Corollary 2

Let Φ be a filter on G invariant under translation. Let $f\in L^1(G)$ such that the Fourier transform of $f$ does not vanish and such that $\int_Gf(x)dx= 1$. Let $g\in L^{\infty}(G)$. Suppose that $f*g$ has a finite limit $\alpha$ following Φ.

a) For every function $h\in L^1(G)$ such that $\int_Gh(x)dx= 1$, the limit of $h*g$ following Φ is equal to $\alpha$;

b) Suppose in addition that $g$ is slowly oscillating following Φ. Then $g$ tends to $\alpha$ following Φ.

Replacing $g$ by $g-\alpha$, we reduce to the case where $\alpha = 0$. Let I be the set of functions $h\in L^1(G)$ such that $h*g$ tends to 0 following Φ. The set I is a vector subspace of $L^1(G)$ invariant under translation. It is a closed space. Indeed, let $h\in \overline{I}$. For every function $h_0\in L^1(G)$ and every $x\in G$, we have

For every $\varepsilon >0$, there exists $h_0\in I$ such that $\|h-h_0\|_1\|g\|_{\infty}< \varepsilon$. Let $M\in$ Φ such that $|(h_0*g)(x)|< \varepsilon$ for every $x\in M$. We then have $|(h*g)(x)|<2\varepsilon$ for every $x\in M$, hence $h*g$ converges to 0 following Φ. This shows that $h\in I$.

The space I is therefore a closed ideal of $L^1(G)$. We have $f\in I$ by assumption, hence $I = L^1(G)$ by th. 1 since the Fourier transform of $f$ does not vanish. This implies a).

Let us place ourselves under the hypotheses of b). Let $\varepsilon >0$. Since $g$ is slowly oscillating according to Φ, there exists $M\in \Phi$ and a compact neighbourhood V of $e$ such that

$x\in M$ and $y\in V =\Rightarrow |g(y^{-1}x)-g(x)|\leqslant \varepsilon$.

Let $\varphi$ be the characteristic function of V and $\mu=\int\varphi (x)dx$. For every $x\in G$, we have

Thus for every $x\in M$, we have

Since, by a), the limit of $\varphi *g$ according to Φ is zero, we have lim sup$_{\Phi}|g|\leqslant \varepsilon$. Since $\varepsilon$ is arbitrary, we conclude that the limit of $g$ according to Φ is zero, which proves b).