An elementary operator and generalized Weyl’s theorem

Abstract

A Hilbert space operator T belongs to class A if $|T^2|-{|T|}^2\ge 0$. Let $d_{AB}$ denote either ${\delta }_{AB}$ or ${\mathrm{△}}_{AB}$, where ${\delta }_{AB}$ and ${\mathrm{△}}_{AB}$ denote the generalized derivation and the elementary operator on a Banach space $B(\mathcal{H})$ defined by ${\delta }_{AB}X=AX-XB$ and ${\mathrm{△}}_{AB}X=AXB-X$ respectively. If A and $B^{\ast }$ are class A operators, we show that $d_{AB}$ is polaroid and generalized Weyl’s theorem holds for $f(d_{AB})$, generalized a-Weyl’s theorem holds for $f({(d_{AB})}^{\ast })$ for every $f\in H(\sigma (d_{AB}))$ and f is not constant on each connected component of the open set U containing $\sigma (d_{AB})$, where $H(\sigma (d_{AB}))$ denotes the set of all analytic functions in a neighborhood of $\sigma (d_{AB})$.

MSC:47B20, 47A63.

1 Introduction

Let ℋ be a complex Hilbert space and ℂ be the set of complex numbers. Let $B(\mathcal{H})$ and $K(\mathcal{H})$ denote the $C^{\ast }$-algebra of all bounded linear operators and the ideal of compact operators acting on ℋ respectively. For operators $A,B\in B(\mathcal{H})$, let ${\delta }_{AB}\in B(B(\mathcal{H}))$ denote the generalized derivation on a Banach space $B(\mathcal{H})$ defined by ${\delta }_{AB}X=AX-XB$; let ${\mathrm{△}}_{AB}\in B(B(\mathcal{H}))$ denote the elementary operator on a Banach space $B(\mathcal{H})$ defined by ${\mathrm{△}}_{AB}=AXB-X$. Let $d_{AB}$ denote either ${\delta }_{AB}$ or ${\mathrm{△}}_{AB}$. $d_{AB}$ has been studied by a number of authors [1–4]. Also let $\alpha (T)=dimkerT$, $\beta (T)=dimker{T}^{\ast }$, and let $\sigma (T)$, ${\sigma }_a(T)$ denote the spectrum and approximate point spectrum of T. An operator $T\in B(\mathcal{H})$ is called upper (resp. lower) semi-Fredholm if ranT is closed and $\alpha (T)<\mathrm{\infty }$ (resp. $\beta (T)<\mathrm{\infty }$). In the sequel, let $S{F}_{+}(\mathcal{H})$ denote the set of all upper semi-Fredholm operators. If both $\alpha (T)$ and $\beta (T)$ are finite, then T is called a Fredholm operator. An operator $T\in B(\mathcal{H})$ is called Weyl if it is Fredholm of index zero and Browder if it is Fredholm of finite ascent and descent. Let ${\sigma }_e(T)$, ${\sigma }_w(T)$ and ${\sigma }_b(T)$ denote the essential spectrum, the Weyl spectrum and the Browder spectrum of $T\in B(\mathcal{H})$. Let $iso\mathcal{K}$ denote the isolated points of $\mathcal{K}\subseteq \mathbb{C}$. We write ${\pi }_{00}(T)=\{\lambda \in iso\sigma (T):0<\alpha (T-\lambda )<\mathrm{\infty }\}$, ${\pi }_{00}^a(T)=\{\lambda \in iso{\sigma }_a(T):0<\alpha (T-\lambda )<\mathrm{\infty }\}$, and $p_{00}(T)=\sigma (T)\setminus {\sigma }_b(T)$. It is evident that ${\sigma }_e(T)\subseteq {\sigma }_w(T)\subseteq {\sigma }_b(T)={\sigma }_e(T)\bigcup acc\sigma (T)$ and $p_{00}(T)\subseteq {\pi }_{00}(T)\subseteq {\pi }_{00}^a(T)$, where $acc\sigma (T)=\sigma (T)\setminus iso\sigma (T)$.

