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general >> complex analysis >> Sometimes unbounded entire implies constant
(Message started by: Aryabhatta on Apr 15th, 2007, 8:32pm)

Title: Sometimes unbounded entire implies constant
Post by Aryabhatta on Apr 15th, 2007, 8:32pm
Let f be an entire function such that |f(z)| <= 1 + sqrt(|z|) for all z.

Show that f must be constant.

Title: Re: Sometimes unbounded entire implies constant
Post by Obob on Apr 19th, 2007, 9:46am
[hide]Apply Cauchy's inequality to f on a disk of radius r.  Let r go to infinity to see that f' = 0.[/hide]

What can you say about f if it is entire and |f(z)| <= (1+|z|^k)?

Title: Re: Sometimes unbounded entire implies constant
Post by Aryabhatta on Apr 19th, 2007, 10:20am
if k < 1, then your proof works in showing that f is constant.

if k >=1, then we have  a counterexample, f(z) = z.

Title: Re: Sometimes unbounded entire implies constant
Post by Eigenray on Apr 19th, 2007, 10:24am
More generally, if an entire function f is everywhere bounded by a polynomial (of degree k), then f is itself a polynomial (of degree no more than k).

Title: Re: Sometimes unbounded entire implies constant
Post by Aryabhatta on Apr 20th, 2007, 5:06pm
Yes. I guess it can be proved using Cauchy's estimate for the nth derivative and the fact that the polynomial is bounded by it a finite power of z...



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