Further Properties of an Operator Commuting with an Injective Quasi-Nilpotent Operator

Abstract

In (Aiena et al., Math. Proc. R. Irish Acad. 122A(2):101–116, 2022), it has been shown that a bounded linear operator $T\in L(X)$, defined on an infinite-dimensional complex Banach space X, for which there exists an injective quasi-nilpotent operator that commutes with it, has a very special structure of the spectrum. In this paper, we show that we have much more: if a such quasi-nilpotent operator does exist, then some of the spectra of T originating from B-Fredholm theory coalesce. Further, the spectral mapping theorem holds for all the B-Weyl spectra. Finally, the generalized version of Weyl type theorems hold for T assuming that T is of polaroid type. Our results apply to the operators that belong to the commutant of Volterra operators.

1 Introduction

This paper concerns operators $T\in L(X)$ that belong to the commutant of an injective quasi-nilpotent operator Q. The spectral properties of such operators, from the point of view of the classical Fredholm theory, have been studied in [7]. In this paper, we consider the spectra originating from the B-Fredholm theory in the sense of Berkani et al., and we show, if a such operator Q does exist, that the upper B-Weyl spectrum $\sigma _{\textrm{ubw}}(T)$ coincides with the left Drazin spectrum $\sigma _{\textrm{ld}}(T)$, while the B-Weyl spectrum $\sigma _{\textrm{bw}}(T)$ coincides with the Drazin spectrum $\sigma _{\textrm{d}}(T)$. These spectral equalities have, as a consequence, that the B-Weyl spectrum, as well as the upper B-Weyl spectrum, obeys to the spectral mapping theorem. Dually, assuming that the dual $Q^*$ of a quasi-nilpotent operator is injective, then the lower B-Weyl spectrum $\sigma _{\textrm{lbw}}(T)$ coincides with the right Drazin spectrum $\sigma _{\textrm{rd}}(T)$. In this case, both the lower B-Weyl spectrum and the B-Weyl spectrum satisfy the spectral mapping theorem.

The second section of this article concerns some generalized version of Weyl type theorems, as the generalized version of a-Weyl’s theorem and the generalized version of property $(\omega )$. If T commutes with an injective quasi-nilpotent operator, then these generalized versions hold for T assuming that T is left polaroid, or that T is a-polaroid. Note that the generalized versions of Weyl type theorems are stronger than the classical versions, see [1, Chapter 5]. Another stronger variant of Weyl’s theorem, the so-called S-Weyl’s theorem, is also discussed in the last part.

Our results find a natural application to the operators which belong to the commutant of the Volterra V operator in $L^2[a,b]$, since, as it is well known, both V and its adjoint $V'$ are quasi-nilpotent and injective. Since the Volterra operator is also compact, our results complement each other with the celebrate Lomonosov result [15] that V admits a non-trivial closed invariant subspace.

2 Preliminaries and Definitions

Let L(X) denote the Banach algebra of all bounded linear operators acting on a complex Banach space X. If $T\in L(X)$, by $\alpha (T): =\text{ dim }\, \ker \, T$ and $\beta (T): = \text{ codim }\, T(X)$, we denote the defects of T. The class of all upper semi-Fredholm operators is defined by

$$\begin{aligned} \Phi _+(X): =\{T \in L(X): \alpha (T)< \infty \ \text{ and }\ T(X) \ \text{ is } \text{ closed }\}, \end{aligned}$$

while the class all lower semi-Fredholm operators is defined by

$$\begin{aligned} \Phi _-(X): =\{T \in L(X): \beta (T)< \infty \}. \end{aligned}$$

The class of all semi-Fredholm operators is defined by $\Phi _\pm (X): = \Phi _+(X) \cup \Phi _-(X)$. For every $T\in \Phi _{\pm }(X)$, the index of T is defined by $\text{ ind }\, T =\alpha (T) - \beta (T)$. The upper semi-Fredholm spectrum is defined by

$$\begin{aligned} \sigma _{\textrm{usf}}(T): =\{\lambda \in {\mathbb {C}}: \lambda I -T \notin \Phi _+(X)\}, \end{aligned}$$

and similarly, it is defined the lower semi-Fredholm spectrum $\sigma _{\textrm{lsf}}(T)$. Recall that an operator $T\in L(X)$ is said to be bounded below if is injective and has closed range. The classical approximate point spectrum is defined by

$$\begin{aligned} \sigma _{\textrm{ap}}(T): =\{\lambda \in {\mathbb {C}}: \lambda I-T \ \text{ is } \text{ not } \text{ bounded } \text{ below }\}, \end{aligned}$$

while the surjectivity spectrum is defined as

$$\begin{aligned} \sigma _{{\textrm{s}}}(T): =\{\lambda \in {\mathbb {C}}: \lambda I-T \ \text{ is } \text{ not } \text{ onto }\}. \end{aligned}$$

If $T^*$ denotes the dual of T, it is well known that $\sigma _{\textrm{ap}}(T)=\sigma _{\textrm{s}}(T^*)$ and $\sigma _{\textrm{s}}(T)=\sigma _{\textrm{ap}}(T^*)$.

Denote by $\Phi (X):= \Phi _+(X)\cap \Phi _-(X)$ the class of all Fredholm operators. An operator $T\in L(X)$ is said to be a Weyl operator if $T\in \Phi (X)$ and $\text{ ind }\, T =0$, $T\in L(X)$ is said to be upper semi-Weyl if $T\in \Phi _+(X)$ and $\text{ ind }\, T\le 0$, while $T\in L(X)$ is said to be lower semi-Weyl if $T\in \Phi _-(X)$ and $\text{ ind }\, T\ge 0$. Denote by $\sigma _{\textrm{w}}(T)$, $\sigma _{\textrm{uw}}(T)$ and $\sigma _{\textrm{lw}}(T)$, the Weyl spectrum, the upper semi-Weyl spectrum, and the lower semi-Weyl spectrum, respectively. Obviously, the inclusions

$$\begin{aligned} \sigma _{\textrm{usf}}(T) \subseteq \sigma _{\textrm{uw}}(T)\subseteq \sigma _{\textrm{ap}}(T) \quad \text{ and } \quad \sigma _{\textrm{lsf}}(T)\subseteq \sigma _{\textrm{lw}}(T)\subseteq \sigma _{\textrm{s}}(T) \end{aligned}$$

hold for every $T\in L(X)$, and there is a duality

$$\begin{aligned}\sigma _{\textrm{uw}}(T)= \sigma _{\textrm{lw}}(T^*) \quad \text{ and }\quad \sigma _{\textrm{lw}}(T)= \sigma _{\textrm{uw}}(T^*). \end{aligned}$$

Recall that the ascent of $T\in L(X)$ is the smallest positive integer $p = p(T)$, whenever it exists, such that $\ker \ T^{p}$ = $\ker \ T^{p+1}$. If such p does not exist, we set $p(T)= \infty$. Analogously, the descent of T is defined to be the smallest integer $q = q(T)$, whenever it exists, such that $T^{q+1}(X)$ = $T^{q}(X)$. If such q does not exist, we set $q(T)= \infty$. Note that if p(T) and q(T) are both finite, then $p(T) = q(T)$, see Chapter 1 of [1] Moreover, $\lambda$ is a pole of the resolvent if and only if $0<p(\lambda I-T)= q(\lambda I-T)< \infty$; see [14, Proposition 50.2].

