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

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 $$\{ \lambda \in\mathbb{C} : \mbox{dist}(\lambda,N) > 0\} \subseteq\rho(A) \implies \sigma(A)\subseteq N^{c}.$$
• Do you mean $\lambda \notin N$, right? – Benzio Jul 25 '14 at 10:39
• 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! – Benzio Jul 25 '14 at 11:05
• @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 $\delta > 0$. – DisintegratingByParts Jun 25 '16 at 14:32