Property (k) and commuting Riesz-type perturbations

Abstract

We study the stability of the spectral property (k) introduced and studied by Kaushik and Kashyap (Int J Math Arch 12:167–171, 2014), under commuting perturbations by Riesz operators, and we give generalizations of some known results.

1 Introduction and basic definitions

For T in the Banach algebra $\mathcal {B}(X)$ of bounded linear operators acting on a Banach space X,  we will denote by $\sigma (T)$ the spectrum of T,  by $\sigma _a(T)$ the approximate point spectrum of T,  by $\mathcal {N}(T)$ the null space of T, by n(T) the nullity of T,  by $\mathcal {R}(T)$ the range of T and by d(T) its defect. If $n(T)<\infty$ and $d(T)<\infty$, then T is called a Fredholm operator and its index is defined by $\text{ ind }(T)=n(T)-d(T).$ A Weyl operator $T\in \mathcal {B}(X)$ is a Fredholm operator of index zero and the Weyl spectrum is defined by $\sigma _{w}(T)=\{\lambda \in \mathbb {C}\, |\,T-\lambda I \text{ is } \text{ not } \text{ a } \text{ Weyl } \text{ operator }\}.$ $T\in \mathcal {B}(X)$ is called an upper (resp., a lower) semi-Fredholm if $\mathcal {R}(T)$ is closed and $n(T)<\infty$ (resp., $d(T)<\infty ).$ The respective semi-Fredholm spectrum and semi-Weyl spectrum of T are defined respectively, by $\sigma _{sf}(T)=\{\lambda \in \mathbb {C}\, |\,T-\lambda I \text{ is } \text{ not } \text{ a } \text{ semi-Fedholm } \text{ operator }\},$ and $\sigma _{sf_+^-}(T)=\{\lambda \in \mathbb {C}\, |\,T-\lambda I \text{ is } \text{ not } \text{ an } \text{ upper } \text{ semi-Fredholm } \text{ operator } \text{ with } \text{ index } \text{ less } \text{ or } \text{ equal } \text{ than } \text{ zero }\}.$

For a bounded linear operator T and $n\in \mathbb {N},$ let $T_{[n]}: \mathcal {R}(T^n)\rightarrow \mathcal {R}(T^n)$ be the restriction of T to $\mathcal {R}(T^n).$ $T \in \mathcal {B}(X)$ is said to be a semi B-Fredholm if for some integer $n \ge 0,$ the range $\mathcal {R}(T^n)$ is closed and $T_{[n]}$ is a semi-Fredholm; its index is defined as the index of the semi-Fredholm operator $T_{[n]}.$ The respective semi B-Fredholm spectrum of T is defined by $\sigma _{sbf}(T)=\{\lambda \in \mathbb {C}\, |\, T-\lambda I \text{ is } \text{ not } \text{ a } \text{ semi } \text{ B-Fredholm } \text{ operator }\}.$

The ascent of an operator T is defined by $a(T)=\text{ inf } \{ n\in \mathbb {N}\, |\,\mathcal {N}(T^n)=\mathcal {N}(T^{n+1})\}$, and the descent of T is defined by $\delta (T)= \text{ inf } \{ n \in \mathbb {N}\, |\, \mathcal {R}(T^n)= \mathcal {R}(T^{n+1})\},$ with $\text{ inf }\, \emptyset = \infty .$ According to Heuser [8], a complex number $\lambda \in \sigma (T)$ is a pole of the resolvent of T if $T-\lambda I$ has finite ascent and finite descent, and in this case they are equal. An operator $T\in \mathcal {B}(X)$ is said to be Browder operator if it is a Fredholm with finite ascent and descent, and is said to be an upper Browder operator if it is an upper semi-Fredholm operator with finite ascent. The respective Browder spectrum and upper Browder spectrum of T are defined respectively, by $\sigma _{b}(T)=\{\lambda \in \mathbb {C}\, |\, T-\lambda I \text{ is } \text{ not } \text{ a } \text{ Browder } \text{ operator }\},$ and $\sigma _{ub}(T)=\{\lambda \in \mathbb {C}\, |\, T-\lambda I \text{ is } \text{ not } \text{ an } \text{ upper } \text{ Browder } \text{ operator }\}.$

In the following, we recall the definition of a property which has a relevant role in local spectral theory. For more details about this property see the monographs of Laursen and Neumann [10] and Aiena [1].

