For if $d$ is a metric on $Y$ compatible with its uniform structure, the structure of uniform convergence on $\mathscr{F}(X; Y)$ is defined by the single pseudometric

in general this pseudometric is not finite, but it is equivalent to a finite one (Chapter IX, § 1, no. 2), and since the uniformity of uniform convergence is Hausdorff (§ 1, no. 2, Proposition 1), it is metrizable.

Since the $K_n$ cover $X$, $\mathscr{F}_c(X; Y)$ is isomorphic to a uniform subspace of $\prod_n \mathscr{F}_u(K_n; Y)$ ($§ 1$, no. 2, Remark 3), and the corollary therefore follows from Proposition 1 (Chapter IX, § 2, no. 4, Theorem 1, Corollary 2).

Note that this corollary applies in particular if $X$ is locally compact and σ-compact (Chapter I, § 9, no. 9, Proposition 15, Corollary 1).

Now let $Y$ be a metric space and let $d$ be its metric. If $X$ is any set and $\mathscr{S}$ any set of subsets of $X$, we shall denote by $\mathscr{B}_{\mathscr{S}}(X; Y)$ the set of all mappings $u : X \to Y$ such that $u(A)$ is bounded for each $A \in \mathscr{S}$. Unless the contrary is expressly stated we shall regard $\mathscr{B}_{\mathscr{S}}(X; Y)$ as endowed with the uniformity of $\mathscr{S}$-convergence, which is defined by the following family of pseudometrics on $\mathscr{B}_{\mathscr{S}}(X; Y)$:

which are finite by hypothesis. When $\mathscr{S} = \{X\}$, we write $\mathscr{B}(X; Y)$ in place of $\mathscr{B}_{\mathscr{S}}(X; Y)$. A mapping $u : X \to Y$ is said to be bounded if it belongs to $\mathscr{B}(X; Y)$, i.e. if $u(X)$ is a bounded subset of $Y$.

If $u$ is bounded, then every mapping $v : X \to Y$ such that for all $x \in X$, we have $d(u(x), v(x)) \leq 1$ is bounded, because

hence $\mathscr{B}(X; Y)$ is open. On the other hand, if $u$ lies in the closure of $\mathscr{B}(X; Y)$ in $\mathscr{F}_u(X; Y)$, there is a mapping $u_0 \in \mathscr{B}(X; Y)$ such that $d(u(x), u_0(x)) \leq 1$ for all $x \in X$; hence $u$ is bounded.

For $\mathscr{B}_\mathfrak{S}(X; Y)$ is the inverse image of the subset $\prod_{A \in \mathfrak{S}} \mathscr{B}(A; Y)$ of the product $\prod_{A \in \mathfrak{S}} \mathscr{F}_u(X; Y)$ under the canonical mapping of $\mathscr{F}_\mathfrak{S}(X; Y)$ into $\prod_{A \in \mathfrak{S}} \mathscr{F}_u(A; Y)$; the first assertion therefore follows from § 1, no. 2, Remark 3, and the second follows from the first, if we take account of Theorem 1 of § 1, no. 5.

The space in question is $\mathscr{B}(X; Y) \cap C_u(X; Y)$; the first assertion follows from Proposition 2; the second follows from the first ($§ 1$, no. 6, Theorem 2, Corollary 1).

For since $\mathscr{B}(X; Y)$ is complete (no. 1, Proposition 2, Corollary 1), the result follows from Chapter IX, § 3, no. 6, Proposition 11 and the definition of a uniformly convergent series.

This set consists of all $u \in F(X; Y)$ which satisfy all the relations

For if $\mathscr{G}$ is the set of all bounded subsets of $X$, $\mathscr{L}(X_1, \ldots, X_n; Y)$ is the intersection of the set of all multilinear mappings of $X$ into $Y$ and the set $\mathscr{B}_\mathscr{G}(X; Y)$ (Chapter IX, § 3, no. 5, Theorem 1); the result thus follows from Proposition 4 and Proposition 2, Corollary 1.

For the remainder of this subsection, $K$ denotes a non-discrete valued field.

