© Wuhan University 2025

0 Introduction

The Weyl type theorem of bounded linear operator can well reflect the structure characteristics and distribution of operator's spectral^[1-5]. Therefore, the Weyl type theorem is an important topic in spectral theory. In recent years, the research scope of Weyl type theorems has been extended from general operators to operator functions, operator matrices, etc. Numerous significant results have been obtained^[6-9]. Property $(W_E)$ and property $(R)$ are the latest variations of Weyl type theorems, which have attracted the attention and research of operator theorists^[10-12]. In this paper, by decomposing and constructing the operator spectrum and using the topological uniform scaling property, we give a new method for bounded linear operators and operator functions to satisfy the property $(W_E)$ and property $(R)$. Furthermore, the perturbation of property $(W_E)$ and property $(R)$ is characterized, and the necessary and sufficient conditions for the operator function to have both the property $(W_E)$ and property $(R)$ are studied.

Throughout this paper, $H$ denotes a complex separable infinite dimensional Hilbert space. Let $B(H)$ be the algebra of all bounded linear operators on $H$. For an operator $T\in B(H)$ we shall denote by $n(T)$ the dimension of the kernel $N(T)$, and by $d(T)$ the codimension of the range $R(T)$. We call $T\in B(H)$ is an upper semi-Fredholm operator if $n(T)<\mathrm{\infty }$ and $R(T)$ is closed. If $T\in B(H)$ is an upper semi-Fredholm operator and $n(T)=\mathrm{0}$, we call $T$ a bounded below operator. If $d(T)$$<\mathrm{\infty }$, $T$ is a lower semi-Fredholm operator. An operator $T\in B(H)$ is said to be Fredholm if $R(T)$ is closed and both $n(T)$ and $d(T)$ are finite. If $T\in B(H)$ is an upper (or a lower) semi-Fredholm operator, the index of $T$, $\mathrm{i}\mathrm{n}\mathrm{d}(T)$, is defined to be $\mathrm{i}\mathrm{n}\mathrm{d}(T)=$$n(T)-d(T)$. The ascent of $T$, $\mathrm{a}\mathrm{s}\mathrm{c}(T)$, is the least non-negative integer $n$ such that $N(T^n)=$$N(T^{n+\mathrm{1}})$ and the descent, $\mathrm{d}\mathrm{e}\mathrm{s}(T)$, is the least non-negative integer $n$ such that $R(T^n)$$=$$R(T^{n+\mathrm{1}})$. We call $T$ a Drazin invertible operator if $\mathrm{a}\mathrm{s}\mathrm{c}(T)$$=\mathrm{d}\mathrm{e}\mathrm{s}(T)<\mathrm{\infty }$. The operator $T$ is Weyl if it is Fredholm of index zero, and $T$ is said to be Browder if it is Fredholm "of finite ascent and descent". Let $\sigma (T)$ be the spectrum of $T$ and ${\sigma }_a(T)$ be the approximate point spectrum of $T$. We write ${\sigma }_w(T)$, ${\sigma }_b(T)$, ${\sigma }_e(T)$, ${\sigma }_{SF}(T)$, ${\sigma }_{ea}(T)$ and ${\sigma }_{ab}(T)$ for the Weyl spectrum of $T$, the Browder spectrum of $T$, the essential spectrum of $T$, the semi-Fredholm spectrum of $T$, the essential approximate point spectrum of $T$ and the Browder essential approximate point spectrum of $T$. Let $\rho (T)=$$\mathbb{C}\backslash \sigma (T)$, ${\rho }_a(T)=$$\mathbb{C}\backslash {\sigma }_a(T)$, ${\rho }_w(T)=$$\mathbb{C}\backslash {\sigma }_w(T)$, ${\rho }_b(T)=$$\mathbb{C}\backslash {\sigma }_b(T)$, ${\rho }_e(T)=$$\mathbb{C}\backslash {\sigma }_e(T)$, ${\rho }_{ab}(T)=$$\mathbb{C}\backslash {\sigma }_{ab}(T)$ ($\mathbb{C}$ denotes the set of complex numbers). $T\in B(H)$ is called an isoloid operator if $\mathrm{i}\mathrm{s}\mathrm{o}\sigma (T)$$\subseteq {\sigma }_p(T)$ where ${\sigma }_p(T)$$=\{\lambda \in \mathbb{C}:n(T-\lambda I)>\mathrm{0}\}$. For a set $X\subseteq \mathbb{C}$, we write $\mathrm{i}\mathrm{s}\mathrm{o}X$, $\mathrm{a}\mathrm{c}\mathrm{c}X$ and $\partial X$ for the set of isolated points, accumulation points and boundary points set of $X$. We denote by ${\sigma }_{\mathrm{0}}(T)$ the set of all normal eigenvalues of $T$, thus ${\sigma }_{\mathrm{0}}(T)=\sigma (T)\backslash {\sigma }_b(T)$. $T\in B(H)$ satisfies property $(W_E)$ if $\sigma (T)\backslash {\sigma }_w(T)=E(T)$, where $E(T)=\{\lambda \in \mathrm{i}\mathrm{s}\mathrm{o}\sigma (T):n(T-\lambda I)>$$\mathrm{0}\}$^[10]. $T\in B(H)$ satisfies property $(R)$ if ${\sigma }_a(T)\backslash {\sigma }_{ab}(T)={\pi }_{\mathrm{00}}(T)$, where ${\pi }_{\mathrm{00}}(T)=\{\lambda \in \mathrm{i}\mathrm{s}\mathrm{o}\sigma (T):\mathrm{0}<n(T-\lambda I)<+\mathrm{\infty }\}$^[10].

