Complex Analysis - Normal Families

[EC92]

Homework Statement

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?

Homework Equations

Montel's Theorem: A family of analytic functions on a domain is normal if it is uniformly bounded on compact subsets of the domain.

The Attempt at a Solution

I'm not sure that this is a normal family; if it's not, how would I prove it? Would I have to produce an explicit sequence of functions from the family which has no subsequence which converges uniformly (on compact subsets)?
If it is normal, I presumably need to show that it is uniformly bounded on compact subsets, which I don't know how to do either.

Thanks.

 

[mighty2000]

Thank you for your question. To determine whether \mathcal{F} is a normal family, we can use Montel's Theorem as you suggested. To show that the family is not normal, we would need to produce a sequence of functions from \mathcal{F} that is not uniformly bounded on compact subsets of U.

To do this, let's consider the function f_n(z) = -\frac{1}{n}z^2, which is analytic on U and satisfies f_n(z_0) = -1 for all n. Additionally, for any z \in U, we have f_n(z) = -\frac{1}{n}z^2 \notin \mathbb{Q}_{\geq 0}, since \mathbb{Q}_{\geq 0} does not contain any negative numbers.

Therefore, the family \mathcal{F} is not uniformly bounded on compact subsets of U, since for any compact subset K \subset U, there exists an n \in \mathbb{N} such that f_n(z) = -\frac{1}{n}z^2 is not bounded on K. This shows that \mathcal{F} is not a normal family.

I hope this helps. Good luck with your studies!