© Wuhan University 2024

0 Introduction

In this paper, let $\mathbb{C}$ denote the set of complex numbers, $\mathbb{N}$ the set of nonnegative integer, $\mathbb{D}$ the closed unit disc and $\mathrm{\Gamma }$ the unit circle. Let $B(H)(F(H))$ denote the algebra of all bounded linear operators (finite rank operators) acting on a complex, separable, infinite dimensional Hilbert space $H$. For an operator $T\in B(H)$, we denote by $\sigma (T),$$N(T)$ and $R(T)$ the spectrum, the kernel and the range of $T$, respectively. Also, we write $\rho (T)=\mathbb{C}\backslash \sigma (T)$,$n(T)=$dim$N(T)$ and $d(T)=$codim$R(T).$ If $R(T)$ is closed and $n(T)$ ($d(T)$) is finite, then $T$ is called an upper (a lower) semi-Fredholm operator. And the upper semi-Fredholm operators and the lower semi-Fredholm operators are called semi-Fredholm operators. If $T$ is a semi-Fredholm operator, the index of $T$ is written as $\mathrm{i}\mathrm{n}\mathrm{d}(T)=n(T)-d(T).$ In particular, if $T$ is a semi-Fredholm operator with $n(T)=\mathrm{0}$, then $T$ is said to be a bounded below operator. Moreover, if $T$ is upper semi-Fredholm with $\mathrm{i}\mathrm{n}\mathrm{d}(T)\le \mathrm{0}$, then we call that $T$ is an upper Weyl operator. The semi-Fredholm spectrum ${\sigma }_{\mathrm{S}\mathrm{F}}(T)$,the approximate point spectrum ${\sigma }_a(T)$ and the essential approximate point spectrum ${\sigma }_{ea}(T)$ are defined^[1,2] by

$\begin{array}{l}{\sigma }_{\mathrm{S}\mathrm{F}}(T)=\{\lambda \in \mathbb{C}:T-\lambda I\mathrm{i}\mathrm{s}\mathrm{n}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{s}\mathrm{e}\mathrm{m}\mathrm{i}\mathrm{‑}\mathrm{F}\mathrm{r}\mathrm{e}\mathrm{d}\mathrm{h}\mathrm{o}\mathrm{l}\mathrm{m}\mathrm{o}\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{o}\mathrm{r}\},\\ {\sigma }_a(T)=\{\lambda \in \mathbb{C}:T-\lambda I\mathrm{i}\mathrm{s}\mathrm{n}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{b}\mathrm{o}\mathrm{u}\mathrm{n}\mathrm{d}\mathrm{e}\mathrm{d}\mathrm{b}\mathrm{e}\mathrm{l}\mathrm{o}\mathrm{w}\mathrm{o}\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{o}\mathrm{r}\},\\ {\sigma }_{ea}(T)=\{\lambda \in \mathbb{C}:T-\lambda I\mathrm{i}\mathrm{s}\mathrm{n}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{u}\mathrm{p}\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{W}\mathrm{e}\mathrm{y}\mathrm{l}\mathrm{o}\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{o}\mathrm{r}\},\end{array}$

and let ${\rho }_{\mathrm{S}\mathrm{F}}(T)=\mathbb{C}\backslash {\sigma }_{\mathrm{S}\mathrm{F}}(T),$${\rho }_a(T)=\mathbb{C}\backslash {\sigma }_a(T),$${\rho }_{ea}(T)=\mathbb{C}\backslash {\sigma }_{ea}(T).$

If $n(T)$ and $d(T)$ are finite, then we call that $T$ is a Fredholm operator (Ref.[1], Theorem 1.53). Particularly, if $T$ is a Fredholm operator with $n(T)=d(T)$, then $T$ is called a Weyl operator. Moreover, if $T$ is a Fredholm operator of finite ascent asc(T) and descent des(T), then $T$ is called a Browder operator (Ref.[2], Chapter III, Definition 6), where

$\mathrm{a}\mathrm{s}\mathrm{c}(T)=\mathrm{i}\mathrm{n}\mathrm{f}\{n\in \mathbb{N}:N(T^n)=N(T^{n+\mathrm{1}})\},$

$\mathrm{d}\mathrm{e}\mathrm{s}(T)=\mathrm{i}\mathrm{n}\mathrm{f}\{n\in \mathbb{N}:R(T^n)=R(T^{n+\mathrm{1}})\}.$

As we all know, if $T$ is not invertible, then $T$ is Browder if and only if $T$ is semi-Fredholm and $\mathrm{0}$ is the boundary point of $\sigma (T)$ (Ref.[3], Theorem 7.9.3). The classes of operators defined above generate the following spectra: the Weyl spectrum ${\sigma }_w(T)$ and the Browder spectrum ${\sigma }_b(T)$ are defined by

$\begin{array}{l}{\sigma }_w(T)=\{\lambda \in \mathbb{C}:T-\lambda I\mathrm{i}\mathrm{s}\mathrm{n}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{W}\mathrm{e}\mathrm{y}\mathrm{l}\mathrm{o}\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{o}\mathrm{r}\},\\ {\sigma }_b(T)=\{\lambda \in \mathbb{C}:T-\lambda I\mathrm{i}\mathrm{s}\mathrm{n}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{B}\mathrm{r}\mathrm{o}\mathrm{w}\mathrm{d}\mathrm{e}\mathrm{r}\mathrm{o}\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{o}\mathrm{r}\},\end{array}$

and let ${\rho }_w(T)=\mathbb{C}\backslash {\sigma }_w(T)$, ${\rho }_b(T)=\mathbb{C}\backslash {\sigma }_b(T)$.

