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I'm trying to understand the following inequality. Let $f$ be holomorphic, such that $\mathrm{Im}f(z)\geq 0$ when $\mathrm{Im}(z)>0$. Why is it that $$ \displaystyle\frac{|f(z)-f(z_0)|}{|f(z)-\overline{f(z_0)}|}\leq\frac{|z-z_0|}{|z-\bar{z}_0|}? $$

Since $f$ maps the upper half plane to itself, I was thinking of mapping the plane to the unit disk by some linear fractional, and then attempt to use Schwarz' lemma somehow. I haven't been able to execute a good plan.

Does anyone have any hints and/or solutions to show this inequality? Thank you.

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  • $\begingroup$ Surely there must be some assumptions about $z$ and $z_0$. Otherwise a counterexample is $f: w \mapsto i + e^{iw}$ with $z_0 = i$ and $z = 2 - i$. $\endgroup$ Commented Mar 14, 2012 at 1:45
  • $\begingroup$ I'm not sure if there's a much cleaner way of writing the title, but right now it parses as 'Why does $(f-f_0/f-\bar{f_0} \leq z-z_0/z-\bar{z_0} \mathrm{when}\ \mathrm{Im}(z) \gt 0)$ imply that $\mathrm{Im} f(z)\geq 0$'? $\endgroup$ Commented Mar 14, 2012 at 17:13

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I assume you mean this to hold for $\Im(z) \geq 0$ and also that $\Im(z_0) > 0$ and $\Im(f(z)) > 0$ for all $z \in \mathbb{R}$. The functions

$$ P(z) = \frac{f(z)-f(z_0)}{f(z)-\overline{f(z_0)}} $$

and

$$ Q(z) = \frac{z-z_0}{z-\overline{z_0}} $$

are holomorphic on the closed upper half plane and satisfy:

  • $P(z_0)=Q(z_0)=0$.
  • If $\Im(z) \geq 0$ then $|P(z)| \leq 1$ and $|Q(z)| \leq 1$.

Moreover, $Q$ has a single simple zero at $z_0$ and if $z \in \mathbb{R}$ then $|Q(z)|=1$. Therefore

$$ \frac{P(z)}{Q(z)} $$

is holomorphic on the closed upper half plane and for all $z \in \mathbb{R}$

$$ \left| \frac{P(z)}{Q(z)} \right| = |P(z)| \leq 1. $$

By the maximum modulus principle this equality must hold on all of the upper half plane. In other words, $|P(z)| \leq |Q(z)|$ for all $z$ with $\Im(z) \geq 0$.

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