线性算子谱理论是算子理论的重要组成部分, 在许多应用实践中具有举足轻重的作用. 自伴算子的谱理论是量子力学的基础, 非自伴算子的谱理论在实际应用中扮演重要角色. 非自伴算子的谱分析比自伴算子复杂, 比如在由非自伴算子决定的系统中, 当受迫振动频率的值离算子谱远的时候就可能发生共振现象^[1]. 所以说, 对于非自伴算子, 仅仅用谱去刻画非自伴系统是不科学的. 又由于线性算子谱对扰动的敏感性, 相关数值算法不能保证它的精确性, 因此经典的谱分析在应用实践中出现很大的弊端. 为了克服这样的缺点, 后来人们提出了拟谱的概念. 拟谱是谱的延伸与推广, 能够更好地刻画非正规系统的稳定性^[2–4]以及分析非正规性动力系统的瞬时状态^{[3, 5]}. 学者们已从不同角度对其进行了研究, 并得到较好的结论. 文献[6, 7]研究了一些特殊算子的拟谱和线性算子的和、差与乘积之拟谱的保持问题. 文献[8]研究了两个矩阵在一定条件下具有相同的拟谱. 文献[9]讨论了矩阵的分数阶拟谱. 文献[10]研究了Banach空间中有界线性算子的本质拟谱和条件本质拟谱. 文献[11, 12]剖析了分块矩阵拟谱的上下界. 文献[13]讨论了在一定条件下分块算子矩阵对角元的本质拟谱与其在对角扰动下的本质拟谱之间的关系. 对于拟谱的更多结论, 请参阅[14–16]及其引用文献.

分块算子矩阵是以Hilbert空间或Banach空间中线性算子为元素的特殊矩阵, 它广泛出现于系统理论、非线性分析以及发展方程等领域, 在应用偏微分方程求解问题、弹性力学、流体力学、量子力学等领域有重要应用. 近二十年来, 许多文献主要集中于对$ 2\times2 $上三角算子矩阵和$ 2\times2 $算子矩阵的的可逆性与正则性以及它们的谱分析和谱补问题的研究, 均获得了丰富的成果, 参见[17–22]和塞尔维亚学者Dragana S. 最近出版的专著[23]. 为了更好地了解拟谱的特性, 文献[24]对拟谱作了精细划分, 讨论了闭线性算子与其共轭算子的精细拟谱之间的关系, 进而研究了Hamilton算子的拟谱结构. 在此基础上, 用对角元的拟谱刻画了整体算子矩阵的拟谱分布. 文献[25]对$ 2\times2 $分块对角算子矩阵在对角有界扰动情形下的拟谱作了精细描述, 同时也探讨了上三角有界扰动情形下的拟点谱分布.

本文在文献[24, 25]的基础上, 提出了稠定闭线性算子的$ \varepsilon $-单射, $ \varepsilon $-稠密以及固有拟谱、固有拟点谱、固有拟剩余谱和固有拟连续谱的概念, 研究了分块对角算子矩阵在上三角有界扰动情形下的$ \varepsilon $-单射性, 同时也得到了分块对角算子矩阵在上三角有界扰动情形下的拟剩余谱和拟连续谱. 最后, 给出了$ M_{0} $在上三角有界扰动情形下的固有拟点谱、固有拟剩余谱和固有拟连续谱的描述.

在下文中, 始终用$ X, \; Y $表示无穷维可分的复Hilbert空间. 符号$ \mathcal{B}(X, Y), \; \mathcal{C}(X, Y) $分别表示从$ X $到$ Y $的有界线性算子全体构成的集合和从$ X $到$ Y $的稠定闭线性算子全体构成的集合. 为了简便, 记$ \mathcal{B}(X)=\mathcal{B}(X, X), \; \mathcal{C}(X)=\mathcal{C}(X, X). $设$ T $是$ X $中的稠定线性算子, 用$ T^{\ast} $, $ \mathcal{D}(T) $, $ \mathcal{N}(T) $, $ \mathcal{R}(T), $ $ \overline{\mathcal{R}(T)} $和$ \mathcal{R}(T)^{\bot} $分别表示$ T $的共轭算子、定义域、零子空间、值域、值域的闭包和值域的正交补. 用$ T\mid_{\mathcal{M}} $表示$ T $在空间$ \mathcal{M} $上的限制. 统一用$ I $表示给定空间上的单位算子. 若$ T\in\mathcal{C}(X) $, 我们用$ \sigma(T) $, $ \rho(T) $, $ \sigma_{p}(T) $, $ \sigma_{r}(T) $和$ \sigma_{c}(T) $分别表示$ T $的谱集、预解集、点谱、剩余谱和连续谱.

下面给出稠定闭线性算子的拟谱、拟点谱、拟剩余谱与拟连续谱定义.

定义2.1(参见[3])  令$ T\in \mathcal{C}(X) $. 对于任意的$ \varepsilon>0 $, 算子$ T $的拟谱$ \sigma_{\varepsilon}(T) $定义如下:

$ \sigma_{\varepsilon}(T)=\sigma(T)\cup\{\lambda\in\rho(T): \|(T-\lambda I)^{-1}\|>\frac{1}{\varepsilon}\}. $

利用扰动原理, 等价地可得到^[2],

$ \begin{align*} \sigma_{\varepsilon}(T)=\bigcup\limits_{\substack{E\in\mathcal{B}(X)\\\| E\|<\varepsilon}}{\sigma(T+E)} =\{\lambda\in\mathbb{C}:\text{存在}\; E\in\mathcal{B}(X), \; \| E\|<\varepsilon, \; \text{使得}\; \lambda\in\sigma(T+E)\}. \end{align*} $

