Proposition 5

Let $p\in [1,+\infty [$. The map $m:g\mapsto \widetilde{m}_g$ of $L^{\infty}(X, \mu)$ into $\mathscr{L}(L^p(X, \mu))$ is an isometric morphism of unital Banach algebras.

For $a,b$ in $\mathbf{C}$ and $g_1,g_2$ in $L^{\infty}(X, \mu)$, one verifies at once that $\widetilde{m}_{ag_1+bg_2}=a\widetilde{m}g_1+b\widetilde{m}_{g_2}$, hence the mapping $\widetilde{m}$ is linear. Moreover, one has $\widetilde{m}_1= 1_{L^p(X,\mu)}$ and $\widetilde{m}_{g_1g_2}=\widetilde{m}_{g_1}\widetilde{m}_{g_2}$, hence the mapping $\widetilde{m}$ is a unital algebra morphism of $L^{\infty}(X, \mu)$ into $\mathscr{L}(L^p(X, \mu))$.

Formula (1) above proves that $\widetilde{m}$ has norm $\leqslant 1$.

Let $g\in \mathscr{L}^{\infty}(X, \mu)$ and $\widetilde{g}$ its class in $L^{\infty}(X, \mu)$. Let $\varepsilon >$ 0. The set Y of the $x\in X$ such that $|g(x)|>\|\widetilde{g}\|_{\infty}-\varepsilon$ is not locally $\mu$-negligible. There therefore exists a compact subset C of X such that $\mu(Y\cap C)>0$. Let $\varphi$ be the characteristic function of $Y\cap C$. Since

it follows that $\|\widetilde{m}_{\widetilde{g}}\|\geqslant \|\widetilde{g}\|_{\infty}-\varepsilon$. Since $\varepsilon$ is arbitrary, one deduces that $\|\widetilde{m}_{\widetilde{g}}\|\geqslant \|\widetilde{g}\|_{\infty}$, which proves that $\widetilde{m}$ is isometric.