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Let $K$ be a compact Hausdorff space and suppose $\langle \cdot, \cdot \rangle$ is an inner product on $C(K)$ such that $\langle f, g \rangle \ge 0$ whenever $f(t),g(t)\ge 0$ for all $t \in K$. It follows that the inner product is bounded, so it induces a bounded positive map $T \colon C(K \times K) \to \mathbb{R}$ satisfying $T(f \otimes g) = \langle f, g \rangle$ via identifying $C(K \times K)$ with the projective tensor product $C(K)\otimes_\pi C(K)$. By the Riesz representation theorem there is a positive Borel measure $\mu$ on $K\times K$ such that $$ \langle f, g \rangle = \int_{K\times K}f(x)g(y)\,\mathrm{d}\mu(x,y) $$ if I'm not mistaken. My question is:

  • Is $\mu$ concentrated on the diagonal $D := \{(x,x) \colon x \in K\}$, because $\langle \cdot, \cdot \rangle$ is an inner product on $C(K)$?

I would like to add that I'm not an expert in measure theory.

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  • $\begingroup$ You cannot identify $C(K\times K)$ with the projective tensor product; it identifies with the injective tensor product. So I think one needs to go back to the drawing board $\endgroup$
    – Yemon Choi
    Commented Aug 2 at 19:44
  • $\begingroup$ Hint: What is wrong with inner product $\iint_{[0,1]\times[0,1]} f(x) g(y)\,dx\,dy$ ? $\endgroup$ Commented Aug 2 at 20:22
  • $\begingroup$ @Yemon Choi Thank you for pointing that out! The drawing board it is. $\endgroup$ Commented Aug 2 at 20:56
  • $\begingroup$ @Gerald Edgar Yes, I suppose you can integrate nonzero functions to zero. $\endgroup$ Commented Aug 2 at 20:57

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