We may suppose that $\mu \geqslant 0$. Let $F \in \mathscr{K}_+(\mathbf{G} \times \mathbf{X})$. Then

Now, the function $(s, x) \mapsto F(s, x)\chi(s, x)$ has compact support and is $(\mu \otimes \beta)$-measurable. By Ch. V, §8, No. 3, Prop. 7, the preceding equality proves that this function is $(\mu \otimes \beta)$-integrable and that

This proves at the same time both assertions of Lemma 1.

We may assume that $f \geqslant 0$ and $\mu \geqslant 0$. Let $h \in \mathscr{K}_+(\mathbf{X})$. We are to prove that the function $(s, x) \mapsto h(sx)$ is essentially integrable for $\mu \otimes (f \cdot \beta) = (1 \otimes f) \cdot (\mu \otimes \beta)$ (Ch. V, §8, No. 5, Prop. 10), that is, that $\iint^\bullet h(sx)f(x)\varphi_K(s) d\mu(s) d\beta(x) < +\infty$ (Ch. V, §5, No. 3, Prop. 3); it will clearly suffice to prove that there exists an $a > 0$ such that for every compact subset $K$ of $\mathbf{G}$,

By Lemma 1, Now, the function $(s, x) \mapsto h(x)f(s^{-1}x)\varphi_K(s)\chi(s^{-1}, x)$ is $(\mu \otimes \beta)$-measurable (Lemma 1) and has compact support. The preceding expression is therefore equal (Ch. V, §8, No. 3, Prop. 7) to

where S denotes the support of h. Whence the proposition.

Let $h \in \mathscr{H}(X)$. Since $\mu$ and f are convolvable, the function $(s, x) \mapsto h(sx)f(x)$ is essentially $(\mu \otimes \beta)$-integrable. By Lemma 1, the function $(s, x) \mapsto h(x)f(s^{-1}x)\chi(s^{-1}, x)$ is essentially $(\mu \otimes \beta)$-integrable. Under hypothesis (i) or (ii) of the statement, one then deduces that this function is $(\mu \otimes \beta)$-integrable; for, in the first case it is continuous and one applies Prop. 3 of Ch. V, §1, No. 1, and in the second case it is zero outside a countable union of compact sets, and one applies Prop. 7, 2) of No. 2, loc. cit. By the Lebesgue–Fubini theorem,

the function $x \mapsto g(x) = \int f(s^{-1}x)\chi(s^{-1}, x)\, d\mu(s)$ being moreover locally $\beta$-integrable. One thus sees that

whence $g = \mu *_{\beta} f$.

Suppose now that $\mu$ is carried by the union S of a sequence of compact sets. The function

is essentially $(\mu \otimes \beta)$-integrable, and zero outside a countable union of compact sets, hence $(\mu \otimes \beta)$-integrable. Since $\mu = \varphi_S \cdot \mu$, the argument concludes as before.

We know that $\mu$ and $f$ are convolvable (§3, No. 2, Prop. 8 (i)). On the other hand, with the notations of §2, we have

since $\mathscr{C}(X)$ is complete. In particular, for every $x \in X$, This, combined with Prop. 2 (i), and §2, No. 6, proves (ii). Finally, if $f \in \mathscr{K}(X)$ then $\mu * (f \cdot \beta)$ has compact support (§3, No. 2, Prop. 9), therefore $\mu *_{\beta} f \in \mathscr{K}(X)$. For, let us consider the continuous representation $U$ of $G$ in the completion $\mathscr{K}(X)^{\sim}$ obtained by extending by continuity the continuous operators $\gamma_\chi(s)$ in $\mathscr{K}(X)$ (§ 2, No. 1, Remark 3). Let $S$ be the support of $\mu$. The functions $\gamma_\chi(s)f$, for $s \in S$, have their support contained in a fixed compact set $K$. The set $\mathscr{K}(X, K)$ is a complete linear subspace of $\mathscr{K}(X)$. Therefore $U(\mu)f \in \mathscr{K}(X)$. One sees as before that $U(\mu)f = \mu *^\beta f$, and (iii) again follows from § 2, No. 6.

(i) $\mu$ and $f$ are convolvable relative to $\beta$.

(ii) Formula (3) of No. 1 defines for every $x \in X$ a convolution product $\mu *^\beta f$ that is continuous.

