Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 4 down vote accepted

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

share|cite|improve this answer

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 $(S S^* h, h) = (S^* h, S^* h) = ||S^* h||^2 \ge 0$.

share|cite|improve this answer
+1 good point! I missed that, obviously. – t.b. Oct 6 '11 at 3:16
oh wow. the wikipedia article needs to be read very carefully, because they have both definitions in the same section. thanks, I will edit the post. luckily, my application can still be rescued :) – user12014 Oct 6 '11 at 3:59

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.