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This is exercise 5, page 236 from Remmert, Theory of complex functions

For each of the following properties produce a function which is holomorphic in a neighborhood of $ 0 $ or prove that no such function exists:

i) $ f (\frac{1}{n}) = (-1)^{n}\frac{1}{n} \ $ for almost all $ n \in \mathbb{N}\ , n \neq 0 $

ii) $ f (\frac{1}{n}) = \frac{1}{n^{2} - 1 } $ for almost all $ n \in \mathbb{N}\ , n \neq 0, n \neq 1 $

iii) $|f^{(n)}(0)|\geq (n!)^{2} $ for almost all $ n \in \mathbb{N} $

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And what is your question? –  Jonas Meyer Feb 12 '12 at 7:58
    
I would like to know which argument or theorem I can use –  WLOG Feb 12 '12 at 8:12
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up vote 2 down vote accepted

Your title is misleading, as you cannot determine a holomorphic function from its values on $\mathbb{N}$. However, in this case you can determine it, using the uniqueness theorem for analytic functions: if $f$ and $g$ are two analytic functions and there is a convergent series $a_n$ such that $f(a_n)=g(a_n)$ for all $n$ then $f=g$.

Thus, for example, if your (i), putting $g(z)=z$. we see that $f(1/2n)=g(1/2n)$, so if $f$ is analytic we must have $f=g$. But then $f(1/(2n+1)) = 1/(2n+1)$, contradicting the fact that $f(1/(2n+1)) = -1/(2n+1)$, so no such analytic function exists.

(ii) is similar.

as for (iii), try to think on the Taylor expansion of $f$ near $0$. What will be its radius of convergence?

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for iii) by Cauchy - Hadamard the radius of convergence is $ 0 \ $, a contradiction because f is holomorphic in a neighborhood of $ 0 $ –  WLOG Feb 12 '12 at 8:48
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