Property $(UW{\scriptstyle \Pi })$ under perturbations

Abstract

Property $(UW {\scriptstyle \Pi })$, introduced in Berkani and Kachad (Bull Korean Math Soc 49:1027–1040, 2015) and studied more recently in Aiena and Kachad (Acta Sci Math (Szeged) 84:555–571, 2018) may be thought as a variant of Browder’s theorem, or Weyl’s theorem, for bounded linear operators acting on Banach spaces. In this article we study the stability of this property under some commuting perturbations, as quasi-nilpotent perturbation and, more in general, under Riesz commuting perturbations. We also study the transmission of property $(UW {\scriptstyle \Pi })$ from T to f(T), where f is an analytic function defined on a neighborhood of the spectrum of T. Furthermore, it is shown that this property is transferred from a Drazin invertible operator T to its Drazin inverse S.

1 Introduction and preliminaries

Let $T\in L(X)$ be a bounded linear operator on an infinite-dimensional complex Banach spaces X,  and denote by $\alpha (T)$ and $\beta (T)$ the dimension of the kernel N(T) and the codimension of the range $R(T)=T(X),$ respectively. Let $\Phi _+(X):=\{ T\in L(X:\alpha (T)<\infty , T(X) \ \text{ is } \text{ closed } \}$ be the class of all upper semi-Fredholm operators and $\Phi _-(X):=\{ T\in L(X): \beta (T)<\infty \}$ the class of all lower semi-Fredholm operators. If $T\in \Phi _-^+(X):=\Phi _+(X) \cup \Phi _-(X),$ the index of T is defined by $ind(T):=\alpha (T)-\beta (T).$ If $\Phi (X):=\Phi _+(X) \cap \Phi _-(X),$ denotes the set of all Fredholm operators, the class of Weyl operators is defined by

$$\begin{aligned} W(X)=\{ T\in \Phi (X): ind(T)=0\}, \end{aligned}$$

the class of upper semi-Weyl operators is defined by

$$\begin{aligned} W_+(X)=\{ T\in \Phi _+(X): ind(T)\leqslant 0\}, \end{aligned}$$

while the class lower semi-Weyl operators is defined by

$$\begin{aligned} W_-(X)=\{ T\in \Phi _-(X): ind(T)\geqslant 0\}. \end{aligned}$$

Evidently, $W(X)= W_-(X)\cap W_+(X)$. If $T^*$ denotes the dual of $T\in L(X)$, it is well known that $T\in W_+(X)$ (respectively, $T\in W_-(X)$) if and only if $T^*\in W_-(X^*)$ (respectively, $T^*\in W_+(X^*)$). The classes of operators above defined generate the following spectra: the Weyl spectrum, defined by

$$\begin{aligned} \sigma _{w}(T):= \{ \lambda \in {\mathbb {C}}: \lambda I -T \not \in W(X)\}, \end{aligned}$$

the upper semi-Weyl spectrum, defined by

$$\begin{aligned} \sigma _{uw}(T):= \{ \lambda \in {\mathbb {C}}: \lambda I-T \not \in W_+(X)\}, \end{aligned}$$

and the lower semi-Weyl spectrum, defined by

$$\begin{aligned} \sigma _{lw}(T):= \{ \lambda \in {\mathbb {C}}: \lambda I-T \not \in W_-(X)\}. \end{aligned}$$

Two classical quantities in operator theory are defined as follows. The ascent of an operator T,  is the smallest non-negative integer $p: = p(T)$ such that $N( T^p)=N( T^{p+1}).$ If such integer does not exist we put $p(T)=\infty .$ Analogously, the descent of T,  is the smallest non-negative integer $q:=q(T)$ such that $R(T^q)=R(T^{q+1}),$ and if such integer q does not exist we put $q(T)=\infty .$ It is well known that if p(T) and q(T) are both finite then $p(T)=q(T),$ see [1, Theorem 1.20]. Moreover, $0<p(\lambda I-T) = q(\lambda I-T)< \infty$ exactly when $\lambda$ is a pole of the resolvent of T, see [14, Proposition 50.2]. The class of all Browder operators is defined

$$\begin{aligned} B(X):=\{ T\in \Phi (X): p(T)=q(T)<\infty ,\}, \end{aligned}$$

while the class of all upper semi-Browder operators is defined

$$\begin{aligned} B_+(X):=\{ T\in \Phi _+(X): p(T)<\infty ,\}. \end{aligned}$$

Obviously, $B(X)\subseteq W(X)$ and $B_+(X)\subseteq W_+(X),$ see [1, Chapter 3]. If $\sigma _{b}(T)$ and $\sigma _{ub}(T)$ denote the Browder spectrum and the upper semi-Browder spectrum, respectively, then $\sigma _{w}(T)\subseteq \sigma _{b}(T)$ and $\sigma _{uw}(T)\subseteq \sigma _{ub}(T)$.

In [6, 8, 10] Berkani et al. generalize semi-Fredholm operators 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$=T_{[0]}$). $T\in L(X)$ is said to be semi B-Fredholm,(respectively, B-Fredholm, upper semi B-Fredholm, lower semi B-Fredholm) if, for some integer n, the range $T^n(X)$ is closed and $T_{[n]}$ is a semi-Fredholm operator (resp. Fredholm, upper semi-Fredholm, lower semi-Fredholm). If $T_{[n]}$ is a semi-Fredholm operator, then $T_{[n]}$ is a semi-Fredholm operator for all $m\geqslant n$ ( [10]), with the same index of $T_{[n]}.$ This enables one to define the index of a semi-B-Fredholm as $ind(T)=ind(T_{[n]}).$ Analogously, 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 $n\geqslant 0$ the range $T^n(X)$ is closed and $T_{[n]}$ is Weyl (respectively, upper semi-Weyl, lower semi-Weyl ). Analogous definitions are given for semi B-Browder operators. The B-Weyl spectrum is defined as

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

and analogously, the upper semi B-Weyl spectrum of T is defined by

$$\begin{aligned} \sigma _{ubw}(T):= \{ \lambda \in {\mathbb {C}}: \lambda I -T\ is \ not \ upper \ semi \ B-Weyl \}. \end{aligned}$$

The concept of Drazin invertibility has been introduced in a more abstract setting than operator theory. In the case the Banach algebra L(X),  an operator $T\in L(X)$ is said to be Drazin if $p(T)=q(T)<\infty$. Evidently, if T is Drazin invertible then either $\lambda I-T$ is invertible or $\lambda$ is a pole of the resolvent of T. An operator $T\in L(X)$ is said to be left Drazin invertible if $p=p(T)<\infty$ and $R(T^{p+1})$ is closed. The Drazin spectrum is then defined as

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

while the left Drazin spectrum is defined as

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

In the sequel we denote by $\sigma _a(T)$ the approximate point spectrum, defined by $\sigma _a(T)=\{ \lambda \in {\mathbb {C}}:\lambda I-T$ is not bounded below}, where an operator is said to be bounded below if it is injective and has closed range. The classical surjective spectrum of T is denoted by $\sigma _s(T).$

The following property has a fundamental role in local spectral theory. An operator $T\in L(X)$ 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 $T\in L(X)$ is said to have the SVEP if T has this property at every $\lambda \in {\mathbb {C}}$. (See [1] and [15] for more details about this concept). Evidently, an operator $T\in L(X)$ has SVEP at every point of the resolvent $\rho (T)={\mathbb {C}}\setminus \sigma (T),$ and both T and $T^*$ have SVEP at the isolated points of the spectrum. Note that

$$\begin{aligned} p(\lambda I- T)<\infty \Longrightarrow T \ \text{ has } \text{ SVEP } \text{ at } \ \lambda , \end{aligned}$$

(1)

and dually,

$$\begin{aligned} q(\lambda I- T)<\infty \Longrightarrow T^*\ \text{ has } \text{ SVEP } \text{ at } \ \lambda , \end{aligned}$$

(2)

see [1, Chapter 2]. Furthermore, if $\text{ acc } \, F$ denote the set of all cluster points of $F\subseteq {\mathbb {C}}$, we have:

$$\begin{aligned} \lambda \notin \text{ acc }\,\sigma _a(T)\Longrightarrow T \ \text{ has } \text{ SVEP } \text{ at } \ \lambda , \end{aligned}$$

(3)

and dually,

$$\begin{aligned} \lambda \notin \text{ acc }\,\sigma _s(T)\Longrightarrow T^*\ \text{ has } \text{ SVEP } \text{ at } \ \lambda , \end{aligned}$$

(4)

see [1, Chapter 2].

Remark 1.1

In [1, Chapter 2] it is shown that the implications above are equivalences if $\lambda I-T$ semi B-Fredholm, in particular semi-Fredholm.