We say that Weyl’s theorem holds for $T\in B(\mathcal{H})$ if

$$\sigma (T)\setminus {\sigma }_w(T)={\pi }_{00}(T),$$

and that Browder’s theorem holds for $T\in B(\mathcal{H})$ if

$$\sigma (T)\setminus {\sigma }_w(T)=p_{00}(T).$$

By definition, ${\sigma }_{ea}(T)=\bigcap \{{\sigma }_a(T+K):K\in K(\mathcal{H})\}$ is the essential approximate point spectrum of T, and ${\sigma }_{ab}(T)=\bigcap \{{\sigma }_a(T+K):K\in K(\mathcal{H})\text{ and }KT=TK\}$ is the Browder approximate point spectrum of T.

We say that a-Weyl’s theorem holds for $T\in B(\mathcal{H})$ if

$${\sigma }_a(T)\setminus {\sigma }_{ea}(T)={\pi }_{00}^a(T),$$

and that a-Browder’s theorem holds for $T\in B(\mathcal{H})$ if

$${\sigma }_{ea}(T)={\sigma }_{ab}(T).$$

For a bounded linear operator T and a nonnegative integer n, define $T_n$ to be the restriction of T to $ran{T}^n$ viewed as a map from $ran{T}^n$ into $ran{T}^n$ (in particular $T_0=T$). If for some integer n, the range space $ran{T}^n$ is closed and $T_n$ is a Fredholm operator, then T is called a B-Fredholm operator. If T is a B-Fredholm operator of index zero, then T is called a B-Weyl operator. The B-Fredholm spectrum ${\sigma }_{\mathrm{BF}}(T)$ and B-Weyl spectrum ${\sigma }_{\mathrm{BW}}(T)$ of T are defined by ${\sigma }_{\mathrm{BF}}(T)=\{\lambda \in \mathbb{C}:T-\lambda \text{ is not a B-Fredholm operator}\}$ and ${\sigma }_{\mathrm{BW}}(T)=\{\lambda \in \mathbb{C}:T-\lambda \text{ is not a B-Weyl operator}\}$. An operator $T\in B(\mathcal{H})$ satisfies generalized Weyl’s theorem [[5], Definition 2.13] if

$${\sigma }_{\mathrm{BW}}(T)=\sigma (T)\setminus {\pi }_0(T),$$

where ${\pi }_0(T)$ is the set of all isolated eigenvalues of T, and $T\in B(\mathcal{H})$ satisfies generalized Browder’s theorem [[5], Definition 2.13] if

$${\sigma }_{\mathrm{BW}}(T)=\sigma (T)\setminus p_0(T),$$

where $p_0(T)$ is the set of all poles of the resolvent of T.

Let ${SBF}_{+}(\mathcal{H})$ be the class of all the upper semi-B-Fredholm operators and ${SBF}_{+}^{-}(\mathcal{H})$ be the class of all $T\in {SBF}_{+}(\mathcal{H})$ such that $ind(T)\le 0$. Let

$${\sigma }_{{\mathrm{SBF}}_{+}^{-}}(T)=\{\lambda \in \mathbb{C}:T-\lambda \notin {SBF}_{+}^{-}(\mathcal{H})\}$$

be called the semi-B-essential approximate point spectrum of T. We say that $T\in B(\mathcal{H})$ satisfies generalized a-Weyl’s theorem [[5], Definition 2.13] if

$${\sigma }_{{\mathrm{SBF}}_{+}^{-}}(T)={\sigma }_a(T)\setminus {\pi }_0^a(T),$$

where ${\pi }_0^a(T)$ is the set of all eigenvalues of T which are isolated points of ${\sigma }_a(T)$.

