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.

  • $\begingroup$ There is an analogous result for normal bounded operators on a Hilbert space converging in norm topology. A proof is given e.g. in Aupetit's A primer on spectral theory. Aupetit also lists a couple of similar results, but I don't recall whether he treats your problem, too. $\endgroup$
    – lvb
    May 23, 2011 at 5:29
  • $\begingroup$ I don't think there's much chance for a positive result to hold without much stronger hypotheses. For instance, choose an orthonormal basis and let $P_{n}$ be the orthogonal projection onto the span of the first $n$ basis vectors and let $I$ be the identity operator. Of course, each $P_{n}$ is a bounded self-adjoint operator, and so is $I$. Then $P_{n} \to I$ in the strong operator topology by Bessel's inequality. However, the spectrum of each $P_{n}$ is $\{0,1\}$ and the spectrum of $I$ is $\{1\}$. $\endgroup$
    – t.b.
    May 23, 2011 at 7:05
  • $\begingroup$ Could you elaborate a bit on what kinds of conditions you expect? The question, as it is, is a bit too open-ended for my taste. $\endgroup$
    – t.b.
    May 23, 2011 at 7:09
  • $\begingroup$ @lvb: Thanks for your comment. Will look it up. @Theo: I know that this isn't true in general, and I do expect some strong (but hopefully still usable) conditions. I think my problem can be reduced to the following: With the hypothesis of the original problem, suppose $x\in\sigma_n$ for all $n$. Under what conditions can we conclude that $x\in \sigma$? Here's something I dug up in that direction: jstor.org/stable/2156538. Are you aware of other results in this direction? Thank you! $\endgroup$
    – user2093
    May 23, 2011 at 7:27
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    $\begingroup$ I don't know if this helps at all: www2.im.uj.edu.pl/actamath/PDF/34-153-163.pdf $\endgroup$
    – cfh
    Apr 16, 2015 at 8:08

3 Answers 3


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 :-)


  • $\begingroup$ Hi Tim. Thank you for your answer. It isn't obvious to me how one could apply this to the general question. Perhaps the question is far too general to begin with. I think I have some partial answers in special cases, which should suffice for my means. Thanks again! $\endgroup$
    – user2093
    May 23, 2011 at 11:04

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.

  • $\begingroup$ Yes, norm convergence does imply convergence of spectra in the Hausdorff metric. It is of course strong convergence that raises the nontrivial question. By the way, there might not even be "honest" eigenvalues. Anyway, that doesn't affect the result you posted. $\endgroup$
    – user2093
    Apr 15, 2015 at 19:59
  • $\begingroup$ @William: Maybe this result is trivial to you; it's not to me. It certainly gives one possible answer to the question "which additional assumptions do I need to make this true". $\endgroup$
    – cfh
    Apr 16, 2015 at 7:55

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


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.

  • $\begingroup$ This is nice! Thanks! $\endgroup$
    – user2093
    Aug 19, 2015 at 2:12

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