Proposition 5

Let $u$ be a normal partial operator on a Hilbert space E.

a) If $f\in \mathscr{L}_u^{\infty}$(Sp($u$)), then $f(u)\in \mathscr{L}(E)$;

b) If $f\in \mathscr{L}_u^{\infty}$(Sp($u$)) and $g\in \mathscr{L}_u$(Sp($u$)), then $f(u)\circ g(u)\subset (f g)(u)$;

c) The mapping $f\mapsto f(u)$ of $\mathscr{L}_u^{\infty}$(Sp($u$)) into $\mathscr{L}(E)$ is a continuous unital morphism of involutive algebras. In particular, we have $\|f(u)\|\leqslant \|f\|_{\infty}$ for $f\in \mathscr{L}_u^{\infty}$(Sp($u$));

d) If $u\in \mathscr{L}(E)$, then for every $f\in \mathscr{L}_u^{\infty}$(Sp($u$)), the endomorphism $f(u)$ belongs to the bicommutant of $u$ in $\mathscr{L}(E)$.

Let $f\in \mathscr{L}_u^{\infty}$(Sp($u$)). We have $D_f= E$ and $f(u)$ is a continuous endomorphism of E by formula (8), p. 272, whence assertion a).

Let $g\in \mathscr{L}_u$(Sp($u$)). Let $y\in D_g$. We have $y\in D_{f g}$ and, for every $x\in E$, there follows $\langle \overline{f}(u)x|g(u)y\rangle =\langle x|(f g)(u)y\rangle$ by formula (7), p. 271, whence $f(u)(g(u)y) = (f g)(u)(y)$, which proves b).

From a) and b), the mapping $f\mapsto f(u)$ of $\mathscr{L}_u^{\infty}$(Sp($u$)) into $\mathscr{L}(E)$ is a unital morphism of involutive algebras; consequently, it is a continuous morphism of norm $\leqslant 1$(I, p. 104, prop. 2), whence assertion c).

Suppose that $u$ is an endomorphism of E. Let $v\in \mathscr{L}(E)$ be permutable with $u$. Then one has $v\circ f(u) =f(u)\circ v$ for $f\in \mathscr{C}$ (Sp($u$)) by the properties of the continuous functional calculus (I, p. 110, remark). Let $x$ and $y$ be in E. The formulas

valid for every function $f\in \mathscr{C}$ (Sp($u$)), mean that the spectral measures of $(v^*(x), y)$ and of $(x, v(y))$ relative to $u$ are equal. By formula (6), p. 271, this equality implies that formula (9) is valid for every $f\in \mathscr{L}_u^{\infty}$(Sp($u$)). Hence $v\circ f(u) =f(u)\circ v$.