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.

Careful, you have the wrong sense of "positive".
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 $(S S^* h, h) = (S^* h, S^* h) = ||S^* h||^2 \ge 0$.
