Convergence of spectrum along with the convergence of the Operator.

Question

This seems to be very interesting result , ie

If operators $\{A_n\}$ in Banach space $B(X)$ and if $A_n \to A$ , $A \in B(X)$in operator norm then $\lambda_n \in \sigma(A_n)$ ie spectrum of $A_n$ then $\lambda_n$ converges to $\lambda$ and $\lambda$ is spectrum of $A$ .

Can you help me to prove that the above statements is infact true ? Thank you .

Answer 1

• If $\lVert A_n-A\rVert_{\mathcal B(X)}\to 0$ and $\lambda_n\to \lambda$, then $\lVert (A_n-\lambda_nI)-(A-\lambda I)\rVert_{\mathcal B(X)}\to 0$.
• The set of non-invertible operators is closed for the norm topology of $\mathcal B(X)$.