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Suppose that $f$ is analytic on the open unit disk and there is a constant $M > 1$ such that $|f(1/k)| \leq M^{-k}$ for $k \geq 1$. Show that $f$ is identically zero.

I see that $f(0) = 0$, that $f'(0) = 0$, and that the zero at $0$ must be isolated if $f$ isn't identically zero on the disk. Any hint to move forward from here would be appreciated.

Context: I'm studying for a qual, so just a hint at this point would be most helpful.

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2 Answers 2

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Hint: if $f$ is not identically zero, $f(z) = z^m g(z)$ for some positive integer $m$ and $g(0) \neq 0$. Use the given inequality to obtain a contradiction.

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Suppose $f(z)=\sum_{j=l}^\infty a_j z^j$ with $a_l\neq 0$. Consider $g(z)=\frac{f(z)}{z^l}$. Show that there is still some constant $\bar{M}>1$ (most likely $\bar{M}<M$) such that $|g(1/k)|\leq \bar{M}^{-k}$ for big enough $k$.

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