Resolvent algebra of finite rank operators

Abstract

Let ${\mathscr {H}}$ be a Hilbert space. Suppose that $A\in {\mathbb {B}}({\mathscr {H}})$ and the operators $I+mA$ are invertible for all integers $m \ge 1$. We characterize the resolvent algebra

$$\begin{aligned} R_A:= \left\{ T \in {\mathbb {B}}({\mathscr {H}}) : \sup _{m \ge 1}\Vert (I+mA)T(I+mA)^{-1}\Vert < \infty \right\} , \end{aligned}$$

when A is a finite rank operator with $\mathrm{cov}(A)\ne 0$. Moreover, we determine the elements of $\{A\}'$ and $R_A^{c_0}$ and prove that $R_A = R_A^c = \{A\}'\oplus R_A^{c_0}$, where $\{A\}'$ is the commutant A and both $R_A^c$ and $R_A^{c_0}$ are subclasses of $R_A$ defined by

$$\begin{aligned} R_A^c = \left\{ T \in R_A : \lim _{m \rightarrow \infty } \Vert (I+mA)T(I+mA)^{-1}\Vert ~ \mathrm {exists} \right\} \end{aligned}$$

and

$$\begin{aligned} R_A^{c_0} = \left\{ T \in R_A^c : \lim _{m \rightarrow \infty } \Vert (I+mA)T(I+mA)^{-1}\Vert =0 \right\} . \end{aligned}$$

We provide a counterexample showing that $R_A^c= \{A\}'\oplus R_A^{c_0}$ is not true for some compact operators.