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Topic: Sometimes unbounded entire implies constant (Read 3004 times) 

Aryabhatta
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Sometimes unbounded entire implies constant
« on: Apr 15^{th}, 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 19^{th}, 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 19^{th}, 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 19^{th}, 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 20^{th}, 2007, 5:06pm » 
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Yes. I guess it can be proved using Cauchy's estimate for the n^{th} derivative and the fact that the polynomial is bounded by it a finite power of z...


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