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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?

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  • $\begingroup$ 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. $\endgroup$ Apr 28, 2018 at 9:32

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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.

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  • $\begingroup$ Simply and beautiful proof. $\endgroup$
    – user92646
    Apr 28, 2018 at 23:21
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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 $\overline{D(AX)}=H$.

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