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.



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


J.B. Garnett, D.E. Marshall, Harmonic measure. New Mathematical Monographs 2. Cambridge University Press, Cambridge, 2005.


