Quasi-Fredholm spectrum and compact perturbations

TL;DR

This paper investigates the properties of the quasi-Fredholm spectrum of operators on Banach spaces and examines the stability of the SVEP under compact perturbations in Hilbert spaces.

Contribution

It introduces new insights into the quasi-Fredholm spectrum and characterizes operators preserving the SVEP under compact perturbations.

Findings

Characterization of the quasi-Fredholm resolvent set.

Conditions for SVEP stability under compact perturbations.

Identification of operators maintaining SVEP in Hilbert spaces.

Abstract

In this paper we explore some characteristics of the quasi-Fredholm resolvent set $\rho_{qf}(T)$ of an operator $T$ defined on an infinite dimensional Banach space $X$. Moreover, in the case of Hilbert space $H$, we study the stability of the SVEP and describe the operators for which the SVEP is preserved under compact perturbations using quasi-Fredholm spectrum and $\rho_{qf}(T)$.

Equations

$${\sigma }_p(T):$$

$${\sigma }_a(T):$$

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

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

$${\sigma }_w(T):$$

$${\Psi }_n:T^n(X)/T^{n+1}(X)⟶T^{n+1}(X)/T^{n+2}(X)$$

$${\Psi }_n(y+T^{n+1}(X)):=Ty+T^{n+1}(X),y\in T^n(X).$$

$$\Vert y{\Vert }_n:=\underset{x\in X}{\inf }\{\Vert x\Vert :y=T^{n}x\}.$$

$${\sigma }_{\Gamma }(T):=\{\lambda \in \mathbb{C}:\lambda I-T\text{mbox}doesnothavetopologicaluniformdescent\}.$$

$$\Delta (T):=\{n\in \mathbb{N}:m\ge n,m\in \mathbb{N}\text{mbox}impliesthat{T}^n(X)\cap N(T)\subset T^m(X)\cap N(T)\}.$$

$${\sigma }_{qf}(T):=\{\lambda \in \mathbb{C}:\lambda I-T\text{mbox}isnotquasi-Fredholm\}.$$

$${\sigma }_{sbf}(T):=\{\lambda \in \mathbb{C}:\lambda I-T\text{mbox}isnotsemiB-Fredholm\}.$$

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

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

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

$${\sigma }_d(T):$$

$$p(\lambda I-T)<\infty \text{mbox}impliesthatT\text{mbox}hasSVEPat\lambda$$

$$q(\lambda I-T)<\infty \text{mbox}impliesthat{T}^{\ast }\text{mbox}hasSVEPat\lambda .$$

$${\rho }_{qf}(T)=\bigcup _{n=1}^{\infty }{\Omega }_{qf}^n=\bigcup _{n=1}^{\infty }{\Omega }_{sbf}^n\cup E.$$

$${\rho }_{qf}(T)\subset \text{mbox}iso(\partial P)\cup (\partial P{)}^c\subset \overline{{\rho }_{qf}(T)}=\mathbb{C}.$$

$${\sigma }_{\ast }(T)={\sigma }_{\ast }(T+K),$$

$$\sigma (T+K)={\sigma }_{qf}(T+K)\cup \Pi (T+K)\cup ({\sigma }_{sbf}(T+K)\setminus {\sigma }_{qf}(T+K)).$$

$${\Omega }_{qf}={\rho }_{sbf}(T+K)\cup E,$$

$${\rho }_{qf}(T+K)=\Omega \cup ({\rho }_{qf}(T+K)\cap {\Omega }^c).$$

$${\rho }_{qf}(T+K)=\rho (T+K)\cup \Pi (T+K)\cup ({\sigma }_{sbf}(T+K)\setminus {\sigma }_{qf}(T+K))$$

$${\rho }_{sbf}^{+}(T)=\{\lambda \in {\rho }_{sbf}:\text{mbox}ind(\lambda I-T)>0\}$$

$${\rho }_{sf}^{+}(T)=\{\lambda \in {\rho }_{sf}:\text{mbox}ind(\lambda I-T)>0\}.$$

$$T+K+K_{1}=\left({\begin{array}[]{cc}A+K_{2}&*\\ 0&B\\ \end{array}}\right)\left(\begin{array}[]{c}H_{1}\\ H_{1}^{\perp}\\ \end{array}\right)$$

$$\Omega \subset {\rho }_a(T+K)\cup \text{mbox}iso{\sigma }_a(T+K).$$

$${\Omega }^{{}^{\prime }}\subset {\rho }_a(T+K+K_1)\cup \text{mbox}iso{\sigma }_a(T+K+K_1).$$

$$\sigma (T+K)=\text{mbox}iso{\sigma }_{sbf}(T+K)\cup \Pi (T+K)$$

$${\sigma }_{qf}(T+K)={\sigma }_{sbf}(T+K)=\text{mbox}iso{\sigma }_{sbf}(T+K).$$

Full text

Mathematics Subject Classification: 47A10, 47A55, 47B15.

Keywords: Quasi-Fredholm spectrum, topological uniform descent, SVEP, compact perturbation.

Introduction and Preliminaries

Throughout this paper, denote by $B(X)$ the Banach algebra of all bounded linear operators defined on an infinite dimensional complex Banach space $X$. For $A\subset\mathbb{C},$ $\mbox{iso}\thinspace A$, $\mbox{int}\thinspace A$, $\overline{A}$ and $\mbox{acc}\thinspace A$ denote the set of isolated points of $A$, interior points of $A$, closure of $A$ and accumulation points of $A$, respectively. For $\lambda\in\mathbb{C}$ and $r>0$, $B(\lambda,r)$ denotes the open disc of radius $r$ centred at $\lambda.$ For $T\in B(X)$, the null space of $T$, range of $T$, spectrum of $T$ and adjoint of $T$ are denoted by $N(T)$, $T(X)$, $\sigma(T)$ and $T^{*}$, respectively. Let $\alpha(T)=$ dim $N(T)$ and $\beta(T)=$ codim $T(X)$ be the nullity of $T$ and deficiency of $T,$ respectively. An operator $T\in B(X)$ is called bounded below if $T$ is injective and $T(X)$ is closed. A bounded linear operator $T$ is said to be an upper semi-Fredholm operator if $\alpha(T)<\infty$ and $T(X)$ is closed. An operator $T\in B(X)$ is said to be a lower semi-Fredholm operator if $\beta(T)<\infty$. An operator $T\in B(X)$ is called a semi-Fredholm operator if it is either upper semi-Fredholm or lower semi-Fredholm. For a semi-Fredholm operator $T$, the index of $T$ is defined by $\mbox{ind}(T):=\alpha(T)-\beta(T)$. The point spectrum, approximate point spectrum and semi-Fredholm spectrum are defined by

