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It was shown in [2] that if T is a contraction in a Hilbert space with finite defect (‖T‖ ≤ 1, rank(I−T ∗T ) <∞), and its spectrum σ(T ) doesn’t coincide with the closed unit disk D, then the following Linear Resolvent Growth condition ‖(λI − T )−1‖ ≤ C dist(λ, σ(T )) , λ ∈ C\σ(T ), implies that T is similar to a normal operator. The condition rank(I − T ∗T ) < ∞ characterizes how close is T to a unitary operator. A natural question arises about relaxing this condition. For example, it was conjectured in [2] that one can replace the condition rank(I − T ∗T ) <∞ by I − T ∗T ∈ S1, where S1 denotes the trace class. In this note we show that this conjecture is not true, moreover it is impossible to replace the condition rank(I − T ∗T ) < ∞ by any reasonable condition of closedness to a unitary operator. For example, we construct a contraction T (i. e. ‖T‖ ≤ 1), σ(T ) 6= D, satisfying I − T, I − T ∗T, I − TT ∗ ∈ S := ∩p>0Sp, where Sp stands for the Schatten–von-Neumann class, satisfying the above Linear Resolvent Growth condition but not similar to a normal operator.

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