Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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: – copper.hat Jan 18 at 19:24
up vote 9 down vote accepted

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.

share|cite|improve this answer
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.

share|cite|improve this answer

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.