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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.2011-08-22
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    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?2011-08-22
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    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 $\infty$).2011-08-22
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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 $\mathbb C$.2012-01-24

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