How to prove that the function $$f(r)=(1 - r^2) \int_0^{2\pi}|Re[\frac{e^{i a} (2 - e^{-i s} r)}{(-e^{i s} + r)^2}]|ds$$ for real $a$ and $r\in[0,1]$ attains its maximum for $r=0$ with $f(0)=8$.
8
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$\begingroup$ Could you give some more background to this problem? Surely this is not a randomly chosen integral; is this an exercise, or a step in the proof of a larger result? $\endgroup$– Yemon ChoiCommented Jul 16, 2014 at 19:51
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$\begingroup$ A step on the proof of the following inequality $|f'(z)-f'(0)|\le \frac{8|z|}{\pi(1-|z|^2)} \max|f(z)|$, where $f$ is analytic in the unit disk. $\endgroup$– MarjoCommented Jul 16, 2014 at 19:54
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1$\begingroup$ OK, so this looks like an exercise from a book. If so, then I think math.stackexchange.com would be a more suitable site for this question. $\endgroup$– Yemon ChoiCommented Jul 16, 2014 at 20:00
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$\begingroup$ Not an exercise, I am guessing that this is true but probably a problem which no one can solve it. $\endgroup$– MarjoCommented Jul 16, 2014 at 20:23
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1$\begingroup$ It should be $|f'(z)-f'(0)|\le \frac{4|z|}{\pi(1-|z|^2)} \max|f(z)|$ instead. $\endgroup$– MarjoCommented Jul 16, 2014 at 20:45
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