Using Weyl sequences to prove relation between quadratic form and spectral radius

Question

I know that the formula $$\lVert A\lVert=\sup_{\lVert x\lVert=1} \langle x,Ax\rangle$$ holds true for self adjoint operators. While reading Teschl's book I saw a comment that on can prove this formula for normal operators using Weyl sequences. It's clear that $\lVert A\lVert=\sup_{\lambda\in\sigma(A)}|\lambda|$ for a normal operator. Since one of the inequalities is trivial it would remain to show that for a given $\lambda\in\sigma(A)$ one has $$|\lambda|\leq\sup_{\|x\|=1}\langle x,Ax\rangle.$$ By Weyl's criterion I get a sequence $x_n$ with $\|x_n\|=1$ and $\|(A-\lambda)x_n\|\to0$. Are those $x_n$'s sufficient for the inequality above? Any ideas about how to use this sequence to prove above's inequality? I tried to expand the norm but then didn't no what to do about the $\|A x_n\|^2$ term.

Answer 1

Note that $\langle x_n,x_n\rangle=1$. So $$ |\langle Ax_n,x_n\rangle-\lambda|=|\langle (A-\lambda)x_n,x_n\rangle|\leq\|(A-\lambda)x_n\|\to 0 $$