Proposition 10

*Let $F$ be a closed subset of $X$. The following properties are equivalent: a) $F$ contains $\check{S}_{\mathscr{H}}(X)$. b) For every function $f \in \mathscr{H}$, $F$ intersects the set of points of $X$ where $|f|$ attains its supremum. c) For every point $x \in X$, there exists a positive measure $\mu$ of total mass 1 on $X$ such that $\mathrm{Supp}(\mu) \subset F$ and $f(x) = \int f d\mu$ for every function $f \in \mathscr{H}$. Let us prove the equivalence of the conditions a) and c): let $f = f_1 + i f_2$ with $f_1, f_2$ in $\mathscr{H}_r$; the relation $f(x) = \int f d\mu$ is equivalent to the two relations $f_1(x) = \int f_1 d\mu$ and $f_2(x) = \int f_2 d\mu$; it thus suffices to apply to $\mathscr{H}_r$ the equivalence of the conditions a) and c) of Prop. 7 of No. 3. The fact that a) implies b) follows from Prop. 9. Let us show that b) implies a); this is a matter of seeing that if b) is verified, then, for every $h \in \mathscr{H}_r$, F intersects the set of points where $h$ attains its infimum in X. The condition b) implies that F is nonempty; since F is compact, there exists $a \in F$ such that $h(a) \leq h(y)$ for all $y \in F$. Let $f \in \mathscr{H}$ be such that $h = \mathscr{R}(f)$; for every $\varepsilon > 0$, the function $g = f - h(a) + \varepsilon$ belongs to $\mathscr{H}$, and

for all $y \in F$. Let $c$ be the supremum of $|g|$ in X, and set $b = c^2 / 2\varepsilon$; for every $y \in F$, in other words, the supremum in F of the function $|g - b|$ is $\leq b$. Since $g - b \in \mathscr{H}$, the hypothesis on F implies that $|g - b| \leq b$, whence and so $\mathscr{R}(g) \geq |g|^2 / 2b \geq 0$; since $\mathscr{R}(g) = h - h(a) + \varepsilon$, and $\varepsilon > 0$ is arbitrary, we have $h \geq h(a)$, and $h(a)$ is the infimum of $h$ in X, which completes the proof.