This should be a trivial question for mathematicians but not for typical physicists.


I know that the spectrum of a linear operator on a Banach space splits into the so-called "point," "continuous" and "residual" parts [I gather that no boundedness assumption is needed but I could be wrong]. I further know that the point spectrum coincides with the set of eigenvalues of the operator. It seems from the terminology that the point spectrum is a discrete set of isolated point and that the eigenvalues cannot form a continuum. But I haven't been able to find a clear statement in a math reference about this.


Actually, I'm mostly interested in self-adjoint operators on a Hilbert space; so a simpler version of my question would be: Can a self-adjoint operator have a continuous set of eigenvalues? And if yes, under what conditions do the eigenvalues have to be discrete?


I appreciate any help.



Can a self-adjoint operator have a continuous set of eigenvalues?
Eigenvectors for different eigenvalues of a self-adjoint operator are orthogonal. In a separable Hilbert space, any orthogonal set is countable. So a self-adjoint operator on separable Hilbert space has only countably many eigenvalues. (As noted, this does not mean the spectrum is countable.)




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Proof for "Eigenvectors for different eigenvalues of a self-adjoint operator are orthogonal".  

Let $\alpha \ne \beta$ be eigenvalues of the self-adjoint operator $A$. They are real. There exist nonzero eigenvectors $x, y$ so that $Ax=\alpha x, Ay=\beta y$. We must show $x \perp y$.


Compute
$$
\alpha\langle x,y\rangle =
\langle \alpha x, y\rangle =
\langle Ax, y\rangle =
\langle x, Ay\rangle =
\langle x, \beta y\rangle =
\beta\langle x,y\rangle
$$
Then
$$
(\alpha-\beta)\langle x,y\rangle = 0
$$
But $\alpha \ne \beta$, so $\alpha - \beta \ne 0$ and thus
$$
\langle x,y\rangle = 0,
$$
so $x \perp y$ as claimed.


