Lemma 3

Let $u$ be a normal partial operator on E. One has

Sp($b(u)$)$\cap D =\beta$(Sp($u$)).

One has $b(u) =u\circ W(u)^{1/2}$ (Prop. 1 of IV, p. 262). The endomorphism $W(u)^{1/2}$ is injective and its image is the domain of $u$ (formula (1) of IV, p. 262).

Let $\lambda \in \mathbf{C}$. The number $\lambda$ belongs to the resolvent set of $u$ if and only if $(u-\lambda 1_E)\circ W(u)^{1/2}$ is a bijective linear mapping of E into E (Remark 3 of IV, p. 245). Now, by formula (3) of IV, p. 262, one has

where $f_{\lambda}$ is the continuous function defined on $\overline{D}$ by $z\mapsto z-\lambda (1- |z|^2)^{1/2}$. The endomorphism $f_{\lambda}(b(u))$ is bijective if and only if its spectrum does not contain 0. Since Sp($f_{\lambda}(b(u))$) $=f_{\lambda}$(Sp($b(u)$)) (Corollary 2 of Proposition 7 of I, p. 111), this is so if and only if 0 does not belong to the set $f_{\lambda}$(Sp($b(u)$)). Thus, $\lambda \in$ Sp($u$) if and only if there exists $z\in$ Sp($b(u)$)$\cap D$ such that

This equality, if it is valid, implies that $z\in D$, and means that $\lambda =\beta^{-1}(z)$. It follows that Sp($u$) $=\beta^{-1}$(Sp($b(u)$)$\cap D$), as stated.

Let $u$ be a normal partial operator on E. Let $f\in \mathscr{K}$ (Sp($u$)). The mapping $z\mapsto f(\beta^{-1}(z))$ of Sp($b(u)$)$\cap D$ into $\mathbf{C}$ is continuous and has compact support; the unique mapping $f_{\beta}$ of Sp($b(u)$) into $\mathbf{C}$ which extends it by zero is therefore continuous.