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[parent] proof of extended Liouville's theorem (Proof)

This is a proof of the second, more general, form of Liouville's theorem given in the parent article.

Let $f:\Complex\to\Complex$ be a holomorphic function such that $$|f(z)| < c \cdot |z|^n$$ for some $c\in\Reals$ and for $z\in\Complex$ with $|z|$ sufficiently large. Consider

$\displaystyle g(z)=\begin{cases}\frac{f(z)-f(0)}{z}&z\neq 0\ f'(0)&z=0 \end{cases}$
Since $f$ is holomorphic, $g$ is as well, and by the bound on $f$ , we have $$ |g(z)|< c_1 + c_2 \cdot |z|^{n-1} < c'\cdot |z|^{n-1 $$ again for $|z|$ sufficiently large.

By induction, $g$ is a polynomial of degree at most $n-1$ , and thus $f$ is a polynomial of degree at most $n$ .




"proof of extended Liouville's theorem" is owned by rm50.
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Cross-references: degree, polynomial, induction, bound, holomorphic function, Liouville's theorem, proof

This is version 5 of proof of extended Liouville's theorem, born on 2006-10-08, modified 2008-03-31.
Object id is 8431, canonical name is ProofOfExtendedLiouvillesTheorem.
Accessed 2391 times total.

Classification:
AMS MSC30D20 (Functions of a complex variable :: Entire and meromorphic functions, and related topics :: Entire functions, general theory)

Pending Errata and Addenda
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I think there is a mistake by futurebird on 2008-03-04 09:21:44
Just because f(z) < |z|^n we don't know that f'(z) < c' |z|^n

I don't see how you got around this in this proof.
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