TITLE: A bounded holomorphic function
QUESTION [15 upvotes]: If $\Omega$ is a region which is dense in $\mathbb{C}$, $f\in H(\Omega)$ and is continuous on $\mathbb{C}$, moreover $f$ is bounded on $\mathbb{C}$, can we claim that $f$ is a constant?

REPLY [9 votes]: The answer is no. Indeed, let $K$ be a nowhere dense compact subset of the plane with Hausdorff dimension strictly greater than one (for example, one can take some Julia set appearing in the following list) and let $\Omega$ be the complement of $K$. The fact that every $f \in H(\Omega)$ continuous and bounded in $\mathbb{C}$ is constant is equivalent to saying that the continuous analytic capacity of $K$ is zero (see my answer to this question). However, as remarked by  Emil Jeřábek in my answer, every compact set $K$ with zero continuous analytic capacity must have Hausdorff dimension at most $1$.
EDIT : As remarked in the very nice comment by user85506, since the Hausdorff dimension of $K$ is strictly greater than one, $K$ supports a nontrivial measure with growth $\mu(B(z,r)) \leq C r^{1+\epsilon}$, by Frostman's Lemma. The growth of $\mu$ implies that the Cauchy Transform
$$f(z):=\int_{K} \frac{1}{z-\zeta}d\mu(\zeta)$$
is continuous on $\mathbb{C}$ and holomorphic on $\mathbb{C} \setminus K$. Furthermore, $f(z)\rightarrow 0$ as $z \rightarrow \infty$ and so in particular, $f$ is bounded. However, $f$ is non-constant because $zf(z) \rightarrow \mu(K) \neq 0$.