Adjoint of multiplication operator To keep it simple, let $\phi : I\to\mathbb C$ be a measurable function on a finite interval $I\subset\mathbb R$. The multiplication operator $M_\phi$ is defined as $M_\phi f = \phi\cdot f$, $f\in\operatorname{dom}M_\phi$, where
$$
\operatorname{dom}M_\phi = \{f\in L^2(I) : \phi\cdot f\in L^2(I)\}.
$$
I want to show that $M_\phi^* = M_{\bar\phi}$, where $\bar\phi$ is the complex conjugate of $\phi$. My first question: why is $M_\phi$ densely defined?
It is easy to see that $M_{\bar\phi}\subset M_\phi^*$, but I cannot prove the opposite inclusion. For this, let $g\in\operatorname{dom}M_\phi^*$. Then $\int f\overline{\bar{\phi}g}\,dx = (\phi f,g) = (f,h)$ for all $f\in\operatorname{dom}M_\phi$, where $h = M_\phi^*g$. How can I infer from here that $\bar\phi g\in L^2(I)$?
 A: Let $A_n = \{x \in I: |\phi(x)| \le n \}$.  Note that
$\bigcup_{n=1}^\infty A_n = I$.  Let $V_n$ be the subspace of $L^2(I)$ consisting of functions that are $0$ outside $A_n$.
Then $\bigcup_{n} V_n \subset \text{dom} M_\phi$ and is dense.
A: Suppose $g\perp \mathcal{D}(M_{\phi})$. Then $\frac{1}{|\phi|^2+1}g\in\mathcal{D}(M_{\phi})$ because $\frac{\phi}{|\phi|^2+1}g \in L^2$, owing to the fact that $|\phi| = |\phi|\cdot 1 \le \frac{1}{2}(|\phi|^2+1)$. Therefore, $g\perp \frac{1}{|\phi|^2+1}g$, which gives
$$
   0= \langle g,\frac{1}{|\phi|^2+1}g\rangle = \int |g|^2\frac{1}{|\phi|^2+1}  \implies g=0\; a.e..
$$
So $M_{\phi}$ is densely-defined.
A: As already shown in DisintegratingByParts' answer, for all $h\in L^2(I)$ we have that $\tfrac{h}{1+|\phi|^2}\in\operatorname{dom}M_\phi$. Now, let $g\in\operatorname{dom}M_\phi^*$. Then for all $f\in\operatorname{dom}M_\phi$ we have $(\phi f,g) = (f,M_\phi^*g)$. Hence, for all $h\in L^2(I)$ we get
$$
\left(\frac{\phi h}{1+|\phi|^2},g\right) = \left(\frac{h}{1+|\phi|^2},M_\phi^*g\right),
$$
that is,
$$
\left(h,\frac{\overline{\phi}g}{1+|\phi|^2}\right) = \left(h,\frac{M_\phi^*g}{1+|\phi|^2}\right).
$$
This implies $\overline\phi g = M_\phi^*g\in L^2(I)$.
