Lemma 4

There exists a unique measure $\nu$ on $\widehat{G}$ such that

for every $f\in A(G)$. For $f\in A(G)$, one has $\mathscr{F}(f)\in L^1(\widehat{G}, \nu )\cap L^2(\widehat{G}, \nu )$.

Let $f\in A(G)$. Let $\Omega_f$ denote the open set in $\widehat{G}$ formed by the $\chi \in \widehat{G}$ such that $\mathscr{F}(f)(\chi )\not= 0$. Let $\varphi$ be the characteristic function of $\widehat{G}-\Omega_f$. Then for every $g\in A(G)$, one has

in view of formula (18).

By corollary 1 of II, p. 212, the image $\widehat{A}(G)$ of A(G) under the Fourier transform is dense in $\mathscr{C}_0(\widehat{G})$. One then deduces from the preceding formula that $\varphi \cdot \mu_f= 0$, hence that $\mu_f$ is concentrated on $\Omega_f$ (INT, IV, §4, n$^o7$, def. 4). Let $\nu_f$ be the measure on $\Omega_f$ with density $\mathscr{F}(f)^{-1}$ with respect to $\mu_f|\Omega_f$.

The sets $\Omega_f$, for $f\in A(G)$, form an open covering of $\widehat{G}$ (cor. 2). For all $f$ and $g$ in A(G), formula (18) shows that $\nu_f|(\Omega_f\cap \Omega_g) =\nu_g|(\Omega_f\cap \Omega_g)$. Consequently, there exists a unique measure $\nu$ on $\widehat{G}$ such that one has $\nu_f=\nu |\Omega_f$ for every $f\in A(G)$ (INT, III, §2, No.$^o1$, prop. 1).

If $f\in A(G)$, the measures $\mu_f$ and $\mathscr{F}(f)\cdot \nu$ are concentrated on $\Omega_f$, and their restrictions to $\Omega_f$ are equal to $\mathscr{F}(f)\cdot \nu_f$; these measures are therefore equal.

Since $\mu_f$ is a bounded measure, the Fourier transform $\mathscr{F}(f)$ belongs to the space $L^1(\widehat{G}, \nu )$. Since moreover $\mathscr{F}(f)$ belongs to $\mathscr{C}_0(\widehat{G})$, one has also $\mathscr{F}(f)\in L^2(\widehat{G}, \nu )$.

Formula (17) now takes the form, for $\varphi \in$ Stell(G) and $f\in A(G)$: (20) $(\varphi *f)(e) =\int_{\widehat{G}}\mathscr{F}(\varphi )\mathscr{F}(f)d\nu$.

In particular, for $f$ and $g$ in A(G), one has