Linear resolvent growth test for similarity of a weak contraction to a normal operator
In order to describe the main result of the paper under review, consider a complex separable Hilbert space \(H\) and a bounded linear operator \(T\) on \(H\) with \(\|T\|\leq 1\). Denote by \({\mathbf D}\) the open unit disc in the complex plane and note that for any \(\mu\in{\mathbf D}\) one can define the bounded linear operator \(b_\mu(T)=(1-\bar\mu T)^{-1}(T-\mu)\). Then the main theorem of the present paper says that \(T\) is similar to a normal operator provided that the spectrum \(\sigma(T)\) is different from the closure of \({\mathbf D}\) and the following conditions are satisfied: \[ \sup_{\lambda\in{\mathbf C}\setminus\sigma(T)}\|(T-\lambda)^{-1}\|\text{dist} (\lambda,\sigma(T))<\infty \] and \[ \sup_{\mu\in{\mathbf D}}\text{tr} (1-b_\mu(T)^*b_\mu(T))<\infty, \] where \(\text{tr}(\cdot)\) stands for the usual trace supported by the ideal of trace-class operators on \(H\). The proof is accomplished firstly under the additional assumption that \(T\) has a complete set of eigenvectors. (A basic tool used at this stage is the Sz-Nagy-Foiaş model for completely nonunitary contractions.) The author then derives the result in the general case described above. We conclude by mentioning that the paper is clearly written and to a large extent self-contained.
scientific article; zbMATH DE number 1987644