Remark

Suppose that $u$ is a normal endomorphism of E, so that Sp($u$) is a compact subset of $\mathbf{C}$. Since the bounded transform $b(u)$ coincides with $\beta (u)$ (formula (5)), and Sp($b(u)$) $=\beta$(Sp($u$)) (Corollary 2 of Proposition 7 of I, p. 111), the spectrum of $b(u)$ is a compact subset of D. For every continuous function $f\in \mathscr{C}$ (Sp($u$)), the function $f_{\beta}$ coincides with the restriction of $f\circ \beta^{-1}$ to Sp($u$). Consequently, $f_{\beta}(\beta (u))$ coincides with the endomorphism $f(u)$ defined by the continuous functional calculus of $u$. The above definition is therefore compatible with that of the continuous functional calculus of the involutive algebra $\mathscr{L}(E)$.

Let $f\in \mathscr{K}$ (Sp($u$)). We have

so that, for all $x$ and $y$ in E, one obtains the inequality $|\langle x|f(u)y\rangle |\leqslant \|x\| \|y\| \|f\|_{\infty}$. The mapping $f\mapsto \langle x|f(u)y\rangle$ is therefore a bounded measure on Sp($u$), of total mass $\leqslant \|x\|\|y\|$ (INT, IV, p. 154, § 4, n$^o7$). If $x=y$, it is a positive measure, since $f(u)$ is positive when $f\geqslant 0$.