Corollary 2

Let $X$ be a compact space, $Y$ a uniform space. Then every equicontinuous subset of $\mathscr{F}(X; Y)$ is uniformly equicontinuous.

Now suppose we have a set $T$, a topological space $X$, a uniform space $Y$ and a mapping $f : T \times X \to Y$. Let $\tilde{f}$ denote the mapping $x \to f(., x)$ of $X$ into $\mathscr{F}_u(T; Y)$, and let us consider the canonical mapping $\theta : (t, g) \to g(t)$ of $T \times \mathscr{F}_u(T; Y)$ into $Y$. It is clear that the diagram

(where $i_T$ is the identity mapping of T) is commutative. Suppose now that T is endowed with a topology and that, for each $x \in X$, the mapping $f(., x) : t \to f(t., x)$ is continuous; we can then replace $\mathscr{F}_u(T; Y)$ by $\mathscr{C}_u(T; Y)$ in the above diagram. But we know that $\theta$ is continuous from § 1, no. 6, Proposition 9; hence if $\tilde{f}$ is continuous it follows that $f$ is continuous. Since the continuity of $\tilde{f}$ can be expressed with the help of Proposition 1, we obtain the following result: