I'm trying to do the following exercise:
Let $M: \mathcal{D}(M) \subset L^2(\mathbb{R}) \rightarrow L^2(\mathbb{R})$ be the multiplication operator $M f=x f$ with $$ \mathcal{D}(M)= \{f \in L^2(\mathbb{R}) : x f \in L^2(\mathbb{R})\} $$ Show that $M$ is self-adjoint.
After a fair amount of searching, I've been able to locate this answer: A multiplicative operator is self-adjoint
However, I have failed to understand the "how that $A_f\pm i$ is invertible. Just verify that $A_{(f\pm i)^{-1}}$ is the inverse" part, including why the invertibility of $A_f\pm i$ implies the self-adjointness of $M$. Could anyone help me understand this, or would there be another approach? I think it is straightforward to show that $M$ is symmetric, but I am having trouble showing $\mathcal{D}(M^*) \subset \mathcal{D}(M)$ and have been unable to make any progress except write out definitions and stare blankly at it.
Assume $g\in D(M^*).$ Then for any $f\in D(M)$ there is $h\in L^2$ such that $$\langle xf,g\rangle=\langle Mf,g\rangle =\langle f,h\rangle$$ In particular this holds for $f:=xg\mathbf{1}_{[-n,n]}.$ Thus $$\int\limits_{-n}^nx^2|g(x)|^2\,dx =\langle f,h\rangle\le \|f\|_2\|h\|_2\\ =\|h\|_2\left (\int\limits_{-n}^nx^2|g(x)|^2\,dx\right )^{1/2}$$ Hence $$\left (\int\limits_{-n}^nx^2|g(x)|^2\,dx\right )^{1/2}\le \|h\|_2$$ As $n$ is arbitrary, we get $xg\in L^2,$ i.e. $g\in D(M).$