Consider first the particular case where $\mathscr{S} = \{ T \}$, i.e. $\mathscr{F}_\mathscr{S}(T; Y) = \mathscr{F}_u(T; Y)$. For each entourage $V$ of $Y$, the condition $(f(., x), f(., x')) \in W(V)$ signifies that $(f(t, x), f(t, x')) \in V$ for all $t \in T$. To say that $x \to f(., x)$ is continuous at $x_0$ (resp. is uniformly continuous) is therefore equivalent to saying that, for each entourage $V$ of $Y$, there is a neighbourhood $U$ of $x_0$ in $X$ (resp. an entourage $M$ of $X$) such that the relation $x \in U$ [resp. $(x, x') \in M$] implies $(f(t, x), f(t, x_0)) \in V$ [resp. $(f(t, x), f(t, x')) \in V$] for all $t \in T$, and the proposition follows from Definitions 1 and 2. In the general case, we have to express that, for each $A \in \mathscr{S}$, the mapping $x \to f(., x)|A$ of $X$ into $\mathscr{F}_u(A; Y)$ is continuous at $x_0$ (resp. uniformly continuous), by virtue of § 1, no. 2; from what has been said, this is equivalent to saying that, for each $A \in \mathscr{S}$, $H_A$ is equicontinuous at $x_0$ (resp. uniformly equicontinuous).

Proposition 1 allows us to translate Definitions 1 and 2 into forms which are sometimes useful, by applying it to the case where $T = H$ and $f$ is the mapping $(h, x) \to h(x)$ of $H \times X$ into $Y$; since $f(., x)$ is the mapping $h \to h(x)$ of $H$ into $Y$, we see that:

In particular, if $X$ is compact, every continuous mapping of $X$ into $\mathscr{F}_u(H; Y)$ is uniformly continuous (Chapter II, § 4, no. 1, Theorem 2). Therefore:

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:

1) For each $x \in X$, the partial mapping $t \to f(t, x)$ is continuous.

2) As t runs through T, the partial mappings $x \to f(t, x)$ form an equicontinuous subset of $\mathscr{F}(X; Y)$.

In particular, take T to be a subset H of $\mathscr{F}(X; Y)$ and f to be the canonical mapping $(h, x) \to h(x)$ of $H \times X$ into Y; condition 1) of Corollary 3 means that H is endowed with a topology finer than that of pointwise convergence, and condition 2) means that H is equicontinuous. Hence:

More intuitively, this expresses the fact that if $h \in H$ converges pointwise to $h_0 \in H$ and if $x \in X$ converges to $x_0$, then $h(x)$ converges to $h_0(x_0)$.

We have to show that, for each $x \in X$, the mapping $(u, v) \to u(v(x))$ of $H \times \mathscr{C}(X; Y)$ into Z is continuous. Now $v \to v(x)$ is continuous on H (§ 1, no. 2, Remark 6), and it follows from Corollary 4 that $(u, y) \to u(y)$ is a continuous mapping of $H \times Y$ into Z; since $(u, v) \to u(v(x))$ is the composition of $(u, y) \to u(y)$ and $(u, v) \to (u, v(x))$, the result is proved.

The following proposition and its corollary are the analogues of Corollaries 3 and 4 of Proposition 1 for uniformly equicontinuous sets of mappings:

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:

This is an immediate consequence of Definitions 1 and 2 and the definition of the entourages of $Y$.

We recall that the diagram

is commutative, where $i$ and $j$ are the canonical mappings; moreover, the uniformity of $X$ (resp. $Y$) is the inverse image under $i$ (resp. $j$) of that of $\hat{X}$ (resp. $\hat{Y}$). Hence $H$ is uniformly equicontinuous if and only if its image under the mapping $u \to j \circ u$ is uniformly equicontinuous (Proposition 3), and we may already restrict ourselves to the case where $Y$ is Hausdorff and complete; moreover, if $\tilde{H}$ is uniformly equicontinuous then so is $H$, because it is the image of $\tilde{H}$ under the mapping $\hat{u} \to \hat{u} \circ i$; thus it remains to prove the converse when $Y = \hat{Y}$. Let $V$ be a closed entourage of $Y$; by hypothesis there is an entourage $U$ of $X$ such that the relations $(x, x') \in U$ and $u \in H$ imply that $(u(x), u(x')) \in V$. Now, if $U'$ is the image of $U$ under $i \times i$, the closure $\overline{U'}$ of $U'$ in $\hat{X} \times \hat{X}$ is an entourage of $\hat{X}$ (Chapter II, § 3, no. 7, Proposition 12); the hypothesis implies that, whenever $(z, z') \in U'$ and $u \in H$, we have $(\hat{u}(z), \hat{u}(z')) \in V$. Since $V$ is closed and $\hat{u}$ is continuous, we have also $(\hat{u}(t), \hat{u}(t')) \in V$ for all $(t, t') \in \overline{U'}$ and all $u \in H$; this completes the proof.