Definition 1.1

An operator $T\in \mathcal {B}(X)$ is said to have the single valued extension property at $\lambda _{0}\in \mathbb {C}$ (abbreviated SVEP at $\lambda _{0}$), if for every open neighborhood ${\mathcal U}$ of $\lambda _{0}$, the only analytic function $f:{\mathcal U} \longrightarrow X$ which satisfies the equation $(T-\lambda I)f(\lambda )=0$ for all $\lambda \in {\mathcal U}$ is the function $f\equiv 0$. An operator $T\in \mathcal {B}(X)$ is said to have SVEP if T has SVEP at every point $\lambda \in \mathbb {C}$.

Evidently, $T\in \mathcal {B}(X)$ has SVEP at every isolated point of the spectrum. We summarize in the following list the usual notations and symbols needed later.

Notations and symbols

${\mathcal F}(X)$ :

The ideal of finite rank operators in $\mathcal {B}(X),$

${\mathcal K}(X)$ :

The ideal of compact operators in $\mathcal {B}(X),$

${\mathcal N}(X)$ :

The class of nilpotent operators on X, 

${\mathcal Q}(X)$ :

The class of quasi-nilpotent operators on X, 

$\text{ iso }\,A$ :

Isolated points of a subset $A\subset \mathbb {C},$

$\text{ acc }\,A$ :

Accumulations points of a subset $A\subset \mathbb {C},$

D(0, 1):

The closed unit disc in $\mathbb {C},$

C(0, 1):

The unit circle of $\mathbb {C},$

$\Pi (T)$ :

poles of T, 

$\Pi ^0(T)$ :

Poles of T of finite rank,

$\Pi _a(T)$ :

Left poles of T, 

$\Pi _a^0(T)$ :

Left poles of T of finite rank,

$\sigma _{p}(T)$ :

Eigenvalues of T, 

$\sigma _{p}^f(T)$ :

Eigenvalues of T of finite multiplicity,

$E^0(T)$ :

$\text{ iso }\,\sigma (T)\cap \sigma _{p}^f(T),$

E(T):

$\text{ iso }\,\sigma (T)\cap \sigma _{p}(T),$

$E_a^0(T)$ :

$\text{ iso }\,\sigma _a(T)\cap \sigma _{p}^f(T),$

$\sigma _{b}(T)= \sigma (T){\setminus }\Pi ^0(T)$ :

Browder spectrum of T, 

$\sigma _{ub}(T)= \sigma _a(T){\setminus }\Pi _a^0(T)$ :

Upper-Browder spectrum of T, 

$\sigma _{w}(T)$ :

Weyl spectrum of T, 

$\sigma _{sf_+^-}(T)$ :

Semi-Weyl spectrum of T, 

$\sigma _{sbf}(T)$ :

Semi B-Fredholm spectrum of T.

Definition 1.2

[2, 6, 9, 12] Let $T\in \mathcal {B}(X).$ T is said to satisfy

• (i) a-Browder’s theorem if $\sigma (T){\setminus }\sigma _{sf_+^-}(T)=\Pi _a^0(T),$ or equivalently $\sigma _{ub}(T)=\sigma _{sf_+^-}(T).$

• (ii) Browder’s theorem if $\sigma (T){\setminus }\sigma _{w}(T)=\Pi ^0(T),$ or equivalently $\sigma _{b}(T)=\sigma _{w}(T).$

• (iii) Weyl’s theorem if $\sigma (T){\setminus }\sigma _{w}(T)=E^0(T).$

• (iv) Property (k) if $\sigma (T){\setminus }\sigma _{w}(T)=E(T).$

Definition 1.3

Let $T\in \mathcal {B}(X)$ and $S\in \mathcal {B}(X).$ We will say that T and S have a shared stable sign index if for each $\lambda \not \in \sigma _{sbf}(T)$ and $\mu \not \in \sigma _{sbf}(S)$, $\text{ ind }(T-\lambda I)$ and $\text{ ind }(S-\mu I)$ have the same sign.

For examples we have:

• 1. It is easily verified that if $T\in \mathcal {B}(X)$ has SVEP then $\text{ ind }(T-\mu I)\le 0$ for every $\mu \not \in \sigma _{sbf}(T).$ So if S and T have SVEP, then they have a shared stable sign index.

