For a multiplication operator $M_f$ on $L^2$ with $f\geq 0$, is $SM_fS^{*}$ positive? I have the following problem. Let $\Omega \subset R^n$ have finite measure, let $H = L^2(\Omega)$ and let $S: H \to H$ be a bounded linear operator. Then it is well known that $P = SS^*$ is a positive operator, i.e., $(Px, x) \geq 0$. for all $x \in H$. Now let $M_f:H \to H$ be the multiplication operator induced by $f: \Omega \to R$ where $f \geq 0$ (or even $f \geq c > 0$). Is it true that $P_1 = SM_fS^*$ is also positive? If so, do you have a proof or reference?
Thanks in advance.

Edit: this answer addresses positivity in the sense of $\langle Px,x\rangle \geq 0$ for all $x$, not positivity in the sense $Pg \geq 0$ for all $g \geq 0$ (there was some confusion in an earlier version of the question). See Nate's answer for a discussion.

Yes that's true.
Since $f$ has a square root $g$, we have $f= g^2 = gg^\ast$, and thus 
$$SM_{f}S^\ast = SM_{gg^\ast}S^\ast = SM_g M_{g}^{\ast}S^\ast = (SM_g)(SM_g)^\ast \geq 0,$$ as desired.

Added: Of course this is true for every positive operator: Every positive operator $P$ can be written as $P = Q^2$ with $Q$ self-adjoint and positive. Hence $SPS^\ast = SQQ^\ast S^\ast = (SQ)(SQ)^\ast \geq 0$.