The following implications are known to hold:

$$\begin{array}{c}\text{generalized }a\text{-Weyl’s theorem}\hfill \\ ⟹\text{generalized Weyl’s theorem}⟹\text{Weyl’s theorem}\hfill \\ ⟹\text{Browder’s theorem};\hfill \\ \text{generalized }a\text{-Weyl’s theorem}\hfill \\ ⟹a\text{-Weyl’s theorem}⟹a\text{-Browder’s theorem}\hfill \\ ⟹\text{Browder’s theorem}.\hfill \end{array}$$

In this paper, we shall study the generalized Weyl’s theorem for the elementary operator and the generalized derivation with class A operators as entries. Recall that $T\in B(\mathcal{H})$ is called p-hyponormal for $p>0$ if ${(T^{\ast }T)}^p-{(T{T}^{\ast })}^p\ge 0$ [6]; when $p=1$, T is called hyponormal. And T is called paranormal if ${\parallel Tx\parallel }^2\le \parallel T^{2}x\parallel \parallel x\parallel$ for all $x\in \mathcal{H}$ [7, 8]. In order to discuss the relations between paranormal and p-hyponormal and log-hyponormal operators (T is invertible and $log{T}^{\ast }T\ge logT{T}^{\ast }$), Furuta, Ito and Yamazaki [9] introduced a very interesting class of operators: class A defined by $|T^2|-{|T|}^2\ge 0$, where $|T|={(T^{\ast }T)}^{\frac{1}{2}}$ which is called the absolute value of T, and they showed that class A is a subclass of paranormals and contains p-hyponormal and log-hyponormal operators.

Definition 1.1 An operator $T\in B(\mathcal{H})$ is said to have the single valued extension property (SVEP) at $\lambda \in \mathbb{C}$ if for every open neighborhood $\mathcal{G}$ of λ, the only function $f\in H(\mathcal{G})$ such that $(T-\mu )f(\mu )=0$ on G is $0\in H(\mathcal{G})$, where $H(\mathcal{G})$ means the space of all analytic functions on G. When T has SVEP at each $\lambda \in \mathbb{C}$, say that T has SVEP.

The single valued extension property dates back to the early days of local spectral theory; see the recent monograph of Laursen and Neumann [10] or Aiena [11]. In addition to the definition of SVEP, there are notions of property (β), property (δ) and condition (C). The interested reader is referred to [10] for more details.

2 The main results

For $T\in B(\mathcal{H})$, let $L_T$ and $R_T$ denote the operators of left and right multiplication by T respectively.

Chō and Yamazaki proved that class A operators have property β in [12] Theorem 3.1; unfortunately, there are some mistakes in the proof of this theorem; see details in [13]. So, Theorem 3.1 in [12] is still an open problem.

Lemma 2.1 Let A and $B^{\ast }$ be class A operators satisfying property (β), then $d_{AB}$ has SVEP.

Proof By assumption and [10] Theorem 2.5.5, A satisfies property (β) and B satisfies property (δ). Hence, both $L_A$ and $R_B$ satisfy condition (C) by [10] Corollary 3.6.11. Clearly, $L_A$ and $R_B$ commute. By Theorem 3.6.3 and Note 3.6.19 on p.283 of [10], $L_A-R_B$ and $L_A{R}_B$ have SVEP, which implies that $d_{AB}$ has SVEP. □

It is well known that the isolated points of the spectrum of a class A (indeed, paranormal) operator T are poles of the resolvent of the operator (hence, eigenvalues of the operator), the restriction of T to an invariant subspace is again of class A (resp., paranormal), and that if T has countable spectrum then T is normal. (We shall use this information freely in the following without any further reference.)

Recall, [14], that $\sigma ({\delta }_{AB})=\{\lambda :\lambda \in \sigma (A)-\sigma (B)\}$ and $\sigma ({\mathrm{△}}_{AB})=\{\lambda :\lambda \in \sigma (A)\sigma (B)-1\}$. If $\lambda \in iso\sigma (d_{AB})$, then we have one of the following two cases:

• (1) $\lambda \ne -1$ if $d_{AB}={\mathrm{△}}_{AB}$. Then there exist finite sequences ${\{{\alpha }_i\}}_{i=1}^m$ and ${\{{\beta }_i\}}_{i=1}^m$, where ${\alpha }_i\in iso\sigma (A)$ and ${\beta }_i\in iso\sigma (B)$ respectively, such that $\lambda ={\alpha }_i-{\beta }_i$ if $\lambda \in iso\sigma ({\delta }_{AB})$ and $\lambda ={\alpha }_i{\beta }_i-1$ if $\lambda \in iso\sigma ({\mathrm{△}}_{AB})$, for all $1\le i\le m$.