• 2. Here and elsewhere, $\mathcal {H}$ denotes a Hilbert space. It is well known that every hyponormal operator T acting on $\mathcal {H}$ has property $(H_1)$ (for the definition of $(H_1),$ see the end of the second section) and hence has SVEP. As a consequence of the first point, every two hyponormal operators have a shared stable sign index. Recall that $T\in \mathcal {B}(\mathcal {H})$ is said to be hyponormal if $T^*T-TT^*\ge 0$ (or equivalently $\Vert T^*x\Vert \le \Vert Tx\Vert$ for all $x\in \mathcal {H}$). The class of hyponormal operators includes also subnormal operators and quasinormal operators, see Conway [4].

After giving an introduction and some preliminaries in the first section, we study in the second section the preservation of property (k) introduced and studied by Kaushik and Kashyap [9], under several commuting Riesz-type perturbations. We prove in particular that if T is an isoloid operator acting on a Banach space and satisfies property (k),  then $T+S$ satisfies property (k) for every finite rank power operator S which commutes with T. Moreover, we give generalization of some perturbation results to commuting Riesz operators such as Theorems 2.3 and 2.8 of Berkani and Zariouh [3]. In the end of this paper, we prove that if S and T are isoloid bounded operators acting on Banach spaces and satisfy property (k),  then $S\oplus T$ satisfies property (k) if and only if $\sigma _{w}(S\oplus T)=\sigma _{w}(S)\cup \sigma _{w}(T),$ extending a result of Kaushik and Kashyap [9]. Some crucial examples are also given.

2 Property (k) and perturbations

We recall that an operator $R\in \mathcal {B}(X)$ is said to be Riesz if $R-\mu I$ is Fredholm for every non-zero complex $\mu ,$ that is, $\pi (R)$ is quasi-nilpotent in the Calkin algebra $C(X)=\mathcal {B}(X)/{\mathcal K}(X)$ where $\pi$ is the canonical mapping of $\mathcal {B}(X)$ into C(X). We denote by $\mathcal {R}(X)$ the class of Riesz operators and by ${\mathcal F^0}(X) ,$ the class of finite rank power operators as follows:

$$\begin{aligned} {\mathcal F^0}(X) =\left\{ S\in \mathcal {B}(X) : S^n \in {\mathcal F}(X) \text{ for } \text{ some } n\in \mathbb {\mathbb {N}}\right\} . \end{aligned}$$

Clearly, ${\mathcal F}(X) \cup {\mathcal N}(X) \subset {\mathcal F^0}(X) \subset {\mathcal R}(X) ,$ and ${\mathcal K}(X) \cup {\mathcal Q}(X) \subset {\mathcal R}(X) .$

According to Oberai [11], Rakocěvi$\acute{c}$ [14] and Tylli [15], we know that for every $T\in \mathcal {B}(X)$ and $R\in {\mathcal R}(X)$ such that $TR=RT,$ $\sigma _{*}(T+R)=\sigma _{*}(T);$ where $\sigma _{*}\in \{\sigma _{sf_+^-}, \sigma _{w}, \sigma _{b}, \sigma _{ub}\}.$ From this, we give the following known lemma which we need in the proof of the next main results.

Lemma 2.1

Let $T\in \mathcal {B}(X)$ and let $R\in {\mathcal R}(X)$ be a commuting operator with T. The following statements hold:

• (i) T satisfies Browder’s theorem if and only if $T+R$ satisfies Browder’s theorem.

• (ii) T satisfies a-Browder’s theorem if and only if $T+R$ satisfies a-Browder’s theorem.

We start this section by the following nilpotent perturbation result.

Proposition 2.2

Let $T\in \mathcal {B}(X)$ and let $N\in {\mathcal N}(X)$ which commutes with T. Then T satisfies property (k) if and only if $T+N$ satisfies property (k).

Proof

Since N is nilpotent and commutes with T,  we know that $\sigma (T+N)=\sigma (T),$ and it is easily seen that $0<n(T+N)\Longleftrightarrow 0<n(T).$ Since $\sigma _{w}(T)=\sigma _{w}(T+N),$ it follows that T satisfies property (k) if and only if $T+N$ satisfies property (k). $\square$

• Note that the assumption of commutativity in the Proposition 2.2 is crucial. Let T and N be defined on $\ell ^2(\mathbb {N})$ by $T(x_1, x_2, \ldots )=(0, \frac{x_1}{2}, \frac{x_2}{3}, \ldots )$ and $N(x_1, x_2, \ldots )=(0, \frac{-x_1}{2}, 0, 0, \ldots ).$ Clearly N is nilpotent and does not commute with T. The property (k) is satisfied by T,  since $\sigma (T)=\{0\}=\sigma _{w}(T)$ and $E(T)=\emptyset .$ But $T+N$ does not satisfy property (k),  because $\sigma (T+N)=\sigma _{w}(T+N)=\{0\}$ and $\{0\}= E(T+N).$

