Lemma 7

Let $g\in \mathscr{L}^{\infty}(X, \mu)$. The endomorphism $\widetilde{m}_g$ of $L^p(X, \mu)$ is injective if and only if the set of $x\in X$ such that $g(x) = 0$ is locally $\mu$-negligible.

Let A denote the set of $x\in X$ such that $g(x) = 0$.

Suppose that A is locally $\mu$-negligible. Let $f\in \mathscr{L}^p(X, \mu)$ and $\widetilde{f}$ its class in $L^p(X, \mu)$. Suppose that $\widetilde{m}_g(\widetilde{f}) = 0$. By definition, this is the case if and only if the function $f g$ is $\mu$-negligible, so that the set B of $x\in X$ such that $f(x)g(x)\not = 0$ is $\mu$-negligible. The function $f$ is zero outside $A\cup B$, hence is $\mu$-negligible. This implies that the endomorphism $\widetilde{m}_g$ is injective.

Conversely, suppose that A is not locally $\mu$-negligible. Let C be a compact subset of X such that $A\cap C$ is not $\mu$-negligible, and let $\varphi$ be the characteristic function of $A\cap C$. The class of the function $\varphi$ in $L^p(X, \mu)$ is not zero, but that of $m_g(\varphi )$ is, hence $m_g$ is not injective.

Since the product of a locally $\mu$-negligible function and a $\mu$-negligible function is $\mu$-negligible, the endomorphism $\widetilde{m}_g$ depends only on the class of $g$ in $L^{\infty}(X, \mu)$. For $\widetilde{g}\in L^{\infty}(X, \mu)$, one also denotes by $\widetilde{m}_g$ the endomorphism $\widetilde{m}_g$ of $L^p(X, \mu)$ for every function $g\in \mathscr{L}^{\infty}(X, \mu\widetilde{)}$ whose class is $\widetilde{g}$. One says that $\widetilde{m}_g$ is the endomorphism of multiplication by $\widetilde{g}$ in $L^p(X, \mu)$.