Suppose $X,Y$ to be compact spaces and $f\in C(X\times Y)$ different from zero. Is there two probability measures $\mu_f\in C(X)_+^*$, $\nu_f\in C(Y)_+^*$ such that $\mu_f\otimes\nu_f(|f|^2)>0$. In other words, do the product measure separate the elements of $C(X\times Y)$? If the answer is yes (as I suppose), I need a reference where such a proof is exhibited. If the answer is no, I need a counterexample.

  • $\begingroup$ Not sure if I am understanding your question correctly, but assuming that $X$ and $Y$ are Hausdorff, how about taking $(x_0, y_0) \in X \times Y$ such that $f(x_0, y_0) \neq 0$ and then setting $\mu_f = \delta_{x_0}$ and $\nu_f = \delta_{y_0}$? $\endgroup$ – Sangchul Lee Oct 7 '18 at 11:20
  • $\begingroup$ for "points" I mean the (continuous) functions in $C(X\times Y)$ $\endgroup$ – francesco fidaleo Oct 7 '18 at 11:24
  • $\begingroup$ What I mean is that, for $\mu_f$ and $\nu_f$ chosen as above, we should have $ (\mu_f \otimes \nu_f)(|f|^2) = |f(x_0, y_0)|^2 > 0$. I am not sure if this answers your question. And of course, if $X$ or $Y$ is no longer Hausdorff, I have no good idea... $\endgroup$ – Sangchul Lee Oct 7 '18 at 11:26
  • $\begingroup$ Perfect! This seems the answer. Indeed, I was searching for the noncommutative version of this problem: Let $A$ and $B$ be two unital $C^*$-algebras and $x\in A\otimes B$ (the algebraic tensor product), different from $0$. Is there $\omega\in A^*_+$ and $\varphi\in B^*_+$ such that $\omega\times \varphi(x^*x)>0$. You are suggesting to look at pure states. $\endgroup$ – francesco fidaleo Oct 7 '18 at 11:33

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