Spectrums of converging operators

Question

Let $T_n, T$ be self-adjoint and compact operators on a Hilbert space $H$ s.t. $T_n \rightarrow T$.

When can we say that for any $\lambda \in \sigma(T)$ there exists a sequence $\lambda_n \in \sigma(T_n)$ s.t. $\lambda_n \rightarrow \lambda$ as $n \rightarrow \infty$?

Any reference for these type of results and proofs? Thank you.

Answer 1

Assume that $H$ is not finite dimensional. If $\lambda = 0$, you can take $\lambda_n \equiv 0$. If $\lambda \neq 0$, it is an isolated point of $\sigma(T)$. Let $\delta > 0$ and assume by contradiction that $\sigma(T_n) \cap B(\lambda,\delta) \neq \emptyset$ only for finitely many $n \in \mathbb{N}$. Then for large enough $n$ we have

$$ \rho((T_n - \lambda I)^{-1}) = \sup_{\mu \in \sigma(T_n)} \left| \frac{1}{\mu - \lambda} \right| \leq \frac{1}{\delta}. $$

Let us write

$$ T - \lambda I = (T_n - \lambda I) + (T - T_n). $$

We see that $T - \lambda I$ is a small perturbation of the invertible operator $T_n - \lambda I$. If $n$ is large enough so that $\| T_n - T \| < \delta$ then we have $$\|T_n - T\| < \delta \leq \frac{1}{\rho(T_n - \lambda I)^{-1})} = | (T_n - \lambda I)^{-1} \|^{-1}$$

and so the regular Neumann series argument shows that $T - \lambda I$ is invertible, a contradiction.