \lambda \in {\bf{C}}:n( {T - \lambda I} ) > d( T $$\left. {\left. { - \lambda I} \right)} \right\} \cup \left\{ {\lambda \in {{\bf{C}}}:n\left( {T - \lambda I} \right){\rm{ = }}\infty } \right\} \cup \left[ {{\rho _a}\left( T \right) \cap \sigma \left( T \right)} \right] \subseteq $         ${\sigma _b}\left( T \right). $因此     ${\sigma _b}\left( T \right) = \left[ {{\rho _1}\left( T \right) \cap {\rm{acc}}\sigma \left( T \right)} \right] \cup \{ \lambda \in {{\bf{C}}}: $$n( {T - \lambda I} ) > d( {T - \lambda I} ) \} \cup \{ \lambda \in {{\bf{C}}}:n( T - \lambda I) = \infty \} \cup [ {\rho _a}( T ) \cap \sigma ( T ) ]$.

    反之,由于有    ${\rho _1}\left( T \right) \cap {\rm{acc}}\sigma \left( T \right) = \left[ {\partial {\rho _1}\left( T \right) \cap {\rm{acc}}\sigma \left( T \right)} \right] \cup \left[ {\operatorname{int} {\rho _1}\left( T \right) \cap {\rm{acc}}\sigma \left( T \right)} \right] \subseteq $    $\partial {\sigma _1}\left( T \right) \cup\left[ {{\rho _w}} \right.\left. {\left( T \right) \cap {\rm{acc}}\sigma \left( T \right)} \right] \cup \left[ {{\sigma _d}\left( T \right) \!\cap \!{\rm{acc}}\sigma \left( T \right)} \right] \cup \left\{ {\lambda \!\in\! {\rm{acc}}\sigma \left( T \right)\backslash {\sigma _d}\left( T \right):n\left( {T \!-\! \lambda I} \right) \!=\! d\left( {T \!-\! \lambda I} \right) \!=\! \infty } \right\}\subseteq $    $ \partial {\sigma _1}\left( T \right) \cup \left[ {{\rho _w}\left( T \right) \cap {\rm{acc}}\sigma \left( T \right)} \right] \cup {\sigma _d}\left( T \right) \cup \left\{ {\lambda \in {\bf{C}}:} \right.$$d\left( {{T^ * } \!-\! \bar \lambda I} \right)\left. { \!=\! \infty } \right\} $,    $\left\{ {\lambda \!\in\! {\bf{C}}\!:\!n\left( {T - \lambda I} \right) \!=\! \infty } \right\} \!=\! \left\{ \lambda \right.\left. { \notin {\sigma _d}\left( T \right)\!:\!d\left( {{T^ * } \!-\! \bar \lambda I} \right) \!=\! \infty } \right\} \cup \left\{ {\lambda \!\in \!{\sigma _d}\left( T \right)\!:\!} \right.n\left( {T \!-\! \lambda I} \right)\left. { \!=\! \infty } \right\} \subseteq $    $\left\{ {\lambda \in {\bf{C}}\!:\!} \right.d\left( {{T^ * } \!-\! \bar \lambda I} \right)=\left. { \infty } \right\} \cup {\sigma _d}\left( T \right)$,并且    $\left\{ {\lambda \in {\bf{C}}:} \right.n\left( {\left. {T \!-\! \lambda I} \right)} \right.\left. { > d\left( {T \!-\! \lambda I} \right)} \right\} \!=\! \left\{ {\lambda \in {\bf{C}}:n\left( {T \!-\! \lambda I} \right)} \right. > \left. {d\left( {T \!-\! \lambda I} \right) \!=\! 0} \right\} \cup \left\{ {\lambda \in {\bf{C}}} \right.:n\left( {T \!-\! \lambda I} \right) > d\left( T \right. \!-\!\left. {\left. {\lambda I} \right) > 0} \right\} = $    $ \left[ {{\rho _s}\left( T \right) \cap \sigma \left( T \right)} \right] \cup \left\{ {\lambda \in {\bf{C}}:n\left( {T - \lambda I} \right)} \right. > \left. {d\left( {T - \lambda I} \right) > 0} \right\}$. 由 ${T^ * }$ 滿足 $\left( {{\omega _1}} \right)$ 性質,可得     $\{ \lambda \in {\bf{C}}:n\left( T- { \lambda I} \right) > $$ d\left( {T - \lambda I} \right) > 0\} \cap {\sigma _b}\left( T \right) = \emptyset$ 且     $\left[ {{\rho _w}\left( T \right) \cap {\rm{acc}}\sigma \left( T \right)} \right] \cap {\sigma _b}\left( T \right) = \emptyset .$ 因此,由所給條件可得     ${\sigma _b}\left( T \right) = \{ {{[ \rho }_1}\left( T \right) \cap $${\rm{acc}}\sigma ( T ) ] \cup \{ \lambda \in {\bf{C}}:n( {T - \lambda I} ) > d( {T - \lambda I} ) \} \cup \{ {\lambda \in } {\bf{C}}: {n( {T - \lambda I} ) = \infty } \} \cup \left[ {{\rho _a}} \right.\left( T \right)\left. {\left. { \cap \sigma \left( T \right)} \right] } \right\} \cap $        ${\sigma _b}\left( T \right) =\partial {\sigma _1}\left( T \right) \cup \left\{ {\lambda \in {\bf{C}}:} \right. d( {T^ * } - \bar \lambda I )\left. { = \infty } \right\} \cup \left[ {{\rho _a}\left( T \right) \cap \sigma \left( T \right)} \right] \cup \left[ {{\rho _s}\left( T \right) \cap } \right.$$\left. {\sigma \left( T \right)} \right] \cup {\sigma _d}\left( T \right)$. 則由推論2(4)可得 $T$ 滿足 $\left( {{\omega _1}} \right)$ 性質.

    并且,假設 ${\lambda _0} \in {{\rm{\pi }}_{00}}\left( T \right)$. 若 ${\lambda _0} \!\in\! \left\{ {\lambda \!\in\! {\bf{C}}\!:\!n\left( {T \!-\! \lambda I} \right)} \right. \! > \! \left. {d\left( {T \!-\! \lambda I} \right)} \right\}$,則 ${\lambda _0} \in {\rho _b}\left( T \right)$. 若 ${\lambda _0} \notin \left\{ {\lambda \in } \right.$${\bf{C}}:n\left( {T - \lambda I} \right) > \left. {d\left( {T - \lambda I} \right)} \right\}$,則 ${\lambda _0} \notin \left. {{{\left[ \rho \right.}_1}\left( T \right) \cap {\rm{acc}}\sigma \left( T \right)} \right] \cup \left\{ {\lambda \in {\bf{C}}:n\left( {T \!-\! \lambda I} \right)} \right. > \left. {d\left( {T \!-\! \lambda I} \right)} \right\} \cup \left\{ \lambda \right. \in $${\bf{C}}:\left. {n\left( {T \!-\! \lambda I} \right) = \infty } \right\} \cup \left[ {{\rho _a}\left( T \right) \cap \sigma \left( T \right)} \right]$,故 ${\lambda _0} \in {\rho _b}\left( T \right) \subseteq $${\;\rho _{ea}}\left( T \right) $. 因此 ${{\rm{\pi }}_{00}}\left( T \right) \subseteq {\sigma _a}\left( T \right)\backslash {\sigma _{ea}}\left( T \right)$.

    $\left\{ {\lambda \in {\rm{iso}}\sigma{ \left( T \right):n\left( {T \!-\! \lambda I} \right) = 0}} \right\} \cap \left\{ {\left. {{{\left[ \rho \right.}_1}\left( T \right) \cap {\rm{acc}}\sigma \left( T \right)} \right] \cup \left\{ {\lambda \in {\bf{C}}:n\left( {T \!-\! \lambda I} \right)} \right. > \left. {d\left( {T - \lambda I} \right)} \right\}} \right. \cup \{ {\lambda \in } {\bf{C}}: n( {T \!-\! \lambda I} ) \!=\! \infty \} \cup $$[ {\rho _a}( T ) \cap \sigma ( T ) ]\} = \emptyset $,可得 ${\rm{iso}}\sigma \left( T \right) \subseteq {\sigma _p}\left( T \right)$. 這就證明了 $T$ 滿足 $\left( \omega \right)$ 性質且為isoloid算子. 證畢.

    注1?。?)若 $T$ 滿足 $\left( \omega \right)$ 性質且為isoloid算子,則有     ${\sigma _b}\left( T \right) = {\left[ \rho \right._1}\left( T \right)\left. { \cap {\rm{acc}}\sigma \left( T \right)} \right] \cup \left\{ {\lambda \in {\bf{C}}:} \right.n\left( {T - \lambda I} \right) > $$\left. {d\left( {T - \lambda I} \right)} \right\} \cup \left. {\left\{ {\lambda \in {\bf{C}}:n\left( {T - \lambda I} \right)} \right. = \infty } \right\} \cup \left[ {{\rho _a}\left( T \right) \cap \sigma \left( T \right)} \right]$.

    (2)定理2中條件“${T^ * }$ 滿足 $\left( {{\omega _1}} \right)$ 性質”是本質的. 例如,設 ${T_1},{T_2} \in B\left( {{\ell ^2}} \right)$ 定義為:

    $\qquad{T_1}\left( {{x_1},{x_2},{x_3}, \cdots } \right) = \left( {0,{x_1},{x_2},{x_3}, \cdots } \right),$

    $\qquad{T_2}\left( {{x_1},{x_2},{x_3}, \cdots } \right) = \left( {{x_2},{x_3},{x_4}, \cdots } \right),$

    并令 $T = {T_1} \oplus {T_2}$,則    ${\sigma _b}\left( T \right) = {\left[ \rho \right._1}\left( T \right)\left. { \cap {\rm{acc}}\sigma \left( T \right)} \right] \cup \left\{ {\lambda \in {\bf{C}}:} \right.n\left( {T - \lambda I} \right) > \left. {d\left( {T - \lambda I} \right)} \right\} \cup \left\{ {\lambda \in {\bf{C}}:} \right.n( {T - \lambda I} ) = \infty \} \cup \left[ {\rho _a}( T ) \cap\right.$$\left. \sigma ( T ) \right]$${T^ * }$ 不滿足 $\left( {{\omega _1}} \right)$ 性質,但 $T$ 不滿足 $\left( \omega \right)$ 性質.

