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),
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
- ℂ+ :
-
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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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