Corollary 2

Let G be a compact topological group endowed with a Haar measure $\mu$. Let $p$ be a real number such that $1< p <+\infty$ and $q$ the conjugate exponent of $p$. Let $f\in \mathscr{L}^q(G)$. Put $k(x, y) =f(x^{-1}y)$ for all $(x, y)\in G\times G$.

a) One has $k\in \mathscr{L}^{q,p}(G,G)$;

b) For $\varphi \in L^p(G)$, the convolution $\varphi *f$ belong to $L^p(G)$ and the linear mapping $v_f:\varphi \mapsto \varphi *f$ of $L^p(G)$ into itself is continuous. It coincides with the endomorphism $u_k$. In particular, the linear mapping $v_f$ is compact.

Since G is compact, the function $f$ belong to $\mathscr{L}^1(G)$. The linear mapping $v_f$ is therefore defined and continuous by INT, VIII, p. 167, § 4, No.$^o5$, Prop. 13.

The mapping $k$ is $(\mu\otimes \mu$)-measurable, and one has $\|k_y\|_q=\|f\|_q$ for every $y\in G$. Since the measure $\mu$ is bounded, it follows that

hence $k\in \mathscr{L}^{q,p}(G,G)$. The linear mapping $u_k$ is therefore a compact linear mapping of $L^p(G)$ into $L^p(G)$ (Prop. 7).

Let $\varphi \in \mathscr{K}(G)$. By INT, VIII, p. 166, § 4, No.$^o5$, Prop. 11 and Prop. 4 of III, p. 29, one then has

for almost every $y\in G$. This implies that $v_f=u_k$.