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Topic: Sometimes unbounded entire implies constant (Read 2794 times) |
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Aryabhatta
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Sometimes unbounded entire implies constant
« on: Apr 15th, 2007, 8:32pm » |
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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 » |
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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 » |
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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 » |
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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 » |
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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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