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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Silvia Ghinassi
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Dontknowanything
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Think about $\mathbb{C}\setminus { 0}$ and removable singularities. – Jose27 Jan 18 '16 at 18:48
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If $U$ is simply connected the Riemann mapping theorem guarantees the existence of an analytic function $f:>U\to D$. – Christian Blatter Jan 18 '16 at 19:09
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This question has some relevance: http://math.stackexchange.com/q/432810/27978. – copper.hat Jan 18 '16 at 19:24
3 Answers
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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.
quid
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4Yes. As you can use Riemann mapping theorem, in that case there always exists such a function. – Silvia Ghinassi Jan 18 '16 at 18:58
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2@MartinR I think RMT applies to non-empty proper simply connected open subsets of $\mathbb C$. – Silvia Ghinassi Jan 18 '16 at 19:06
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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.
Silvia Ghinassi
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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.
copper.hat
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Are you asking to show this for an arbitrary $U$ (that satisfies the conditions)? – copper.hat Jan 18 '16 at 18:41
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