• (2) $\lambda =-1$ and $d_{AB}={\mathrm{△}}_{AB}$. Then either $0\in iso\sigma (A)$ and $0\in iso\sigma (B)$ or $0\in iso\sigma (A)$ and $0\notin \sigma (B)$ or $0\in iso\sigma (B)$ and $0\notin \sigma (A)$.

Theorem 2.2 Let A and $B^{\ast }$ be class A operators, then $H_0(d_{AB}-\lambda )=ker(d_{AB}-\lambda )$ for all $\lambda \in iso\sigma (d_{AB})$.

Proof We consider the case $d_{AB}={\delta }_{AB}$ and $d_{AB}={\mathrm{△}}_{AB}$ respectively.

• (1) We consider the case $d_{AB}={\delta }_{AB}$. The idea comes from [1]. If $\lambda \in iso\sigma ({\delta }_{AB})$, then there exist finite sets $\{{\alpha }_1,{\alpha }_2,\dots ,{\alpha }_m\}$ and $\{{\beta }_1,{\beta }_2,\dots ,{\beta }_m\}$, where ${\alpha }_i\in iso\sigma (A)$ and ${\beta }_i\in iso\sigma (B)$ such that $\lambda ={\alpha }_i-{\beta }_i$ for all $1\le i\le m$. Let

  $$M_1=\underset{i=1}{\overset{m}{\bigvee }}ker(A-{\alpha }_i),M_2=\mathcal{H}\ominus M_1$$

and

$$N_1=\underset{i=1}{\overset{m}{\bigvee }}ker(B^{\ast }-\overline{{\beta }_i}),N_2=\mathcal{H}\ominus N_1.$$

Then A and B have representations $A=\left(\begin{array}{cc}{A}_{11}& A_{12}\\ 0& A_{22}\end{array}\right)$ on $M_1\oplus M_2$ and $B=\left(\begin{array}{cc}{B}_{11}& 0\\ B_{21}& B_{22}\end{array}\right)$ on $N_1\oplus N_2$ respectively, where $A_{11}$ and $B_{11}$ are normal, $\sigma (A)=\sigma (A_{11})\cup \sigma (A_{22})$, $\sigma (B)=\sigma (B_{11})\cup \sigma (B_{22})$ and $\lambda \notin \sigma ({\delta }_{A_{ii}{B}_{jj}})$ for all $1\le i$, $j\le 2$ other than $i=j=1$. Consider an $X\in H_0({\delta }_{AB}-\lambda )$. Letting $X:N_1\oplus N_2⟶M_1\oplus M_2$ have the matrix representation $X={[X_{ij}]}_{i,j=1}^2$, we have

$${({\delta }_{AB}-\lambda )}^{n}X=\left(\begin{array}{cc}\ast & \ast \\ \ast & {({\delta }_{A_{22}{B}_{22}}-\lambda )}^n{X}_{22}\end{array}\right)$$

for some yet to be determined entries ∗.

Since

$$\underset{n⟶\mathrm{\infty }}{lim}{\parallel {({\delta }_{AB}-\lambda )}^{n}X\parallel }^{\frac{1}{n}}=0,$$

we have that

$$\underset{n⟶\mathrm{\infty }}{lim}{\parallel {({\delta }_{A_{22}{B}_{22}}-\lambda )}^n{X}_{22}\parallel }^{\frac{1}{n}}=0.$$

Since $\lambda \notin \sigma ({\delta }_{A_{22}{B}_{22}})$, we have that ${\delta }_{A_{22}{B}_{22}}-\lambda$ is invertible. Hence, $X_{22}=0$. So, we have that

$${({\delta }_{AB}-\lambda )}^{n}X=\left(\begin{array}{cc}\ast & {({\delta }_{A_{11}{B}_{22}}-\lambda )}^n{X}_{12}\\ {({\delta }_{A_{22}{B}_{11}}-\lambda )}^n{X}_{21}& 0\end{array}\right).$$

