Lemma 6

Let $\varphi \in L^1(\widehat{G})\cap L^2(\widehat{G})$. Then $f=\overline{\mathscr{F}}_{\widehat{G}}(\varphi )\circ \eta$ belongs to $L^2(G)$ and $\mathscr{F}_G(f) =\varphi$ in $L^2(\widehat{G})$.

The function $f$ is continuous and bounded on G since $\varphi \in L^1(\widehat{G})$. For every function $g\in L^1(G)\cap L^2(G)$, one has

applying the Lebesgue-Fubini theorem (INT, V, §8, n$^o4$, th. 1, a)) to the function $(x, \chi )\mapsto g(x)\varphi (\chi )\langle \chi , x\rangle$ which is integrable on $G\times \widehat{G}$ with respect to the product measure $dx\otimes d\widehat{x}$. It follows that

by Plancherel's formula. The linear form $g\mapsto \int_Gf g$ is therefore continuous on $L^1(G)\cap L^2(G)$, and since $L^1(G)\cap L^2(G)$ is dense in the Hilbert space $L^2(G)$, it follows that $f$ belongs to $L^2(G)$.

Applying then Theorem 1 of II, p. 215, one obtains on the other hand

for every $g\in L^2(G)$. Comparing with (28), one concludes that $\varphi =\mathscr{F}_G(f)$ in $L^2(\widehat{G})$, since A(G) is contained in $L^1(G)\cap L^2(G)$ and $\widehat{A}(G)$ is dense in $L^2(\widehat{G})$ (Lemma 5 of II, p. 215).

Proposition 12 (Fourier inversion formula)

Let $f\in L^2(G)$ be such that $\mathscr{F}_G(f)\in L^1(\widehat{G})$. Then one has $f$ = $\overline{\mathscr{F}}_G(\mathscr{F}_G(f))\circ \eta$ in $L^2(G)$. In other words, for almost every $x\in G$, one has

The function $\varphi =\mathscr{F}_G(f)$ belongs to $L^1(\widehat{G})\cap L^2(\widehat{G})$, and the desired formula is obtained by applying the lemma.