Let $\mathscr{F}$ be a collection of holomorphic functions $f:U\to \mathbb{C}\smallsetminus [0,1] $, for some domain $U$. Suppose there exists a $z_0\in U$, such that $f(z_0)=g(z_0)$, for all $f,g\in\mathscr{F}$. Prove that every sequence in $\mathscr F$ has a locally uniformly convergent subsequence.
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Should that last word say "subsequence?" – T.J. Gaffney Jan 11 '14 at 07:02
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1Are you sure that there exists a uniformly convergent subsequence? How about a locally uniformly convergent subsequence? – Yiorgos S. Smyrlis Jan 11 '14 at 07:24
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@YiorgosS.Smyrlis Year. I do not know why $f(z_0)=g(z_0)$ is needed. – XLDD Jan 11 '14 at 08:58
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Not exactly the same but very similar. If you don't place any extra assumption you can't expect convergent subsequences: take for example $f_n(z) = n$ (for $n \ge 2$). – mrf Jan 11 '14 at 09:27
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Consider $$\mathscr{G} = \left\lbrace \sqrt{1 - \frac{1}{f(z)}} : f\in \mathscr{F} \right\rbrace,$$ where the used branch of the square root is the principal branch. – Daniel Fischer Jan 11 '14 at 12:21
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The existence of a locally uniformly (and not uniformly) convergent subsequence is guaranteed by Montel's Theorem.
In this exercise, nevertheless, we have stronger assumption: (1) The functions miss a whole interval and (2) They agree on a point.
Yiorgos S. Smyrlis
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How can I dominate the ${f_n}$ by a $M>0$ under these two assumptions? Using linear transformation? I have no idea. – XLDD Jan 11 '14 at 11:56