Proposition 12

— Let T be a completely regular space, and let B be the unit ball of the normed space $\mathscr{C}^b(T; \mathbf{C})$. Let I be a linear form on $\mathscr{C}^b(T; \mathbf{C})$. In order that there exist a bounded complex measure $\theta$ on T such that $I(f) = \theta(f)$ for all $f \in \mathscr{C}^b(T; \mathbf{C})$, it is necessary and sufficient that the restriction of I to B be continuous for the topology of compact convergence. The measure $\theta$ is then unique.

Let us show that the condition in the statement is necessary. Let $\theta$ be a bounded complex measure on T, $\varepsilon$ a number $> 0$, and K a compact subset of T such that $|\theta|^*(T - K) < \varepsilon$. Let $f \in B$; we denote by U the neighborhood of $f$ in B for the topology of compact convergence, formed by the functions $g \in B$ such that $\sup_{x \in K} |g(x) - f(x)| \leq \varepsilon$. Then, for every $g \in U$,

because $|g - f|$ is bounded above by $\varepsilon$ on K and by 2 on $T - K$.

Conversely, consider a linear form I on $\mathscr{C}^b(T; \mathbf{C})$ whose restriction to B is continuous for the topology of compact convergence. Then, for every number $\varepsilon > 0$, there exist a number $a > 0$ and a compact subset K of T such that the relations $f \in B, \sup_{x \in K} |f(x)| \leq a$ imply $|I(f)| \leq \varepsilon$. Prop. 5 of No. 2 then implies the existence of a unique bounded complex measure $\theta$ such that $I(f) = \theta(f)$ for all $f \in \mathscr{C}^b(T; \mathbf{C})$.