Then $\mathscr{L}(X_1, \ldots, X_n; Y)$ is a vector subspace of $\mathscr{F}(X; Y)$. Let $B$ be the unit ball in $X$, the set of all $(x_i)_{1 \leq i \leq n}$ such that $\sup_{1 \leq i \leq n} \|x_i\| \leq 1$.

Then the mapping $u \to u|B$ of $\mathscr{L}(X_1, \ldots, X_n; Y)$ into $\mathscr{B}(B; Y)$ is injective; moreover, the inverse image, under this mapping, of the uniformity of uniform convergence on $\mathscr{B}(B; Y)$ is the uniformity of bounded convergence on $\mathscr{L}(X_1, \ldots, X_n; Y)$. For every bounded subset of $X$ is contained in a set of the form $\mu B$ (for some $\mu \in K^*$), and if $u$ is an element of $\mathscr{L}(X_1, \ldots, X_n; Y)$, to say that $\|u(z)\| \leq a$ for all $z \in \mu B$ is equivalent to saying that $\|u(z)\| \leq a/|\mu|^n$ for all $z \in B$. It is easily verified that the number

Unless the contrary is expressly stated, whenever $\mathscr{L}(X_1, \ldots, X_n; Y)$ is considered as a normed space, it is the norm defined above which is in question.

This is an immediate consequence of the inequality (2) (Chapter IX, § 3, no. 5, Theorem 1).

This follows immediately from the definitions and the relation

For if $u \in \mathscr{L}(X; Y)$ and $v \in \mathscr{L}(Y; Z)$ we have

In particular, on the set $\mathscr{L}(X)$ of continuous endomorphisms of a normed space $X$ over $K$, the norm $||u||$ is compatible with the $K$-algebra structure of $\mathscr{L}(X)$.

a) Let $d$ (resp. $d'$) be a metric compatible with the topology of $X$ (resp. with the uniformity of $Y$); then $\delta(f, g) = \sup_{x \in X} d'(f(x), g(x))$ is a metric defining the uniformity of uniform convergence on the space $C(X; Y)$, the functions of $C(X; Y)$ being bounded because $X$ is compact (no. 1). For each pair of integers $m > 0, n > 0$, let $G_{mn}$ be the set of functions $f \in \mathscr{C}(X; Y)$ such that the relation $d(x, x') \leq 1/m$ implies $d(f(x), f(x')) \leq 1/n$. Every function $f \in \mathscr{C}(X; Y)$ is uniformly continuous (Chapter II, § 4, no. 1, Theorem 2) and therefore, for each $n > 0$, $\mathscr{C}(X; Y)$ is the union of the sets $G_{mn}$ ($m > 0$). Let $\{a_1, \ldots, a_{p(m)}\}$ be a finite subset of $X$ such that the open balls with centres $a_i$ and radii $1/m$ cover $X$ ($1 \leq i \leq p(m)$); and let $(b_r)_{r \in \mathbf{N}}$ be a countable sequence which is dense in $Y$. For each mapping $\varphi : [1, p(m)] \to \mathbf{N}$, let $H_\varphi$ be the set of those $f \in G_{mn}$ such that $d'(f(a_k), b_{\varphi(k)}) \leq 1/n$ for $1 \leq k \leq p(m)$. By the definition of the $b_r$, $G_{mn}$ is the union of the sets $H_\varphi$ for $\varphi \in \mathbf{N}^{p(m)}$; let $C_{mn}$ be the set of mappings $\varphi \in \mathbf{N}^{p(m)}$ such that $H_\varphi \neq \emptyset$, and for each $\varphi \in C_{mn}$ let $g_\varphi$ be an element of $H_\varphi$; finally, let $L_{mn}$ denote the countable set of $g_\varphi$ for $\varphi \in C_{mn}$. Let $f \in G_{mn}$, and let $\varphi$ be an element of $C_{mn}$ such that $f \in H_\varphi$; then it follows immediately from the definitions that we have $d'(f(x), g_\varphi(x)) \leq 4/n$ for all $x \in X$, i.e. $\delta(f, g_\varphi) \leq 4/n$. Hence the union of the sets $L_{mn}$ is dense in $\mathscr{C}_u(X; Y)$, because for every integer $n > 0$ and every $f \in \mathscr{C}(X; Y)$ there exists $m$ such that $f \in G_{mn}$, and we have just seen that the distance from $f$ to $L_{mn}$ is $\leq 4/n$.

