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Spring 2016 - Problem 8

conformal mappings, normal families

Let C+={z∈C∣Im⁡z>0}\C_+ = \set{z \in \C \mid \Im{z} > 0} and let fn ⁣:C+→C+\func{f_n}{\C_+}{\C_+} be a sequence of holomorphic functions. Show that unless ∣fn∣→∞\abs{f_n} \to \infty uniformly on compact subsets of C+\C_+, there exists a subsequence converging uniformly on compact subsets of C+\C_+.
Solution.

Let φ ⁣:C+→D\func{\phi}{\C_+}{\D} be a conformal mapping, e.g., φ(z)=z−iz+i\phi\p{z} = \frac{z - i}{z + i}. Thus, ∣φ∘fn∣≤1\abs{\phi \circ f_n} \leq 1, so the sequence {φ∘fn}n\set{\phi \circ f_n}_n forms a normal family. Hence, there exists a subsequence {φ∘fnk}k\set{\phi \circ f_{n_k}}_k which converges locally uniformly to some holomorphic g ⁣:C+→D\func{g}{\C_+}{\D}.

Let K⊆C+K \subseteq \C_+ be a compact set. Notice that because φ−1\phi^{-1} is holomorphic, it is bounded on KK, hence Lipschitz on KK with some constant CKC_K. Thus, if we set f=φ−1∘gf = \phi^{-1} \circ g

∣fnk−f∣=∣φ−1(φ∘fnk)−φ−1(g)∣≤CK∣φ∘fnk−g∣,\abs{f_{n_k} - f} = \abs{\phi^{-1}\p{\phi \circ f_{n_k}} - \phi^{-1}\p{g}} \leq C_K \abs{\phi \circ f_{n_k} - g},

so fnkf_{n_k} converges locally uniformly to ff, which completes the proof.