It is enough to show that a) implies c). Let $V'$ be a neighbourhood of the identity element $e'$ of $G'$; then, by hypothesis, there is a neighbourhood $V$ of $e$ in $G$ such that $u(V) \subset V'$ for all $u \in H$; since the elements of $H$ are homomorphisms, the relation $x^{-1} y \in V$ implies that we have

The condition is sufficient, trivially. To show that it is necessary, consider an entourage $V$ of $Y$ which is closed in $Y \times Y$; by hypothesis, there is a neighbourhood $U$ of $x_0$ in $X$ (resp. an entourage $M$ of $X$) such that the relation $x \in U$ (resp. $(x', x'') \in M$) implies $(h(x_0), h(x)) \in V$ [resp. $(h(x'), h(x'')) \in V$] for all $h \in H$. Since $V$ is closed, the mappings $h \in \mathscr{F}(X; Y)$ which satisfy the relation $(h(x_0), h(x)) \in V$ for all $x \in U$ [resp. the relation $(h(x'), h(x'')) \in V$ for all $(x', x'') \in M$] form a closed subset of $\mathscr{F}_s(X; Y)$ (§ 1, no. 2, Remark 6); since this closed subset contains $H$, it contains $\overline{H}$. Hence the result, since the closed entourages of $Y$ form a fundamental system of entourages (Chapter II, § 1, no. 2, Proposition 2, Corollary 2).

It is enough to show that the last uniformity on $H$ is finer than the first; in other words that, given an entourage $V$ of $Y$ and a compact (resp. precompact) subset $A$ of $X$, there exists an entourage $W$ of $Y$ and a finite subset $F$ of $D$ such that the relation

implies

Suppose first that $A$ is compact and $H$ equicontinuous. Given a symmetric entourage $W$ of $Y$, every point $x \in X$ has a neighbourhood $U(x)$ such that the relation $x' \in U(x)$ implies $(u(x),\ u(x')) \in W$ for all $u \in H$. We can therefore cover the compact set $A$ by a finite number of open sets $U_i$ such that, for each pair of points $x',\ x''$ of the same set $U_i$, we have $(u(x'),\ u(x'')) \in \hat{W}$ for all $u \in H$. Let $a_i$ be a point of $D \cap U_i$, let $F$ be the set of the $a_i$, and suppose that (2) is true; then for each $x \in A$ there exists an index $i$ such that $a_i$ and $x$ belong to the same set $U_i$, so that we have $(u(x),\ u(a_i)) \in \hat{W}$ and $(v(a'),\ v(x)) \in \hat{W}$; thus (2) implies (3) provided that $W$ is chosen so that $\hat{W} \subset V$.

If $A$ is precompact and $H$ uniformly equicontinuous, we use Proposition 4 of no. 2; it is enough to note that $i(A)$ is compact in $\hat{X}$, $i(D)$ dense in $\hat{X}$ and that the entourages of $Y$ are the inverse images of those of $\hat{Y}$ under the mapping $j \times j$.

For the set $\overline{H}$ is equicontinuous (resp. uniformly equicontinuous) by Proposition 6 of no. 3, and hence is contained in $\mathscr{C}(X; Y)$; the result follows immediately from the fact that, on $\overline{H}$, the two topologies under consideration are the same, by virtue of Theorem 1.

a) For each $A \in \mathscr{S}$, the set $H|A \subset \mathscr{F}(A; Y)$ of restrictions to $A$ of functions of $H$ is equicontinuous (resp. uniformly equicontinuous).

b) For each $x \in X$, the set $H(x) \subset Y$ of points $u(x)$ ($u \in H$) is precompact.

