Prove that for a bounded self adjoint operator, $\langle Tx,x\rangle \geq 0$ is equivalent to $\sigma(T)\subset [0,\infty)$

Question

Prove that for a bounded self adjoint operator, the following are equivalent:

A: $\langle Tx,x\rangle \geq 0$

B: $\sigma(T)\subset [0,\infty)$

What I have said so far:

Since $T$ is self adjoint, $\sigma_r(T)=\emptyset$ and $\sigma(T)\subset\mathbb R$

$A\implies B$

Assume $\langle Tx,x\rangle \geq 0$

Then for any $\lambda \in\sigma_p(T)$ with eigenvector $x$,

$$0\leq \langle Tx,x\rangle =\langle \lambda x,x\rangle =\lambda\|x\| \implies 0\leq \lambda $$

If $\lambda\in \sigma_c(T)$, then

$$ \overline{ R(T-\lambda I)} = \mathcal H \implies N(T-\lambda I)=\{0\} $$

But I'm not sure where to go from here.

$B \implies A$

I have no clue. Any ideas?

Answer 1

Assume $T=T^{\star}\in\mathcal{L}(H)$, where $H$ is a complex Hilbert Space.

Implication 1: Show $(Tx,x) \ge 0$ for all $x \in H$ implies $\sigma(T)\subseteq [0,\infty)$.
To do this, assume that $(Tx,x) \ge 0$ for all $x \in H$, and let $\lambda < 0$. Then $$ 0 \le -\lambda(x,x) \le ((T-\lambda I)x,x) $$ implies $$ |\lambda|\|x\|^{2} \le \|(T-\lambda I)x\|\|x\|,\\ |\lambda|\|x\| \le \|(T-\lambda I)x\|. $$ Therefore $T-\lambda I$ is injective for $\lambda < 0$, and its range is closed because the inverse is bounded. But the range is dense because $\mathcal{R}(T-\lambda I)^{\perp}=\mathcal{N}(T-\lambda I)=\{0\}$. Therefore $\lambda \in \rho(A)$ for $\lambda < 0$.

Implication 2: Show $\sigma(T)\subseteq [0,\infty)$ implies $(Tx,x) \ge 0$ for all $x \in H$.
There are many techniques for proving this implication, but I'll assume you know that the norm $\|T\|$ is the same as the spectral radius $r_{\sigma}(T)$ for a bounded selfadjoint operator $T$. Therefore, $\sigma(T)\subseteq [0,\|T\|]$. So, $\sigma(T-\|T\|/2)\subseteq [-\|T\|/2,\|T\|/2]$ which then implies that $$ \left\|T-\frac{\|T\|}{2}I\right\| \le \frac{\|T\|}{2}. $$ Consequently, $$ \left|\left(\left(T-\frac{\|T\|}{2}I\right)x,x\right)\right| \le \frac{\|T\|}{2}\|x\|^{2},\\ %% \left|(Tx,x)-\frac{\|T\|}{2}\|x\|^{2}\right| \le \frac{\|T\|}{2}\|x\|^{2},\\ %% -\frac{\|T\|}{2}\|x\|^{2} \le (Tx,x) -\frac{\|T\|}{2}\|x\|^{2} %% \le \frac{\|T\|}{2}\|x\|^{2},\\ \implies 0 \le (Tx,x) \le \|T\|\|x\|^{2}. $$