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Suppose $f_n:\mathbb{D} \to \mathbb{C} $ are injective holomorphic functions such that $ f(0) = 0, |f^\prime(0)|= 1 $. Show that there exists an injective holomorphic map $f_\infty:\mathbb{D} \to \mathbb{C} $ such that $f_{n_k} \to f_\infty$ locally uniformly for some subsequence $f_{n_k}$ of $f_n$.

I want to use Schwartz lemma, but the codomain is not $\mathbb{D}$. So I can only know that each $f_n$ behaves like a rotation near the origin. I want to mimic the strategy as in the proof of Riemann mapping theorem, and instead consider $g_n = f_n^{-1}$ such that $ g(0) = 0, |g^\prime(0)|= 1 $ and $g_n: U_n \to \mathbb{D}$ for open $U_n = f_n(\mathbb{D})$. And suppose I can show that $U_n$ are uniformly bounded, then I can extend $g_n$ to have a common domain and the result follows by uniformly boundedness. But I wonder why it is so as I think $f_n(\mathbb{D})$ can grow infinitely. I think there is something to do with $|f^\prime(0)|= 1$, like mean value theorem or something, but I'm not sure if this is correct. Any suggestion would be much appreciated.

The One
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1 Answers1

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The hypotesis of injectivity here is fundamental. Indeed given that $f$ is injective, $f(0)=0$ and $|f'(0)|=1$, we know from Koebe distortion theorem that $$|f(z)|\le \frac{|z|}{(1-|z|)^2}$$ This gives uniform boundedness of the family on $\{|z|\le r\}$ for every $r\in [0,1)$. Since this is an exaustion of $\mathbb{D}$, we have by Montel's theorem that the family is normal.