An operator $T\in L(X)$ is said to be Browder if $T\in \Phi (X)$ and $p(T)=q(T)<\infty$. $T\in L(X)$ is said to be upper semi-Browder if $T\in \Phi _+(X)$ and $p(T)< \infty$, while $T\in L(X)$ is said to be lower semi-Browder if $T\in \Phi _-(X)$ and $q(T)< \infty$. Every Browder operator is Weyl, and every upper semi-Browder (respectively, lower semi-Browder) operator is upper semi-Weyl (respectively, lower semi-Weyl); see [1, Chapter 3].

The Browder spectrum, the upper semi-Browder spectrum, and the lower semi-Browder spectrum are denoted by $\sigma _{\textrm{b}}(T)$, $\sigma _{\textrm{ub}}(T)$, and $\sigma _{\textrm{lb}}(T)$, respectively. Note that if $\lambda$ is a spectral point for which $\lambda I-T$ is Browder, then $\lambda$ is an isolated point of $\sigma (T)$. Recall that $R\in L(X)$ is said to be a Riesz operator if $\lambda I-T\in \Phi (X)$ for each $\lambda \ne 0$. Quasi-nilpotent operators and compact operators are examples of Riesz operators. By a well-known result of Rakočević [16] (see also [3]), the Browder spectra are invariant under Riesz commuting perturbations R, that is

$$\begin{aligned} \sigma _{\textrm{ub}}(T)= \sigma _{\textrm{ub}}(T+R), \quad \sigma _{\textrm{lb}}(T)= \sigma _{\textrm{lb}}(T+R), \quad \sigma _{\textrm{b}}(T)= \sigma _{\textrm{b}}(T+R). \end{aligned}$$

(1)

Semi-Fredholm operators have been generalized by Berkani [8, 9] in the following way: for every $T\in L(X)$ and a nonnegative integer n, let us denote by $T_{[n]}$ the restriction of T to $T^n(X)$, viewed as a map from the space $T^n(X)$ into itself (we set $T_{[0]} = T$). $T\in L(X)$ is said to be semi B-Fredholm (resp. B-Fredholm, upper semi B-Fredholm, lower semi B-Fredholm), if, for some integer $n\ge 0$, the range $T^n (X)$ is closed and $T_{[n]}$ is a semi-Fredholm operator (resp. Fredholm, upper semi-Fredholm, lower semi-Fredholm). In this case, $T_{[m]}$ is a semi-Fredholm operator for all $m\ge n$ [8] with the same index of $T_{[n]}$. This enables one to define the index of a semi B-Fredholm as $\text{ ind } \ T = \text{ ind }\ T_{[n]}$. The upper semi B-Fredholm spectrum is defined

$$\begin{aligned} \sigma _{\textrm{ubf}}(T): = \{\lambda \in {\mathbb {C}}: \lambda I-T \ \text{ is } \text{ not } \text{ upper } \text{ semi } \text{ B-Fredholm }\}, \end{aligned}$$

and analogously, it may be defined the lower semi B-Fredholm spectrum $\sigma _{\textrm{lbf}}(T)$.

A bounded operator $T\in L(X)$ is said to be B-Weyl (respectively, upper semi B-Weyl, lower semi B-Weyl), if, for some integer $n\ge 0$, the range $T^n (X)$ is closed and $T_{[n]}$ is Weyl (respectively, upper semi-Weyl, lower semi-Weyl). The B-Weyl spectrum is defined by

$$\begin{aligned} \sigma _{\textrm{bw}}(T): =\{\lambda \in {\mathbb {C}}: \lambda I-T \ \text{ is } \text{ not } \text{ B-Weyl }\}, \end{aligned}$$

and analogously may be defined the upper semi B-Weyl spectrum $\sigma _{\textrm{ubw}}(T)$ and the lower semi B-Weyl spectrum $\sigma _{\textrm{lbw}}(T)$.

Recall that an operator $T\in L(X)$ is said to be Drazin invertible if $p(T)=q(T)< \infty$, i.e., 0 is a pole of the resolvent or T is invertible. An operator $T\in L(X)$ is said to be left Drazin invertible if $p: =p(T)< \infty$ and $T^{p+1}(X)$ is closed. A scalar $\lambda \in {\mathbb {C}}$ is said to be a left pole if $\lambda I-T$ is left Drazin invertible and $\lambda \in \sigma _{\textrm{ap}}(T)$. A left pole $\lambda$ for which $\alpha (\lambda I-T)< \infty$ is said to have finite rank. Dually, $T\in L(X)$ is said to be right Drazin invertible if $q: = q(T)< \infty$ and $T^q(X)$ is closed. A scalar $\lambda \in {\mathbb {C}}$ is said to be a right pole if $\lambda I-T$ is right Drazin invertible and $\lambda \in \sigma _{\textrm{s}}(T)$.

The Drazin spectrum is defined by

$$\begin{aligned} \sigma _{\textrm{d}}(T): = \{\lambda \in {\mathbb {C}}: \lambda I-T \ \text{ is } \text{ not } \text{ Drazin } \text{ invertible }\}, \end{aligned}$$

and analogously are defined the left Drazin spectrum $\sigma _{\textrm{ld}}(T)$ and the right Drazin spectrum $\sigma _{\textrm{rd}}(T)$. It should be noted that there is a perfect duality, i.e., $\sigma _{\textrm{ld}}(T)= \sigma _{\textrm{rd}}(T^*)$ and $\sigma _{\textrm{ld}}(T^*)= \sigma _{\textrm{rd}}(T)$; see [1, Chapter 1]. The following inclusions hold for every operator $T\in L(X)$:

$$\begin{aligned} \sigma _{\textrm{ubf}}(T) \subseteq \sigma _{\textrm{ubw}}(T) \subseteq \sigma _{\textrm{ld}}(T)\subseteq \sigma _{\textrm{ap}}(T), \end{aligned}$$

(2)

and

$$\begin{aligned} \sigma _{\textrm{lbf}}(T) \subseteq \sigma _{\textrm{lbw}}(T) \subseteq \sigma _{\textrm{rd}}(T)\subseteq \sigma _{\textrm{s}}(T); \end{aligned}$$

(3)

see [1, Chapter 1]. The following lemma has been proved in [4, Lemma 3.5].

Lemma 2.1

Suppose that $\rho _{uw}(T)$ is connected. Then, $\sigma _{\textrm{uw}}(T)= \sigma _{\textrm{w}}(T)$.

A proof of the following theorem can be found in [1, Theorem 1.140].

Theorem 2.2

Let $T\in L(X)$. Then, T is left Drazin invertible (respectively, right Drazin invertible, Drazin invertible) if and only if there exists $n\in {\mathbb {N}}$, such that $T^n(X)$ is closed and the restriction $T|T^n(X)$ is bounded below (respectively, onto, invertible).

3 Injective Quasi-Nilpotent Operators

Recall that an operator $Q\in L(X)$ is said to be quasi-nilpotent if $\sigma (Q)= \{0\}$. In the sequel by $comm\,(T)$, we denote the commutant of T. A proof of the following result may be found in [7].

Theorem 3.1

Let $T\in L(X)$ and suppose that $Q\in \text{ comm }(T)$ is a quasi-nilpotent operator.

