Proposition 2

(i) Any two elements of $\mathscr{M}^\rho(G)$ are convolvable. (ii) For convolution, and for the norm $\| \lambda \|$, $\mathscr{M}^\rho(G)$ is a complete normed algebra, admitting $\varepsilon_e$ as unity element. (iii) $\mathscr{C}'(G)$ is a subalgebra of $\mathscr{M}^\rho(G)$.

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.