定义2.2(参见[24])  令$ T\in \mathcal{C}(X) $. 对于任意的$ \varepsilon>0 $, 定义

$ \begin{align*} \sigma_{\varepsilon, p}(T):&=\bigcup\limits_{\substack{E\in\mathcal{B}(X)\\\| E\|<\varepsilon}}\sigma_{p}(T+E) =\{\lambda\in\mathbb{C}: \text{存在}\; E\in\mathcal{B}(X), \; \| E\|<\varepsilon, \; \text{使得}\; \lambda\in\sigma_{p}(T+E)\};\\ \sigma_{\varepsilon, r}(T):&=\bigcup\limits_{\substack{E\in\mathcal{B}(X)\\\| E\|<\varepsilon}}\sigma_{r}(T+E) =\{\lambda\in\mathbb{C}: \text{存在}\; E\in\mathcal{B}(X), \; \| E\|<\varepsilon, \; \text{使得}\; \lambda\in\sigma_{r}(T+E)\};\\ \sigma_{\varepsilon, c}(T):&=\bigcup\limits_{\substack{E\in\mathcal{B}(X)\\\| E\|<\varepsilon}}\sigma_{c}(T+E) =\{\lambda\in\mathbb{C}: \text{存在}\; E\in\mathcal{B}(X), \; \| E\|<\varepsilon, \; \text{使得}\; \lambda\in\sigma_{c}(T+E)\}, \end{align*} $

称$ \sigma_{\varepsilon, p}(T) $, $ \sigma_{\varepsilon, r}(T) $和$ \sigma_{\varepsilon, c}(T) $分别为$ T $的拟点谱、拟剩余谱和拟连续谱.

注2.1  容易看出

$ \begin{align*} &\sigma(T)\subseteq\sigma_{\varepsilon}(T), \; \sigma_{\alpha}(T)\subseteq\sigma_{\varepsilon, \alpha}(T), \; \alpha\in\{p, \; r, \; c\}, \\ & \sigma_{\varepsilon}(T)=\sigma_{\varepsilon, p}(T)\cup\sigma_{\varepsilon, r}(T)\cup\sigma_{\varepsilon, c}(T). \end{align*} $

当$ \varepsilon\rightarrow0 $时, 拟点谱、拟剩余谱和拟连续谱分别退化为点谱、剩余谱和连续谱; 但对较大的正数$ \varepsilon $, 拟点谱、拟剩余谱、拟连续谱之间可能是有交集的(参见[24]).

定义2.3(参见[25])  设$ T\in\mathcal{C}(X) $且$ \varepsilon>0 $. 定义$ T $的$ \varepsilon $-零子空间$ \mathcal{N}_{\varepsilon}(T) $和$ \varepsilon $-值域$ \mathcal{R}_{\varepsilon}(T) $分别为:

$ \begin{align*} \mathcal{N}_{\varepsilon}(T)&=\bigcup\limits_{\substack{E\in\mathcal{B}(X)\\\| E\|<\varepsilon}}\mathcal{N}(T+E) =\{x\in\mathcal{D}(T):\; \text{使得}\; (T+E)x=0, \; \text{其中}\; E\in\mathcal{B}(X), \; \| E\|<\varepsilon\};\\ \mathcal{R}_{\varepsilon}(T)&=\bigcup\limits_{\substack{E\in\mathcal{B}(X)\\\| E\|<\varepsilon}}\mathcal{R}(T+E) =\{(T+E)x:\; x\in \mathcal{D}(T), \; \text{其中}\; E\in\mathcal{B}(X)\; \text{且}\; \| E\|<\varepsilon\}. \end{align*} $

并称$ \alpha_{\varepsilon}(T)=\dim\mathcal{N}_{\varepsilon}(T) $为$ T $的$ \varepsilon $-零度, $ \; \beta_{\varepsilon}(T)=\dim\mathcal{R}_{\varepsilon}(T)^{\perp} $为$ T $的$ \varepsilon $-余维数.

定义2.4  设$ T\in\mathcal{C}(X) $且$ \varepsilon>0 $. 如果存在$ E\in\mathcal{B}(X), \; \| E\|<\varepsilon $使得$ T+E $是单射, 则称线性算子$ T $是$ \varepsilon $-单射, 如果$ \overline{\mathcal{R}(T+E)}=X $, 那么称线性算子$ T $是$ \varepsilon $-稠密.

定义2.5  令$ T\in \mathcal{C}(X) $. 对于任意给定的$ \varepsilon>0 $, 定义

$ \begin{align*} \sigma^{I}_{\varepsilon}(T):&=\bigcap\limits_{\substack{E\in\mathcal{B}(X)\\\| E\|<\varepsilon}}\sigma(T+E)\\ & =\{\lambda\in\mathbb{C}: \text{对于所有的}\; E\in\mathcal{B}(X), \; \| E\|<\varepsilon, \; \text{都有}\; \lambda\in\sigma(T+E)\}, \\ \sigma^{I}_{\varepsilon, p}(T):&=\bigcap\limits_{\substack{E\in\mathcal{B}(X)\\\| E\|<\varepsilon}}\sigma_{p}(T+E)\\ & =\{\lambda\in\mathbb{C}: \text{对于所有的}\; E\in\mathcal{B}(X), \; \| E\|<\varepsilon, \; \text{都有}\; \lambda\in\sigma_{p}(T+E)\};\\ \sigma^{I}_{\varepsilon, r}(T):&=\bigcap\limits_{\substack{E\in\mathcal{B}(X)\\\| E\|<\varepsilon}}\sigma_{r}(T+E)\\ & =\{\lambda\in\mathbb{C}: \text{对于所有的}\; E\in\mathcal{B}(X), \; \| E\|<\varepsilon, \; \text{都有}\; \lambda\in\sigma_{r}(T+E)\};\\ \sigma^{I}_{\varepsilon, c}(T):&=\bigcap\limits_{\substack{E\in\mathcal{B}(X)\\\| E\|<\varepsilon}}\sigma_{c}(T+E)\\ & =\{\lambda\in\mathbb{C}: \text{对于所有的}\; E\in\mathcal{B}(X), \; \| E\|<\varepsilon, \; \text{都有}\; \lambda\in\sigma_{c}(T+E)\}, \end{align*} $

称$ \sigma^{I}_{\varepsilon}(T), \; \sigma^{I}_{\varepsilon, p}(T) $, $ \sigma^{I}_{\varepsilon, r}(T) $和$ \sigma^{I}_{\varepsilon, c}(T) $分别为$ T $的固有拟谱、固有拟点谱、固有拟剩余谱和固有拟连续谱.

