Let $E$ be a [[Hilbert Space|Hilbert space]], and $u \in K(E, E)$ such that $u = u^*$ be a [[Compact Operator|compact]] [[Self-Adjoint Operator|self-adjoint]] operator. > [!theorem] > > Either $\norm{u}$ or $-\norm{u}$ is in $\sigma(u)$. > > *Proof*. Let $\seq{x_n} \subset E$ such that $\norm{x_n} = 1$ for all $n \in \nat$, and $\angles{ux_n, x_n} \to \norm{u}$. Assume without loss of generality that $\angles{ux_n, x_n} \to \alpha$ where $\alpha = \pm\norm{u}$. From here, > $ > \begin{align*} > 0 \le \norm{ux_n - \alpha x_n}^2 &= \angles{ux_n - \alpha x_n, ux_n - \alpha x_n} \\ > &= \norm{ux_n}^2 - 2\alpha\angles{ux_n, x_n} + \alpha^2 \norm{x_n}^2 \\ > &\le \alpha^2 - 2\alpha\angles{ux_n, x_n} + \alpha^2\norm{x_n}^2 > \end{align*} > $ > which converges to $0$ as $n \to \infty$. Taking a further subsequence such that $\seq{ux_n}$ converges. Since $\norm{ux_n - \alpha x_n} \to 0$, $\seq{x_n}$ converges as well, yielding the desired eigenvector.