Let $B$ be an unbounded closed operator on a Hilbert space $H$. If $B=\int \lambda d E\_\lambda $ is positive self-adjoint and a positive bounded operator $X$ commutes with every $E\_\lambda $, then why $BX$ is positive and self-adjoint?


I am struggling in dealing with unbounded operators...


see page 48, line +6 (just consider $p=1$) in [link](https://www.cambridge.org/core/services/aop-cambridge-core/content/view/7ECF9AE5715F6C8D8A7BA8EC248B28B7/S0305004100058515a.pdf/div-class-title-isometries-of-non-commutative-span-class-italic-l-span-class-sup-p-span-span-spaces-div.pdf). I want to understand from line 5 to line 8.


I know it is symmetric, but I have no idea why it is self-adjoint.



For $B=\int \lambda d E_\lambda $ and $X$ commutes with every $E_\lambda $, why $BX$ is positive and self-adjoint?
If $B$ is positive self-adjoint then $B=A^2$ with $A$ positive self-adjoint. If $X$ is bounded non-negative and commutes with $B$, it commutes as well with a function of $B$ such as $A=\sqrt B$. Then we have 
$$
XB=BX=AAX=AXA\ge 0.
$$
The domain of $BX$ can be taken as the domain of $B$ with $XB u$ well-defined for $u\in D\_B$. Now $Y=XB$ is self-adjoint since, with $\mathbb H$ the ambient Hilbert space, 
$$
D\_{Y^\*}=\{v\in \mathbb H,\forall u\in D\_Y, \exists C, \vert\langle v, Yu\rangle\vert\le C\Vert u\Vert\},
$$
and if for all $u\in D\_B$, $\exists C$, $\vert\langle v, XBu\rangle\vert\le C\Vert u\Vert$,
then $\vert\langle Xv, Bu\rangle\vert\le C\Vert u\Vert$ and thus 
$X\nu\in D\_{B^\*}=D\_B$.