注2.2  易知

$ \begin{equation*} \sigma^{I}_{\varepsilon}(T)\subseteq\sigma(T), \sigma^{I}_{\varepsilon, \ast}(T)\subseteq\sigma_{\ast}(T), \; \ast\in\{p, \; r, \; c\}. \end{equation*} $

本文后续结果显示: 对于给定的$ \varepsilon>0 $, 上述包含关系是严格的.

设$ A\in\mathcal{C}(X), \; B\in\mathcal{C}(Y) $, 下文中我们始终用

$ M_{0}=\left( \begin{array} {cc}{A} & {0}\\ {0}& {B} \end{array} \right) $

表示$ 2\times2 $对角分块算子矩阵. 同时我们考虑扰动算子为上三角有界扰动情形, 即

$ E=\left( \begin{array} {cc}{E_{11}} & {E_{12}}\\ {0}& {E_{22}} \end{array} \right)\in\mathcal{B}(X\oplus Y), $

其中$ E_{11}\in\mathcal{B}(X), \; E_{12}\in\mathcal{B}(Y, X), \; E_{22}\in\mathcal{B}(Y), \; \|E_{ij}\|<\varepsilon, \; i, \; j=1, \; 2 $且$ \|E\|=\max\limits_{i, j=1, 2}\{\|E_{ij}\|\}<\varepsilon $. 并分别用$ \sigma^{T}_{\varepsilon, \ast}(M_{0}), \; \sigma^{I T}_{\varepsilon, \ast}(M_{0}) (\ast\in\{p, r, c\}) $表示$ M_{0} $在上三角有界扰动情形下的精细拟谱和固有精细拟谱.

在本部分中, 首先我们讨论了$ M_{0} $在上三角有界扰动情形下的$ \varepsilon $-单射性, 在此基础上, 刻画了$ M_{0} $的拟剩余谱与拟连续谱分布.

在下文, 取常数$ \gamma $满足$ 0<\gamma<\varepsilon, $并且$ \gamma J $表示有界线性算子$ J $的数乘. 为了表述方便, 记

$ \begin{eqnarray*} \Delta_{0}&=&\{\lambda\in\mathbb{C}:\alpha_{\varepsilon}(B-\lambda I)>\beta_{\varepsilon}(A-\lambda I)\};\\ \Delta_{1}&=&\{\lambda\in\mathbb{C}:1\leq\alpha_{\varepsilon}(B-\lambda I)=\beta_{\varepsilon}(A-\lambda I)\};\\ \Delta_{2}&=&\{\lambda\in\mathbb{C}:0<\alpha_{\varepsilon}(B-\lambda I)<\beta_{\varepsilon}(A-\lambda I)\}. \end{eqnarray*} $

引理3.1.1 ^[25]  若$ M_{0} $是$ 2\times2 $对角分块算子矩阵, 则

$ \begin{eqnarray*} \sigma^{T}_{\varepsilon, p}(M_{0})=\sigma_{\varepsilon, p}(A)\cup\Delta_{0}\cup\Delta_{1}\cup\Delta_{2}. \end{eqnarray*} $

引理3.1.2 ^[26]  设$ T\in\mathcal{B}(X, Y), \; \mathcal{R}(T) $不闭, 那么存在无穷维子空间$ X_{0}\subseteq \overline{\mathcal{R}(T)} $使得$ X_{0}\cap\mathcal{R}(T)=\{0\}. $

定理3.1.3  设$ \varepsilon>0, $算子$ M_{0} $是$ \varepsilon $-单射的充要条件$ A $是$ \varepsilon $-单射且满足如下条件之一:

$ (1) $ $ \mathcal{R}_{\varepsilon}(A) $闭的且$ \alpha_{\varepsilon}(B)\leq\beta_{\varepsilon}(A) $;

$ (2) $ $ \mathcal{R}_{\varepsilon}(A) $不闭.

证  (充分性.) 设$ A $是$ \varepsilon $-单射, 则存在算子$ E_{11}\in\mathcal{B}(X), \; \|E_{11}\|<\varepsilon $使得$ A+E_{11} $是单射. 如果$ \mathcal{R}_{\varepsilon}(A) $闭且$ \alpha_{\varepsilon}(B)\leq\beta_{\varepsilon}(A), $那么定义一下方有界算子$ J:\mathcal{N}_{\varepsilon}(B)\rightarrow \mathcal{R}_{\varepsilon}(A)^{\perp}, \; \|J\|=1. $令

$ \begin{equation} E_{12}=\left( \begin{array} {cc}{\gamma J} & {0} \\ {0}& {0} \end{array} \right):\left(\begin{array} {c}{\mathcal{N}_{\varepsilon}(B)} \\ {\mathcal{N}_{\varepsilon}(B)^{\perp}}\end{array} \right)\rightarrow \left(\begin{array} {c}{\mathcal{R}_{\varepsilon}(A)^{\perp}} \\ {\mathcal{R}_{\varepsilon}(A)} \end{array} \right). \end{equation} $

(3.1)

如前所述$ \gamma $表示一常数且$ 0<\gamma<\varepsilon. $显然, $ \|E_{12}\|=\|\gamma J\|<\varepsilon $. 取

$ \begin{equation} E=\left( \begin{array} {cc}{E_{11}} & {E_{12}}\\ {0}& {E_{22}} \end{array} \right)\in\mathcal{B}(X\oplus Y), \; \|E\|=\max\limits_{i, j}\{\|E_{ij}\|\}<\varepsilon, \end{equation} $

(3.2)

其中$ E_{22}\in\mathcal{B}(Y), \; \|E_{22}\|<\varepsilon $.

