Say $\left\{A_n\right\}$ is a sequence of bounded self-adjoint operators on a separable Hilbert space, converging in strong operator topology to a (bounded, self-adjoint) operator $A$. Denote the spectrum of $A_n$ by $\sigma_n$, and the spectrum of $A$ by $\sigma$. Under what conditions does it follow that $\sigma_n\rightarrow\sigma$ in Hausdorff metric? Any references will be appreciated.
Functional Analysis – Convergence of Spectra Under Strong Convergence of Operators
functional-analysisspectral-theory
Best Answer
FWIW: The best result that comes close to what you seek, that I know, is theorem 50.16 in
which is an extension of a theorem of Rellich that you can also find in
It says: For a smooth curve of unbounded self-adjoint operators in a Hilbert space $t \to A$, with common domain of definition and compact resolvent, the eigenvalues of $A(t)$ may be arranged increasingly ordered in such a way that they become $C^1-$ functions. If the curve is real analytic, then the eigenvalues and eigenvectors can be chosen smoothly in t.
A smooth curve of unbounded operators means that $t \to (A(t)u, v)$ is smooth for all $u, v \in H$ vectors in the Hilbert space, and $u$ in the domain of definition of $A(t)$, of course.
On the other hand, there is a theorem that approaches the problem from a different angle in
chapter X.7 "Perturbation Theory", corollary 3: For $E_n, E$ being the resolutions of the identity of the normal operators $T_n, T$ with $T_n \to T$ in the strong operator topology, we have: If $E$ vanishes on the boundary of the Borel set $\sigma$, then $E_n(\sigma) \to E(\sigma)$ in the strong operator topology.
I haven't thought if it is possible to use this result to get closer to an answer to your question, though :-)
HTH.