b) Let $I = [0, 1]$. Since $\mathscr{C}_u(X; I)$ is a uniform subspace of $\mathscr{C}_u(X; \mathbf{R})$, it is of countable type. Let $(f_n)$ be a sequence which is dense in $\mathscr{C}_u(X; I)$. Consider the product space $K = \mathbf{N}$ and the mapping $\psi : x \to (f_n(x))$ of $X$ into $K$, which is obviously continuous. The mapping $\psi$ is injective; for, by definition of the sequence $(f_n)$, the relation $f_n(x) = f_n(x')$ for all $n$ implies, on passing to the limit, $f(x) = f(x')$ for every function $f \in \mathscr{C}(X; I)$; but this is impossible if $x \neq x'$ by virtue of Axiom (OIV) applied to the point $x$ and to a neighbourhood $V$ of $x$ which does not contain $x'$ (Chapter IX, § 1, no. 5, Theorem 2). It follows that the compact space $X$ is homeomorphic to the subspace $\psi(X)$ of $K$ (Chapter I, § 9, no. 4, Theorem 2, Corollary 2); since $K$ is metrizable and of countable type, so is $\psi(X)$ and therefore so is $X$.

Q.E.D.

a) The space $\mathscr{L}$ of continuous mappings of $X$ into $Y$ which have a limit at infinity, endowed with the topology of uniform convergence in $X$, is a metrizable space of countable type.

b) The space $\mathscr{C}_c(X; Y)$ of continuous mappings of $X$ into $Y$, endowed with the topology of compact convergence, is a metrizable space of countable type.

a) Let $X'$ be the compact space obtained by adjoining a point at infinity to $X$ (Chapter I, § 9, no. 8, Theorem 4); by definition, every function $f \in \mathscr{L}$ can be uniquely extended to a continuous function $\overline{f} : X' \to Y$, and $f \to \overline{f}$ is therefore a bijection of L onto $C(X'; Y)$; and this bijection is a homeomorphism of the space L onto $C_u(X'; Y)$ by Proposition 6 of § 1, no. 6. Since $X'$ is metrizable (Chapter IX, § 2, no. 9, Proposition 16, Corollary) the result follows from Theorem 1, applied to $X'$ and Y.

b) Let $(U_n)$ be a covering of X by relatively compact open sets, such that every compact subset of X is contained in some $U_n$ (Chapter I, § 9, no. 9, Proposition 15, Corollary 1). If $S$ is the set of the $\overline{U}_n$, the topology of compact convergence on $C(X; Y)$ is the same as the topology of $S$-convergence. Consequently ($§ 1,$ no. 2, Remark 3) the space $C_c(X; Y)$ is homeomorphic to a subspace of the product $\prod_n C_u(\overline{U}_n; Y)$; since each of the compact spaces $\overline{U}_n$ has a countable base, it is metrizable (Chapter IX, § 2, no. 9, Proposition 16); each of the $C_u(\overline{U}_n; Y)$ is therefore metrizable and of countable type by Theorem 1, and hence so is $C_c(X; Y)$.

Note that the space of all bounded continuous real-valued functions on $\mathbf{R}$, endowed with the topology of uniform convergence, is not of countable type (Exercise 4).