设$ \left(\begin{array} {c}{x} \\ {y} \end{array} \right)\in \mathcal{D}(M_{0}+E) $, 令

$ \begin{eqnarray} (M_{0}+E)\left(\begin{array} {c}{x} \\ {y} \end{array} \right)&=&\left( \begin{array} {cc}{A+E_{11}} & {E_{12}}\\ {0}& {B+E_{22}} \end{array} \right)\left(\begin{array} {c}{x} \\ {y} \end{array} \right)\\ &=&\left( \begin{array} {c}{(A+E_{11})x+E_{12}y}\\ {(B+E_{22})y} \end{array} \right)\\&=&\textbf{0}. \end{eqnarray} $

(3.3)

可以得到

$ \begin{equation} y\in\mathcal{N}(B+E_{22})\subseteq\mathcal{N}_{\varepsilon}(B)\; \text{且}\; (A+E_{11})x=-E_{12}y. \end{equation} $

(3.4)

进而, 有

$ \begin{equation} (A+E_{11})x=-\gamma Jy=0. \end{equation} $

(3.5)

结合$ J $的定义得到$ y=0 $. 又因为$ A+E_{11} $是单射, 所以$ x=0. $这表明$ M_{0}+E $是单射. 即$ M_{0} $是$ \varepsilon $-单射.

如果$ \mathcal{R}_{\varepsilon}(A) $不闭, 那么有$ \beta_{\varepsilon}(A)=\infty. $于是, 根据引理3.1.2知, 必存在闭子空间$ X_{0}\subset \overline{\mathcal{R}_{\varepsilon}(A)} $满足$ X_{0}\cap\mathcal{R}_{\varepsilon}(A)=\{0\}. $现定义一下方有界算子

$ \begin{equation} E^{(1)}_{12}:\mathcal{N}_{\varepsilon}(B)\rightarrow \overline{\mathcal{R}_{\varepsilon}(A)} \end{equation} $

(3.6)

使得$ \mathcal{R}(E^{(1)}_{12})\subseteq X_{0}. $令

$ \begin{equation} E_{12}=\left( \begin{array} {cc}{E^{(2)}_{12}} & {E^{(3)}_{12}} \\ {E^{(1)}_{12}}& {E^{(4)}_{12}} \end{array} \right):\left(\begin{array} {c}{\mathcal{N}_{\varepsilon}(B)} \\ {\mathcal{N}_{\varepsilon}(B)^{\perp}}\end{array} \right)\rightarrow \left(\begin{array} {c}{\mathcal{R}_{\varepsilon}(A)^{\perp}} \\ {\overline{\mathcal{R}_{\varepsilon}(A)}} \end{array} \right). \end{equation} $

(3.7)

这里$ E^{(i)}_{12}\in\mathcal{B}(Y, X) $且$ \|E^{(i)}_{12}\|<\varepsilon. $

根据$ E^{(1)}_{12} $定义, 有

$ \begin{equation} \mathcal{N}(E^{(2)}_{12})\cap\mathcal{N}(E^{(1)}_{12})=\{0\}\; \text{且}\; \mathcal{R}(E^{(1)}_{12})\cap\mathcal{R}_{\varepsilon}(A)=\{0\}, \end{equation} $

(3.8)

从而$ E_{12}|_{\mathcal{N}_{\varepsilon}(B)} $是单射且

$ \begin{equation} \mathcal{R}(E_{12}\mid_{\mathcal{N}_{\varepsilon}(B)})\cap\mathcal{R}_{\varepsilon}(A)=\{0\}. \end{equation} $

(3.9)

又由于$ A $是$ \varepsilon $-单射, 故存在算子$ E_{11}\in\mathcal{B}(X), \; \|E_{11}\|<\varepsilon $使得$ A+E_{11} $是单射. 取

$ \begin{equation*} E=\left( \begin{array} {cc}{E_{11}} & {E_{12}}\\ {0}& {E_{22}} \end{array} \right)\in\mathcal{B}(X\oplus Y), \; \|E\|=\max\limits_{i, j}\{\|E_{ij}\|\}<\varepsilon, \end{equation*} $

其中$ E_{22}\in\mathcal{B}(Y), \; \|E_{22}\|<\varepsilon $.

设$ \left(\begin{array} {c}{x_{1}} \\ {y_{1}} \end{array} \right)\in \mathcal{D}(M_{0}+E) $, 令$ (M_{0}+E)\left(\begin{array} {c}{x_{1}} \\ {y_{1}} \end{array} \right)=\textbf{0}. $如同等式$ 3.3-3.5 $, 易得$ M_{0} $是$ \varepsilon $-单射.

(必要性.) 假设算子$ M_{0} $是$ \varepsilon $-单射, 则存在

$ \begin{equation*} E=\left( \begin{array} {cc}{E_{11}} & {E_{12}}\\ {0}& {E_{22}} \end{array} \right)\in\mathcal{B}(X\oplus Y), \; \|E\|=\max\limits_{i, j}\{\|E_{ij}\|\}<\varepsilon, \end{equation*} $

使得算子$ M_{0}+E $是单射, 这表明$ A $是$ \varepsilon $-单射且$ \mathcal{N}(E_{12})\cap\mathcal{N}(B+E_{22})=\{0\}. $进而,

$ \begin{equation} E_{12}(\mathcal{N}(B+E_{22}))\cap\mathcal{R}(A+E_{11})=\{0\}. \end{equation} $

(3.10)

如果$ \mathcal{R}_{\varepsilon}(A) $不闭, 则结论自然成立. 以下讨论$ \mathcal{R}_{\varepsilon}(A) $为闭的情形, 考虑$ E_{12} $的下述分解:

$ \begin{equation} E_{12}=\left( \begin{array} {cc}{E^{(2)}_{12}} & {E^{(3)}_{12}} \\ {E^{(1)}_{12}}& {E^{(4)}_{12}} \end{array} \right):\left(\begin{array} {c}{\mathcal{N}_{\varepsilon}(B)} \\ {\mathcal{N}_{\varepsilon}(B)^{\perp}}\end{array} \right)\rightarrow \left(\begin{array} {c}{\mathcal{R}_{\varepsilon}(A)^{\perp}} \\ {\mathcal{R}_{\varepsilon}(A)} \end{array} \right), \end{equation} $

