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If $f:\mathbb{D}\rightarrow\mathbb{D}$ is holomorphic then prove that $$\frac{|f(0)|-|z|}{1 + |f(0)||z|} \leq|f(z)| \leq\frac{|f(0)| + |z|}{1 - |f(0)||z|} $$

I have been wracking my brain for hours and am out of ideas. The hypothesis seems to suggest an application of the Schwarz lemma but I cannot find a suitable function - I've tried functions of the form $$\frac{f(z) - f(0)}{\mbox{scale factor}}$$ and none have worked. I also tried working backwards to obtain an inequality and got $$-|z|\leq|f(z)||1\pm f(0)z| - |f(0)|\leq |z|\;\;\;\mbox{(*) }$$ If I could prove $$|f(z)[1\pm f(0)z] - f(0)|\leq 1$$ then I could obtain (*), but with so few conditions on $f$ I have no idea how to proceed.

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You can consider the function, $$g(z)=\frac{f(z)-f(0)}{1-\overline{f(0)}f(z)}$$ You have $g(0)=\frac{f(0)-f(0)}{1-\overline{f(0)}f(0)}=0$. Now consider $|g(z)|=\frac{|f(z)-f(0)|}{|1-\overline{f(0)}f(z)|}$, we have $|g(z)| \le 1$, it is a consequence of the fact that $ |f(z)| < 1 $ and that the Möbius transformation is an automorphism of the unit disk (see lemma (21.3) for a proof http://www.personal.psu.edu/jxr57/501-03/lecture21.pdf).

Now you can apply Schwarz lemma. Can you go from there? For all $|z|<1$, we have:

$$\begin{align} |g(z)|\le |z| &\Rightarrow |f(z)-f(0)|\le |z||1-\overline{f(0)}f(z)| \\ &\Rightarrow ||f(z)|-|f(0)|| \le |f(z)-f(0)| \le |z||1-\overline{f(0)}f(z)| \le |z|+|z||f(0)||f(z)| \\ &\Rightarrow ||f(z)|-|f(0)|| \le |z|+|z||f(0)||f(z)| \\ &\Rightarrow -|z|-|z||f(0)||f(z)| \le |f(z)|-|f(0)| \le |z|+|z||f(0)||f(z)| \\ & \Rightarrow -|z|+|f(0)| \le|f(z)|[1+|z||f(0)|] \ \text{and} \space |f(z)|[1-|z||f(0)|] \le |z|+|f(0)| \\ & \Rightarrow \frac{|f(0)| - |z|}{1 + |f(0)||z|} \leq|f(z)| \leq\frac{|f(0)| + |z|}{1 - |f(0)||z|} \end{align} $$

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  • $\begingroup$ Sorry, I am still having trouble. You can apply the Schwarz lemma to $g$ to obtain $$|g(z)|\leq|z|$$ and I eventually get $$|f(0)| - |z||1 - \overline{f(0)}f(z)| \leq |f(z)|\leq |f(0)| + |z||1 - \overline{f(0)}f(z)|$$ No idea where to go from here. Unless I am missing something patently obvious this seems like a hard question for a timed exam ... $\endgroup$ – matty_k_walrus May 29 '16 at 19:13
  • $\begingroup$ Ok I will edit to finish the proof :). $\endgroup$ – Jennifer May 29 '16 at 19:33
  • $\begingroup$ @matty_k_walrus Done $\endgroup$ – Jennifer May 29 '16 at 19:50
  • $\begingroup$ Merci, c'est gentille de ta part ! Ça me fait un peu peur qu'on est censé deviner tout ça en 30 minutes ... $\endgroup$ – matty_k_walrus Jun 4 '16 at 22:14
  • $\begingroup$ Si tu n'as jamais vu l'exercice avant, ca me semble difficile de le faire seul - If you never had done this exercise in the past, it seems very hard to find a solution from your own $\endgroup$ – Jennifer Jun 5 '16 at 6:56

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