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Let $A$ be a bounded linear operator in a Hilbert space $H$.

I had the misconception that the continuous spectrum of $A$ would necessarily have some "continuous" appearance: an interval, a union of intervals, or something like that. This is false as the following operator shows:

$Vf(t)=\int_0^t f(s)\, ds,\qquad f \in L^2([0, 1])$

(see Wikipedia) the spectrum of this operator is purely continuous and is reduced to $\{0\}$.

However this operator is not self-adjoint. This leads to the question:

Question Is it true that, if $A$ is self-adjoint and $0$ is an isolated spectral value, then $0$ is not in the continuous spectrum (and so it is an eigenvalue)?

This is true if $A$ is compact, I believe: in that case, if $0$ is an isolated spectral value then $A$ only has a finite number of eigenvalues and so, by Hilbert-Schmidt theorem, it is a finite rank operator. Then any spectral value is an eigenvalue. But what about the general case? I suspect that in the general setting things are not so easy.

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    Yes, sorry for the deletion of the comment and thanks for clarifying, but you had me confused for a moment (and I now really must run, but I wanted to clear it up): The point is of course that eigenvectors to distinct eigenvalues are orthogonal, so you can choose an orthonormal system of eigenvectors, and an orthonormal system can at most be countable. As I said before, the names *point spectrum* and *continuous spectrum* derive from how these parts of the spectrum *typically* look like. It does not describe the spectrum accurately, as you show with your examples.2011-09-10

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