I am reading through Chapter XIII.16 of Reed and Simon's Methods of Modern Mathematical Physics IV: Analysis of Operators about Schrödinger operators with periodic potentials. I think it is a fairly general statement I do not know why it is true.
The question occurs in the proof Theorem XIII.89(d):
The part that is not so obvious to me is the last sentence. Why is $\tilde{E}$ an eigenvalue of $H(\epsilon)$? I know that if the resolvent $H(\theta_n)$ converges in norm to $H(\epsilon)$ this should be true by a general result, but I do not know whether this is true here or how one would derive this from the properties of $H(\theta)$.
Remark: It was already shown in part (a) of the theorem that the mapping $\theta \mapsto H(\theta)$ is real-analytic on $(0, \pi)$. I assume that this means that $\theta \mapsto \langle \phi, H(\theta) \psi \rangle$ for all $\phi, \psi$. I don't know if this is necessary for the above or if continuity suffices.
Thanks for any help in advance!