Convergence of spectrum with multiplicity under norm convergence

Question

This article by Joachim Weidmann claims that, if a sequence $A_n$ of bounded operators in a Hilbert space converges in norm topology, i.e., $\|A_n - A\| \rightarrow 0$, then "isolated eigenvalues $\lambda$ of $A$ of finite multiplicity are exactly the limits of eigenvalues of $A_n$ (including multiplicity)".

Unfortunately, no reference is given. Can somebody give a source for this claim? I have checked Kato, "Perturbation theory for linear operators", where the convergence of the spectrum is given, but I could not find a statement that the eigenvalues converge with the proper multiplicities.

Answer 1

It's not quite stated precisely. This should be more-or-less in Kato.

The spectral projection for the isolated eigenvalue $\lambda$ is $$ P = \dfrac{1}{2\pi i} \oint_\Gamma (z I-A)^{-1}\; dz $$ where $\Gamma$ is a small circle centred at $\lambda$. By assumption, this is a projection of finite rank. It is the limit (in operator norm) of the corresponding integrals with $A$ replaced by $A_n$, which are spectral projections for the part of the spectrum of $A_n$ inside $\Gamma$, and for sufficiently large $n$ those projections will have the same rank as $P$. So $\lambda$ is indeed the limit of eigenvalues of $A_n$ (i.e. for every $\epsilon > 0$, all $A_n$ for $n$ sufficiently large will have eigenvalues within $\epsilon$ of $\lambda$, with total multiplicity the same as $\lambda$).

Conversely, given $\lambda \in \mathbb C$, suppose for every $\epsilon>0$, all $A_n$ for $n$ sufficiently large have eigenvalues within $\epsilon$ of $\lambda$, with total multiplicity $r$ (i.e. the rank of the spectral projection is $r$). Then $\lambda$ is an isolated eigenvalue of $A$ with multiplicity $r$.

Answer 2

I'll give here another proof which makes the (admittedly, strong) additional assumption that the operators are compact and self-adjoint. For such operators, we have a very useful tool in the Courant-Fischer min-max principle, which we will use here in the form $$ \lambda_k(A) = \min_{\dim V=k-1} \max_{x \in V^\bot, \|x\|=1} \langle A x, x \rangle. $$ Here $\lambda_k(A)$ is the $k$-th largest eigenvalue of $A$, and $V$ ranges over all $k-1$-dimensional subspaces of our Hilbert space $H$.

Assume that two operators $A$ and $B$ satisfy $$ \| A - B \| \le \epsilon. $$ Then we see using the Cauchy-Schwarz inequality that $$ | \langle (A-B) x,x \rangle | \le \|(A-B) x\| \|x\| \le \epsilon \|x\|^2, $$ and hence for any $x$ with norm 1, $$ \langle B x,x \rangle - \epsilon \le \langle A x,x \rangle \le \langle B x,x \rangle + \epsilon. $$ By applying the min-max principle from above, we obtain $$ |\lambda_k(A) - \lambda_k(B)| \le \epsilon. $$