$\lambda$ is an eigenvalue iff spectral measure of $\lambda$ is nonzero

Question

Let $M$ be a normal operator on a Hilbert space and let $E$ be the spectral measure of $\sigma(M)$ (the spectrum of $M$). Show that $\lambda$ is an eigenvalue to $M$ $\iff E(\{\lambda\})\not = 0$.

Hints?

Answer 1

The operator $M$ can be written as $\int_{\sigma}\lambda dE(\lambda)$. Using properties to the spectral measure, $$ ME\{\mu\}=\int_{\sigma}\lambda dE(\lambda)E\{\mu\}=\int_{\sigma\cap\{\mu\}}\lambda dE(\lambda)=\mu E\{\mu\}, \\ E\{\mu\}(M-\mu I)=(M-\mu I)E\{\mu\} =0. $$ If $E\{\mu\} \ne 0$, then $\mathcal{N}(M-\mu I)\ne \{0\}$.

Conversely, suppose $Mx=\mu x$ for some $x\ne 0$. For any $\epsilon > 0$, define $$ S_{\epsilon} = \{ \lambda\in\mathbb{C} : |\lambda-\mu| \ge \epsilon \}. $$ Then $0=E(S_{\epsilon})(M-\mu I)x=(M-\mu I)E(S_{\epsilon})x$ forces $E(S_{\epsilon})x=0$ because \begin{align} 0 &= \|(M-\mu I)E(S_{\epsilon})x\|^2 \\ & = \int_{|\lambda-\mu|\ge\epsilon}|\lambda-\mu|^2d\|E(\lambda)x\|^2 \\ & \ge \int_{|\lambda-\mu|\ge\epsilon}\epsilon^2d\|E(\lambda)x\|^2 \\ & =\epsilon^2 \|E(S_{\epsilon})x\|^2. \end{align} Therefore $$ \int_{|\lambda-\mu|> 0}d\|E(\lambda)x\|^2=\lim_{\epsilon\downarrow 0}\int_{S_{\epsilon}}d\|E(\lambda)x\|^2 = 0. $$ Hence, $$ E\{\mu\}x = E\{\mu\}x+E\{\lambda :|\lambda-\mu|>0\}x=x. $$ Therefore, if $\mu$ is an eigenvalue of $M$, then $E\{\mu\} \ne 0$.