• The stability of property (k) showed in Proposition 2.2 cannot be extended to commuting quasi-nilpotent operators, as we can see in the next example:

Example 2.3

Let T be the operator defined on $\ell ^2(\mathbb {N})$ by $T(x_1,x_2, \ldots )=(0, \frac{x_1}{2},\frac{x_2}{3}, \ldots ).$ Put $R=-T,$ clearly R is quasi-nilpotent, compact and commutes with T. As it is already mentioned, T satisfies property (k). But $T+R=0$ does not satisfy this property. Indeed, $\sigma (T+R)=\{0\}=\sigma _{w}(T+R),$ $E(T+R)=\{0\}.$ Note also that $\Pi ^ 0(T+R)=\emptyset ,$ $\Pi ^0(T)=\emptyset .$

Theorem 2.4

Let $R\in {\mathcal R}(X)$ and let $T\in \mathcal {B}(X)$ be a commuting operator with T. If T satisfies property (k),  then the following statements are equivalent:

• (i) $T+R$ satisfies property (k); 

• (ii) $\Pi ^0(T+R)=E(T+R);$

• (iii) $E(T+R)\cap \sigma (T)\subset \Pi ^0(T)$.

Proof

• (i) $\Longleftrightarrow$ (ii) If $T+R$ satisfies property (k) then from [9, Theorem 2.5], $\Pi ^0(T+R)=E(T+R).$ Conversely, since T satisfies property (k),  then it satisfies Browder’s theorem, and from Lemma 2.1, $T+R$ satisfies Browder’s theorem too. So $\sigma (T+R){\setminus }\sigma _{w}(T+R)=\Pi ^0(T+R).$ Thus $T+R$ satisfies property (k).

• (ii) $\Longleftrightarrow$ (iii) Suppose that $\Pi ^0(T+R)=E(T+R)$ and let $\lambda _0\in E(T+R)\cap \sigma (T)$ be arbitrary. Then $\lambda _0\in \Pi ^0(T+R)\cap \sigma (T)$ and hence $\lambda _0\in \sigma (T){\setminus }\sigma _{b}(T)=\Pi ^0(T).$ Consequently, $E(T+R)\cap \sigma (T)\subset \Pi ^0(T).$ Conversely, suppose that $E(T+R)\cap \sigma (T)\subset \Pi ^0(T)$ and let $\mu _0\in E(T+R)$ be arbitrary. We distinguish two cases: the first is $\mu _{0}\in \sigma (T).$ Then $\mu _{0}\in E(T+R)\cap \sigma (T)\subset \Pi ^0(T).$ It follows that $\mu _{0}\not \in \sigma _{b}(T+R)$ and since $\mu _{0}\in \sigma (T+R),$ then $\mu _{0}\in \Pi ^0(T+R).$ The second case is $\mu _{0}\not \in \sigma (T).$ This implies that $\mu _{0}\not \in \sigma _{b}(T+R)$ and then $\mu _{0}\in \Pi ^0(T+R).$ In the two cases we have $\Pi ^0(T+R)\supset E(T+R)$ and as the opposite inclusion is always true, then $\Pi ^0(T+R)=E(T+R).$ Remark that the statements (ii) and (iii) are always equivalent without the assumption that T satisfies property (k). $\square$

As an application of Theorem 2.4 to commuting isoloid operators, we give the following corollary. Recall that an operator $T\in \mathcal {B}(X)$ is said to be isoloid (resp., polaroid) if $\text{ iso }\,\sigma (T)=E(T)$ (resp., $\text{ iso }\,\sigma (T)=\Pi (T)$).

Corollary 2.5

Let $S\in {\mathcal F^0}(X)$ and let $T\in \mathcal {B}(X)$ be an isoloid operator commuting with S. If T satisfies property (k) then $T+S$ satisfies property (k).