[TABLE]

Clearly, $\sigma_{sf}(T)\subset\sigma_{a}(T)$. Let $\rho_{a}(T)=\mathbb{C}\setminus\sigma_{a}(T)$ and $\rho_{sf}(T)=\mathbb{C}\setminus\sigma_{sf}(T)$. An operator $T\in B(X)$ is called said to be an upper semi-Weyl (lower semi-Weyl, respectively) operator if it is upper semi-Fredholm (lower semi-Fredholm, respectively) and $\mbox{ind}(T)\leq 0$ ($\mbox{ind}(T)\geq 0$, respectively). A bounded linear operator $T$ is called Weyl if it is semi-Fredholm and $\mbox{ind}(T)=0$. The Weyl essential approximate point spectrum and Weyl spectrum are defined by

[TABLE]

Let $\rho_{uw}(T)=\mathbb{C}\setminus\sigma_{uw}(T)$ and $\rho_{w}(T)=\mathbb{C}\setminus\sigma_{bw}(T)$. Let $T\in B(X)$, then for each non negative integer $n$, $T$ induces a linear transformation

[TABLE]

defined by

[TABLE]

Clearly, $\Psi_{n}$ is surjective for each $n$. For each $n$, let $k_{n}(T)=\alpha(\Psi_{n})$. Define a norm $\|.\|_{n}$ on $T^{n}(X)$ by

[TABLE]

The topology induced by this norm is called operator range topology on $T^{n}(X)$. An operator $T\in B(X)$ is said to have uniform descent for $n\geq d$ if there exists a non negative integer $d$ such that $k_{n}(T)=0$ for $n\geq d$. In addition, if $T^{n}(X)$ is closed in the operator range topology of $T^{d}(X)$ for $n\geq d$, then $T$ is said to have topological uniform descent for $n\geq d$. The topological uniform descent spectrum is defined by

[TABLE]

Let $\rho_{\Gamma}(T)=\mathbb{C}\setminus\sigma_{\Gamma}(T)$ be the topological uniform descent resolvent of $T$.

For $T\in B(X)$ consider the set

[TABLE]

The degree of stable iteration is defined by $\mbox{dis}(T):=\inf\Delta(T)$ whenever $\Delta(T)\neq\emptyset$. If $\Delta(T)=\emptyset$, set $\mbox{dis}(T)=\infty$. Let $T\in B(X)$. An operator $T\in B(X)$ is said to be quasi-Fredholm of degree $d$ if there exists a $d\in\mathbb{N}$ such that

(i) $\mbox{dis}(T)=d$,

(ii) $T^{n}(X)$ is a closed subspace of $X$ for each $n\geq d$,

(iii) $T(X)+N(T^{d})$ is a closed subspace of $X$.

For $T\in B(X)$, the quasi-Fredholm spectrum is defined by

[TABLE]

Let $\rho_{qf}(T)=\mathbb{C}\setminus\sigma_{qf}(T)$ be the quasi-Fredholm resolvent of $T$. By [2, Theorem 1.96] we know that $\sigma_{\Gamma}(T)\subset\sigma_{qf}(T)\subset\sigma_{sf}(T)$. For a bounded linear operator $T$ and a non negative integer $n$, denote by $T_{[n]}$ the restriction of $T$ to $T^{n}(X)$. An opeartor $T\in B(X)$ is said to be B-Fredholm (an upper semi B-Fredholm, a lower semi B-Fredholm, respectively) if for some non negative integer $n$, $T^{n}(X)$ is closed and $T_{[n]}$ is Fredholm (an upper semi B-Fredholm, a lower semi B-Fredholm, respectively). In this case, the $\mbox{ind}(T)$ is defined to be the $\mbox{ind}T_{[n]}$ (see [4]). An operator is said to be a semi B-Fredholm operator if is a lower semi B-Fredholm or an upper semi B-Fredholm operator. The * semi B-Fredholm spectrum* is defined by

[TABLE]

Clearly, $\sigma_{sbf}(T)\subset\sigma_{sf}(T)$. Let $\rho_{sbf}(T)=\mathbb{C}\setminus\sigma_{sbf}(T)$. By [2, Theorem 1.116] we know that every semi B-Fredholm operator is quasi-Fredholm. Therefore, $\sigma_{\Gamma}(T)\subset\sigma_{qf}(T)\subset\sigma_{sbf}(T)\subset\sigma_{sf}(T)$. An operator $T\in B(X)$ is called an upper semi B-Weyl (B-Weyl, respectively) operator if $T$ is an upper semi B-Fredholm (B-Fredholm, respectively) having $\mbox{ind}(T)\leq 0$ ($\mbox{ind}(T)=0$, respectively). The upper semi B-Weyl spectrum and B-Weyl spectrum are defined by

[TABLE]

Let $\rho_{bw}(T)=\mathbb{C}\setminus\sigma_{bw}(T)$ and $\rho_{w}(T)=\mathbb{C}\setminus\sigma_{w}(T)$. For an operator $T\in B(X)$, the ascent of $T$ denoted by $p(T)$ is the smallest non negative integer $p$ such that $N(T^{p})=N(T^{p+1})$. If no such integer exists, set $p(T)=\infty$. For an operator $T\in B(X)$, the descent of $T$ denoted by $q(T)$ is the smallest non negative integer $q$ such that $T^{q}(X)=T^{q+1}(X)$. If no such integer exists, set $q(T)=\infty$. Evidently, $p(T)=0$ if and only if $T$ is injective and $q(T)=0$ if and only if $T$ is surjective. By [2, Theorem 1.20] we know that if both $p(T)$ and $q(T)$ are finite, then $p(T)=$ $q(T)$. An operator $T\in B(X)$ is called left Drazin invertible if $p(T)<\infty$ and $T^{p+1}(X)$ is closed. We say that $\lambda\in\mbox{iso}\thinspace\sigma_{a}(T)$ is a left pole of the resolvent of $T$ if $\lambda I-T$ is left Drazin invertible. An operator $T\in B(X)$ is called right Drazin invertible if $q(T)<\infty$ and $T^{q}(X)$ is closed. An operator $T\in\mathcal{B}(X)$ is called Drazin invertible if $p(T)=q(T)<\infty$. We say that $\lambda\in\mbox{iso}\thinspace\sigma(T)$ is a pole of the resolvent of $T$ if $\lambda I-T$ is Drazin invertible. The left Drazin spectrum and Drazin spectrum are defined by

