Local Spectral Theory for R and S Satisfying RⁿSRⁿ = R^j

Abstract

In this paper, we analyze local spectral properties of operators $R,$ S and $RS$ which satisfy the operator equations $R^{n}S{R}^n=R^j$ and $S^{n}R{S}^n=S^j$ for same integers $j\ge n\ge 0.$ We also continue to study the relationship between the local spectral properties of an operator R and the local spectral properties of $S.$ Thus, we investigate the transmission of some local spectral properties from R to S and we illustrate our results with an example. The theory is exemplified in some cases.

1. Introduction

In this paper, we continue the analysis undertaken in [1,2,3,4,5,6] on the general problem of study the local spectral properties for $R,$ $S,RS$ and $SR\in L\left(X\right)$ in the case R and S satisfy the operator equations $R^{n}S{R}^n=R^j$ for same integers $j\ge n\ge 0.$ Following the procedure of [1], we study the relationship of Dunford property (C) for products $R^{n}S$ and $S{R}^n$ for operator $R^{j-n}\in L\left(X\right)$ which satisfy the operator equations

$$S{R}^n=R^{j-n}\mathrm{for}\mathrm{same}\mathrm{integers}j\ge n\ge 0,$$

(1)

and hence

$$R^{n}S{R}^n=R^j\mathrm{for}\mathrm{same}\mathrm{integers}j\ge n\ge 0.$$

(2)

The paper is organized as follows.

In Section 2, to keep the paper sufficiently self-contained, we collect some preliminary definitions and propositions that are used in what follows. In Section 3, we show some results concerning the transmission of some local spectral properties from R to $S.$ In Section 4, we give an example that plays a crucial role for the theory. The final considerations are given in Section 5.

2. Notation and Complementary Results

A bounded operator $T\in L\left(X\right)$ on a complex infinite dimensional Banach space X is said to have the single valued extension property at ${\lambda }_o\in \mathbf{C}$. In short, T has the SVEP at ${\lambda }_o$, if for every open disc ${\mathbf{D}}_{{\lambda }_o}$ centered at ${\lambda }_o$ the only analytic function $f:{\mathbf{D}}_{{\lambda }_o}\to X$ which satisfies the equation

$$(\lambda I-T)f\left(\lambda \right)=0$$

(3)

is the constant function $f\equiv 0$.

T is said to have the SVEP if T has the SVEP for every $\lambda \in \mathbf{C}$.

To facilitate the reader, we remember that the SVEP is a typical tool of the local spectral theory. If ${\rho }_T\left(x\right)$ denote the local resolvent set of T at the point $x\in X$, defined as the union of all open subsets $\mathcal{U}$ of $\mathbf{C}$ for which there exists an analytic function $f:\mathcal{U}\to X$ that satisfies

$$(\lambda I-T)f\left(\lambda \right)=x\mathrm{for}\mathrm{all}\lambda \in \mathcal{U}$$

(4)

then the local spectrum ${\sigma }_T\left(x\right)$ of T at x is defined by

$${\sigma }_T\left(x\right):=\mathbf{C}\backslash {\rho }_T\left(x\right),$$

and, obviously, ${\sigma }_T\left(x\right)\subseteq \sigma \left(T\right)$, where $\sigma \left(T\right)$ denotes the spectrum of T.

Remark 1.

Let $\lambda \in {\rho }_T\left(x\right)$ and $\mathcal{U}$ denotes an open neighborhood of λ. If $f:\mathcal{U}\to X$ satisfies the equation $(\lambda I-T)f\left(\mu \right)=x$ on $\mathcal{U}$, then ${\sigma }_T\left(f\left(\lambda \right)\right)={\sigma }_T\left(x\right)$ for all $\lambda \in \mathcal{U}$ (see [7], Lemma 1.2.14). Moreover, $0\in {\sigma }_{\lambda I-T}\left(x\right)$ if and only if $\lambda \in {\sigma }_T\left(x\right)$.

Theorem 1.

Let $T\in L\left(X\right)$, X a Banach space. Then, T has SVEP if and only if every $0\ne x\in X$ the local spectrum ${\sigma }_T\left(x\right)$ is non-empty.

Proof. 

See ([7], Proposition 1.2.16). □

The SVEP has a decisive role in local spectral theory it has a certain interest to find conditions for which an operator has the SVEP.

Definition 1.

Let T is a linear operator on a vector space X. The hyperrange of T is the subspace

$$T^{\infty }\left(X\right):=\bigcap _{n\in \mathbf{N}}{T}^n\left(X\right).$$

Generally, $T\left(T^{\infty }\left(X\right)\right)\subseteq T^{\infty }\left(X\right)$, thus we are interested in finding conditions for which $T\left(T^{\infty }\left(X\right)\right)=T^{\infty }\left(X\right)$. For every linear operator T on a vector space X, there corresponds the two chains:

$$\left\{0\right\}=\ker T^0\subseteq \ker T\subseteq \ker T^2\cdots .$$

and

$$X=T^0\left(X\right)\supseteq T\left(X\right)\supseteq T^2\left(X\right)\cdots .$$

The ascent of T is the smallest positive integer $p=p\left(T\right)$, whenever it exists, such that $\ker T^p$ = $\ker T^{p+1}$. If such p does not exist, we let $p=+\infty$. Analogously, the descent of T is defined to be the smallest integer $q=q\left(T\right)$, whenever it exists, such that $T^{q+1}\left(X\right)$ = $T^q\left(X\right)$. If such q does not exist, we let $q=+\infty$.

It is possible to prove that, if $p\left(T\right)$ and $q\left(T\right)$ are both finite, then $p\left(T\right)=q\left(T\right).$ Note that $p\left(T\right)=0$ means that T is injective, and $q\left(T\right)=0$ that T is surjective.

Theorem 2.

If $T\in L\left(X\right)$ and X is a Banach space, then

$$TdoesnothavetheSVEPat0\Rightarrow p\left(T\right)=\infty .$$

(5)

As noted in [1] (Lemma 1.1), the local spectrum of $Tx$ and x may differ only at $0,$ i.e.,

For every $T\in L\left(X\right)$ and $x\in X$, we have

$${\sigma }_T\left(Tx\right)\subseteq {\sigma }_T\left(x\right)\subseteq {\sigma }_T\left(Tx\right)\cup \left\{0\right\}.$$

(6)

Moreover, if T is injective, then

$${\sigma }_T\left(Tx\right)={\sigma }_T\left(x\right)\mathrm{for}\mathrm{all}x\in X.$$

(7)

For every subset $F\subseteq \mathbf{C},$ the analytic spectral subspace of T associated with F is the set

$$X_T\left(F\right):=\{x\in X:{\sigma }_T\left(x\right)\subseteq F\}$$

For every subset $F\subseteq \mathbf{C},$ the global spectral subspace ${\mathfrak{X}}_T\left(F\right)$ consists of all $x\in X$ for which there exists an analytic function $f:\mathbf{C}\backslash F\to X$ that satisfies

$$(\lambda I-T)f\left(\lambda \right)=x\mathrm{for}\mathrm{all}\lambda \in \mathbf{C}\backslash F.$$

(8)

In general, ${\mathfrak{X}}_T\left(F\right)\subseteq X_T\left(F\right)$ for every closed sets $F\subseteq \mathbf{C}.$ The identity $X_T\left(F\right)={\mathfrak{X}}_T\left(F\right)$ holds for all closed sets $F\subseteq \mathbf{C}$ whenever T has SVEP, precisely. T has SVEP if and only if $X_T\left(F\right)={\mathfrak{X}}_T\left(F\right)$ holds for all closed sets $F\subseteq \mathbf{C}.$

Definition 2.

The analytical core $K(\lambda I-T)$ of $\lambda I-T$ is the set

$$K(\lambda I-T):=X_T(\mathbf{C}\backslash \left\{\lambda \right\})=\{x\in X:\lambda \notin {\sigma }_T\left(x\right)\}$$

(9)

The analytic core of an operator T is an invariant subspace, which, in general, is not closed [8].

