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Aryabhatta
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 Sometimes unbounded entire implies constant   « on: Apr 15th, 2007, 8:32pm » Quote Modify

Let f be an entire function such that |f(z)| <= 1 + sqrt(|z|) for all z.

Show that f must be constant.
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Obob
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 Re: Sometimes unbounded entire implies constant   « Reply #1 on: Apr 19th, 2007, 9:46am » Quote Modify

Apply Cauchy's inequality to f on a disk of radius r.  Let r go to infinity to see that f' = 0.

What can you say about f if it is entire and |f(z)| <= (1+|z|^k)?
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Aryabhatta
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 Re: Sometimes unbounded entire implies constant   « Reply #2 on: Apr 19th, 2007, 10:20am » Quote Modify

if k < 1, then your proof works in showing that f is constant.

if k >=1, then we have  a counterexample, f(z) = z.
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Eigenray
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 Re: Sometimes unbounded entire implies constant   « Reply #3 on: Apr 19th, 2007, 10:24am » Quote Modify

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).
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Aryabhatta
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 Re: Sometimes unbounded entire implies constant   « Reply #4 on: Apr 20th, 2007, 5:06pm » Quote Modify

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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