Remark 2

Let $f\in L^2(G)$. One then has the Plancherel formula

Moreover, by Theorem 2$,f$ is the sum in $L^2(G)$ of the family $(f_{\pi})_{\pi\in\widehat{G}}$ where

$=\langle \pi (x)|\pi (f)\rangle =$ dim($\pi$ )$\int_Gf(g)\chi_{\pi}(x^{-1}g)d\mu(g)$

for every $x\in G$ and every $\pi \in \widehat{G}$.

Suppose that G is commutative. Since the irreducible representations of G are of dimension 1 (cor. 7 of V, p. 390) and the dual group $\widehat{G}$ is discrete (prop. 18 of II, p. 233), the algebras $F_b(\widehat{G})$ and $F_0(\widehat{G})$ are identified, respectively, with the algebra $\mathscr{C}_b(\widehat{G})$ of bounded continuous functions on $\widehat{G}$ and the algebra $\mathscr{C}_0(\widehat{G})$ of continuous functions tending to 0 at infinity on $\widehat{G}$. Since G is compact, the Haar measure on $\widehat{G}$ dual to $\mu$ is the counting measure $\widehat{\mu}$ (prop. 18 of II, p. 233). The Hilbert sum $F^2(\widehat{G})$ of the spaces End(E$_{\pi}$) for $\pi \in \widehat{G}$ is identified with the Hilbert space $L^2(\widehat{G},\widehat{\mu})$.

Let $\nu \in \mathscr{M}^1(G)$. Then, for every unitary character $\chi \in \widehat{G}$, one has

which proves that the Fourier cotransform defined above coincides with the Fourier cotransform of G defined in II, p. 206.

Every $f\in \mathscr{L}^2(G)$ is integrable and the formula $\|\overline{\mathscr{F}}_G(f)\|_2=\|f\|$ is the Plancherel formula (II, p. 215, Theorem 1).