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

1$\begingroup$ Where is the statement that you are reading? $\endgroup$– LSpiceMay 3, 2020 at 1:18

$\begingroup$ It is actually an exercise asking one to prove the statement above. $\endgroup$– Jasper LiangMay 3, 2020 at 1:39

1$\begingroup$ Where is the exercise? $\endgroup$– LSpiceMay 3, 2020 at 1:40

$\begingroup$ Sorry it is on a set of notes from a class I am taking. I am not sure where the exercise is originally from. $\endgroup$– Jasper LiangMay 3, 2020 at 1:42
1 Answer
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 nonconstant 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 nonconstant 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
J.B. Garnett, D.E. Marshall, Harmonic measure. New Mathematical Monographs 2. Cambridge University Press, Cambridge, 2005.

$\begingroup$ Thank you so much for your answer! Would you mind elaborating a bit on why $K$ is regular for the Dirichlet problem? $\endgroup$ May 3, 2020 at 3:45

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$\begingroup$ I have checked out the paper but not the book yet. Thank you for your help and your references! $\endgroup$ May 3, 2020 at 4:25