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I am trying to prove the following:

Let $n\in \mathbb{N}$. Prove that $\not \exists$ a holomorphic function $f$ on the open unit disk satisfying:

$f\left(\displaystyle \frac{1}{n}\right) = 2^{-n}$

Any help will be deeply appreciated!

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What have you tried? –  Antonio Vargas Nov 20 '12 at 3:12
    
The problem is that I cannot think of a strategy to solve this. I am assuming its contradiction but I cannot see how to proceed if we assume that such holomorphic function exists. –  user44069 Nov 20 '12 at 3:54

3 Answers 3

up vote 2 down vote accepted

Any function $f$ analytic in the unit disk and not identically $0$ can be written in the form $$f(z)=z^r\ g(z)\ ,$$ where $r\in{\mathbb N}_{\geq0}$, $\ g$ is analytic in the unit disk, and $g(0)=:c\ne0$. Assume that for such an $f$ we have $$2^{-n}=f\left({1\over n}\right)=n^{-r}\ g\left({1\over n}\right)\qquad (n\geq1)\ ,$$ or $g\bigl({1\over n}\bigr)=\ n^r\ 2^{-n}$. Letting $n\to \infty$ we obtain $$c=\lim_{n\to\infty} g\left({1\over n}\right)=\lim_{n\to\infty}\bigl(n^r\ 2^{-n}\bigr)=0\ ,$$ a contradiction.

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Hint: Assume that $f$ is analytic on the open unit disk, then think about power series. Show that $f(z)/z^k \to 0$ as $z \to 0$ for all fixed integers $k \geq 0$.

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$\bf{Hint}$: Suppose we had such an analytic function $f$ on the open unit disk. Then in order for $f$ to even be continuous, $f(0) = 0$. Since $f$ is analytic we can express $f$ as a power series centered around zero $$f(z) = \sum_{n = 0}^{\infty}{\frac{f^{n}(0)}{n!}z^n}$$ Show that $f^n(0) = 0$ for all $n\in\mathbb{N}$. Deduce that $f$ cannot be analytic

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