Proof

Let $\lambda _{0}\in E(T+S)\cap \sigma (T)$ be arbitrary. Then $\lambda _{0}\not \in \text{ acc }\,\sigma (T+S)=\text{ acc }\,\sigma (T),$ see [16, Theorem 2.2]. As $\lambda _{0}\in \sigma (T),$ then $\lambda _{0}\in \text{ iso }\,\sigma (T)=E(T).$ The property (k) for T implies that $E(T)=\Pi ^0(T)$ and hence $E(T+S)\cap \sigma (T)\subset \Pi ^0(T).$ But this is equivalent from Theorem 2.4, to say that $T+S$ satisfies property (k). $\square$

The next theorem extends [3, Theorem 2.3] to commuting Riesz perturbations which are not necessary nilpotent or compact. According to Rakocěvić [13], we recall that an operator $T\in \mathcal {B}(X)$ is said to satisfy a-Weyl’s theorem if $\sigma _a(T){\setminus }\sigma _{sf_+^-}(T)=E_a^0(T).$

Theorem 2.6

Let $R\in {\mathcal R}(X) .$ If $T\in \mathcal {B}(X)$ satisfies a-Weyl’s theorem and commutes with R,  then the following statements are equivalent:

• (i) $T+R$ satisfies a-Weyl’s theorem;

• (ii) $\Pi _a^0(T+R)=E_a^0(T+R);$

• (iii) $E_a^0(T+R)\cap \sigma _a(T)\subset E_a^0(T)$.

Proof

• (i) $\Longleftrightarrow$ (ii) Suppose that $T+R$ satisfies a-Weyl’s theorem and let $\mu _0\in E_a^0(T+R)$ be arbitrary. Then $\mu _0\in E_a^0(T+R)\Longleftrightarrow \mu _0\in \text{ iso }\,\sigma _a(T+R)\cap \,\sigma _{sf_+^-}(T+R)^C \Longleftrightarrow \mu _0\in \Pi _a^0(T+R),$ where $\sigma _{sf_+^-}(T+R)^C$ is the complement of the semi-Weyl spectrum of $T+R.$ Thus $\Pi _a^0(T+R)=E_a^0(T+R).$ For the converse, since T satisfies a-Weyl’s theorem, then it satisfies a-Browder’s theorem and therefore $T+R$ satisfies a-Browder’s theorem too, see Lemma 2.1. So $\sigma _a(T+R){\setminus }\sigma _{sf_+^-}(T+R)=\Pi _a^0(T+R)=E_a^0(T+R).$

• (ii) $\Longleftrightarrow$ (iii) Suppose that $\Pi _a^0(T+R)=E_a^0(T+R)$ and let $\lambda _0\in E_a^0(T+R)\cap \, \sigma _a(T)$ be arbitrary. Then $\lambda _0\in \Pi _a^0(T+R)\cap \, \sigma _a(T).$ Hence $\lambda _0\in \sigma _a(T){\setminus }\sigma _{ub}(T)=\Pi _a^0(T).$ Consequently, $E_a^0(T+R)\cap \sigma _a(T)\subset E_a^0(T).$ Conversely, suppose that $E_a^0(T+R)\cap \sigma _a(T)\subset E_a^0(T)$ and let $\mu _0\in E_a^0(T+R)$ be arbitrary. We distinguish two cases: the first is $\mu _{0}\in \sigma _a(T).$ Then $\mu _{0}\in E_a^0(T+R)\cap \sigma _a(T)\subset E_a^0(T).$ It follows that $\mu _{0}\not \in \sigma _{sf_+^-}(T+R)$ and since $\mu _{0}\in \text{ iso }\,\sigma _a(T+R),$ then $\mu _{0}\in \Pi _a^0(T+R).$ The second case is $\mu _{0}\not \in \sigma _a(T).$ This implies that $\mu _{0}\not \in \sigma _{ub}(T+R)$ and so $\mu _{0}\in \Pi _a^0(T+R).$ In the two cases we have $\Pi _a^0(T+R)\supset E_a^0(T+R)$ and as the opposite inclusion is always true, then $\Pi _a^0(T+R)=E_a^0(T+R)$. $\square$

The next theorem extends [3, Theorem 2.8] to commuting Riesz perturbations which are not necessary nilpotent or compact.

Theorem 2.7

Let $R\in {\mathcal R}(X) .$ If $T\in \mathcal {B}(X)$ satisfies Weyl’s theorem and commutes with R,  then the following statements are equivalent:

• (i) $T+R$ satisfies Weyl’s theorem;

• (ii) $\Pi ^0(T+R)=E^0(T+R);$

• (iii) $E^0(T+R)\cap \sigma (T)\subset E^0(T)$.

