Proposition 6

For every point $x \in X$, the following properties are equivalent:

a) $x$ is $\mathscr{H}$*-extremal*.

b) For every open neighborhood $U$ of $x$ in $X$ and every $\varepsilon > 0$, there exists a function $h \geq 0$ in $\mathscr{H}$ such that $h(x) \leq \varepsilon$ and $h(y) \geq 1$ for every $y \in X - U$.

Let $x$ be any point of $X$, $f$ a function in $\mathscr{C}(X; \mathbf{R})$; it is known (TVS, II, §3, No. 1, Prop. 1) that the infimum of the numbers $\lambda(f)$, for all the positive measures on $X$ such that $\lambda(h) = h(x)$ for every function $h \in \mathscr{H}$, is equal to the supremum of the numbers $h(x)$, where $h$ runs over the set of functions $h \in \mathscr{H}$ such that $h \leq f$. Suppose that $x$ is $\mathscr{H}$-extremal; it then follows from Prop. 4, (ii) that for every function $f \in \mathscr{C}(X; \mathbf{R})$,

To show that a) implies b), we take for $f$ a continuous mapping of $X$ into $[0, 1]$, with support contained in $U$, such that $f(x) = 1$; then, by (11), there exists a function $h' \in \mathscr{H}$ such that $h' \leq f$ and $h'(x) \geq 1 - \varepsilon$. Since $1 \in \mathscr{H}$, the function $h = 1 - h'$ meets the conditions of b).

Conversely, suppose that the condition b) is verified; this condition implies that $1 - h \leq \varphi_U$; for every positive measure $\lambda$ on $X$ satisfying the condition (10), we therefore have

Since, by hypothesis, this relation holds for every $\varepsilon > 0$ and every open neighborhood $U$ of $x$, it follows that

for every $\varepsilon > 0$, therefore $\lambda(\{x\}) = 1$. Since $\lambda$ is positive and of total mass 1, necessarily $\lambda = \varepsilon_x$, which proves that $x$ is $\mathscr{H}$-extremal, by virtue of Prop. 4, (ii).