Corollary

Let $u$ be an endomorphism of the Hilbert space $L^2(X, \mu)$ commuting with $\widetilde{m}_g$ for every function $g\in \mathscr{K}(X)$. Then there exists a unique element $f\in L^{\infty}(X, \mu)$ such that $u=\widetilde{m}_f$.

After Prop. 7, it is enough to prove that $u$ commutes with $\widetilde{m}_g$ for every $g\in \mathscr{L}^{\infty}(X, \mu)$. Let $h_1$ and $h_2$ be elements of $\mathscr{L}^2(X, \mu)$ with classes $\widetilde{h}_1$ and $\widetilde{h}_2$ in $L^2(X, \mu)$. Let $k_1$ (resp. $k_2$) be a function in $\mathscr{L}^2(X, \mu)$ whose class is $u(\widetilde{h}_1)$ (resp. $u^*(\widetilde{h}_2)$).

Let us set $h=h_1\overline{k}_2-k_1\overline{h}_2$; then $h\in \mathscr{L}^1(X, \mu)$. Define the measure $\nu =h\cdot \mu$ on X; it is bounded. For every $g\in \mathscr{L}^{\infty}(X, \mu)$, one has

Hence by hypothesis $\nu (g) = 0$ for every function $g\in \mathscr{K}$ (X), that is, $\nu = 0$. Consequently, one obtains

for every $g\in \mathscr{L}^{\infty}(X, \mu)$. Since this is the case for all $h_1$ and $h_2$ in $\mathscr{L}^2(X, \mu)$, one has $u\circ \widetilde{m}_g=\widetilde{m}_g\circ u$.