(iii) The mapping $(\mu, f) \mapsto \mu *^\beta f$ of $\mathscr{M}(G) \times \mathscr{K}(X)$ into $\mathscr{C}(X)$ is hypocontinuous relative to the bounded subsets of $\mathscr{M}(G)$ and the compact subsets of $\mathscr{K}(X)$ that are contained in some subspace $\mathscr{K}(X, L)$ (where $L$ is a variable compact subset of $X$).

We know that $\mu$ and $f$ are convolvable (§ 3, No. 2, Prop. 8 (ii)), and it is clear that the integrals occurring in (3) exist for every $x \in X$. Let $K$ and $L$ be two compact subsets of $X$. There exists a compact subset $H$ of $G$ such that the relations $x \in K$ and $s^{-1}x \in L$ imply $s \in H$; let $\varphi \in \mathscr{K}_+(G)$ with $\varphi(s) = 1$ for $s \in H$. Then, for $f \in \mathscr{K}(X, L)$ and $x \in K$,

Consequently $\int f(s^{-1}x)\chi(s^{-1}, x)\, d\mu(s)$ is a continuous function of $x$ and defines a convolution product $\mu *^\beta f \in \mathscr{C}(X)$. Moreover, the mapping $\mu \mapsto \varphi \cdot \mu$ of $\mathscr{M}(G)$ into $\mathscr{C}'(G)$ is continuous for the topologies of compact convergence. Prop. 3 (iii) therefore implies that the mapping $(\mu, f) \mapsto \mu *^\beta f$ of $\mathscr{M}(G) \times \mathscr{K}(X, L)$ into $\mathscr{C}(X)$ is, for every compact subset $L$ of $X$, hypocontinuous relative to the compact subsets of $\mathscr{K}(X, L)$. In particular, the mapping $(\mu, f) \mapsto \mu *^\beta f$ of $\mathscr{M}(G) \times \mathscr{K}(X)$ into $\mathscr{C}(X)$ is separately continuous. Since $\mathscr{K}(X)$ is barreled, this mapping is hypocontinuous relative to the bounded subsets of $\mathscr{M}(G)$ (TVS, III, § 5, No. 3, Prop. 6).

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).

We have

by §2, No. 5, formula (5). On the other hand, the mapping $s \mapsto \gamma_\chi(s)f$ of $G$ into $L^p(X, \beta)$ is continuous (§2, No. 5, Prop. 9). Therefore this mapping is $\mu$-integrable. Let We have $\|g\|_p \leq (\int \rho^{1/q}(s) d|\mu|(s)) \|f\|_p$. Applying the preceding remarks to $|f|$, one sees that the mapping $s \mapsto \varepsilon_s * |f|$ of $G$ into $L^p(X, \beta)$ is $\mu$-integrable, therefore that, for every $h \in \mathscr{H}(X)$, the mapping $s \mapsto \langle h, \varepsilon_s * (|f| \cdot \beta) \rangle$ is $\mu$-integrable. Prop. 7 of §1, No. 5 then proves that $\mu$ and $f \cdot \beta$ are convolvable. Moreover, and this last integral is equal to $\langle h, \mu * (f \cdot \beta) \rangle$ by Prop. 7 of §1, No. 5. One therefore sees that $g$ is a convolution product of $\mu$ and $f$. This convolution product is given locally $\beta$-almost everywhere by (3), by Prop. 2 and the Remark that follows it.

2For a function $f$, the expression " $f \in L^p(X, \beta)$ " is an abuse of notation signifying that, depending on the context, the symbol $f$ is to be interpreted either as a function defined $\beta$-almost everywhere, or as the equivalence class of such a function for the relation of equality $\beta$-almost everywhere. Thus $f \in L^p$ can symbolize either $f \in \mathscr{L}^p$ or $\dot{f} \in L^p$.

By an abuse of notation, it is often one of the functions $g$ of the statement that is denoted $\mu *^\beta f$, which permits writing

If $X$ is countable at infinity, this style of notation is, moreover, entirely justified.

This follows from Props. 5 and 6 and the associativity of the convolution product.

Since $g(s, x)$ is integrable for $\mu \otimes \nu$, the function $f(sx)$ is essentially integrable for $(\chi \cdot \check{\mu}) \otimes \nu$ and $f$ is essentially integrable for $(\chi \cdot \check{\mu}) * \nu$. By the Lebesgue–Fubini theorem, $\mu *_{\beta} f = \int g(s, x) d\mu(s)$ is $\nu$-integrable and