A bounded holomorphic function

Solution 1:

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 $zf(z) \rightarrow \mu(K) \neq 0$.

Solution 2:

This answer is incorrect.

You can consider the function $g(x):=\sum_{k=0}^{\infty}\frac{x^{2^k}}{2^{k^2}}$. This function, and many examples like it, is going to be analytic on the unit disc, and continuous and even differentiable infinitely many times on the unit circle (but not analytic there!). We can have a biholomorphism between the unit disc and the plane minus the non-negative real line. Composing $g$ with this biholomorphism we get a function analytic in the whole plane minus the non-negative real line, and continuous on the non-negative real line.