Showing that $\inf_{\|x\|=1}\langle Tx,x\rangle$ and $\sup_{\|x\|=1}\langle Tx,x\rangle$ are eigenvalues of $T$ (in particular when they are $0$)

Question

Let $T$ be a compact self-adjoint operator on a Hilbert space $H$. I'd already proved that $m=\inf_{\|x\|=1}\langle Tx,x\rangle$ and $M=\sup_{\|x\|=1}\langle Tx,x\rangle$ are elements of $\sigma(T)$ and that $m=\inf\sigma(T)$ and $M=\sup\sigma(T)$. Since $T$ is compact, we have $$\tag{$\star$} \sigma(T)=\{0\}\cup \{\lambda\in\Bbb K:\lambda~\text{is an eigenvalue of}~T\}.$$ I aim to show that $m$ and $M$ are the smallest and highest eigenvalues of $T$, respectively. When $m\neq0$ and $M\neq0$, we are done by $\star$. How can I show that these values are eigenvalues of $T$ when one of them is $0$?

Answer 1

Assume that $$ m\|x\|^2 \le \langle Tx,x\rangle \le M\|x\|^2. $$ Define $$ [x,y] = \langle (MI-T)x,y\rangle. $$ This is a pseudo inner product on $H$ because $\langle (MI-T)x,x\rangle$ is non-negative, but not necessarily strictly positive for $x\ne 0$. As such, the Cauchy-Schwarz inequality holds: $$ |[x,y]|^2 \le [x,x][y,y]. $$ Let $y=(MI-T)x$: \begin{align} \|(MI-T)x\|^4 &\le [x,x]\langle (MI-T)^2x,(MI-T)x\rangle \\ &\le [x,x](M-m)\langle(MI-T)x,(MI-T)x\rangle \\ & = (M-m)[x,x]\|(MI-T)x\|^2. \end{align} Therefore, $$ \|(MI-T)x\|^2 \le (M-m)[x,x]. $$ Suppose there is a sequence of unit vectors $\{ x_n \}$ such that $\langle Tx_n,x_n\rangle \rightarrow M$. Then $[x_n,x_n]\rightarrow 0$, which forces $(MI-T)x_n\rightarrow 0$. By compactness, there is a subsequence $\{ x_{n_k} \}$ such that $\{ Tx_{n_k} \}$ converges. Therefore, assuming $M \ne 0$, it follows that $\{ x_{n_k} \}$ converges to some $x$ because $(MI-T)x_{n_k}\rightarrow 0$, $Tx_{n_k}$ converges, and $M\ne 0$. And $\|x\|=1$ because $\|x_n\|=1$. Finally, $(MI-T)x=\lim_n (MI-T)x_{n_k}=0$.