Question about positive operators on a Hilbert space

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?

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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 $SPS^\ast = SQQ^\ast S^\ast = (SQ)(SQ)^\ast \geq 0$.

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An operator of the form $A = S S^*$ on $L^2$ need not have the property that $Ah \ge 0$ for any $h \ge 0$ (this is sometimes called "positivity-preserving"). Even in finite dimensions this may fail: take $$S = \begin{pmatrix} 1 & 2 \\ 1 & -2 \end{pmatrix}, \quad x = \begin{pmatrix}1 \\ 0 \end{pmatrix}.$$ Then $x \ge 0$, but $S S^* x = \begin{pmatrix}5 \\ -3 \end{pmatrix}$.
Instead, $A = S S^*$ is positive in the sense that $(A h, h) \ge 0$ for any $h$ in the Hilbert space. We might say that $A$ is "positive semidefinite". This latter property is obvious in this case because $(S S^* h, h) = (S^* h, S^* h) = ||S^* h||^2 \ge 0$.