2 Property $( UW {\scriptstyle \Pi })$

Denote by $p_{00}(T):=\sigma (T)\setminus \sigma _b(T)$ the set of all poles of T having finite rank, by $\Pi (T)=\sigma (T)\setminus \sigma _d(T)$ the set of all poles of T, and by $p_{00}^a(T)=\sigma _a(T)\setminus \sigma _{ub}(T)$ the set of all left poles of T having finite rank. It is easy to check that $p_{00}(T)\subseteq p_{00}^a(T)$ for all $T\in L(X),$ and obviously every point of $p_{00}(T)$ is an isolated point of $\sigma (T)$, and hence an isolated point of $\sigma _a(T)$ (since every isolated point of the spectrum belongs to $\sigma _a (T)$). Set

$$\begin{aligned} \Delta (T): = \sigma (T)\setminus \sigma _{w}(T) \quad \text{ and }\quad \Delta _a (T): = \sigma _a(T)\setminus \sigma _{uw}(T). \end{aligned}$$

Let $\Pi _a(T): = \sigma _a(T)\setminus \sigma _{ld}(T)$ be the set of all left poles of the resolvent of T. Obviously, $p_{00}(T) \subseteq \Delta _a (T)$, since each point of $p_{00}(T)$ is an eigenvalue of T and every Browder operator is upper semi-Weyl.

Hereafter, the symbol $\bigsqcup$ stands for disjoint union.

Definition 2.1

A bounded operator $T\in L(X)$ is said to satisfy:

• (i) Browder’s theorem if $\sigma _w(T)=\sigma _b(T)$ or equivalently $\sigma (T)\setminus \sigma _w(T)= p_{00}(T)$, ([13]).

• (ii) a-Browder’s theorem if $\sigma _{uw}(T)=\sigma _{ub}(T)$ or equivalently $\Delta _a(T)= p_{00}^a(T)$, ([13, 1, Chapter 5]).

• (iii) Property $(UW {\scriptstyle \Pi })$ if $\Delta _a(T)= \Pi (T),$ or equivalently $\sigma _a (T)= \sigma _{uw}(T) \bigsqcup \Pi (T)$, ( [2, 9]).

Remark 2.2

a-Browder’s theorem entails Browder’s theorem. Furthermore, a-Browder’s theorem is equivalent to generalized a-Browder’s theorem, i.e., a-Browder’s theorem is equivalent to saying that the equality $\sigma _{ubw}(T)= \sigma _{ld}(T)$ holds, where $\sigma _{ubw}(T)$ denotes the upper semi B-Weyl spectrum, see [5].

Property $(UW {\scriptstyle \Pi })$ may be characterized as follows ([2]):

Theorem 2.3

Let $T\in L(X).$ Then the following assertions are equivalent:

• (i) T satisfies property $(UW {\scriptstyle \Pi });$

• (ii) T satisfies a-Browder’s theorem, and $p_{00}^a(T)=\Pi (T)$;

• (iii) $T^*$ has SVEP at every $\lambda \in \Delta _a(T)$ and $\sigma _b (T)= \sigma _d (T)$.

Property $(UW {\scriptstyle \Pi })$ entails some relevant equalities between parts of the spectrum:

Theorem 2.4

Suppose that T satisfies property $( UW {\scriptstyle \Pi })$. Then we have:

• (i) $\sigma _{w}(T) \setminus \sigma _{uw}(T)= \sigma (T)\setminus \sigma _a(T)$.

• (ii) $\sigma _{uw}(T) \setminus \sigma _{ubw}= \Pi _a(T)\setminus \Pi (T)$.

• (iii) $\sigma (T)\setminus \sigma _w(T)= \Pi (T)$.

Proof

If $\lambda \in \sigma _{w}(T) \setminus \sigma _{uw}(T)$ then $\alpha (\lambda I-T)< \infty$ and $(\lambda I-T)(X)$ is closed. Suppose that $0< \alpha (\lambda I-T)$. Then $\lambda \in \sigma _{a}(T) \setminus \sigma _{uw}(T) = \Pi (T)$. Hence, $p(\lambda I-T)= q(\lambda I-T)< \infty$ and consequently, by [1, Theorem 1.22], $\lambda I -T$ is Browder, in particular $\lambda \notin \sigma _{w}(T)$, a contradiction. Hence $\alpha (\lambda I-T)= 0$, so $\lambda \notin \sigma _a (T)$, and consequently $\lambda \in \sigma (T)\setminus \sigma _a(T)$.

Conversely, if $\lambda \in \sigma (T)\setminus \sigma _a(T)$ then $\lambda \notin \sigma _{uw}(T)$, since $\sigma _{uw}(T)\subseteq \sigma _a (T)$. Suppose that $\lambda \notin \sigma _w(T)$. Then $\alpha (\lambda I-T)= \beta (\lambda I-T)$, and since $\alpha (\lambda I-T)=0$ we have $\lambda \notin \sigma (T)$, a contradiction. Hence $\lambda \in \sigma _{w}(T) \setminus \sigma _{uw}(T)$.

(ii) Suppose that T has property $( UW {\scriptstyle \Pi })$. Then a-Browder’s theorem holds for T, or equivalently generalized a-Browder’s theorem holds for T. If $\lambda \in \sigma _{uw}(T) \setminus \sigma _{ubw}$ then $\lambda \notin \sigma _{ubw} (T) =\sigma _{ld}(T)$, and $\lambda \in \sigma _a (T)$, since $\sigma _{uw}(T)\subseteq \sigma _a(T)$. Hence, $\lambda \in \Pi _a (T)$. Since a-Browder’s theorem holds for T we then have $\lambda \in \sigma _{ub}(T)$, hence $\lambda \notin \sigma _a (T)\setminus \sigma _{ub}(T)= \pi _{00}^a$, and hence $\lambda \notin \Pi (T)$, by Theorem 2.3.

Conversely, suppose that $\lambda \in \Pi _a(T)\setminus \Pi (T)$. Then $\lambda \in \Pi _a(T)$ and hence $\lambda \notin \sigma _{ld}(T)= \sigma _{ubw}(T)$, since generalized a-Browder’s theorem holds for T. On the other hand, $\lambda \notin \Pi (T)= p_{00}^a (T)$, by Theorem 2.3, and since $\lambda \in \sigma _a(T)$, then $\lambda \in \sigma _{ub}(T)= \sigma _{uw}(T)$.

(iii) Write $\sigma (T)= \sigma (T)\setminus \sigma _a(T) \bigsqcup \sigma _a (T)$. Since $\sigma _a (T)= \sigma _{uw}(T) \bigsqcup \Pi (T)$, by part (i) we then have

$$\begin{aligned} \sigma (T)= & {} \sigma _{w}(T) \setminus \sigma _{uw}(T) \bigsqcup \sigma _a (T) \\= & {} \sigma _{w}(T) \setminus \sigma _{uw}(T) \bigsqcup \sigma _{uw}(T) \bigsqcup \Pi (T) = \sigma _{w}(T)\bigsqcup \Pi (T), \end{aligned}$$

so $\sigma (T)\setminus \sigma _w(T)= \Pi (T)$. $\square$

The following variant of Weyl’s theorem and Browder’s theorem has been introduced recently by Zariouh [19]. An operator $T\in L(X)$ is said to satisfies property$(Z_{\Pi _a})$ if $\Delta (T): = \sigma (T)\setminus \sigma _w(T)$ coincides with the set $\Pi _a(T)$ of left poles of T.

The next theorem improves a result of [2], which was proved by assuming that $T^*$ has SVEP.