Proof

• (i) $\Longleftrightarrow$ (ii) Suppose that $T+R$ satisfies Weyl’s theorem and let $\mu _0\in E^0(T+R)$ be arbitrary. Then $\mu _0\in E^0(T+R)\Longleftrightarrow \mu _0\in \text{ iso }\,\sigma (T+R)\cap \,\sigma _{w}(T+R)^C \Longleftrightarrow \mu _0\in \Pi ^0(T+R),$ where $\sigma _{w}(T+R)^C$ is the complement of the Weyl spectrum of $T+R.$ Thus $\Pi ^0(T+R)=E^0(T+R).$ For the converse, since T satisfies Weyl’s theorem, then it satisfies Browder’s theorem and therefore $T+R$ satisfies Browder’s theorem too. So $\sigma (T+R){\setminus }\sigma _{w}(T+R)=\Pi ^0(T+R)=E^0(T+R).$

• (ii) $\Longleftrightarrow$ (iii) Suppose that $\Pi ^0(T+R)=E^0(T+R)$ and let $\lambda _0\in E^0(T+R)\cap \, \sigma (T)$ be arbitrary. Then $\lambda _0\in \Pi ^0(T+R)\cap \, \sigma (T)$ and hence $\lambda _0\in \sigma (T){\setminus }\sigma _{b}(T)=\Pi ^0(T).$ Consequently, $E^0(T+R)\cap \,\sigma (T)\subset E^0(T).$ Conversely, suppose that $E^0(T+R)\cap \, \sigma (T)\subset E^0(T)$ and let $\mu _0\in E^0(T+R)$ be arbitrary. We distinguish two cases: the first is $\mu _{0}\in \sigma (T).$ Then $\mu _{0}\in E^0(T+R)\cap \,\sigma (T)\subset E^0(T).$ It follows that $\mu _{0}\in \sigma _{w}(T+R)^C\cap \,\text{ iso }\,\sigma (T+R) \Longleftrightarrow \mu _{0}\in \Pi ^0(T+R).$ The second case is $\mu _{0}\not \in \sigma (T).$ This implies that $\mu _{0}\not \in \sigma _{b}(T+R).$ Thus $\mu _{0}\in \Pi ^0(T+R).$ In the two cases we have $\Pi ^0(T+R)\supset E^0(T+R)$ and as the opposite inclusion is always true, then $\Pi ^0(T+R)=E^0(T+R)$. $\square$

Corollary 2.8

Let $S\in {\mathcal F^0}(X)$ and let $T\in \mathcal {B}(X)$ be a bounded operator commuting with S. The following assertions hold:

• (i) If T satisfies a-Weyl’s theorem and $\text{ iso }\,\sigma _a(T)=E_a^0(T),$ then $T+S$ satisfies a-Weyl’s theorem.

• (ii) If T satisfies Weyl’s theorem and $\text{ iso }\,\sigma (T)=E^0(T),$ then $T+S$ satisfies Weyl’s theorem.

Proof

• (i) Let $\lambda _{0}\in E_a^0(T+S)\cap \sigma _a(T)$ be arbitrary. Then $\lambda _{0}\not \in \text{ acc }\,\sigma _a(T+S)=\text{ acc }\,\sigma _a(T),$ see [16, Theorem 2.2]. As $\lambda _{0}\in \sigma _a(T)$ then $\lambda _{0}\in \text{ iso }\,\sigma _a(T)=E_a^0(T).$ So $E_a^0(T+S)\cap \sigma _a(T)\subset E_a^0(T).$ But this is equivalent from Theorem 2.6, to say that $T+S$ satisfies a-Weyl’s theorem.

• (ii) Goes similarly with the proof of the first assertion, as an application of Theorem 2.7. $\square$

In the next, we explore conditions on $S\in \mathcal {B}(X)$ and $T\in \mathcal {B}(Y)$ so that $S\oplus T$ satisfies property (k). The motivation for this work has come from Duggal and Kubrusly [7]. We begin with an example which shows that even if two operators S and T satisfy property (k),  yet there direct sum may fail to satisfy property (k).