(3.11)

由

$ E_{12}(\mathcal{N}(B+E_{22}))\cap\mathcal{R}(A+E_{11})=\{0\}, $

得$ \mathcal{N}(E^{(2)}_{12})\subseteq\mathcal{N}(E^{(1)}_{12}) $. 又由$ E_{12}|_{\mathcal{N}(B+E_{22})} $是单射, 故$ \mathcal{N}(E^{(2)}_{12})\cap\mathcal{N}(E^{(1)}_{12})=\{0\}. $从而$ \mathcal{N}(E^{(2)}_{12})=\{0\}. $进一步推出, 算子$ E^{(2)}_{12} $是单射. 从而, $ \alpha_{\varepsilon}(B)\leq\beta_{\varepsilon}(A). $

通过对定理3.1.3的逆否命题的分析, 得到如下推论.

推论3.1.4  设$ \varepsilon>0 $, 则有

$ \begin{equation} \sigma^{IT}_{\varepsilon, p}(M_{0})=\sigma^{I}_{\varepsilon, p}(A)\cup\{\lambda\in\mathbb{C}:\alpha_{\varepsilon}(B-\lambda I)>\beta_{\varepsilon}(A-\lambda I)\}. \end{equation} $

(3.12)

接下来, 我们来讨论上三角有界扰动情形下的拟剩余谱.

为了表述的方便, 在下文记

$ \begin{eqnarray*} \Delta_{3}&=&\{\overline{\lambda}\in\mathbb{C}:\alpha_{\varepsilon}(A^{\ast}-\overline{\lambda}I)>\beta_{\varepsilon}(B^{\ast}-\overline{\lambda} I)\};\\ \Delta_{4}&=&\{\overline{\lambda}\in\mathbb{C}:1\leq\alpha_{\varepsilon}(A^{\ast}-\overline{\lambda}I)=\beta_{\varepsilon}(B^{\ast}-\overline{\lambda} I)\};\\ \Delta_{5}&=&\{\overline{\lambda}\in\mathbb{C}:0<\alpha_{\varepsilon}(A^{\ast}-\overline{\lambda}I)<\beta_{\varepsilon}(B^{\ast}-\overline{\lambda} I)\}. \end{eqnarray*} $

定理3.1.5  设$ \; \lambda\in\mathbb{C} $, $ \lambda\in\sigma^{T}_{\varepsilon, r}(M_{0}) $的充要条件为$ A-\lambda I\; $是$ \varepsilon $-单射且满足以下条件之一:

$ (1) $ $ \mathcal{R}_{\varepsilon}(A-\lambda I) $是闭的, $ \alpha_{\varepsilon}(B-\lambda I)\leq\beta_{\varepsilon}(A-\lambda I) $且$ \{\lambda:\overline{\lambda}\in\sigma_{\varepsilon, p}(B^{\ast})\cup\Delta_{3}\cup\Delta_{4}\cup\Delta_{5}\} $;

$ (2) $ $ \mathcal{R}_{\varepsilon}(A-\lambda I) $不闭且$ \{\lambda:\overline{\lambda}\in\sigma_{\varepsilon, p}(B^{\ast})\cup\Delta_{3}\cup\Delta_{4}\cup\Delta_{5}. \}$.

证  假设$ \lambda\in\sigma^{T}_{\varepsilon, r}(M_{0}) $, 那么存在

$ E=\left( \begin{array} {cc}{E_{11}} & {E_{12}}\\ {0}& {E_{22}} \end{array} \right)\in\mathcal{B}(X\oplus Y), \; \|E\|=\max\limits_{i, j}\{\|E_{ij}\|\}<\varepsilon, $

使得算子$ M_{0}+E-\lambda I $是单射且$ \overline{\mathcal{R}(M_{0}+E-\lambda I)}\neq X\oplus Y. $这样, 我们根据定理3.1.3得, 算子$ M_{0}+E-\lambda I $是$ \varepsilon $-单射的充要条件为算子$ A-\lambda I $是$ \varepsilon $-单射且满足以下条件之一: $ \mathcal{R}_{\varepsilon}(A-\lambda I) $是闭的且$ \alpha_{\varepsilon}(B-\lambda I)\leq\beta_{\varepsilon}(A-\lambda I) $或$ \mathcal{R}_{\varepsilon}(A-\lambda I) $不闭.

另一方面, 由$ \; \overline{\mathcal{R}(M_{0}+E-\lambda I)}\neq X\oplus Y $知, $ \mathcal{N}(M_{0}^{\ast}+E^{\ast}-\overline{\lambda }I)\neq\{0\}. $故$ \overline{\lambda }\in\sigma^{T}_{\varepsilon, p}(M_{0}^{\ast}). $根据引理3.1.1得,

$ \begin{equation} \overline{\lambda}\in\sigma_{\varepsilon, p}(B^{\ast})\cup\Delta_{3}\cup\Delta_{4}\cup\Delta_{5}. \end{equation} $

(3.13)

综上所述, 有$ \lambda\in\sigma^{T}_{\varepsilon, r}(M_{0}) $的充要条件为$ A-\lambda I\; $是$ \varepsilon $-单射且满足以下条件之一:

$ (1) $ $ \mathcal{R}_{\varepsilon}(A-\lambda I) $是闭的, $ \alpha_{\varepsilon}(B-\lambda I)\leq\beta_{\varepsilon}(A-\lambda I) $且$ \{\lambda:\overline{\lambda}\in\sigma_{\varepsilon, p}(B^{\ast})\cup\Delta_{3}\cup\Delta_{4}\cup\Delta_{5}. \}$;

$ (2) $ $ \mathcal{R}_{\varepsilon}(A-\lambda I) $不闭且$ \{\lambda:\overline{\lambda}\in\sigma_{\varepsilon, p}(B^{\ast})\cup\Delta_{3}\cup\Delta_{4}\cup\Delta_{5}. \}$.

根据上述已有结论和方法, 下面将研究上三角有界扰动情形下的拟连续谱.

定理3.1.6  设$ \; \lambda\in\mathbb{C} $, $ \lambda\in\sigma_{\varepsilon, c}^{T}(M_{0})\backslash\sigma_{\varepsilon, \delta}(A) $的充要条件为$ A-\lambda I $是$ \varepsilon $-单射, $ B-\lambda I $是$ \varepsilon $-稠密, $ \lambda\in\sigma_{\varepsilon, \delta}(B) $且满足以下四种情形之一:

$ (1) $ $ \mathcal{R}_{\varepsilon}(A-\lambda I) $闭, $ \mathcal{R}_{\varepsilon}(B-\lambda I) $闭, $ \alpha_{\varepsilon}(B-\lambda I)\leq\beta_{\varepsilon}(A-\lambda I) $, $ \alpha_{\varepsilon}(A^{\ast}-\overline{\lambda} I)\leq\beta_{\varepsilon}(B^{\ast}-\overline{\lambda} I); $

$ (2) $ $ \mathcal{R}_{\varepsilon}(A-\lambda I) $闭, $ \mathcal{R}_{\varepsilon}(B-\lambda I) $不闭, $ \alpha_{\varepsilon}(B-\lambda I)\leq\beta_{\varepsilon}(A-\lambda I) $;

$ (3) $ $ \mathcal{R}_{\varepsilon}(A-\lambda I) $不闭, $ \mathcal{R}_{\varepsilon}(B-\lambda I) $闭, $ \alpha_{\varepsilon}(A^{\ast}-\overline{\lambda} I)\leq\beta_{\varepsilon}(B^{\ast}-\overline{\lambda} I) $;

$ (4) $ $ \mathcal{R}_{\varepsilon}(A-\lambda I) $不闭, $ \mathcal{R}_{\varepsilon}(B-\lambda I) $不闭.

证  设$ \lambda\in\sigma_{\varepsilon, c}^{T}(M_{0}) $, 那么存在算子$ E=\left( \begin{array} {cc}{E_{11}} & {E_{12}}\\ {0}& {E_{22}} \end{array} \right)\in\mathcal{B}(X\oplus Y), \; \|E\|=\max\limits_{i, j}\{\|E_{ij}\|\}<\varepsilon, $使得$ \mathcal{N}(M_{0}+E-\lambda I)=\{0\}, \; \overline{\mathcal{R}(M_{0}+E-\lambda I)}=X\oplus Y $且$ \mathcal{R}(M_{0}+E-\lambda I)\neq X\oplus Y. $首先根据定理3.1.3知: 算子$ M_{0}+E-\lambda I $是单射的充要条件, 算子$ A-\lambda I $是$ \varepsilon $-单射且满足以下条件之一:

(ⅰ) $ \mathcal{R}_{\varepsilon}(A-\lambda I) $是闭的且$ \alpha_{\varepsilon}(B-\lambda I)\leq\beta_{\varepsilon}(A-\lambda I) $;

(ⅱ) $ \mathcal{R}_{\varepsilon}(A-\lambda I) $不闭.

其次, $ \overline{\mathcal{R}(M_{0}+E-\lambda I)}=X\oplus Y $意味着$ \mathcal{N}(M^{\ast}_{0}+E^{\ast}-\overline{\lambda} I)=\{0\}, $这表明$ M^{\ast}_{0}-\overline{\lambda} I $是$ \varepsilon $-单射. 根据定理3.1.3得, 算子$ B^{\ast}-\overline{\lambda} I $是$ \varepsilon $-单射且满足以下条件之一:

(a) $ \mathcal{R}_{\varepsilon}(B^{\ast}-\overline{\lambda} I) $是闭的且$ \alpha_{\varepsilon}(A^{\ast}-\overline{\lambda} I)\leq\beta_{\varepsilon}(B^{\ast}-\overline{\lambda} I); $

(b) $ \mathcal{R}_{\varepsilon}(B^{\ast}-\overline{\lambda} I) $不闭.

然而, 由$ B^{\ast}-\overline{\lambda} I $是$ \varepsilon $-单射, 得$ B-\lambda I $是$ \varepsilon $-稠密.

由$ \mathcal{R}(M_{0}+E-\lambda I)\neq X\oplus Y $, $ \lambda\not\in\sigma_{\varepsilon, \delta}(A) $知, $ \; \mathcal{R}(B+E_{22}-\lambda I)\neq Y. $因此$ \lambda\in\sigma_{\varepsilon, \delta}(B) $.

综上所述, $ \lambda\in\sigma_{\varepsilon, c}^{T}(M_{0})\backslash\sigma_{\varepsilon, \delta}(A) $的充要条件为$ A-\lambda I $是$ \varepsilon $-单射, $ B-\lambda I $是$ \varepsilon $-稠密, $ \lambda\in\sigma_{\varepsilon, \delta}(B) $且以下四种情形之一成立:

$ (1) $ $ \mathcal{R}_{\varepsilon}(A-\lambda I) $闭, $ \mathcal{R}_{\varepsilon}(B-\lambda I) $闭, $ \alpha_{\varepsilon}(B-\lambda I)\leq\beta_{\varepsilon}(A-\lambda I) $, $ \alpha_{\varepsilon}(A^{\ast}-\overline{\lambda} I)\leq\beta_{\varepsilon}(B^{\ast}-\overline{\lambda} I); $

$ (2) $ $ \mathcal{R}_{\varepsilon}(A-\lambda I) $闭, $ \mathcal{R}_{\varepsilon}(B-\lambda I) $不闭, $ \alpha_{\varepsilon}(B-\lambda I)\leq\beta_{\varepsilon}(A-\lambda I) $;

$ (3) $ $ \mathcal{R}_{\varepsilon}(A-\lambda I) $不闭, $ \mathcal{R}_{\varepsilon}(B-\lambda I) $闭, $ \alpha_{\varepsilon}(A^{\ast}-\overline{\lambda} I)\leq\beta_{\varepsilon}(B^{\ast}-\overline{\lambda} I) $;

$ (4) $ $ \mathcal{R}_{\varepsilon}(A-\lambda I) $不闭, $ \mathcal{R}_{\varepsilon}(B-\lambda I) $不闭.

