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Given $\{f_{n}(z)\}$, a sequence of analytic functions in the upper half plane $\mathbb C^{+}$, where each $f_{n}(z)$ has continuous extension to the real line, and $|f_{n}(z)|\leq 1$ for all $z\in \mathbb C^{+}\cup \mathbb R$ (i.e., uniformly bounded). How to prove that there is a subsequence which converges to some analytic function $f$ (whatever the convergence is; uniformly or pointwise)?

(I think there is a theorem called Montel's theorem, but I'm not sure if we can apply it here directly!)

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    Use Cauchy's integral formula and the mean value inequality to show that $\{ f_n|_K \}$ is an equicontinuous family for each compact $K\subset \mathbb{C}^+$. Use Arzela-Ascoli to prove convergence on compact sets and Weierstrass' theorem to prove that the limit is analytic.2012-06-13
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    @Jose27: Is there a known proof in some books? reference?2012-06-13
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    What prevents you from applying Montel's theorem? This is the easy version of Montel: a uniformly bounded family is normal. http://en.wikipedia.org/wiki/Montel's_theorem2012-06-13
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    @Leonid: So you mean we can consider $\mathbb C^{+}$ as the open subset of $\mathbb C$? and then apply the theorem directly!2012-06-13
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    Well, $\mathbb C^+$ is an open subset of $\mathbb C$, is it not? By the way, the theorem applies just as well for open subsets of the sphere $\mathbb C\cup\{\infty\}$.2012-06-13
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    @Leonid: What about $\mathbb C^{+}\cup \mathbb R$, it is not open! So the convergence above will be just in $\mathbb C^{+}$, what about convergent on the real line? Does the same subsequence will converge uniformly to the same limit (found in $\mathbb C^{+}$) on the real line?2012-06-13
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    No. Here is an example of functions on the closed disk instead of closed halfplane: $f_n(z)=z^n$. The functions are bounded by $1$ on the unit disk. According to Montel, they converge uniformly on compact subsets of $\{|z|<1\}$; the limit is $0$. They don't converge to anything on the boundary, and certainly not to zero. You can transplant this example to the half-plane: $f_n(z)=((z-i)/(z+i))^n$.2012-06-13
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    Do we care about the boundary? the sequence of functions $x^{n}$ still converge to 0 on $[-1,1]$, right! and this is what we need, on the real line.2012-06-13

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