• (i) If Q is injective, then $\alpha (T) < \infty$ if and only if T is injective.

• (ii) If the dual $Q^*$ is injective, then $\beta (T) < \infty$ if and only if T is onto.

In the sequel, we set

$$\begin{aligned} \textbf{Q}_i (X):=&\{ T\in L(X)&: \text{ there } \text{ exists } \text{ an } \text{ injective } \text{ quasi-nilpotent } \text{ operator }\ \\&Q\in L(X)&\text{ such } \text{ that } \ TQ=QT\}. \end{aligned}$$

Several examples of operators that commute with an injective quasi-nilpotent operator are given in [7]. Theorem 3.1 has some important consequences:

Theorem 3.2

Let $T\in L (X)$ and $Q\in L(X)$ a quasi-nilpotent operator that commutes with T.

• (i) If Q is injective, then

  $$\begin{aligned} \sigma _{\textrm{usf}}(T) =\sigma _{\textrm{uw}}(T) =\sigma _{\textrm{ub}}(T) = \sigma _{\textrm{ap}}(T) \quad \text{ and } \quad \sigma _{\textrm{b}}(T)= \sigma _{\textrm{w}}(T) = \sigma (T). \end{aligned}$$

  (4)

• (ii) If $Q^*$ is injective, then

  $$\begin{aligned} \sigma _{\textrm{lsf}}(T) =\sigma _{\textrm{lw}}(T) =\sigma _{\textrm{lb}}(T) = \sigma _{\textrm{s}}(T) \quad \text{ and } \quad \sigma _{\textrm{b}}(T)= \sigma _{\textrm{w}}(T) = \sigma (T). \end{aligned}$$

  (5)

Proof

(i) Part (i) has been proved in [7, Corollary 3.7 and Theorem 3.8].

(ii) We have $\sigma _{\textrm{lsf}}(T) \subseteq \sigma _{\textrm{lw}}(T) \subseteq \sigma _{\textrm{lb}}(T) \subseteq \sigma _{\textrm{s}}(T)$, and so, to show the first equalities in (5), it suffices to prove the inclusion $\sigma _{\textrm{s}}(T) \subseteq \sigma _{\textrm{lsf}}(T)$. Let $\lambda \notin \sigma _{\textrm{lsf}}(T)$ be arbitrary given. Then, $\beta (\lambda I-T)< \infty$. Since $T^*Q^*= Q^*T^*$, by Theorem 3.1, we have $\beta (\lambda I-T)=0$, so $\lambda \notin \sigma _{\textrm{s}} (T)$, and hence, $\sigma _{\textrm{s}}(T) \subseteq \sigma _{\textrm{lw}}(T)$.

The equalities $\sigma _{\textrm{b}}(T)= \sigma _{\textrm{w}}(T) = \sigma (T)$ may be proved in a similar way. $\square$

An operator $T\in L(X)$, X a Banach space, is said to have the single-valued extension property at $\lambda _0 \in {{\mathbb {C}}}$, in short T has the SVEP at $\lambda _0$, if for every open disc ${{\textbf{D}}}_{\lambda _0}$ centered at $\lambda _0$ the only analytic function $f: {{\textbf{D}}}_{\lambda _0} \rightarrow X$ which satisfies the equation

$$\begin{aligned} (\lambda I - T) f(\lambda ) = 0 \quad \text{ for } \text{ all } \ \lambda \in {{\textbf{D}}}_{\lambda _0} \end{aligned}$$

is the constant function $f \equiv 0$. T is said to have the SVEP if T has the SVEP for every $\lambda \in {\mathbb {C}}$. Evidently, both T and $T^*$ have SVEP at the points $\lambda \notin \sigma (T)$.

Remark 3.3

Let $\lambda _0 \in {\mathbb {C}}$ and suppose that T has SVEP at the points $\lambda$ of a punctured open disc ${\mathbb {D}}(\lambda _0, \varepsilon ){\setminus }\{\lambda _0\}$. Then, T has SVEP at $\lambda _0$. Indeed, suppose that $f: {\mathbb {D}}(\lambda _0, \varepsilon ) \rightarrow X$ is an analytic function, such that $(\lambda I - T) f(\lambda ) = 0$ holds for every $\lambda \in \mathbb D(\lambda _0, \varepsilon )$. Choose $\mu \in {\mathbb {D}}(\lambda _0, \varepsilon ){\setminus } \{\lambda _0\}$ and let ${\mathbb {D}}(\mu , \delta )$ be an open disc contained in ${\mathbb {D}}(\lambda _0, \varepsilon ){\setminus } \{\lambda _0\}$. The SVEP for T at $\mu$ entails $f(\lambda )=0$ on ${\mathbb {D}}(\mu , \delta )$. Since f is continuous at $\lambda _0$, we then conclude that $f(\lambda _0) = 0$. Hence, $f \equiv 0$ on ${\mathbb {D}}(\lambda _0, \varepsilon )$, and thus, T has the SVEP at $\lambda _0$.

We now consider the Weyl spectra relative to the B-Fredholm theory.

Theorem 3.4

Let $T\in L(X)$ and suppose that $Q\in L(X)$ is a quasi-nilpotent operator, such that $TQ=QT$.

• (i) If Q is injective, then

  $$\begin{aligned} \sigma _{\textrm{ubf}}(T) = \sigma _{\textrm{ubw}}(T) =\sigma _{\textrm{ld}}(T) \quad \text{ and } \quad \sigma _{\textrm{bw}}(T) = \sigma _{\textrm{d}}(T). \end{aligned}$$

  (6)

• (ii) If $Q^*$ is injective, then

  $$\begin{aligned} \sigma _{\textrm{lbf}}(T) = \sigma _{\textrm{lbw}}(T) =\sigma _{\textrm{rd}}(T) \quad \text{ and } \quad \sigma _{\textrm{bw}}(T) = \sigma _{\textrm{d}}(T). \end{aligned}$$

  (7)

Proof

(i) To show the first equality it suffices, by (2), to prove the inclusion $\sigma _{\textrm{ld}}(T) \subseteq \sigma _{\textrm{ubf}}(T)$. Let $\lambda \notin \sigma _{\textrm{ubf}}(T)$ be arbitrarily given. There is no harm if we assume that $\lambda =0$. Then, T is upper semi B-Fredholm, so $T^n(X)$ is closed for some $n\in {\mathbb {N}}$ and $T_{[n]} =T|T^n(X)$ is upper semi-Fredholm, and hence, $\alpha ( T_{[n]}) < \infty$ and $T_{[n]}$ has a closed range. Now, from

$$\begin{aligned}Q(T^n(X)) = T^n (Q(X))\subseteq T^n(X),\end{aligned}$$

we see that $T^n(X)$ is invariant under Q, so we can consider the restriction $Q_{[n]} = Q|T^n(X)$, and, since T and Q commutes, we have the following:

$$\begin{aligned}T_{[n]} Q_{[n]} = Q_{[n]} T_{[n]}. \end{aligned}$$

Clearly, $Q_{[n]}$ is injective and quasi-nilpotent. By Theorem 3.1, then $\alpha (T_{[n]})=0$. Since $T_{[n]}$ has closed range, then $T_{[n]}$ is bounded below, so, by Theorem 2.2, T is left Drazin invertible, i.e., $0\notin \sigma _{\textrm{ld}}(T)$. Therefore, the first equalities in (6) are proved.