Let $Y'$ be the Hausdorff uniform space associated with Y (Chapter II, § 3, no. 8) and let $i : Y \to Y'$ be the canonical mapping of Y onto $Y'$. The topology of compact convergence is the coarsest topology for which the mappings $u \to (i \circ u)|K$ of $C(X; Y)$ into $C_u(K; Y')$ are continuous, as K runs through the set of all compact subsets of X ($§ 1,$ no. 4, Proposition 4). Hence we obtain a subbase of the topology of $C_c(X; Y)$ by taking a subbase of the topology of $C_c(K; Y')$ for each compact subset K of X and then taking the union [in $\mathfrak{B}(C(X; Y))$] of the inverse images of these subbases in $C(X, Y)$. On the other hand, every open subset of Y is of the form $\overline{i}^{-1}(U')$, where $U'$ is open in $Y'$ (Chapter II, § 3, no. 7, Proposition 12); hence, for each compact subset $K' \supset K$, $T(K, \overline{i}^{-1}(U'))$ is the inverse image of $T(K, U')$ under the mapping

It thus remains for us to prove the theorem when X is compact and Y is Hausdorff; we shall make these assumptions from now on.

Let us first show that $T(K, U)$ is open in $C_c(X; Y)$. Let $u_0$ be a point of this set; since $u_0(K)$ is compact (Chapter I, § 9, no. 4, Theorem 2, Corollary 1) and contained in the open set $U$, there exists a symmetric entourage $V$ of $Y$ such that $V(u_0(K)) \subset U$ (Chapter II, § 4, no. 3, Proposition 4, Corollary). Let $W$ be the neighbourhood of $u_0$ in $C_c(X; Y)$ consisting of all continuous mappings $u : X \to Y$ such that $(u(x), u_0(x)) \in V$ for all $x \in K$. For such mappings we clearly have $u(K) \subset V(u_0(K)) \subset U$; hence $u \in T(K, U)$ and therefore $W \subset T(K, U)$, which proves our assertion.

Conversely, if $W$ is a neighbourhood of a point $u_0 \in C_c(X; Y)$, let us show that $W$ contains the intersection of a finite number of neighbourhoods of the form $T(K, U)$. We may suppose that $W$ is the set of all $u \in C(X; Y)$ such that $(u(x), u_0(x)) \in V$ for all $x \in X$, $V$ being a given entourage of $Y$. Since $u_0$ is continuous on $X$, it is uniformly continuous (Chapter II, § 4, no. 1, Theorem 2). Let $V_1$ be a symmetric entourage of $Y$, open in $Y \times Y$ and such that $V_1^2 \subset V$. $X$ can be covered by a finite number of compact sets $K_i$ ($1 \leq i \leq n$) such that each $u_0(K_i)$ is $V_1$-small ($1 \leq i \leq n$). Let $U_i$ be the open set $V_1(u_0(K_i))$, and let $u : X \to Y$ be a continuous mapping contained in the intersection of the $n$ sets $T(K_i, U_i)$ (which are neighbourhoods of $u_0$). Then, for every $x \in K_i$, $u(x)$ belongs to $U_i$ and therefore $u_0(x)$ and $u(x)$ are $V_1^2$-close, hence $V$-close. Since each $x \in X$ belongs to some $K_i$, we have $u \in W$ and the proof is complete.

This result leads us to make the following definition:

Suppose that $f$ is continuous. To show that $\tilde{f}$ is continuous we have to prove that, for each compact subset $K$ of $Y$ and each open subset $U$ of $Z$, the inverse image $V$ of $T(K, U)$ under $\tilde{f}$ is open in $X$. Let $x_0 \in V$; for each $y \in K$, we have $f(x_0, y) \in U$, and since $f$ is continuous there is a neighbourhood $V_y$ of $x_0$ in $X$ and a neighbourhood $W_y$ of $y$ in $Y$ such that $f(V_y \times W_y) \subset U$. Since $K$ is compact, there exists a finite number of points $y_i \in K$ such that the sets $W_{y_i} (1 \leq i \leq n)$ cover $K$. Let $V'$ be the intersection of the neighbourhoods $V_{y_i}$ of $x_0$, which is a neighbourhood of $x_0$; if $x \in V'$ and $y \in K$, we have $f(x, y) \in U$, since $y$ is contained in some one of the $W_{y_i}$ and $x$ is contained in each $V_{y_i}$; hence $V' \subset V$, and therefore $V$ is a neighbourhood of each of its points and consequently is open in $X$.