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.
Best Answer
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, $\sigma(U)=\sigma_{\rm ap}(U)$.