The proof of the equality $\sigma _{\textrm{bw}}(T)= \sigma _{\textrm{d}}(T)$ is analogous: if $0 \notin \sigma _{\textrm{bw}}(T)$, then T is semi B-Weyl, so $T^n(X)$ is closed for some $n\in {\mathbb {N}}$ and the restriction $T_{[n]}$ is Weyl, in particular $\alpha ( T_{[n]} )= \beta ( T_{[n]})< \infty$. The restriction $Q_{[n]}= Q|T^n(X)$ is injective and commutes with $T_{[n]}$, so, always by Theorem 3.1, $\alpha (T_{[n]})=\beta (T_{[n]} )=0$, and hence, $T_{[n]}$ is invertible. By Theorem 2.2, then T is Drazin invertible, so $0\notin \sigma _{\textrm{d}}(T)$. Therefore, $\sigma _{\textrm{bw}}(T)= \sigma _{\textrm{d}}(T)$.

(ii) According the inclusions (3), to show the first equalities in (7), we need only to prove that $\sigma _{\textrm{rd}}(T)\subseteq \sigma _{\textrm{lbf}}(T)$. Let $\lambda _0 \notin \sigma _{\textrm{lbf}}(T)$. Then, $\lambda _0 I-T$ is lower semi B-Fredholm, so, by [1, Theorem 1.117], there exists an open disc ${\mathbb {D}}_\varepsilon (\lambda _0)$ centered at $\lambda _0$ and radius $\varepsilon >0$, such that $\lambda I-T$ is lower semi-Fredholm for all $\lambda \in {\mathbb {D}}_\varepsilon (\lambda _0){\setminus } \{\lambda _0\}$. Since $\lambda I-T$ commutes with Q, by part (ii) of Theorem 3.2, then $\lambda I-T$ is lower semi-Browder, and hence, $q(\lambda I-T)< \infty$ for all $\lambda \in {\mathbb {D}}_\varepsilon (\lambda _0){\setminus } \{\lambda _0\}$, and this implies, by [1, Theorem 2.65], that $T^*$ has SVEP at every $\lambda \in {\mathbb {D}}_\varepsilon (\lambda _0){\setminus } \{\lambda _0\}$. From Remark 3.3, we then conclude that $T^*$ has SVEP also at $\lambda _0$. Since $\lambda _0 I-T$ has topological uniform descent (see Chapter 1 of [1] for details), then, by Theorem 2.98 of [1], $\lambda _0 I-T$ is right Drazin invertible, and hence, $\lambda _0 \notin \sigma _{\textrm{rd}}(T)$. Therefore, $\sigma _{\textrm{rd}}(T) \subseteq \sigma _{\textrm{lbf}}(T)$.

The proof of the second equality in (7) is similar: we have only to prove the inclusion $\sigma _{\textrm{d}}(T) \subseteq \sigma _{\textrm{bw}}(T)$. Let $\lambda _0 \notin \sigma _{\textrm{bw}}(T)$. Then, $\lambda _0 I-T$ is semi-B-Weyl and, by [1, Theorem 1.117], there exists an open disc ${\mathbb {D}}_\varepsilon (\lambda _0)$ centered at $\lambda _0$ and radius $\varepsilon >0$, such that $\lambda I-T$ is Weyl for all $\lambda \in {\mathbb {D}}_\varepsilon (\lambda _0){\setminus } \{\lambda _0\}$. From part (ii) of Theorem 3.2, it then follows that $\lambda I-T$ is Browder; hence, $p(\lambda I-T)=q(\lambda I-T)< \infty$ for all $\lambda \in \mathbb D_\varepsilon (\lambda _0){\setminus } \{\lambda _0\}$. This implies, see [1, Theorem 2.65], that both T and $T^*$ have SVEP at every $\lambda \in {\mathbb {D}}_\varepsilon (\lambda _0){\setminus } \{\lambda _0\}$. From Remark 3.3, we then deduce that both T and $T^*$ have SVEP at $\lambda _0$. Since $\lambda _0 I-T$ has topological uniform descent, from Theorem 2.97 and Theorem 2.98 of [1], we then conclude that $\lambda _0 I-T$ is both left and right Drazin invertible, i.e., Drazin invertible; hence, $\lambda _0 \notin \sigma _{\textrm{d}}(T)$. $\square$

Example 3.5

An important example of injective quasi-nilpotent operator is provided by the classical Volterra operator V, on the Banach space X, where $X: =C[0,1]$, the space of all continuous functions on the closed interval [0, 1], or $X: =L^2[0,1]$, the Hilbert space of all complex-valued square-integrable functions on the interval [0, 1]. The operator V is defined by means

$$\begin{aligned} (Vf)(x): = \int _0^x f(t){\textrm{d}}t \quad \text{ for } \text{ all } \ f \in X \quad \text{ and } \ x\in [0,1]. \end{aligned}$$

The class of operators which commute with the Volterra operators is large; for instance, examples of operators which commute with V have been studied in the framework of supercyclic operators [17]. Note that the adjoint of the Volterra operator V on $L^2[0,1]$ is given by

$$\begin{aligned} (V'f)(x): = \int _x^1 f(t){\textrm{d}}t. \end{aligned}$$

Evidently, also $V'$ is injective and quasi-nilpotent; moreover, $V'$ commutes with the adjoint of every operator which belongs to the commutant of V.

For operators T defined on a Hilbert spaces, H is better to consider the adjoint $T'$ instead of the dual $T^*$. We recall now the relationship between the Hilbert adjoint $T'$ of an operator T defined on a Hilbert space and the dual $T^*$. By the Frechét–Riesz representation theorem, there exists a conjugated-linear isometry $U: H\rightarrow H^*$, $H^*$ the dual of H, that associates to every $y\in H$ the linear form defined

$$\begin{aligned}f_y(x):= <x, \,y> \quad \text{ for } \text{ every } \ x \in H.\end{aligned}$$

Moreover

$$\begin{aligned} ( {{\bar{\lambda }}} I - T') = U^{-1}(\lambda I -T^*)U \quad \text{ for } \text{ every }\ \lambda \in {\mathbb {C}}. \end{aligned}$$

Hence, for a Hilbert space operator T

$$\begin{aligned}T' \ \text{ is } \text{ injective } \Leftrightarrow \ T^*\ \text{ is } \text{ injective }. \end{aligned}$$

Since the adjoint $V'$ of the Volterra operator V in $L^2[0,1]$ is also injective and quasi-nilpotent, from Theorems 3.2 and 3.4, we then conclude the following:

Corollary 3.6

The equalities (4), (5), (6), and (7) hold for every bounded linear operator $T:L^2[0,1]\rightarrow L^2[0,1]$ that belongs to the commutant of the Volterra operator V.

