TITLE: Existence of bounded analytic function on unbounded domain?
QUESTION [5 upvotes]: Given any proper open connected unbounded set $U$  in $\mathbb C$.Does there always exist a non constant bounded analytic function $ f\colon U \to \mathbb C$ ?

Edit: $U$ is any arbitrary domain. I don't have idea to do it. Please help.

REPLY [9 votes]: No not always. Take $ U= \mathbb{C} \setminus \{0\}$. Take a bounded analytic function on $U$. As it is bounded it can only have a removeable singularity at $0$. Thus it extends to an entire function, which must be constant.
On the other hand if the closure of $U$ is not all of $\mathbb{C}$ take a $z_0$ outside the closure of $U$  and consider $(z-z_0)^{-1}$. 
This is not a full classification of all $U$ though, but you did not ask for this.

REPLY [6 votes]: No. Take $U=\mathbb{C}\setminus \{p\}$, and take $f$ bounded holomorphic on $U$. Then we can extend $f$ to the whole complex plane (a point is removable), but being bounded and entire, $f$ has to be constant.

REPLY [5 votes]: Take $f(z) = {1 \over z} $ on $U=\{z \mid |z|>1 \}$.
This example can be extended to any $U$ such that $U^c$ contains an open set.