It is easy to prove ${\sigma }_w(T)\subseteq {\sigma }_b(T)$ and ${\sigma }_w(T)={\sigma }_w(T+F)$, where $F\in F(H)$. From Ref.[4], we know if $F\in F(H)$ commutes with $T$, then ${\sigma }_b(T)={\sigma }_b(T+F)$. ${\sigma }_{\mathrm{0}}(T)$ is denoted by the set of all normal eigenvalues, that is ${\sigma }_{\mathrm{0}}(T)=\sigma (T)\backslash {\sigma }_b(T)$. Then we have ${\sigma }_{\mathrm{0}}(T)\subseteq \mathrm{i}\mathrm{s}\mathrm{o}\sigma (T)$, where $\mathrm{i}\mathrm{s}\mathrm{o}\sigma (T)$ denotes the set of all isolated points of $\sigma (T)$.

Recall that Weyl's theorem holds for $T\in B(H)$ when the complement of the Weyl spectrum in the spectrum coincide with the set ${\pi }_{\mathrm{00}}(T)$, where ${\pi }_{\mathrm{00}}(T)$ is denoted by all the isolated points of $\sigma (T)$ and these points are eigenvalues with finite multiplicity. In 1909, Weyl^[5] discovered that for each Hermitian operator $T\in B(H)$, $\sigma (T)\backslash {\sigma }_w(T)={\pi }_{\mathrm{00}}(T)$ and since then the variations of Weyl's theorem have received a considerable attention^[6-11]. Property $(\omega )$ as one variant was given by Rakočević^[7]. In 2010, using the variant of essential approximate point spectrum, Sun et al^[11] characterized property $({\omega }_{\mathrm{1}})$, which was a necessary condition of property $(\omega )$, and discussed the relation between property $({\omega }_{\mathrm{1}})$ and hypercyclic (or supercyclic) property. $T\in B(H)$ has the Property $(\omega )$ if

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

where ${\pi }_{\mathrm{00}}(T)=\{\lambda \in \mathrm{i}\mathrm{s}\mathrm{o}\sigma (T):\mathrm{0}<n(T-\lambda I)<\mathrm{\infty }\}.$ In 2021, by the spectrum originated from the single-valued extension property, Dai et al^[12] characterized property $(\omega )$, and discussed the relation between property $(\omega )$ and hypercyclic (supercyclic) property.

In this paper, using the new spectrum originating from CFI property around an operator, we continue to study property $(\omega )$ and hypercyclic (or supercyclic) property. In Section 1, we conclude the necessary and sufficient conditions for $T\in B(H)$ satisfying the property $(\omega )$. In addition, the stability of the property $(\omega )$ is also studied. In Section 2, we considered the relations between property $(\omega )$ and hypercyclic (or supercyclic) property.

1 Property $(\mathit{\omega })$ and Its Perturbation

Now we begin with a definition and a lemma.

Definition 1   (Ref.[13], Definition 3.1) We say that $T\in B(H)$ is Consistence in Fredholm and Index (CFI for short), if for each $S\in B(H)$, one of the following cases occurs:

1) Both $TS$ and $ST$ are Fredholm, and $\mathrm{i}\mathrm{n}\mathrm{d}(TS)=\mathrm{i}\mathrm{n}\mathrm{d}(ST)=\mathrm{i}\mathrm{n}\mathrm{d}(S)$;

2) Both $TS$ and $ST$ are not Fredholm.

Lemma 1   (Ref.[13], Theorem 3.2) Let $T\in B(H)$. Then $T$ is a CFI operator if and only if one of the following three mutually disjoint cases occurs:

1) $T$ is Weyl;

2) $R(T)$ is closed and $n(T)=d(T)=\mathrm{\infty }$;

3) $R(T)$ is not closed.

It follows from Lemma 1 that $T\in B(H)$ is not CFI if and only if $T$ is semi-Fredholm and $\mathrm{i}\mathrm{n}\mathrm{d}(T)\ne \mathrm{0}$.

Following we define the new spectrum set. Let

${\rho }_{\mathrm{1}}(T)=\{\lambda \in \mathbb{C}:n(T-\lambda I)<\mathrm{\infty }$

$T-\mu I\mathrm{i}\mathrm{s}\mathrm{C}\mathrm{F}\mathrm{I}\mathrm{i}\mathrm{f}\mathrm{0}<|\mu -\lambda |<ϵ\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{s}\mathrm{o}\mathrm{m}\mathrm{e}ϵ>\mathrm{0}\}$

and let ${\sigma }_{\mathrm{1}}(T)=\mathbb{C}\backslash {\rho }_{\mathrm{1}}(T)$. From the perturbation of semi-Fredholm operators, if $\lambda \notin {\sigma }_w(T)$, then $n(T-\lambda I)<\mathrm{\infty }$, and $T-\mu I$ is Weyl if $\mathrm{0}<|\mu -\lambda |<ϵ$ for some $ϵ>\mathrm{0}$. It easily follows from Lemma 1 and the definition of ${\rho }_{\mathrm{1}}(T)$ that ${\sigma }_{\mathrm{1}}(T)\subseteq {\sigma }_w(T)\subseteq {\sigma }_b(T)\subseteq \sigma (T)$.