Theorem 2.5

If $T\in L(X)$ satisfies property $(UW {\scriptstyle \Pi }),$ then T satisfies property $(Z{\scriptstyle \Pi _a}).$

Proof

Suppose that T satisfies property $(UW {\scriptstyle \Pi })$, i.e. $\sigma _a (T)= \sigma _{uw}(T) \bigsqcup \Pi (T)$. By part (i) and part (iii) of Theorem 2.4 we have

$$\begin{aligned} \sigma (T)= & {} \sigma _w (T)\bigsqcup \Pi (T)= (\sigma _w (T)\setminus \sigma _{uw}(T)) \bigsqcup \sigma _{uw}(T) \bigsqcup \Pi (T)\\= & {} (\sigma (T)\setminus \sigma _{a}(T)) \bigsqcup \sigma _{uw}(T) \bigsqcup \Pi (T)\\= & {} (\sigma (T)\setminus \sigma _{a}(T)) \bigsqcup [(\sigma _{uw}(T)\setminus \sigma _{ubw}(T))\bigsqcup \sigma _{ubw}(T)] \bigsqcup \Pi (T). \end{aligned}$$

From part (ii) of Theorem 2.4 we then obtain

$$\begin{aligned} \sigma (T)= & {} (\sigma _w (T)\setminus \sigma _{uw}(T)) \bigsqcup (\Pi _a (T)\setminus \Pi (T)) \bigsqcup \sigma _{ubw}(T) \bigsqcup \Pi (T)\\= & {} (\sigma _w (T)\setminus \sigma _{uw}(T)) \bigsqcup \Pi _a (T)\bigsqcup \sigma _{ubw}(T). \end{aligned}$$

Since $\sigma _{ubw}(T)\subseteq \sigma _{uw}(T)$ then

$$\begin{aligned} \sigma (T) \subseteq (\sigma _w (T)\setminus \sigma _{uw}(T)) \bigsqcup \Pi _a (T) \bigsqcup \sigma _{uw}(T) = \sigma _w (T) \bigsqcup \Pi _a (T). \end{aligned}$$

Trivially, $\sigma _w (T) \bigsqcup \Pi _a (T) \subseteq \sigma (T)$ Consequently, $\sigma (T)= \sigma _w (T) \bigsqcup \Pi _a (T)$ and hence property $(Z{\scriptstyle \Pi _a})$ holds for T. $\square$

The converse of Theorem 2.5 holds if $T^*$ has SVEP, see [2]. The precise relationship between property $(UW {\scriptstyle \Pi })$ and property $(Z{\scriptstyle \Pi _a})$ is described by the following theorem.

Theorem 2.6

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

• (i) T satisfies property $(UW {\scriptstyle \Pi });$

• (ii) T satisfies property $(Z{\scriptstyle \Pi _a})$ and $\sigma _{w}(T)\setminus \sigma _{uw}(T)=\sigma (T)\setminus \sigma _{a}(T);$

• (iii) T satisfies property $(Z {\scriptstyle \Pi _a})$ and $\Delta _a (T)\cap \sigma _w (T)= \emptyset$.

Proof

(i) $\Longleftrightarrow$ (ii) Suppose that T satisfies property $(UW {\scriptstyle \Pi }).$ Then T has property $(Z{\scriptstyle \Pi _a})$ and, by Theorem 2.4, we have $\sigma _{w}(T)\setminus \sigma _{uw}(T)=\sigma (T)\setminus \sigma _{a}(T).$ Conversely, if T has property $(Z{\scriptstyle \Pi _a})$ and $\sigma _{w}(T)\setminus \sigma _{uw}(T)=\sigma (T)\setminus \sigma _{a}(T)$, we have $\sigma (T)=\sigma _w(T)\bigsqcup \Pi _a(T).$ Our assumption $\sigma _{w}(T)\setminus \sigma _{uw}(T)=\sigma (T)\setminus \sigma _{a}(T)$ entails that $\sigma (T) =[\sigma (T)\setminus \sigma _a(T)] \bigsqcup [\sigma _{uw}(T) \bigsqcup \Pi _a(T)],$ hence $\sigma _a(T)=\sigma _{uw}(T) \bigsqcup \Pi _a(T).$ From [19, Lemma 2.9] we have that $\Pi _a(T)=\Pi (T),$ so $\sigma _a(T)= \sigma _{uw}(T) \bigsqcup \Pi (T),$ hence T satisfies property $(UW {\scriptstyle \Pi }).$

(i) $\Longleftrightarrow$ (iii) Suppose that T satisfies property $(UW {\scriptstyle \Pi }).$ Then T has property $(Z{\scriptstyle \Pi _a})$. Let $\lambda \in \Delta _a (T)=\Pi (T)$. Then $\lambda I-T\in W_+(X)$ and since $p(\lambda I-T)= q(\lambda I-T)< \infty$ it follows, from [1, Theorem 1.22] that $\lambda I-T \in B(X)$ and in particular $\lambda I-T\in W(X)$, so $\lambda \notin \sigma _w(T)$.

Conversely, suppose that T satisfies property $(Z {\scriptstyle \Pi _a})$ and $\Delta _a (T)\cap \sigma _w (T)= \emptyset$. If $\lambda \in \Delta _a (T)$ then $\lambda \notin \sigma _w(T)$ and hence $\lambda \notin \sigma _{uw} (T)$. This implies that $\lambda \in \Delta _a (T)= \Pi _a (T)= \Pi (T)$, always by [19, Lemma 2.9]. Therefore T has property $(UW {\scriptstyle \Pi }).$$\square$

3 Property $(UW {\scriptstyle \Pi })$ and perturbations

An operator $R\in L(X)$ is called a Riesz operator if $\lambda I-R$ is Fredholm for all nonzero $\lambda \in \mathbb {C}$. Evidently, quasi-nilpotent operators and compact operators are Riesz operators. Also every operator K for which, for some $n\in {\mathbb {N}}$, $K^n$ is finite-dimensional is a Riesz operator. Indeed, $K^n$ is a Riesz operator and hence K is a Riesz operator, see [14]. The spectrum of a Riesz operator is either a finite set or a sequence of eigenvalues which clusters at 0. It is known that if $T\in L(X)$ then $\sigma _b (T+R)= \sigma _b (T)$ for every Riesz operator R commuting with T, see [1, Chapter 2], and $\sigma _{uw}(T+R)= \sigma _{uw}(T)$ see [17].

• Every Riesz operator T having infinite spectrum has property $(UW {\scriptstyle \Pi })$. To see this, observe first that $\sigma _a (T)= \sigma (T)$, since $T^*$ has SVEP. Further, $\sigma _{uw} (T)= \sigma _d (T)= \{0\}$, so $\Delta _a(T)= \Pi (T)$.

• An operator $T\in L(X)$ is said to be algebraic if there exists a nontrivial complex polynomial h such that h(T)=0. Examples of algebraic operators are idempotent operators, nilpotent operators and every operator K such that $K^n$ is finite-dimensional for some $n\in {\mathbb {N}}$.If T is algebraic then $\sigma (T)$ is a finite set of poles [1, Chapter 3], say $\{\lambda _1,\dots , \lambda _n\}$. Therefore, $\sigma _d (T)= \emptyset$ and hence $\Pi (T)= \{\lambda _1,\dots , \lambda _n\}$. Moreover, $\sigma _a (T) = \sigma (T)$, since a pole is always an eigenvalue and $\sigma _{uw}(T) \ne \emptyset$ for every operator $T\in L(X)$. Therefore, for an algebraic operator we have $\Delta _a (T) \ne \Pi (T)$, i.e., for every algebraic operator property $(UW {\scriptstyle \Pi })$ fails.

• A Riesz operator which has finite spectrum may fail property $(UW {\scriptstyle \Pi })$. Indeed, every finite-dimensional operator K does not satisfy this property, since K is algebraic.

• If Q be a quasi-nilpotent operator, then

  $$\begin{aligned} Q \ \text{ has } \text{ property } \ (UW {\scriptstyle \Pi }) \Longleftrightarrow \ 0\notin \sigma _d(Q). \end{aligned}$$

  Indeed, if 0 is not a pole then $\{0\}=\sigma (Q)= \sigma _{ uw}(Q)= \sigma _{ a}(Q)= \sigma _d (Q)$, since both $\sigma _{ uw}(Q)$ and $\sigma _{ a}(Q)$ are non-empty. Therefore $\Delta _a (Q) = \Pi (Q)= \emptyset$, so Q has property $(UW {\scriptstyle \Pi })$. Conversely, if Q has property $(UW {\scriptstyle \Pi })$ then, we have $\emptyset =\sigma _a (Q)\setminus \sigma _{uw}(Q)= \Pi (T)$ and since $\sigma (Q)= \{0\}$ it then follows that $\sigma _{d}(Q)=\{0\}$.

• It should be noted that if $K^n$ is finite-dimensional for some $n\in {\mathbb {N}}$ and $KT=TK$ then

  $$\begin{aligned} \sigma _d (T)= \sigma _d(T+K), \end{aligned}$$

  see [21], or [1, Chapter 3] for an alternative proof. Moreover, if $\text{ acc }\, F$ denotes the set of all cluster points of $F\subseteq {\mathbb {C}}$, then

  $$\begin{aligned} \text{ acc }\,\sigma (T)= \text{ acc }\, \sigma (T+K) \quad \text{ and } \quad \text{ acc }\,\sigma _a (T)= \text{ acc } \, \sigma _a (T+K) \end{aligned}$$

  see [20].

