prove $Ax_n - \lambda x_n \to 0$

Question

Given a Hilbert space $H$ over $\mathbb{C}$ and a linear self-adjoint operator $A: H \to H$.

Let $(x_n) \subset H$ with $\|x_n \|= 1$ such that $\langle Ax_n,x_n \rangle \to \lambda$ where $|\lambda| = \| A \|$.

I need to prove $Ax_n - \lambda x_n \to 0$

I started by $|\lambda| = \| A \|=\sup_{\|x\|=1}\langle Ax,x \rangle \geq \langle Ax_n, x_n \rangle = \langle x_n,Ax_n \rangle$ but not sure how it helps. How can we prove this?

Answer 1

We have

$$\|Ax_n-\lambda x_n\|^2=\|Ax_n\|^2+|\lambda|^2-2Re\langle Ax_n,\lambda x_n\rangle$$

Now $\langle Ax_n,\lambda x_n\rangle=\overline{\lambda}\langle Ax_n,x_n\rangle\to|\lambda|^2$. Also, $\|Ax_n\|^2\leq\|A\|^2=|\lambda|^2$.

So, $$0\leq\|Ax_n-\lambda x_n\|^2\leq2|\lambda|^2-2Re(\overline{\lambda}\langle Ax_n,x_n\rangle)\to2|\lambda|^2-2|\lambda|^2=0. $$