Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have encountered the following problem while reading in complex analysis.

Let $U$ be a domain in $\mathbb{C}$ with $z_0 \in U$. Let $\mathcal{F}$ be the family of analytic functions $f$ in $U$ such that $f(z_0) = -1$ and $f(U) \cap \mathbb{Q}_{\geq 0} = \emptyset$, where $ \mathbb{Q}_{\geq0}$ denotes the set of non-negative rational numbers. Is $\mathcal{F}$ a normal family?

[A normal family of functions is a family such that every sequence of functions from the family has a subsequence which converges uniformly on compact subsets of the domain.]

I know I'm supposed to show my work here, but I don't know how to begin to solve this problem. I know that Montel's Theorem says that if $\mathcal{F}$ is locally uniformly bounded, then it is a normal family, but there doesn't seem to be much to work with to show bounds here. And if it's not a normal family, would I have to produce an explicit sequence which has no uniformly convergent subsequence?

Any guidance would be appreciated. Thanks.

share|improve this question
You do know that analytic functions are open mappings, right? Meaning they map open sets to open sets. So the rational numbers seems like a red herring in this problem. –  Harald Hanche-Olsen Jun 26 '12 at 19:30
It's not a red herring. I've left this comment on lots of questions, but the trick is so simple and powerful that I'll do it again. Consider the points $0,1,2$. Since the image of the family omits them, the family omits three points and hence must be normal. –  Matt Jun 26 '12 at 19:51
Thanks - I wasn't aware of that version of Montel's Theorem. None of the books I've looked at progressed that far. –  ec92 Jun 26 '12 at 23:40

Your Answer


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

Browse other questions tagged or ask your own question.