In this paper, we mainly study property $(W_E)$ and property $(R)$ for bounded linear operators and its functions. Some meaningful conclusions are obtained.

1 Judgement of Property $({\mathit{W}}_{\mathit{E}})$ and Property $(\mathit{R})$ for Bounded Linear Operators

Although both property $(W_E)$ and property $(R)$ are variations of Weyl's theorem based on their definitions, there is no necessary connection between them.

Remark 1   (i) $T\in B(H)$ satisfies property $(W_E)$$\nRightarrow$$T$ satisfies property $(R)$.

Let $A,B\in B({{l}}^{\mathrm{2}})$ be defined by $A(x_{\mathrm{1}},x_{\mathrm{2}},x_{\mathrm{3}},\cdots )=(\mathrm{0},x_{\mathrm{1}},x_{\mathrm{2}},\cdots )$, $B(x_{\mathrm{1}},x_{\mathrm{2}},x_{\mathrm{3}},\cdots )=(\mathrm{0},x_{\mathrm{2}},x_{\mathrm{3}},\cdots )$. Suppose that $T=\left(\begin{array}{cc}A& \mathrm{0}\\ \mathrm{0}& B\end{array}\right)$. Then $\sigma (T)=\{\lambda \in \mathbb{C}:\left|\lambda \right|\le \mathrm{1}\}={\sigma }_w(T)$, $E(T)={\pi }_{\mathrm{00}}(T)=\mathrm{\varnothing }$, ${\sigma }_a(T)=\{\lambda \in \mathbb{C}:|\lambda |=\mathrm{1}\}\bigcup \{\mathrm{0}\}$, ${\sigma }_{ab}(T)=\{\lambda \in \mathbb{C}:|\lambda |=\mathrm{1}\}$. $T$ satisfies property $(W_E)$, but property $(R)$ does not hold for $T$.

(ii) $T\in B(H)$ satisfies property $(R)$$\nRightarrow$$T$ satisfies property $(W_E)$.

Let $A,B\in B({{l}}^{\mathrm{2}})$ be defined by $A(x_{\mathrm{1}},x_{\mathrm{2}},x_{\mathrm{3}},\cdots )=(\mathrm{0},x_{\mathrm{1}},x_{\mathrm{2}},\cdots )$, $B(x_{\mathrm{1}},x_{\mathrm{2}},x_{\mathrm{3}},\cdots )=(x_{\mathrm{2}},x_{\mathrm{3}},\cdots )$, $T=\left(\begin{array}{cc}A& \mathrm{0}\\ \mathrm{0}& B\end{array}\right)$. We have that $\sigma (T)=\{\lambda \in \mathbb{C}:\left|\lambda \right|\le \mathrm{1}\}\ne {\sigma }_w(T)$, ${\sigma }_a(T)={\sigma }_{ab}(T)=\{\lambda \in \mathbb{C}:|\lambda |\le \mathrm{1}\}$. ${\pi }_{\mathrm{00}}(T)=E(T)=\mathrm{\varnothing }$. So $T$ satisfies property $(R)$ and $T$ does not have property $(W_E)$.

(iii) $T\in B(H)$ has property $(R)$ and property $(W_E)$$\iff$${\sigma }_w(T)={\sigma }_{ab}(T)\bigcup [{\rho }_a(T)\bigcap \sigma (T)]$ and ${\pi }_{\mathrm{00}}(T)=E(T)={\sigma }_{\mathrm{0}}(T)$.

Topological uniform descent is an important property of operators, which is widely used in spectral theory. If $T\in B(H)$, then for each nonnegative integer $n$, $T$ induces a linear transformation from the vector space $R(T^n)/R(T^{n+\mathrm{1}})$ to $R(T^{n+\mathrm{1}})/R(T^{n+\mathrm{2}})$. We denote $k_n(T)$ the dimension of the null space of the induced map and let $k(T)={\displaystyle \sum _{n=\mathrm{0}}^{\mathrm{\infty }}}{k}_n\left(T\right)$. The following definition was introduced by Grabiner^[13]. Let $T\in B(H)$, if there is a nonnegative integer $d$ for which $k_n(T)=\mathrm{0}$ for $n\ge d$ and $R(T^n)$ is closed in the operator range topology of $R(T^d)$ for $n\ge d$, then we say that $T$ has topological uniform descent.

