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I need to show that $\mathcal{F}=\{f\in \mathcal{O}(\mathbb{H}): |f(z)|\neq 5 \forall z \in \mathbb{H}\}$ is a normal family. Here $\mathbb{H}$ is the upper half plane and $\mathcal{O}(\mathbb{H})$ is all of the holomorphic functions on $\mathbb{H}$. Since $\mathbb{H}$ is connected and each $f\in \mathcal{F}$ is continuous we see that either $f(\mathbb{H})\subset \{z\in \mathbb{C}: |z|<5\}$ or $f(\mathbb{H})\subset \{z\in \mathbb{C}: |z|>5\}$. Let's split $\mathcal{F}$ into the two sub families just mentioned: $\mathcal{F_{1}}$ and $\mathcal{F_{2}}$, respectively. I'm aiming to use Montel's theorem so I've got to establish a uniform bound on each compact $K\subset\mathbb{H}$. This trivial for $\mathcal{F_{1}}$ but I don't know how to do this for $\mathcal{F_{2}}$. Any hints or answers would be greatly appreciated.

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The functions $f_n\colon z\mapsto n$ where $n\geq 6$ are in $\mathcal F$, but it seems there is no uniform bound. –  Davide Giraudo Aug 22 '11 at 21:04
    
It's been two years since I've thought about complex analysis at all, but I seem to recall Montel saying something like if the family omits more than three values then it must be normal, and as you've stated $\mathcal{F}$ omits an entire circle. Maybe these had to be defined on all of $\mathbb{C}$ or something? –  Matt Aug 22 '11 at 22:18
    
Indeed, if $f: \mathbb H \to \mathbb P^1 \setminus \{0, 1, \infty\}$ is holomorphic, then by simple connectedness of $\mathbb H$, it lifts to the open unit disk. Therefore $\mathcal F$ forms a normal family (in the extended sense where one allows local uniform convergence to $\infty$). –  Henri Aug 22 '11 at 22:31
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Note that ${\cal F}_2$ is a normal family of functions into the Riemann sphere ${\mathbb C}_\infty$, but not as functions into $\mathbb C$. –  Robert Israel Jan 24 '12 at 18:54

1 Answer 1

I assume by Montel's theorem, you here mean its simplest form: a family of functions that is (locally) bounded is normal.

(This is the "normal families version" of Liouville's theorem, and the easiest to prove.)

As you mention, the family $\mathcal{F}_1$ is normal by this theorem.

To show that $\mathcal{F}_2$ is also normal, consider the family of functions $1/f$, for $f\in\mathcal{F}_2$.

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