Convergence of spectrum with multiplicity under norm convergence 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.
 A: 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 $r$.
A: 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
$$
  |\lambda_k(A) - \lambda_k(B)| \le \epsilon.
$$
