Let $D := \{z \in \mathbb{C}: |z| < 1\}$ and $f\colon D \rightarrow \mathbb{C}$ be holomorphic. Suppose $\lvert f(z)\rvert \leq 1$ on $D$, show that $$\frac{|f(0)| - |z|}{1 + |f(0)||z|} \leq |f(z)| \leq \frac{|f(0)| + |z|}{1 - |f(0)||z|} \ \forall z \in D.$$

I have tried using Cauchy Integral Formula to $f$ and expanding $f$ at $0$, but I have no idea why $f(0)$ appears in the inequality in the former case. Both answers and hints are appreciated and thanks in advance!

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    $\begingroup$ Just a guess...have you tried something like $g = B \circ f$, where $B$ is some well-chosen fractional linear transformation? This is a common trick in problems like these. I'd try to choose $B$ so that $g(0) = 0$, so that the Schwarz lemma applies... $\endgroup$ – bryanj Jul 8 '13 at 14:21
  • $\begingroup$ Actually I have not learn Schwarz lemma, so I have not try using your method at all...... Nevertheless, it becomes clear to me about why $f(0)$ appears in the inequality after you mentioned it since we need $g(0) = 0$. Thank you! $\endgroup$ – user43378 Jul 8 '13 at 14:58

Following the comment from bryanj, if $f(0)f(z) \neq 1$, consider $$g(z) := \frac{f(z) - f(0)}{1 - f(0)f(z)},$$ then $g(0) = 0$ and $g$ is holomorphic on $D$. Schwarz lemma then gives $$|g(z)| \leq |z|$$ and the result follows.

Otherwise, $f(z) = \frac{1}{f(0)}$, then the desired inequality becomes $$(|f(z)| \pm |z|)^2 \leq 2.$$ But it is true since $|f(z)| \leq 1 \ \forall z \in D.$


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