Theorem 3.1

Suppose that $T\in L(X)$ has property $(UW {\scriptstyle \Pi })$ and $T^*$ has SVEP. If $K\in L(X)$ commutes with T and $K^n$ is finite-dimensional for some $n\in {\mathbb {N}}$, then $T+K$ has property $(UW {\scriptstyle \Pi }).$

Proof

Since T has property $(UW {\scriptstyle \Pi })$, by Theorem 2.3 we have $\sigma _b (T) = \sigma _d(T)$. The spectrum $\sigma _b (T)$ is invariant under Riesz commuting perturbations, and being K a Riesz operator, we then have $\sigma _b (T+K)= \sigma _b(T)$, while the equality $\sigma _d (T+K) = \sigma _d (T)$, has been already observed. Therefore, $\sigma _b (T+K)= \sigma _d (T+K)$ . Now, K is a Riesz operator and hence also $K^*$ is a Riesz operator. Consequently, $T^*+K^*$ has SVEP, see [1, Chapter 2]. By Theorem 2.3 it then follows that $T+K$ has property $(UW {\scriptstyle \Pi }).$

$\square$

Corollary 3.2

Suppose that $Q \in L(X)$ is a quasi-nilpotent such that $\alpha (Q)< \infty$. If $K \in L(X)$ commutes with Q and $K^n$ is finite-dimensional for some $n\in {\mathbb {N}}$, then $Q+K$ satisfies property $(UW {\scriptstyle \Pi })$.

Proof

If $\alpha (Q) < \infty$ then $0\in \sigma _d (Q)$, otherwise, by [1, Theorem 1.22], we would have $0 \notin \sigma _b(Q)$, hence $\sigma _b (Q)= \emptyset$, and this is impossible. Therefore, 0 is not a pole of the resolvent of T and, as noted before, Q has property $(UW {\scriptstyle \Pi })$. Theorem 3.1 then applies, since $Q^*$ has SVEP. $\square$

Property $(UW {\scriptstyle \Pi })$ is also transmitted to $T+K$ if we assume that the approximate-point spectrum $\sigma _a(T)$ has no isolated points:

Theorem 3.3

Suppose that $T, K\in L(X)$ commute, and $K^n$ is finite-dimensional for some $n\in {\mathbb {N}}$. If $\text{ iso }\, \sigma _a (T)= \emptyset$ and T has property $(UW {\scriptstyle \Pi })$, then $T+K$ has property $(UW {\scriptstyle \Pi }).$

Proof

Observe that if $\text{ iso }\, \sigma _a (T)= \emptyset$ then $\text{ iso }\, \sigma (T)= \emptyset$, since $\text{ iso } \, \sigma (T)\subseteq \text{ iso }\,\sigma _a(T)$. Therefore,

$$\begin{aligned} \sigma (T)= \text{ iso }\, \sigma (T) \cup \text{ acc }\, \sigma (T)=\text{ acc }\, \sigma (T), \end{aligned}$$

and, similarly, $\sigma _a (T)= \text{ acc }\, \sigma _a (T)$. Since $\text{ acc }\, \sigma _{a}(T)=\text{ acc }\ \sigma _{a}(T+K)$, and $\text{ acc }\, \sigma _{}(T)=\text{ acc }\, \sigma _{}(T+K)$ see [20], we then have $\sigma _a (T+K)= \sigma _a(T)$ and $\sigma (T+K)= \sigma (T)$. Also the spectrum $\sigma _{uw}(T)$ is invariant under Riesz commuting perturbations, so $\sigma _{uw}(T+K)= \sigma _{uw}(T)$. Hence,

$$\begin{aligned} \Delta _a (T+K)= \sigma _a (T+K)\setminus \sigma _{uw}(T+K)= \sigma _a (T)\setminus \sigma _{uw}(T)=\Delta _a (T). \end{aligned}$$

As observed before, $\sigma _d (T+K) = \sigma _d (T)$, so

$$\begin{aligned} \Pi (T+K)= \sigma (T+K)\setminus \sigma _d(T+K)= \sigma (T)\setminus \sigma _d(T)= \Pi (T). \end{aligned}$$

Therefore, $\Delta _a (T+K)= \Pi (T+K) = \Pi (T)= \Delta _a(T)$. $\square$

Theorem 3.3 applies to nilpotent commuting perturbations. However, in this case no assumption on the spectrum is required. It is well-known that $\sigma (T)$ and $\sigma _a(T)$ are invariant under commuting quasi-nilpotent perturbations.

Theorem 3.4

Let $T\in L(X)$ and let N be a nilpotent operator which commutes with T. If T has property $(UW {\scriptstyle \Pi })$, then $T+N$ has property $(UW {\scriptstyle \Pi }).$

Proof

We have that $\sigma (T+N)= \sigma (T)$, $\sigma _a (T+N)= \sigma _a (T)$ and $\sigma _d(T+N)= \sigma _d(T)$, so $\Delta _a (T+N) = \Delta _a (T)= \Pi (T)= \Pi (T+N)$. $\square$

The following example shows that the result of Theorem 3.4 cannot be extended to quasi-nilpotent commuting perturbations:

Example 3.5

Let $Q\in L(X)$ denote an injective quasi-nilpotent operator (for instance the Volterra integral operator defined on the space of all continuous function C[a, b]). Then $\Delta _a (Q)= \sigma _a (Q)\setminus \sigma _{uw}(Q)= \emptyset$. Since $\alpha (Q)= 0$, the point 0 cannot be a pole of the resolvent, i.e. $\sigma _d(Q) = \{0 \}$, otherwise the condition $p(Q)= q(Q)< \infty$ would implies $\alpha (Q) = \beta (Q)= 0$ and hence $0\notin \sigma (Q)$). Therefore, Q has property $(UW {\scriptstyle \Pi }).$ On the other hand, $0=Q-Q$ does not have property $(UW {\scriptstyle \Pi })$ since $\Delta _a (0)= \emptyset$, while $\sigma _d(0) = \emptyset$ and $\sigma (0)= \{0\}$, hence $\Pi (0) = \{0\}$.

Theorem 3.6

Suppose that $T\in L(X)$ has property $(UW {\scriptstyle \Pi })$. If $\text{ iso }\, \sigma _b(T)= \emptyset$, or $\text{ iso }\, \sigma _{ub}(T)= \emptyset$, then $T+Q$ has property $(UW {\scriptstyle \Pi })$ for every commuting quasi-nilpotent operator Q.

Proof

The spectra $\sigma (T)$, $\sigma _a (T)$ and $\sigma _{uw}(T)$ are invariant under a commuting quasi-nilpotent perturbation Q, so, if T has property $(UW {\scriptstyle \Pi })$, then $T+Q$ has property $(UW {\scriptstyle \Pi })$ exactly when $\sigma _d(T+Q) = \sigma _d(T)$. Each one of the assumptions $\text{ iso }\, \sigma _b(T)= \emptyset$, or $\text{ iso }\, \sigma _{ub}(T)= \emptyset$, entails the equality $\sigma _d(T+Q) = \sigma _d(T)$, see [11, Proposition 2.7]. $\square$

The result of Corollary 3.2 may be strongly improved if we consider injective quasi-nilpotent perturbations. To see this we need a preliminary lemma.

Lemma 3.7

Let $T\in L(X)$ be such that $\alpha (T)< \infty$. Suppose that there exists an injective quasi-nilpotent operator $Q\in L(X)$ such that $TQ=QT$. Then T is injective.

Proof

Set $Y: = \ker \, T$. Clearly, Y is invariant under Q and the restriction $(\lambda I-Q)|Y$ is injective for all $\lambda \ne 0$. Since Y is finite-dimensional then $(\lambda I-Q)|Y$ is also surjective for all $\lambda \ne 0$, thus $\sigma (Q|Y)\subseteq \{0\}$. On the other hand, from assumption we know that Q|Y is injective and hence Q|Y is surjective, so $\sigma (Q|Y)= \emptyset$, from which we conclude that $Y= \{0\}$. $\square$

Recall that $T\in L(X)$ is said to be finite-isoloid (respectively, finite a-isoloid) if every isolated point of $\sigma (T)$ (respectively, $\sigma _a (T)$) is an eigenvalue having finite rank. It is easily seen that if T is finite-isoloid then $\sigma _b (T)= \sigma _d (T)$. Indeed, if $\lambda \notin \sigma _d(T)$ then $\lambda \in \text{ iso } \, \sigma (T)$, so $\alpha (\lambda I-T)< \infty$. Since $p(\lambda I-T)= q(\lambda I-T)< \infty$, then $\lambda I-T$ is Browder, by [1, Chapter 1], so $\lambda \notin \sigma _b(T)$. Therefore, $\sigma _b(T)\subseteq \sigma _d (T)$, and being the opposite inclusion true for every operator, we have $\sigma _b (T)= \sigma _d (T)$.

