I am doing the following problem: Let $(M,\mathcal{A},\mu)$ be a general measure space, $f:M\to \mathbb{R}$ be a measurable function. Define the operator $A_f:\mathrm{dom}(A_f)\to L^2(\mu)$ by $$\mathrm{dom}(A_f):=\{\psi\in L^2(\mu)|f\psi\in L^2(\mu)\}$$ $$A_f\psi=f\psi,\forall\psi\in\mathrm{dom}(A_f)$$ Show this operator is self-adjoint for any given measurable function $f$.
It is clear that this operator is symmetric, but I can't think of any method to argue the domain of $A^*_f$ is the same as that of $A_f$ (I know it's at least larger so we only need to prove $\mathrm{dom}(A^*_f)\subset\mathrm{dom}(A_f)$). By definition I can have $$\left<A_f^*\psi,\phi\right>=\left<\psi,A_f\phi\right>=\left<\psi,f\phi\right>=
\left<f\psi,\phi\right>,\forall \phi\in\mathrm{dom}(A_f)$$
so $A^*_f\psi=f\psi,\forall \psi\in\mathrm{dom}(A^*_f)$.
However it doesn't seem to help much.
Best Answer
If you can show $A_f^\ast \psi=f\psi$, you are done (since $A_f^\ast \psi\in L^2$ by definition). However, this does not follow immediately from the equality $$ \langle A_f^\ast \psi,\phi\rangle=\int f\psi\phi\,d\mu. $$ because $f\psi$ is a priori only a measurable function and not in $L^2$ (that's what you want to prove).
I think the easiest way to prove self-adjointness here is to show that $A_f\pm i$ is invertible. Just verify that $A_{(f\pm i)^{-1}}$ is the inverse.