Show that the point spectrum is non empty. Let, $H$ be a non zero complex Hilbert Space. $U: H \rightarrow H$ be a compact self adjoint linear operator on $H$. Prove that the point spectrum contains atleast one point.
Things I know so far,
Let $ s = \underset{\Vert x \Vert = 1}{\inf}\langle Ux, x \rangle$ and $S = \underset{\Vert x  \Vert = 1}{\sup}\langle Ux, x \rangle$. Then we have spectrum is a proper subset of $[s, S]$. Further, $s,S \in \sigma(U)$.
I am not sure how to proceed with this information. I believe It can be shown that either $s$ or $S$ is in $\sigma_p(U)$.
Kindly help me out. Thanks !

This depends a bit on what background information you have available. Here is one possible path:

*

*The spectrum is nonempty.

*If $\sigma(U)=\{0\}$, then $s=S=0$; this gives you $\langle Ux,x\rangle=0$ for all $x$, which forces $U=0$. Then $0$ is in the point spectrum.

*Otherwise, there exists $\lambda\in\sigma(U)\setminus \{0\}$.

*One shows that $U-\lambda I$ has closed range.

*Then one shows that $\ker(U-\lambda I)$ is nontrivial.

The arguments I know to prove the last two steps are not super complicated, but not trivial either.
