I'm having a bit of difficulty following the proof. It would be great if someone could give me a bit of advice. I get the feeling that my question is rather trivial so sorry in advance. Many thanks.

For $\mathcal{H}$, a complex Hilbert Space and $S\in B(\mathcal{H})$ self adjoint, at least one of $\|S\|$ or -$\|S\|$ must lie in the spectrum of $S$.

The proof I have is from Linear Functional Analysis by Rynne and Youngson. It goes as follows:


This is true if $S=0$, so by working with $\|S\|^{-1}S$ if necessary we may assume that $\|S\|=1$. From the definition of the norm of $S$, there exists a sequence $\{x_n\}$ in $\mathcal{H}$ such that $\|x_n\|=1$ and $\lim_{n\rightarrow\infty}\|Sx_n\|=1$. Then

\begin{align} \|(I-S^2)x_n\|^2&=((I-S^2)x_n,(I-S^2)x_n) \\ &=\|x_n\|^2+\|Sx_n\|^2-2(S^2x_n,x_n) \\ &\leq2-2(Sx_n,Sx_n) \\ &=2-2\|Sx_n\|^2 \end{align} Where $(\cdot,\cdot)$ is the inner product. So $\lim_{n\rightarrow\infty}\|(I-S^2)x_n\|^2=0$ and hence $1\in\sigma(S^2)$. Thus $1\in(\sigma(S))^2$ and hence $1$, or $-1$ is in $\sigma(S)$.

What I don't understand is why by the definition of the norm we can pick such a sequence $\{x_n\}$ in $\mathcal{H}$ such that $\|x_n\|=1$ and $\lim_{n\rightarrow\infty}\|Sx_n\|=1$.

I would really appreciate it if someone could explain why this is true or even offer some hints as to how I could see this for myself.

  • $\begingroup$ There are a few subtle points in this proof. $\endgroup$ – copper.hat Apr 8 '17 at 0:35

The definition of $\| S \|$ is

$$ \| S \| = \sup_{\| x \| = 1} \| Sx \| $$

so using the definition of $\sup$ we see that we must have a sequence $x_n \in \mathcal{H}$ with $\| x_n \| = 1$ and $\| Sx_n \| \to \| S \|$.

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