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I was given the following exercise:

Let $f(z)$ be analytic in an open and connected set $U$ containing the point $z=0$ and assume $|f(1/n)| < \frac{1}{2^n}$ for $n \in \mathbb{N}_{> 0}$. Prove that $f(z)=0$ in $U$.

My idea was to show that there exists an open neighborhood $A$ around $z=0$ such that $f(z)=0$ in $A$ since it is easy to show that $f(0)=0$. From this the result follows due to connectedness of $U$. I have however no idea how to do this.

Another idea was to consider the set $B = \{ 1/n | n \in \mathbb{N} \} \cup \{ 0 \}$ (which is not discrete) and show that $f(x)=0$ for all $x \in B$.

Can someone please help me out? Thanks in advanced!

Mathemagician
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    I don't understand, isn't $f(z) = e^{-z}$ a counter-example ? –  May 22 '16 at 19:46
  • I don't think so because $e^{1/n}$ is not necessarily smaller than $1/2^n$ I think. – Mathemagician May 22 '16 at 19:54
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    See http://math.stackexchange.com/questions/1564554 (Not exactly a duplicate, but almost.) – mrf May 22 '16 at 19:58
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    clearly $f(0) = 0$. suppose that $f(z)$ has a zero of order $k$ at $z=0$, hence $|f(z)| \sim C |z|^k$ and $|f(1/n)| \sim \frac{C }{n^k}$, i.e. it is not $< 2^{-n}$ for every $n$ – reuns May 22 '16 at 20:01
  • @Mathemagician : yes of course, sorry for this comment. –  May 22 '16 at 20:02
  • @mrf : no finally it is not a duplicate since unfortunately the answer uses that $f(1/n) = e^{-n }$ not $|f(1/n)| \le e^{-n}$ – reuns May 22 '16 at 20:06
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    @user1952009 That doesn't matter much. – mrf May 22 '16 at 20:24

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