Let $A_n$ and $A$ be self-adjoint and $A_n\rightarrow A$ in norm resolvent sense i.e. $R_{A_n}(z)\rightarrow R_A(z)$ converges in norm for some $z\in\mathbb{C}/\mathbb{R}$. Also let $A$ to be bounded. I want to show $|| A_n ||\rightarrow ||A||$.
My Attempt: There is a theorem that states (with above conditions) $$\sigma(A)=\lim_{n\rightarrow\infty} \sigma(A_n) $$ $\lim_{n\rightarrow\infty} \sigma(A_n)$ denotes the set of all $\lambda$ which has a sequence $\lambda_n\in\sigma(A_n)$ converging to $\lambda$. There is another theorem which states for self-adjoint operator $T$ $$\textrm{spectral radius }r(T)=\sup_{\lambda\in\sigma(T)}|\lambda|=\lim_{k\rightarrow\infty}||T^k||^{1/k}=||T||$$ combining these two theorems gave me \begin{align} \lim_{n\rightarrow\infty}||A_n||&=\lim_{n\rightarrow\infty}\sup_{\lambda\in\sigma(A_n)}|\lambda|\stackrel{?}{=}\sup_{\lambda\in\sigma(A)}|\lambda|=||A|| \\ &=\lim_{n\rightarrow\infty}\lim_{k\rightarrow\infty}||A_n^k||^{1/k}\stackrel{?}{=}\lim_{k\rightarrow\infty}||A^k||^{1/k}=||A|| \end{align} I put "?" into the places that i'm not sure.
I have 2 questions related to this: Is my proof correct, if it is why? Is there another way? For example, with using norm resolvent convergence directly instead of using it indirectly.
Thank you for your answer.