Let $f\in {\mathcal {H}}(\sigma (T))$ be an analytic function defined on an open neighborhood U which contains the spectrum, and let f(T) be defined by the classical functional calculus

$$\begin{aligned} f(T): = \frac{1}{2\pi i}\int _\Gamma f(\lambda )(\lambda I-T)^{-1} \, {\textrm{d}} \lambda , \end{aligned}$$

where $\Gamma$ is a contour that surrounds $\sigma (T)$ in U. It is known that, in general, the spectral mapping theorem does not hold for $\sigma _{\textrm{ubw}}(T)$ and $\sigma _{\textrm{bw}}(T)$; indeed, we have only the inclusions

$$\begin{aligned}\sigma _{\textrm{ubw}}(f(T)) \subseteq f(\sigma _{\textrm{ubw}}(T)) \quad \text{ and }\quad \sigma _{\textrm{bw}}(f(T)) \subseteq f(\sigma _{\textrm{bw}}(T));\end{aligned}$$

see [1, Theorem 3.24], and these inclusions may be strict.

Let ${\mathcal {H}}_{\textrm{nc}}(\sigma (T))$ denote the subset of ${\mathcal {H}}(\sigma (T))$ of all functions nonconstant on each component of its domain of definition.

Theorem 3.7

Let $T\in L(X)$ and suppose that there exists a quasi-nilpotent operator Q that commutes with T.

• (i) If Q is injective, then the spectral mapping theorem holds for $\sigma _{\textrm{ubw}}(T)$ and $\sigma _{\textrm{bw}}(T)$, i.e.,

  $$\begin{aligned}\sigma _{\textrm{ubw}}(f(T))= f(\sigma _{\textrm{ubw}}(T)) \quad \text{ for } \text{ all }\ f \in {\mathcal {H}}_{\textrm{nc}}(\sigma (T)),\end{aligned}$$

  and

  $$\begin{aligned}\sigma _{\textrm{bw}}(f(T))= f(\sigma _{\textrm{bw}}(T)) \quad \text{ for } \text{ all }\ f \in {\mathcal {H}}_{\textrm{nc}}(\sigma (T)).\end{aligned}$$

• (ii) If $Q^*$ is injective, then the spectral mapping theorem holds for $\sigma _{\textrm{lbw}}(T)$ and $\sigma _{\textrm{bw}}(T)$, i.e.,

  $$\begin{aligned}\sigma _{\textrm{lbw}}(f(T))= f(\sigma _{\textrm{lbw}}(T)) \quad \text{ for } \text{ all } \ f \in {\mathcal {H}}_{\textrm{nc}}(\sigma (T)),\end{aligned}$$

  and

  $$\begin{aligned}\sigma _{\textrm{bw}}(f(T))= f(\sigma _{\textrm{bw}}(T)) \quad \text{ for } \text{ all } \ f \in {\mathcal {H}}_{\textrm{nc}}(\sigma (T)).\end{aligned}$$

Proof

(i) By Theorem 3.4, we know that $\sigma _{\textrm{ubw}}(T)=\sigma _{\textrm{ld}}(T)$. The spectral mapping theorem holds for the left Drazin spectrum, since the set of all left Drazin invertible operators is a regularity; see [1, Theorem 3.109], hence

$$\begin{aligned} f(\sigma _{\textrm{ubw}}(T))= f(\sigma _{\textrm{ld}}(T))= \sigma _{\textrm{ld}}(f(T)). \end{aligned}$$

Let Q be an injective quasi-nilpotent operator Q which commutes with T. Evidently, if $\lambda \in \rho (T):= {\mathbb {C}}{\setminus } \sigma (T)$, then Q commutes also with $(\lambda I-T)^{-1}$, and consequently, Q commutes with f(T). Therefore, always by Theorem 3.4, we have $\sigma _{\textrm{ld}}(f(T))= \sigma _{\textrm{ubw}}(f(T))$, so the spectral mapping theorem holds for $\sigma _{\textrm{ubw}}(T)$.

The spectral mapping theorem for the B-Weyl spectrum follows similarly, taking into account that, by Theorem 3.4, we have $\sigma _{\textrm{bw}}(T)=\sigma _{\textrm{d}}(T)$ and that the spectral mapping theorem holds for $\sigma _{\textrm{d}}(T)$.

(ii) Evidently, $Q^*$ commutes with $T^*$, and hence, with $f(T^*)=f(T)^*$, so by Theorem 3.4, we have $\sigma _{\textrm{rd}}(f(T))= \sigma _{\textrm{lbw}}(f(T))$. The spectral mapping theorem also holds for the right Drazin spectrum, since the set of all right Drazin invertible operators is a regularity, see [1, Theorem 3.109], hence

$$\begin{aligned} f(\sigma _{\textrm{lbw}}(T))= f(\sigma _{\textrm{rd}}(T))= \sigma _{\textrm{rd}}(f(T))= \sigma _{\textrm{lbw}}(f(T)). \end{aligned}$$

The spectral mapping theorem for $\sigma _{\textrm{bw}}(T)$ is proved similarly using the equality $\sigma _{\textrm{d}}(f(T))= \sigma _{\textrm{bw}}(f(T))$, proved in part (ii) of Theorem 3.4. $\square$

Corollary 3.8

If a bounded linear operator $T:L^2[0,1] \rightarrow L^2[0,1] )$ commutes with the Volterra operator V, then the spectral mapping theorem holds for all B-Weyl spectra.

4 Weyl Type Theorems

An operator $T\in L(X)$ is said to verify Weyl’s theorem if $\sigma (T)\setminus \sigma _{\textrm{w}} (T)=\pi _{00}(T)$, where $\pi _{00} (T): = \{\lambda \in \text{ iso }\, \sigma (T): 0< \alpha (\lambda I-T) < \infty \}$. The operator $T\in L(X)$ is said to verify a-Weyl’s theorem if $\sigma _{\textrm{ap}}(T)\setminus \sigma _{\textrm{uw}} (T)=\pi _{00}^a (T)$, where $\pi _{00}^a (T): = \{\lambda \in \text{ iso }\, \sigma _{\textrm{ap}}(T): 0< \alpha (\lambda I-T) < \infty \}$, while $T\in L(X)$ is said to verify property $(\omega )$ if $\sigma _{\textrm{ap}}(T){\setminus } \sigma _{\textrm{uw}} (T)=\pi _{00} (T)$. Note that

$$\begin{aligned} \text{ either } a\text {-Weyl's theorem or property} \ (\omega ) \ \Rightarrow \text {Weyl' s theorem;} \end{aligned}$$

see [1, Chapter 6]. An operator $T\in L(X)$ is said to verify generalized Weyl’s theorem (shortly, (gWt)), if $\sigma _\mathrm{}(T){\setminus } \sigma _{\textrm{bw}} (T)=E(T)$, where $E(T): = \{\lambda \in \text{ iso }\, \sigma (T): 0< \alpha (\lambda I-T) \}$. The operator $T\in L(X)$ is said to verify generalized a-Weyl’s theorem (shortly, (gaWt)) if $\sigma _{\textrm{ap}}(T){\setminus } \sigma _{\textrm{ubw}} (T)=E_a (T)$, where $E_a (T): = \{\lambda \in \text{ iso }\, \sigma _{\textrm{ap}}(T): 0< \alpha (\lambda I-T) \}$, while $T\in L(X)$ is said to verify generalized property $(\omega )$ (shortly, $(g\omega )$) if $\sigma _{\textrm{ap}}(T){\setminus } \sigma _{\textrm{ubw}} (T)=E (T)$. Note that (gaWt) entails a-Weyl’s theorem and generalized property $(\omega )$ entails property $(\omega )$. Furthermore

$$\begin{aligned} \text{ either }\ (gaWt)\ \text{ or } \ (g\omega ) \ \Rightarrow \ (gWt); \end{aligned}$$

see [1, Chapter 6]. In [7], it has been proved that the existence of an injective quasi-nilpotent operator that commutes with T ensures that a-Weyl’s theorem or property $(\omega )$ hold for T. It is a natural question if the generalized versions of Weyl type theorems hold for $T\in \textbf{Q}_i (X)$. In the sequel, we shall prove that this is true under some additional conditions.

