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Let $\Omega$ be a domain in $\mathbb{C}$ and let $\{f_n\}_{n \in \mathbb{N}}$ be a sequence of injective functions that converge in $O(\Omega)$ to $f$ . Show that $f$ is either injective or a constant function.

How does the conclusion change if, instead of a domain, we allow $\Omega$ to be an arbitrary open set ?

I know that $f$ is holomorphic as an almost uniform limit. But I dont know how to proceed.

Rusty
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  • Hint: the derivative of an analytic function can be recovered from the function's values via the Cauchy integral formula. What can you say about the derivatives of all the functions involved? – Greg Martin Apr 11 '16 at 04:24
  • @GregMartin: I don't see how that would work. Even if you show that the derivative of the limit function vanishes nowhere, that would not be sufficient to prove that an analytic function is injective. Or what did you have in mind? – Martin R Apr 12 '16 at 08:07

1 Answers1

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Assume that the limit function $f$ is neither constant nor injective. Then $f$ takes some value $a$ in disjoint disks $B_1, B_2 \subset \Omega$. As in

it follows from Rouché's theorem that there are $n_1, n_2$ such that for $n \ge n_j$, $f_n$ takes the value $a$ in $B_j$ (at least once). For $n \ge \max(n_1, n_2)$ this is a contradiction to $f_n$ being injective.

Rouché's theorem is applied separately to the two disks in $\Omega$, so the same conclusion holds if $\Omega$ is a (not necessarily connected) open set.

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Martin R
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