Since

$$\underset{n⟶\mathrm{\infty }}{lim}{\parallel {({\delta }_{A_{ii}{B}_{jj}}-\lambda )}^n{X}_{ij}\parallel }^{\frac{1}{n}}=0$$

for all $1\le i$, $j\le 2$ other than $i=j=1$ and since $\lambda \notin \sigma ({\delta }_{A_{11}{B}_{22}})$ and $\lambda \notin \sigma ({\delta }_{A_{22}{B}_{11}})$, we have that $X_{12}=X_{21}=0$. Hence,

$${({\delta }_{AB}-\lambda )}^{n}X=\left(\begin{array}{cc}{({\delta }_{A_{11}{B}_{11}}-\lambda )}^n{X}_{11}& 0\\ 0& 0\end{array}\right).$$

Since $A_{11}$ and $B_{11}$ are normal,

$$\underset{n⟶\mathrm{\infty }}{lim}{\parallel {({\delta }_{A_{11}{B}_{11}}-\lambda )}^n{X}_{11}\parallel }^{\frac{1}{n}}=0$$

if and only if

$$({\delta }_{A_{11}{B}_{11}}-\lambda )X_{11}=0$$

by [15] Lemma 2. Hence, we have $H_0({\delta }_{AB}-\lambda )\subseteq ker({\delta }_{AB}-\lambda )$. Since $ker({\delta }_{AB}-\lambda )\subseteq H_0({\delta }_{AB}-\lambda )$ is always true, we have

$$H_0({\delta }_{AB}-\lambda )=ker({\delta }_{AB}-\lambda ).$$

• (2) We consider the case $d_{AB}={\mathrm{△}}_{AB}$. When $\lambda \ne -1$, the proof is similar to the proof of the first part. We omit the proof. When $\lambda =-1$, then either $0\in iso\sigma (A)$ and $0\in iso\sigma (B)$ or $0\in iso\sigma (A)$ and $0\notin \sigma (B)$ or $0\in iso\sigma (B)$ and $0\notin \sigma (A)$. If 0 is both in $iso\sigma (A)$ and $iso\sigma (B)$, then let $M_1=ker(A)$, $M_2=\mathcal{H}\ominus M_1$ and $N_1=ker(B^{\ast })$, $N_2=\mathcal{H}\ominus N_1$.

Then we have $A=\left(\begin{array}{cc}0& C_1\\ 0& A_2\end{array}\right)$ on $\mathcal{H}=M_1\oplus M_2$ and $B=\left(\begin{array}{cc}0& 0\\ C_2& B_2\end{array}\right)$ on $\mathcal{H}=N_1\oplus N_2$ for some operators $C_1$, $A_2$ and $C_2$, $B_2$ respectively. Here both $A_2$ and $B_2$ are invertible. So, we have that ${\mathrm{△}}_{A_2{B}_2}-\lambda =L_{A_2}{R}_{B_2}$ is invertible. Let $X:N_1\oplus N_2⟶M_1\oplus M_2$ have the matrix representation $X={[X_{ij}]}_{i,j=1}^2$. If $X\in H_0({\mathrm{△}}_{AB}-\lambda )=H_0(L_A{R}_B)$, it follows that $X_{22}=0$ as in the proof of the first part. Hence, $L_A{R}_{B}X=0$ for every $X\in H_0({\mathrm{△}}_{AB}-\lambda )=H_0(L_A{R}_B)$. So, we have $H_0({\mathrm{△}}_{AB}-\lambda )\subseteq ker({\mathrm{△}}_{AB}-\lambda )$. Since $ker({\mathrm{△}}_{AB}-\lambda )\subseteq H_0({\mathrm{△}}_{AB}-\lambda )$ is always true, we have $H_0({\mathrm{△}}_{AB}-\lambda )=ker({\mathrm{△}}_{AB}-\lambda )$. The proofs of the other remaining cases are similar, we consider $0\in iso\sigma (A)$ and $0\notin \sigma (B)$. Here ${\mathrm{△}}_{AB}-\lambda =L_A{R}_B$. In the following, we shall prove that $H_0(L_A{R}_B)=H_0(L_A)$. If $X\in H_0(L_A{R}_B)$, then