    根據定理2,我們可以得到以下推論:

    推論3 若 $T$ 滿足 $\left( {{\omega _1}} \right)$ 性質. 則 ${T^ * }$ 滿足 $\left( \omega \right)$ 性質且為isoloid算子 $ \Leftrightarrow {\sigma _b}\left( T \right) = $    ${\left[ \rho \right._1}\left( T \right) \cap \left. {{\rm{acc}}\sigma \left( T \right)} \right] \cup \left\{ {\lambda \in {\bf{C}}:n\left( {{T^ * } - \bar \lambda I} \right)} \right. > \left. {d\left( {{T^ * } - \bar \lambda I} \right)} \right\} \cup \left. {\left\{ {\lambda \in {\bf{C}}:n\left( {{T^ * } - \bar \lambda I} \right)} \right. = \infty } \right\} \cup \left[ {{\rho _s}\left( T \right) \cap \sigma \left( T \right)} \right]$.

    推論4 設 $T \in B\left( H \right)$. ${T^ * }$ 滿足 $\left( {{\omega _1}} \right)$ 性質,$T$ 滿足 $\left( \omega \right)$ 性質且 $T$ 為isoloid算子 $ \Leftrightarrow {\sigma _b}\left( T \right) = $    ${\left[ \rho \right._1}\left( T \right) \cap {\sigma _w}\left( T \right)\cap\left. { {\rm{acc}}\sigma \left( T \right)} \right] \cup \left\{ {\lambda \in {\rm{iso}}\sigma \left( T \right):n\left( {T - \lambda I} \right)} \right.\left. { = \infty } \right\} \cup \left[ {{\rho _a}\left( T \right) \cap \sigma \left( T \right)} \right] \cup \left[ {{\rho _s}\left( T \right) \cap \sigma \left( T \right)} \right]$.

    證明 因為     $\left\{ {\lambda \in {\bf{C}}:n\left( {T - \lambda I} \right)} \right.\left. { = \infty } \right\} = \left\{ {\lambda \in {\rm{iso}}\sigma \left( T \right):n\left( {T - \lambda I} \right)} \right.\left. { = \infty } \right\} \cup $              $\left\{ {\lambda \in {\rm{acc}}\sigma \left( T \right):d\left( {T - \lambda I} \right)} \right. < n( T - \lambda I)=\left. {\infty } \right\} \cup $              $\left\{ {\lambda \in {\rm{acc}}\sigma} \right. \left( T \right): d\left( {T - \lambda I} \right)=n\left( {T - \lambda I} \right)\left. { = \infty } \right\} $,根據定理2可得     ${\sigma _b}\left( T \right) \!=\! {\left[ \rho \right._1}\left( T \right) \cap \left. {{\rm{acc}}\sigma \left( T \right)} \right] \cup \left\{ {\lambda \in {\bf{C}}:n\left( {T \!\!-\! \lambda I} \right)} \right. \!>\!\left. {d\left( {T \!\!-\!\! \lambda I} \right)} \right\} \cup \left. {\left\{ {\lambda \in {\bf{C}}:n\left( {T - \lambda I} \right)} \right. \!=\! \infty } \right\} \cup\left[ {{\rho _a}\left( T \right) \cap \sigma \left( T \right)} \right] \subseteq $        $ {\left[ \rho \right._1}\left( T \right)\left. { \cap {\rm{acc}}\sigma \left( T \right)} \right] \cup $        $\left\{ {\lambda \in {\bf{C}}:n\left( {T \!\!-\!\! \lambda I} \right)} \right. > \left. {d\left( {T \!-\! \lambda I} \right)} \right\} \cup\left\{ {\lambda \in {\rm{iso}}\sigma \left( T \right):} \right. \left.n\left( {T - \lambda I} \right)= { \infty } \right\} \cup \left[ {{\rho _a}\left( T \right) \cap \sigma \left( T \right)} \right] \subseteq $        ${\left[ \rho \right._1}\left( T \right) \cap {\sigma _w}\left( T \right)\left. { \cap {\rm{acc}}\sigma \left( T \right)} \right] \cup {\left[ \rho \right._1}\left( T \right) \cap {\rho _w}\left( T \right)\left. { \cap {\rm{acc}}\sigma \left( T \right)} \right] \cup \left\{ {\lambda \in {\bf{C}}:n\left( {T - \lambda I} \right)} \right.> $$\left. { d\left( {T - \lambda I} \right) > 0} \right\} \cup $        $\left\{ {\lambda \in {\rm{iso}}\sigma \left( T \right):n\left( {T - \lambda I} \right)} \right.\left. { = \infty } \right\} \cup \left[ {{\rho _a}\left( T \right) \cap \sigma \left( T \right)} \right] \cup \left[ {{\rho _s}\left( T \right) \cap \sigma \left( T \right)} \right]$. 由于 $T$${T^ * }$ 滿足 $\left( {{\omega _1}} \right)$ 性質,可得     ${\;\rho _1}\left( T \right) \cap {\rho _w}\left( T \right) \cap {\rm{acc}}\sigma \left( T \right) = \emptyset $ 以及     $\left\{ {\lambda \in {\bf{C}}:n\left( {T - \lambda I} \right)} \right.\left. { > d\left( {T - \lambda I} \right) > 0} \right\} = \emptyset $. 從而    ${\sigma _b}\left( T \right) \subseteq {\left[ \rho \right._1}\left( T \right) \cap {\sigma _w}\left( T \right)\cap $$\left. { {\rm{acc}}\sigma \left( T \right)} \right] \cup \left\{ {\lambda \in {\rm{iso}}\sigma \left. {\left( T \right):n\left( {T - \lambda I} \right) = \infty } \right\}} \right. \cup \left[ {{\rho _a}\left( T \right) \cap \sigma \left( T \right)} \right] \cup \left[ {{\rho _s}\left( T \right) \cap \sigma \left( T \right)} \right].$ 其反包含顯然成立,因此等式成立.

