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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.

am_11235...
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    (1) Doesn't $c$ depend on $\lambda$? (2) How does this contradict $T$ being a bounded operator? – Nate Eldredge Feb 27 '20 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 '20 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 '20 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 '20 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 '20 at 16:40
  • @am_11235..., didn’t you be embarrassed that you have never used self-adjointness in your proof? In addition, why should lambda and $c$ be directly related? Why not there be an inverse relationship? – thing Feb 27 '20 at 16:46
  • @thing, the line I have wrote above "..we have a $c>0$ s.t. $||T_\lambda(x)||\geq c||x||$.." holds only for bounded self-adjoint linear operators. It's a standard theorem for resolvent sets of bounded self-adjoint linear operators. That is, if the operator weren't bounded self-adjoint, I could never write that line. – am_11235... Feb 27 '20 at 17:16
  • Take a look at $T=\begin{pmatrix}1&0\ 0 &-1\end{pmatrix}$, for this $T$ you have a bound $|T_\lambda|≥ c(\lambda)$, where here you may set $c(\lambda)=1$ for all $\lambda$, this is not contradiction to the boundedness of $T$. – s.harp Feb 27 '20 at 23:50
  • A way to achieve the result is to consider the sub-algebra of $B(H)$ generated by $\Bbb1$ and $T$. Every element of this algebra except for $0$ is invertible, because $\sigma(p(T))=p(\sigma(T))=\emptyset$ for any polynomial $p$. Further it is a commutative algebra. Hence you have here a normed field containing $\Bbb C$, Compare this with the Gelfand-Mazur theorem to get a contradiction. – s.harp Feb 27 '20 at 23:54
  • @am_11235...If $\lambda$ is a point of the resolvent set, then the statement $|T_\lambda x|\geq c|x|$ is true for any bounded operator (in fact, there is no even boundedness needed, linearity enough). Self-adjointness is needed here if this statement were applied in the opposite direction. – thing Feb 28 '20 at 01:14

2 Answers2

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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.

thing
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  • If $A$ happens to be a complex self adjoint operator. Can we show that either $m$ or $M$ is in the point spectrum. – cabmetric May 13 '21 at 16:47
  • You mean to show that these numbers are eigenvalues? This don't follow from anywhere unless the operator is compact. – thing May 15 '21 at 01:00
  • The last equality in this sentence "Moreover, $|Ax_n| \leq |A| |x_n| = M$" does not follow without proof. What follows is $|Ax_n| \leq |A| |x_n| = |A|$. But then the proof fails. – travelingbones Jan 05 '24 at 13:48
  • In this case everything is correct and follows from the equality $|A|=\sup\limits_{|x|=1}|(Ax,x)|=M$. – thing Jan 06 '24 at 02:33
  • concur, if we suppose that $|A| = \sup_{|x|= 1} \langle Ax, x\rangle$ (which is true), then I believe your proof holds. My proof of $|A| = \sup_{|x|= 1} \langle Ax, x\rangle$ uses the fact that the spectrum is nonempty. John Conway's Functional Analysis book has a proof of this fact in Proposition 2.13 https://www.datascienceassn.org/sites/default/files/A%20Course%20in%20Functional%20Analysis%20-%20Conway.pdf. Conway's proof does not use the spectrum, it uses the parallelogram equality. – travelingbones Jan 09 '24 at 15:23
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Suppose, for contradiction, that $(L-\lambda I)^{-1} \in \mathcal{L}(H)$ for every $\lambda \in \mathbb{C}$. Then $f:\mathbb{C} \to \mathcal{L}(H)$ by $f(\lambda) = (\lambda I-L)^{-1}$ is a non-constant, analytic function on (all of) $\mathbb{C}$. This function is uniformly bounded on $\{|\lambda| \leq \|L\| + 1 \}$ because it is continuous on a compact set, and for $|\lambda| > \|L\| + 1$, we see $\|f(\lambda)\| = \|(L/\lambda - I)^{-1}\|/|\lambda| \leq 1$. But a uniformly bounded analytic function on $\mathbb{C}$ is constant, which is a contradiction.

travelingbones
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