Recall that $T$ is isoloid if $\mathrm{i}\mathrm{s}\mathrm{o}\sigma (T)\subseteq {\sigma }_p(T)$, where ${\sigma }_p(T)$ is the point spectrum of $T$. And let ${\sigma }_d(T)=\{\lambda \in \mathbb{C}:R(T-\lambda I)\mathrm{i}\mathrm{s}\mathrm{n}\mathrm{o}\mathrm{t}\mathrm{c}\mathrm{l}\mathrm{o}\mathrm{s}\mathrm{e}\mathrm{d}\}$. Then we have the following theorem.

Theorem 1   Let $T\in B(H)$. Then $T$ is isoloid and has property $(\omega )$ if and only if

$\begin{array}{l}\sigma (T)=[{\sigma }_{\mathrm{1}}(T)\bigcap \{\lambda \in \mathbb{C}:n(T-\lambda I)\ge d(T-\lambda I)\}]\\ \bigcup [\mathrm{a}\mathrm{c}\mathrm{c}\sigma (T)\bigcap {\sigma }_d(T)]\bigcup {\sigma }_{\mathrm{0}}(T)\bigcup [{\rho }_a(T)\bigcap \sigma (T)],\end{array}$

where $\mathrm{a}\mathrm{c}\mathrm{c}\sigma (T)$ denotes the set of accumulation points of $\sigma (T)$.

Proof   Suppose $T$ is isoloid and has property $(\omega )$.. Let ${\lambda }_{\mathrm{0}}\notin [{\sigma }_{\mathrm{1}}(T)\bigcap \{\lambda \in \mathbb{C}:n(T-\lambda I)\ge d(T-\lambda I)\}]\bigcup [\mathrm{a}\mathrm{c}\mathrm{c}\sigma (T)\bigcap {\sigma }_d(T)]$$\bigcup {\sigma }_{\mathrm{0}}(T)\bigcup [{\rho }_a(T)\bigcap \sigma (T)]$. We claim that $n(T-{\lambda }_{\mathrm{0}}I)\ge d(T-{\lambda }_{\mathrm{0}}I)$. In fact, assume that $n(T-{\lambda }_{\mathrm{0}}I)<d(T-{\lambda }_{\mathrm{0}}I)$, then $n(T-{\lambda }_{\mathrm{0}}I)<\mathrm{\infty }$. Note that ${\lambda }_{\mathrm{0}}\notin \mathrm{a}\mathrm{c}\mathrm{c}\sigma (T)\bigcap {\sigma }_d(T)$. If ${\lambda }_{\mathrm{0}}\notin \mathrm{a}\mathrm{c}\mathrm{c}\sigma (T)$, then ${\lambda }_{\mathrm{0}}\in \mathrm{i}\mathrm{s}\mathrm{o}\sigma (T)\bigcup \rho (T)$. Since $T$ is isoloid and has property $(\omega )$, it follows from Chapter XI, Property 6.9 of Ref. [14] that ${\lambda }_{\mathrm{0}}\in {\pi }_{\mathrm{00}}(T)\bigcup \rho (T)\subseteq {\sigma }_{\mathrm{0}}(T)\bigcup \rho (T)$. Then $\mathrm{i}\mathrm{n}\mathrm{d}(T-{\lambda }_{\mathrm{0}}I)=\mathrm{0}$, which is a contradiction. If ${\lambda }_{\mathrm{0}}\notin {\sigma }_d(T)$, then $T-{\lambda }_{\mathrm{0}}I$ is an upper semi-Fredholm operator and $\mathrm{i}\mathrm{n}\mathrm{d}(T-{\lambda }_{\mathrm{0}}I)<\mathrm{0}$. Observing that ${\lambda }_{\mathrm{0}}\notin {\rho }_a(T)\bigcap \sigma (T)$, we know ${\lambda }_{\mathrm{0}}\in [{\sigma }_a(T)\backslash {\sigma }_{ea}(T)]\bigcup \rho (T)\subseteq {\sigma }_{\mathrm{0}}(T)\bigcap \rho (T)$, a contradiction. Therefore $n(T-{\lambda }_{\mathrm{0}}I)\ge d(T-{\lambda }_{\mathrm{0}}I)$, which implies ${\lambda }_{\mathrm{0}}\notin {\sigma }_{\mathrm{1}}(T)$. Hence $n(T-{\lambda }_{\mathrm{0}}I)<\mathrm{\infty }$ and $T-\lambda I$ is CFI if $\mathrm{0}<|\lambda -{\lambda }_{\mathrm{0}}|<ϵ$ for some $ϵ>\mathrm{0}$. Again note that ${\lambda }_{\mathrm{0}}\notin \mathrm{a}\mathrm{c}\mathrm{c}\sigma (T)\bigcap {\sigma }_d(T)$. Similarly, we know that if ${\lambda }_{\mathrm{0}}\notin \mathrm{a}\mathrm{c}\mathrm{c}\sigma (T)$, then ${\lambda }_{\mathrm{0}}\in {\sigma }_{\mathrm{0}}(T)\bigcup \rho (T)$. Since ${\lambda }_{\mathrm{0}}\notin {\sigma }_{\mathrm{0}}(T)$, it follows that ${\lambda }_{\mathrm{0}}\in \rho (T)$, i.e.,${\lambda }_{\mathrm{0}}\notin \sigma (T)$. If ${\lambda }_{\mathrm{0}}\notin {\sigma }_d(T)$, then $T-{\lambda }_{\mathrm{0}}I$ is an upper semi-Fredholm operator. By the punctured neighborhood theorem of semi-Fredholm operators and the definition of ${\rho }_{\mathrm{1}}(T)$, we conclude that $T-\lambda I$ is both semi-Fredholm and CFI if $\mathrm{0}<|\lambda -{\lambda }_{\mathrm{0}}|<ϵ$ for some $ϵ>\mathrm{0}$. Therefore $T-\lambda I$ is Weyl operator if $\mathrm{0}<|\lambda -{\lambda }_{\mathrm{0}}|<ϵ$. Now we can obtain that $T-{\lambda }_{\mathrm{0}}I$ is Weyl. The fact that property $(\omega )$ holds for $T$ tells us that ${\lambda }_{\mathrm{0}}\in {\rho }_b(T)$. Since ${\lambda }_{\mathrm{0}}\notin {\sigma }_{\mathrm{0}}(T)$, it follows that ${\lambda }_{\mathrm{0}}\in \rho (T)$, which means ${\lambda }_{\mathrm{0}}\notin \sigma (T)$. The opposite conclusion is clear. So we have