    反之,由所給條件可得     ${\sigma _b}\left( T \right) = {\left[ \rho \right._1}\left( T \right) \cap {\sigma _w}\left( T \right)\left. { \cap {\rm{acc}}\sigma \left( T \right)} \right] \cup \left\{ {\lambda \in {\rm{iso}}\sigma \left. {\left( T \right):n\left( {T - \lambda I} \right) = \infty } \right\}} \right. \cup \left[ {{\rho _a}\left( T \right) \cap \sigma \left( T \right)} \right] \cup\left[ {{\rho _s}\left( T \right) \cap \sigma \left( T \right)} \right] \subseteq$        $ {\left[ {\partial \rho } \right._1}\left( T \right) \cap {\sigma _w}\left( T \right)\left. { \cap {\rm{acc}}\sigma \left( T \right)} \right] \cup {\left[ {\operatorname{int} \rho } \right._1}\left( T \right) \cap {\sigma _w}\left( T \right) \cap {\rm{acc}}\left. {\sigma \left( T \right)} \right] \cup $        $\{ \lambda \in {\rho _1} ( T ):n( {T - \lambda I} ) =d\left( {T - \lambda I} \right) = \infty \} \cup \left[ {{\rho _a}\left( T \right) \cap \sigma \left( T \right)} \right] \cup \left[ {{\rho _s}\left( T \right) \cap \sigma \left( T \right)} \right] \subseteq $        $\partial {\rho _1}\left( T \right) \cup \{ \lambda \in \operatorname{int} {\rho _1} ( T ):d( {T - \lambda I} ) = \infty \} \cup [ {{\rho _a}( T ) \cap \sigma ( T )} ] \cup [ {\rho _s}( T ) \cap \sigma ( T ) ] \subseteq {\sigma _b}\left( T \right). $根據推論2中(1)即可得 $T$${T^ * }$ 滿足 $\left( {{\omega _1}} \right)$ 性質. 再者,由     ${{\rm{\pi }}_{00}}\!\left( T \right) \!\cap\! \left\{ {{{\left[ \rho \right.}_1}\!\!\left( T \right) \!\cap\! {\sigma _w}\!\left( T \right)\left. { \!\cap {\rm{acc}}\sigma \!\left( T \right)} \right] \!\cup\! } \right.\left\{ {\lambda \in {\rm{iso}}\sigma } \right.\!\left( T \right)\!:$$\left. {n\left( T \right. \!\!-\!\! \left. {\lambda I} \right) = \infty } \right\} \cup \left[ {{\rho _a}\left( T \right) \cap \sigma \left( T \right)} \right] \cup \left. {\left[ {{\rho _s}\left( T \right) \cap \sigma \left( T \right)} \right]} \right\} \!=\! \emptyset $,即 ${{\rm{\pi }}_{00}}\left( T \right) \subseteq {\rho _b}\left( T \right) \subseteq {\rho _{ea}}\left( T \right)$. 因此可得 $T$ 滿足 $\left( \omega \right)$ 性質.

    ${\lambda _0} \in {\rm{iso}}\sigma \left( T \right)$. 若 $n\left( {T - {\lambda _0}I} \right) = 0$,則 ${\lambda _0} \notin {\left[ \rho \right._1}\left( T \right) \cap {\sigma _w}\left( T \right)\left. { \cap {\rm{acc}}\sigma \left( T \right)} \right] \cup \left\{ {\lambda \in } \right.{\rm{iso}}\sigma \left( T \right):n\left( T \right.\left. {\left. { - \lambda I} \right) = \infty } \right\} \cup \left[ {{\rho _a}\left( T \right) \cap } \right.$$\left. {\sigma \left( T \right)} \right] \left[ {{\rho _a}\left( T \right) \cap \sigma \left( T \right)} \right] \cup \left[ {{\rho _s}\left( T \right) \cap \sigma \left( T \right)} \right]$. 故有 ${\lambda _0} \in {\rho _b}\left( T \right) \subseteq {\rho _w}\left( T \right)$,即 ${\lambda _0} \in \rho \left( T \right)$,這與 ${\lambda _0} \in {\rm{iso}}\sigma \left( T \right)$ 矛盾. 因此,${\rm{iso}}\sigma \left( T \right) \subseteq $${\sigma _p}\left( T \right)$,即 $T$ 為isoloid算子. 由此,${T^ * }$ 滿足 $\left( {{\omega _1}} \right)$ 性質,$T$ 滿足 $\left( \omega \right)$ 性質且 $T$ 為isoloid算子得證. 證畢.