$$\underset{n⟶\mathrm{\infty }}{lim}{\parallel {(L_A)}^{n}X\parallel }^{\frac{1}{n}}=\underset{n⟶\mathrm{\infty }}{lim}{\parallel {(L_A{R}_B)}^{n}X{B}^{-n}\parallel }^{\frac{1}{n}}\le \parallel B^{-1}\parallel \underset{n⟶\mathrm{\infty }}{lim}{\parallel {(L_A{R}_B)}^{n}X\parallel }^{\frac{1}{n}}=0.$$

On the other hand, if $X\in H_0(L_A)$, then

$$\underset{n⟶\mathrm{\infty }}{lim}{\parallel {(L_A{R}_B)}^{n}X\parallel }^{\frac{1}{n}}\le \parallel B\parallel \underset{n⟶\mathrm{\infty }}{lim}{\parallel {(L_A)}^{n}X\parallel }^{\frac{1}{n}}=0.$$

Hence, $H_0(L_A{R}_B)=H_0(L_A)$. Next, we shall prove that $H_0(L_A)=ker(L_A)$. Let $M_1=kerA$, $M_2=\mathcal{H}\ominus M_1$. We have $A=\left(\begin{array}{cc}0& C_1\\ 0& A_2\end{array}\right)$ on $\mathcal{H}=M_1\oplus M_2$, where $A_2$ is invertible. Let X have the following matrix representation: $X={[X_{ij}]}_{i,j=1}^2$ on $\mathcal{H}=M_1\oplus M_2$. If $X\in H_0(L_A)$, as in the proof above, we have that $X_{21}=X_{22}=0$. So, we have that $L_{A}X=0$. Hence, $H_0(L_A)\subseteq ker(L_A)$. Since $ker(L_A)\subseteq H_0(L_A)$ is always true, we have that $H_0(L_A)=ker(L_A)$. Since B is invertible, we have $ker(L_A{R}_B)=ker(L_A)$. Therefore, we have that

$$H_0(L_A{R}_B)=H_0(L_A)=ker(L_A)=ker(L_A{R}_B),$$

hence

$$H_0(L_A{R}_B)=ker(L_A{R}_B).$$

That is,

$$H_0({\mathrm{△}}_{AB}-\lambda )=ker({\mathrm{△}}_{AB}-\lambda ).$$

This completes the proof. □

An operator $T\in B(\mathcal{H})$ is said to be isoloid if every isolated point of $\sigma (T)$ is an eigenvalue of T and polaroid if every isolated point of $\sigma (T)$ is a pole of the resolvent of T. In general, if T is polaroid then it is isoloid.

Lemma 2.3 Let A and $B^{\ast }$ be class A operators, then $d_{AB}$ and $d_{AB}^{\ast }$ are polaroid. In particular, $d_{AB}$ and $d_{AB}^{\ast }$ are isoloid.

Proof We only need to prove that $d_{AB}$ is polaroid. Let $\mathcal{X}=B(\mathcal{H})$ and $\lambda \in iso\sigma (d_{AB})$. Then we have that $H_0(d_{AB}-\lambda )=ker(d_{AB}-\lambda )$ by Theorem 2.2. Hence,

$$\mathcal{X}=H_0(d_{AB}-\lambda )\oplus K(d_{AB}-\lambda )=ker(d_{AB}-\lambda )\oplus K(d_{AB}-\lambda ).$$

So, we have

$$(d_{AB}-\lambda )\mathcal{X}=0\oplus (d_{AB}-\lambda )(K(d_{AB}-\lambda ))=K(d_{AB}-\lambda ).$$

Therefore,

$$\mathcal{X}=ker(d_{AB}-\lambda )\oplus (d_{AB}-\lambda )\mathcal{X}.$$

Thus, isolated points of $\sigma (d_{AB})$ are simple poles of the resolvent of $d_{AB}$. Hence, $d_{AB}$ is polaroid. So, we have that $d_{AB}$ and $d_{AB}^{\ast }$ are polaroid. Since polaroid operators are always isoloid, we have that $d_{AB}$ and $d_{AB}^{\ast }$ are isoloid. □

Theorem 2.4 Let A and $B^{\ast }$ be class A operators satisfying property (β). Then generalized Weyl’s theorem holds for $f(d_{AB})$ for every $f\in H(\sigma (d_{AB}))$ and f is not constant on each connected component of the open set U containing $\sigma (d_{AB})$.