Theorem 3.8

Suppose that $T \in L(X)$ is finite-isoloid. If Q is an injective quasi-nilpotent operator commuting with T, then both T and $T+Q$ have property $(UW {\scriptstyle \Pi })$.

Proof

Let $\lambda \in \Delta _a (T)= \sigma _a (T)\setminus \sigma _{uw}(T)$. Since $\lambda I-T \in W_+(X)$ then $\alpha (\lambda I-T)< \infty$ and $\lambda I-T$ has closed range. Since $\lambda I-T$ commutes with Q it then follows, by Lemma 3.7, that $\lambda I-T$ is injective, so $\lambda \notin \sigma _{ a} (T)$, a contradiction. Therefore, $\Delta _a (T)$ is empty. Also $\Pi (T)$ is empty. Indeed, suppose that $\lambda \in \Pi (T)=\sigma (T)\setminus \sigma _d(T)$. Then $\lambda$ is an isolated point of $\sigma (T)$, and since T is finite-isoloid then $0<\alpha (\lambda I-T)< \infty$. But by Lemma 3.7 we have $\alpha (\lambda I-T)=0$, a contradiction. Therefore $\Pi (T) = \emptyset$, and hence T has property $(UW {\scriptstyle \Pi })$.

To show that $T+Q$ has property $(UW {\scriptstyle \Pi })$, observe first that

$$\begin{aligned} \Delta _a (T+Q)= \sigma _a (T+Q)\setminus \sigma _{uw} (T+Q)= \sigma _a (T)\setminus \sigma _{uw}(T)=\Delta _a (T), \end{aligned}$$

so, from the first part, $\Delta _a (T+Q) = \emptyset$. Also $\Pi (T+Q)$ is empty. Indeed, suppose that there exists $\lambda \in \Pi (T+Q)$. Then $\lambda \in \text{ iso }\, \sigma (T+Q)= \text{ iso }\, \sigma (T)$, and hence, since T is finite-isoloid, $0<\alpha (\lambda I-T)< \infty$. Again by Lemma 3.7 we have $\alpha (\lambda I-T)=0$, a contradiction. $\square$

The assumption that T is finite-isoloid in Theorem 3.8 is crucial. For instance, if T is an injective quasi-nilpotent operator Q then T is not finite-isoloid, and, as observed before, T has property $(UW {\scriptstyle \Pi })$, while $T-Q=0$ does not have property $(UW {\scriptstyle \Pi })$.

In the next result, we give a necessary and sufficient condition under which property $(UW {\scriptstyle \Pi })$ is preserved under commuting Riesz perturbations.

Theorem 3.9

Let $T\in L(X)$ and let R be a Riesz operator commuting with T. If T satisfies property $(UW {\scriptstyle \Pi })$ then the following assertions are equivalent:

• (i) $T+R$ satisfies property $(UW {\scriptstyle \Pi });$

• (ii) $p_{00}^a(T+R)=\Pi (T+R)$.

Proof

(i)$\Rightarrow$ (ii) Suppose that $T+R$ satisfies property $(UW {\scriptstyle \Pi }).$ By Theorem 2.3 we then have $\ p_{00}^a(T+R)=\Pi (T+R).$

(ii) $\Rightarrow$ (i) Assume hat $\ p_{00}^a(T+R)=\Pi (T+R)$. Since T satisfies property $(UW {\scriptstyle \Pi })$ then T satisfies a-Browder’s theorem. Hence $\sigma _{uw}(T)=\sigma _{ub}(T)$. Since $\sigma _{uw}(T+R)=\sigma _{uw}(T)$ and $\sigma _{ub}(T+R)=\sigma _{ub}(T),$ then so $\sigma _{ub}(T+R)=\sigma _{uw}(T+R)$, and $T+R$ satisfies a-Browder’s theorem i.e $\Delta _a(T+R)= p_{00}^a(T+R)$. Since $p_{00}^a(T+R)=\Pi (T+R)$, then $\Delta _a(T+R)=\Pi (T+R).$ Therefore $T+R$ satisfies property $(UW {\scriptstyle \Pi })$.$\square$

Property $(UW {\scriptstyle \Pi }))$ is transmitted under commuting Riesz perturbations in a very special case. Recall that an operator $T\in L(X)$ is said to be finitely a-polaroid if every isolated point of $\sigma _a (T)$ is a pole of T having finite rank. Recall that, in general, the equality $\sigma _a (T+R) = \sigma _a (T)$, for a Riesz commuting perturbation, does not hold, even if T is a finite-rank operator.

Theorem 3.10

Suppose that $T\in L(X)$ is finitely a-polaroid and let $R\in L(X)$ be a commuting Riesz operator such that $\sigma _a (T+R)= \sigma _a (T)$. If T has property $(UW {\scriptstyle \Pi }))$ then $T+R$ has property $(UW {\scriptstyle \Pi })$.

Proof

By Theorem 3.9 it suffices to show the equality $\Pi (T+R)= p_{00}^a (T+R)$. Let $\lambda \in \Pi (T+R)$. Then $\lambda \in \text{ iso } \,\sigma _a (T+R)= \text{ iso }\,\sigma _a (T)$. Since T is finitely a-polaroid then $\lambda I-T$ is Drazin invertible and $\alpha (\lambda I-T)< \infty$. Then $\lambda I-T\in B(X)$, see [1, Chapter 1], in particular is upper semi-Browder. This implies that $\lambda I- (T+R) \in B_+(X)$, so $\lambda \in \sigma _a (T+R)\setminus \sigma _{ub}(T+R) = p_{00}^a (T+R)$.

Conversely, suppose that $\lambda \in p_{00}^a (T+R)$. Then

$$\begin{aligned} \lambda\in & {} \sigma _a (T+R)\setminus \sigma _{ub}(T+R) \subseteq \sigma _a (T+R)\setminus \sigma _{uw}(T+R)\\= & {} \sigma _a (T)\setminus \sigma _{uw}(T) = \Delta _a (T)=\Pi (T). \end{aligned}$$

In particular, $\lambda$ is an isolated point of $\sigma _a (T)$, and consequently, a pole of T having finite rank. This implies that $\lambda I-T \in B(X)$, see [1, Chapter 3], and hence $\lambda I- (T+R) \in B(X)$, so $\lambda \in \Pi (T+R)$. $\square$

4 Property $(UW {\scriptstyle \Pi })$ under functional calculus

In this section we study the preservation of property $(UW {\scriptstyle \Pi })$ under functional calculus. Furthermore, we show that this property is transferred from a Drazin invertible operator to its Drazin inverse. Let ${\mathcal {H}}(\sigma (T))$ be the set of all analytic functions defined on a neighborhood of $\sigma (T)$, and for every $f\in {\mathcal {H}}(\sigma (T))$ let f(T) be defined by means of the classical functional calculus. The spectral theorem for the spectrum asserts that $\sigma (f(T))= f (\sigma (T))$ for every $f\in {\mathcal {H}}(\sigma (T))$ and a similar equality holds for the approximate point spectrum.

Lemma 4.1

Let $T \in L(X)$ and let $\{\lambda _1, \cdots , \lambda _k \}$ be a finite subset of ${\mathbb {C}}$ such that $\lambda _i \ne \lambda _j$ for $i\ne j$. Assume that $\{\nu _1, \cdots , \nu _n\}\subset {\mathbb {N}}$ and set $h(\lambda ) : = \prod _{i=1}^{n}(\lambda _i - \lambda )^{\nu _i}$. Then, for the operator $h(T) : = \prod _{i=1}^{n}(\lambda _i I - T)^{\nu _i}$ we have

$$\begin{aligned} \mathrm{ker}\ h(T) = \bigoplus _{i=1}^n \mathrm{ker} \ (\lambda _i I - T)^{\nu _i} \end{aligned}$$

(5)

and

$$\begin{aligned} h(T)(X) = \bigcap _{i=1}^n (\lambda _i I - T)^{\nu _i}(X). \end{aligned}$$

(6)

Furthermore,

$$\begin{aligned} 0\in \Pi (h(T)) \Leftrightarrow \lambda _i \in \Pi (T) \ \text{ for } \text{ all } \ i=1, \dots , n. \end{aligned}$$

Proof

A proof of the equalities (5) and (6) may be found in [1]. Suppose that $\lambda _i \in \Pi (T)$ for all $i=1, \dots , n$, i.e., $\lambda _i \in \sigma (T)$ and $\lambda _i I-T$ Drazin invertible, for all $i=1, \dots , n$. Then, for k large enough, we have $\ker (\lambda _i I-T)^{\nu _i k}= \ker (\lambda _i I-T)^{\nu _i (k+1)}$ and $(\lambda _i -T)^{\nu _i k} (X)= (\lambda _i -T)^{\nu _i (k+1)} (X)$. This implies, by (5) and (6), that $\ker \, (h(T)^k)= \ker \, (h(T)^{k+1})$ and $h(T)^k (X)= h(T)^{k+1}(X)$, so h(T) is Drazin invertible. Since $0 =h(\lambda _i)\in h(\sigma (T)) = \sigma (h(T))$ it then follows that $0\in \Pi (h(T))$.