    在定理2中,如果我們去掉條件“${T^ * }$ 滿足 $\left( {{\omega _1}} \right)$ 性質”,則可得到以下結論.

    定理3 設 $T \in B\left( H \right)$. $T$ 滿足 $\left( \omega \right)$ 性質且為isoloid算子 $ \Leftrightarrow {\sigma _b}\left( T \right) = $    ${\left[ \rho \right._1}\left( T \right) \cap {\sigma _w}\left( T \right) \cap \left. {{\rm{acc}}\sigma \left( T \right)} \right] \cup \{ \lambda \in {\bf{C}}:n\left( {T - \lambda I} \right) \left. { > d\left( {T - \lambda I} \right)} \right\} \cup \left\{ {\lambda \in {\bf{C}}:n\left( {T - \lambda I} \right)} \right.\left. { = \infty } \right\} \cup \left[ {{\rho _a}\left( T \right) \cap \sigma \left( T \right)} \right]$.

    證明 由定理2可得     $ {\sigma _b}\left( T \right) = {\left[ \rho \right._1}\left( T \right) \cap \left. {{\rm{acc}}\sigma \left( T \right)} \right] \cup \left\{ {\lambda \in {{\bf{C}}}:n\left( {T - \lambda I} \right)} \right. > \left. {d\left( {T - \lambda I} \right)} \right\} \cup $        $ \left\{ \lambda \right. \in \left. {{\bf{C}}:n\left( {T - \lambda I} \right) = \infty } \right\} \cup\left[ {{\rho _a}\left( T \right) \cap \sigma \left( T \right)} \right] =$        $ {\left[ \rho \right._1}\left( T \right) \cap {\sigma _w}\left( T \right)\left. { \cap {\rm{acc}}\sigma \left( T \right)} \right] \cup {\left[ \rho \right._1}\left( T \right) \cap {\rho _w}\left( T \right)\left. { \cap {\rm{acc}}\sigma \left( T \right)} \right] \cup \left\{ {\lambda \in {\bf{C}}:n\left( {T - \lambda I} \right)} \right.\left. { > d\left( {T - \lambda I} \right)} \right\} \cup $        $\left\{ \lambda \in {\bf{C}}: \right.n( {T - \lambda I} ) = \infty \} \cup [ {\rho _a}( T ) \cap \sigma ( T ) ]$. 由 $T$ 滿足 $\left( \omega \right)$ 性質,可得 ${\rho _1}\left( T \right) \cap {\rho _w}\left( T \right) \cap {\rm{acc}}\sigma \left( T \right) = \emptyset $. 因此    ${\sigma _b}\left( T \right) = $$ {\left[ \rho \right._1}\left( T \right) \cap {\sigma _w}\left( T \right) \cap \left. {{\rm{acc}}\sigma \left( T \right)} \right] \cup \left\{ {\lambda \in {\bf{C}}:} \right.n$$\left. {\left( {T - \lambda I} \right) > d\left( {T - \lambda I} \right)} \right\} \cup \left\{ {\lambda \in {{\bf{C}}}:n\left( {T - \lambda I} \right)} \right.\left. { = \infty } \right\} \cup $        $\left[ {{\rho _a}\left( T \right) \cap \sigma \left( T \right)} \right] $.

