Fall 2012 - Problem 12

Solution.

1. Let $M' = \sup_{z \in \Omega} u\p{z}$ and let $\set{z_k}_k$ be a sequence such that $u\p{z_k} \to M'$. Since $\Omega$ is bounded, $\cl{\Omega}$ is compact, so passing to a subsequence if necessary, we may assume without loss of generality that $\set{z_k}_k$ converges to some $z_0 \in \cl{\Omega}$.

   If $z_0 \in \Omega$, then $u$ attains its maximum $u\p{z_0} = M'$ in $\Omega$, so by the maximum principle, $u \equiv M'$ on the connected component containing $z_0$. Furthermore, $u \leq M$, since if $z \in \partial\Omega$, then $M' = \limsup_{w \to z} u\p{z} \leq M$, so by definition of $M'$, $u\p{z} \leq M' \leq M$ for all $z \in \Omega$.

   If $z_0 \in \partial\Omega$, then by definition,

   $$M' = \lim_{n\to\infty} u\p{z_n} \leq \limsup_{z \to z_0} u\p{z} \leq M,$$

   and the claim follows since $M'$ was the supremum of $u$ on $\Omega$.

2. Write $F = \set{p_1, \ldots, p_n}$ and consider the function

   $$v\p{z} = \sum_{k=1}^n \log\,\abs{\frac{z - p_k}{d}},$$

   where $d = 2\sup_{z,w\in\Omega} \abs{z - w} < \infty$, since $\Omega$ is bounded. In particular, $\abs{z - p_k} \leq \frac{d}{2} < d$ for all $z \in \cl{\Omega}$ and $1 \leq k \leq n$. Thus, $v\p{z}$ is non-positive, and it is also a harmonic function on $\Omega$: notice that $f\p{z} \coloneqq \prod_{k=1}^n \frac{z - p_k}{d}$ does not vanish on $\Omega$, so it has a holomorphic logarithm $g$ on all of $\Omega$. Thus, $v\p{z} = \Re{g\p{z}}$, hence harmonic.

   Let $\epsilon > 0$ and consider $\phi_\epsilon = u + \epsilon v$. Notice that as $z \to p_k$, $v\p{z} \to -\infty$. Thus, because $u$ is bounded on $\Omega$ by, say, $M'$, we have $\phi_\epsilon \leq M' + \epsilon v$. Thus,

   $$\limsup_{z \to p_k} \phi_\epsilon\p{z} \leq M' + \limsup_{z \to p_k} \epsilon v\p{z} = -\infty \leq M.$$

   Moreover, since $v \leq 0$, we see that $\phi_\epsilon \leq u$ so in particular, $\limsup_{z \to z_0} \phi_\epsilon\p{z} \leq M$ for all $z_0 \in \partial\Omega \setminus F$. In other words, $\phi_\epsilon$ satisfies the hypotheses of (1) and so $\phi_\epsilon\p{z} = u\p{z} + \epsilon v\p{z} \leq M$ for all $z \in \Omega$. Letting $\epsilon \to 0$, we get $u\p{z} \leq M$ on $\Omega$, which completes the proof.