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?


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, $$ \left|\left(\left(T-\frac{\|T\|}{2}I\right)x,x\right)\right| \le \frac{\|T\|}{2}\|x\|^{2},\\ %% \left|(Tx,x)-\frac{\|T\|}{2}\|x\|^{2}\right| \le \frac{\|T\|}{2}\|x\|^{2},\\ %% -\frac{\|T\|}{2}\|x\|^{2} \le (Tx,x) -\frac{\|T\|}{2}\|x\|^{2} %% \le \frac{\|T\|}{2}\|x\|^{2},\\ \implies 0 \le (Tx,x) \le \|T\|\|x\|^{2}. $$

  • $\begingroup$ Why does $\sigma(T - \| T \| / 2) \subset [- \| T \| / 2, \| T \| / 2 ]$ follow from $\sigma(T) \subset [0, \| T \| ]$? $\endgroup$ – Viktor Glombik Oct 9 '19 at 17:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.