Proposition 4

Let $f$ be a universally measurable function on the spectrum of $u$.

a) The set $D_f$ is a dense vector subspace of E. The mapping $f(u) :y\mapsto$ ev$_y(f)$ of $D_f$ into E is linear and coincides with $1_E$ if $f= 1$;

b) Let $g\in \mathscr{L}_u$(Sp($u$)). For every $y\in D_f$ and every $x\in D_g$, one has $f g\in \mathscr{L}^1$(Sp($u$)$, \mu_{x,y}$), and

Sp($u$)

c) Suppose that $E = L^2(X, \mu)$ where X is a locally compact topological space and $\mu$ a positive measure on X, and that $u=m_h$, where $h: X\rightarrow \mathbf{C}$ is $\mu$-measurable. One has $f(m_h) =m_{f\circ h}$.

According to Theorem 1 of IV, p. 266, one may suppose that one is in the situation of assertion c), that is to say that $E = L^2(X, \mu)$ and $u=m_h$, where X is a locally compact topological space, $\mu$ a positive measure on X and $h: X\rightarrow \mathbf{C}$ is $\mu$-measurable. We shall denote by $\widetilde{\varphi}$ the class in $L^2(X, \mu)$ of a function $\varphi \in \mathscr{L}^2(X, \mu)$.

Let S = Sp($m_h$). We denote by $\mu_{\widetilde{\varphi}_1,\widetilde{\varphi}_2}$ (resp. $\mu_{\widetilde{\varphi}}$) the spectral measure of $(\widetilde{\varphi}_1,\widetilde{\varphi}_2)$ (resp. of $\widetilde{\varphi}$) relative to $m_h$ for every $(\varphi_1, \varphi_2)\in \mathscr{L}^2(X, \mu)^2$ (resp. every $\varphi \in \mathscr{L}^2(X, \mu)$).

Let $\varphi \in \mathscr{L}^2(X, \mu)$. The spectral measure $\mu_{\widetilde{\varphi}}$ is equal to the image measure $h(|\varphi |^2\cdot \mu)$ on S (Lemma 4 of IV, p. 269). Since the function $f$ is $\mu_{\varphi}$-measurable, one has $\varphi \in D_f$ if and only if the integrals

are finite (INT, V, p. 70, § 6, n$^o2$, prop. 2). This means that $D_f$ is the domain of the multiplication operator $m_{f\circ h}$, which is a dense subspace of E (prop. 5 of IV, p. 232).

The restriction of ev$_{\widetilde{\varphi}}$ to classes of functions $g\in \mathscr{K}(S)$ is the mapping which to a class $\widetilde{g}$ associates the element $g(m_h)(\widetilde{\varphi}) =m_{g\circ h}(\widetilde{\varphi})$ (lemma 4 of IV, p. 269). The mapping ev$_{\varphi}$ therefore coincides with the isometric mapping of $L^2(S, \mu_{\varphi})$ into $L^2(X, \mu)$ deduced by passing to quotients from the mapping $g\widetilde{\mapsto}(g\circ h)\cdot \varphi$ of $\mathscr{L}^2(S, \mu_{\widetilde{\varphi}})$ into $\mathscr{L}^2(X, \mu)$.

In particular, therefore, we have ev$_{\varphi}(f) =m_{f\circ h}(\widetilde{\varphi})$, so that the mapping $f(m_h)$ of $D_f$ into E coincides with the partial operator $m_{f\circ h}$. This proves assertion c); if $f= 1$, we find ev$_{\varphi}(1) =\widetilde{\varphi}$, which also concludes the proof of a).

Let us prove assertion b). Let $\varphi_1$ and $\varphi_2$ be functions such that $\widetilde{\varphi}_1\in D_f$ and $\widetilde{\varphi}_2\in D_g$. We then have (INT, V, loc. cit.)

which is finite; hence $f g\in \mathscr{L}^1(S, \mu_{\widetilde{\varphi}_1,\widetilde{\varphi}_2})$. Moreover, we then have

which concludes the proof.