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There is a generalized version of Liouville: Let $f:\mathbb{C}\to \mathbb{C} $ be holomorphic function such that $|f(z)|\le M |z|^n$ for all $z\in \mathbb{C}$ with $|z|\ge R$ for certain $n\in \mathbb{N}$ and $R,M>0$. Then $f$ is a polynomial of degree $\le n$.

My question now is how to get the original Liouville (bounded implies constant) from this? I see if I set $n=0$ I get that $f$ is bounded, but only for $|z|\ge R$. The assumption in the original Liouville is, however, that $f$ is bounded on all of $\mathbb{C}$.

Do I miss an assumption here or can the total boundedness be concluded? Thanks in advance!

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    $\begingroup$ Welcome to math.stackexchange! Remember that a holomorphic function is automatically continuous and continuous functions are bounded on compact sets. $\endgroup$ – Hans Engler May 30 '16 at 0:28
  • $\begingroup$ You stated no theorem. $\endgroup$ – zhw. May 30 '16 at 0:28
  • $\begingroup$ Tsemo, that's not the point of my question. Thank you Hans! Now I see that the assumption ensures boundedness where it is not bounded automatically as you kindly explained :) And I added the implication, thanks for pointing it out zhw. $\endgroup$ – user829100 May 30 '16 at 0:44
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The key point is that the set $\{z:|z|\leq R\}$ is compact. Thus since $f$ is continuous, it must be bounded on this set. Since $f$ is bounded on $\{z:|z|\leq R\}$ and also on $\{z:|z|\geq R\}$, it is bounded on all of $\mathbb{C}$ (just take the larger of the two bounds).

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If $f$ is bounded by $M$ on $|z|\geq R$, then $f$ is bounded by $$\sup\left(M,\sup_{|z|\leq R} |f|\right)$$ on $\mathbb{C}$.

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