Example

— 1) Suppose that $E$ is finite-dimensional over $\mathbf{C}$ and Hausdorff, so that if $(e_i)_{1 \leq i \leq n}$ is a basis of $E$ then the mapping

is an isomorphism of $\mathbf{C}^n$ onto $E$. We then know that every linear form on $E$ is continuous, in other words that $E'$ is identical to the algebraic dual $E^*$ of $E$, and ${E'}^*$ may be canonically identified with $E$. Let $(e'_i)_{1 \leq i \leq n}$ be the basis of $E'$ dual to $(e_i)$; for a mapping $f$ of $X$ into $E$ to be weakly continuous and scalarly of compact support, it is necessary and sufficient that the functions $f_i = e'_i \circ f$ be continuous with compact support; we then have $f(x) = \sum_{i=1}^n f_i(x) e_i$ for all $x \in X$, and

2) Let us take for $E$ the space $\mathscr{M}(X; \mathbf{C})$ of measures on $X$, equipped with the vague topology (§ 1, No. 9); the dual $E'$ of $E$ may then be canonically identified with the space $\mathscr{K}(X; \mathbf{C})$ (TVS, II, § 6, No. 2, Prop. 3). The mapping $x \mapsto \varepsilon_x$ of $X$ into $E$ is continuous (§ 1, No. 9, Prop. 13), but its support is not compact if $X$ is not compact; however, it is scalarly of compact support, because for every function $f \in E'$ the function $x \mapsto \langle \varepsilon_x, f \rangle = f(x)$ by definition has compact support. Moreover,

for every function $f \in \mathscr{K}(X; \mathbf{C}) = E'$, which proves that

for every measure $\mu$ on $X$.

3) If $E$ is Hausdorff then, for every point $y \in X$ and every function $f \in \widetilde{\mathscr{K}}(X; E)$, we have

because $\int \langle f, z' \rangle d\varepsilon_y = \langle f(y), z' \rangle$ by definition.