Definition 3.

An operator $T\in L\left(X\right)$ is said to be upper semi-Fredholm, $T\in {{\Phi }}_{+}\left(X\right),$ if T(X) is closed and the kernel $kerT$ is finite-dimensional. An operator $T\in L\left(X\right)$ is said to be lower semi-Fredholm, $T\in {{\Phi }}_{-}\left(X\right)$ if the range $T\left(X\right)$ has finite codimension.

Definition 4.

An operator $T\in L\left(X\right)$ is said to be Drazin invertible if there exist $C\in \mathsf{Ł}\left(X\right)$ such that

1. 

$T^m\left(X\right)=T^{m+1}C$ for some integer $m\ge 0;$

2. 

$C=T{C}^2$; and

3. 

$TC=CT$

In this case, C is called Drazin inverse of T and the smallest $m\ge 0$ in (4) is called the index $i\left(T\right)$ of T.

3. Operator Equation ${\mathit{R}}^{\mathit{n}}{\mathit{SR}}^{\mathit{n}}={\mathit{R}}^{\mathit{j}}$

As mentioned in the Introduction, in this section, we show some results concerning the transmission of some local spectral properties from R to $S.$

We study the relationship between the local spectral properties of an operator R and the local spectral properties S, if this exists. In particular, we study a reciprocal relationship, analogous to that of (2). We also show that many local spectral properties, such as SVEP and Dunford property (C), are transferred from operator R to S somehow through a bond. While these properties are, in general, not preserved under sums and products of commuting operators, we obtain positive results in the case of our perturbations.

We suppose that $R,S\in \mathsf{Ł}\left(X\right)$ satisfy $R^{n}S{R}^n=R^j$ for some integers $j\ge n\ge 0.$ The case $n=2$ and $j=1$ is studied in [1,9,10]; if $n=j=1$, the operators A and B are relatively regular.

Moreover, if $T\in \mathsf{Ł}\left(X\right)$ is Drazin invertible operator with $i\left(T\right)=k$, then, by (4),

$$T^{2k+1}=T^{k+1}C{T}^{k+1}.$$

Therefore, in this case, $j=2k+1$ and $n=k+1.$

Lemma 1.

For every $x\in X$, we have

$${\sigma }_{R^{j-n}}\left(R^{n}x\right)\subseteq {\sigma }_{S{R}^n}\left(x\right).$$

(10)

Moreover,

$${\sigma }_{S{R}^n}\left(S{R}^{n}x\right)\subseteq {\sigma }_{R^{j-n}}\left(x\right),{\sigma }_{S{R}^n}\left(S{R}^{n}x\right)\subseteq {\sigma }_{R^{j-n}}\left(R^{n}x\right)$$

(11)

Proof. 

Suppose that ${\lambda }_0\in {\rho }_{S{R}^n}\left(x\right)$; then, there exists an open neighborhood ${\mathcal{U}}_0$ if ${\lambda }_0$ and an analytic function $f:{\mathcal{U}}_o\to X$ such that

$$(\lambda I-S{R}^n)f\left(\lambda \right)=x\mathrm{for}\mathrm{all}\lambda \in {\mathcal{U}}_0.$$

(12)

From this, it then follows that

$$\begin{array}{ccc}\hfill R^{n}x& =& R^n(\lambda I-S{R}^n)f\left(\lambda \right)=(\lambda R^n-R^{n}S{R}^n)f\left(\lambda \right)\hfill \\ & =& (\lambda R^n-R^j)f\left(\lambda \right)=(\lambda I-R^{j-n})R^{n}f\left(\lambda \right),\hfill \end{array}$$

for all $\lambda \in {\mathcal{U}}_0.$ Hence, ${\lambda }_o\in {\rho }_{R^{j-n}}\left(R^{n}x\right)$; thus,

$${\sigma }_{R^{j-n}}\left(R^{n}x\right)\subseteq {\sigma }_{S{R}^n}\left(x\right).$$