[TABLE]

By [2, Theorem 1.142] we know that $\sigma_{qf}(T)\subset\sigma_{ld}(T)\subset\sigma_{d}(T)$. The set of all the poles of the resolvent of $T$ and all left poles of the resolvent of $T$ are denoted by $\Pi(T)=\sigma(T)\setminus\sigma_{d}(T)$ and $\Pi^{a}(T)=\sigma_{a}(T)\setminus\sigma_{ld}(T)$, respectively.

An operator $T\in B(X)$ is said to have the single-valued extension property (SVEP) at $\lambda_{0}\in\mathbb{C}$, if for every neighborhood $V$ of $\lambda_{0}$ the only analytic function $f:V\rightarrow X$ which satisfies the equation $(\lambda I-T)f(\lambda)=0$ is the function $f=0$. An operator $T\in B(X)$ is said to have SVEP if $T$ has SVEP at every $\lambda\in\mathbb{C}$. It is known that if $\mbox{int}\thinspace\sigma_{p}(T)=\emptyset$, then $T$ has SVEP. Recall that

[TABLE]

and

[TABLE]

Zeng et al. [9] studied the components of quasi-Fredholm resolvent and characterized them by means of localized SVEP. Shi [8] considered the topological uniform descent and studied how topological uniform descent resolvent is distributed in $\rho_{sf}(T)$. As we know that for an operator $T\in B(X)$, topological uniform descent, quasi-Fredholmness, semi-Fredholmness and semi B-Fredholness are closely related to each other. Motivated by them we study the distribution of $\rho_{qf}(T)$ in $\rho_{sbf}(T)$. Zhu and Li [10] obtained results for non commuting compact perturbations of an operator $T\in B(X)$ using semi-Fredholm spectrum. Recently, for $T\in B(X)$ various authors (see [3, 6, 7]) discussed various spectral properties under compact (not necessarily commuting) perturbations. Motivated by them we obtain results for compact perturbations of an operator $T\in B(X)$ using quasi-Fredholm spectrum.

In this paper we discuss some characteristics of quasi-Fredholm resolvent set $\rho_{qf}(T)$ for $T\in B(X)$. We give results regarding the distribution of semi B-Fredholm domain $\rho_{sbf}(T)$ in $\rho_{qf}(T)$. We prove that if $\mbox{int}\thinspace\sigma_{sbf}(T)=\emptyset$, then there is one-to-one correspondence between the bounded components of $\rho_{sbf}(T)$ and the bounded components of $\rho_{qf}(T)$. In the last section we discuss the permanence of SVEP under(small) compact perturbations using quasi-Fredholm resolvent set and quasi-Fredholm spectrum. Also, we describe those operators for which SVEP is stable under compact perturbations by means of quasi-Fredholm resolvent.

Main Results

It is known that the sets $\rho_{sf}(T),\rho_{sbf}(T),\rho_{qf}(T)$ and $\rho_{sf}(T)$ are nonempty open sets of $\mathbb{C},$ they can be decomposed into (pairwise disjoint, maximal, open, connected) non-empty components.

Lemma 2.1**.**

Let $T\in B(X)$. Then $\sigma_{sf}(T)=\sigma_{sbf}(T)\cup\mbox{iso}\thinspace\sigma_{sf}(T)$.

Proof.

Let $\lambda_{0}\in\sigma_{sf}(T)\setminus\sigma_{sbf}(T)$. Then $\lambda_{0}I-T$ is semi B-Fredholm. By [2, Theorem 1.117] there exists an $\epsilon>0$ such that $\lambda I-T$ is semi-Fredholm for all $\lambda\in B(\lambda_{0},\epsilon)\setminus\{\lambda_{0}\}.$ Therefore, $\lambda_{0}\in\mbox{iso}\thinspace\sigma_{sf}(T)$. Thus, $\sigma_{sf}(T)\subset\sigma_{sbf}(T)\cup\mbox{iso}\thinspace\sigma_{sf}(T)$. The reverse inclusion always holds. ∎

Recall that a hole of a compact set $\sigma\subset\mathbb{C}$ is a bounded component of $\mathbb{C}\setminus\sigma$. It is known that $\mathbb{C}\setminus\sigma$ has always an unbounded component. Therefore, $\mathbb{C}\setminus\sigma$ is connected if and only if $\sigma$ has no holes.

Theorem 2.2**.**

Let $T\in B(X)$, then $\rho_{sbf}(T)$ is connected if and only if $\rho_{sf}(T)$ is connected.

Proof.

Suppose that $\rho_{sbf}(T)$ is connected. Since $\sigma_{sf}(T)=\sigma_{sbf}(T)\cup\mbox{iso}\thinspace\sigma_{sf}(T)$, $\rho_{sf}(T)=\rho_{sbf}(T)\setminus\mbox{iso}\thinspace\sigma_{sf}(T)$. As $\rho_{sbf}(T)$ is connected and $\mbox{iso}\thinspace\sigma_{sf}(T)$ is at most countable we deduce that $\rho_{sf}(T)$ is connected.