Proof Suppose that A and $B^{\ast }$ are class A operators. By Lemma 2.1 and Lemma 2.3, we have that $d_{AB}$ has SVEP and $d_{AB}$ is polaroid. So, we have that generalized Weyl’s theorem holds for $d_{AB}$ by [[16], Theorem 3.10(ii)]. Since $d_{AB}$ has SVEP and $d_{AB}$ is isoloid, we have that generalized Weyl’s theorem holds for $f(d_{AB})$ for every $f\in H(\sigma (d_{AB}))$ by [[17], Theorem 2.2]. □

Corollary 2.5 Let A and $B^{\ast }$ be class A operators satisfying property (β). Then Weyl’s theorem holds for $f(d_{AB})$ for every $f\in H(\sigma (d_{AB}))$ and f is not constant on each connected component of the open set U containing $\sigma (d_{AB})$.

A bounded linear operator $T\in B(\mathcal{H})$ is called a-isoloid if every isolated point of ${\sigma }_a(T)$ is an eigenvalue of T. Note that every a-isoloid operator is isoloid and the converse is not true in general.

Lemma 2.6 Let A and $B^{\ast }$ be class A operators satisfying property (β), then $d_{AB}^{\ast }$ is a-isoloid.

Proof Let λ be an isolated point of ${\sigma }_a(d_{AB}^{\ast })$. Suppose that A and $B^{\ast }$ are class A operators satisfying property (β). By Lemma 2.1 and Lemma 2.3, we have that $d_{AB}$ has SVEP and $d_{AB}^{\ast }$ is isoloid. Hence, ${\sigma }_a(d_{AB}^{\ast })=\sigma (d_{AB}^{\ast })$ by [[18], Corollary 7]. We have that λ is an isolated point of $\sigma (d_{AB}^{\ast })$. Since $d_{AB}^{\ast }$ is isoloid, we have that λ is an eigenvalue of $d_{AB}^{\ast }$. Hence, $d_{AB}^{\ast }$ is a-isoloid. □

Theorem 2.7 Let A and $B^{\ast }$ be class A operators satisfying property (β). Then generalized a-Weyl’s theorem holds for $f(d_{AB}^{\ast })$ for every $f\in H(\sigma (d_{AB}))$, and f is not constant on each connected component of the open set U containing $\sigma (d_{AB})$.

Proof Suppose that A and $B^{\ast }$ are class A operators satisfying property (β). By Lemma 2.1 and Lemma 2.3, we have that $d_{AB}$ has SVEP and $d_{AB}^{\ast }$ is polaroid. By Corollary 2.5, Weyl’s theorem holds for $d_{AB}$. Hence, $d_{AB}^{\ast }$ satisfies Weyl’s theorem by [[19], Proposition 2.1]. Since $d_{AB}$ has SVEP and $d_{AB}^{\ast }$ is polaroid, generalized a-Weyl’s theorem holds for $d_{AB}^{\ast }$ by [[16], Theorem 3.10]. T is a-isoloid by Lemma 2.6, hence generalized a-Weyl’s theorem holds for $f(T)$ for every $f\in H(\sigma (T))$ by [[17], Theorem 2.4]. □

Corollary 2.8 Let A and $B^{\ast }$ be class A operators satisfying property (β). Then a-Weyl’s theorem holds for $f(d_{AB}^{\ast })$ for every $f\in H(\sigma (d_{AB}))$, and f is not constant on each connected component of the open set U containing $\sigma (d_{AB})$.