# bounded self-adjoint operator has non-empty spectrum

Show that the spectrum of a bounded self-adjoint linear operator on a complex Hilbert space $$H\neq\{0\}$$ is not empty.

If possible, let the spectrum $$\sigma(T)=\emptyset$$. So its resolvent set $$\rho(T)$$ equals $$\mathbb{C}$$. So for all $$\lambda\in\mathbb{C}$$ we have a $$c>0$$ such that $$||T_\lambda(x)||\geq c||x||$$ where $$T_\lambda=T-\lambda I$$. Dividing both sides by $$||x||$$ and taking supremum, we have $$||T_\lambda||\geq c \ \ \ \ \ \forall \ \ \ \lambda\in\mathbb{C}$$ which contradicts that $$T$$ is a bounded linear operator. So $$\sigma(T)\neq\emptyset$$. Is my proof correct? Any help is appreciated.

• (1) Doesn't $c$ depend on $\lambda$? (2) How does this contradict $T$ being a bounded operator? – Nate Eldredge Feb 27 at 15:32
• Yes $c$ depends on $\lambda$, but since $\lambda$ takes value in all of $\mathbb{C}$, so $||T_\lambda||\geq c$ for all $\lambda\in\mathbb{C}$, so eventually unbounded on whole $\mathbb{C}$. – am_11235... Feb 27 at 15:36
• $c$ can for example be $0$ or just $1$, you will not get a contradiction in this way. – s.harp Feb 27 at 16:02
• In fact, every bounded linear operator on a complex (!) Hilbert space has a non-empty spectrum. However, even a self adjoint one may have empty point spectrum. – G. Chiusole Feb 27 at 16:38
• @s.harp I have already written that $c>0$, so how can $c$ be zero I don't understand. Also why $c=1$ is not contradiction? – am_11235... Feb 27 at 16:40

Let $$M=\sup\limits_{\|x\|=1}(Ax,x)$$, $$m=\inf\limits_{\|x\|=1}(Ax,x)$$ and let for definiteness $$0\leq m\leq M$$, then $$\|A\|=\sup\limits_{\|x\|=1}|(Ax,x)|=M$$. We show that $$M$$ is point of the spectrum. By the property of the supremum, there exists a sequence $$x_n\in H$$ such that $$\|x_n\|=1$$ and $$(Ax_n,x_n)\to M$$. Moreover, $$\|Ax_n\|\leq\|A\|\|x_n\|=M$$.
Further, $$\|Ax_n-Mx_n\|^2=\|Ax_n\|^2-2M(Ax_n,x_n)+M^2\|x_n\|^2\leq 2M^2-2M(Ax_n,x_n)$$. By $$n\to\infty$$ we have $$\|Ax_n-Mx_n\|\to0$$, so $$M$$ -- is the point of spectrum.