Conversely, suppose that $\rho_{sf}(T)$ is connected. Assume that $\rho_{sbf}(T)$ is not connected then there exists a bounded component $\Omega$ of $\rho_{sbf}(T)$. Then either $\Omega\cap\rho_{sf}(T)=\emptyset$ or $\Omega\cap\rho_{sf}(T)\neq\emptyset$. If $\Omega$ $\cap$ $\rho_{sf}(T)=\emptyset$, then $\Omega\subset\sigma_{sf}(T)$ which implies that $\Omega\subset\mbox{iso}\thinspace\sigma_{sf}(T)$ which is not possible. Therefore, $\Omega\cap\rho_{sf}(T)\neq\emptyset$. Then there exists $\lambda_{0}$ such that $\lambda_{0}\in\Omega\cap\rho_{sf}(T)$. Let $\Omega^{{}^{\prime}}$ be the component of $\rho_{sf}(T)$ containing $\lambda_{0}$. Therefore, $\Omega^{{}^{\prime}}$ is an open connected subset of $\rho_{sbf}(T)$ such that $\Omega\cap\Omega^{{}^{\prime}}\neq\emptyset$. This implies that $\Omega^{{}^{\prime}}\subset\Omega$. Thus, $\Omega^{{}^{\prime}}$ is a bounded component of $\rho_{sf}(T)$, a contradiction. Hence, $\rho_{sbf}(T)$ is connected. ∎

By [5, Lemma 2.2] we have $\sigma_{uw}(T)=\sigma_{usbw}(T)\cup\mbox{iso}\thinspace\sigma_{uw}(T)$ and $\sigma_{w}(T)=\sigma_{bw}(T)\cup\mbox{iso}\thinspace\sigma_{w}(T)$. Following the lines of the proof of Theorem 2.2 we have the following result:

Theorem 2.3**.**

Let $T\in B(X)$, then

(i) $\rho_{bw}(T)$ is connected if and only if $\rho_{w}(T)$ is connected.

(ii) $\rho_{usbw}(T)$ is connected if and only if $\rho_{uw}(T)$ is connected.

Theorem 2.4**.**

Let $T\in B(X)$ and $\Omega_{qf}$ be a connected component of $\rho_{qf}(T)$. If $\Omega_{qf}\cap\rho_{sbf}(T)\neq\emptyset$, then there exists a unique connected component $\Omega_{sbf}$ of $\rho_{sbf}(T)$ such that $\Omega_{qf}=\Omega_{sbf}\cup E$, where $E_{0}\subset\mbox{iso}\thinspace\sigma_{sf}(T)$.

Proof.

As $\Omega_{qf}$ is a connected component of $\rho_{qf}(T)$ and $\rho_{qf}(T)\subset\rho_{\Gamma}(T)$, there exists a component $\Omega_{\Gamma}$ of $\rho_{\Gamma}(T)$ such that $\Omega_{qf}\subset\Omega_{\Gamma}$. Since $\Omega_{qf}\cap\rho_{sbf}(T)\neq\emptyset$ and $\rho_{sbf}(T)\subset\rho_{qf}(T)$, proceeding as in the proof of Theorem 2.2 there exists a component $\Omega_{sbf}$ of $\rho_{sbf}(T)$ such that $\Omega_{sbf}\subset\Omega_{qf}$. By the proof of Theorem 2.2 we get a component $\Omega_{sf}$ of $\rho_{sf}$ such that $\Omega_{sf}\subset\Omega_{sbf}\subset\Omega_{qf}\subset\Omega_{\Gamma}$. Using [8, Theorem 1] we have $\Omega_{\Gamma}=\Omega_{sf}\cup E$, where $E\subset\mbox{iso}\thinspace\sigma_{sf}(T)$. This gives $\Omega_{sbf}\subset\Omega_{qf}\subset\Omega_{sf}\cup E\subset\Omega_{sbf}\cup E$. Therefore, there exists $E_{0}\subset E\subset\mbox{iso}\thinspace\sigma_{sf}(T)$ such that $\Omega_{qf}=\Omega_{sbf}\cup E_{0}$.

Assume that there exist connected components $\Omega_{sbf}$ and $\Omega_{sbf}^{{}^{\prime}}$ of $\rho_{sbf}(T)$ such that $\Omega_{qf}=\Omega_{sbf}\cup E$ and $\Omega_{qf}=\Omega_{sbf}^{{}^{\prime}}\cup F$, where $E,F\subset\mbox{iso}\thinspace\sigma_{sf}(T)$. Then $\Omega_{sbf}\cup E=\Omega_{sbf}^{{}^{\prime}}\cup F$ which implies that $\Omega_{sbf}\subset F$, a contradiction. ∎

Corollary 2.5**.**

Let $T\in B(X)$ and $\Omega_{sbf}$ be a connected component of $\rho_{sbf}(T)$. Then there exists a unique connected component $\Omega_{qf}$ of $\rho_{qf}(T)$ such that $\Omega_{qf}=\Omega_{sbf}\cup E,$ where $E\subset\mbox{iso}\thinspace\sigma_{sf}(T).$

Proof.

Since $\Omega_{sbf}\subset\rho_{sbf}(T)\subset\rho_{qf}(T),$ there exists a connected component $\Omega_{qf}$ of $\rho_{qf}(T)$ such that $\Omega_{sbf}\subset\Omega_{qf}$. By Theorem 2.4 we get $\Omega_{qf}=\Omega_{sbf}\cup E_{0}$, where $E_{0}\subset\mbox{iso}\thinspace\sigma_{sf}(T)$. Assume that there exists another connected component $\Omega_{qf}^{{}^{\prime}}$ of $\rho_{qf}(T)$ such that $\Omega_{qf}^{{}^{\prime}}=\Omega_{sbf}\cup F$, where $F\subset\mbox{iso}\thinspace\sigma_{sf}(T)$. This gives $\Omega_{sbf}\subset\Omega_{qf}\cap\Omega_{qf}^{{}^{\prime}}$, a contradiction. ∎

Corollary 2.6**.**

Let $T\in B(X)$ and $\mbox{int}\thinspace\sigma_{sbf}(T)=\emptyset.$ Then $\rho_{qf}(T)\setminus\rho_{sbf}(T)$ is at most countable and $\sigma_{sbf}(T)=\sigma_{qf}(T)\cup\mbox{iso}\thinspace\sigma_{sbf}(T)$.

Proof.

