A question around Liouville's theorem

Question

Liouville's theorem of complex analysis states that a bounded entire function is constant. I am trying to understand if a sort of converse holds in the following sense: consider a closed set $S \subset \mathbb{C}$ ($S$ may be unbounded) such that $\mathbb{C} \setminus S$ is unbounded. Can one construct a non-constant entire function $f$ such that $f$ is bounded on $S$ ?

More specifically, I am interested in understanding what types of unbounded closed subsets $S$ of $\mathbb{C}$ (if any) satisfy the following:

1. the complement $\mathbb{C} \setminus S$ is unbounded, and

2. All non-constant entire functions are unbounded on $S$.

Comment: I asked this question on MSE, but haven't got any answers there. With explicit examples involving trigonometric and exponential functions, one can construct rather "large" subsets of $\mathbb{C}$ where entire functions can be bounded (but my stock of examples is rather limited).

Answer 1

This is a classical problem of complex analysis which is usually stated as a problem on removable singularities. By definition, a Riemann surface (for example a plane domain) belongs to the class $O_{HB}$ if all bounded holomorphic functions on it are constant. For results on general Riemann surfaces, the reference is Tsuji, Potential theory in modern function theory, (which is old but very good), and more recent book Ahlfors and Sario, Riemann surfaces.

For the case of plain regions of the form $C\backslash E$, the complement $E$ in this case is called "removable for bounded functions". For example, if $E$ is of zero logarithmic capacity, it is removable, but this is very crude statement. To characterize the removable sets, an Analytic capacity was introduced. (Analytic capacity is zero if and only if the set is removable). It is difficult to characterize the sets of zero analytic capacity in metric terms. See the Wikipedia article on Analytic capacity, to begin with; it contains some references. The problem of characterization of removable sets in the plane is called "Painleve problem". This is a hot research area, and the number of results is enormous.

Answer 2

Edit: the answer is given by user1952009 in the comments when $\infty$ is locally accessible from $\mathbb{C}\setminus{S}$, but I'll leave my original incomplete answer here for illustration.

Note that if $\mathbb{C}\setminus{S}$ is not required to be connected, you can take $S$ to be any close union of simple closed curves $\bigcup_{i\in \mathbb{N}}\gamma_i$ such that $\gamma_i$ tends to $\infty$ as $i\to \infty$. More generally, if $\mathbb{C}\setminus{S}$ is unbounded but has only bounded connected components then $S$ satisifes 1. and 2. in virtue of the maximum principle.

Therefore only the case of $\mathbb{C}\setminus{S}$ with an unbounded connected component from which $\infty$ is not accessible remains open, as far as I know.

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The family of Mittag-Leffler functions $$E_a(z):=\sum_{n=0}^{\infty}\frac{z^n}{\Gamma(an+1)}$$ for $a>0$ has growth order $\rho=\frac{1}{a}$. It is well-known (see e.g. [1,p25-26]) that $$\lim_{r\to \infty}E(r\exp(\mathrm{i}\theta))=0$$ uniformly in compact subsectors of $\{\frac{a\pi}{2} < \theta <\frac{(4-a)\pi}{2} \}$ as soon as $0<a<2$. This shows the existence of bounded entire functions on any sector of aperture $2\pi-\varepsilon$.

Hence for a $S$ to exist satisfying 1. and 2. no member of its tail at $\infty$, the family $(S_r)_r:=(S\setminus{r\mathbb D})_r$, can be included in any angular sector, which seriously constraints its shape. In particular if $\infty$ is accessible from $\mathbb{C}\setminus S$ by a path $\gamma$ then $\arg\gamma$ cannot converge.

[1] Gorenflo, R., Kilbas, A.A., Mainardi, F., Rogosin, S.V. , "Mittag-Leffler Functions, Related Topics and Applications", Springer, 2014

(can be found on Google books)