(i) If $\lambda, \mu, \nu$ are convolvable, then so are $\lambda$ and $\mu$, $|\lambda| * |\mu|$ and $\nu$, $\mu$ and $\nu$, $\lambda$ and $|\mu| * |\nu|$, and one has

(ii) If $\lambda$ and $\mu$ are convolvable, as well as $|\lambda| * |\mu|$ and $\nu$, then $\lambda, \mu, \nu$ are convolvable. Similarly if $\mu$ and $\nu$ are convolvable, as well as $\lambda$ and $|\mu| * |\nu|$.

This follows from Prop. 1 of §1, No. 2.

There can exist measures $\lambda, \mu, \nu$ on G such that the convolution products $\lambda * \mu$, $(\lambda * \mu) * \nu$, $\mu * \nu$, $\lambda * (\mu * \nu)$ are all defined, and yet $(\lambda * \mu) * \nu \neq \lambda * (\mu * \nu)$ (cf. Exer. 4).

Let $\rho$ be a lower semi-continuous finite function $> 0$ on $G$ such that $\rho(st) \leq \rho(s)\rho(t)$ for all $s, t$ in $G$. We denote by $\mathscr{M}^\rho(G)$ the vector space of measures $\lambda$ on $G$ such that $\rho$ is $\lambda$-integrable, and by $\| \lambda \|_\rho$ (or simply $\| \lambda \|$) the norm $\int_G \rho(s) d|\lambda|(s)$ on this space. When $\rho = 1$, one recovers the set $\mathscr{M}^1(G)$ of bounded measures on $G$.

Let $\lambda, \mu$ be in $\mathscr{M}^\rho(G)$, and let us show that $\lambda$ and $\mu$ are convolvable. Let $f \in \mathscr{K}_+(G)$. Since $\rho$ is $> 0$ and lower semi-continuous, there exists a constant $k > 0$ such that $f \leq k\rho$. Then

(Ch. V, §8, No. 3, Cor. 1 of Prop. 8). Therefore $(s,t) \mapsto f(st)$ is $(\lambda \otimes \mu)$-integrable, so that $\lambda$ and $\mu$ are convolvable. On the other hand, using Ch. V (§1, Prop. 4, §6, Prop. 2, §8, Cor. 1 of Prop. 8) and the fact that $(s,t) \mapsto \rho(s)\rho(t)$ is lower semi-continuous in $G \times G$, one has

One sees that $\lambda * \mu \in \mathscr{M}^\rho(G)$ and that $\| \lambda * \mu \| \leq \| \lambda \| \cdot \| \mu \|$. In view of Prop. 1, $\mathscr{M}^\rho(G)$ is an algebra. The mapping $\lambda \mapsto \rho \cdot \lambda$ is an isometric linear mapping $\theta$ of $\mathscr{M}^\rho(G)$ into $\mathscr{M}^1(G)$; if $\mu \in \mathscr{M}^1(G)$ then $1/\rho$, which is locally bounded and upper semi-continuous, is locally $\mu$-integrable, and $\rho$ is $(1/\rho) \cdot \mu$-integrable, thus $(1/\rho) \cdot \mu \in \mathscr{M}^\rho(G)$; this proves that $\theta$ is surjective; therefore $\mathscr{M}^\rho(G)$ is a complete normed algebra. Finally, it is clear that $\varepsilon_e$ is a unity element for $\mathscr{M}^\rho(G)$ and that $\mathscr{C}'(G)$ is a subalgebra of $\mathscr{M}^\rho(G)$ (§1, No. 4, Cor. of Prop. 5).

If $\rho = 1$, Prop. 2, (i) and (ii) also follow from §1, Prop. 2.

For, let $S_i$ be the support of $\mu_i$, and suppose that $S_i$ is compact for $i \neq i_0$. Let K be a compact subset of G. The set of $(x_1, \ldots, x_n) \in \prod_i S_i$ such that $x_1 x_2 \cdots x_n \in K$ is compact, because the conditions $x_i \in S_i$ for all $i$, $x_1 x_2 \cdots x_n \in K$ imply

Therefore the $\mu_i$ are convolvable (§1, No. 4, Prop. 4).

