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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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  • $\begingroup$ Think about $\mathbb{C}\setminus \{ 0\}$ and removable singularities. $\endgroup$ – Jose27 Jan 18 '16 at 18:48
  • $\begingroup$ If $U$ is simply connected the Riemann mapping theorem guarantees the existence of an analytic function $f:\>U\to D$. $\endgroup$ – Christian Blatter Jan 18 '16 at 19:09
  • $\begingroup$ This question has some relevance: math.stackexchange.com/q/432810/27978. $\endgroup$ – copper.hat Jan 18 '16 at 19:24
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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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    $\begingroup$ Does this change if we assume simple connectedness? $\endgroup$ – Cameron Williams Jan 18 '16 at 18:56
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    $\begingroup$ Yes. As you can use Riemann mapping theorem, in that case there always exists such a function. $\endgroup$ – Silvia Ghinassi Jan 18 '16 at 18:58
  • $\begingroup$ @SilviaGhinassi: ... if the complement has at least two elements. $\endgroup$ – Martin R Jan 18 '16 at 19:03
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    $\begingroup$ @MartinR I think RMT applies to non-empty proper simply connected open subsets of $\mathbb C$. $\endgroup$ – Silvia Ghinassi Jan 18 '16 at 19:06
  • $\begingroup$ @SilviaGhinassi: In that case you are right. $\endgroup$ – Martin R Jan 18 '16 at 19:07
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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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  • $\begingroup$ But $U$ is any arbitrary domain? $\endgroup$ – Dontknowanything Jan 18 '16 at 18:40
  • $\begingroup$ Are you asking to show this for an arbitrary $U$ (that satisfies the conditions)? $\endgroup$ – copper.hat Jan 18 '16 at 18:41
  • $\begingroup$ Yeah,I'm asking for arbitrary $U$ $\endgroup$ – Dontknowanything Jan 18 '16 at 18:42

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