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The Liouville theorem state:

Let $f$ be an entire function for which there exists a positive number M such that $|f(z)|\leq M$ for all z in $\mathbb{C}$, the $f$ must be a constant.

more general form of this theorem is :

Let $f$ be an entire function for which there exists a positive number M and a polynomial $P$ such that $|f(z)|\leq M |P(z)|$ for all z in $\mathbb{C}$ then $f(z)=k P(z)$ for $k$ a constant.

who know a good proof of this generalized form ?

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    I don't think your generalized form is correct. Consider $f=z^2$, $P=z^2+1$, $M=1$. Then surely $|f(z)|=|z^2|\leq 1\cdot(|z^2|+1)$, but $z^2$ is not a constant multiple of $z^2+1$. I believe the correct statement should be $|f(z)\leq M|P(z)|$ implies $f(z)=kP(z)$. Then the proof would consist of showing that $\frac fP$ is entire, and using the usual form of Liouville's theorem.2012-03-17
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    yes, maybe $|f(z)|\leq M|P(z)|$ for all $z\in\mathbb{C}$. thans for the remarks :)2012-03-17
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    I think the theorem is that if $|f(z)|\leq C\cdot |z|^k$ for some constant $C\in \Bbb C$, then $f$ is a polynomial of degree at most $k$.2014-11-09

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