Proposition 10

The Fourier cotransform is an injective morphism of involutive Banach algebras of $\mathscr{M}^1(G)$ into $F_b(\widehat{G})$ which maps $L^1(G)$ into $F_0(\widehat{G})$.

Since G is compact, the space $\mathscr{C}(G)$ is included in and dense in $\mathscr{L}^1(G)$ and $\mathscr{L}^2(G)$.

Let $\nu \in \mathscr{M}^1(G)$ be such that $\overline{\mathscr{F}}_G(\nu ) = 0$. Then for every continuous function $f$ on G, one has $\overline{\mathscr{F}}_G(\nu *f) = 0$. Now $\nu *f$ belongs to $\mathscr{C}(G)$ (INT, VIII, p. 152, § 4, No.$^o2$, prop. 3). Since, by Theorem 2, the Fourier cotransform is injective on $L^2$(G), and a fortiori on the space $\mathscr{C}$ (G), one has $\nu *f= 0$ for $f\in \mathscr{C}(G)$. In particular, it follows that

for every function $f\in \mathscr{C}(G)$ (INT, VIII, loc. cit.), hence the measure $\nu$ is zero.

The image of $\mathscr{C}(G)$ under the Fourier cotransform is contained in $F^2(\widehat{G})$, and a fortiori in $F_0(\widehat{G})$. Since $F_0(\widehat{G})$ is closed in $F_b(\widehat{G})$ and $\mathscr{C}(G)$ is dense in $L^1$(G), the image of $L^1(G)$ under the Fourier cotransform is likewise contained in $F_0(\widehat{G})$.