Let $A$ be a bounded self-adjoint operator on a Hilbert space $\mathcal H$ and let $$a = \inf_{\|x\|=1} \left \langle Ax,x \right \rangle,\ \ b = \sup\limits_{\|x\| = 1} \left \langle Ax, x \right \rangle$$
Then the spectrum $\sigma (A)$ of $A$ is contained in the interval $[a,b]$ and contains the points $a$ and $b.$
I have tried to prove it but failed to proceed further. Could anybody give me some suggestion?
Thanks!
There is a very clever, elementary trick that you might not know about, which uses the Cauchy-Schwarz inequality for an inner product or pseudo inner product such as $$ [x,y] = \langle (A-aI)x,y\rangle. $$ This form has all of the properties of an inner product, except possibly strict positivity. Cauchy-Schwarz still holds: $$ |[x,y]|^2 \le [x,x][y,y]. $$ Now, set $y=(A-aI)x$: \begin{align} \|(A-aI)x\|^4 &\le \langle(A-aI)x,x\rangle\langle (A-aI)^2x,(A-aI)x\rangle \\ \|(A-aI)x\|^4 &\le \langle(A-aI)x,x\rangle \|(A-aI)^2x\|\|(A-aI)x\| \\ &\le \langle (A-aI)x,x\rangle\|(A-aI)\|\|(A-aI)x\|^2\\ \|(A-aI)x\|^2&\le\|(A-aI)\|\langle(A-aI)x,x\rangle. \end{align} Now you see that, if $\{ x_n \}$ is a sequence of unit vectors chosen so that $\langle(A-aI)x_n,x_n\rangle\rightarrow 0$, then $(A-aI)x_n\rightarrow 0$. That's enough to guarantee that $a\in\sigma(A)$. Similarly, $b\in\sigma(A)$.
Without using this Cauchy-Schwarz inequality for general inner products or pseudo inner products, I don't know how you would come up with this. It's a trick worth knowing because it proves that, for a bounded self-adjoint $A$, both $\sup_{\|x\|=1}\langle Ax,x\rangle$ and $\inf_{\|x\|=1}\langle Ax,x\rangle$ are in the spectrum of $A$. And it proves that the spectrum is non-empty as well. If $A$ is compact, you get eigenvectors with eigenvalues equal to the $\sup$ and $\inf$, and that ends up giving you a full spectral decomposition for the compact $A$ in terms of eigenvectors.