It can be shown that if $T$ is upper semi-Fredholm, then $T$ has topological uniform descent. Let ${\rho }_{\tau }(T)=\{\lambda \in \mathbb{C}:T-\lambda I$ has topological uniform descent}, ${\sigma }_{\tau }(T)=\mathbb{C}\backslash {\rho }_{\tau }(T)$. Grabiner discovered many properties of topological uniform descent. We will use the following property (Ref. [13], Corollary 4.9): Suppose that $T\in B(H)$, $\lambda \in \partial \sigma (T)$, if $T-\lambda I$ has topological uniform descent, then $\lambda$ is a pole of $T$. Next, we will discuss property $(W_E)$ and property $(R)$ by using the topological uniform descent.

Lemma 1   Let $T\in B(H)$, the following statements are equivalent:

(1) $T$ satisfies property $(W_E)$;

(2) ${\sigma }_b(T)=[{\sigma }_{\tau }(T)\bigcap \mathrm{a}\mathrm{c}\mathrm{c}\sigma (T)]\bigcup \mathrm{a}\mathrm{c}\mathrm{c}{\sigma }_w(T)\bigcup \{\lambda \in \sigma (T):n(T-\lambda I)=\mathrm{0}\}$;

(3) ${\sigma }_b(T)=[{\sigma }_{\tau }(T)\bigcap \mathrm{a}\mathrm{c}\mathrm{c}{\sigma }_a(T)]\bigcup \mathrm{a}\mathrm{c}\mathrm{c}{\sigma }_{ea}(T)\bigcup \{\lambda \in \sigma (T):n(T-\lambda I)=\mathrm{0}\}\bigcup \mathrm{a}\mathrm{c}\mathrm{c}\{\lambda \in \mathbb{C}:n(T-\lambda I)<d(T-\lambda I)\}.$

Proof   (1)$\Rightarrow$(2). Suppose that $T$ satisfies property $(W_E)$. If ${\lambda }_{\mathrm{0}}\notin [{\sigma }_{\tau }(T)\bigcap \mathrm{a}\mathrm{c}\mathrm{c}\sigma (T)]\bigcup \mathrm{a}\mathrm{c}\mathrm{c}{\sigma }_w(T)\bigcup$$\{\lambda \in \sigma (T):n(T-\lambda I)=\mathrm{0}\}$, then there exists a deleted neighborhood $B^{°}({\lambda }_{\mathrm{0}})$ centered on ${\lambda }_{\mathrm{0}}$ such that for any $\lambda \in B^{°}({\lambda }_{\mathrm{0}})$, $T-\lambda I$ is a Weyl operator. Since $T$ has property $(W_E)$, we know that $T-\lambda I$ is a Browder operator. Then we get that ${\lambda }_{\mathrm{0}}\in \rho (T)\cup \partial \sigma (T)$. We assume that ${\lambda }_{\mathrm{0}}\in \partial \sigma (T)$. If ${\lambda }_{\mathrm{0}}\in {\rho }_{\tau }(T)$, we get that $T-{\lambda }_{\mathrm{0}}I$ is Drazin invertible. According to $n(T-{\lambda }_{\mathrm{0}}I)>\mathrm{0}$, we have ${\lambda }_{\mathrm{0}}\in E(T)$. Since $T$ satisfies property $(W_E)$, we can get that $T-{\lambda }_{\mathrm{0}}I$ is a Browder operator. If ${\lambda }_{\mathrm{0}}\notin \mathrm{a}\mathrm{c}\mathrm{c}\sigma (T)$, we can also get that ${\lambda }_{\mathrm{0}}\in E(T)$ and $T-{\lambda }_{\mathrm{0}}I$ is Browder operators. The inclusion "$\supseteq$" is obviously true.

(2)$\Rightarrow$(1). Since $\sigma (T)\backslash {\sigma }_w(T)]\bigcap [{\sigma }_{\tau }(T)\bigcap \mathrm{a}\mathrm{c}\mathrm{c}{\sigma }_a(T)]=\mathrm{\varnothing }$ and $[\sigma (T)\backslash {\sigma }_w(T)]\bigcap [\mathrm{a}\mathrm{c}\mathrm{c}{\sigma }_w(T)\bigcup$$\{\lambda \in \sigma (T):n(T-\lambda I)=\mathrm{0}\}]=\mathrm{\varnothing }$, we know that $\sigma (T)\backslash {\sigma }_w(T)\subseteq E(T)$. Similarly, $E(T)\subseteq$$\sigma (T)\backslash {\sigma }_w(T)$. Hence $T$ satisfies property $(W_E)$.

