Proposition 9

For every function $f \in \mathscr{H}$, $\mathrm{Ch}_{\mathscr{H}}(X)$ intersects the set of points where $|f|$ attains its supremum.

We may limit ourselves to the case that $f$ is not the constant 0. Let $a$ be a point of $X$ where $|f|$ attains its supremum, and set $g = f/f(a)$; then $g(a) = 1$ and $|g(x)| \leq 1$ for all $x \in X$, whence

By Prop. 5 of No. 3 applied to $\mathscr{H}_r$, there exists $b \in \mathrm{Ch}_{\mathscr{H}}(X)$ where $\Re((g(x))$ attains its supremum 1, whence $|g(b)| = 1$ since $|g(b)| \leq 1$; it follows that $|f(b)| = |(f(a)| \geq |f(x)|$ for all $x \in X$.