(i) The locally convex space $\mathscr{K}(X; E)$ is Hausdorff. For every compact subset $K$ of $X$, the topology on $\mathscr{K}(X, K; E)$ induced by that of $\mathscr{K}(X; E)$ is the topology of uniform convergence in $K$, and each of the subspaces $\mathscr{K}(X, K; E)$ is closed in $\mathscr{K}(X; E)$.

(ii) If $E$ is the product of a finite number of locally convex spaces $E_i$ ($1 \leq i \leq n$), then the mapping $f \mapsto (\mathrm{pr}_i \circ f)$ is an isomorphism of the space $\mathscr{K}(X; E)$ onto the product space $\prod_{1 \leq i \leq n} \mathscr{K}(X; E_i)$.

(iii) If $X$ is the sum of a family $(X_\lambda)_{\lambda \in L}$ of locally compact spaces, then the mapping $f \mapsto (f|X_\lambda)_{\lambda \in L}$ is an isomorphism of the space $\mathscr{K}(X; E)$ onto the topological direct sum space of the family $(\mathscr{K}(X_\lambda; E))_{\lambda \in L}$.

(i) Note that, on $\mathscr{K}(X; E)$, the topology of uniform convergence in $X$ is compatible with the vector space structure of $\mathscr{K}(X; E)$ because, for every $f \in \mathscr{K}(X; E)$, with (compact) support $S$, the set $f(X) = f(S) \cup \{0\}$ is compact, hence bounded in $E$ (TVS, III, §3, No. 1, Prop. 1). Since this topology $\mathscr{T}_0$ is locally convex and induces on each $\mathscr{K}(X, K; E)$ the topology of uniform convergence in $K$, the same is true of the direct limit topology $\mathscr{T}$ on $\mathscr{K}(X; E)$ (TVS, II, §4, No. 4, Remark); moreover, $\mathscr{T}$ is finer than $\mathscr{T}_0$ and $\mathscr{T}_0$ is Hausdorff, therefore $\mathscr{T}$ is Hausdorff. Finally, suppose that a function $f \in \mathscr{K}(X; E)$ belongs to the closure of $\mathscr{K}(X, K; E)$; by the definitions, there exists a compact subset $K' \supset K$ of $X$ such that $f \in \mathscr{K}(X, K'; E)$. By the foregoing, $f$ belongs to the closure of $\mathscr{K}(X, K; E)$ in the space $\mathscr{K}(X, K'; E)$, hence belongs to $\mathscr{K}(X, K; E)$.

(ii) The criterion for continuity in a direct limit (TVS, II, §4, No. 4, Prop. 5) shows at once that the mapping $f \mapsto (\mathrm{pr}_i \circ f)$ is continuous and that the same is true of the inverse mapping (for the latter, it suffices to note that if, for every function $f_i \in \mathscr{K}(X; E_i)$, one denotes by $f'_i$ the mapping

² Limite inductive, translated as "direct limit" in S, A and GT.

of X into E such that $\mathrm{pr}_i \circ f'_i = f_i$ and $\mathrm{pr}_j \circ f'_i = 0$ for $j \neq i$, then each of the mappings $f_i \mapsto f'_i$ is continuous).

(iii) Each compact subset K of X intersects only the $X_\lambda$ of a finite sub-family $(X_\lambda)_{\lambda \in H}$ of $(X_\lambda)_{\lambda \in L}$, and it is immediate that if one sets $K_\lambda = K \cap X_\lambda$ for $\lambda \in H$, the mapping $f \mapsto (f|X_\lambda)_{\lambda \in H}$ is an isomorphism of $\mathscr{K}(X, K; E)$ onto $\prod_{\lambda \in H} \mathscr{K}(X_\lambda, K_\lambda; E)$. Conversely, for every function $f_\lambda \in \mathscr{K}(X_\lambda; E)$, let $f''_\lambda$ be the mapping of X into E such that $f''_\lambda|X_\lambda = f_\lambda$ and $f''_\lambda|X_\mu = 0$ for $\mu \neq \lambda$; it is immediate that the mapping $f_\lambda \mapsto f''_\lambda$ of $\mathscr{K}(X_\lambda; E)$ into $\mathscr{K}(X; E)$ is continuous. The assertion (iii) follows from these remarks and the criterion for continuity in direct limits (TVS, II, §4, No. 4, Prop. 5).

(i) If E is a Fréchet space, then the space $\mathscr{K}(X; E)$ is barreled.