Let $\{\Omega_{qf}^{n}\}_{n=1}^{\infty}$ be an enumeration of connected components of $\rho_{qf}(T)$. Since $\mbox{int}\thinspace\sigma_{sbf}(T)=\emptyset$, for every connected component $\Omega_{qf}^{n}$ of $\rho_{qf}(T)$ we have $\Omega_{qf}^{n}\cap\rho_{sbf}(T)\neq\emptyset$. Using Theorem 2.4 for $\Omega_{qf}^{n}$, there exists a unique connected component $\Omega_{sbf}^{n}$ of $\rho_{sbf}(T)$ such that $\Omega_{qf}^{n}$=$\Omega_{sbf}^{n}\cup E^{n},$ where $E^{n}\subset\mbox{iso}\thinspace\sigma_{sf}(T)$. Let $E=\bigcup\limits_{n=1}^{\infty}E^{n}$, then $E$ is at most countable and $E\subset\mbox{iso}\thinspace\sigma_{sf}(T)$. Also,

[TABLE]

Since $\rho_{sbf}(T)\subset\rho_{qf}(T)$, $\rho_{qf}(T)=\rho_{sbf}(T)\cup E$. Let $E^{{}^{\prime}}=E\cap\sigma_{sbf}(T)$. Then $E^{{}^{\prime}}\subset\mbox{iso}\thinspace\sigma_{sbf}(T)$ and $\rho_{qf}(T)=\rho_{sbf}(T)\cup E^{{}^{\prime}}$. This gives $\sigma_{sbf}(T)=\sigma_{qf}(T)\cup E^{{}^{\prime}}$ which implies that $\sigma_{sbf}(T)=\sigma_{qf}(T)\cup\mbox{iso}\thinspace\sigma_{sbf}(T)$. ∎

Let $W_{1}$, $W_{2}$, $W_{3}$ and $W_{4}$ be the set of all bounded components of $\rho_{sbf}(T)$, $\rho_{qf}(T),\rho_{sf}(T)$ and $\rho_{\Gamma}(T)$, respectively.

Theorem 2.7**.**

Let $T\in B(X)$. Then there exists an injective mapping $f:W_{1}\rightarrow W_{2}$. Moreover, if $\mbox{int}\thinspace\sigma_{sbf}(T)=\emptyset$, then $f$ is also surjective.

Proof.

Suppose that $\Omega\in W_{1}$. Using Corollary 2.5 we get a unique connected component $\Omega^{{}^{\prime}}$ of $\rho_{qf}(T)$ such that $\Omega^{{}^{\prime}}=\Omega\cup E$, where $E\subset\mbox{iso}\thinspace\sigma_{sf}(T)$. Since $\mbox{iso}\thinspace\sigma_{sf}(T)\subset\sigma_{sf}(T)$, $\Omega^{{}^{\prime}}$ is bounded component of $\rho_{qf}(T)$ which implies that $\Omega^{{}^{\prime}}\in W_{2}$. Define $f:W_{1}\rightarrow W_{2}$ by $f(\Omega)=\Omega^{{}^{\prime}}$. Then $f$ is a well defined mapping. We prove that $f$ is an injective mapping. Let $\Omega_{1}$ and $\Omega_{2}$ be two distinct elements of $W_{1}$ such that $f(\Omega_{1})=f(\Omega_{2})$. This implies that there exists a component $\Omega^{{}^{\prime}}$ of $\rho_{qf}(T)$ such that $\Omega^{{}^{\prime}}=\Omega_{1}\cup E=\Omega_{2}\cup F$, where $E,F\subset\mbox{iso}\thinspace\sigma_{sf}(T)$. As $\Omega_{1}\cap\Omega_{2}=\emptyset$, $\Omega_{1}\subset F$, a contradiction. Therefore, $f$ is an injective mapping.

Suppose that $\tau\in W_{2}$. Since $\mbox{int}\thinspace\sigma_{sbf}(T)=\emptyset$, $\tau\cap\rho_{sbf}(T)\neq\emptyset$. Using Theorem 2.4 there exists a unique component $\tau^{{}^{\prime}}$ of $\rho_{sbf}(T)$ such that $\tau=\tau^{{}^{\prime}}\cup E$, where $E\subset\mbox{iso}\thinspace\sigma_{sf}(T)$. Therefore, $f(\tau^{{}^{\prime}})=\tau$. ∎

Similarly, using [8, Theorem 1, Corollary 1] we establish the following result:

Theorem 2.8**.**

Let $T\in B(X)$. Then there exists an injective mapping $g:W_{3}\rightarrow W_{4}$. Moreover, if $\mbox{int}\thinspace\sigma_{sf}(T)=\emptyset$, then $g$ is also surjective.

Theorem 2.9**.**

Let $T\in B(X)$. Then every non isolated boundary point of $\sigma_{sbf}(T)$ belongs to $\sigma_{qf}(T)$.

Proof.

Let $\lambda$ be a non isolated boundary point of $\sigma_{sbf}(T)$. Let $\lambda\in\rho_{qf}(T)$ and $\Omega_{qf}$ be the component of $\rho_{qf}(T)$ containing $\lambda$. Then there exists $\epsilon>0$ such that $B(\lambda,\epsilon)\subset\Omega_{qf}$. Since $\lambda$ is the boundary point of $\sigma_{sbf}(T)$, $B(\lambda,\epsilon)\cap\rho_{sbf}(T)\neq\emptyset$ which implies that $\Omega_{qf}\cap\rho_{sbf}(T)\neq\emptyset$. Therefore, by Theorem 2.4 there exists a component $\Omega_{sbf}$ of $\rho_{sbf}(T)$ such that $\Omega_{qf}=\Omega_{sbf}\cup E$, where $E\subset\mbox{iso}\thinspace\sigma_{sf}(T)$. Since $\lambda\in\Omega_{qf}\cap\mbox{acc}\thinspace\sigma_{sbf}(T)\subset\Omega_{qf}\cap\mbox{acc}\thinspace\sigma_{sf}(T)$ we deduce that $\lambda\in\Omega_{sbf}$, a contradiction. Therefore, $\lambda\in\sigma_{qf}(T)$. ∎

Remark 2.10**.**

It is observed that if $P$ is a closed subset of $\mathbb{C}$ such that $\mbox{int}\thinspace P\neq\emptyset$ and $\mbox{int}\thinspace P^{c}\neq\emptyset$, then $(\partial P)^{c}$ is disconnected.**

Lemma 2.11**.**

Let $T\in B(X)$, $\rho_{qf}(T)$ be connected and $\mbox{int}\thinspace\sigma_{qf}(T)=\emptyset$. Suppose that $P$ is a closed set contained in $\sigma(T)$. Then $\mbox{int}\thinspace P=\emptyset$.

