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Let $U$ be the bilateral shift operator in $l^2(Z)$, and let $V$ stand for a rotation on an irrational angle $\alpha$ in $L^2(T, \mu)$, where $T$ is a circle with a rotation-invariant Lebesgue measure $\mu$. The spectra of both operators coincide as sets, but the spectrum of $U$ is absolutely continuous, while that of $V$ is pure point. Hence, no unitary equivalence between two operators is possible.

However, intuitively $U$ and $V$ are related in the followng way. Let's extend the action of $V$ to the space of tempered distributions ${\cal S}'(T)$. Then, $V$ acts on the set of vectors $e_n=V^n \delta$ exactly by a bilateral shift (here $\delta$ stands for the Dirac delta at some point of $T$). Unfortunately, there is no obvious way to identify the closure of $\mathrm {span} (e_n)$ in ${\cal S}'(T)$ with $l^2(Z)$. Thus the question is: is it possible make this naive argument rigorous, say by extending the action of $U$ and $V$?

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One difficulty here is that the spectrum of the bilateral shift really is more than countable, and every point has "approximate eigenvectors". – paul garrett May 16 '13 at 19:31
In what sense this is a difficulty? The spectrum of an irrational rotation is also uncountable (it is a unit circle, as well as that of the bilateral shift). It is pure point though, since it coincides with the closure of the (countable) set of eigenvalues. – Pavel Kalouguine May 20 '13 at 18:19

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