Let $\text{ iso }\, F$ denote the isolated points of $F\subseteq {\mathbb {C}}$. Recall that if Q is any quasi-nilpotent operator that commutes with T, then

$$\begin{aligned} \sigma _{\textrm{ap}}(T) = \sigma _{\textrm{ap}}(T+Q) \quad \text{ and } \quad \sigma _{\textrm{s}}(T) = \sigma _{\textrm{s}}(T+Q); \end{aligned}$$

(8)

see [1, Corollary 3.24]. Since $\sigma (T)= \sigma _{\textrm{ap}}(T) \cup \sigma _{\textrm{s}}(T)$, this easily implies that $\sigma (T)= \sigma (T+Q)$.

Theorem 4.1

Let $T\in L(X)$ and suppose that Q is an injective quasi-nilpotent operator that commutes with T.

• (i) If $\text{ iso }\, \sigma (T)= \emptyset$, then both T and $T+Q$ satisfy generalized Weyl’s theorem.

• (ii) If $\text{ iso }\, \sigma _{\textrm{ap}} (T)= \emptyset$, then both T and $T+Q$ satisfy generalized a-Weyl’s theorem and generalized property $(\omega )$.

Proof

 

• (i) We show first that $\sigma _{\textrm{d}}(T)=\sigma (T)$. It suffices to prove that $\sigma (T)\subseteq \sigma _{\textrm{d }}(T)$. Let $\lambda \notin \sigma _{\textrm{d}}(T)$. Then, $\lambda I-T$ is Drazin invertible, and hence, $\lambda I-T$ is either invertible or $\lambda$ is a pole of T. If $\lambda$ is a pole, then $\lambda$ is an isolated point of $\sigma (T)$, but this is impossible, since by assumption $\text{ iso }\, \sigma (T)= \emptyset$. Hence, $\lambda \notin \sigma (T)$, so $\sigma _{\textrm{d}}(T)=\sigma (T)$. To show that T satisfies generalized Weyl’s theorem, observe that, by Theorem 3.4, $\sigma (T){\setminus } \sigma _{\textrm{bw}}(T) =\sigma (T){\setminus } \sigma _{\textrm{d}}(T) = \emptyset$, and obviously, $E(T)= \emptyset$, since there are no isolated points in $\sigma (T)$. Since $T+Q$ commutes with Q and $\text{ iso }\, \sigma (T+Q)= \text{ iso }\,\sigma (T)= \emptyset$, the argument above proves that also $T+Q$ satisfies generalized Weyl’s theorem.

• (ii) We show first that $\sigma _{\textrm{ld}}(T)=\sigma _{\textrm{ap }}(T)$. Evidently, every bounded below operator is left Drazin invertible, so $\sigma _{\textrm{ld}}(T)\subseteq \sigma _{{\textrm{ap }}}(T)$. To prove the opposite inclusion, let $\lambda \notin \sigma _{\textrm{ld}}(T)$ be arbitrary given. Then, $\lambda I-T$ is left Drazin invertible. There are two possibilities: $\lambda \in \sigma _{\textrm{ap}} (T)$ or $\lambda \notin \sigma _{\textrm{ap}} (T)$. If $\lambda \in \sigma _{\textrm{ap}} (T)$, then $\lambda$ is a left pole of T, and hence, by [1, Theorem 4.3], an isolated point of $\sigma _{\textrm{ap}} (T)$, but this is impossible since $\text{ iso }\, \sigma _{\textrm{ap}} (T)= \emptyset$. Hence, $\lambda \notin \sigma _{\textrm{ap}} (T)$, so the equality $\sigma _{\textrm{ld}}(T)=\sigma _{\textrm{ap }}(T)$ holds. Now, by Theorem 3.4,

  $$\begin{aligned}\sigma (T)\setminus \sigma _{\textrm{bw}}(T) =\sigma (T)\setminus \sigma _{\textrm{d}}(T) = \emptyset , \end{aligned}$$

  and obviously, $E_a(T)= \emptyset$, since there is no isolated point in $\sigma _{\textrm{ap }}(T)$; hence, T satisfies generalized Weyl’s theorem. Since $T+Q$ commutes with Q and

  $$\begin{aligned}\text{ iso }\,\sigma _{\textrm{ap}} (T+Q)= \text{ iso }\, \sigma _{\textrm{ap}} (T)= \emptyset ,\end{aligned}$$

  the argument above proves that $T+Q$ satisfies generalized a-Weyl’s theorem.

To prove the generalized property $(\omega )$ for T, observe, as above, that $\sigma (T){\setminus } \sigma _{\textrm{bw}}(T) = \emptyset$. Furthermore, also $E(T)= \emptyset$, since every isolated point $\lambda$ of the spectrum belongs to $\sigma _{\textrm{ap}} (T)$; see [1, Theorem 1.12], and hence, $\lambda \in \text{ iso } \, \sigma _{\textrm{ap}} (T)$ and this is impossible. Therefore, also generalized property $(\omega )$ holds for T.

Since $T+Q$ commutes with Q and $\text{ iso }\, \sigma _{\textrm{ap}} (T+Q)= \emptyset$, then the previous reasoning shows that both generalized a-Weyl’s theorem and generalized property $(\omega )$ hold for $T+Q$. $\square$

An operator $T\in L(X)$ is said to be polaroid if every$\lambda \in \text{ iso }\, \sigma (T)$ is a pole of the resolvent. $T\in L(X)$ is said to be left polaroid if every $\lambda \in \text{ iso }\, \sigma _{\textrm{ap}} (T)$ is left pole of the resolvent, while $T\in L(X)$ is said to be a-polaroid if every $\lambda \in \text{ iso }\, \sigma _{\textrm{ap}} (T)$ is pole of the resolvent. Note that

$$\begin{aligned} T \ a\text{-polaroid }\ \Rightarrow T \ \text{ left } \text{ polaroid } \Rightarrow T \ \text{ polaroid }; \end{aligned}$$

see [1, Corollary 4.13]. Recall that the spectral theorem holds for $\sigma _{\textrm{ap}}(T)$.

Lemma 4.2

Suppose that $T\in L(X)$ is polaroid (respectively, left polaroid, a-polaroid). Then, f(T) is polaroid (respectively, left polaroid, a-polaroid) for every $f\in {\mathcal {H}}_{\textrm{nc}}(\sigma (T))$.

