A question regarding normal and adjoint operators

Question

Let $(X,A,\mu)$ be a measurable space.

For all $f\in L^{\infty}(X,A,\mu)$ define $M_f\in B(L^2(X,A,\mu))$ by: $M_fg:=fg$ for all $g\in L^2(X,A,\mu)$. So, $M_f^*=M_{\overline{f}}$ and then $M_f$ is normal ($ff^*=f^*f$). And if $f=\overline{f}$ then f is self-adjoint ($f=f^*$).

I am trying to show this theorem. I'm not getting why is $M_f^*=M_{\overline{f}}$. However, if i did succeed to show this, then using the fact that: $M_{f1}M_{f2}=M_{f1f2}$ we get: $M_fM_f^*=M_fM_\overline{f}=M_{f\overline{f}}=M_{\overline{f}f}=M_\overline{f}M_f=M_f^*M_f$ so $M_f$ is normal (by definition). For the second part: If $f=\overline{f}$ then by first part we have: $M_f^*=M_\overline{f}=M_f$ so $M_f$ is self-adjoint (by defnition).

Appreciate any help with this.

Answer 1

We show $M_f^*= M_{\overline{f}}$. I'll assume that the inner product $\langle \cdot,\cdot\rangle$ is linear in the first component, i.e. $$\langle g,h\rangle = \int_X g\overline{h}d \mu.$$

Indeed, let $g,h \in L^2(X, \mathcal{A}, \mu)$ and note that $$\langle g, M_{\overline{f}}h\rangle = \langle g, \overline{f}h\rangle = \int_X g f\overline{h}d\mu.$$ On the other hand $$\langle M_f g, h\rangle = \langle fg, h\rangle = \int_X fg \overline{h}d\mu$$ and thus $$\langle g, M_{\overline{f}}h\rangle = \langle M_f g,h\rangle, \quad g,h \in L^2(X, \mathcal{A}, \mu)$$ so by definition of adjoint $M_f^* = M_{\overline{f}}.$

Answer 2

$ \langle M_f^{*}g , h \rangle=\langle g , M_fh \rangle=\int g\overline {fh}$. This is true for all $h \in L^{2}$ so $M_f^{*}g=\overline f g=M_{\overline f} g$.