Complex plane minus Cantor set admits non-constant bounded harmonic function

Let $$K\subset [0,1]$$ denote the usual 1/3 Cantor set. I know that $$\mathbb{C}\backslash K$$ has no non-constant bounded analytic function, since the singularity $$K$$ can be removed. However, a statement I am reading says that $$\mathbb{C}\backslash K$$ admits a non-constant bounded harmonic function. Why is this true? Any help would be appreciated.

• Where is the statement that you are reading? May 3, 2020 at 1:18
• It is actually an exercise asking one to prove the statement above. May 3, 2020 at 1:39
• Where is the exercise? May 3, 2020 at 1:40
• Sorry it is on a set of notes from a class I am taking. I am not sure where the exercise is originally from. May 3, 2020 at 1:42

Since the Cantor set $$K$$ has Hausdorff dimension $$\log2/\log 3<1$$, it is a removable set for bounded analytic functions, and so, as you say, there is no bounded analytic function outside of $$K$$. But it does not mean that there are no bounded harmonic functions outside of $$K$$. Actually, any point of $$K$$ is regular for the Dirichlet problem. So, choosing any non-constant continuous function $$u$$ on $$K$$, there exists a unique harmonic function $$f$$ in $$\overline{\mathbb C}\setminus K$$ such that $$\forall\zeta\in K,~\lim_{z\to\zeta}f(z)=u(\zeta).$$ Then $$f$$ is a non-constant bounded harmonic function in $$\overline{\mathbb C}\setminus K$$. Note that $$K$$ is of positive capacity so there is no reason that $$f$$ can be extended to $$\mathbb C$$ (and actually it cannot be extended).
A proof that any point of $$K$$ is regular for the Dirichlet problem, that uses Wiener criterion, is given p.100 of
• Thank you so much for your answer! Would you mind elaborating a bit on why $K$ is regular for the Dirichlet problem? May 3, 2020 at 3:45