Lemma 3

For every $f\in A(G)$, there exists a unique bounded measure $\mu_f$ on $\widehat{G}$ such that

for every $\varphi \in$ Stell(G).

Moreover, for every $f$ and $g$ in A(G), we have the equality

between the measure with density $\mathscr{F}(f)$ with respect to $\mu_g$ and the measure with density $\mathscr{F}(g)$ with respect to $\mu_f$.

Let $f,g$ be elements of $L^1(G)\cap L^2(G)$. For every $\varphi \in$ Stell(G), one has $\varphi *f\in L^2(G)$ and $\|\varphi *f\|_2\leqslant \|\varphi \|_*\|f\|_2($I, p. 126, formula (8)). Moreover, one has $\varphi *(f*g) = (\varphi *f)*g($loc. cit., formula (9)). This latter function belongs to the closure $\mathscr{C}_0(G)$ of $\mathscr{K}(G)$ in $\mathscr{C}(G)$ (INT, VIII, §4, n$^o5$, prop. 15). Moreover, one has

Since the functions $f*g$ for $f$ and $g$ in $L^1(G)\cap L^2(G)$ generate A(G), it follows that $\varphi *f\in \mathscr{C}_0(G)$ for all $f\in A(G)$ and $\varphi \in$ Stell(G), and that the mapping $\varphi \mapsto (\varphi *f)(e)$ is a continuous linear form on Stell(G). Since $\mathscr{F}$ is an isomorphism of Stell(G) onto $\mathscr{C}_0(\widehat{G})$ (prop. 5 of II, p. 209), the first assertion follows.

Now let $f$ and $g$ be in A(G). For $\varphi \in L^1(G)$, one has

Since $(\varphi *f)*g= (\varphi *g)*f$ and since the image of $L^1(G)$ under the Fourier transform is dense in $\mathscr{C}_0(\widehat{G})$ (cor. of II, p. 210), one deduces from formula (19) that formula (18) is satisfied for all $f$ and $g$ in A(G).