定理3.2.1  设$ \varepsilon>0 $, 则有

$ \begin{equation} \sigma^{IT}_{\varepsilon, r}(M_{0})=(\sigma^{I}_{\varepsilon, r}(B)\setminus\sigma_{\varepsilon, p}(A))\cup\{\lambda:\; \lambda\in\Gamma\}\backslash(\sigma_{\varepsilon, p}(A)\cup\sigma_{\varepsilon, p}(B)). \end{equation} $

(3.14)

在这里, $ \Gamma=\{\lambda: \; \dim(\bigcap\limits_{\substack{E_{11}\in\mathcal{B}(X)\\\| E_{11}\|<\varepsilon}}\overline{\mathcal{R}(A+E_{11}-\lambda I)})^{\bot}>0\} $.

证  假设$ \lambda\in\sigma^{IT}_{\varepsilon, r}(M_{0}) $, 那么对于所有的

$ E=\left( \begin{array} {cc}{E_{11}} & {E_{12}}\\ {0}& {E_{22}} \end{array} \right)\in\mathcal{B}(X\oplus Y), \; \|E\|=\max\limits_{i, j}\{\|E_{ij}\|\}<\varepsilon, $

算子$ M_{0}+E-\lambda I $是单射且$ \overline{\mathcal{R}(M_{0}+E-\lambda I)}\neq X\oplus Y. $算子$ \; M_{0}+E-\lambda I $是单射等价于$ \lambda\not\in\sigma^{T}_{\varepsilon, p}(M_{0}). $从而, 我们根据引理3.1.1可得$ \lambda\not\in\sigma_{\varepsilon, p}(A)\cup\Delta_{0}\cup\Delta_{1}\cup\Delta_{2} $. 经进一步分析得$ \lambda\not\in\sigma_{\varepsilon, p}(A)\cup\sigma_{\varepsilon, p}(B). $而由$ \overline{\mathcal{R}(M_{0}+E-\lambda I)}\neq X\oplus Y $知, $ \; \mathcal{N}(M_{0}^{\ast}+E^{\ast}-\overline{\lambda }I)\neq\{0\}, $即

$ \overline{\lambda }\in\sigma^{IT}_{\varepsilon, p}(M^{\ast}_{0}). $

根据推论3.1.4知,

$ \begin{equation} \overline{\lambda}\in\sigma^{I}_{\varepsilon, p}(B^{\ast})\cup\{\lambda\in\mathbb{C}:\alpha_{\varepsilon}(A^{\ast}-\overline{\lambda} I)>\beta_{\varepsilon}(B^{\ast}-\overline{\lambda }I)\}. \end{equation} $

(3.15)

由$ \overline{\lambda}\in\sigma^{I}_{\varepsilon, p}(B^{\ast}) $可得, $ \; \mathcal{N}(B^{\ast}+E_{22}^{\ast}-\overline{\lambda }I)\neq\{0\}. $从而, 有$ \overline{\mathcal{R}(B+E_{22}-\lambda I)}\neq X. $由于

$ \begin{align} \alpha_{\varepsilon}(A^{\ast}-\overline{\lambda}I)=&\dim(\bigcup\limits_{\substack{E_{11}\in\mathcal{B}(X)\nonumber\\\| E_{11}\|<\varepsilon}}\mathcal{N}(A^{\ast}+E_{11}^{\ast}-\overline{\lambda} I)) \\ =&\dim(\bigcup\limits_{\substack{E_{11}\in\mathcal{B}(X)\nonumber\\\| E_{11}\|<\varepsilon}}\overline{\mathcal{R}(A+E_{11}-\lambda I)}^{\bot}) \\ =&\dim(\bigcap\limits_{\substack{E_{11}\in\mathcal{B}(X)\\\| E_{11}\|<\varepsilon}}\overline{\mathcal{R}(A+E_{11}-\lambda I)})^{\bot}, \end{align} $

(3.16)

$ \begin{align} \beta_{\varepsilon}(B^{\ast}-\overline{\lambda}I)=&\dim(\bigcup\limits_{\substack{E_{22}\in\mathcal{B}(Y)\nonumber\\ \| E_{22}\|<\varepsilon}}\mathcal{R}(B^{\ast}+E_{22}^{\ast}-\overline{\lambda }I))^{\bot} \\ =&\dim(\bigcap\limits_{\substack{E_{22}\in\mathcal{B}(Y)\nonumber\\ \| E_{22}\|<\varepsilon}}\mathcal{R}(B^{\ast}+E_{22}^{\ast}-\overline{\lambda} I)^{\bot}) \\ =&\dim(\bigcap\limits_{\substack{E_{22}\in\mathcal{B}(Y)\\\| E_{22}\|<\varepsilon}}\mathcal{N}(B+E_{22}-\lambda I)). \end{align} $

(3.17)

故当$ \lambda\not\in\sigma_{\varepsilon, p}(A)\cup\sigma_{\varepsilon, p}(B) $时,

$ \begin{equation} \{\lambda:\dim(\bigcap\limits_{\substack{E_{11}\in\mathcal{B}(X)\\\| E_{11}\|<\varepsilon}}\overline{\mathcal{R}(A+E_{11}-\lambda I)})^{\bot}>\beta_{\varepsilon}(B^{\ast}-\overline{\lambda}I)\} \end{equation} $

(3.18)

等价于

$ \begin{equation} \{\lambda:\dim(\bigcap\limits_{\substack{E_{11}\in\mathcal{B}(X)\\\| E_{11}\|<\varepsilon}}\overline{\mathcal{R}(A+E_{11}-\lambda I)})^{\bot}>0\}. \end{equation} $

(3.19)