Conversely, assume that $\lambda _j \notin \Pi (T)$ for some $j\in \{1,\dots , n\}$. Let $p_j$ and $q_j$ denote the ascent and the descent of $\lambda _j I-T$, respectively. We have either $\lambda _j \notin \sigma (T)$ or $\lambda _j \in \sigma (T)$. In the case where $\lambda _j \notin \sigma (T)$, then $0= h(\lambda _j)\notin h(\sigma (T))= \sigma (h(T))$, so $0\notin \Pi (T)$. Consider the other case $\lambda _j \in \sigma (T)$. Since $\lambda _j \notin \Pi (T)$ then $\lambda _j I-T$ is not Drazin invertible, so either $p_j=\infty$, i.e. $\ker \, (\lambda _j I-T)^k$ is properly contained in $\ker \, (\lambda _j I-T)^{k+1}$ for every $k\in {\mathbb {N}}$, or $q_j=\infty$, i.e., $(\lambda _j I-T)^{k+1} (X)$ is properly contained in $(\lambda _j I-T)^{k} (X)$ for every $k\in {\mathbb {N}}$. If $p_j=\infty$, being $\ker \, (\lambda _i I -T)^{\nu _i} \cap \ker \, (\lambda _j I -T)^{\nu _j}= \emptyset$ for all $i\ne j$, it then follows that $p(h(T))= \infty$. Analogously, if $q_j= \infty$ then $q(h(T))= \infty$. Therefore, also in this case $0\notin \Pi (T)$. $\square$

Remark 4.2

If $T\in L(X)$ is invertible and $S\in L(X)$ commutes with T then $N(S^n)= N((TS))^n$ and $(TS)^n (X)= S^n (X)$ for all $n\in {\mathbb {N}}$. Consequently, TS is Drazin invertible if and only if S is Drazin invertible, while $0\in \Pi (TS)$ if and only if $0\in \Pi (S)$.

In the sequel we need the following lemma.

Lemma 4.3

For every $T\in L(X)$, X a Banach space, and $f\in {\mathcal {H}} (\sigma (T))$ we have

$$\begin{aligned} \sigma _a (f(T))\setminus \Pi (f(T)) \subseteq f \left( \sigma _a (T)\setminus \Pi (T) \right) . \end{aligned}$$

(7)

Furthermore, if T is a-isoloid then

$$\begin{aligned} \sigma _a (f(T))\setminus \Pi (f(T)) = f\left( \sigma _a (T) \setminus \Pi (T)\right) . \end{aligned}$$

(8)

Proof

Suppose that $\lambda _0 \in \sigma _a (f(T))\setminus \Pi (f(T))$. We consider two cases.

Case (I). Suppose first that $\lambda _0 \notin \text{ iso } \, f(\sigma _a (T))= \text{ iso } \, \sigma _a (f(T))$. In this case $\lambda _0 \notin \Pi (f(T))$, since each point of $\Pi (f(T))$ is an isolated point of $\sigma (f(T))$ and hence an isolated point of $\sigma _a (f(T))$. Let $(\lambda _n)\subset f(\sigma _a (T))= \sigma _a (f(T))$ be a sequence such that $\lambda _n \rightarrow \lambda _0$ and let $(\mu _n)\in \sigma _a (T)$ be such that $f(\mu _n) =\lambda _n$. Then $(\mu _n)$ admits a subsequence which converge to some $\mu _0 \in \sigma _a (T))$. There is no harm if we assume that $\mu _n \rightarrow \mu _0$. Then $f(\mu _n)\rightarrow f(\mu _0)= \lambda _0$. Evidently, $\mu _0 \notin \Pi (T)$, since $\mu _0$ is a cluster point of $\sigma _a (T)$. Hence

$$\begin{aligned} \lambda _0 =f(\mu _0) \in f\left( \sigma _a (T) \setminus \Pi (T)\right) . \end{aligned}$$

Case (II). Suppose that $\lambda \in \text{ iso } \, f(\sigma _a (T)) =\text{ iso } \, \sigma _a (f(T))$. Define $g(\lambda ): = \lambda _0 -f(\lambda )$, Since $g(\lambda )$ is analytic then g has only a finite number of zeros in $\sigma (T)$, say $\{\lambda _1, \dots , \lambda _n\}$. Write

$$\begin{aligned} g(\lambda ) = h(\lambda ) k(\lambda ) \quad \text{ and } \quad h(\lambda ) = \prod _{i=1}^{k}(\lambda _i - \lambda )^{\nu _i}, \end{aligned}$$

(9)

where $k(\lambda )$ has no zero in $\sigma (T)$. Then

$$\begin{aligned} g(T)= \lambda _0 I- f(T)= h(T) k(T), \end{aligned}$$

where k(T) invertible, and $h(T): = \prod _{i=1}^{k}(\lambda _i I - T)^{\nu _i}$.

Now, $\lambda _0 \notin \Pi (f(T)$ implies that $\lambda _0 I-f(T)$ is invertible, or $\lambda _0 I-f(T)$ is not Drazin invertible. In the first case g(T) is invertible, so $0\notin \Pi (g(T))$. If $\lambda _0 I-f(T)$ is not Drazin invertible, we have either $p(g(T))= p(\lambda _0 I-f(T))= \infty$ or $q(g(T))= q(\lambda _0 I-f(T))= \infty$, so also in this case $0\notin \Pi (g(T))$. According Remark 4.2 then $0\notin \Pi (h(T))$. By Lemma 4.1 it then follows that $\lambda _j \notin \Pi (T)$ for some j. Thus, $\lambda _j \in \sigma _a (T)\setminus \Pi (T)$, and hence $\lambda _0 = f(\lambda _j)\in f\left( \sigma _a (T) \setminus \Pi (T)\right) .$

To show the equality (8), assume that T is a-isoloid.

To show the equality (8) we need only to prove the inclusion $\supseteq$. Let $\lambda _0 \in f\left( \sigma _a(T) \setminus \Pi (T)\right)$ be arbitrary given. Since $\lambda _0 \in f(\sigma _a (T))= \sigma _a (f(T)) \subseteq \sigma (f (T))$ it then suffices to prove that $\lambda _0 \notin \Pi (f(T))$. To do this, suppose that $\lambda _0 \in \Pi (f(T))$. Then $0 \in \Pi (h(T)$, and by Lemma 4.1 it then follows that $\lambda _i \in \Pi (T)$ for all $i=1, \dots , n$. Thus, $\lambda _i \notin \sigma _a (T)\setminus \Pi (T)$ for all i. Since $0= g(\lambda _i)= \lambda _0-f(\lambda _i)$, we have $\lambda _0 = f(\lambda _i) \notin f\left( \sigma _a(T) \setminus \Pi (T)\right)$, a contradiction. Therefore, $\lambda _0 \notin \Pi (f(T))$, and hence

$$\begin{aligned} \lambda _0 \in \sigma _a (f(T))\setminus \Pi (f(T)), \end{aligned}$$

as desired. $\square$

Recall that, in general, the spectral theorem does not hold for $\sigma _{uw} (T)$, see [1, Chapter 3].

Theorem 4.4

Let $T\in L(X)$ be a-isoloid and $f\in {\mathcal {H}} (\sigma (T))$. If T satisfies property $(UW {\scriptstyle \Pi }))$ then the following are equivalent:

• (i) f(T) satisfies property $(UW {\scriptstyle \Pi })$;

• (ii) $f(\sigma _{uw}(T))= \sigma _{uw} (f(T))$.

Proof

(i) $\rightarrow$ (ii) Observe first that the spectral mapping theorem holds for $\sigma _{\mathrm{ub}}(T)$, i.e., $f(\sigma _{\mathrm{ub}}(T))= \sigma _{\mathrm{ub}} (f(T))$ for every $f\in {\mathcal {H}} (\sigma (T))$, see [1, Chapter 3]. Since f(T) satisfies property $(UW {\scriptstyle \Pi })$, then both T and f(T) satisfies a-Browder’s theorem, by Theorem 2.3, so $\sigma _{\mathrm{uw}}(T)= \sigma _{\mathrm{ub}} (T)$ and $\sigma _{\mathrm{uw}}(f(T))= \sigma _{\mathrm{ub}} (f(T))$. Then we have

$$\begin{aligned} f(\sigma _{\mathrm{uw}}(T))= f(\sigma _{\mathrm{ub}}(T))= \sigma _{\mathrm{ub}} (f(T))= \sigma _{\mathrm{uw}}(f(T)). \end{aligned}$$

(ii) $\rightarrow$ (i) Since T satisfies property $(UW {\scriptstyle \Pi }))$ we have $\sigma _{uw}(T) = \sigma _a (T)\setminus \Pi (T)$. By Lemma 4.3 it then follows that

$$\begin{aligned} \sigma _{uw} (f(T))= f(\sigma _{uw}(T))= f ( \sigma _a (T)\setminus \Pi (T))= \sigma _a (f(T))\setminus \Pi (f(T)), \end{aligned}$$

so f(T) satisfies property $(UW {\scriptstyle \Pi })$. $\square$

It is known that the spectral theorem for $\sigma _{uw}(T)$ holds if T or $T^*$ has SVEP ( [1, Chapter 3], or if f is injective ( [18]. Consequently, property $(UW {\scriptstyle \Pi }))$ is transmitted from T to f(T) if T is a-isoloid and T or $T^*$ has SVEP. In the case, that $T^*$ has SVEP, then we can require that T is isoloid, since in this case $\sigma _a(T)= \sigma (T)$ and hence the properties of being a-isoloid and isoloid coincide.

Lemma 4.5

Suppose that for a bounded operator $T\in L(X)$ there exists $\lambda _0 \in {\mathbb {C}}$ such that $K(\lambda _0 I-T) = \{0\}$ and $\ker \, (\lambda _0 I-T)= \{0\}$. Then $\sigma _{p}(T)= \emptyset$.

Proof

For all complex $\lambda \ne \lambda _0$ we have $\ker \, (\lambda I-T) \subseteq K(\lambda _0 I-T)$, so that $\ker \, (\lambda I-T)= \{0\}$, for $\lambda \ne \lambda _0$. Since $\ker \, (\lambda _0 I-T)= \{0\}$ we then conclude that $\ker \, (\lambda I-T)= \{0\}$ for all $\lambda \in {\mathbb {C}}$. $\square$

Theorem 4.6

Let $T\in L(X)$ be such that there exists $\lambda _0 \in {\mathbb {C}}$ such that

$$\begin{aligned} K(\lambda _0 I-T) = \{0\} \quad \text{ and } \quad \ker \, (\lambda _0 I-T)= \{0\}. \end{aligned}$$

(10)

Then property $(UW {\scriptstyle \Pi })$ holds for f(T) for all $f\in {\mathcal {H}}(\sigma (f(T))$.

Proof

We know from Lemma 4.5 that $\sigma _{p}(T)= \emptyset$, so T has SVEP. We show that also $\sigma _{p}(f(T))= \emptyset$. Let $\mu \in \sigma (f(T))$ and write $\mu -f(\lambda ) = p (\lambda ) g(\lambda )$, where g is analytic on an open neighborhood ${\mathcal {U}}$ containing $\sigma (T)$ and without zeros in $\sigma (T)$, p a polynomial of the form $p(\lambda ) = \Pi _{k=1}^n (\lambda _k- \lambda )^{\nu _k},$ with distinct roots $\lambda _1, \dots , \lambda _n$ lying in $\sigma (T)$. Then

$$\begin{aligned} \mu I- f(T) = \Pi _{k=1}^n (\lambda _k I - T)^{\nu _k} g(T). \end{aligned}$$

Since g(T) is invertible, $\sigma _{\mathrm{p}}(T)= \emptyset$ implies that $\ker \, (\mu I - f(T))= \{0\}$ for all $\mu \in {\mathbb {C}}$, so $\sigma _{p}(f(T))= \emptyset$. Since T has SVEP then f(T) has SVEP, see [1, Chapter 2], so that a-Browder’s theorem holds for f(T) ( [1, Chapter 5]). To prove that property $(UW {\scriptstyle \Pi })$ holds for f(T), by Theorem 2.3 it then suffices to prove that

$$\begin{aligned} p_{00}^a (f(T))= \Pi (f(T)). \end{aligned}$$

Obviously, the condition $\sigma _{p}(f(T))= \emptyset$ entails that $\Pi (f(T))= \emptyset$, since every point of $\Pi (f(T))$ is an eigenvalue of f(T). On the other hand, if $\lambda \in p_{00}^a (f(T))$ then $\lambda \in \sigma _a (f(T))$ and $\lambda \notin \sigma _{ub}(f(T)$, so $\lambda I-f(T)$ has closed range, since it is a upper semi-Browder operator. This implies that $\alpha (\lambda I-f(T)) >0$ and we get a contradiction, since $\sigma _{p}(f(T))= \emptyset$. $\square$

The conditions of Theorem 4.6 are satisfied by any injective operator for which the hyperrange$T^\infty (X): = \bigcap T^n (X)$ is $\{0\}$. In fact, $K(T)\subseteq T^\infty (X)$ for all $T\in L(X)$, so that $K(T)= \{0\}$. In particular, the conditions of Theorem 4.6 are satisfied by a semi-shift T, i.e. T is an isometry for which $T^\infty (X)= \{0\},$ see [15] for details on this class of operators. Clearly, a semi-shift T on a non-trivial Banach space is a non-invertible isometry.

The following result applies to several operator, in particular to shift operators.

Lemma 4.7

Suppose that for $T\in L(X)$ we have $\text{ iso }\, \sigma _{\mathrm{a}}\, (T)= \emptyset$. If $K\in L(X)$ commutes with T and $K^n$ is finite-dimensional for some $n\in {\mathbb {N}}$, then $\text{ iso }\, \sigma _{\mathrm{a}} (T+K)= \emptyset$. Consequently, $\sigma _{\mathrm{a}} (T+K)= \sigma _{\mathrm{a}} (T)$.

Proof

By Lemma 4.7 we have

$$\begin{aligned} \sigma _{\mathrm{a}} (T)= & {} \text{ iso }\, \sigma _{\mathrm{a}}\, (T) \cup \text{ acc }\, \sigma _{\mathrm{a}}\, (T)= \text{ acc }\, \sigma _{\mathrm{a}}\, (T)\\= & {} \text{ acc }\, \sigma _{\mathrm{a}}\, (T+K )\subseteq \sigma _{\mathrm{a}} (T+K). \end{aligned}$$

On the other hand, if $\sigma _ \mathrm{a} (K)= \{\lambda _1, \lambda _2, \dots \lambda _n\}$ we have

$$\begin{aligned} \text{ iso }\, \sigma _{\mathrm{a}} (T+K)\subseteq \text{ iso }\,(\sigma _\mathrm{a} (T) + \sigma _ \mathrm{a} (K)) = \text{ iso }\,\bigcup _{k=1}^n (\lambda _k + \sigma _{\mathrm{a}} (T))= \emptyset , \end{aligned}$$

hence,

$$\begin{aligned} \sigma _{\mathrm{a}} (T+K)= & {} \text{ iso }\, \sigma _{\mathrm{a}} (T+K) \cup \text{ acc }\, \sigma _{\mathrm{a}} (T+K)= \text{ acc }\, \sigma _{\mathrm{a}} (T+K)\\= & {} \text{ acc }\, \sigma _{\mathrm{a}} (T) = \sigma _{\mathrm{a}} (T), \end{aligned}$$

so $\sigma _{\mathrm{a}} (T+K)= \sigma _{\mathrm{a}} (T)$ holds.

$\square$

Theorem 4.8

Let $T\in L(X)$ be a semi-shift and suppose that $K \in L(X)$ is such that $K^n$ is finite-dimensional for some $n\in {\mathbb {N}}$ and that $TK=KT$. Then property $(UW {\scriptstyle \Pi })$ holds for $f(T)+K$ for all $f\in {\mathcal {H}}(\sigma (f(T))$.

