Proposition 3

The algebra $L^1(G)$ satisfies the Ditkin condition (I, p. 92, def. 2).

Let $\chi$ be a character of $L^1(G)$. Distinguish two cases according as $\chi$ is zero or nonzero. If $\chi$ is zero, it must be verified that for every function $f\in L^1(G)$, there exists a sequence $(f_n)_{n\geqslant 1}$ in $L^1(G)$ such that $\mathscr{F}(f_n)$ vanishes outside a compact subset of $\widehat{G}$ and such that $f_n*f$ tends to $f$ in $L^1(G)$. The existence of such a sequence follows from Lemma 1 above and Proposition 8 of II, p. 211.

Suppose now that $\chi$ is nonzero, hence $\chi \in \mathsf{X}(L^1(G)) =\widehat{G}$ (Proposition 1 of II, p. 202). Let $f\in L^1(G)$ be such that $\mathscr{G}_{L^1(G)}(f)(\chi ) =$ $\overline{\mathscr{F}}(f)(\chi ) = 0$. It is a question of proving the existence of a sequence $(f_n)_{n\geqslant 1}$ in $L^1(G)$ such that $f*f_n$ converges to $f$ in $L^1(G)$ and such that $\overline{\mathscr{F}}(f_n)$ vanishes in a neighbourhood of $\chi$. We may suppose that $\|f\|_1= 1$. By translation in $\widehat{G}$, we reduce to the case where $\chi =e$.

Let $K_n$ be a compact subset of G such that

Let $u_n\in L^1(G)$ be a function $\geqslant 0$ such that $\|u_n\|_1= 1$ and

(cf. Proposition 8 of II, p. 211, (iii)). According to Lemma$\surd$ 2, there exists a function $j_n$ in $L^1(G)$ such that $\|j_n\|_1\leqslant \overline{2}$, whose Fourier cotransform is equal to 1 in a neighbourhood of $e$, and moreover such that $\|j_n-j_n*\varepsilon_x\|_1\leqslant n^{-1}$ for every $x\in K_n$. Put

We shall prove that the sequence $(f_n)_{n\geqslant 1}$ has the required properties. First of all, we have

hence the Fourier transform of $f_n$ vanishes in a neighbourhood of $\chi =e$. On the other hand,

1

Now, for almost every $y\in G$, we have

since, by assumption, we have $\mathscr{F}(f)(e) =\int_Gf(x)dx= 0$. Hence

$G-K_n$

$1\int\int$ 5 $\leqslant |f(x)|dx+ 4|f(x)|dx\leqslant$.

$n_{K_n}G-K_{_n}n$

Finally, $\|f*f_n-f\|_1\leqslant 6n^{-1}$ and therefore $f*f_n$ converges in $f$ in $L^1(G)$, as desired.

Applying prop. 5, we obtain the following result: