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Suppose $\{f_n\}$ is sequence of analytic functions that is uniformly bounded in the open unit disk and for every positive integer $k$, $f_n(\frac{1}{1+k})\to 0$ pointwise. Then, $\{f_n\}$ converges uniformly to $0$ on every compact subset of the open unit disk.

I do not know how to get started on this problem. It seems as though either Montel's or Arzela-Ascoli's theorem could be used but I fail to see how. I will appreciate any help.

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    $\begingroup$ if you knew that $a_{n}(k) = f_n(\frac1{1+k})$ converges to $0$ uniformly, could you prove that $f_n \to 0$ uniformly on $|z| < 1-\epsilon$ ? $\endgroup$ – reuns May 12 '16 at 23:53
  • $\begingroup$ I think I see it now. If no one posts an answer, I will write up one. Thanks for the hint. $\endgroup$ – dezdichado May 13 '16 at 3:42
  • $\begingroup$ I'd like to know your arguments, I start writing an answer, but then I stopped : I used 4 times different arguments for showing that $a_n(k) \to 0$ uniformly, that the finite differences $\to 0$, that $f_n^{(k)}(0) \to 0$, that $f_n(z_0) \to 0$, but I'm sure there is much simpler. $\endgroup$ – reuns May 13 '16 at 3:45
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Use Montel's theorem. They form a normal family, so $(f_n)_{n\ge1 }$ has a subsequence $f_{n_r}$ converging uniformly on compact subsets to say, $g$. Then $g(\frac{1}{1+k}) = 0$ for all $k \ge 1$. So basically the zero set of $g$ has a limit point, so $g \equiv 0$.

Use the above argument to in fact prove that, if you take any subsequence $f_{n_p}$ of $(f_n)_{n\ge1 }$, then it has a further subsequence $f_{n_{p_r}}$ which converges uniformly on compact subsets to the identically zero function.

I claim that it implies that $f_n$ converges to the zero function. For this use the metric $d$ on $H(\Omega)$. (Which is defined as an infinite sum over a countable family of compact subsets) Assume $f_n \not \to 0$. Then there is a subsequence $f_{n_k}$ such that $d(f_{n_k}, 0) \geq \epsilon$. But then $f_{n_k}$ can not have a subsequence converging to $0$.

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