Proof

The case where T is polaroid is proved in [1, Theorem 4.19], while for the case where T is left polaroid; see [1, Remark 4.20]. Now, let T be a-polaroid and suppose that $\lambda _0 \in \text{ iso }\, \sigma _{\textrm{ap}}(f(T))= \text{ iso }\, f(\sigma _{\textrm{ap}}(T))$. We show that $\lambda _0$ is a pole of the resolvent of f(T). Let us first to show that $\lambda _0 \in f(\text{ iso }\, \sigma _{\textrm{ap}} (T))$. Select $\mu _0\in \sigma _{\textrm{ap}}(T)$, such that $f(\mu _0)= \lambda _0$. Let $\Omega$ be the connected component of the domain of f which contains $\mu _0$ and suppose that $\mu _0$ is not isolated in $\sigma _{\textrm{ap}}(T)$. Then, there exists a sequence $(\mu _n) \subset \sigma _{\textrm{ap}}(T)\cap \Omega$ of distinct scalars, such that $\mu _n \rightarrow \mu _0$. The set $\Gamma : =\{\mu _0, \mu _1, \mu _2, \dots \}$ is a compact subset of $\Omega$, so, by the Principle of isolated zeros of analytic functions, the function f may assume the value $\lambda _0= f(\mu _0)$ at only a finite number of points of $\Gamma$. Consequently, for n sufficiently large $f(\mu _n)\ne f(\mu _0)= \lambda _0$, and since $f(\mu _n)\rightarrow f(\mu _0)= \lambda _0$, it then follows that $\lambda _0$ is not an isolated point of $f(\sigma _{\textrm{ap}} (T))$, a contradiction.

Hence, $\lambda _0 =f(\mu _0)$, with $\mu _0 \in \text{ iso }\ \sigma _{\textrm{ap}}(T)$. Since T is a-polaroid, then $\mu _0$ is a pole of the resolvent of T, and hence, by [1, Theorem 4.16], $\lambda _0$ is a pole of the resolvent of f(T). Thus, f(T) is a-polaroid. $\square$

Theorem 4.3

Let $T\in \textbf{Q}_i (X)$. Then, we have the following:

• (i) If T is polaroid, then f(T) satisfies generalized Weyl’s theorem for every $f\in {\mathcal {H}}_{\textrm{nc}}(\sigma (T))$.

• (ii) If T is left polaroid, then f(T) satisfies generalized a Weyl’s theorem theorem for every $f\in \mathcal H_{\textrm{nc}}(\sigma (T))$.

• (iii) If T is a-polaroid, then f(T) satisfies generalized property $(\omega )$ for every $f\in {\mathcal {H}}_{\textrm{nc}}(\sigma (T))$.

Proof

 

• (i) Let T be polaroid. By Theorem 3.4, we have $\sigma _{\textrm{bw}} (T)= \sigma _{\textrm{d}} (T)$. Hence

  $$\begin{aligned} \sigma (T){\setminus } \sigma _{\textrm{bw}} (T) = \sigma (T){\setminus } \sigma _{\textrm{d}} (T) = \Pi (T),\end{aligned}$$

  so every $\lambda \in \sigma (T){\setminus } \sigma _{\textrm{bw}} (T)$ is a pole of the resolvent, hence an isolated point of $\sigma (T)$ and an eigenvalue of T. Therefore, $\lambda \in E (T)$ and this show that $\sigma (T){\setminus } \sigma _{\textrm{bw}} (T) \subseteq E(T)$.

  The opposite inclusion is also true, if $\lambda \in E(T)$ then $\lambda$ is a pole, since T is polaroid, and so, by Theorem 3.2, $\lambda \in \sigma (T){\setminus } \sigma _{\textrm{d}} (T)= \sigma (T){\setminus } \sigma _{\textrm{bw}} (T)$, and hence, (gW) holds for T. By Lemma 4.2, also f(T) is polaroid for every $f\in {\mathcal {H}}_{\textrm{nc}}(\sigma (T))$, and since f(T) commutes with Q, then (gW) holds for f(T), by the first part of the proof.

• (ii) Let T be left polaroid. If $\lambda \in E_a(T)$, then $\lambda \in \text{ iso }\, \sigma _{\textrm{ap}}(T)$, and hence, $\lambda$ is a left pole, since T is left polaroid; in particular, $\lambda I-T$ is left Drazin invertible.Taking into account Theorem 3.2

  $$\begin{aligned} \lambda \in \sigma _{\textrm{ap}}(T)\setminus \sigma _{\textrm{ld}}(T)= \sigma _{\textrm{ap}}(T)\setminus \sigma _{\textrm{ubw}}(T),\end{aligned}$$

  so $E_a(T)\subseteq \sigma _{\textrm{ap}}(T)\setminus \sigma _{\textrm{ubw}}(T)$.

  On the other hand, the opposite inclusion also holds. Indeed, if

  $$\begin{aligned}\lambda \in \sigma _{\textrm{ap}}(T){\setminus } \sigma _{\textrm{ubw}}(T)= \sigma _{\textrm{ap}}(T){\setminus } \sigma _{\textrm{ld}}(T)= \Pi _a(T),\end{aligned}$$

  then $\lambda \in \text{ iso }\, \sigma _{\textrm{ap}}(T)$. Furthermore, $\alpha (\lambda I-T)>0$; otherwise, if were $\alpha (\lambda I-T) =0$, we would have $p(\lambda I-T)= 0$, and hence, since $\lambda \notin \sigma _{\textrm{ld}}(T)$, $(\lambda I-T)^{0+1}(X)= (\lambda I-T)(X)$ would be closed, contradicting the assumption that $\lambda \in \sigma _{\textrm{ap}}(T)$. Therefore, $\lambda \in E_a(T)$, so $\sigma _{\textrm{ap}}(T)\setminus \sigma _{\textrm{ubw}}(T)=E_a(T)$; thus, generalized a-Weyl theorem holds for T.

  As above, the same argument shows that generalized a-Weyl theorem holds for f(T), since f(T) commutes with Q and f(T) is left polaroid, by Lemma 4.2.

• (iii) Observe first that E(T) is contained in $\sigma _{\textrm{ap}}(T){\setminus } \sigma _{\textrm{ubw}}(T)$. Indeed, every isolated point of the spectrum belongs to its boundary and hence to $\sigma _{\textrm{ap}}(T)$, by [1, Theorem 1.12]. Therefore, every $\lambda \in E(T)$ is an isolated point of $\sigma _{\textrm{ap}}(T)$, and hence, a pole of the resolvent, since by assumption T is a-polaroid; in particular, $\lambda$ is a left pole. By Theorem 3.4, then

  $$\begin{aligned}\lambda \in \sigma _{\textrm{ap}}(T)\setminus \sigma _{\textrm{ld}}(T)= \sigma _{\textrm{ap}}(T)\setminus \sigma _{\textrm{ubw}}(T).\end{aligned}$$

To show the opposite inclusion $\sigma _{\textrm{ap}}(T){\setminus } \sigma _{\textrm{ubw}}(T)\subseteq E(T)$, observe that if $\lambda \in \sigma _{\textrm{ap}}(T){\setminus } \sigma _{\textrm{lbw}}(T)= \sigma _{\textrm{ap}}(T){\setminus } \sigma _{\textrm{ld}}(T)$, then $\lambda$ is a left pole, hence an isolated point of $\sigma _{\textrm{ap}}(T)$. The assumption that T is a-polaroid then entails that $\lambda$ is a pole of T, hence an isolated point of $\sigma (T)$. Since $\lambda$ is also an eigenvalue of T, we then conclude that $\lambda \in E(T)$, so property $(g\omega )$ holds for T. Since f(T) is a-polaroid then, as above, we conclude that property $(g\omega )$ holds for f(T). $\square$

Corollary 4.4

Let $T\in L(L^2[0,1])$ commute with the Volterra operator and $f\in {\mathcal {H}}_{\textrm{nc}} (\sigma (T))$. If T is polaroid, then f(T) satisfies generalized Weyl’s theorem. If T is left polaroid, then generalized a-Weyl’s theorem holds for f(T). If T is a-polaroid, then generalized property (w) holds for f(T).