Example 2.9

Let R and L be the operators defined on $\ell ^2(\mathbb {\mathbb {N}})$ by $R(x_1,x_2, x_3, \ldots )=(0, x_1, x_2, x_3, \ldots )$ and $L(x_1,x_2, x_3, \ldots )=( x_2, x_3, \ldots ).$ Then property (k) holds for R and L,  since $\sigma (R)= \sigma _{w}(R)= D(0, 1),$ $E(R)=\emptyset ,$ $\sigma (L)=\sigma _{w}(L)=D(0, 1)$ and $E(L)=\emptyset$. But it does not hold for $R\oplus L.$ In fact $\sigma (R\oplus L)=D(0, 1),$ and as $n(R\oplus L)=d(R\oplus L)=1$ then $0\not \in \sigma _{w}(R\oplus L)$. So $\sigma _{w}(R\oplus L)\subsetneq \sigma (R\oplus L).$ We also remark that $E(R\oplus L)=\emptyset .$ Thus $\sigma (R\oplus L){\setminus }\sigma _{w}(R\oplus L)\ne E(R\oplus L).$ Note that S and T are isoloid and $\sigma _{w}(R\oplus L)\subsetneq \sigma _{w}(R)\cup \sigma _{w}(L)=D(0, 1).$

Nonetheless, and under the assumption that S and T are isoloid, we give in the following result a characterization of the stability of property (k) under direct sum.

Theorem 2.10

Let $S\in \mathcal {B}(X)$ and let $T\in \mathcal {B}(Y),$ X and Y are Banach spaces. If S and T satisfy property (k) and are isoloid, then the following assertions are equivalent:

• (i) $S\oplus T$ satisfies property (k);

• (ii) $\sigma _{w}(S\oplus T)=\sigma _{w}(S)\cup \sigma _{w}(T)$.

Proof

• (i) $\Longrightarrow$ (ii) The property (k) for $S\oplus T$ implies with no other restriction on either S or T that $\sigma _{w}(S\oplus T)=\sigma _{w}(S)\cup \sigma _{w}(T)$. Indeed, as $S\oplus T$ satisfies property (k) then it satisfies Browder’s theorem and so $\sigma _{w}(S\oplus T)=\sigma _{b}(S\oplus T).$ Since $\sigma _{b}(S\oplus T)=\sigma _{b}(S)\cup \sigma _{b}(T),$ then $\sigma _{w}(S\oplus T)=\sigma _{b}(S)\cup \sigma _{b}(T),$ and as $\sigma _{w}(S)\cup \sigma _{w}(T)\subset \sigma _{b}(S)\cup \sigma _{b}(T)$, we then have $\sigma _{w}(S)\cup \sigma _{w}(T)\subset \sigma _{w}(S\oplus T).$ Hence $\sigma _{w}(S\oplus T)=\sigma _{w}(S)\cup \sigma _{w}(T).$

• (ii) $\Longrightarrow$ (i) Suppose that $\sigma _{w}(S\oplus T)=\sigma _{w}(S)\cup \sigma _{w}(T)$. As S and T are isoloid then

  $$\begin{aligned} E(S\oplus T)= & {} \left[ E(S)\cap \rho (T)\right] \cup \left[ E(T)\cap \rho (S)\right] \cup \left[ E(S)\cap E(T)\right] , \end{aligned}$$

  where $\rho (.)=\mathbb {\mathbb {C}}{\setminus }\sigma (.)$. On the other hand, since S and T satisfy property (k), i.e. $\sigma (S){\setminus }\sigma _{w}(S)=E(S)$ and $\sigma (T){\setminus }\sigma _{w}(T)=E(T),$ we then have

  $$\begin{aligned}&\left[ \sigma (S)\cup \sigma (T)\right] {\setminus } \left[ \sigma _{w}(S)\cup \sigma _{w}(T)\right] \\= & {} \left[ (\sigma (S){\setminus }\sigma _{w}(S))\cap \rho (T)\right] \cup \left[ (\sigma (T){\setminus }\sigma _{w}(T))\cap \rho (S)\right] \\&\cup \left[ (\sigma (S){\setminus }\sigma _{w}(S))\cap (\sigma (T){\setminus }\sigma _{w}(T))\right] \\= & {} \left[ E(S)\cap \rho (T)\right] \cup \left[ E(T)\cap \rho (S)\right] \cup \left[ E(S)\cap E(T)\right] . \end{aligned}$$

Hence $E(S\oplus T)=[\sigma (S)\cup \sigma (T)]{\setminus }[\sigma _{w}(S)\cup \sigma _{w}(T)]= \sigma (S\oplus T){\setminus }\sigma _{w}(S\oplus T),$ and this shows that property (k) is satisfied by $S\oplus T$. $\square$

Remark 2.11

• 1. Theorem 2.10 extends [9, Theorem 3.2] which proves that if $T\in \mathcal {B}(\mathcal {H})$ and $S\in \mathcal {B}({\mathcal K} )$ are isoloid operators acting on Hilbert spaces $\mathcal {H}$ and ${\mathcal K}$ satisfying property (k) with the supplementary condition $\sigma _{w}(S\oplus T)=\sigma _{w}(S)\cup \sigma _{w}(T),$ then $S\oplus T$ satisfies property (k).