综上所述, 我们有

$ \begin{equation*} \sigma^{IT}_{\varepsilon, r}(M_{0})=(\sigma^{I}_{\varepsilon, r}(B)\setminus\sigma_{\varepsilon, p}(A))\cup\{\lambda:\; \lambda\in\Gamma\}\backslash(\sigma_{\varepsilon, p}(A)\cup\sigma_{\varepsilon, p}(B)). \end{equation*} $

定理3.2.2  设$ \varepsilon>0 $, 则

$ \begin{equation} \sigma^{IT}_{\varepsilon, c}(M_{0})=(\sigma^{I}_{\varepsilon, c}(A)\cap\sigma^{I}_{\varepsilon, c}(B)) \cup(\sigma^{I}_{\varepsilon, c}(A)\cap\rho_{\varepsilon}(B))\cup(\sigma^{I}_{\varepsilon, c}(B)\cap\rho_{\varepsilon}(A)). \end{equation} $

(3.20)

证  设$ \lambda\in\sigma^{IT}_{\varepsilon, c}(M_{0}) $, 那么对于所有的算子$ E=\left( \begin{array} {cc}{E_{11}} & {E_{12}}\\ {0}& {E_{22}} \end{array} \right)\in\mathcal{B}(X\oplus Y), \; \|E\|=\max\limits_{i, j}\{\|E_{ij}\|\}<\varepsilon, $使得$ \mathcal{N}(M_{0}+E-\lambda I)=\{0\}, \; \overline{\mathcal{R}(M_{0}+E-\lambda I)}=X\oplus Y $且$ \mathcal{R}(M_{0}+E-\lambda I)\neq X\oplus Y. $这样, 由$ \mathcal{N}(M_{0}+E-\lambda I)=\{0\} $得, $ \lambda\not\in\sigma^{T}_{\varepsilon, p}(M_{0}). $即

$ \lambda\not\in\sigma_{\varepsilon, p}(A)\cup\Delta_{0}\cup\Delta_{1}\cup\Delta_{2}. $

这进一步说明了$ \lambda\not\in\sigma_{\varepsilon, p}(A)\cup\sigma_{\varepsilon, p}(B). $故对于所有的$ E_{11}\in\mathcal{B}(X), \; E_{22}\in\mathcal{B}(Y), $有

$ \mathcal{N}(A+E_{11}-\lambda I)=\{0\}\; \text{且}\; \mathcal{N}(B+E_{22}-\lambda I)=\{0\}. $

其次, 由$ \overline{\mathcal{R}(M_{0}+E-\lambda I)}=X\oplus Y $知$ \mathcal{N}(M^{\ast}_{0}+E^{\ast}-\overline{\lambda} I)=\{0\}. $从而有$ \overline{\lambda}\not\in\sigma^{T}_{\varepsilon, p}(M^{\ast}_{0}). $即

$ \overline{\lambda}\not\in\sigma_{\varepsilon, p}(B^{\ast})\cup\Delta_{3}\cup\Delta_{4}\cup\Delta_{5}. $

进而表明$ \overline{\lambda}\not\in\sigma_{\varepsilon, p}(A^{\ast})\cup\sigma_{\varepsilon, p}(B^{\ast}). $因此对于所有的$ E_{11}\in\mathcal{B}(X), \; E_{22}\in\mathcal{B}(Y), $有

$ \overline{\mathcal{R}(A+E_{11}-\lambda I)}=X\; \text{且}\; \overline{\mathcal{R}(B+E_{22}-\lambda I)}=Y. $

最后由$ \mathcal{R}(M_{0}+E-\lambda I)\neq X\oplus Y $可知, 对于对应的$ E_{11}\in\mathcal{B}(X), \; E_{22}\in\mathcal{B}(Y) $,

$ \mathcal{R}(A+E_{11}-\lambda I)\neq X\; \text{或}\; \mathcal{R}(B+E_{22}-\lambda I)\neq Y. $

假设上述情形不成立, 那么我们就得到存在$ E_{11}\in\mathcal{B}(X), \; E_{22}\in\mathcal{B}(Y) $使得$ \mathcal{R}(A+E_{11}-\lambda I)=X $且$ \; \mathcal{R}(B+E_{22}-\lambda I)=Y. $这样, 显然有$ \mathcal{R}(M_{0}+E-\lambda I)= X\oplus Y $成立. 这与已知是矛盾的.

综上所述, 得

$ \begin{equation*} \sigma^{IT}_{\varepsilon, c}(M_{0})=(\sigma^{I}_{\varepsilon, c}(A)\cap\sigma^{I}_{\varepsilon, c}(B)) \cup(\sigma^{I}_{\varepsilon, c}(A)\cap\rho_{\varepsilon}(B))\cup(\sigma^{I}_{\varepsilon, c}(B)\cap\rho_{\varepsilon}(A)). \end{equation*} $

注3.1  通过以上研究, 我们不难发现:

$ \begin{equation} \sigma^{IT}_{\varepsilon, \ast}(M_{0})\subseteq\sigma_{\ast}(M_{0}), \; \ast\in\{p, \; r, \; c\}. \end{equation} $

(3.21)

经简单分析, 上述包含关系是严格的, 这进一步印证了注2.2的断言. 另外, $ \varepsilon\rightarrow0 $时, $ \sigma^{IT}_{\varepsilon, \ast}(M_{0}) $退化为$ \sigma_{\ast}(M_{0}), $这说明结果是合理的. 设$ M_{C}=\left( \begin{array} {cc}{A} & {C}\\ {0}& {B} \end{array} \right) $表示$ 2\times2 $上三角分块算子矩阵, 文献[18]给出了$ \bigcap\limits_{\substack{C\in\mathcal{B}(Y, X)}}\sigma_{\ast}(M_{C}) $的描述, 这为本文结果的证明提供了思路. 经进一步分析, 我们有如下结论:

$ \begin{equation} \bigcap\limits_{\substack{C\in\mathcal{B}(Y, X)}}\sigma_{\ast}(M_{C})\subseteq\sigma^{IT}_{\varepsilon, \ast}(M_{0}), \; \ast\in\{p, \; r, \; c\}. \end{equation} $

(3.22)