Spring 2010 - Problem 10

Solution.

We first state Harnack's inequality: Let $u$ be a non-negative harmonic function on the open disk $D\p{z_0, R}$. Then for $z \in D\p{z_0, R}$,

Let $\delta = \frac{1}{2}d\p{K, \Omega^\comp} > 0$, since $K$ and $\Omega^\comp$ are disjoint closed sets. Since $K$ is totally bounded, this implies that we can cover $K$ with finitely many balls $B\p{z_1, \delta}, \ldots, B\p{z_n, \delta}$. By construction, the corresponding closed balls $\cl{B\p{z_j, \delta}} \subseteq \Omega$ also. Set $E = \bigcup_{j=1}^n \cl{B\p{z_j, \delta}} \subseteq \Omega$. Since $E$ is a finite union of compact balls, $E$ is itself compact and has finitely many connected components (it has at most $N$, one for each ball). Hence, we may connect all the components together with finitely many curves to form a new set $F$, since $\Omega$ is connected, hence path connected. Because $F$ is a union of a compact set $E$ and finitely many compact curves, $F$ is also compact, and by construction, $E$ is connected. Since $K \subseteq F$, if we find an upper bound $M$ for $F$, this gives us an upper bound on $K$. Thus, by replacing $K$ with $F$, we may assume without loss of generality that $K$ is a connected, compact set in $\Omega$.

With $\delta$ as before, compactness implies that we can cover $K$ with finitely many balls $B\p{z_1, \frac{\delta}{3}}, \ldots, B\p{z_N, \frac{\delta}{3}}$ such that $B\p{z_j, \delta} \subseteq \Omega$ as well. By Harnack's inequality, for $z \in B\p{z_j, \frac{2\delta}{3}}$, we have

for any $u \in \mathcal{U}$. Notice that these balls have the following property: if $B\p{z_j, \frac{\delta}{3}} \cap B\p{z_k, \frac{\delta}{3}} \neq \emptyset$, then $z_k \in B\p{z_j, \frac{2\delta}{3}}$. Indeed, given $w \in B\p{z_j, \frac{\delta}{3}} \cap B\p{z_k, \frac{\delta}{3}}$, we have

Now let $z \in K$ and let $\gamma$ be a curve in $K$ connecting $z$ and $z_0$. Then $\gamma$ traverses through the $B\p{z_j, \frac{\delta}{3}}$ in some order. Let $A = \set{w_j}_j \subseteq \set{z_1, \ldots, z_N}$ be the sequence of the centers of these balls, which is finite since $\gamma$ is compact and these balls cover $\gamma$. For notation, let $\abs{A} = M$.

Notice that $B\p{w_j, \frac{\delta}{3}} \cap B\p{w_{j+1}, \frac{\delta}{3}} \neq \emptyset$ for all $1 \leq j < M$. Indeed, because $j < M$, there exists $t_0$ so that $\abs{\gamma\p{t_0} - w_j} = \frac{\delta}{3}$, i.e., a time when $\gamma$ leaves $B\p{w_j, \frac{\delta}{3}}$ and enters $B\p{w_{j+1}, \frac{\delta}{3}}$. Since $B\p{w_{j+1}, \frac{\delta}{3}}$, there exists an open neighborhood $\gamma\p{t_0} \in U \subseteq B\p{w_{j+1}, \frac{\delta}{3}}$. Since $\gamma\p{t_0}$ is a boundary point of $B\p{w_j, \frac{\delta}{3}}$, this means that $U \cap B\p{w_j, \frac{\delta}{3}} \neq \emptyset$, and so $B\p{w_j, \frac{\delta}{3}} \cap B\p{w_{j+1}, \frac{\delta}{3}} \neq \emptyset$, as claimed.

Next, we need to remove duplicates from $A$. We do this as follows: let $m$ be the first index so that $w_m = w_k$ for some $1 \leq k < m$, and remove $w_{k+1}, \ldots, w_m$ from $A$. Notice that

so the property we described before still holds with $w_{k+1}, \ldots, w_m$ removed. Renumber $A$ so that $w_{k+\ell} = w_{m+\ell}$ for $\ell \geq 1$.

By induction, we may repeat this process, and this must terminate after finitely many steps since $A$ is finite. When this occurs, $A$ has no more duplicates, which implies that $\abs{A} = M \leq N$. To complete the proof, observe that $w_{j+1} \in B\p{w_j, \frac{2\delta}{3}}$ by construction, and so by (1),

By construction, $w_M = z$ and $w_1 = z_0$, and by induction, we conclude that

Since $z \in K$ was chosen arbitrarily, we see that

which completes the proof.