# Existence of bounded analytic function on unbounded domain?

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.

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Think about $\mathbb{C}\setminus \{ 0\}$ and removable singularities. – Jose27 Jan 18 at 18:48
If $U$ is simply connected the Riemann mapping theorem guarantees the existence of an analytic function $f:\>U\to D$. – Christian Blatter Jan 18 at 19:09
This question has some relevance: math.stackexchange.com/q/432810/27978. – copper.hat Jan 18 at 19:24

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.

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Does this change if we assume simple connectedness? – Cameron Williams Jan 18 at 18:56
Yes. As you can use Riemann mapping theorem, in that case there always exists such a function. – Silvia Ghinassi Jan 18 at 18:58
@SilviaGhinassi: ... if the complement has at least two elements. – Martin R Jan 18 at 19:03
@MartinR I think RMT applies to non-empty proper simply connected open subsets of $\mathbb C$. – Silvia Ghinassi Jan 18 at 19:06
@SilviaGhinassi: In that case you are right. – Martin R Jan 18 at 19:07

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.

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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.

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But $U$ is any arbitrary domain? – Dontknowanything Jan 18 at 18:40
Are you asking to show this for an arbitrary $U$ (that satisfies the conditions)? – copper.hat Jan 18 at 18:41
Yeah,I'm asking for arbitrary $U$ – Dontknowanything Jan 18 at 18:42