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Let $G,H$ be disjoint open subsets of $\mathbb{C}$ and $f_n:G\to H$ be analytic functions. If $f_n(z)\to f(z)$ for all $z\in G$, then prove that $f$ is analytic and $f(G)\subset H$. Any help is appreciated.

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I think the pointwise limit $f(z)$ could be a constant, equal to a value on the boundary of $H$, so contradicting $f(G) \subset H$. –  GEdgar Dec 19 '12 at 2:41
    
I must ask: What is the alternative? –  copper.hat Dec 19 '12 at 3:06
    
@copper.hat en.wikipedia.org/wiki/Tits_alternative –  user53153 Dec 21 '12 at 5:50
    
Thanks @PavelM! Breast answer so far today... –  copper.hat Dec 21 '12 at 5:54

2 Answers 2

up vote 1 down vote accepted

The claim that $f(G) \subset H$ is not true in general, it is easy to make up examples where $(f_n)$ converges to a constant in $\partial H$. For the other (true) part of the claim, you need to show that the convergence is not only pointwise, but locally uniform. The crucial step here is to show that the family $(f_n)$ is normal, and the tool to do it is Montel's theorem, using that all $(f_n)$ omit all values in $G$. If you need more details, let me know.

Added details: Let $z_0 \in G$ be arbitrary. Then $g_n(z) = \frac{1}{f_n(z)-z_0}$ is analytic for all $n$, and the family $(g_n)$ is locally bounded, because each $f_n$ omits a fixed disk about $z_0$. So $(g_n)$ is a normal family, and then $f_n(z) = z_0 + \frac1{g_n(z)}$ is normal as well.

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That is the strategy I have been trying, but have been unable to prove that family $f_n$ is locally bounded, which will imply the result. It would be helpful if you could give some details. Thanks. –  Tits_Alternative Dec 19 '12 at 2:48

Let's the series expansion of $f_n$ $$ f_{n}(x)=\lim_{N\to \infty}\sum_{k=1}^{N}a_k(n)x^k. $$ A suficient condition is:

1) the convergence $f_n\to f$ is uniform,

2) exist $\lim_{t\to\infty}\lim_{N\to \infty}\sum_{k=1}^{N}a_k(n)x^k$ and $\lim_{N\to \infty}\lim_{n\to\infty}\sum_{k=1}^{N}a_k(n)x^k$.

This result is a consequence of

Theorem. Let $\{ F_t ; t\in T\}$ be a family of functions $F_t : X \rightarrow \mathbb{C}$ depending on a parameter t; let $\mathcal{B}_X$ be a base $X$ and $\mathcal{B}_{T}$ a base in $T$. If the family converges uniformly on $X$ over the base $\mathcal{B}_{T}$ to a function $F : X \rightarrow \mathbb{C}$ and the limit $\lim_{\mathcal{B}_{T}} F_t(x)=A_t$ exists for each $t\in T$, the both repeated limits $\lim_{\mathcal{B}_{X}}(\lim_{\mathcal{B}_{T}}F_t(x))$ and $\lim_{\mathcal{B}_{T}}(\lim_{\mathcal{B}_{X}}F_t(x))$ exist and the equality

$$ \lim_{\mathcal{B}_{X}}(\lim_{\mathcal{B}_{T}}F_t(x))=\lim_{\mathcal{B}_{T}}(\lim_{\mathcal{B}_{X}}F_t(x)) $$ holds.

The proof can be found in books of Zorich ([Mathematical Analysis II][1] p. 381).

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