The fact that $\mathrm{a}\mathrm{c}\mathrm{c}{\sigma }_{ea}(T)\bigcup \mathrm{a}\mathrm{c}\mathrm{c}\{\lambda \in \mathbb{C}:n(T-\lambda I)<d(T-\lambda I)\}=\mathrm{a}\mathrm{c}\mathrm{c}{\sigma }_w(T)$ implies that (2)$\iff$(3).

The closeness of operator range is very important in spectral theory. According to the closeness of operator range, the following spectral set is defined: ${\sigma }_c(T)=\{\lambda \in \mathbb{C}:R(T-\lambda I)$ is not closed}.

Lemma 2   Let $T\in B(H)$, the following statements are equivalent:

(1) $T$ satisfies property $(R)$;

(2) ${\sigma }_b(T)=[{\sigma }_{\tau }(T)\bigcap \mathrm{a}\mathrm{c}\mathrm{c}\sigma (T)]\bigcup \mathrm{a}\mathrm{c}\mathrm{c}{\sigma }_{ab}(T)\bigcup \{\lambda \in \sigma (T):n(T-\lambda I)=\mathrm{0}\}\bigcup \{\lambda \in \mathbb{C}:n(T-\lambda I)=\mathrm{\infty }\}\bigcup [\mathrm{a}\mathrm{c}\mathrm{c}\{\lambda \in \mathbb{C}:n(T-\lambda I)<d(T$

$d(T-\lambda I)\}\bigcap {\sigma }_c(T)]$;

(3) ${\sigma }_b(T)=[{\sigma }_{\tau }(T)\bigcap \mathrm{a}\mathrm{c}\mathrm{c}{\sigma }_a(T)]\bigcup \mathrm{a}\mathrm{c}\mathrm{c}{\sigma }_{ab}(T)\bigcup \{\lambda \in \sigma (T):n(T-\lambda I)=\mathrm{0}\}\bigcup \{\lambda \in \mathbb{C}:n(T-\lambda I)=\mathrm{\infty }\}\bigcup [\mathrm{a}\mathrm{c}\mathrm{c}\{\lambda \in \mathbb{C}:n(T-\lambda I)<-\lambda$

$I)\}\bigcap {\sigma }_c(T)]$

Proof   (1)$\Rightarrow$(2). Suppose ${\lambda }_{\mathrm{0}}\notin [{\sigma }_{\tau }(T)\bigcap \mathrm{a}\mathrm{c}\mathrm{c}\sigma (T)]\bigcup \mathrm{a}\mathrm{c}\mathrm{c}{\sigma }_{ab}(T)\bigcup \{\lambda \in \sigma (T):n(T-\lambda I)=\mathrm{0}\}\bigcup \{\lambda \in \mathbb{C}:n(T-\lambda I)=\mathrm{\infty }\}\bigcup [\mathrm{a}\mathrm{c}\mathrm{c}\{\lambda$

$\in \mathbb{C}:n(T-\lambda I)<d(T-\lambda I)\}\bigcap {\sigma }_c(T)]$. Without loss of generality, we assume that ${\lambda }_{\mathrm{0}}\in \sigma (T)$. Then we have $\mathrm{0}<n(T-{\lambda }_{\mathrm{0}}I)<\mathrm{\infty }$. There exists a deleted neighborhood $B^{°}({\lambda }_{\mathrm{0}};\epsilon )$ centered on ${\lambda }_{\mathrm{0}}$ such that for any $\lambda \in B^{°}({\lambda }_{\mathrm{0}};\epsilon )$, $\lambda \in {\rho }_{ab}(T)$. If ${\lambda }_{\mathrm{0}}\notin \mathrm{a}\mathrm{c}\mathrm{c}\sigma (T)$, then ${\lambda }_{\mathrm{0}}\in {\pi }_{\mathrm{00}}(T)$. Since $T$ satisfies property $(R)$, we have that $T-{\lambda }_{\mathrm{0}}I$ is a Browder operator. Suppose that ${\lambda }_{\mathrm{0}}\in {\rho }_{\tau }(T)$, if ${\lambda }_{\mathrm{0}}\in {\rho }_c(T)$, then $T-{\lambda }_{\mathrm{0}}I$ is an upper semi-Fredholm operator. By $\forall \lambda \in B^{°}({\lambda }_{\mathrm{0}};\epsilon )$, $\lambda \in {\rho }_{ab}(T)$ and ${\lambda }_{\mathrm{0}}\in \mathrm{i}\mathrm{s}\mathrm{o}{\sigma }_a(T)$, we know that ${\lambda }_{\mathrm{0}}\in {\sigma }_a(T)\backslash {\sigma }_{ab}(T)$. Because $T$ has property $(R)$, we can get that ${\lambda }_{\mathrm{0}}\in {\rho }_b(T)$. If ${\lambda }_{\mathrm{0}}\notin \mathrm{a}\mathrm{c}\mathrm{c}\{\lambda \in \mathbb{C}:n(T-\lambda I)<d(T-\lambda I)\}$, then exists a deleted neighborhood $B^{°}({\lambda }_{\mathrm{0}})\subseteq B^{°}({\lambda }_{\mathrm{0}};\epsilon )$ centered on ${\lambda }_{\mathrm{0}}$ such that for any $\lambda \in B^{°}({\lambda }_{\mathrm{0}})$, $\lambda \in {\rho }_b(T)$. Therefore ${\lambda }_{\mathrm{0}}\in \partial \sigma (T)$. Since ${\lambda }_{\mathrm{0}}\in {\rho }_{\tau }(T)$ and $n(T-{\lambda }_{\mathrm{0}}I)<\mathrm{\infty }$, we can also get that ${\lambda }_{\mathrm{0}}\in {\rho }_b(T)$.