$\begin{array}{l}\sigma (T)=[{\sigma }_{\mathrm{1}}(T)\bigcap \{\lambda \in \mathbb{C}:n(T-\lambda I)\ge d(T-\lambda I)\}]\\ \bigcup [\mathrm{a}\mathrm{c}\mathrm{c}\sigma (T)\bigcap {\sigma }_d(T)]\bigcup {\sigma }_{\mathrm{0}}(T)\bigcup [{\rho }_a(T)\bigcap \sigma (T)].\end{array}$

Conversely, suppose $\sigma (T)=[{\sigma }_{\mathrm{1}}(T)\bigcap \{\lambda \in \mathbb{C}:n(T-\lambda I)$

$\ge d(T-\lambda I)\}]\bigcup [\mathrm{a}\mathrm{c}\mathrm{c}\sigma (T)\bigcap {\sigma }_d(T)]\bigcup {\sigma }_{\mathrm{0}}(T)\bigcup [{\rho }_a(T)\bigcap \sigma (T)].$ Since ${\sigma }_a(T)\backslash {\sigma }_{ea}(T)=\{\lambda \in {\sigma }_a(T)\backslash {\sigma }_{ea}(T):\mathrm{i}\mathrm{n}\mathrm{d}(T-\lambda I)<\mathrm{0}\}\bigcup$

$\{\lambda \in {\sigma }_a(T)\backslash$ ${\sigma }_{ea}(T):\mathrm{i}\mathrm{n}\mathrm{d}(T-\lambda I)=\mathrm{0}\}$, it follows from $\{\lambda \in {\sigma }_a(T)\backslash {\sigma }_{ea}(T):\mathrm{i}\mathrm{n}\mathrm{d}(T-\lambda I)<\mathrm{0}\}\bigcap \{[{\sigma }_{\mathrm{1}}(T)\bigcap \{\lambda \in \mathbb{C}:$$n(T-\lambda I)\ge d(T-\lambda I)\}]\bigcup [\mathrm{a}\mathrm{c}\mathrm{c}\sigma (T)\bigcap {\sigma }_d(T)]\bigcup {\sigma }_{\mathrm{0}}(T)\bigcup$

$[{\rho }_a(T)\bigcap \sigma (T)]\}=\mathrm{\varnothing }$ that $\{\lambda \in {\sigma }_a(T)\backslash {\sigma }_{ea}(T):\mathrm{i}\mathrm{n}\mathrm{d}(T-\lambda I)<\mathrm{0}\}$$\bigcap \sigma (T)=\mathrm{\varnothing }$.Then $\{\lambda \in {\sigma }_a(T)\backslash {\sigma }_{ea}(T):\mathrm{i}\mathrm{n}\mathrm{d}(T-\lambda I)<\mathrm{0}\}=\mathrm{\varnothing }$. Thus ${\sigma }_a(T)\backslash {\sigma }_{ea}(T)\subseteq {\rho }_w(T)$. Since ${\rho }_w(T)\bigcap \{[{\sigma }_{\mathrm{1}}(T)\bigcap$$\{\lambda \in \mathbb{C}:n(T-\lambda I)\ge d(T-\lambda I)\}]\bigcup [\mathrm{a}\mathrm{c}\mathrm{c}\sigma (T)\bigcap {\sigma }_d(T)]\bigcup {\sigma }_{\mathrm{0}}(T)$$\bigcup [{\rho }_a(T)\bigcap \sigma (T)]\}={\sigma }_{\mathrm{0}}(T)$, we obtain that ${\sigma }_a(T)\backslash {\sigma }_{ea}(T)\subseteq {\sigma }_{\mathrm{0}}(T)$. Then ${\sigma }_a(T)\backslash {\sigma }_{ea}(T)$$\subseteq {\pi }_{\mathrm{00}}(T)$. From the fact that

