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Prove that any holomorphic function , but not one-to-one, must satisfy .

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    What is $\Delta(1)$? – 2017-02-02
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    It is the unit disk in the complex plane. – 2017-02-02

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Assuming $\Delta(1)$ is the unit disc, then it suffices to prove the Schwarz-Pick lemma which states \begin{align} \left|\frac{f(z_1)-f(z_2)}{1-\overline{f(z_1)}f(z_2)}\right| \leq \left|\frac{z_1-z_2}{1-\overline{z_1}z_2}\right| \end{align} which means \begin{align} \frac{|f'(z)|}{1-|f(z)|^2} \leq \frac{1}{1-|z|^2}. \end{align} Hence it follows when $z=0$ that \begin{align} |f'(0)| \leq 1-|f(0)|^2<1. \end{align}

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    Why is $f(0) \neq 0$? – 2017-02-02
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    @copper.hat Suppose $f(0) = 0$, then it's clear that you recover Schwarz's lemma which states that $|f'(0)| = 1$ means $f(z) =az$ which is injective. – 2017-02-02
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    +1: Thanks! I have no intuition whatsoever for applications of the Schwarz lemma :-(. – 2017-02-02