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To keep it simple, let $phi : Itomathbb C$ be a measurable function on a finite interval $Isubsetmathbb R$. The multiplication operator $M_phi$ is defined as $M_phi f = phicdot f$, $finoperatornamedomM_phi$, where
$$
operatornamedomM_phi = fin L^2(I) : phicdot fin L^2(I).
$$
I want to show that $M_phi^* = M_barphi$, where $barphi$ is the complex conjugate of $phi$. My first question: why is $M_phi$ densely defined?

It is easy to see that $M_barphisubset M_phi^*$, but I cannot prove the opposite inclusion. For this, let $ginoperatornamedomM_phi^*$. Then $int foverlinebarphig,dx = (phi f,g) = (f,h)$ for all $finoperatornamedomM_phi$, where $h = M_phi^*g$. How can I infer from here that $barphi gin L^2(I)$?

Suppose $gperp mathcalD(M_phi)$. Then $frac1phiginmathcalD(M_phi)$ because $fracphiphig in L^2$, owing to the fact that $|phi| = |phi|cdot 1 le frac12(|phi|^2+1)$. Therefore, $gperp frac1phig$, which gives
$$
0= langle g,frac1phigrangle = int |g|^2frac1phi implies g=0; a.e..
$$
So $M_phi$ is densely-defined.

Let $A_n = x in I: $. 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 $bigcup_n V_n subset textdom M_phi$ and is dense.

As already shown in DisintegratingByParts' answer, for all $hin L^2(I)$ we have that $tfrach1+inoperatornamedomM_phi$. Now, let $ginoperatornamedomM_phi^*$. Then for all $finoperatornamedomM_phi$ we have $(phi f,g) = (f,M_phi^*g)$. Hence, for all $hin L^2(I)$ we get
$$
left(fracphi h1+,gright) = left(frach1+,M_phi^*gright),
$$
that is,
$$
left(h,fracoverlinephig1+right) = left(h,fracM_phi^*g1+right).
$$
This implies $overlinephi g = M_phi^*gin L^2(I)$.