(ii) If X is paracompact then, for every bounded set B in $\mathscr{K}(X; E)$, there exists a compact subset K of X such that $B \subset \mathscr{K}(X, K; E)$.

Suppose E is a Fréchet space. Then, for every compact subset K of X, $\mathscr{K}(X, K; E)$ is a Fréchet space, hence is barreled, and one knows that a direct limit of barreled spaces is barreled (TVS, III, §4, No. 1, Cor. 3 of Prop. 3), whence (i).

If X is paracompact, one knows (GT, I, §9, No. 10, Th. 5) that X is the sum of a family $(X_\lambda)_{\lambda \in L}$ of locally compact spaces that are countable at infinity; thus (Prop. 1, (iii)), $\mathscr{K}(X; E)$ is the topological direct sum of the family of subspaces $\mathscr{K}(X_\lambda; E)$ ($\lambda \in L$). By virtue of the characterization of the bounded sets in a topological direct sum (TVS, III, §1, No. 4, Prop. 5), every bounded set in $\mathscr{K}(X; E)$ is contained in the sum of a finite number of subspaces $\mathscr{K}(X_\lambda; E)$, and it will suffice to prove that every bounded set in $\mathscr{K}(X_\lambda; E)$ is contained in a subspace $\mathscr{K}(X_\lambda, K_\lambda; E)$, with $K_\lambda$ compact in $X_\lambda$. We are thus reduced to the case that X is countable at infinity, in other words is the union of a sequence of relatively compact open sets $U_n$ such that $\overline{U}_n \subset U_{n+1}$ (GT, I, §9, No. 9, Prop. 15). But then $\mathscr{K}(X; E)$ is the strict direct limit of the sequence of spaces $\mathscr{K}(X, \overline{U}_n; E)$, whence the assertion (ii) (TVS, III, §1, No. 4, Prop. 6).

We shall say that a subset H of $\mathscr{K}(X; E)$ is strictly compact if it is compact and if there exists a compact subset K of X such that $H \subset \mathscr{K}(X, K; E)$. It follows at once from Proposition 2 that if X is a paracompact locally compact space and if E is Hausdorff, then every compact set in $\mathscr{K}(X; E)$ is strictly compact. One can give examples of locally compact

3 Dénombrable à l'infini, also translated as "σ-compact" (GT, I, §9, No. 9, Def. 5).

spaces $X$ (not paracompact) such that there exist sets in $\mathscr{K}(X; \mathbf{R})$ that are compact but not strictly compact (Exercises 3 and 4).

We recall that, by virtue of Ascoli’s theorem (GT, X, §2, No. 5, Cor. 3 of Th. 2), a strictly compact subset $H$ of $\mathscr{K}(X; E)$ contained in $\mathscr{K}(X, K; E)$ is characterized by the following conditions: $1^\circ$ it is closed; $2^\circ$ it is equicontinuous; $3^\circ$ for every $x \in K$, the set $H(x)$ is relatively compact in $E$.

It suffices, by virtue of Proposition 2, (ii), to note that for every compact subset $K$ of $X$, $\mathscr{K}(X, K; E)$ is a closed subspace of $C(K; E)$, which is quasi-complete since every bounded subset of $C(K; E)$ consists of functions taking values in a same bounded subset of $E$.

For, let $X'$ be the compact space obtained by adjoining to $X$ a point at infinity $\omega$ (GT, I, §9, No. 8, Th. 4); the sets $V_0 = X' - K$ and $V_k$ ($1 \leq k \leq n$) form an open covering of $X'$. Let $(f_k)_{0 \leq k \leq n}$ be a continuous partition of unity subordinate to this covering of $X'$ (GT, IX, §4, No. 3, Prop. 3); the functions $f_k$ with index $k \geq 1$ satisfy the conditions of the lemma.

For every $y$ belonging to the boundary of $K$, one has $f(y) = 0$ for all $f \in \Phi$, therefore there exists an open neighborhood $V_y$ of $y$ in $X$ such that, for every $z \in V_y$ and every $f \in \Phi$, one has $q(f(z)) \leq \varepsilon/2$. Let $K'$ be the set of points of $K$ not belonging to any of the $V_y$ as $y$ runs over the boundary of $K$; $K'$ is compact and is contained in the interior of $K$. The set $\Phi$ is uniformly equicontinuous in $K$; therefore, there exists a finite open covering $(U_j)_{1 \leq j \leq n}$ of $K'$ consisting of open sets in $X$ contained in $K$, such that for every pair of points $x, y$ of a same $U_j$, one has $q(f(x) - f(y)) \leq \varepsilon/2$ for every $f \in \Phi$. By Lemma 1, there exist $n$ continuous mappings $\varphi_j$ of $X$ into $[0, 1]$ ($1 \leq j \leq n$) such that $\operatorname{Supp}(\varphi_j) \subset U_j$ and such that