Proof.

Suppose that $\mbox{int}\thinspace P\neq\emptyset$. First we prove that $\rho_{qf}(T)\cap\mbox{acc}\thinspace(\partial P)\neq\emptyset$. If $\rho_{qf}(T)\cap\mbox{acc}\thinspace(\partial P)=\emptyset$, then

[TABLE]

Since $\rho_{qf}(T)$ is connected, $\mbox{iso}\thinspace(\partial P)\cup(\partial P)^{c}$ is connected. Let $S=\mbox{iso}\thinspace(\partial P)\cup(\partial P)^{c}.$ This implies that $(\partial P)^{c}=S\setminus\mbox{iso}\thinspace(\partial P)$ which gives $(\partial P)^{c}$ is connected. As $\mbox{int}\thinspace P\neq\emptyset$ then by Remark 2.10 we get a contradiction. Therefore, there exists $\lambda$ such that $\lambda\in\rho_{qf}(T)\cap\mbox{acc}\thinspace(\partial P)$. As $\rho_{qf}(T)$ is connected and $\rho(T)\subset\rho_{qf}(T)$, by [9, Theorems 3.6, 3.7] $p(\lambda I-T)=q(\lambda I-T)<\infty$ for all $\lambda\in\rho_{qf}(T)$. Therefore, $(\lambda I-T)$ is drazin invertible for all $\lambda\in\rho_{qf}(T)$. This gives $\rho_{qf}(T)=\rho(T)\cup\Pi(T)$, where $\Pi(T)$ denotes the set of poles of the resolvent of $T$. Since $\lambda\in\mbox{acc}\thinspace(\partial P)\subset\sigma(T)$ which implies that $\lambda\in\Pi(T)\subset\mbox{iso}\thinspace\sigma(T)$. Then there exists an $\epsilon>0$ such that $B(\lambda,\epsilon)\setminus\{\lambda\}\subset\rho(T)$. Since $\lambda\in\mbox{acc}\thinspace(\partial P)$, there exists $\mu\in B(\lambda,\epsilon)\cap\partial P\subset\rho(T)\cap\partial P$, a contradiction. Hence, $\mbox{int}\thinspace P=\emptyset$. ∎

If $\rho_{\Gamma}(T)$ is connected, then by [8, Proposition 2] we know that $\rho_{\Gamma}(T)=\rho(T)\cup\Pi(T)$. Then proceeding likewise as in Lemma 2.11 we have the following result:

Lemma 2.12**.**

Let $T\in B(X)$, $\rho_{\Gamma}(T)$ be connected and $\mbox{int}\thinspace\sigma_{\Gamma}(T)=\emptyset$. Suppose that $P$ is closed set contained in $\sigma(T)$. Then $\mbox{int}\thinspace P=\emptyset$.

Theorem 2.13**.**

Let $T\in B(X)$. Then following statements are equivalent:

(i) $\rho_{sf}(T)$ is connected and int $\sigma_{sf}(T)=\emptyset$,

(ii) $\rho_{sbf}(T)$ is connected and int $\sigma_{sbf}(T)=\emptyset$,

(iii) $\rho_{qf}(T)$ is connected and int $\sigma_{qf}(T)=\emptyset$,

(iv) $\rho_{\Gamma}(T)$ is connected and $\mbox{int}\thinspace\sigma_{\Gamma}(T)=\emptyset$.

Proof.

Since $\sigma_{\Gamma}(T)\subset\sigma_{qf}(T)\subset\sigma_{sbf}(T)\subset\sigma_{sf}(T)$, (i)$\Rightarrow$(ii)$\Rightarrow$(iii)$\Rightarrow$(iv) is obvious. Now suppose that $\rho_{\Gamma}(T)$ is connected and int $\sigma_{\Gamma}(T)=\emptyset$. Using Lemma 2.12 we deduce that $\mbox{int}\thinspace\sigma_{sf}(T)=\emptyset$. It remains to prove that $\rho_{sf}(T)$ is connected. As $\rho_{\Gamma}(T)$ is connected, by [8, Theorem 1] there exists a unique connected component $\Omega_{sf}$ of $\rho_{sf}(T)$ such that $\rho_{\Gamma}(T)=\Omega_{sf}\cup E_{0}$, where $E_{0}\subset\mbox{iso}\thinspace\sigma_{sf}(T)$. This gives $\rho_{\Gamma}(T)=\rho_{sf}(T)\cup E_{0}$. This implies that $rho_{sf}(T)=\rho_{\Gamma}(T)\setminus E_{0}$. Therefore, $\rho_{sf}(T)$ is connected. ∎

Lemma 2.14**.**

Let $T\in B(X)$. Then if $\rho_{sf}(T)$ consists of finite bounded components, then $\rho_{sbf}(T)$ consists of finite bounded components.

Proof.

Suppose that $\Omega_{1}$ and $\Omega_{2}$ are two distinct bounded components of $\rho_{sbf}(T)$. Then by the proof of Theorem 2.2 we get bounded components $\Omega_{1}^{{}^{\prime}}$, $\Omega_{2}^{{}^{\prime}}$ of $\rho_{sf}(T)$ such that $\Omega_{1}^{{}^{\prime}}\subset\Omega_{1}$ and $\Omega_{2}^{{}^{\prime}}\subset\Omega_{2}$. This gives $\Omega_{1}^{{}^{\prime}}\cap\Omega_{2}^{{}^{\prime}}=\emptyset$ since if $\Omega_{1}^{{}^{\prime}}=\Omega_{2}^{{}^{\prime}}$, then $\Omega_{1}\cap\Omega_{2}\neq\emptyset$ which is a contradiction. ∎

Remark 2.15**.**

If $\mbox{int}\thinspace\sigma_{qf}(T)=\emptyset$ and $\Omega$ is bounded component of $\rho_{\Gamma}(T)$, then $\Omega\cap\rho_{qf}(T)\neq\emptyset$. Therefore, there exists a component $\Omega^{{}^{\prime}}$ of $\rho_{qf}(T)$ such that $\Omega^{{}^{\prime}}\subset\Omega$. From this we can conclude that for any two bounded distinct components of $\rho_{\Gamma}(T)$ we get two distinct component of $\rho_{qf}(T)$. Hence, if $\rho_{qf}(T)$ consists of finite bounded components, then $\rho_{\Gamma}(T)$ consists of bounded components.**