    反之,由于     $[ {{\sigma _a}( T )\backslash {\sigma _{ea}}( T )} ] \cap \{ { {{[ \rho }_1}( T ) \cap {\sigma _w}( T ) \cap {{\rm{acc}}\sigma ( T )} ]} \cup \{ {\lambda \in {{\bf{C}}}:n( {T - \lambda I} )} > d( T - { {\lambda I} )} \} \cup $    $\{ {\lambda \in {\bf{C}}:n( {T - \lambda I} )= }{ \infty } \} \cup [ {{\rho _a}( T ) \cap \sigma ( T )} ] \} = \emptyset $,故 ${\sigma _a}\left( T \right)\backslash {\sigma _{ea}}\left( T \right) \subseteq {\rho _b}\left( T \right)$,即 ${\sigma _a}\left( T \right)\backslash {\sigma _{ea}}\left( T \right) \subseteq {{\rm{\pi }}_{00}}\left( T \right)$. 此外,設 ${\lambda _0} \in {{\rm{\pi }}_{00}}\left( T \right)$. 若 ${\lambda _0} \in \left\{ {\lambda \in {\bf{C}}:n\left( {T - \lambda I} \right)}> \right. d\left( {T - } \right. $$\left. {\left. {\lambda I} \right)} \right\}$,則 ${\lambda _0} \in {\rho _b}\left( T \right)$. 若 ${\lambda _0} \notin \left\{ {\lambda \in {\bf{C}}:n\left( {T - \lambda I} \right)} \right.\left. { > d\left( {T - \lambda I} \right)} \right\}$,則     $ {\lambda _0} \notin {[ \rho _1}( T ) \cap {\sigma _w}( T ) \cap $$ {{\rm{acc}}\sigma ( T )} ] \cup \left\{ {\lambda \in {\bf{C}}:n\left( {T \!-\! \lambda I} \right)} \right.$$\left. { > d\left( {T \!-\! \lambda I} \right)} \right\} \cup \left\{ {\lambda \in {\bf{C}}:n\left( {T\! -\! \lambda I} \right)} \right.\left. { = \infty } \right\} \cup \left[ {{\rho _a}\left( T \right) \cap \sigma \left( T \right)} \right]$,即 ${\lambda _0} \in {\rho _b}\left( T \right) \subseteq {\rho _{ea}}\left( T \right)$. 因此 ${{\rm{\pi }}_{00}}\left( T \right)$$ \subseteq {\sigma _a}\left( T \right)\backslash {\sigma _{ea}}\left( T \right)$. 并且,我們知道     $\left\{ {\lambda \!\in\! {\rm{iso}}\sigma \left( T \right):n\left( {T \!-\! \lambda I} \right) \!=\! 0} \right\} \!\cap\! \left\{ {{{\left[ \rho \right.}_1}\left( T \right) \!\cap\! {\sigma _w}\left( T \right) \!\cap\! \left. {{\rm{acc}}\sigma \left( T \right)} \right]} \right. \!\cup\! \left\{ {\lambda \!\in\! {{\bf{C}}}:n\left( {T \!-\! \lambda I} \right)} \right. \! >\left. { d\left( {T - \lambda I} \right)} \right\} \cup $    $ \left\{ {\lambda \in {{\bf{C}}}:n\left( {T - \lambda I} \right)} \right.\left. { = \infty } \right\}\left. { \cup \left[ {{\rho _a}\left( T \right) \cap \sigma \left( T \right)} \right]} \right\}{\rm{ = }}\emptyset $,由此即可得 $\left\{ {\lambda \in {\rm{iso}}\sigma \left( T \right):n\left( {T - \lambda I} \right) = 0} \right\} \subseteq {\rho _b}\left( T \right)$. 在這種情況下,$\left\{ {\lambda \in {\rm{iso}}\sigma \left( T \right):n\left( {T - \lambda I} \right) = 0} \right\} = \emptyset $. 因此 ${\rm{iso}}\sigma \left( T \right) \subseteq {\sigma _p}\left( T \right)$. 這就證明了 $T$ 滿足 $\left( \omega \right)$ 性質且為isoloid算子. 證畢.

    對于 $T$ 的共軛算子 ${T^ * }$,類似定理3的證明,可得以下推論.

    推論 5 設 $T \in B\left( H \right)$. ${T^ * }$ 滿足 $\left( \omega \right)$ 性質且為isoloid算子 $ \Leftrightarrow {\sigma _b}\left( T \right) = $    $ {\left[ \rho \right._1}\left( T \right) \cap {\sigma _w}\left( T \right) \cap \left. {{\rm{acc}}\sigma \left( T \right)} \right] \cup \left\{ {\lambda \in {\bf{C}}:n\left( {T^ * }\!-\! \right.} \right.\left.\bar \lambda I \right)\left. {> d\left( {{T^ * } \!-\! \bar \lambda I} \right)} \right\} \cup \left\{ {\lambda \in {\bf{C}}:n\left( {{T^ * } \!-\! \bar \lambda I} \right)} \right.\left. { \!=\! \infty } \right\} \cup \left[ {{\rho _s}\left( T \right) \cap \sigma \left( T \right)} \right] $.

    推論 6 設 $T \in B\left( H \right)$. $T$${T^ * }$ 均滿足 $\left( \omega \right)$ 性質且均為isoloid算子 $ \Leftrightarrow {\sigma _b}\left( T \right) = $    ${\left[ \rho \right._1}\left( T \right) \cap \left. {{\sigma _w}\left( T \right) \cap {\rm{acc}}\sigma \left( T \right)} \right] \cup \left\{ \lambda \in {\bf{C}}:n\left( {T - \lambda I} \right) = n\left. {\left( {{T^ * } - \bar \lambda I} \right) = \infty } \right\} \right. \cup \left[ {{\rho _a}\left( T \right) \cap \sigma \left( T \right)} \right] \cup \left[ {{\rho _s}\left( T \right) \cap \sigma \left( T \right)} \right] $.

    推論 7 設 $T \in B\left( H \right)$. $T$ 滿足 $\left( \omega \right)$ 性質 $ \Leftrightarrow {\sigma _b}\left( T \right) = {\left[ \rho \right._1}\left( T \right) \cap {\sigma _w}\left( T \right) \cap \left. {{\rm{acc}}\sigma \left( T \right)} \right] \cup \left\{ {\lambda \in {{\bf{C}}}:} \right.\left. {n\left( {T - \lambda I} \right) > d\left( {T - \lambda I} \right)} \right\} \cup $$\left\{ {\lambda \in {\bf{C}}:n\left( {T - \lambda I} \right)} { = \infty } \right\} \cup \left\{ {\lambda \in \sigma \left( T \right):n\left( {T - \lambda I} \right) = 0} \right\} $.

