Proposition 21

Let E be a complex Banach space and $u$ a closed partial operator on E. Let $\varepsilon \in \mathbf{R}^*_+$. Every bounded connected component of PSp$_{\varepsilon}(u)$ meets the spectrum of $u$.

Let U be a connected component of PSp$_{\varepsilon}(u)$ which does not meet Sp($u$). The set U is open and closed in PSp$_{\varepsilon}(u)$, and its closure U in $\mathbf{C}$ therefore satisfies $U\cap$ PSp$_{\varepsilon}(u) = U$.

Since Sp($u$) is contained in PSp$_{\varepsilon}(u)$ and U does not meet Sp($u$), this equality proves that the set U is disjoint from Sp($u$), hence contained in the resolvent set of $u$. Moreover, it implies that the set $\overline{U}-$ U does not meet PSp$_{\varepsilon}(u)$.

One therefore has $\|R(u, \lambda )\|\leqslant \varepsilon^{-1}$ for every $\lambda$ in $\overline{U}-$ U, whereas $\|R(u, \lambda )\|> \varepsilon^{-1}$ for $\lambda \in U$. If the set U is bounded, its closure $\overline{U}$ is compact and there exists $\lambda_0\in \overline{U}$ such that $\|R(u, \lambda )\|\leqslant \|R(u, \lambda_0)\|$ for $\lambda \in U$. The foregoing implies that $\lambda_0\in U$, which contradicts the maximum principle (VAR, R1, p. 29, 3.3.7) since the resolvent of $u$ is holomorphic in the resolvent set of $u$ (Prop. 14 of IV, p. 246).