Proposition 2

Let T, X, Y be uniform spaces and let f be a mapping of T × X into Y. Then f is uniformly continuous if and only if the following two conditions are satisfied:

1) The mappings x → f(t, x) (t ∈ T) form a uniformly equicontinuous subset of $\mathscr{F}(X; Y)$.

2) The mappings t → f(t, x) (x ∈ X) form a uniformly equicontinuous subset of $\mathscr{F}(T; Y)$.

It is easily seen that the conditions are necessary. Conversely, suppose that they are satisfied. Let W be an entourage of Y; then there exists an entourage U of T and an entourage V of X such that:

1) $(t', t'') \in U$ implies that, for each $x \in X$,

2) $(x', x'') \in V$ implies that, for each $t \in T$,

It is now clear that the relation "$(t', t'') \in U$ and $(x', x'') \in V$" implies that $(f(t', x'), f(t'', x'')) \in \hat{W}$, whence the result.

In particular, take T to be a subset H of $\mathscr{F}(X; Y)$, endowed with the uniformity of uniform convergence, and take f to be the canonical mapping $(h, x) \to h(x)$; then condition 2) of Proposition 2 is automatically satisfied because, for each entourage W of Y, the set of pairs $(h', h'')$ such that $(h'(x), h''(x)) \in W$ for all $x \in X$ is by definition an entourage of the uniform structure of H. Hence only condition 1) has to be expressed; in other words: