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.

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    $\begingroup$ (1) Doesn't $c$ depend on $\lambda$? (2) How does this contradict $T$ being a bounded operator? $\endgroup$ – Nate Eldredge Feb 27 at 15:32
  • $\begingroup$ 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 $\mathbb{C}$. $\endgroup$ – am_11235... Feb 27 at 15:36
  • $\begingroup$ $c$ can for example be $0$ or just $1$, you will not get a contradiction in this way. $\endgroup$ – s.harp Feb 27 at 16:02
  • $\begingroup$ 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. $\endgroup$ – G. Chiusole Feb 27 at 16:38
  • $\begingroup$ @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? $\endgroup$ – 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.

Other cases are treated similarly.

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