Convergence of the spectrum under norm resolvent convergence

Question

Suppose $\{A_n\}$ is a sequence of self-adjoint operators in a Hilbert space $\mathcal H$, and $A$ is a self-adjoint operator, with $A_n \to A$ in norm resolvent sense.

Since $A_n \to A$ in strong resolvent sense also, if $\lambda \in \sigma(A)$ then there exists $\lambda_n \in \sigma(A_n)$ for all $n$, such that $\lambda_n \to \lambda$.

Since $A_n \to A$ in norm resolvent sense, if $\rho \not \in \sigma(A)$ then there exists $N$ such that $\rho \not \in \sigma(A_n)$ for all $n > N$.

Suppose now that we have $\lambda_n \in \sigma(A_n)$ for all $n$, and $\lambda_n \to \lambda$ for some $\lambda$. How can we see that $\lambda \in \sigma(A)$?

Answer 1

Assuming you mean that $\|A_n - A\| \to 0$, suppose that $\lambda \notin \sigma(A)$, then $(A-\lambda I)$ is invertible. The set of invertible elements is open, so $\exists \epsilon > 0$ such that if $$ \|B - (A-\lambda I)\| < \epsilon $$ then $B$ must also be invertible. Now choose $B$ to be of the form $A_n - \lambda_n I$ for large enough $n$. This would contradict the fact that $\lambda_n \in \sigma(A_n)$.

Answer 2

Note that for a self-adjoint operator the norm of the resolvent equals the distance to the spectrum: $\|(A-z)^{-1}\|=\mathrm{dist}(z,\sigma(A))$. Hence we have $\lambda\not\in\sigma(A)$ if and only if $\|(A-(\lambda+\mathrm{i}))^{-1}\|<1$. Now if you have a sequence $\lambda_n\in\sigma(A_n)$ such that $\lambda_n\to\lambda \not\in\sigma(A)$, then $\|(A_n-(\lambda+\mathrm{i}))^{-1}\| \to \|(A-(\lambda+\mathrm{i}))^{-1}\|<1$. Implying $\|(A_n-(\lambda_n+\mathrm{i}))^{-1}\|<1$ for $n$ sufficiently large contradicting $\lambda_n\in \sigma(A_n)$.

See Chapter 6 in my book for more on this topic: https://www.mat.univie.ac.at/~gerald/ftp/book-schroe/