Convergence of spectra under strong convergence of operators

Question

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.

Answer 1

FWIW: The best result that comes close to what you seek, that I know, is theorem 50.16 in

• Kriegl, Michor: "The Convenient Setting of Global Analysis",

which is an extension of a theorem of Rellich that you can also find in

• Kato: "Perturbation Theory of Linear Operators", chapter 7, theorem 3.9

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

• Dunford, Schwartz: "Linear Operators, Part II"

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.

Answer 2

Here is another partial answer. From the book by Kato, "Perturbation theory for linear operators", this is Theorem 4.10 in Chapter 5 (p. 291); I'm paraphrasing a bit:

Let $T$ be selfadjoint and $A$ be selfadjoint and bounded operators in a Hilbert space. Then $$ \operatorname{dist}(\Sigma(T + A), \Sigma(T)) \le \| A \|. $$

Here $\Sigma(T)$ denotes the spectrum of $T$.

In other words, if you replace the strong operator topology by the norm topology, so you have $\| T - T_n \| \le \epsilon$, then you know that for each element $\sigma$ of the spectrum of $T$, there is at least one element $\mu$ of the spectrum of $T_n$ such that $|\sigma - \mu| \le \epsilon$ - and vice versa.

Answer 3

Even if it might be to late for answering. The following work provides a characterization of the convergence of the spectrum with respect to the Hausdorff metric. The continuity is satisfied whenever the norms of all polynomials (up to degree 2) of the operator behave continuous. One could do much better: In particular quantitative estimates are provided. Even there it can be read as a characterization.

"Continuity of the spectrum of a field of self-adjoint operators"

by Siegfried Beckus and Jean Bellissard

http://arxiv.org/abs/1507.04641

A tool to prove this continuity is the theory about continuous fields of C*-algebras. In the paper the result can be read also as follows: The spectrum of a field of bounded self-adjoint operators behaves continuous if and only if the related field of C*-algebras is continuous.