Proposition 18

In order that $\widehat{G}$ be compact, it is necessary and sufficient that G be discrete. If G is discrete, the dual measure of the counting measure on G is the normalised Haar measure on $\widehat{G}$. If G is compact, the dual measure of the normalised Haar measure is the counting measure on $\widehat{G}$.

Suppose G discrete, and let $\alpha$ be the counting measure on G. Let $\varphi$ be the characteristic function of $e\in G$. One has $\mathscr{F}_G(\varphi ) = 1$ on $\widehat{G}$. Since $\mathscr{F}_G(\varphi )$ tends to 0 at infinity, the group $\widehat{G}$ is compact.

Moreover, for the dual measure $\widehat{\alpha}$ of $\alpha$, the function $\mathscr{F}_G(\varphi )$ must have integral $\varphi (e) = 1$ (prop. 12 of II, p. 219). Hence $\widehat{\alpha}(\widehat{G}) = 1$.

Suppose G compact. Then the Haar measure $dx$ belong to $\mathscr{M}^1(G)$. Its Fourier transform is strictly positive at $\chi =e$ and zero for $\chi \not= 0$ (prop. 6 of II, p. 210). Since it is continuous on $\widehat{G}$, the group $\widehat{G}$ is discrete. If the measure of G is 1, one deduces by duality from the preceding case that the dual measure of the measure $dx$ is the counting measure on $\widehat{G}$.