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Let $G$ be a discrete Abelian group and denote by $\widehat G$ the (compact) Pontryagin-Van Kampen dual of $G$. I was reading in a paper of Justin Peters that Fourier Transform induces a bijection between the following sets of functions:

(1) $L^1(G)^+\cap \mathcal P(G)$ (continuous, non-negative, positive-definite and absolutely integrable functions $G\to \mathbb C$);

(2) $L^1(\widehat G)^+\cap \mathcal P(\widehat G)$ (continuous, non-negative, positive-definite and absolutely integrable functions $\widehat G\to \mathbb C$).

Peters says that this is easy to deduce from Fourier Inversion Theorem but it does not seem so elementary to me. Can anyone help me? Is this true for any LCA group?

Is there any analog if we start with a compact non-commutative group and we use the Tannaka-Krein duality?

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Maybe I should add some motivation to my question. When we have a duality we can pass from one side (read "category") to the other any kind of categorical statements. Anyway, sometimes one has non-categorical statements, this happens for example when speaking about finite subsets of a discrete Abelian group G. The idea of Peters is to associate to a given finite subset $S\subseteq G$ the function $\chi_S*\chi_{-S}\in L^{1}(G)^+\cap\mathcal P(G)$, in this way, using the correspondence I'm asking about, he passes non-categorical information from the discrete to the compact side. – Simone Virili Jul 5 '12 at 12:14
It's not really my area, but I think the point is that a function is pointwise positive if and only if its Fourier transform is positive definite (Bochner's theorem). Also the transform of any positive definite function is a finite measure, but if you start with an $L^1$ function then its transform is a function; and if it's both a function and a finite measure then it's an $L^1$ function. So any $L^1$ function which is positive and positive definite will transform to an $L^1$ function which is positive and positive definite. Now invoke the inversion theorem. – Nik Weaver Jul 5 '12 at 15:11
Ok, I see that my problem is that I'm not completely understand Bochner's Theorem. How can you reformulate it as in your comment? – Simone Virili Jul 5 '12 at 17:26
Maybe the point is that you only need to prove one inclusion (Fourier maps first space into second), because the Fourier inversion theorem says that Fourier inverse is the same as Fourier, up to a flip (when applied to an integrable function with integrable Fourier transform). This implies the other inclusion by duality (exchange $G$ with its dual). But now Bochner easily implies the first inclusion. And note that this applies to any LCA group $G$. – BS. Jul 6 '12 at 10:24
up vote 1 down vote accepted

I try to write down in full detail the answers of Nik and BS (many thanks to both!).

The references in the proof are to Folland's book "A Course in Abstract Harmonic Analysis" and Rudin's book "Fourier Analysis on Groups".

Let $G$ be an LCA group, then {$\widehat\phi:\phi\in L^1(G)^+\cap \mathcal P(G) $}$=L^1(\widehat G)^+\cap \mathcal P(\widehat G)$.

Proof. Let us start proving the inclusion {$\widehat\phi:\phi\in L^1(G)^+\cap \mathcal P(G)$}$\subseteq L^1(\widehat G)^+\cap \mathcal P(\widehat G)$. Indeed, consider $\phi \in L^1(G)^+\cap \mathcal P(G)$, then $\widehat\phi\in L^1(\widehat G)$ by the Fourier Inversion Theorem (see [Rudin, page 22, line 1]) and $\widehat\phi\geq 0$ by [Folland, Corollary 4.23]. Let now $\mu_\phi$ be the non-negative and bounded (as $\phi\in L^1(G)^+$) regular measure defined on a generic Borel subset $E$ of $G$ by $\mu_\phi(E)=\int_{x\in E}\phi(x)d\mu(x)$ (here $\mu$ is a fixed Haar measure on $G$). One can show that $$\widehat\phi(\gamma)=\int_{x\in G}\phi(x)\gamma(-x)d\mu(x)=\int_{x\in G}\gamma(-x)d\mu_\phi(x)\, .$$ By Bochner's Theorem (see [Rudin, page 19]), $\widehat\phi\in \mathcal P(\widehat G)$.

On the other hand, let $\phi\in L^1(\widehat G)^+\cap \mathcal P(\widehat G)$ and $\psi$ be the function defined by $\psi(\gamma)=\phi(-\gamma)$ for all $\gamma\in \widehat G$. It is not difficult to see that $\psi\in L^1(\widehat G)^+\cap \mathcal P(\widehat G)$. By the first part of the proof, $\widehat\psi\in L^1( G)^+\cap \mathcal P( G)$ and, using Fourier Inversion Theorem (see [Folland, Theorem 4.32]), one obtains that $\widehat{\widehat\psi}=\phi$, which is therefore the Fourier transform of a function in $ L^1( G)^+\cap \mathcal P( G)$.\\\

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