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This is probably something basic that I am missing. I am reading the article Normal Families: New Perspectives by Lawrence Zalcman, and in one of his examples he makes the following assertion (I am paraphrasing, not quoting, for brevity - I hope I didn't ruin the correctness of the statement):

Let $f_n : D \to \mathbb C$ be a sequence of analytic univalent functions on a domain $D$. Suppose the sequence of first derivatives $g_n := f_n\prime$ converges locally uniformly to an analytic function $g$. Then $g$ is also the first derivative of a univalent function on $D$, or zero.

Why is this true? It looks like a twist on Hurwitz's Theorem, but I don't get it.

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Yes, this is a twist on Hurwitz's theorem: the limit of non-vanishing functions $f_n'$ is either identically zero, or nowhere zero. The first case is clear. In the second case, fix a point $z_0\in D$ and consider the functions $\tilde f_n=f_n-f_n(z_0)$. Since $\tilde f_n$ is pinned down at $z_0$, and the derivatives $\tilde f_n'=f_n'$ converge locally uniformly, the functions $\tilde f_n$ converge locally uniformly. Let $f$ be their limit. Since $f'=\lim \tilde f_n' = g$ is nowhere zero, $f$ is locally univalent.

To prove that $f$ is univalent in $D$, argue by contradiction: suppose there is $w\in \mathbb C$ and points $z_1\ne z_2$ such that $f-w$ vanishes at $z_1,z_2$. Pick $r>0$ such that the $r$-neighborhoods of $z_1,z_2$ are disjoint. By Hurwitz's theorem, for all sufficiently large $n$ the function $\tilde f_n-w$ vanishes in the aforementioned neighborhoods. This is a contradiction, because $\tilde f_n$ is univalent.

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  • $\begingroup$ Thanks, that helps a lot! Just one more thing I've yet to understand: You claim that since $\tilde f_n$ is pinned down at $z_0$, and the derivatives converge locally uniformly, the functions converge locally uniformly. Why is this true in a general domain? I've managed to prove it in a simply connected domain, where I can express $\tilde f_n$ as an integral of $\tilde f_n '$. $\endgroup$ – Yoni Rozenshein Mar 2 '13 at 16:23
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    $\begingroup$ @Yoni You can always express a function in terms of its derivative, as long as the domain of definition is connected. Just use any curve from $z_0$ to another point. You may be thinking of whether there exists a function with given derivative, but this is not an issue here: our functions do exist. $\endgroup$ – user53153 Mar 2 '13 at 16:35
  • $\begingroup$ Ah yes, the fundamental theorem of calculus for connected domains is a much easier result than Cauchy's theorem. Thanks! :) $\endgroup$ – Yoni Rozenshein Mar 2 '13 at 16:38

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