(2)$\Rightarrow$(1). Suppose that ${\lambda }_{\mathrm{0}}\in {\sigma }_a(T)\backslash {\sigma }_{ab}(T)$. It follows that ${\lambda }_{\mathrm{0}}\notin [{\sigma }_{\tau }(T)\bigcap \mathrm{a}\mathrm{c}\mathrm{c}\sigma (T)]\bigcup \mathrm{a}\mathrm{c}\mathrm{c}{\sigma }_{ab}(T)\bigcup \{\lambda \in \sigma (T):n(T-\lambda I)=\mathrm{0}\}\bigcup \{\lambda \in \mathbb{C}:n(T-\lambda I)=\mathrm{\infty }\}\bigcup [\mathrm{a}\mathrm{c}\mathrm{c}\{\lambda \in \mathbb{C}:n(T-\lambda I)<$$d(T-\lambda I)\}\bigcap {\sigma }_c(T)]$. So $T-{\lambda }_{\mathrm{0}}I$ is a Browder operator. Then we have ${\sigma }_a(T)\backslash {\sigma }_{ab}(T)\subseteq {\pi }_{\mathrm{00}}(T)$. Similarly, we can get the inclusion "$\supseteq$". Hence $T$ satisfies property $(R)$.

If (2) holds, then $T$ has property $(R)$. This implies that $\{\lambda \in \mathbb{C}:\mathrm{0}<n(T-\lambda I)<\mathrm{\infty }\}\bigcap {\rho }_c(T)\bigcap \mathrm{a}\mathrm{c}\mathrm{c}[{\rho }_a(T)\bigcap \sigma (T)]=\mathrm{\varnothing }$, where ${\rho }_c(T)=\mathbb{C}\backslash {\sigma }_c(T)$. It follows that $\mathrm{a}\mathrm{c}\mathrm{c}\sigma (T)=\mathrm{a}\mathrm{c}\mathrm{c}{\sigma }_a(T)\bigcup \{\mathrm{a}\mathrm{c}\mathrm{c}[{\rho }_a(T)\bigcap \sigma (T)]\bigcap {\sigma }_c(T)\}\bigcup \{\mathrm{a}\mathrm{c}\mathrm{c}[{\rho }_a(T)\bigcap \sigma (T)]\bigcap {\rho }_c(T)\bigcap \{\lambda$

$\in \sigma (T):n(T-\lambda I)=\mathrm{0}\}\}\bigcup \{\mathrm{a}\mathrm{c}\mathrm{c}[{\rho }_a(T)\bigcap \sigma (T)]\bigcap {\rho }_c(T)\bigcap \{\lambda \in \mathbb{C}:n(T-\lambda I)=\mathrm{\infty }\}\}$. Hence $[{\sigma }_{\tau }(T)\bigcap \mathrm{a}\mathrm{c}\mathrm{c}\sigma (T)]\subseteq \{[{\sigma }_{\tau }(T)\bigcap \mathrm{a}\mathrm{c}\mathrm{c}{\sigma }_a(T)]$

$\bigcup$ $\{\lambda \in \sigma (T):n(T-\lambda I)=\mathrm{0}\}\bigcup \{\lambda \in \mathbb{C}:n(T-\lambda I)=\mathrm{\infty }\}\bigcup [\mathrm{a}\mathrm{c}\mathrm{c}\{\lambda \in \mathbb{C}:$ $n(T-\lambda I)<d(T-$ $\lambda I)\}\bigcap {\sigma }_c(T)]\}$. Then we have that ${\sigma }_b(T)\subseteq [{\sigma }_{\tau }$

$(T)\bigcap \mathrm{a}\mathrm{c}\mathrm{c}\sigma (T)]\bigcup \mathrm{a}\mathrm{c}\mathrm{c}{\sigma }_{ab}(T)\bigcup \{\lambda \in \sigma (T):n(T-\lambda I)=\mathrm{0}\}\bigcup \{\lambda \in \mathbb{C}:n(T-\lambda I)=\mathrm{\infty }\}\bigcup [\mathrm{a}\mathrm{c}\mathrm{c}\{\lambda \in \mathbb{C}:n(T-\lambda I)<d(T-\lambda I)\}\bigcap {\sigma }_c(T)]$

The inclusion "$\supseteq$" is obviously true. This implies that (3) holds. For the converse, if (3) holds, by $\mathrm{a}\mathrm{c}\mathrm{c}{\sigma }_a(T)\subseteq \mathrm{a}\mathrm{c}\mathrm{c}\sigma (T)$ we know that (2) holds. Therefore (2)$\iff$(3).