• 2. The assumption “S and T are isoloid” is essential in Theorem 2.10. For this define on $\mathbb {\mathbb {C}}^n\oplus \ell ^2(\mathbb {N})$ the operator $U=0\oplus S$ where S is defined by $S(x_1, x_2, \ldots )=(0, \frac{x_1}{2}, \frac{x_2}{3}, \ldots ).$ It is clear that the null operator satisfies property (k). Also the operator S satisfies property (k),  since $\sigma (S)=\sigma _{w}(S)=\{0\}$ and $E(S)=\emptyset .$ But U does not satisfy this property, since $\sigma (U)=\sigma _{w}(U)=\{0\}$ and $E(U)=\{0\}.$ Here $\sigma _{w}(0\oplus S)=\sigma _{w}(0)\cup \sigma _{w}(S)=\{0\},$ the null operator is isoloid, but S is not isoloid.

Before we state our next corollary as an application of Theorem 2.10 to the class of (H)-operators, we recall the definition of this class and definitions of some classes of operators which are contained in the class (H).

According to the monograph of Aiena [1], the quasinilpotent part $H_0(T)$ of $T\in \mathcal {B}(X)$ is defined as the set $H_{0}(T)=\{x\in X: \displaystyle \lim _{n\rightarrow \infty }\Vert T^{n}(x)\Vert ^{\frac{1}{n}}=0\}.$ Note that generally, $H_0(T)$ is not closed and from [1, Theorem 2.31] we have if $H_0(T-\lambda I)$ is closed then T has SVEP at $\lambda$. We also recall that T is said to belong to the class (H) if for all $\lambda \in \mathbb {C}$ there exists $p:=p(\lambda )\in \mathbb {N}$ such that $H_{0}(T-\lambda I)=\mathcal {N}((T-\lambda I)^p),$ see Aiena [1] for more details about this class of (H)-operators. Of course, every operator T which belongs the class (H) has SVEP, since $H_0(T-\lambda I)$ is closed. Observe also that $a(T-\lambda I)\le p,$ for every $\lambda \in \mathbb {C}.$ The class of operators having the property (H) is rather large. Obviously, it contains every operator having the property $(H_1).$ Recall that an operator $T\in \mathcal {B}(X)$ is said to have the property $(H_1)$ if $H_{0}(T-\lambda I)=\mathcal {N}(T-\lambda I)$ for all $\lambda \in \mathbb {C}.$ Although the property $(H_1)$ seems to be rather strong, the class of operators having the property $(H_1)$ is considerably large. In the sequel we give some important classes of operators which satisfy property $(H_1).$ Every totally paranormal operator has property $(H_1),$ and in particular every hyponormal operator has property $(H_1).$ Also every transaloid operator or log-hyponormal has the property $(H_1).$ Some other operators satisfy property (H);  for example M-hyponormal operators, p-hyponormal operators, algebraically p-hyponormal operators, algebraically M-hyponormal operators, subscalar operators and generalized scalar operators. For more details about these definitions and comments which we cited above, we refer the reader to Aiena [1], Curto and Han [5] and, Laursen and Neumann [10].

Corollary 2.12

Let $S\in \mathcal {B}(X)$ and $T\in \mathcal {B}(Y)$ be isoloid operators and have a shared stable sign index. If S and T satisfy property (k), then $S\oplus T$ satisfies property (k). In particular, if S and T are (H)-operators satisfying property (k) then $S\oplus T$ satisfies property (k).

Proof

Assume that S and T are isoloid and satisfy property (k). Since S and T have a shared stable sign index, then it is easily seen that $\sigma _{w}(S\oplus T)=\sigma _{w}(S)\cup \sigma _{w}(T).$ But this is equivalent by Theorem 2.10, to say that property (k) holds for $S\oplus T.$ In particular if S and T are (H)-operators, then they are polaroid and so isoloid. But every (H)-operator has SVEP. Hence $\text{ ind }(T-\lambda I)$ and $\text{ ind }(S-\mu I)$ are less or equal than zero, for each $\lambda \in \rho _{sbf}(T)$ and $\mu \in \rho _{sbf}(S)$. Hence $S\oplus T$ satisfies property (k). $\square$