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Linear resolvent growth of a weak contraction does not imply its similarity to a normal operator.

S. Kupin and S. Treil

It was shown by N. Nikolski and N. Benamara that if T is a contraction in a Hilbert space with finite defect ($\Vert T\Vert\le 1$, $\operatorname{rank} (I- T^*T) <\infty$), and its spectrum $\sigma(T)$ doesn't coincide with the closed unit disk $\overline{\mathbb {D}}$, then the following Linear Resolvent Growth condition

$$\Vert(\lambda I - T)^{-1} \Vert\le\frac{C}{\operatorname{dis... ...da,\sigma(T))},\ \lambda\in\mathbb {C}\backslash \sigma(T),$$

implies that T is similar to a normal operator.

The condition $\operatorname{rank} (I- T^*T) <\infty$ characterizes how close is T to a unitary operator. A natural question arises about relaxing this condition. For example, it was conjectured that one can replace the condition $\operatorname{rank} (I- T^*T) <\infty$ by $I - T^*T\in \mathfrak {S}_1$, where $\mathfrak {S}_1$ denotes the trace class.

In this note we show that this conjecture is not true, moreover it is impossible to replace the condition $\operatorname{rank} (I- T^*T) <\infty$ by any reasonable condition of closedness to a unitary operator. For example, we construct a contraction T (i. e. $\Vert T\Vert\le 1$), $\sigma(T)\ne \overline{\mathbb {D}}$, satisfying $I-T,\,I-T^*T, \, I- TT^*\in \mathfrak {S}:= \cap_{p\gt} \mathfrak {S}_p$, where $\mathfrak {S}_p$ stands for the Schatten-von-Neumann class, satisfying the above Linear Resolvent Growth condition but not similar to a normal operator.
 

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• About this document ...

9/17/1999