# Entire bounded function must be constant.

• Mar 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?
• Mar 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?
• Mar 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 \$\displaystyle \exp(f(z)) = e^{u+iv} = e^ue^{iv}\$ and therefore \$\displaystyle |\exp(f(z))| = e^u\$ (since \$\displaystyle e^u>0\$ and \$\displaystyle |e^{iv}|=1\$). Then apply Liouville's theorem.
• Mar 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 \$\displaystyle \exp(f(z)) = e^{u+iv} = e^ue^{iv}\$ and therefore \$\displaystyle |\exp(f(z))| = e^u\$ (since \$\displaystyle e^u>0\$ and \$\displaystyle |e^{iv}|=1\$). Then apply Liouville's theorem.

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