${\pi }_{\mathrm{00}}(T)\bigcap [{\sigma }_{\mathrm{1}}(T)\bigcap \{\lambda \in \mathbb{C}:n(T-\lambda I)\ge d(T-\lambda I)\}]=\mathrm{\varnothing }$

and

$\begin{array}{l}{\pi }_{\mathrm{00}}(T)\bigcap [\mathrm{a}\mathrm{c}\mathrm{c}\sigma (T)\bigcap {\sigma }_d(T)]=\mathrm{\varnothing },\\ {\pi }_{\mathrm{00}}(T)\bigcap [{\rho }_a(T)\bigcap \sigma (T)]=\mathrm{\varnothing },\end{array}$

We have that ${\pi }_{\mathrm{00}}(T)={\pi }_{\mathrm{00}}(T)\bigcap \sigma (T)={\pi }_{\mathrm{00}}(T)\bigcap {\sigma }_{\mathrm{0}}(T)$$\subseteq {\sigma }_{\mathrm{0}}(T)\subseteq {\sigma }_a(T)\backslash {\sigma }_{ea}(T)$. Thus ${\sigma }_a(T)\backslash {\sigma }_{ea}(T)$$={\pi }_{\mathrm{00}}(T)$, that is, property $(\omega )$ holds for $T$. Since $\{\lambda \in \mathrm{i}\mathrm{s}\mathrm{o}\sigma (T):n(T-\lambda I)=\mathrm{0}\}\bigcap \{[{\sigma }_{\mathrm{1}}(T)\bigcap \{\lambda \in \mathbb{C}:n(T-\lambda I)\ge d(T-\lambda I)\}]\bigcup [\mathrm{a}\mathrm{c}\mathrm{c}\sigma (T)\bigcap$${\sigma }_d(T)]\bigcup {\sigma }_{\mathrm{0}}(T)\bigcup [{\rho }_a(T)\bigcap \sigma (T)]\}=\mathrm{\varnothing }$, then $\{\lambda \in \mathrm{i}\mathrm{s}\mathrm{o}\sigma (T):$$n(T-\lambda I)=\mathrm{0}\}=\{\lambda \in \mathrm{i}\mathrm{s}\mathrm{o}\sigma (T):n(T-\lambda I)=\mathrm{0}\}\bigcap \sigma (T)=\mathrm{\varnothing }$. It follows that $T$ is isoloid.

Remark 1   From the proof of Theorem 1, we know that if $T$ is isoloid and has property $(\omega )$, then $\sigma (T)={\sigma }_{\mathrm{1}}(T)\bigcup [\mathrm{a}\mathrm{c}\mathrm{c}\sigma (T)\bigcap {\sigma }_d(T)]\bigcup {\sigma }_{\mathrm{0}}(T)\bigcup [{\rho }_a(T)\bigcap \sigma (T)].$ But the converse is not true.

For example, let $T_{\mathrm{1}},T_{\mathrm{2}}\in B({{l}}^{\mathrm{2}})$ be defined as

$\begin{array}{l}{T}_{\mathrm{1}}(x_{\mathrm{1}},x_{\mathrm{2}},x_{\mathrm{3}},\cdots )=(\mathrm{0,0},x_{\mathrm{1}},x_{\mathrm{2}},x_{\mathrm{3}},\cdots )\\ T_{\mathrm{2}}(x_{\mathrm{1}},x_{\mathrm{2}},x_{\mathrm{3}},\cdots )=(x_{\mathrm{2}},x_{\mathrm{3}},x_{\mathrm{4}},\cdots )\end{array}$

and let $T=\left(\begin{array}{cc}{T}_{\mathrm{1}}& \mathrm{0}\\ \mathrm{0}& T_{\mathrm{2}}\end{array}\right)$. Then $\sigma (T)={\sigma }_{\mathrm{1}}(T)=\mathbb{D}$, thus $\sigma (T)={\sigma }_{\mathrm{1}}(T)\bigcup [\mathrm{a}\mathrm{c}\mathrm{c}\sigma (T)\bigcap {\sigma }_d(T)]\bigcup {\sigma }_{\mathrm{0}}(T)\bigcup [{\rho }_a(T)$$\bigcap \sigma (T)]$. But since ${\sigma }_a(T)=\mathbb{D},{\sigma }_{ea}(T)=\mathrm{\Gamma },{\pi }_{\mathrm{00}}(T)=\mathrm{\varnothing }$, it follows that property $(\omega )$ does not hold for $T$.

