I want to prove the following are equivalent.
(a) For any $p \in U$ and $r>0$ with $\overline{D(p, r )} \subseteq U$,
if $h$ is harmonic on a neighborhood of $\overline{D(p, r)}$ satisfying $f \le h $ on $\partial D(p, r)$, then $f \le h$ on $D(p, r)$.
(b) For any $p \in U$ and $r>0$ s.t. $\overline{D(p, r )} \subseteq U$,
if $h$ is continuous on $\overline{D(p, r)}$ and harmonic on $D(p, r)$,satisfying $f \le h $ on $\partial D(p, r)$, then $f \le h$ on $D(p, r)$.
(a) $\Leftarrow $ (b) is obvious.
But I don't know how to prove the converse. The author says that I should use a 'limiting argument', but I'm not sure what 'limiting argument' refers to.
Any help will be appreciated. Thanks.
You didn't say what is $f$ so I assume it is a continuous function on $U$.
For sake of simplicity, let assume that $r = 1$. If $h$ is harmonic on $D(p, 1)$ then for any $t \in (0, 1)$, the function $h$ is harmonic on a neighbourhood of $\overline{D(p, t)}$. Since $h$ and $f$ are continuous on the compact set $\overline{D(p, 1)}$, then there are uniformly continuous on it. Thus, for any $\varepsilon > 0$ there exists $t_{\varepsilon} \in (0, 1)$ such that $$ \forall t \in (t_{\varepsilon}, 1),\ \forall z \in \partial D(p, 1),\ |h(tz)-h(z)| \leqslant \varepsilon\ \mbox{ and }\ |f(tz)-f(z)| \leqslant \varepsilon. $$ In particular, we have $f - 2\varepsilon \leqslant h$ on $\partial D(p, t)$ for any $t \in (t_{\varepsilon}, 1)$. According to (a), we got $f-2\varepsilon \leqslant h$ on $D(p, t)$ for all $t \in (t_{\varepsilon}, 1)$. Thus $f-2\varepsilon \leqslant h$ on $D(p, 1)$. Let $\varepsilon$ tends towards zero to conclude.