Real eigenvalues of continuum spectrum of a self-adjoint operator

Question

Is my understanding that if you assume eigenvectors of a self-adjoint operator are in Hilbert space, then is easy to prove that the eigenvalues must be real. However, it could happen that such eigenvectors and eigenvalues do not exist. For instance it could happen that the spectrum is only continuous.Then the solution to the differential equation Af=cf (A is a linear self-addjoint operator, c is a constant and f is a function) would not be in Hilbert space and c would be a continuous variable. In such a case would c still be part of the real numbers? Does the spectral theorem guarantees it?

Answer 1

See https://proofwiki.org/wiki/Eigenvalues_of_Self-Adjoint_Operator_are_Real.

There are two (in a sense) generalizations of the finite-dimensional spectral theorem that gives us real eigenvalues. Let $A$ be a self-adjoint operator on a Hilbert space $H$. The fully general theorem states that there is a projection-valued measure $E$ such that $E(A)$ is orthogonal to $E(B)$ if $A \cap B = 0$, a sort of orthogonalization criterion.

If we know that $A$ is a compact operator, then we get something nicer: we get a sequence of real eigenvalues $\lambda_n$ converging to zero with a corresponding orthonormal eigenbasis.