Remark 2   If $T\in B(H)$ is isoloid and has property $(\omega )$, the spectrum $\sigma (T)$ can be decomposed as four blocks, and each block cannot be avoided.

1) "${\sigma }_{\mathrm{1}}(T)\bigcap \{\lambda \in \mathbb{C}:n(T-\lambda I)\ge d(T-\lambda I)\}$" cannot be avoided.

For example, let $T_{\mathrm{1}}\in B({{l}}^{\mathrm{2}})$ be defined as

$T_{\mathrm{1}}(x_{\mathrm{1}},x_{\mathrm{2}},x_{\mathrm{3}},\cdots )=(\mathrm{0},x_{\mathrm{1}},\mathrm{0},\frac{x_{\mathrm{3}}}{\mathrm{3}},\mathrm{0},\frac{x_{\mathrm{5}}}{\mathrm{5}},\mathrm{0},\cdots ),$

and let $T=I-T_{\mathrm{1}}$. Then

$(\mathrm{a}){\sigma }_a(T)={\sigma }_{ea}(T)=\{\mathrm{1}\},{\pi }_{\mathrm{00}}(T)=\mathrm{\varnothing }.$

$(\mathrm{b})\sigma (T)=\{\mathrm{1}\},\mathrm{a}\mathrm{c}\mathrm{c}\sigma (T)\bigcap {\sigma }_d(T)=\mathrm{\varnothing },{\sigma }_{\mathrm{0}}(T)=\mathrm{\varnothing },{\rho }_a(T)\bigcap \sigma (T)=\mathrm{\varnothing }.$

Therefore $T$ is isoloid and has property $(\omega )$, but $\sigma (T)\ne [\mathrm{a}\mathrm{c}\mathrm{c}\sigma (T)\bigcap {\sigma }_d(T)]\bigcup {\sigma }_{\mathrm{0}}(T)\bigcup [{\rho }_a(T)\bigcap \sigma (T)]$.

2) "$\mathrm{a}\mathrm{c}\mathrm{c}\sigma (T)\bigcap {\sigma }_d(T)$" cannot be avoided.

For example, let $T_{\mathrm{1}},T_{\mathrm{2}}\in B({{l}}^{\mathrm{2}})$ be defined as

$\begin{array}{l}{T}_{\mathrm{1}}(x_{\mathrm{1}},x_{\mathrm{2}},x_{\mathrm{3}},\cdots )=(\mathrm{0},x_{\mathrm{1}},x_{\mathrm{2}},x_{\mathrm{3}},\cdots ),\\ T_{\mathrm{2}}(x_{\mathrm{1}},x_{\mathrm{2}},x_{\mathrm{3}},\cdots )=(\mathrm{0},x_{\mathrm{1}},\frac{x_{\mathrm{2}}}{\mathrm{2}},\frac{x_{\mathrm{3}}}{\mathrm{3}},\cdots ),\end{array}$

and let $T=\left(\begin{array}{cc}{T}_{\mathrm{1}}& \mathrm{0}\\ \mathrm{0}& T_{\mathrm{2}}\end{array}\right)$. Then

$(\mathrm{a})\sigma (T)=\mathbb{D},{\sigma }_a(T)={\sigma }_{ea}(T)=\mathrm{\Gamma }\bigcup \{\mathrm{0}\},{\pi }_{\mathrm{00}}(T)=\mathrm{\varnothing }.$

$(\mathrm{b}){\sigma }_{\mathrm{1}}(T)\bigcap \{\lambda \in \mathbb{C}:n(T-\lambda I)\ge d(T-\lambda I)\}=\mathrm{\varnothing },{\sigma }_{\mathrm{0}}(T)=\mathrm{\varnothing },{\rho }_a(T)\bigcap \sigma (T)=\{\lambda \in \mathbb{C}:\mathrm{0}<|\lambda |<\mathrm{1}\}.$

Therefore $T$ is isoloid and property $(\omega )$ holds for $T$, but

$\sigma (T)\ne [{\sigma }_{\mathrm{1}}(T)\bigcap \{\lambda \in \mathbb{C}:n(T-\lambda I)\ge d(T-\lambda I)\}]\bigcup {\sigma }_{\mathrm{0}}(T)\bigcup$

$[{\rho }_a(T)\bigcap \sigma (T)].$

3) "${\sigma }_{\mathrm{0}}(T)$" cannot be avoided.

