Adjoint of multiplication operator

Question

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 $\bar\phi g\in L^2(I)$?

Answer 1

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.

Answer 2

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.

Answer 3

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 $\overline\phi g = M_\phi^*g\in L^2(I)$.