$A$ is self-adjoint and $X$ is bounded commuting with $A$, why $AX$ is densely defined?

Question

Let $A$ be a self-adjoint linear operator on a Hilbert space and $X$ is a positive bounded linear operator on this Hilbert space. Assume that $X$ strongly commute with $A$ (commute with all the spectral projections of $A$). How to show that $AX$ is densely defined?

Answer 1

If $A = \int \lambda dE(\lambda)$ is the spectral representation, then you are given that $E[a,b]$ commutes with $X$ for all finite $a,b$. $AE[a,b]$ is defined on all of $\mathcal{H}$. So, $AXE[a,b]=AE[a,b]X$ is defined on all of $\mathcal{H}$. The set of all $E[a,b]\mathcal{H}$ is dense in $\mathcal{H}$ because $E[a,b]x\rightarrow x$ as $a,b\rightarrow-\infty,\infty$, respectively.

Answer 2

Another way of seeing this is (you don't even have to assume that $A$ is self-adjoint just a densely defined $A$ will do): Since $XA\subset AX$, we have $D(XA)=D(A)\subset D(AX)\subset H$ and hence passing to the closure in $H$ as $A$ is densely defined yields $\overline{D(AX)}=H$.