For example, let $T_{\mathrm{1}},T_{\mathrm{2}}\in B({{l}}^{\mathrm{2}})$ be defined as

$\begin{array}{l}{T}_{\mathrm{1}}(x_{\mathrm{1}},x_{\mathrm{2}},x_{\mathrm{3}},\cdots )=(x_{\mathrm{2}},x_{\mathrm{3}},x_{\mathrm{4}},\cdots ),\\ T_{\mathrm{2}}(x_{\mathrm{1}},x_{\mathrm{2}},x_{\mathrm{3}},\cdots )=(x_{\mathrm{1}},\mathrm{0,0},\cdots )\end{array}$

and let $T=\left(\begin{array}{cc}{T}_{\mathrm{1}}& \mathrm{0}\\ \mathrm{0}& T_{\mathrm{2}}+I\end{array}\right)$. Then

$(\mathrm{a})\sigma (T)={\sigma }_a(T)=\mathbb{D}\bigcup \{\mathrm{2}\},{\sigma }_{ea}(T)=\mathbb{D},{\pi }_{\mathrm{00}}(T)=\{\mathrm{2}\}.$

$(\mathrm{b}){\sigma }_{\mathrm{1}}(T)\bigcap \{\lambda \in \mathbb{C}:n(T-\lambda I)\ge d(T-\lambda I)\}=\mathbb{D},\mathrm{a}\mathrm{c}\mathrm{c}\sigma (T)\bigcap {\sigma }_d(T)=\mathrm{\Gamma },{\rho }_a(T)\bigcap \sigma (T)=\mathrm{\varnothing }.$

Therefore $T$ is isoloid and has property $(\omega )$, but $\sigma (T)\ne [{\sigma }_{\mathrm{1}}(T)\bigcap \{\lambda \in \mathbb{C}:n(T-\lambda I)\ge d(T-\lambda I)\}]\bigcup [\mathrm{a}\mathrm{c}\mathrm{c}\sigma (T)\bigcap {\sigma }_d(T)]\bigcup [{\rho }_a(T)\bigcap \sigma (T)].$

4) "$[{\rho }_a(T)\bigcap \sigma (T)]$" cannot be avoided.

For example, let $T\in B({{l}}^{\mathrm{2}})$ be defined as

$T(x_{\mathrm{1}},x_{\mathrm{2}},x_{\mathrm{3}},\cdots )=(\mathrm{0},x_{\mathrm{1}},x_{\mathrm{2}},x_{\mathrm{3}},\cdots ).$

Then

$(\mathrm{a})\sigma (T)={\sigma }_{\mathrm{1}}(T)=\mathbb{D},{\sigma }_a(T)={\sigma }_{ea}(T)=\mathrm{\Gamma },{\pi }_{\mathrm{00}}(T)=\mathrm{\varnothing }.$

$(\mathrm{b}){\sigma }_{\mathrm{1}}(T)\bigcap \{\lambda \in \mathbb{C}:n(T-\lambda I)\ge d(T-\lambda I)\}=\mathrm{\varnothing },$

$\mathrm{a}\mathrm{c}\mathrm{c}\sigma (T)\bigcap {\sigma }_d(T)=\mathrm{\Gamma },{\sigma }_{\mathrm{0}}(T)=\mathrm{\varnothing }.$

Therefor $T$ is isoloid and has propert $(\omega )$, but $\sigma (T)\ne [{\sigma }_{\mathrm{1}}(T)\bigcap \{\lambda \in \mathbb{C}:n(T-\lambda I)\ge d(T-\lambda I)\}]\bigcup [\mathrm{a}\mathrm{c}\mathrm{c}\sigma (T)\bigcap {\sigma }_d(T)]\bigcup {\sigma }_{\mathrm{0}}(T).$

Next, we will consider the stability of property $(\omega )$.

Remark 3   1) There exist $T\in B(H)$ and $F\in F(H)$ with $TF=FT$ such that property $(\omega )$ holds for $T$ but not holds for $T+F$.

For example, let $T_{\mathrm{1}}\in B({{l}}^{\mathrm{2}})$ be defined as

$T_{\mathrm{1}}(x_{\mathrm{1}},x_{\mathrm{2}},\cdots )=(\mathrm{0},x_{\mathrm{1}},\frac{x_{\mathrm{2}}}{\mathrm{2}},\frac{x_{\mathrm{3}}}{\mathrm{3}},\cdots ),$

and let $T=\left(\begin{array}{cc}{T}_{\mathrm{1}}& \mathrm{0}\\ \mathrm{0}& I\end{array}\right)$.Then ${\sigma }_a(T)={\sigma }_{ea}(T)=\{\mathrm{0,1}\},{\pi }_{\mathrm{00}}(T)=\mathrm{\varnothing }$. Hence $T$ has property $(\omega )$.

Let $T_{\mathrm{2}}\in B({{l}}^{\mathrm{2}})$ be defined as

$T_{\mathrm{2}}(x_{\mathrm{1}},x_{\mathrm{2}},\cdots )=(x_{\mathrm{1}},\mathrm{0,0},\cdots ),$

and let $F=\left(\begin{array}{cc}\mathrm{0}& \mathrm{0}\\ \mathrm{0}& -T_{\mathrm{2}}\end{array}\right)$. Then $F\in F(H)$, $TF=FT$ and $T+F=\left(\begin{array}{cc}{T}_{\mathrm{1}}& \mathrm{0}\\ \mathrm{0}& I-T_{\mathrm{2}}\end{array}\right)$. Hence ${\sigma }_a(T+F)={\sigma }_{ea}(T+F)=\{\mathrm{0,1}\}$,${\pi }_{\mathrm{00}}(T+F)=\{\mathrm{0}\}$. Therefore, $T+F$ does not have property $(\omega )$.

