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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. |
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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)? |
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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. |
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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). |
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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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