Based on Lemma 1 and Lemma 2, the following results demonstrate that the two properties can be valid at the same time:

Theorem 1   $T\in B(H)$ satisfies property $(R)$ and property $(W_E)$ if and only if $\begin{array}{l}{\sigma }_b(T)=[{\sigma }_{\tau }(T)\bigcap \mathrm{a}\mathrm{c}\mathrm{c}{\sigma }_a(T)]\bigcup [\mathrm{a}\mathrm{c}\mathrm{c}{\sigma }_{ab}\\ (T)\bigcap {\sigma }_w(T)]\bigcup \{\lambda \in \sigma (T):n(T-\lambda I)=\mathrm{0}\}\bigcup \{\lambda \in \mathrm{a}\mathrm{c}\mathrm{c}\sigma (T):n(T-\lambda I)=\mathrm{\infty }\}\bigcup [\mathrm{a}\mathrm{c}\mathrm{c}\{\lambda \in \mathbb{C}:n(T-\lambda I)<d(T-\lambda I)\}\bigcap {\sigma }_c(T)].\end{array}$

Proof   "$\Rightarrow$". Suppose $T$ has property $(R)$. Using Lemma 2, we have that $\begin{array}{l}{\sigma }_b(T)=[{\sigma }_{\tau }(T)\bigcap \mathrm{a}\mathrm{c}\mathrm{c}{\sigma }_a(T)]\bigcup \mathrm{a}\mathrm{c}\mathrm{c}{\sigma }_{ab}(T)\bigcup \\ \{\lambda \in \sigma (T):n(T-\lambda I)=\mathrm{0}\}\bigcup \{\lambda \in \mathbb{C}:n(T-\lambda I)=\mathrm{\infty }\}\bigcup [\mathrm{a}\mathrm{c}\mathrm{c}\{\lambda \in \mathbb{C}:n(T-\lambda I)<d(T-\lambda I)\}\bigcap {\sigma }_c(T)].\end{array}$ Since $T$ has property $(W_E)$, we know that ${\rho }_w(T)={\rho }_b(T)andE(T)={\sigma }_{\mathrm{0}}(T).Hence\mathrm{a}\mathrm{c}\mathrm{c}{\sigma }_{ab}(T)\bigcap {\rho }_w(T)=\mathrm{\varnothing },\{\lambda \in \mathrm{i}\mathrm{s}\mathrm{o}\sigma (T):n(T-\lambda I)=\mathrm{\infty }\}=\mathrm{\varnothing }.$ Thus $\begin{array}{l}\mathrm{a}\mathrm{c}\mathrm{c}{\sigma }_{ab}(T)=[\mathrm{a}\mathrm{c}\mathrm{c}{\sigma }_{ab}(T)\bigcap {\sigma }_w(T)]\bigcup [\mathrm{a}\mathrm{c}\mathrm{c}{\sigma }_{ab}(T)\bigcap {\rho }_w(T)]=\mathrm{a}\mathrm{c}\mathrm{c}{\sigma }_{ab}(T)\bigcap {\sigma }_w(T).\{\lambda \in \mathbb{C}:n(T-\lambda I)=\mathrm{\infty }\}=\{\lambda \in \mathrm{a}\mathrm{c}\mathrm{c}\sigma (T):n(T-\lambda I)=\mathrm{\infty }\}\\ \bigcup \{\lambda \in \mathrm{i}\mathrm{s}\mathrm{o}\sigma (T):n(T-\lambda I)=\mathrm{\infty }\}=\{\lambda \in \mathrm{a}\mathrm{c}\mathrm{c}\sigma (T):n(T-\lambda I)=\mathrm{\infty }\}.Therefore{\sigma }_b(T)=[{\sigma }_{\tau }(T)\bigcap \mathrm{a}\mathrm{c}\mathrm{c}{\sigma }_a(T)]\bigcup [\mathrm{a}\mathrm{c}\mathrm{c}{\sigma }_{ab}(T)\bigcap {\sigma }_w(T)]\bigcup \\ \{\lambda \in \sigma (T):n(T-\lambda I)=\mathrm{0}\}\bigcup \{\lambda \in \mathrm{a}\mathrm{c}\mathrm{c}\sigma (T):n(T-\lambda I)=\mathrm{\infty }\}\bigcup [\mathrm{a}\mathrm{c}\mathrm{c}\{\lambda \in \mathbb{C}:n(T-\lambda I)<d(T-\lambda I)\}\bigcap {\sigma }_c(T)].\end{array}$

