Entire bounded function must be constant.

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March 2nd 2010, 07:50 PM
davismj

Entire bounded function must be constant.

http://i45.tinypic.com/mtbkvs.jpg

I can kind of see whats going on, but not well enough to construct a logical progression towards a proof. Any ideas?
March 2nd 2010, 08:16 PM
davismj

This feels like the solution.

http://i45.tinypic.com/2iu6h5f.jpg

This feels right, but I tend to get things wrong with my ignorance. Verify?
March 3rd 2010, 07:07 AM
Opalg

That basic idea (taking the exponential) is exactly what is needed, but some of the details are a bit dubious. You can use inequalities for real numbers, but not for complex numbers. It would be better to say $\exp(f(z)) = e^{u+iv} = e^ue^{iv}$ and therefore $|\exp(f(z))| = e^u$ (since $e^u>0$ and $|e^{iv}|=1$ ). Then apply Liouville's theorem.
March 3rd 2010, 09:17 AM
davismj

Quote:

Originally Posted by Opalg

That basic idea (taking the exponential) is exactly what is needed, but some of the details are a bit dubious. You can use inequalities for real numbers, but not for complex numbers. It would be better to say $\exp(f(z)) = e^{u+iv} = e^ue^{iv}$ and therefore $|\exp(f(z))| = e^u$ (since $e^u>0$ and $|e^{iv}|=1$ ). Then apply Liouville's theorem.

Of course. This makes a lot of sense to me. Thank you.

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