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.