Maximum Modulus Principle Problem

dark_dingo
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Homework Statement



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


Homework Equations



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

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

The Attempt at a Solution



Not exactly sure how to start.
 
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Ok, so basically the problem is that g(z) might be something like \frac{1}{z} 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?
 
that the maximum of g(z) is reached on the boundary of g(z0) ?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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