"$\Leftarrow$". By the condition, we get that ${\sigma }_b(T)\subseteq [{\sigma }_{\tau }(T)\bigcap \mathrm{a}\mathrm{c}\mathrm{c}{\sigma }_a(T)]\bigcup \mathrm{a}\mathrm{c}\mathrm{c}{\sigma }_{ab}(T)\bigcup \{\lambda \in \sigma (T):n(T-\lambda I)=\mathrm{0}\}\bigcup \{\lambda \in \mathbb{C}:n(T-\lambda I)=\mathrm{\infty }\}\bigcup [\mathrm{a}\mathrm{c}\mathrm{c}\{\lambda \in \mathbb{C}:n(T-\lambda I)<d(T-\lambda I)\}\bigcap {\sigma }_c(T)].$ The inclusion "$\supseteq$" is obviously true. From Lemma 2, we know that $T$ satisfies property $(R)$.

If ${\lambda }_{\mathrm{0}}\in \sigma (T)\backslash {\sigma }_w(T)$, then ${\lambda }_{\mathrm{0}}\notin [{\sigma }_{\tau }(T)\bigcap \mathrm{a}\mathrm{c}\mathrm{c}{\sigma }_a(T)]\bigcup [\mathrm{a}\mathrm{c}\mathrm{c}{\sigma }_{ab}(T)\bigcap {\sigma }_w(T)]\bigcup \{\lambda \in \sigma (T):n(T-\lambda I)=\mathrm{0}\}\bigcup \{\lambda \in \mathrm{a}\mathrm{c}\mathrm{c}\sigma (T):n(T-\lambda I)=\mathrm{\infty }\}\bigcup [\mathrm{a}\mathrm{c}\mathrm{c}\{\lambda \in \mathbb{C}:n(T-\lambda I)<d(T-\lambda I)\}\bigcap {\sigma }_c(T)].ThusT-{\lambda }_{\mathrm{0}}I$ is a Browder operator, $\sigma (T)\backslash {\sigma }_w(T)\subseteq E(T)$. The converse is similar. Therefore $T$ satisfies property $(W_E)$.

Remark 2   (i) In Theorem 1, suppose $T\in B(H)$ satisfies property $(R)$ and property $(W_E)$, then each part of the decomposition of ${\sigma }_b(T)$ cannot be deleted.

($\mathrm{a}$) Let $T\in B({{l}}^{\mathrm{2}})$ be defined by $T(x_{\mathrm{1}},x_{\mathrm{2}},x_{\mathrm{3}},\cdots )=(\mathrm{0},\frac{x_{\mathrm{2}}}{\mathrm{2}},\frac{x_{\mathrm{3}}}{\mathrm{3}},\cdots )$. Hence $T$ has property $(R)$ and property $(W_E)$. But ${\sigma }_b(T)\ne [\mathrm{a}\mathrm{c}\mathrm{c}{\sigma }_{ab}(T)\bigcap {\sigma }_w(T)]\bigcup \{\lambda \in \sigma (T):n(T-\lambda I)=\mathrm{0}\}\bigcup \{\lambda \in \mathrm{a}\mathrm{c}\mathrm{c}\sigma (T):n(T-\lambda I)=\mathrm{\infty }\}\bigcup [\mathrm{a}\mathrm{c}\mathrm{c}\{\lambda \in \mathbb{C}:n(T-\lambda I)<d(T-\lambda I)\}\bigcap {\sigma }_c(T)].$Thus $[{\sigma }_{\tau }(T)\bigcap \mathrm{a}\mathrm{c}\mathrm{c}{\sigma }_a(T)]$ cannot be deleted.

($\mathrm{b}$) Let $T\in B({{l}}^{\mathrm{2}})$ be defined by $T(x_{\mathrm{1}},x_{\mathrm{2}},x_{\mathrm{3}},\cdots )=(x_{\mathrm{2}},x_{\mathrm{3}},x_{\mathrm{4}},\cdots )$. Then $T$ has property $(R)$ and property $(W_E)$. But ${\sigma }_b(T)\ne [{\sigma }_{\tau }(T)\bigcap \mathrm{a}\mathrm{c}\mathrm{c}{\sigma }_a(T)]\bigcup \{\lambda \in \sigma (T):n(T-\lambda I)=\mathrm{0}\}$$\bigcup \{\lambda \in \mathrm{a}\mathrm{c}\mathrm{c}\sigma (T):$$n(T-\lambda I)=\mathrm{\infty }\}\bigcup [\mathrm{a}\mathrm{c}\mathrm{c}\{\lambda \in \mathbb{C}:n(T-\lambda I)<d(T-\lambda I)\}\bigcap {\sigma }_c(T)]$. Thus $[\mathrm{a}\mathrm{c}\mathrm{c}{\sigma }_{ab}(T)\bigcap {\sigma }_w(T)]$ cannot be deleted.

