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Let $A$ be a bounded, unitary operator in a Hilbert space $H.$ Is it true that the approximate spectrum of $A$ equals the spectrum of $A$?

My intuition says no. I know that the spectrum lies on the unit circle but I can’t find a way to show they are not equal or show they are equal. Can anyone help? Thanks.

  • wikipedia has a hint how to prove this: https://en.wikipedia.org/wiki/Spectrum_(functional_analysis)#Approximate_point_spectrum – daw Dec 01 '22 at 06:51
  • Every point $\lambda$ which belongs to the boundary of $\sigma(T)$ belongs to the approximate spectrum. In case $U$ is unitary, we have $\sigma(U)=\partial\sigma(U),$ hence the conclusion follows. – Ryszard Szwarc Dec 01 '22 at 15:11

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It is true in general that any point in $\partial \sigma(U)$ is in the approximate point spectrum.

In the case of a unitary, though, things can be done more concretely. A unitary $U$ is normal. Because of this, the residual spectrum of $U$ is empty, which implies that $\sigma(U)=\sigma_{\rm ap}(U)$. Indeed, as $U-\lambda I$ is normal and $\lambda$ is not an eigenvalue, then $$ \{0\}=\ker (U-\lambda I)=\ker(U-\lambda I)^*=\operatorname{ran}(U-\lambda I)^\perp. $$ So the range of $U-\lambda I$ is dense, which implies that $\lambda$ is not in the residual spectrum. As the full spectrum is the union of the approximate and residual spectra, $\sigma(U)=\sigma_{\rm ap}(U)$.

Martin Argerami
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