Proving property of spectrum of compact self-adjoint operator

Question

I need to prove that, if $T$ is a compact, self-adjoint operator on a Hilbert space $\mathsf H$, and $\sigma_p(T)$ is its point spectrum, then $\sigma_p(T) \neq \varnothing$ and $$\sup_{\lambda \in \sigma_p(T)} |\lambda| = \max_{\lambda \in \sigma_p(T)} |\lambda|= \|T \|. $$

Surely $\sup_{\lambda \in \sigma_p(T)} |\lambda| \leq \|T\|$, as if $\psi_\lambda \in \mathsf E_\lambda[T]$ (the $\lambda$-eigenspace of $T$) and $\|\psi_\lambda\| = 1$, then by Cauchy-Schwarz

$$\|T\| \geq |(\psi_\lambda,T\psi_\lambda)| = |(\psi_\lambda,\lambda\psi_\lambda)|= |\lambda| $$

To show (a) equality, and (b) that the $\sup$ is actually a $\max$, it suffices to exhibit $\lambda \in \sigma_p(T)$ such that $|\lambda| = \|T\|$. I know the following facts:

1. $T$ is self-adjoint, so $$ \|T\| = \sup_{\|\psi\|=1} |(\psi,T\psi)|,$$ hence there exists a sequence $\{\psi_n\}\subset \mathsf H$ such that $\lim_{n\to\infty}|(\psi_n,T\psi_n)| = \|T\|$. Fix one such that $\|\psi_n\| = 1$.
2. $T$ is bounded, so that $\|T\psi\| \leq \|T\|\|\psi\|$ for all $\psi\in\mathsf H$.
3. $T$ is compact and $\{\psi_n\}$ is a bounded sequence, so $\{T\psi_n\}$ admits a convergent subsequence $\{T\psi_{n_k}\}$.
4. The sequence $\{T\psi_n - \|T\|\psi_n\}$ converges to $0$: $$\begin{split} \lim_{n\to\infty} \big\| T\psi_n - \| T\| \psi_n\big\|^2 &= \lim_{n\to\infty} \big(T\psi_n - \| T\| \psi_n,T\psi_n - \| T\| \psi_n\big) \\ &= \lim_{n\to\infty} \left(\| T\psi_n\|^2 + \| T\|^2 -2\| T\| (\psi_n, T\psi_n) \right) \\ &\leq \lim_{n\to\infty} \left(2\| T\|^2 - 2\| T\| (\psi_n,T\psi_n) \right) \\ &= 2\| T\|^2 - 2\| T\|^2 = 0. \end{split}$$

These facts allow me to conclude that $\{\psi_{n_k}\}$ is also convergent. Indeed, we may rewrite $$\psi_{n_k} = \frac{T\psi_{n_k} - T\psi_{n_k} + \|T\|\psi_{n_k}}{\|T\|}, $$ so that, by the triangle inequality, the continuity of the norm, and Fact 4, $$\left\|\lim_{k\to\infty}\psi_{n_k}\right\| = \lim_{k\to\infty} \|\psi_{n_k}\| \leq \lim_{k\to\infty}\frac{\|T\psi_{n_k}\|}{\|T\|} + \lim_{k\to\infty}\frac{\|T\psi_{n_k} - \|T\|\psi_{n_k}\|}{\|T\|} = \frac 1 {\|T\|} \left\|\lim_{k\to\infty} T\psi_{n_k}\right\|, $$ where the first limit exists by Facts 3 and 4 combined, and the last limit exists by Fact 3. Setting $\hat\psi := \lim_{k\to\infty}\psi_{n_k}$, we may rewrite, by the continuity of $T$, $$ \|T\| \| \hat \psi\| \leq \|T\hat\psi\|; $$ Fact 2 implies equality. Comparing norms, seeing that $T=T^*$, we have that $$ T\hat \psi = \pm \|T\| \hat \psi, $$ and hence we have found a $\lambda$-eigenvector $\hat\psi$ such that $|\lambda| = \|T\|$. This implies both that $\sigma_p(T) \neq \varnothing$ and the $\sup$ is actually a $\max$.

Is this proof correct? (It is a slight rewriting of the one found in my book.)

Answer 1

A different approach: the spectrum of $T$ is a compact set so there exists $\lambda$ in the spectrum with $|\lambda|=\|T\|$ (because spectral radius is same as the norm). If $T \neq 0$ then $\lambda \neq 0$. For a compact self-adjoint operator non-zero points in the spectrum beliong to the point spectrum so $\lambda \in \sigma_p(T)$.