Fall 2014 - Problem 11

Solution.

First, we will show that we can replace $K$ with a connected compact set. Let $\delta = \frac{1}{2}d\p{K, \Omega^\comp} > 0$ and for each $z \in K$, notice that $B\p{z, \delta} \subseteq \Omega$. By compactness, there exist finitely many balls which cover $K$, and this cover has finitely many connected components (since there are only finitely many balls). Thus, replacing $K$ with the closure of these balls, we may assume without loss of generality that $K$ has finitely many connected components. Since $\Omega$ is (path) connected, we can connect these components with finitely many curves. Replacing $K$ with $K$ union these curves, we may thus assume without loss of generality that $K$ is a connected compact set.

For each $z \in K$, let $R_z > 0$ be such that $\cl{B\p{z, 2R_z}} \subseteq \Omega$. By compactness, there exist $z_1, \ldots, z_N$ such that $\set{B\p{z_n, R_n}}_n$ cover $K$. Fix a $z_0 \in K$. For any $z \in K$, path connectedness of $K$ gives $\gamma$ such that $\gamma\p{0} = z_0$ and $\gamma\p{1} = z$, and so we get a sequence $\set{B\p{z_{n_k}, R_{n_k}}}_{1 \leq k \leq M}$ such that $z_0 \in B\p{z_{n_0}, R_{n_0}}$, $z \in B\p{z_{n_M}, R_{n_M}}$, and $B\p{z_{n_k}, R_{n_k}} \cap B\p{z_{n_{k+1}}, R_{n_{k+1}}} \neq \emptyset$. We may remove duplicates without breaking this property, so we have a strict subset of $A$.

Notice that if $w \in B\p{z_{n_k}, R_{n_k}} \cap B\p{z_{n_{k+1}}, R_{n_{k+1}}}$, then Harnack's inequality gives

so $u\p{z_{n_k}} \leq 9u\p{z_{n_{k+1}}}$. Hence, by induction, we see

Let $C_K = 3 \cdot 9^N$. Since $z$ was arbitrary, we see

Similarly, $z_0$ was arbitrary, so