Generalized derivation, SVEP, finite ascent, range closure

Abstract

Let $\mathcal{X}$ be an infinite complex Banach space and consider two bounded linear operators
$A,B\in L(\mathcal{X})$. Let $L_{A}\in L(L(\mathcal{X}))$ and $R_{B}\in L(L(\mathcal{X}))$ be the left and the right
multiplication operators, respectively. The generalized derivation $\delta_{A,B} \in L(L(\mathcal{X}))$
is defined by $\delta_{A,B}(X)=(L_{A}-R_{B})(X)=AX-XB$.
In this paper we give some sufficient conditions for $\delta_{A,B}$ to satisfy SVEP, in the main result we prove that if $A\in L(\mathcal{X})$, $B^{*}\in L(\mathcal{X}^{*})$ and $L_{A}R_{B}$ have the SVEP, then SVEP holds for
$\delta_{A,B}$, and we prove that $\delta_{A,B}-\lambda I$ has finite ascent for all complex $\lambda$, for general choices of the operators $A$ and $B$, without using the range kernel orthogonality. This information is applied to prove some necessary and sufficient conditions for the range of $\delta_{A,B}-\lambda I$ to be closed. In \cite[Propostion 2.9]{DDK} Duggal, Djordjevic and Kubrusly proved that, if $asc(\delta_{A,B}-\lambda)\leq 1$, for all complex $\lambda$, and if either (i) $A^{*}$ and $B$ have SVEP or (ii) $\delta^{*}_{A,B}$ has SVEP, then $\delta_{A,B}-\lambda$ has closed range for all complex $\lambda$ if and only if $A$ and $B$ are algebraic operators, we prove using the spectral theory that, if $asc(\delta_{A,B}-\lambda)\leq 1$, for all complex $\lambda$, then $\delta_{A,B}-\lambda$ has closed range, for all complex $\lambda$ if and only if $A$ and $B$ are algebraic operators, without the additional conditions (i) or (ii).