Proposition 5

Assume $\chi$ a continuous multiplier and each function $\chi(s, \cdot)$ bounded. (i) The function $s \mapsto \rho(s) = \sup_{x \in X} \chi(s^{-1}, x)$ on $G$ is lower semi-continuous $> 0$ and satisfies $\rho(st) \leq \rho(s)\rho(t)$ for all $s, t$ in $G$. (ii) Let $\mu \in \mathscr{M}^\rho(G)$ and $f \in L^\infty(X, \beta)$.¹ Then $\mu$ and $f$ are convolvable and $\mu *^\beta f$ is given locally almost everywhere by the formula (3) of No. 1. One has $\mu *^\beta f \in L^\infty(X, \beta)$, and $\| \mu *^\beta f \|_\infty \leq \| \mu \|_\rho \| f \|_\infty$.

¹For a function $f$, the expression "$f \in L^\infty(X, \beta)$" is an abuse of notation signifying that, depending on the context, the symbol $f$ is to be interpreted either as a function or the equivalence class of a function. In particular, the symbol $\mu *^\beta f$ can stand for either a function defined locally $\beta$-almost everywhere, or the equivalence class of such a function for the relation of equality locally $\beta$-almost everywhere.

(iii) *If, moreover, $f \in \mathscr{C}^\infty(X)$ (resp. $\overline{\mathscr{K}(X)}$), then formula (3) of No. 1 defines for every $x$ a convolution product $\mu *^\beta f$ that belongs to $\mathscr{C}^\infty(X)$ (resp. $\mathscr{K}(X)$).

(iv) *If $f \in \mathscr{K}(X)$, then the convolution product $\mu *^\beta f$ defined by (3) is none other than the element $\gamma_\chi(\mu)f$ defined by the continuous representation $\gamma_\chi$ of $G$ in $\overline{\mathscr{K}(X)}$.

The identity $\chi(st, x) = \chi(s, tx)\chi(t, x)$ implies at once that $\rho(st) \leq \rho(s)\rho(t)$. On the other hand, $\rho$ is lower semi-continuous, being the upper envelope of continuous functions.

Let $\mu \in \mathscr{M}^\rho(G)$. By Prop. 1 of No. 1, $\mu$ and 1 are convolvable; Prop. 2 (i) shows that $(|\mu| *^\beta 1)(x) \leq \int_G \rho(s) d|\mu|(s)$ locally $\beta$-almost everywhere. Therefore, if $f$ is $\beta$-measurable and $|f| \leq 1$, then $\mu$ and $f$ are convolvable and $N_\infty(\mu *^\beta f) \leq \int \rho(s) d|\mu|(s)$. Moreover, $\mu *^\beta f$ is given locally almost everywhere by formula (3) of No. 1, because condition (iii) of Prop. 2 of No. 1 is satisfied. This implies (ii).

Suppose $f$ continuous and bounded by 1 in absolute value. It is clear that the integrals occurring in (3) exist for all $x \in X$. Let us show that they depend continuously on $x$. We can suppose $\mu \geq 0$. Let $x_0 \in X$ and $\varepsilon > 0$. Let $K$ be a compact subset of $G$ such that $\int_{G - K} \rho(s) d\mu(s) \leq \varepsilon$. There exists a neighborhood $V$ of $x_0$ in $X$ such that $x \in V$ implies

for $s \in K$. Then, for $x \in V$,

whence our assertion. Suppose that in addition $f \in \overline{\mathscr{K}(X)}$. Let $H$ be a compact subset of $X$ such that $|f(y)| \leq \varepsilon$ for $y \notin H$. Let $x \notin KH$. Then $s^{-1}x \notin H$ for $s \in K$, therefore

which completes the proof of (iii).

Finally, if $f \in \mathscr{K}(X)$ then, since $\varepsilon_x \in \mathscr{M}^1(X)$ for all $x \in X$, we have

thus $\gamma_\chi(\mu)f$ is the convolution product $\mu *^\beta f$ defined by (3).