($\mathrm{c}$) Let $T\in B({{l}}^{\mathrm{2}})$ be defined by $T(x_{\mathrm{1}},x_{\mathrm{2}},x_{\mathrm{3}},\cdots )=(\mathrm{0},x_{\mathrm{1}},x_{\mathrm{2}},x_{\mathrm{3}},\cdots )$. We know that $T$ has property $(R)$ and property $(W_E)$. But ${\sigma }_b(T)\ne [{\sigma }_{\tau }(T)\bigcap \mathrm{a}\mathrm{c}\mathrm{c}{\sigma }_a(T)]\bigcup [\mathrm{a}\mathrm{c}\mathrm{c}{\sigma }_{ab}(T)\bigcap {\sigma }_w(T)]\bigcup \{\lambda \in \mathrm{a}\mathrm{c}\mathrm{c}\sigma (T):$$n(T-\lambda I)$$=\mathrm{\infty }\}\bigcup [\mathrm{a}\mathrm{c}\mathrm{c}\{\lambda \in \mathbb{C}:n(T-\lambda I)<d(T-\lambda I)\}\bigcap {\sigma }_c(T)]$. Hence $\{\lambda \in \sigma (T):n(T-\lambda I)=\mathrm{0}\}$ cannot be deleted.

($\mathrm{d}$) Let $A,B\in B({{l}}^{\mathrm{2}})$ be defined by $A(x_{\mathrm{1}},x_{\mathrm{2}},\cdots )=(\mathrm{0},x_{\mathrm{1}},x_{\mathrm{2}},\cdots )$, $B(x_{\mathrm{1}},x_{\mathrm{2}},x_{\mathrm{3}},\cdots )=(x_{\mathrm{1}},\mathrm{0},x_{\mathrm{3}},\mathrm{0},\cdots )$. And suppose $T=\left(\begin{array}{cc}A& \mathrm{0}\\ \mathrm{0}& B\end{array}\right)$. Then $T$ satisfies property $(R)$ and property $(W_E)$. But ${\sigma }_b(T)\ne [{\sigma }_{\tau }(T)\bigcap$$\mathrm{a}\mathrm{c}\mathrm{c}{\sigma }_a(T)]\bigcup$$[\mathrm{a}\mathrm{c}\mathrm{c}{\sigma }_{ab}(T)\bigcap {\sigma }_w(T)]\bigcup \{\lambda \in \sigma (T):n(T-\lambda I)=\mathrm{0}\}\bigcup [\mathrm{a}\mathrm{c}\mathrm{c}\{\lambda \in \mathbb{C}:n(T-\lambda I)<$$d(T-\lambda I)\}$$\bigcap {\sigma }_c(T)]$. Therefore $\{\lambda \in \mathrm{a}\mathrm{c}\mathrm{c}\sigma (T):n(T-\lambda I)=\mathrm{\infty }\}$ cannot be deleted.

($\mathrm{e}$) Let $A,B\in B({{l}}^{\mathrm{2}})$ be defined by $A(x_{\mathrm{1}},x_{\mathrm{2}},\cdots )=(\mathrm{0},x_{\mathrm{1}},x_{\mathrm{2}},\cdots )$, $B(x_{\mathrm{1}},x_{\mathrm{2}},x_{\mathrm{3}},\cdots )=$$(\mathrm{0,0},\frac{x_{\mathrm{2}}}{\mathrm{2}},\frac{x_{\mathrm{3}}}{\mathrm{3}},\cdots )$. And suppose $T=\left(\begin{array}{cc}A& \mathrm{0}\\ \mathrm{0}& B\end{array}\right)$. Thus $T$ satisfies property $(R)$ and property $(W_E)$. But ${\sigma }_b(T)\ne [{\sigma }_{\tau }(T)\bigcap \mathrm{a}\mathrm{c}\mathrm{c}{\sigma }_a(T)]\bigcup [\mathrm{a}\mathrm{c}\mathrm{c}{\sigma }_{ab}(T)\bigcap$${\sigma }_w(T)]\bigcup \{\lambda \in \sigma (T):n(T-\lambda I)=\mathrm{0}\}\bigcup$$\{\lambda \in$$\mathrm{a}\mathrm{c}\mathrm{c}\sigma (T):$$n(T-\lambda I)=\mathrm{\infty }\}$. It follows that $[\mathrm{a}\mathrm{c}\mathrm{c}\{\lambda \in \mathbb{C}:n(T-\lambda I)<d(T-\lambda I)\}\bigcap {\sigma }_c(T)]$ cannot be deleted.