Proposition 3

Let $E$ be a locally convex space. The mapping $\mu \mapsto \mathscr{F}\mu$ of the set of promeasures on $E$ into the set of functions on $E'$ is injective.

The formula (5) permits reducing to the case that $E$ is finite-dimensional; since every finite-dimensional space is isomorphic to a space $\mathbf{R}^n$, we can even suppose that there exists an integer $n \geqslant 0$ such that $E = \mathbf{R}^n$. We therefore have to prove that if $\mu$ is a bounded measure (not necessarily positive) on $\mathbf{R}^n$ and if

for every linear form $y$ on $\mathbf{R}^n$, then $\mu = 0$.

For every integer $m \geqslant 0$, let $G_m$ be the subgroup $m \cdot \mathbf{Z}^n$ of $\mathbf{R}^n$. Denote by $\mathscr{C}_m$ the vector space of continuous functions $f$ on $\mathbf{R}^n$ such that $f(x + a) = f(x)$ for $x \in \mathbf{R}^n$ and $a \in G_m$. By Prop. 8 of GT, X, §4, No. 4, every function in $\mathscr{C}_m$ is the uniform limit of finite linear combinations of functions of the type $x \mapsto e^{2\pi i \langle x, q \rangle}$ with $q \in m^{-1} \cdot \mathbf{Z}^n$. Therefore $\mu(f) = 0$ for every function $f \in \mathscr{C}_m$.

Let $f$ be a continuous function on $\mathbf{R}^n$ with compact support. For every integer $m \geqslant 0$, set $f_m(x) = \sum_{q \in G_m} f(x + q)$. It is immediate that for every $x \in \mathbf{R}^n$, the preceding series has only finitely many terms, and that $f_m$ belongs to $\mathscr{C}_m$. Moreover, it is easy to see that the sequence $(f_m)$ tends to $f$ uniformly on every compact set, and that there exists a constant $C \geqslant 0$ such that $|f_m| \leqslant C$ for all $m$. Consequently, $\mu(f) = \lim_{m \to \infty} \mu(f_m)$ by Prop. 12 of §5, No. 6. Since $f_m \in \mathscr{C}_m$, we have $\mu(f_m) = 0$, whence finally $\mu(f) = 0$. Thus $\mu = 0$.