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I'm stuck on an exercise about the spectrum of a selfadjoint operator on a Hilbert space.

The problem is the following:

Let $(X,\langle \cdot, \cdot\rangle)$ a Hilbert space and let $A \in B(H)$ a selfadjoint operator such that $\langle Ax,x \rangle \geq 0$ for every $x \in X$.

a) If $m=\inf_{\|x\|=1}\langle Ax,x \rangle$, prove that $\lambda<m \Rightarrow \lambda \in \rho(A)$.

b) If $M=\sup_{\|x\|=1}\langle Ax,x \rangle$, show that $\sigma(A) \subseteq [m,M]$

any help is appreciated. Thanks

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Let $N=\{ (Ax,x) : \|x\|=1\}$. Suppose $\lambda \notin N^c$ (closure of $N$) so that there exists $\delta$ such that the following holds whenever $\|x\|=1$: $$ 0 < \delta \le |(Ax,x)-\lambda|=|((A-\lambda I)x,x)|\le \|(A-\lambda I)x\|\|x\| = \|(A-\lambda I)x\| $$ Then $\|(A-\lambda I)x\| \ge \delta \|x\|$ for all $x$. (The same holds for $\overline{\lambda}$ as well.) Use this to argue that $$ \{ \lambda \in\mathbb{C} : \mbox{dist}(\lambda,N) > 0\} \subseteq\rho(A) \implies \sigma(A)\subseteq N^{c}. $$

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    $\begingroup$ Do you mean $\lambda \notin N$, right? $\endgroup$ – Benzio Jul 25 '14 at 10:39
  • $\begingroup$ Well there is the characterization of the Injectives operators with closed range that is just $\|(A-\lambda I)x\| \geq \delta\|x\|$. Then $(A-\lambda I)$ is injective. To show that $\lambda$ is in the resolvent I have to show that $(A-\lambda I)$ is a surjective operator. And now I'm stuck again! $\endgroup$ – Benzio Jul 25 '14 at 11:05
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    $\begingroup$ @ougoah : I started by assuming $\lambda\notin N^c$ where $N^c$ is the closure of the numerical range $N$. That gives the existence of $\delta > 0$. $\endgroup$ – DisintegratingByParts Jun 25 '16 at 14:32
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    $\begingroup$ Ahh thanks, I mistook that for the complement of N. Cheers $\endgroup$ – ougoah Jun 25 '16 at 16:01
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    $\begingroup$ @ougoah : Thanks for pointing out the confusion. I made a parenthetical remark in case someone else sees this answer. It's good to keep the site clean if possible. $\endgroup$ – DisintegratingByParts Jun 25 '16 at 17:02

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