Proof

Since T is a non-invertible isometry, the aproximate point spectrum $\sigma _{\mathrm{a}}(T)$ is the closed unit circle of ${\mathbb {C}}$, see [15, Proposition 1.6.2]. Hence $\text{ iso }\, \sigma _{\mathrm{a}}(T)= \emptyset$. From Lemma 4.7 and Theorem 3.3 then property $(UW {\scriptstyle \Pi })$ holds for $f(T)+K$. $\square$

Recall that an operator $T\in L(X)$ is Drazin invertible if and only if there exists an operator $S\in L(X)$ and $n\in {\mathbb {N}}$ such that

$$\begin{aligned} TS = ST, \quad STS=S, \quad T^n ST=T^n , \end{aligned}$$

(11)

see [16, Chap.3, Theorem 10]. The operator S is called the Drazin inverse of T. From [16, Chap. 3, Theorem 10] we also know that if $T\in L(X)$ is Drazin invertible if and only if there exist two closed invariant subspaces Y and Z such that $X= Y\oplus Z$ and, with respect to this decomposition,

$$\begin{aligned} T= T_1 \oplus T_2, \quad \text{ with } \ T_1: = T|Y \ \text{ nilpotent } \ \text{ and } \ T_2: =T|Z \ \text{ invertible }. \end{aligned}$$

(12)

Note that the Drazin inverse S of an operator, if it exists, is uniquely determined ( [12]), and with respect to the decomposition $X=Y\oplus Z$, the Drazin inverse S may be represented as the directed sum

$$\begin{aligned} S: = 0\oplus S_2 \quad \text{ with } \ S_2: = {T_2}^{-1}. \end{aligned}$$

(13)

From the decomposition (13) it is obvious that the Drazin inverse S is also Drazin invertible. Evidently, if T is invertible then $S= T^{-1}$, while $0 \in \sigma (T)$ if and only if $0\in \sigma (S)$.

The decompositions (12) and (13) are very useful in order to study the spectral properties of a Drazin invertible operator, see [3] and [4], in particular the decomposition (13) shows that the Drazin inverse S is itself Drazin invertible, since is the direct sum of the nilpotent operator 0 and the invertible operator $S_2$. It should be noted that if $0\in \sigma (T)$ then 0 is a pole of the first order of the resolvent of S, see [18]. Furthermore, the following relationship of reciprocity holds for the spectra of S and T:

$$\begin{aligned} \sigma (S)\setminus \{0\}= \left\{ \frac{1}{\lambda }: \lambda \in \sigma (T)\setminus \{0\} \right\} , \end{aligned}$$

(14)

see [1, Chapter 1].

We also have,

$$\begin{aligned} \sigma _{ a}(S)\setminus \{0\}= \left\{ \frac{1}{\lambda }: \lambda \in \sigma _{ a}(T)\setminus \{0\}\right\} , \end{aligned}$$

and

$$\begin{aligned} \sigma _{ uw}(S)\setminus \{0\}= \left\{ \frac{1}{\lambda }: \lambda \in \sigma _{uw}(T)\setminus \{0\}\right\} , \end{aligned}$$

see [4]. By [4, Theorem 2.8] we also have

$$\begin{aligned} \Pi (S) \setminus \{0\} = \left\{ \frac{1}{\lambda }: \lambda \in \Pi (T) \setminus \{0\}\right\} . \end{aligned}$$

Lemma 4.9

Suppose that $T\in L(X)$ is Drazin invertible with Drazin inverse S. Then $T \in W_+(X)$ if and only if $S\in W_+(X)$.

Proof

(i) If $0\notin \sigma (T)$ then T is invertible and the Drazin inverse is $S=T^{-1}$ so the assertion is trivial in this case. Suppose that $0\in \sigma (T)$ and that T is upper semi-Weyl. Since $p(\lambda I-T)= q(\lambda I-T) < \infty$ then, see [1, Chapter 1], T is Browder. Then 0 is a pole of the resolvent of T and is also a pole (of the first order) of the resolvent of S. Let $X=Y\oplus Z$ such that $T= T_1\oplus T_2$, $T_1=T|Y$ nilpotent and $T_2=T|Z$ invertible. Observe that

$$\begin{aligned} \ker \, T = \ker \, T_1 \oplus \ker T_2= \ker T_1 \oplus \{0\}, \end{aligned}$$

(15)

and, analogously, since $S= 0\oplus S_2$ with $S_2= {T_2}^{-1}$, we have

$$\begin{aligned} \ker \, S = \ker \,0 \oplus \ker S_2= Y \oplus \{0\}. \end{aligned}$$

(16)

Since T is upper semi-Weyl we have $\alpha (T)= \text{ dim }\, \ker \, T< \infty$, and from the inclusion $\ker \, T_1 \subseteq \ker \, T$ we obtain $\alpha (T_1)< \infty$. Consequently, $\alpha (T_1^n)< \infty$ for all $n\in {\mathbb {N}}$. Let $T_1^\nu =0$. Since $Y = \ker \, T_1^\nu$ the subspace Y is finite-dimensional, and hence $\ker \, S = Y \oplus \{0\}$ is finite-dimensional, i.e. $\alpha (S)< \infty$. Now, S is Drazin invertible, so $p(S)=q(S)< \infty$ and hence, see [1, Chapter 1], $\alpha (S)= \beta (S) < \infty$. Hence S is Browder, and in particular upper semi-Weyl.

Conversely, suppose that S is upper semi-Weyl. Then $\alpha (S) <\infty$ and hence by (16) the subspace Y is finite-dimensional, from which it follows that also $\ker \, T_1= \ker \, T|Y$ is finite-dimensional. From (15) we then have that $\alpha (T)< \infty$ and since $p(T)= q(T) < \infty$ we then conclude that $\alpha (T)= \beta (T)$, see [1, Chapter 1]. Therefore, T is a Browder operator, in particular upper semi-Weyl. $\square$

In [4] it is shown that several Browder type theorems and Weyl type theorems are transferred from a Drazin invertible operator to its Drazin inverse. We show now that the same happens for property $(UW {\scriptstyle \Pi })$.

Theorem 4.10

Suppose that $T\in L(X)$ is Drazin invertible with Drazin inverse S. Then T satisfies property $(UW {\scriptstyle \Pi })$ if and only if S satisfies property $(UW {\scriptstyle \Pi })$.

Proof

Suppose that T satisfies property $(UW {\scriptstyle \Pi })$. Consider first the case that T is invertible. Then $S= T^{-1}$. Suppose that $\lambda \in \sigma _a (T^{-1})$. Then $\lambda \ne 0$, and hence $\frac{1}{\lambda } \in \sigma _a (T)= \sigma _{uw}(T) \bigsqcup \Pi (T)$. If $\frac{1}{\lambda } \in \sigma _{uw}(T)$ then $\lambda \in \sigma _{uw}(T^{-1}$, while if $\frac{1}{\lambda } \in \Pi (T)$ then $\lambda \in \sigma _{uw}(T^{-1}$. This shows that $\sigma _a (T^{-1})\subseteq \sigma _{uw}(T^{-1}) \bigsqcup \Pi (T^{-1})$, and since the opposite inclusion always holds we then have

$$\begin{aligned} \sigma _a (T^{-1})= \sigma _{uw}(T^{-1}) \bigsqcup \Pi (T^{-1}), \end{aligned}$$

i.e., property $(UW {\scriptstyle \Pi })$ holds for $T^{-1}$. The proof that property $(UW {\scriptstyle \Pi })$ for $T^{-1}$ implies property $(UW {\scriptstyle \Pi })$ for T is similar.

Consider the case that T is not invertible. Then $0\in \sigma (T)$, as well as $0\in \sigma (S)$, and in this case 0 is a pole of both T and S. Let $\lambda \in \sigma _a (S)$. If $\lambda =0$ we have already observed that $0 \in \Pi (S)$ and $0\in \Pi (T)$. Since T has property $(UW {\scriptstyle \Pi })$ then $\sigma _a (T)= \sigma _{uw}(T) \bigsqcup \Pi (T)$, so $0\notin \sigma _{uw} (T)$. But $p(T)= q(T) < \infty$, so, by [1, Chapter 1], T is Browder, and in particular is upper semi-Weyl. By Lemma 4.9, S is upper semi-Weyl, so $0\notin \sigma _{uw} (S)$. Therefore, $0\in \sigma _{uw}(T) \bigsqcup \Pi (T)$. Suppose that $\lambda \ne 0$. Then $\frac{1}{\lambda } \in \sigma _{a}(T)$, and hence $\frac{1}{\lambda } \in \sigma _{uw}(T) \bigsqcup \Pi (T)$. This implies that either $\lambda \in \sigma _{uw}(S)$ or $\lambda \in \Pi (S)$, so $\lambda \in \sigma _{uw}(T) \bigsqcup \Pi (T)$. We have shown that $\sigma _a (S)\subseteq \sigma _{uw}(T) \bigsqcup \Pi (T)$ and since the reverse inclusion holds we then have $\sigma _a (S)\subseteq \sigma _{uw}(T) \bigsqcup \Pi (T)$.

Conversely, suppose that S has property $(UW {\scriptstyle \Pi })$. The Drazin inverse of S is $U: = T^2 S=TST$ and the Drazin inverse of U is T, see [1, Chapter 1]. Therefore, from the first part, U inherits property $(UW {\scriptstyle \Pi })$ from S and property $(UW {\scriptstyle \Pi })$ is then transferred from U to T. $\square$