For a self-adjoint operator $A$ on a Hilbert space $H$, one has $\sup_{|x|=1}\langle Ax,x \rangle = \max\sigma(A)$.
I want to prove this using the spectral theorem.
My idea is: Let $a = \max\sigma(A)$. Choose for $n \in \mathbb{N}$ a unit-length vector $x_n \in \mathbb{E}((a-\frac{1}{n}, a])H$, where $\mathbb{E}$ is the spectral measure. Then $\langle Ax_n,x_n\rangle$ should converge to $\max\sigma(A)$.
More precisely, $\int_{\sigma(A)}\lambda d\mu_{x_n,x_n} = \int_{\sigma(A)}\lambda d\langle\mathbb{E}_\lambda x_n,x_n\rangle$. If $\lambda \in [c,d]$ with $[c,d] \cap (a-\frac{1}{n}, a] = \emptyset$, then $\mathbb{E}_\lambda x_n = \mathbb{E}_\lambda \mathbb{E}((a-\frac{1}{n}, a])x_n = 0$. It follows from this that $\int_{\sigma(A)}\lambda d\langle\mathbb{E}_\lambda x_n,x_n\rangle = \int_{\sigma(A) \cap (a-\frac{1}{n}, a]}\lambda d\langle\mathbb{E}_\lambda x_n,x_n\rangle$
Why does the last integral converge to $a$? One should show that $\langle\mathbb{E}_\lambda x_n,x_n\rangle$ converges to $1$?
Can someone help me?