To show the first inclusion (11), let ${\lambda }_0\in {\rho }_{R^{j-n}}\left(x\right)$; then, there exists an open neighborhood ${\mathcal{U}}_o$ of ${\lambda }_0$ and an analytic function $f:{\mathcal{U}}_o\to X$ such that

$$(\lambda I-R^{j-n})f\left(\lambda \right)=x\mathrm{for}\mathrm{all}{\lambda }_0\in {\mathcal{U}}_0.$$

Consequently,

$$\begin{array}{ccc}\hfill S{R}^{n}x& =& S{R}^n(\lambda I-R^{j-n})f\left(\lambda \right)=(\lambda S{R}^n-S{R}^j)f\left(\lambda \right)\hfill \\ & =& (\lambda S{R}^n-S{R}^{n}S{R}^n)f\left(\lambda \right)=(\lambda SR-{\left[S{R}^n\right]}^2)f\left(\lambda \right)\hfill \\ & =& (\lambda I-S{R}^n)S{R}^{n}f\left(\lambda \right),\hfill \end{array}$$

for all $\lambda \in {\mathcal{U}}_0,$ and since $S{R}^{n}f\left(\lambda \right)$ is analytic, we obtain ${\lambda }_0\in {\rho }_{S{R}^n}\left(S{R}^{n}x\right).$ Hence, this shows the first inclusion of (11). To show the second inclusion, let ${\lambda }_0\in {\rho }_{R^{j-n}}\left(R^{n}x\right)$; then, there exists an open neighborhood ${\mathcal{U}}_o$ of ${\lambda }_0$ and an analytic function $f:{\mathcal{U}}_o\to X$ such that

$$(\lambda I-R^{j-n})f\left(\lambda \right)=R^{n}x\mathrm{for}\mathrm{all}{\lambda }_0\in {\mathcal{U}}_0.$$

Consequently, the argument is similar to that first part. □

Theorem 3.

Suppose that $\mathcal{F}$ is a closed subset of $\mathbf{C}$ and $0\in \mathcal{F}.$ Then, $X_{R^{j-n}}\left(\mathcal{F}\right)$ is closed if and only if $X_{S{R}^n}\left(\mathcal{F}\right)$ is closed.

Proof. 

Suppose that $X_R^j\left(\mathcal{F}\right)$ is closed and let $\left(x_m\right)$ be a sequence of $X_{S{R}^n}\left(\mathcal{F}\right)$ which converges to $x\in X$. Then, for every $m\in \mathbf{N}$, we have ${\sigma }_{S{R}^n}\left(x_m\right)\subseteq \mathcal{F}.$ By (10), we have ${\sigma }_{R^{j-n}}\left(R^n{x}_m\right)\subseteq \mathcal{F}.$ Since $0\in \mathcal{F}$, by (6) where $T=R^{j-n}$, we have ${\sigma }_{R^{j-n}}\left(R^n{x}_m\right)\subseteq {\sigma }_{R^{j-n}}\left(R^j{x}_m\right)\cup \left\{0\right\}\subseteq \mathcal{F}.$ Therefore, $R^j{x}_m\in X_{R^{J-n}}\left(\mathcal{F}\right)$ i.e., $R^{j-n}{R}^n{x}_m\in X_{R^{J-n}}\left(\mathcal{F}\right).$ By [9] (Lemma 2.3), $R^n{x}_m\in X_{R^{J-n}}\left(\mathcal{F}\right)$ and by assumption $X_{R^{j-n}}\left(\mathcal{F}\right)$ is closed. We then have $R^{n}x\in X_{R^{j-n}}\left(\mathcal{F}\right)$, i.e., ${\sigma }_{R^{j-n}}\left(x\right)\subseteq \mathcal{F}.$ By (11),

$${\sigma }_{S{R}^n}\left(S{R}^{n}x\right)\subseteq {\sigma }_{R^{j-n}}\left(x\right)\subseteq \mathcal{F}.$$