This follows from Props. 1 and 3.

We can suppose that $\mu \geqslant 0$. Let $f \in \mathscr{K}_+(G)$. When $\lambda$ is a left Haar measure,

therefore the function $(x, y) \mapsto f(xy)$ is $(\mu \otimes \lambda)$-integrable, and its integral for $\mu \otimes \lambda$ is $\lambda(f)\|\mu\|$. One argues similarly when $\lambda$ is a right Haar measure.

Let $f \in \mathscr{K}(G)$. Then $f\chi \in \mathscr{K}(G)$, therefore the function

on $G \times G$ is integrable for $\mu \otimes \nu$; therefore the function $(x, y) \mapsto f(xy)$ is integrable for $(\chi \cdot \mu) \otimes (\chi \cdot \nu)$; therefore $\chi \cdot \mu$ and $\chi \cdot \nu$ are convolvable. Moreover,

whence $(\chi \cdot \mu) * (\chi \cdot \nu) = \chi \cdot (\mu * \nu)$.

This follows from §1, No. 2, Cor. of Prop. 1.

This follows from Prop. 7 on considering the isomorphism $x \mapsto x^{-1}$ of G onto the opposite group.

With G still a locally compact group, let us assume that G operates continuously on the left in a locally compact space X. This defines (§2, No. 4) a continuous linear representation $\gamma$ of G in $\mathscr{M}(X)$ (equipped with the topology of compact convergence in $\mathscr{H}(X)$).

By Prop. 7 of §1, No. 5,

Now, $\varepsilon_s * \mu = \gamma(s)\mu$ (No. 2, formula (5)) and

by the definition of $\gamma(\lambda)$.

For, $\mathscr{M}(X)$, equipped with the topology of compact convergence in $\mathscr{H}(X)$, is quasi-complete. Therefore the mapping $(\lambda, \mu) \mapsto \gamma(\lambda)\mu$ of $\mathscr{C}'(G) \times \mathscr{M}(X)$ into $\mathscr{M}(X)$ is hypocontinuous relative to the equicontinuous subsets of $\mathscr{C}'(G)$ and the compact subsets of $\mathscr{M}(X)$ (§2, No. 6). It then suffices to apply Prop. 12.

Since $\gamma(s)\mu = \gamma(ss_0^{-1})\gamma(s_0)\mu$, it follows from Remark 1 that it suffices to prove the continuity of the mapping under consideration at a point of the form $(e, \mu_0)$ with $\mu_0 \in \mathscr{M}_+(X)$. Given a function $f \in \mathscr{K}(X)$ and a number $\varepsilon > 0$, it is thus a matter of showing that there exist a neighborhood U of e in G and a neighborhood W of $\mu_0$ in $\mathscr{M}_+(X)$ such that the relations $s \in U, \mu \in W$ imply

Let V be a compact neighborhood of the support K of $f$ in X, and let $\varphi \in \mathscr{K}_+(X)$ be such that $\varphi(x) = 1$ on V; there exists a neighborhood $W_0$ of $\mu_0$ in $\mathscr{M}_+(X)$ such that $a = \sup_{\mu \in W_0} \mu(V)$ is finite: it suffices to take for $W_0$ the set of $\mu \in \mathscr{M}_+(X)$ such that $|\langle \varphi, \mu - \mu_0 \rangle| \leq 1$. Since the mapping $(s, x) \mapsto sx$ is continuous, there is, on the other hand, a compact neighborhood $U_0$ of e in G such that $sK \subset V$ for all $s \in U_0$; the function $(s, x) \mapsto f(sx)$ is then uniformly continuous in $U_0 \times V$ and so there is a neighborhood $U \subset U_0$ of e such that $|f(sx) - f(x)| \leq \varepsilon / 2a$ for all $s \in U$ and $x \in V$. For $s \in U$ and $\mu \in W_0$, we therefore have

if $W \subset W_0$ is the neighborhood of $\mu_0$ in $\mathscr{M}_+(X)$ formed by the measures $\mu \in W_0$ such that $\left| \int f(x)\, d\mu(x) - \int f(x)\, d\mu_0(x) \right| \leq \varepsilon / 2$, U and W meet the requirements.