For $x_j \in \operatorname{Supp}(\varphi_j)$ ($1 \leq j \leq n$) and $f \in \Phi$, we therefore have, for all $x \in U_j$, and this relation remains true if $x \notin U_j$ since then $\varphi_j(x) = 0$. By addition we infer that, for every $x \in X$, where $\varphi_0 = 1 - \sum_{j=1}^n \varphi_j$; whence (1) for $x \in K'$, since then $\varphi_0(x) = 0$; (1) also holds for $x \notin K$, the first member then being zero. Finally, for $x \in K - K'$ we have $q(f(x)\varphi_0(x)) \leq \varepsilon/2$ by the definition of $K'$, therefore this relation and (2) again imply (1) in this case.

Let $X$ be a locally compact space; for every Banach space $E$ (real or complex) we denote by $\mathscr{C}^b(X; E)$ the vector space of continuous and bounded mappings of $X$ into $E$; we know that the topology of uniform convergence in $X$ is compatible with the vector space structure (real, resp. complex) of $\mathscr{C}^b(X; E)$, and it is defined by the norm

Moreover, the normed space thus defined is a Banach space (GT, X, §3, No. 2, and No. 1, Cor. 2 of Prop. 2); the topology defined by this norm on $\mathscr{K}(X; E)$ (in other words, the topology of uniform convergence in $X$) is coarser than the direct limit topology on $\mathscr{K}(X; E)$ defined in No. 1.

Th. 4), and E a Banach space. The closure of $\mathscr{K}(X;E)$ in the normed space $C^b(X;E)$ is the vector space of continuous functions on X, with values in E and tending to 0 at the point $\omega$.

Let $f \in C^b(X;E)$ be a function in the closure of $\mathscr{K}(X;E)$; for every $\varepsilon > 0$, there exists a function $g \in \mathscr{K}(X;E)$ such that $\|f(x) - g(x)\| \leq \varepsilon$ for all $x \in X$; if K is the support of g, it follows that $\|f(x)\| \leq \varepsilon$ for all $x \in \mathbf{C}K$, thus $f(x)$ tends to 0 as $x$ tends to $\omega$. Conversely, if f has this property then, for every $\varepsilon > 0$, there exists a compact set $K \subset X$ such that $\|f(x)\| \leq \varepsilon$ for all $x \in \mathbf{C}K$. By Lemma 1 there exists a continuous mapping h of X into $[0,1]$, with compact support, equal to 1 on K; then $\|f(x)h(x)\| \leq \varepsilon$ on $\mathbf{C}K$ and $f(x) = f(x)h(x)$ on K; since $fh$ has compact support and $\|f(x) - f(x)h(x)\| \leq 2\varepsilon$ for all $x \in X$, the proposition is proved.

We shall denote by $C^0(X;E)$ the subspace of $C^b(X;E)$ formed by the functions tending to zero at the point at infinity $\omega$; it is thus the completion of the normed space $\mathscr{K}(X;E)$.

For every compact set $K \subset X$, there exists a function $h \in \mathscr{K}(X;\mathbf{R})$ equal to 1 on K, by Lemma 1; for every function $f \in C(X;E)$ the function $hf$, which belongs to $\mathscr{K}(X;E)$, is equal to f on K, whence our assertion.

As the second assertion is an obvious consequence of the first, it suffices to prove the latter. We apply Lemma 2 with $\Phi$ reduced to a single element $f$ of $\mathscr{K}(X,K;E)$; then, for every $x \in X$,

where the $\varphi_j$ belong to $\mathscr{K}(X,K;\mathbf{R})$; since the mapping $x \mapsto \sum_{j=1}^n \varphi_j(x)f(x_j)$ may be canonically identified with the element $\sum_{j=1}^n \varphi_j \otimes f(x_j)$, this proves the proposition, by the definition of the topology of $\mathscr{K}(X,K;E)$.

for every function $f \in \mathscr{K}(X, K_\alpha; \mathbf{C})$.

