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Linear resolvent growth of rank one perturbation of a unitary operator does not imply its similarity to a normal operator

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Abstract

The main result of this paper is that for any unitary (selfadjoint) operatorU with non-trivial absolutely continuous part of the spectrum, there exists a rank-one perturbationK = ba* = (a)b such that the operatorT = U + K satisfies the Linear Resolvent Growth condition (LRG),

$$||(\lambda I - T)^{ - 1} || \leqslant C/dist(\lambda ,\sigma (T)), \lambda \in \mathbb{C}\backslash \sigma (T),$$

its spectrum lies on the unit circle T (on the real line ℝ), butT is not similar to a normal operator. This contrasts sharply with the result of M. Benamara and the first author [1] that if a finite rank perturbationT = U + K of a unitary operator is acontraction (¦T¦< 1), then it is similar to a normal operator if and only if it satisfies (LRG) and its spectrum does not cover the unit disc D.

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Abbreviations

D:

Open unit disk in the complex plane ℂ, D := {z ∃ ℂ :¦z¦ < 1

T:

Unit circle, T:=∂D = {z ∃ ℂ : ¦z¦ = 1

+ :

Upper half-plane, ℂ+ := {z ∃ ℂ : Imz > 0

R T λ :

Resolvent of the operatorT, R T λ := (λI-T)-1

a*:

If a is a vector in a Hilbert space, then a* := (·,a) stands for the linear functional (operator)f → (f, a)

ba* :

Stands for the operatorba* := b(·,a), i.e., for the operatorf → (f, a)b

H2, H 2- :

Hardy class of analytic and antianalytic functions in the disk or in the upper half-plane

P+, P- :

Riesz projections, i.e., orthogonal projections onto H2 and Hi, respectively

m:

Normalized (m(T) = 1) Lebesgue measure on T

J:

Length of the arc (interval) I

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Correspondence to Nikolai Nikolski.

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S. Treil is partially supported by the NSF grant DMS 9970395.

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Nikolski, N., Treil, S. Linear resolvent growth of rank one perturbation of a unitary operator does not imply its similarity to a normal operator. J. Anal. Math. 87, 415–431 (2002). https://doi.org/10.1007/BF02868483

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  • DOI: https://doi.org/10.1007/BF02868483

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