Then, $S{R}^{n}x\in X_{S{R}^n}\left(\mathcal{F}\right),$ by [9] (Lemma 2.3) $x\in X_{S{R}^n}\left(\mathcal{F}\right),$ thus $X_{S{R}^n}\left(\mathcal{F}\right)$ is closed. Conversely, suppose that $X_{S{R}^n}\left(\mathcal{F}\right)$ is closed and let $\left(x_m\right)$ be a sequence of $X_{R^{j-n}}\left(\mathcal{F}\right)$ which converges to $x\in X$; then, ${\sigma }_{R^{j-n}}\left(x_m\right)\subseteq \mathcal{F}$ for every $m\in \mathbf{N}.$ By (11), ${\sigma }_{S{R}^n}\left(S{R}^n{x}_m\right)\subseteq \mathcal{F}$, and then $S{R}^n{x}_m\in X_{S{R}^n}\left(\mathcal{F}\right).$ By [9] (Lemma 2.3) $x_m\in X_{S{R}^n}\left(\mathcal{F}\right),$ therefore $x\in X_{S{R}^n}\left(\mathcal{F}\right).$ Hence, ${\sigma }_{S{R}^n}\left(x\right)\subseteq \mathcal{F}.$ Since by (10) ${\sigma }_{R^{j-n}}\left(R^{n}y\right)\subseteq {\sigma }_{S{R}^n}\left(y\right)$ for all $y\in X,$ then, if $y=R^{j-n}x$, we have ${\sigma }_{R^{j-n}}\left(R^{J}x\right)\subseteq {\sigma }_{S{R}^n}\left(R^{j-n}x\right)$. By (6), we have

$$\begin{array}{ccc}\hfill {\sigma }_{R^{j-n}}\left(R^{j-n}x\right)& \subseteq & {\sigma }_{R^{j-n}}\left(R^{j}x\right)\cup \left\{0\right\}\subseteq {\sigma }_{S{R}^n}\left(R^{j-n}x\right)\cup \left\{0\right\}\hfill \\ & \subseteq & {\sigma }_{S{R}^n}\left(S{R}^{j}x\right)\cup \left\{0\right\}\subseteq {\sigma }_{S{R}^n}\left[{\left(S{R}^n\right)}^{2}x\right]\cup \left\{0\right\}\hfill \\ & \subseteq & {\sigma }_{S{R}^n}\left(S{R}^{n}x\right)\cup \left\{0\right\}\subseteq {\sigma }_{S{R}^n}\left(x\right)\cup \left\{0\right\}\subseteq \mathcal{F}\hfill \end{array}$$

i.e., $R^{j-n}x\in X_{R^{j-n}}\left(\mathcal{F}\right).$ Hence, ${\sigma }_{R^{j-n}}\left(R^{j-n}x\right)\subseteq \mathcal{F}$ i.e., $R^{j-n}x\in X_{R^{j-n}}\left(\mathcal{F}\right).$ By [9] (Lemma 2.3) $x\in X_{R^{j-n}}\left(\mathcal{F}\right).$

The following result is inspired by [1] and ([11], Theorem 2.1). □

Lemma 2.

Let $S,R\in L\left(X\right)$ be such that $R^{n}S{R}^n=R^j$ for same integers $j\ge n\ge 0.$ If $R^{j-n}$ has SVEP, then $S{R}^n$ and $R^{n}S$ have SVEP.

Proof. 

