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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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    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 $\mathrm{Im} f(z)\geq 0$'?2012-03-14

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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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    Hi WimC, great answer! I'm asking a similar question - could you please take a look? Thanks! http://math.stackexchange.com/questions/1245940/prove-that-big-fracfz-fwfz-overlinefw-big-le-big-fracz-w2015-04-22