1) Let us show first that conditions a) and b) are necessary. We know (§ 1, no. 2, Remark 6) that the mapping $u \to u(x)$ of $\mathscr{F}_{\mathscr{S}}(X; Y)$ into $Y$ is uniformly continuous; hence, if $H$ is precompact, so is $H(x)$ (Chapter II, § 4, no. 2, Proposition 2), which proves b). To prove a), consider a set $A \in \mathscr{S}$, a point $x_0 \in A$ and an entourage $V$ of $Y$; since $H$ is precompact it can be covered by a finite number of $W(A, V)$-small sets; in other words there is a finite sequence $(u_i)$ of elements of $H$ such that, for each $u \in H$, we have

for at least one index $i$.

Since each of the $u_i|A$ is continuous at $x_0$ (resp. uniformly continuous) there is a neighbourhood $U_i$ of $x_0$ in $A$ (resp. an entourage $M_i$ of $A$) such that

(resp. such that

Let $U$ (resp. $M$) be the intersection of the $U_i$ (resp. $M_i$); it is a neighbourhood of $x_0$ in $A$ (resp. an entourage of $A$). For each $u \in H$ there is an index $i$ for which (4) holds; writing condition (4) for $x_0$ and for $x$ (resp. for $x'$ and $x''$) and taking account of (5) [resp. (6)], we see immediately that the relation $x \in U$ [resp. $(x', x'') \in M$] implies $(u(x),\ u(x_0)) \in \overset{\circ}{V}$ [resp. $(u(x'),\ u(x'')) \in \overset{\circ}{V}$], for each $u \in H$; and this establishes a).

2) Now let us show that the conditions a) and b) are sufficient if the sets $A \in \mathscr{S}$ are compact (resp. precompact). Condition b) implies that $H$ is precompact with respect to the uniformity of pointwise convergence (Chapter II, § 4, no. 2, Proposition 3). But it follows from condition a) and Theorem 1 of no. 4 that on $H|A$ the uniformity of pointwise convergence in $A$ coincides with the uniformity of uniform convergence in $A$; hence $H|A$ is precompact in $\mathscr{F}_u(A; Y)$, which implies that $H$ is precompact with respect to the uniformity of $\mathscr{S}$-convergence (§ 1, no. 2).

Note that condition b) of Theorem 2 is automatically satisfied if $Y$ is a precompact space.

Let $\overline{H}$ be the closure of $H$ in $\mathscr{F}_s(X; Y)$. $\overline{H}$ is equicontinuous (resp. uniformly equicontinuous) (no. 3, Proposition 6). Moreover, we have $\overline{H}(x) \subset \overline{H(x)}$ (§ 1, no. 2, Remark 6) and therefore $\overline{H}(x)$ is also relatively compact; hence Theorem 2 shows that $\overline{H}$ is precompact with respect to $\mathscr{S}$-convergence, where $\mathscr{S}$ denotes the set of all compact (resp. precompact) subsets of $X$. Moreover, since $\overline{H(x)}$ is compact, and therefore complete, $\overline{H}$ is complete with respect to the uniformity of pointwise convergence (Chapter II, § 3, no. 5, Proposition 10 and no. 4, Proposition 8) and therefore also with respect to the uniformity of $\mathscr{S}$-convergence (§ 1, no. 5, Proposition 5, Corollary 2); $\overline{H}$ is therefore compact, since it is precompact, complete and Hausdorff (§ 1, no. 2, Proposition 1).

It is enough to show that $H(x)$ is relatively compact for all $x \in X$, for we can then apply Corollary 1. Since $Y$ is complete it is enough to show that $H(x)$ is precompact for all $x \in X$. Now if $V$ is any symmetric entourage of $Y$, there is a neighbourhood $U$ of $x$ such that $(u(x), u(x')) \in V$ for all $x' \in U$ and all $u \in H$. By hypothesis there exists $x' \in U \cap D$, and since $H(x')$ is relatively compact in $Y$, there exists a finite number of points $y_k \in Y$ such that $H(x')$ is contained in the union of the sets $V(y_k)$; hence $H(x)$ is contained in the union of the sets $V(y_k)$, and the proof is complete.

In view of Corollary 1 it is enough to show that, if $H$ is relatively compact in $C_c(X; Y)$, then $H$ is equicontinuous. Now each point $x \in X$ has a compact neighbourhood $A$, and it follows from Theorem 2 that $H/A$ is equicontinuous; this implies that $H$ is equicontinuous at $x$, and the result is proved.