By ([12], Proposition 2.1), $S{R}^n$ has SVEP if and only if $R^{n}S$ has Svep. Suppose that $R^{j-n}$ has SVEP at ${\lambda }_0$ and let $f:{\mathcal{U}}_0\to X$ be an analytic function for which $(\lambda I-S{R}^n)f\left(\lambda \right)=0$ for all $\lambda {\mathcal{U}}_0.$ Then, $S{R}^{n}f\left(\lambda \right)=\lambda f\left(\lambda \right).$

$$\begin{array}{ccc}\hfill R^n(\lambda I-S{R}^n)f\left(\lambda \right)& =& \\ \hfill (\lambda R^n-R^j)f\left(\lambda \right)& =& (\lambda I-R^{j-n})R^{n}f\left(\lambda \right)=0.\hfill \end{array}$$

The SVEP of $R^{j-n}$ at ${\lambda }_0$ implies that $R^{n}f\left(\lambda \right)=0$ and hence $S{R}^{n}f\left(\lambda \right)=\lambda f\left(\lambda \right)=0.$ Thus, if $0\notin {\mathcal{U}}_0$, then $f\left(\lambda \right)=0$ for $\lambda \ne 0$ and by continuity $f\left(0\right)=0.$ Therefore, $S{R}^n$ has SVEP at ${\lambda }_0.$ □

We now consider the case where $0\notin \mathcal{F}$

Theorem 4.

Let $\mathcal{F}$ be a closed subset of $\mathbf{C}$ such that $0\notin \mathcal{F}.$ Suppose that $R,S\in \mathsf{Ł}\left(X\right)$ satisfy $R^{n}S{R}^n=R^j$ for some integers $j\ge n\ge 0$ and $R^{j-n}$ has SVEP. If $X_{R^{j-n}}\left(\mathcal{F}\right)$ is closed, then $X_{S{R}^n}\left(\mathcal{F}\right)$ is closed.

Proof. 

Let ${\mathcal{F}}_1:=\mathcal{F}\cup \left\{0\right\}$; by assumption, $X_{R^{j-n}}\left({\mathcal{F}}_1\right)$ is closed. By (3), $X_{S{R}^n}\left({\mathcal{F}}_1\right)$ is closed. By (2), $S{R}^n$ has SVEP; therefore, by ([9], Lemma 1.4), $X_{SR}\left(\mathcal{F}\right)$ is closed. □

Definition 5.

An operator $T\in L\left(X\right)$ is said to have Dunford’s property (abbreviated property $\left(C\right)$) if $X_T\left(\mathcal{F}\right)$ is closed for every closed set $\mathcal{F}\subseteq \mathbf{C}$

It is known that Dunford property (C) entails SVEP for T.

Theorem 5.

Let $S,R\in L\left(X\right)$ be such that $R^{n}S{R}^n=R^j$ for some integers $j\ge n\ge 0.$ If $R^{j-n}$ has the property (C), then $S{R}^n$ and $R^{n}S$ have the property (C).

Proof. 

Suppose that $\mathcal{F}$ is a closed set and $R^{j-n}$ has property (C); then, $R^{j-n}$ has SVEP. If $0\in \mathcal{F}$, by (3) and by assumptions $X_{R^{j-n}}\left(\mathcal{F}\right)$ is closed, it follows that $X_{S{R}^n}\left(\mathcal{F}\right)$ is closed. Similarly, if $0\notin \mathcal{F}$, then by (4) we have that $X_{S{R}^n}(\mathcal{F}\cup \left\{0\right\})$ is closed. Therefore, $S{R}^n$ has property (C). □

We prove that somehow there exists a bond, i.e., SR and RS share Dunford’s property (C) when $R^{n}S{R}^n=R^j$ for same integers $j\ge n\ge 0.$

Definition 6.

An operator $T\in L\left(X\right)$ is said to have property $\left(Q\right)$ if the quasi-nilpotent part $H_0(\lambda I-T)$ of $\lambda I-T$ defined by

$$H_0(\lambda I-T):=\{x\in X:\underset{n\to \infty }{lim\; sup}{\parallel (\lambda I-T)}^{n}x{\parallel }^{1/n}=0\}$$

is closed for every $\lambda \in \mathbf{C}.$

It is known that

$$\mathrm{Property}\left(C\right)\Rightarrow \mathrm{Property}\left(Q\right)\Rightarrow SVEP,$$

and moreover for operator T we have ${\mathfrak{X}}_T\left(\lambda \right)=H_0(\lambda I-T).$

Then, if T has SVEP,

$$X_T\left(\lambda \right)={\mathfrak{X}}_T\left(\lambda \right)=H_0(\lambda I-T).$$

(13)

Every multiplier of a semi-simple commutative Banach algebra has property $\left(Q\right)$, see ([13], Theorem 1.8), in particular every convolution operator $T_{\mu }$, $\mu \in M\left(G\right)$, on the group algebra $L_1\left(G\right)$ has property $\left(Q\right)$, but there are convolution operators which do not enjoy property $\left(C\right)$ (see [7], Chapter 4).

Observe that, if T has property $\left(Q\right)$ and f is an injective analytic function defined on an open neighborhood U of $\sigma \left(T\right)$, then $f\left(T\right)$ also has property $\left(Q\right)$. To see this, recall first that the equality

$${\mathcal{X}}_{f\left(T\right)}\left(\mathcal{F}\right)={\mathcal{X}}_T\left(f^{-1}\left(\mathcal{F}\right)\right)$$

(14)

holds for every closed subset of $\mathbb{C}$ and every analytic function f on an open neighborhood U of $\sigma \left(T\right)$, see ([7], Theorem 3.3.6). Now, to show that $f\left(T\right)$ has property $\left(Q\right)$ amd f is injective, we have to prove that $H_0(\lambda I-f\left(T\right))$ is closed for every $\lambda \in \mathbb{C}$. If $\lambda \notin \sigma \left(f\right(T\left)\right)$, then $H_0(\lambda I-f\left(T\right))=\left\{0\right\}$, while, if $\lambda \in \sigma \left(f\right(T\left)\right)=f\left(\sigma \right(T\left)\right)$, then

$$H_0(\lambda I-f\left(T\right))=X_{f\left(T\right)}\left(\left\{\lambda \right\}\right)={\mathcal{X}}_T\left(f^{-1}\left\{\lambda \right\}\right)=H_0(\mu I-T),$$

where $f\left(\lambda \right)=\mu$, and, consequently, $H_0(\lambda I-f\left(T\right))$ is closed. In particular, considering the function $f\left(\lambda \right):=\frac{1}{\lambda }$, we see that, if T is invertible and has property $\left(Q\right)$, then its inverse has property $\left(Q\right)$. Furthermore, property $\left(Q\right)$ for T implies property $\left(Q\right)$ for $T^n$, for every $n\in \mathbb{N}$.

Theorem 6.

Let $S,R\in L\left(X\right)$ be such that $R^{n}S{R}^n=R^j$ for some integers $j\ge n\ge 0.$ If $R^{j-n}$ has the property $\left(Q\right)$, then $S{R}^n$ has the property $\left(Q\right)$.

Proof. 

Suppose that $R^{j-n}$ has property $\left(Q\right)$. Then, $R^{j-n}$ has SVEP, hence by Lemma 2 $S{R}^n$ has SVEP. Therefore, by (13) and by assumption, $H_0(\lambda I-R^{j-n})=X_{R^{j-n}}\left(\left\{\lambda \right\}\right)$ is closed for every $\lambda \in \mathbf{C}.$ By (13) and (3), $H_0\left(S{R}^n\right)=X_{S{R}^n}\left(\left\{0\right\}\right)$ is closed. Following the procedure of [1], let $0\ne \lambda \in \mathbf{C}$; by ([7], Proposition 3.3.1, part (f)) we have

$$X_{R^{j-n}}(\left\{\lambda \right\}\cup \left\{0\right\})=X_{R^{j-n}}\left(\left\{\lambda \right\}\right)+X_{R^{j-n}}\left(\left\{0\right\}\right)=H({\lambda }_{0}I-R^{j-n})+H_0\left(R^{j-n}\right).$$

Since $R^{j-n}$ is upper semi-Fredholm, the SVEP at 0 implies that $H_0\left(R^{j-n}\right)$ is finite-dimensional (see [8], Theorem 3.18). Then, $X_{R^{j-n}}(\left\{\lambda \right\}\cup \left\{0\right\})$ is closed. By Theorem 5, we then have

$$H_0(\lambda I-S{R}^n)=X_{S{R}^n}\left\{\lambda \right\}$$

is closed, therefore $S{R}^n$ has property $Q.$ □

Following the procedure of [1] (Theorem 3), it is possible to prove the following:

Theorem 7.

Let $S,R\in L\left(X\right)$ be such that $R^{n}S{R}^n=R^j$ for same integers $j\ge n\ge 0.$

1. 

(i) If $0\ne \lambda \in \mathbf{C}$, then $K(\lambda I-R^{j-n})$ is closed if and only $K(\lambda I-S{R}^n)$ is closed, or equivalently $K(\lambda I-R^{n}S)$ is closed.

2. 

(ii) If $R^{j-n}$ is injective, then $K(\lambda I-R^{j-n})$ is closed if and only $K(\lambda I-S{R}^n)$ is closed, or equivalently $K(\lambda I-R^{n}S)$ is closed for all $\lambda \in \mathbf{C}.$

Corollary 1.

Suppose $R^{n}S{R}^n=R^j,$ $S^{j}R{S}^J=S^n,$ for some integers $j\ge n\ge 0$ and $\lambda \ne 0.$ Then, the following statements are equivalent:

1. 

$K(\lambda I-R^j)$ is closed.

2. 

$K(\lambda I-S{R}^n)$ is closed.

3. 

$K(\lambda I-R^{n}S)$ is closed.

4. 

$K(\lambda I-S^n)$ is closed.

When R is injective, the equivalence also holds for $\lambda \ne 0.$

Proof. 

The equivalence of (3) and (4) follows from Theorem 3. Since, the injectivity of R is equivalent to the injectivity of S, the equivalence of (1) and (4) also holds for $\lambda =0$. □

We show now that property $\left(Q\right)$ is also transmitted between operators R into $S.$ Let $S,R\in L\left(X\right)$ be such that $R^{n}S{R}^n=R^j$ for some integers $j\ge n\ge 0.$ If R has the property $\left(Q\right)$ and $R^{j-n}$ has the property $\left(Q\right)$, then $S{R}^n$ has the property $\left(Q\right)$, therefore $S^n$ has the property $\left(Q\right)$, thus S has the property $\left(Q\right)$.

4. Example: Drazin Invertible Operators

In this section, we give an example that plays a crucial role for the theory, of operators $R,S\in \mathsf{Ł}\left(X\right)$ that satisfy the equation $R^{n}C{R}^n=R^j$ for some integers $j\ge n\ge 0.$

In the literature, the concept of invertibility admits several generalizations. Another generalization of the notion of invertibility, which satisfies the relationships of “reciprocity” observed above, is provided by the concept of Drazin invertibility.

The concept of Drazin invertibility has been introduced in a more abstract setting than operator theory [14]. In the case of the Banach algebra $L\left(X\right)$, $R\in L\left(X\right)$ is said to be Drazin invertible (with a finite index) if there exists an operator $S\in L\left(X\right)$ and $n\in \mathbb{N}$ such that

$$RS=SR,SRS=S,R^{n}SR=R^n.$$

(15)

The smallest nonnegative integer $\nu$ such that (15) holds is called the index $i\left(R\right)$ of R. In this case, the operator S is called Drazin inverse of R.

Clearly, in this case,

$$R^{n}S{R}^n=R^{n}SR{R}^{n-1}=R^n{R}^{n-1}=R^j\mathrm{for}\mathrm{same}\mathrm{integers}j=2n-1>n\ge 0.$$

(16)

Clearly, any invertible operator or a nilpotent operator R is Drazin invertible.

5. Conclusions

In this paper we give a proof that the operators S and R share property (Q) and in some modes Dunford’s property (C); we prove further results concerning the local spectral theory of R, S, RS and SR, in particular we show several results concerning the quasi-nilpotent parts and the analytic cores of these operators. It should be noted that these results are established in a very general framework. Therefore, we hope to discuss some aspect in a further paper.