    在文獻[15]中,介紹了如下引理.

    引理 1 設 $T \in B\left( H \right)$. 若 $K \in B\left( H \right)$ 是一個可交換的冪有限秩算子,則:

    (1)$K$ 是一個Riesz算子;

    (2)$n\left( {T + K} \right) < \infty \Leftrightarrow n\left( T \right) < \infty $;

    (3)${\rm{iso}}\sigma \left( {T + K} \right) \subseteq {\rm{iso}}\sigma \left( T \right) \cup \rho \left( T \right)$

    (4)${\rm{iso}}{\sigma _a}\left( {T + K} \right) \subseteq {\rm{iso}}{\sigma _a}\left( T \right) \cup {\rho _a}\left( T \right)$.

    顯然,若 $T$ 是一個有限秩算子,則 $T$ 是一個冪有限秩算子.

    定理4 設 $T \in B\left( H \right)$. 則以下命題等價:

    (1) ${\sigma _b}\left( T \right) = {\left[ \rho \right._1}\left( T \right) \cap {\sigma _w}\left( T \right) \cap \left. {{\rm{acc}}\sigma \left( T \right)} \right] \cup \left\{ {\lambda \in {\bf{C}}:} \right.\left. {n\left( {T - \lambda I} \right) > d\left( {T - \lambda I} \right)} \right\} \cup \left\{ {\lambda \in {\bf{C}}:} \right.n( T - \left. {\lambda I) = \infty } \right\}$;

    (2) 對任意可交換的有限秩算子 $F \in B\left( H \right)$,有 ${\sigma _a}\left( T \right) = \sigma \left( T \right)$,并有 $T + F$ 滿足 $\left( \omega \right)$ 性質且為isoloid算子.

    證明 $ (1) \Rightarrow (2)$. 由定理3可知 $T$ 滿足 $\left( \omega \right)$ 性質且為isoloid算子. 設 ${\lambda _0} \in {\rho _a}\left( T \right) \cap \sigma \left( T \right)$,則有 ${\lambda _0} \notin {\left[ \rho \right._1}\left( T \right) \cap {\sigma _w}\left( T \right) \cap \left. {{\rm{acc}}\sigma \left( T \right)} \right] \cup \left\{ {\lambda \in {\bf{C}}:} \right.\left. {n\left( {T - \lambda I} \right) > d\left( {T - \lambda I} \right)} \right\} \cup \left\{ {\lambda \in {\bf{C}}:} \right.n\left( T \right. - \left. {\lambda I} \right) = \left. \infty \right\}$,故 ${\lambda _0} \in {\rho _b}\left( T \right)$. 這與 ${\lambda _0} \in {\rho _a}\left( T \right) \cap \sigma \left( T \right)$ 矛盾,所以 $\sigma \left( T \right) \subseteq {\sigma _a}\left( T \right)$. 又顯然有 $\sigma \left( T \right) \supseteq {\sigma _a}\left( T \right)$,因此 ${\sigma _a}\left( T \right) = \sigma \left( T \right)$. 此外,設 ${\lambda _0} \in {\sigma _a}\left( {T + F} \right)\backslash {\sigma _{ea}}\left( {T + F} \right)$. 若 ${\lambda _0} \in {\rho _a}\left( T \right)$,則由 ${\sigma _a}\left( T \right) = \sigma \left( T \right)$ 可得 ${\lambda _0} \in \rho \left( T \right) \subseteq {\rho _b}\left( T \right) = {\rho _b}\left( {T + F} \right)$,即 ${\lambda _0} \in $${{\rm{\pi }}_{00}}\left( {T + F} \right) $;若 ${\lambda _0} \in {\sigma _a}\left( T \right)$,則 ${\lambda _0} \in {\sigma _a}\left( T \right)\backslash {\sigma _{ea}}\left( T \right)$. 由 $T$ 滿足 $\left( \omega \right)$ 性質,有 ${\lambda _0} \in {\rho _b}\left( T \right) = {\rho _b}\left( {T + F} \right)$,即 ${\lambda _0} \in {{\rm{\pi }}_{00}}\left( {T + F} \right)$. 故 ${\sigma _a}\left( {T + F} \right)\backslash {\sigma _{ea}}\left( {T + F} \right) \subseteq {{\rm{\pi }}_{00}}\left( {T + F} \right)$. 設 ${\lambda _0} \in {{\rm{\pi }}_{00}}\left( {T + F} \right)$,則 ${\lambda _0} \in {\rm{iso}}\sigma \left( {T + F} \right)$$n\left( {T + F - {\lambda _0}I} \right) < \infty $. 由引理1可得 ${\lambda _0} \in {\rm{iso}}\sigma \left( T \right) \cup \rho \left( T \right)$$n\left( {T - {\lambda _0}I} \right) < \infty $. 若 ${\lambda _0} \in \rho \left( T \right)$,則 ${\lambda _0} \in {\rho _b}\left( T \right) \subseteq {\rho _{ea}}\left( T \right) = {\rho _{ea}}\left( {T + F} \right)$;若 ${\lambda _0} \in {\rm{iso}}\sigma \left( T \right)$,則由 $T$ 是isoloid算子可得 ${\lambda _0} \in {{\rm{\pi }}_{00}}\left( T \right)$,即 ${\lambda _0} \in {\rho _{ea}}\left( T \right) = {\rho _{ea}}\left( {T + F} \right)$. 因此 ${{\rm{\pi }}_{00}}\left( {T + F} \right) \subseteq {\sigma _a}\left( T \right.\left. { + F} \right)\backslash {\sigma _{ea}}\left( {T + F} \right)$. 綜上,這就證明了 $T + F$ 滿足 $\left( \omega \right)$ 性質.