The condition being obviously necessary, it suffices to prove that (5) implies (4) for every compact subset $K$ of $X$. Now, $K$ is covered by a finite number of open sets $\overset{\circ}{K}_{\alpha_i}$ ($1 \leq i \leq n$); applying Lemma 1 of No. 2 to $K$ and to the $\overset{\circ}{K}_{\alpha_i}$, there exist continuous functions $g_i \geq 0$ on $X$ such that $\operatorname{Supp}(g_i) \subset K_{\alpha_i}$, $0 \leq \sum_{i=1}^n g_i(x) \leq 1$ for all $x \in X$ and $\sum_{i=1}^n g_i(x) = 1$ for $x \in K$. For every function $f \in \mathscr{K}(X, K; \mathbf{C})$, we can therefore write $f = \sum_{i=1}^n f g_i$ and we have $f g_i \in \mathscr{K}(X, K_{\alpha_i}; \mathbf{C})$ and $\| f g_i \| \leq \| f \|$; if $M_K = \sum_{i=1}^n M_{\alpha_i}$, we then have the relation (4).

We denote by $\mathscr{M}(X; \mathbf{C})$, or simply $\mathscr{M}(X)$ if no confusion can result, the vector space of measures on $X$, in other words, the dual of $\mathscr{K}(X; \mathbf{C})$.

One knows that for every set $\mathscr{S}$ of bounded subsets of $\mathscr{K}(X;\mathbf{C})$, there is defined on $\mathscr{M}(X;\mathbf{C})$ the $\mathscr{S}$-topology, which is locally convex (TVS, III, §3, No. 1, Cor. of Prop. 1). We denote the topological vector space, obtained by equipping $\mathscr{M}(X;\mathbf{C})$ with the $\mathscr{S}$-topology, by $\mathscr{M}_{\mathscr{S}}(X;\mathbf{C})$ or $\mathscr{M}_{\mathscr{S}}(X)$.

This results from the fact that $\mathscr{K}(X;\mathbf{C})$ is barreled (TVS, III, §4, No. 2, Cor. 4 of Th. 1).

Examples of measures. — I. Atomic measures. Let X be a locally compact space, $a$ a point of X; the mapping $f\mapsto f(a)$ of $\mathscr{K}(X;\mathbf{C})$ into $\mathbf{C}$ obviously satisfies the condition (4) with $M_K=1$ for every compact subset K of X containing $a$, hence is a measure on X, which is denoted by $\varepsilon_a$; it is called the Dirac measure at the point $a$, or the measure defined by a unit mass placed at the point $a$.

More generally, let $\alpha$ be a mapping of X into $\mathbf{C}$ such that, for every compact subset K of X,

Then, for every function $f\in\mathscr{K}(X,K;\mathbf{C})$, the sum is defined, being equal to it is clear that $\mu$ is a linear form on $\mathscr{K}(X;\mathbf{C})$ and that, for $f\in\mathscr{K}(X,K;\mathbf{C})$, in other words the condition (4) is satisfied.

A measure $\mu$ on X is said to be atomic if there exists a mapping $\alpha$ of X into $\mathbf{C}$ such that

for every compact subset K of X, and such that $\mu$ is equal to the measure defined as above. If N is the set of $x\in X$ such that $\alpha(x)\neq0$, the condition imposed on $\alpha$ implies that for every compact subset K of X, $K\cap N$ is countable. One also says that $\mu$ is defined by the masses $\alpha(x)$ placed at the points $x\in N$. If one assumes that $N\cap K$ is finite for every compact set $K\subset X$, then obviously it comes to the same to say that N is a closed and discrete subspace of X, because then every point of X has a compact neighborhood containing only a finite number of points of N, and conversely, if this is the case, then every compact subset of X can be covered by a finite number of such neighborhoods. When N is closed and discrete, every atomic measure defined by a function $\alpha$ such that $\alpha(x)=0$ on $\mathbf C N$ is called a discrete measure on X (cf. §2, No. 5).

For, let $\mu$ be a positive linear form on $\mathscr{K}(X; \mathbf{R})$ and let $K$ be a compact subset of $X$. There exists a continuous mapping $f_0$ of $X$ into $[0, 1]$, with compact support, such that $f_0(x) = 1$ on $K$ (No. 2, Lemma 1). For every function $g \in \mathscr{K}(X, K; \mathbf{R})$, we thus have $-\|g\|f_0 \leq g \leq \|g\|f_0$, consequently $|\mu(g)| \leq \|g\| \cdot \mu(f_0)$, which proves the theorem.

We denote by $\mathscr{M}_+(X)$ the pointed convex cone of positive measures on $X$ (or, what amounts to the same thing, the cone of positive linear forms on the Riesz space $\mathscr{K}(X; \mathbf{R})$).