Plancherel theorem for abelian groups

Remarks

The measure $\nu$ is called the dual measure of $\mu$; scaling $\mu$ by $c > 0$ scales $\nu$ by $1/c$. One proof applies Bochner's theorem to functions $f * \tilde{f}$, where $\tilde{f}(x) = \overline{f(-x)}$, whose transform is $\abs{\widehat{f}}^2$. This yields a Haar measure on $\widehat{G}$ for which the inversion formula holds for such functions. Evaluating $f * \tilde{f}$ at $0$ then gives the isometry, and density of $L^1 \cap L^2$ in $L^2$ gives the extension. Example: for $G = \mathbb{T}$ with normalised arc length, $\widehat{G} = \Z$, $\nu$ is counting measure, and the theorem is Parseval's identity $\frac{1}{2\pi}\int_0^{2\pi} \abs{f(\theta)}^2 \, d\theta = \sum_{n \in \Z} \abs{\widehat{f}(n)}^2$. For $G = \R$ with Lebesgue measure and characters $x \mapsto e^{2\pi i \xi x}$, the dual measure is again Lebesgue measure. The general Plancherel theorem for groups extends the result to some non-abelian groups, where the dual is the unitary dual and $\nu$ becomes the Plancherel measure.

Going deeper

For a student

The theorem makes the Fourier transform a unitary operator on $L^2$, so one can transform any square-integrable function, not only integrable ones. It fixes a normalisation of the Haar measure on the dual group, which is the one used in the Fourier inversion formula. Together with Pontryagin duality it shows that $L^2(G)$ is diagonalised by the characters: translation operators become multiplication operators on $L^2(\widehat{G})$.