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

Question

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.

Answer 1

I'm assuming $B=\int \lambda dE(\lambda)$ is the spectral resolution of the positive selfadjoint operator $B$. However, even in this case, your assertion is not true because $X=-B$ commutes with $B$ and, hence, with every $E(\lambda)$, but $XB=-B^2$ is definitely not positive and selfadjoint.