Theorem 2.16**.**

Let $T\in B(H)$ and $int\sigma_{p}(T)=\emptyset$. Then following statements are equivalent:

(i) $\mbox{int}\thinspace\sigma_{sf}(T)=\emptyset$ and $\rho_{sf}(T)$ consists of finite bounded components,

(ii) $\mbox{int}\thinspace\sigma_{sbf}(T)=\emptyset$ and $\rho_{sbf}(T)$ consists of finite bounded components,

(iii) $\mbox{int}\thinspace\sigma_{qf}(T)=\emptyset$ and $\rho_{qf}(T)$ consists of finite bounded components,

(iv) $\mbox{int}\thinspace\sigma_{\Gamma}(T)=\emptyset$ and $\rho_{\Gamma}(T)$ consists of finite bounded components.

Proof.

(i) $\Rightarrow$(ii) Follows directly from Lemma 2.14.

(ii) $\Rightarrow$(iii) Follows from Theorem 2.7.

(iii) $\Rightarrow$(iv) Follows from Remark 2.15.

(iv) $\Rightarrow$(i) From Theorem 2.8 it follows that $\rho_{sf}(T)$ consists of finite bounded components. Also, if $\mbox{int}\thinspace\sigma_{p}(T)=\emptyset$, then $T$ has SVEP. Therefore, by [8, Proposition] we have $\mbox{int}\thinspace\sigma_{sf}(T)=\emptyset$. ∎

Quasi-Fredholm spectrum and compact perturbations

Let $K(X)$ denote the ideal of all compact operators acting on a Banach space $X$. It is known that for $T\in B(X)$ and $K\in K(X)$,

[TABLE]

where $\sigma_{*}=\sigma_{sf}$ or $\sigma_{uw}$ or $\sigma_{w}$. We start the following section with the following theorem:

Theorem 3.1**.**

Let $T\in B(X)$ and $\rho_{qf}(T)$ be connected. Suppose that $K\in K(X)$. Then

[TABLE]

Proof.

Since $\rho_{qf}(T)$ is connected, $\rho_{qf}(T)=\rho(T)\cup\Pi(T)$ which implies that $\rho_{qf}(T)=\rho_{sbf}(T)$. Therefore, $\rho_{sbf}(T)$ is connected. Using Theorem 2.2 we get $\rho_{sf}(T)$ is connected. As $\rho_{sf}(T)=\rho_{sf}(T+K)$, $\rho_{sf}(T+K)$ is connected. Again using Theorem 2.2 $\rho_{sbf}(T+K)$ is connected. By Corollary 2.5 there exists a component $\Omega_{qf}$ of $\rho_{qf}(T+K)$ such that

[TABLE]

where $E\subset\mbox{iso}\thinspace\sigma_{sf}(T+K)$. Let $E_{0}=E\cap\sigma_{sbf}(T+K)$. Then $\Omega_{qf}=\rho_{sbf}(T+K)\cup E_{0}$ and $E_{0}\subset\mbox{iso}\thinspace\sigma_{sbf}(T+K)$. As $\rho(T+K)\subset\rho_{sbf}(T+K)\subset\Omega$, by [9, Theorems 3.6, 3.7] we get $\Omega=\rho(T+K)\cup\Pi(T+K)$. Now

[TABLE]

As $\rho_{qf}(T+K)\cap\Omega^{c}=\sigma_{sbf}(T+K)\setminus\sigma_{qf}(T+K)$,

[TABLE]

which gives $\sigma(T+K)=\sigma_{qf}(T+K)\cup\Pi(T+K)\cup(\sigma_{sbf}(T+K)\setminus\sigma_{qf}(T+K))$. ∎

Denote

[TABLE]

and

[TABLE]

Lemma 3.2**.**

Let $T\in B(X).$ Then $\rho_{sbf}^{+}(T)=\emptyset$ if and only if $\rho_{sf}^{+}(T)=\emptyset$.

Proof.

Evidently, if $\rho_{sbf}^{+}(T)=\emptyset$, then $\rho_{sf}^{+}(T)=\emptyset.$ Conversely, suppose that $\rho_{sf}^{+}(T)=\emptyset$. Without loss of generality, we may assume that $0\in\rho_{sbf}^{+}(T)$ then $0\in\rho_{sbf}(T)$ and $\mbox{ind}(T)>0.$ Now by [2, Theorem 1.117] there exists an $\epsilon>0$ such that $B(0,\epsilon)\setminus\{0\}\subset\rho_{sf}(T)$ and $\mbox{ind}(\lambda I-T)>0$ for all $\lambda\in B(0,\epsilon)$, a contradiction. ∎

The following result is an immediate consequence of Lemma 3.2 and [10, Theorem 1.1]

Theorem 3.3**.**

Let $T\in B(H)$, where $H$ is a Hilbert space. Then the following statements are equivalent:

(i) There exists $K\in K(H)$ such that $T+K$ has SVEP,

(ii) $\rho_{sbf}^{+}(T)=\emptyset$,

(iii) $\rho_{sf}^{+}(T)=\emptyset$.

It is observed that if $T$ has SVEP at every point of $\rho_{qf}(T)$ then $\rho_{qf}(T)$ need not be connected. The following example illustrates this fact:

Example 3.4**.**

Let $R$ be unilateral shift on $l^{2}(\mathbb{N})$. It is known that $\sigma_{a}(T)=S^{1}$, where $S^{1}$ denotes the unit circle. Therefore, $\partial\sigma_{a}(T)\cap\mbox{acc}\thinspace\sigma_{a}(T)=S^{1}$. By [11, Corollary 3.6] we have $\sigma_{qf}(T)=\sigma_{usbb}(T)=S^{1}$. Hence, $T$ has SVEP at every point of $\rho_{qf}(T)$ but $\rho_{qf}(T)$ is not connected.**

Theorem 3.5**.**

Let $T\in B(H)$, where $H$ is a Hilbert space. Then $T+K$ has SVEP at every point of $\rho_{qf}(T+K)$ for any $K\in K(H)$ if and only if $\rho_{qf}(T+K)$ is connected for any $K\in K(H)$.