Denote by $\Pi (T)$ the set of all poles of T, i.e., $\Pi (T):=\sigma (T){\setminus } \sigma _{\textrm{d}}(T)$ and by $\Pi _a(T)$ the set of all left poles. i.e., $\Pi _a(T):=\sigma _{\textrm{ap}} (T){\setminus } \sigma _{\textrm{ld}}(T)$. Note that $\Pi _a(T) \subseteq \text{ iso }\, \sigma _{\textrm{ap}}(T)$ ( [1, Theorem 4.3]. According to [10] and [11], an operator $T\in L(X)$ is said to satisfy property (gb) if $\sigma _{\textrm{ap}}(T){\setminus } \sigma _{ubw}(T)= \Pi (T)$.

Theorem 4.5

Le $T\in \textbf{Q}_i (X)$ be a-polaroid. Then, T satisfies property (gb).

Proof

From Theorem 3.4, we know that

$$\begin{aligned}\sigma _{\textrm{ap}}(T){\setminus } \sigma _{\textrm{ubw}}(T)= \sigma _{\textrm{ap}}(T) {\setminus } \sigma _{\textrm{ld}}(T)= \Pi _a(T).\end{aligned}$$

Obviously, the inclusion $\Pi (T) \subseteq \Pi _a(T)$ holds for every operator. If $\lambda \in \Pi _a(T)$, then $\lambda \in \text{ iso } \, \sigma _{\textrm{ap}}(T)$, and since T is a-polaroid, then $\lambda \in \Pi (T)$; hence, $\Pi _a(T)\subseteq \Pi (T)$, and thus, $\Pi _a(T)=\Pi (T)$. $\square$

To introduce another stronger variant of Weyl’s theorem, we introduce the following property.

Definition 4.6

An operator $T\in L(X)$ is said to verify property (gaz) if $\sigma (T)\setminus \sigma _{\textrm{ubw}}(T)= \Pi _a (T)$.

Theorem 4.7

Let $T\in {\textbf{Q}}_i (X)$. Then, property (gaz) holds for T if and only if $\sigma _{\textrm{ap}} (T)= \sigma (T)$.

Proof

Suppose that property (gaz) holds for T. Then, $\sigma _{\textrm{ap}} (T)= \sigma (T)$, by Theorem 3.2 of [5]. Conversely, suppose that $\sigma _{\textrm{ap}} (T)= \sigma (T)$. Since $T\in \textbf{Q}_i (X)$, then $\sigma _{\textrm{ubw}} (T) = \sigma _{\textrm{ld}} (T)$, by Theorem 3.2, so we have

$$\begin{aligned} \sigma (T){\setminus } \sigma _{\textrm{ubw}}(T)= \sigma _{\textrm{ap }} (T){\setminus } \sigma _{\textrm{ld}}(T)= \Pi _a(T), \end{aligned}$$

i.e., (gaz) holds for T. $\square$

It should be noted that the equality $\sigma _{\textrm{ap}} (T)= \sigma (T)$ holds if $T^*$ has SVEP, and hence, if $T^\star$ has SVEP, then property (gaz) holds for T.

The following variant of Weyl type theorems has been introduced in [6]. An operator $T\in L(X)$ is said to satisfy S-Weyl’s theorem, if $\sigma (T)\setminus \sigma _{\textrm{ubw}}(T)= \pi _{00}(T)$. Note that S-Weyl’s theorem holds for T if and only if T satisfies property (gaz) and the equality $\Pi _a (T)= \pi _{00}(T)$ holds; see [6, Theorem 3.6]. Furthermore, S-Weyl’s theorem entails both a-Weyl’s theorem and property $(\omega )$; see [6, Theorem 3.8].

Observe that for operators $T \in \textbf{Q}_i (X)$, we have

$$\begin{aligned}\pi _{00}(T)= \pi _{00}^a(T)= \emptyset , \end{aligned}$$

since, if $\lambda$ belongs to one of these sets, then $0<\alpha (\lambda I-T)< \infty$ and this is impossible by Theorem 3.1.

Corollary 4.8

Let $T\in \textbf{Q}_i (X)$ be such that $\sigma _{\textrm{ap}} (T)= \sigma (T)$. Then, S-Weyl’s theorem holds for T if and only if T has no left poles. In particular, if $\text{ iso }\, \sigma _{\textrm{ap}}(T)= \emptyset$, then S-Weyl’s theorem holds for T.

Proof

By Theorem 4.3, if $\sigma _{\textrm{ap}} (T)= \sigma (T)$, then T has property (gaz). Furthermore, as noted above, $\pi _{00}(T)= \emptyset$. Hence

$$\begin{aligned} \emptyset = \Pi _a(T)=\pi _{00}(T); \end{aligned}$$

thus, S-Weyl’s theorem holds for T. Conversely, if S-Weyl’s theorem holds for T, then $\Pi _a(T)=\pi _{00}(T) =\emptyset$, so T has no left poles.

The last assertion is clear: every left pole of the resolvent of an operator is an isolated point of $\sigma _{\textrm{ap}}(T)$; see [1, Theorem 4.3]. $\square$

Theorem 4.9

Suppose that $\sigma _{\textrm{ap}}(T)$ has no hole and that $\sigma _{\textrm{ap}}(T)$ has no isolated point. If Q is an injective quasi-nilpotent operator that commutes with T, then S-Weyl’s theorem holds for T and $T+Q$.

Proof

Since $\sigma _{\textrm{ap}}(T)= \sigma _{\textrm{uw}}(T)$, by Theorem 3.2, the set $\rho _{\textrm{uw}} (T):= {\mathbb {C}} {\setminus } \sigma _{\textrm{uw}}(T)$ is connected, so, by Lemma 2.1 and Theorem 3.2, we have

$$\begin{aligned} \sigma _{\textrm{ap}}(T)=\sigma _{\textrm{uw}}(T)= \sigma _{\textrm{w}} (T)= \sigma (T). \end{aligned}$$

Therefore, T has property (gaz), by Theorem 4.3. To prove that S-Weyl’s theorem holds for T, we need to prove that $\Pi _a (T)= \pi _{00}(T)$. As noted before, $\pi _{00}(T)$ is empty and also $\Pi _a(T)= \emptyset$, since $\Pi _a(T)\subseteq \text{ iso }\, \sigma _{\textrm{ap}}(T)$, and thus, T satisfies S-Weyl’s theorem. To show that S-Weyl’s theorem holds also for $T+Q$, note that $T+Q$ commutes with Q, $\text{ iso }\, \sigma _{\textrm{ap}}(T+Q)=\text{ iso }\, \sigma _{\textrm{ap}}(T) = \emptyset$ and $\sigma _{\textrm{ap}}(T+Q)= \sigma _{\textrm{ap}}(T)$ has no hole. $\square$