Fourier inversion on abelian groups

Remarks

Proof idea: one first treats the functions $f$ that are continuous and positive-definite, or finite combinations of them. Bochner's theorem writes such an $f$ as $\int \chi(x)\, d\sigma_f(\chi)$ for a measure $\sigma_f$ on $\widehat{G}$. One shows that $\sigma_f = \widehat{f}\, \nu$ for one Haar measure $\nu$ independent of $f$. The general case follows by approximation with an approximate identity.

Examples with the usual normalisations follow. For $G = \R$ with Lebesgue measure and $\chi_\xi(x) = e^{2\pi i \xi x}$, $\nu$ is Lebesgue measure, which gives the classical Fourier inversion theorem. For $G = \mathbb{T}$ with arc length divided by $2\pi$, $\nu$ is counting measure on $\Z$, and the formula is the Fourier series of $f$. For $G = \Z/n\Z$ with counting measure, $\nu$ is counting measure divided by $n$. This gives $f(a) = \frac{1}{n} \sum_{k} \widehat{f}(k) e^{2\pi i k a / n}$.

The hypothesis $\widehat{f} \in L^1$ is needed for the integral to make sense. For $f = \mathbf{1}_{[-1/2, 1/2]}$ on $\R$, $\widehat{f}(\xi) = \sin(\pi\xi)/(\pi\xi)$ is not integrable.

Going deeper

For a student

The dual measure ties together the normalisations on $G$ and $\widehat{G}$: if $\mu$ is multiplied by $c > 0$, then $\nu$ is divided by $c$. With this choice the Fourier transform extends to a unitary map $L^2(G) \to L^2(\widehat{G})$, the Plancherel theorem. A first consequence is uniqueness: if $f \in L^1(G)$ and $\widehat{f} = 0$, then $f = 0$ almost everywhere. Inversion and the Plancherel theorem are the main steps in the usual proof of the Pontryagin duality theorem.