Lemma 6

Let $\psi \in L^1(G)$. Let B be a bounded subset of the Banach space $L^{\infty}(G)$. The mapping $\varphi \mapsto \psi *\varphi$ of B into $\mathscr{C}_b(G)$ is continuous when B is endowed with the weak topology and $\mathscr{C}_b(G)$ with the topology of compact convergence.

Let $\varphi \in L^{\infty}(G)$ and $\eta \in L^1(G)$. The function $\Delta^{-1}\check{\eta}$ belongs to $L^1(G)$ and $\langle \eta ,\check{\psi}\rangle =\langle \Delta^{-1}\check{\eta , \psi}\rangle ($cf. INT, VII, p. 19, § 1, n$^o3$, formula (22)). The mapping $\varphi \mapsto \check{\varphi}$ is therefore an automorphism of the space $L^{\infty}(G)$ endowed with the weak topology.

Let $\varphi \in L^{\infty}(G)$. By INT, VIII, p. 167, § 4, n$^o5$, prop. 14, the function $\psi *\varphi$ belongs to $\mathscr{C}_b(G)$ and satisfies

$=\int_G\psi (gy)\check{\varphi}(y)d\mu(y) =\langle \check{\varphi ,}\boldsymbol{\gamma }^{(1)}_G(g^{-1})\psi \rangle$ for all $g\in G$. It follows that the linear mapping $u:\varphi \mapsto \psi *\varphi$ is a continuous mapping of the space $L^{\infty}(G)$ endowed with the weak topology into $\mathscr{C}_b(G)$ endowed with the topology of simple convergence.

Let $\varphi \in B$ and $(g, h)\in G\times G$. According to the above formula, we have

$|u(\varphi )(g)-u(\varphi )(h)|\leqslant \|\check{\varphi}\|_{\infty}\|(\boldsymbol{\gamma }^{(1)}_G(g^{-1})-\boldsymbol{\gamma }^{(1)}_G(h^{-1}))\psi \|_1$.

Since B is bounded and the left regular representation of G in $L^1(G)$ is continuous (No. 4 of V, p. 405), this implies that $u(B)$ is an equicontinuous subset of $\mathscr{C}_b(G)$. The assertion then results from the foregoing and from TG, X, p. 16, th. 1.