    接下來,我們證明 $T + F$ 是isoloid算子. 設 ${\lambda _0} \in {\rm{iso}}\sigma \left( {T + F} \right)$,則由引理1可得 ${\lambda _0} \in {\rm{iso}}\sigma \left( T \right)$$ \cup \rho \left( T \right)$. 顯然,$\;\rho \left( T \right) \subseteq {\rho _b}\left( {T + F} \right)$. 因此,若 ${\lambda _0} \in \rho \left( T \right)$,則有 $n\left( {T + F - {\lambda _0}I} \right) > 0$. 若不然,則有 ${\lambda _0} \in \rho \left( {T + F} \right)$,這與 ${\lambda _0} \in $$ {\rm{iso}}\sigma \left( {T + F} \right)$ 矛盾. 若 ${\lambda _0} \in {\rm{iso}}\sigma \left( T \right)$,可以證明 $n\left( {T + F - {\lambda _0}I} \right) > 0$. 實際上,若 $n\left( {T + F - {\lambda _0}I} \right) = 0$,則由 $T$ 是isoloid算子可得 ${\lambda _0} \in {{\rm{\pi }}_{00}}\left( T \right)$. 此外,由 $T$ 滿足 $\left( \omega \right)$ 性質,則有 ${\lambda _0} \in {\rho _b}\left( {T + F} \right)$. 故 $T + F - {\lambda _0}I$ 可逆,矛盾. 因此 $T + F$ 是isoloid算子.

    $(2) \Rightarrow (1) $. 設 ${\lambda _0} \notin {\left[ \rho \right._1}\left( T \right) \cap {\sigma _w}\left( T \right) \cap \left. {{\rm{acc}}\sigma \left( T \right)} \right] \cup \left\{ {\lambda \in {\bf{C}}:} \right.\left. {n\left( {T - \lambda I} \right) > d\left( {T - \lambda I} \right)} \right\} \cup \left\{ {\lambda \in {\bf{C}}:} \right.$$n\left( T \right. - \left. {\lambda I} \right) = \left. \infty \right\}$. 則 $n\left( {T - {\lambda _0}I} \right) < \infty $$n\left( {T - {\lambda _0}I} \right) \leqslant d\left( {T - {\lambda _0}I} \right)$.

    ${\lambda _0} \notin {\rho _1}\left( T \right)$,則有 ${\lambda _0} \in {\rho _{ea}}\left( T \right)$,即 ${\lambda _0} \in {\rho _a}\left( T \right) \cup {{\text{π}}_{00}}\left( T \right)$. 由 ${\sigma _a}\left( T \right) = \sigma \left( T \right)$ 可知 ${\lambda _0} \in {\rho _b}\left( T \right)$. 這與 ${\lambda _0} \in {\sigma _1}\left( T \right)$ 矛盾;若 ${\lambda _0} \notin {\sigma _w}\left( T \right)$,類似以上證明可得 ${\lambda _0} \notin {\sigma _b}\left( T \right)$;若 ${\lambda _0} \notin {\rm{acc}}\sigma \left( T \right)$,則 ${\lambda _0} \in {\rm{iso}}\sigma \left( T \right) \cup \rho \left( T \right)$. 由于 $T$ 是isoloid算子,則可得 ${\lambda _0} \notin {\sigma _b}\left( T \right)$. 證畢.

    推論 8 設 $T \in B\left( H \right)$. 則以下條件等價:

    (1)${\sigma _b}\left( T \right) = {\left[ \rho \right._1}\left( T \right) \!\cap\! {\sigma _w}\left( T \right) \cap \left. {\left\{ {\lambda \in {\rm{acc}}{\sigma _b}\left( T \right):d\left( {T - \lambda I} \right) = \infty } \right\}} \right] \cup \left\{ {\lambda \in {\bf{C}}:} \right.n\left( {T - \lambda I} \right) > d\left( T \right. - {\left. {\lambda I} \right\} \cup }$      $ \left\{ {\lambda \in {\bf{C}}:}n ( T - \right.{\lambda I} )\! =\! \infty\} $;

    (2)對任意可交換的有限秩算子 $F \in B\left( H \right)$,有 ${\sigma _a}\left( T \right) = \sigma \left( T \right)$,且 $\left\{ {\lambda \in {\rm{iso}}} \right.{\sigma _b}\left( T \right):n\left( T \right. - \left. {\lambda I} \right)$$\left. { < \infty } \right\} = \emptyset $,并有 $T + F$ 滿足 $\left( \omega \right)$ 性質且為isoloid算子.

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