Maximum Modulus Principle Problem

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  • #1

Homework Statement



Let g(z) be a function that is analytic and non-constant on D = {|z| < 1}. Suppose that Max |g(z)| [tex]\leq[/tex] [tex]\frac{1}{r}[/tex] for all 0< r <1, |z| = r. Use the Maximum Modulus Principle (or corollary) to prove that |g(z)| < 1 for all z [tex]\in[/tex] D.


Homework Equations



http://hphotos-snc3.fbcdn.net/hs104.snc3/15146_524570240458_58700263_31202039_3403055_n.jpg [Broken]
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Maximum Modulus Principle:

http://hphotos-snc3.fbcdn.net/hs104.snc3/15146_524570060818_58700263_31202038_7869593_n.jpg [Broken]
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http://hphotos-snc3.fbcdn.net/hs084.snc3/15146_524570804328_58700263_31202043_1540731_n.jpg [Broken]

The Attempt at a Solution



Not exactly sure how to start.
 
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Answers and Replies

  • #2
Office_Shredder
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Ok, so basically the problem is that g(z) might be something like [tex] \frac{1}{z}[/tex] But it's analytic so it has to be defined at 0. This means that near 0 it has to be bounded. What can you conclude?
 
  • #3
that the maximum of g(z) is reached on the boundary of g(z0) ?
 

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