# For $B=\int \lambda d E_\lambda$ and $X$ commutes with every $E_\lambda$, why $BX$ is positive and self-adjoint?

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. 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.

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 $X\nu\in D_{B^*}=D_B$.