2) Suppose ${\sigma }_{\mathrm{1}}(T)=\mathrm{\varnothing }$. Then for every $F\in F(H)$ commuting with $T$, $T+F$ is isoloid and has property $(\omega )$ if and only if $T$ is isoloid and has property $(\omega )$.

In fact, we only need to prove the sufficiency. Suppose $T$ is isoloid and has property $(\omega )$. For any finite rank $F$ satisfying $TF=FT$, since ${\sigma }_{\mathrm{1}}(T)=\mathrm{\varnothing }$, it follows that ${\sigma }_{ea}(T)={\sigma }_{\mathrm{S}\mathrm{F}}(T)={\sigma }_w(T)$. Hence

${\sigma }_a(T+F)\backslash {\sigma }_{ea}(T+F)\subseteq {\rho }_w(T)={\rho }_b(T)\subseteq {\rho }_b(T+F).$ Thus

${\sigma }_a(T+F)\backslash {\sigma }_{ea}(T+F)\subseteq {\pi }_{\mathrm{00}}(T+F).$

Conversely, note that ${\pi }_{\mathrm{00}}(T+F)\subseteq \{\lambda \in \mathrm{i}\mathrm{s}\mathrm{o}\sigma (T):n(T-\lambda I)$

$<\mathrm{\infty }\}\bigcup \rho (T)$ and $T$ is isoloid, we know

${\pi }_{\mathrm{00}}(T+F)\subseteq {\pi }_{\mathrm{00}}(T)\bigcup \rho (T)\subseteq {\rho }_b(T)={\rho }_b(T+F).$

Therefore property $(\omega )$ holds for $T+F$.

Observing that $\{\lambda \in \mathrm{i}\mathrm{s}\mathrm{o}\sigma (T+F):n(T+F)=\mathrm{0}\}\subseteq$$\{\lambda \in \mathrm{i}\mathrm{s}\mathrm{o}\sigma (T):n(T-\lambda I)<\mathrm{\infty }\}\bigcup \rho (T)\subseteq {\pi }_{\mathrm{00}}(T)\bigcup \rho (T)\subseteq {\rho }_b(T)$$\subseteq {\rho }_b(T+F)$, we see $\{\lambda \in \mathrm{i}\mathrm{s}\mathrm{o}\sigma (T+F):n(T+F)=\mathrm{0}\}=\mathrm{\varnothing }$, that is, $T+F$ is isoloid.

3) Suppose ${\sigma }_{\mathrm{1}}(T)$ is finite. Then for any $F\in F(H)$ commuting with $T$, $T+F$ is isoloid and has property $(\omega )$ if and only if $T$ is isoloid and has property $(\omega )$.

In fact, since ${\sigma }_{\mathrm{1}}(T)$ is finite, we know that ${\sigma }_{\mathrm{S}\mathrm{F}}(T)={\sigma }_w(T)$. It is similar to the proof of the above statement, we can prove the statement (3) is true.

4) Suppose $T$ is quasinilpotent operator and has property $(\omega )$. Then for any finite rank $F$ commuting with $T$, $T+F$ has property $(\omega )$.

In fact, since $T$ is quasinilpotent and has property $(\omega )$, it follows that ${\sigma }_a(T)={\sigma }_{ea}(T)$, ${\pi }_{\mathrm{00}}(T)=\mathrm{\varnothing }$, then $n(T)=\mathrm{0}$ or $n(T)=\mathrm{\infty }$. As we know, if $n(T)=\mathrm{0}$, then the only finite operator commuting with $T$ is $F=\mathrm{0}$, thus $T+F=T$, which implies that $T+F$ has property $(\omega )$. Next if $n(T)=\mathrm{\infty }$, then $\sigma (T)=\{\mathrm{0}\}$. Note that $T$ is isoloid and has property $(\omega )$, from 3) of Remark 3, we know that for any finite rank operator $F$ commuting with $T$, $T+F$ has property $(\omega )$.

Theorem 2   Suppose $\mathrm{i}\mathrm{n}\mathrm{t}{\sigma }_{\mathrm{1}}(T)=\mathrm{\varnothing }$. Then the following statements are equivalent:

1) $T$ is isoloid and has property $(\omega )$;

2) For any $F\in F(H)$ with $TF=FT$, $T+F$ is isoloid and has property $(\omega )$.

Proof   We only need to prove $\mathrm{1})\Rightarrow \mathrm{2})$. Suppose $T$ is isoloid and has property $(\omega )$. From $\mathrm{i}\mathrm{n}\mathrm{t}{\sigma }_{\mathrm{1}}(T)=\mathrm{\varnothing }$ and the punctured neighborhood theorem of semi-Fredholm operators, we can get that ${\sigma }_{\mathrm{S}\mathrm{F}}(T)={\sigma }_w(T)={\sigma }_b(T)$, then $\begin{array}{l}{\sigma }_a(T+F)\backslash {\sigma }_{ea}(T+F)\subseteq {\rho }_{\mathrm{S}\mathrm{F}}(T+F)\\ ={\rho }_{\mathrm{S}\mathrm{F}}(T)={\rho }_w(T)={\rho }_b(T)\subseteq [\rho b(T+F)].\end{array}$