Proposition 23

Let $g$ be a $\mu$-measurable function on X. The adjoint of the multiplication operator $m_g$ is $m_{\overline{g}}$.

For every integer $n\geqslant$ 1, let $\varphi_n$ be the characteristic function of the set of elements $x\in$ X such that $|g(x)|\leqslant n$, and let $\widetilde{\varphi}_n$ be its class in $L^2(X, \mu)$. Let $f\in \mathscr{L}^2(X, \mu)$ be such that its class $\widetilde{f}$ belongs to dom($m^*_g$), and let $\psi$ be a function whose class is $m^*_g(\widetilde{f})$.

For every $h\in \mathscr{L}^2(X, \mu)$ whose class $\widetilde{h}$ belongs to dom($m_g$), we have also $\widetilde{h}\widetilde{\varphi}_n\in$ dom($m_g$), and therefore $\langle \widetilde{f}|m_g(\widetilde{h}\widetilde{\varphi}_n)\rangle =\langle m^*_g(\widetilde{f})|\widetilde{h}\widetilde{\varphi}_n\rangle$. This yields the equality

Since the domain of $m_g$ is dense in $L^2(X, \mu)$, it follows that $(f g-\psi )\varphi_n$ is zero $\mu$-almost everywhere. Since $n$ is arbitrary, this means that $m^*_g(\widetilde{f})$ is the class in $L^2(X, \mu)$ of $f g$. In particular, since $m^*_g(\widetilde{f})\in L^2(X, \mu)$, one concludes that $f$ belongs to the domain of $m_{\overline{g}}$ and that $m^*_g(\widetilde{f}) =m_{\overline{g}}(\widetilde{f})$.

The adjoint of $m_g$ is therefore an extension of $m_{\overline{g}}$. Moreover, one has

for every $f\in L^2(X, \mu)$ and $h\in$ dom($m_g$), which proves that the linear form $h\mapsto \langle f|m_g(h)\rangle$ is continuous whenever $m_{\overline{g}}(f)g$ belongs to $L^2(X, \mu)$. Consequently the domain of $m_{\overline{g}}$ is contained in that of $m^*_g$, which concludes the proof.