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The problems goes as follows:

Let $f(z)$ be holomorphic in the closed unit disc, with $f(-\log 2)=0$, and $$|f(z)|\leq |e^z|$$ for all $z$ with $|z|=1$, then how large can $|f(\log 2)|$? Find the best possible upper bound.

I am quite sure what the problem is testing. We can possibly get estimate by Cauchy's formula but that is obvious too sparse and did not make use of $f(-\log 2)=0$....

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Take $g(z) = f(z)e^{-z}$, with a view to applying Schwarz lemma. Since $g(0)$ may not be zero, pre-compose it with the map $$ \varphi_{\log(2)} = \frac{z-\log(2)}{1-\log(2)z} $$ which maps $0$ to $-\log(2)$. Now Schwarz lemma applies to $h = g\circ \varphi_{\log(2)}$ from which you get that $$ |h(z)| \leq |z| $$ Now choose $z$ so that $\varphi_{\log(2)}(z) = \log(2)$, ie. $$ z = \frac{2\log(2)}{1+\log(2)^2} $$ Then $$ |h(z)| = |g(\log(2))| = |f(\log(2))| $$ since $exp(\varphi_{\log(2)}(z)) = 1$. Hence, $$ |f(\log(2))| \leq \frac{2\log(2)}{1+\log(2)^2} $$

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  • $\begingroup$ Thank you very much!! I was confused by 'the best possible', and never came up with the Schwarz's lemma! It is so clear! $\endgroup$ – Roy Han Aug 13 '13 at 14:09
  • $\begingroup$ No problem :) Schwarz' lemma is usually the best place to start when dealing with functions on the disc. $\endgroup$ – Prahlad Vaidyanathan Aug 13 '13 at 14:13

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