Proof.

Suppose that $T+K$ has SVEP at every point of $\rho_{qf}(T+K)$ for any $K\in K(H)$. Let $\rho_{qf}(T+K)$ is not connected for some $K\in K(H)$. Then we can find a bounded connected component $\Omega$ of $\rho_{qf}(T+K)$. Now $\partial\Omega\subset\sigma_{qf}(T+K)\subset\sigma_{sf}(T+K)$. Then the proof of [8, corollary 4] shows that there exist compact operators $K_{1}$ and $K_{2}$ such that

[TABLE]

where $A$ is a normal operator and $\lambda I-(A+K_{2})$ is Weyl but not invertible operator for any $\lambda\in\Omega$. Since $T+K$ has SVEP at every point of $\rho_{qf}(T+K)$, $T+K$ has SVEP at $\Omega$. Then by [9, Theorem 3.6] we have

[TABLE]

Therefore, $\Omega\cap\rho_{a}(T+K)\neq\emptyset$ which implies that $\Omega\cap\rho_{sf}(T+K)\neq\emptyset$. Let $\lambda\in\Omega\cap\rho_{sf}(T+K)$ and $\Omega^{{}^{\prime}}$ be a connected component of $\rho_{sf}(T+K)$ contaning $\lambda$. Then $\Omega^{{}^{\prime}}\subset\Omega$. Since $T+K+K_{1}$ has SVEP at every point of $\rho_{qf}(T+K+K_{1})$, $T+K+K_{1}$ has SVEP at every point of $\rho_{sf}(T+K+K_{1})$. Using [1, Theorem 3.36] we have

[TABLE]

This gives $\Omega^{{}^{\prime}}\cap\rho_{a}(T+K+K_{1})\neq\emptyset$. Therefore, there exists $\mu\in\Omega^{{}^{\prime}}\subset\Omega$ such that $\mu I-(T+K+K_{1})$ is bounded below which implies that $\mu I-(A+K_{2})$ is bounded below, a contradiction. Conversely, suppose that $\rho_{qf}(T+K)$ is connected for any $K\in K(H)$. Then by [9, Theorem 3.6], we know that $T+K$ has SVEP at every point of $\rho_{qf}(T+K)$. ∎

The following result follows from [8, Proposition 6, Corollary 4] and Theorem 2.13.

Theorem 3.6**.**

Let $T\in B(H)$, where $H$ is a Hilbert space. Then following statements are equivalent:

(i) $T+K$ has SVEP for any $K\in K(H)$,

(ii) $T^{*}+K$ has SVEP for any $K\in K(H)$,

(iii) $\rho_{sf}(T)$ is connected and $\mbox{int}\thinspace\sigma_{sf}(T)=\emptyset$,

(iv) $\rho_{sbf}(T)$ is connected and $\mbox{int}\thinspace\sigma_{sbf}(T)=\emptyset$,

(v) $\rho_{qf}(T)$ is connected and $\mbox{int}\thinspace\sigma_{qf}(T)=\emptyset$,

(vi) $\rho_{\Gamma}(T)$ is connected and $\mbox{int}\thinspace\sigma_{\Gamma}(T)=\emptyset$,

(vii) $\rho_{sf}(T+K)$ is connected and $\mbox{int}\thinspace\sigma_{sf}(T+K)=\emptyset$ for any $K\in K(H)$,

(viii) $\rho_{sbf}(T+K)$ is connected and $\mbox{int}\thinspace\sigma_{sbf}(T+K)=\emptyset$ for any $K\in K(H)$,

(ix) $\rho_{qf}(T+K)$ is connected and $\mbox{int}\thinspace\sigma_{qf}(T+K)=\emptyset$ for any $K\in K(H)$,

(x) $\rho_{\Gamma}(T+K)$ is connected and $\mbox{int}\thinspace\sigma_{\Gamma}(T+K)=\emptyset$.

The following result is consequence of [10, Theorem 1.2] and Theorem 2.16.

Theorem 3.7**.**

Let $T\in B(H)$, where $H$ is a Hilbert space. If

(i) $\mbox{int}\thinspace\sigma_{p}(T)=\emptyset$,

(ii) $\mbox{int}\thinspace\sigma_{*}(T)=\emptyset$,

*(iii) $\rho_{*}(T)$ consists of finite bounded components,

where $\sigma_{*}$,$\rho_{*}=\sigma_{sf}$, $\rho_{sf}$ or $\sigma_{sbf}$, $\rho_{sbf}$ or $\sigma_{qf}$, $\rho_{qf}$ or $\sigma_{\Gamma}$, $\rho_{\Gamma}$. Then there exists $\delta>0$ such that $T+K$ has SVEP for all $K\in K(H)$ with $\|K\|<\delta$.*

Theorem 3.8**.**

Let $T\in B(H)$, where $H$ is Hilbert space. If $\sigma_{qf}(T)=\emptyset$, then

[TABLE]

for any compact operator $K\in K(X)$.

Proof.

Since $\sigma_{qf}(T)=\emptyset$, $\rho_{qf}(T)=\mathbb{C}$ which implies that $\sigma(T)=\Pi(T)$. As $\mbox{int}\thinspace\sigma_{qf}(T)=\emptyset$, by Theorem 2.16 $\rho_{qf}(T+K)$ is connected. This gives $\rho_{qf}(T+K)=\rho_{sbf}(T+K)$. Therefore, by Theorem 3.1 $\sigma(T+K)=\sigma_{qf}(T+K)\cup\Pi(T+K).$ As $\sigma(T)=\Pi(T)$, $\sigma(T)$ is finite which implies that $\sigma_{sf}(T)=\sigma_{sf}(T+K)$ is finite. This gives

[TABLE]

Therefore, $\sigma(T+K)=\mbox{iso}\thinspace\sigma_{sbf}(T+K)\cup\Pi(T+K).$ ∎

Acknowledgement The corresponding author (Ankit Kumar) is supported by Department of Science and Technology, New Delhi, India (Grant No. DST/INSPIRE Fellowship/[IF170390]).