Proposition 10

Let $\lambda \in \mathbf{C}$. One has $\lambda \in$ Sp($u$) if and only if, for every open neighbourhood V of $\lambda$ in $\mathbf{C}$, the spectral projector $p_V$ of $u$ relative to V is nonzero.

If $\lambda \notin$ Sp($u$), then there exists an open neighbourhood V of $\lambda$ in $\mathbf{C}$ which does not meet Sp($u$), and then $p_V=p_{\emptyset}= 0$.

Conversely, suppose that there exists an open neighbourhood V of $\lambda$ in $\mathbf{C}$ such that $p_V= 0$. Let $c >0$ be such that the disc with center $\lambda$ and radius $c$ is contained in V.

Let $x\in$ dom($u$) and let $\mu_x$ be the spectral measure of $x$ relative to $u$. Since $\mu_x(V) =\langle x|p_V(x)\rangle = 0$, one computes

But, on the other hand, one has

(formula (8), p. 272), whence $\|u(x)-\lambda x\|\geqslant c\|x\|$ and $\|u^*(x)-\lambda x\|\geqslant c\|x\|$. It follows that $\lambda$ belongs to the resolvent set of $u$ (Lemma 5 of IV, p. 248). This concludes the proof.