Theorem 2 (Pontryagin)

The canonical mapping $\eta$ of G into $\widehat{\widehat{G}}$ is an isomorphism of topological groups. It transforms the Haar measure $dx$ into the bidual Haar measure $d\widehat{\widehat{x}}$.

Let us first prove that $\eta$ is injective and strict. For this it is enough to show that for every neighbourhood U of $e$ in G, there exists a neighbourhood W of $e$ in $\widehat{\widehat{G}}$ such that $\overset{-1}{\eta}(W)\subset U$ (Lemma 2 of II, p. 200). Now let V be a compact symmetric neighbourhood of $e$ in G such that $V^2\subset U$, let $f$ be a positive continuous function on G, with support contained in V, and such that $f(e)>0$. Let $g=\widetilde{f}*f$. Then $g$ belongs to A(G), its support is contained in U and $g(e)>0$. Moreover, $\mathscr{F}_G(g)\in L^1(\widehat{G})$ by Prop. 11 of II, p. 217. The set W of the $\xi$ in $\widehat{\widehat{G}}$ such that

is a neighbourhood of $e$ in $\widehat{\widehat{G}}$ since the function $\overline{\mathscr{F}}_{\widehat{G}}(\mathscr{F}_G(g))$ is continuous on $\widehat{\widehat{G}}$. Let $x\in \overset{-1}{\eta}(W)$. By formula (26), we have

and therefore $|g(x)-g(e)|<\frac{1}{2}g(e)$. This implies that $g(x)\not= 0$ and therefore $x\in U$, since the support of $g$ is contained in U. Thus $\overset{-1}{\eta}(W)\subset U$.

Let us prove that the mapping $\eta$ is surjective. Since this mapping is a homeomorphism onto its image, the group $\eta (G)$ is a locally compact subgroup of $\widehat{\widehat{G}}$. It is therefore closed in $\widehat{\widehat{G}}$ (TG, III, p. 22, Cor. 2). Arguing by contradiction, suppose that there exists a character $\xi \in \widehat{\widehat{G}}$ such that $\xi \notin \eta (G)$. Then there exists (Corollary 1 of II, p. 219) a nonzero element $f$ of $L^1(\widehat{G})$ such that $\mathscr{F}_{\widehat{G}}(f)$ is zero on $\eta (G)$. Let $g\in L^1(G)$. The function $(x, \chi )\mapsto g(x)f(\chi )\langle \chi , x\rangle$ belong to $L^1(G\times \widehat{G})$. By Lebesgue-Fubini's theorem (INT, V, §8, n$^o4$, th. 1, a)), there follows therefore

Since the image of the Fourier transform is dense in $\mathscr{C}_0(\widehat{G})$ (corollary of Prop. 5 of II, p. 209), it follows that the measure $f\cdot d\chi$ is zero. This contradicts the fact that $f\not= 0$ in $L^1(\widehat{G})$, and proves that $\eta$ is surjective.

The image measure $\eta (dx)$ and the measure $\nu$ dual to the measure $d\chi$ are Haar measures on $\widehat{\widehat{G}}$. Let $f$ be a nonzero element of A(G) ; in particular $f\in L^2(G)$. By Prop. 12 of II, p. 219$,\mathscr{F}_G(f)\in L^1(\widehat{G})$ and one has

where the second equality follows from two applications of the Plancherel formula, hence the Haar measure dual to $d\chi$ is the measure $\eta (dx)$.

Henceforth we shall identify G and $\widehat{\widehat{G}}$ by means of the isomorphism $\eta$. One then has: