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

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?

• For the adjoint of $A$ itself to be definable, we must have that $A$ is densely defined. Then of course $X$ is defined everywhere, so $AX$ is densely defined. If you have not understood what I have said, then please specify your definition of adjoint of an unbounded linear operator, and the definition of being self adjoint. Apr 28, 2018 at 9:32

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.
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 $$\overline{D(AX)}=H$$.