# Real eigenvalues of continuum spectrum of a self-adjoint operator

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?

• What is your question exactly? Continuous spectrum, self-adjoint operators having real eigenvalues and Diract orthonormality are just 3 different things. Can you define them carefully? – Argyll Mar 24 '18 at 0:49
• Asking about objects that are not in the space being considered sounds more religious than scientific. Can you pin this down a bit more? – DisintegratingByParts Mar 24 '18 at 8:51
• 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? – angel leonardo Mar 24 '18 at 21:42
• @DisintegratingByParts it seems that this is a differential equation context. A differential operator $D$ acts on functions, so there is profit to be had from analysing it with the Hilbert space $L^2$, where it might be self-adjoint. I believe the question asks if there exist solutions to the equation $Df=cf$ with non-real $c$ in the case that $D$ can be restricted to an essentially self-adjoint operator on $L^2$. If the solution is strong or weak is probably up to the preference of the asker or answerer :) – s.harp Mar 25 '18 at 1:42
• As an example: $i\frac{d}{dx}$ is essentially self-adjoint on $L^2(\Bbb R)$ and has continuous spectrum. The solutions to the equation $i\frac{d}{dx} f = c f$ are $f(x) = \text{Const}\,e^{-ic x}$, which works for arbitrary (not necessarily real) $c$. – s.harp Mar 25 '18 at 1:46

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.