Let $f$ be a holomorphic function on $|z|<1$ with $|f(z)|<1$ for all $|z|<1$.

(1) Find necessary and sufficient conditions for equality of $$\frac{|f'(z)|}{1-|f(z)|^2} \leq \frac{1}{1-|z|^2}$$ for all $|z| < 1.$

(2) If $f(\frac{1}{2}) = \frac{1}{3}$, find a sharp upper bound for $|f'(\frac{1}{2})|.$

I know that the inequality is Schwarz-Pick Lemma, and I think that it has something to do with Mobius Transformation (Linear fractional transformation) which I supposed that it will be introduced later in my textbook (Complex Analysis in Spirit of Lipman Bers, second edition). This problem is in chapter 6, and the Mobius should be introduced in chapter 8. So, actually, I do not know much about Mobius Transformation. So I do not have any clear idea for the condition concerning equality of the inequality above. Also, I do not know what is a sharp upper bound. I found the definition on http://en.wikipedia.org/wiki/Upper_and_lower_bounds ,but I do not fully understand what it means, and, according to the question, what I have to do with $|f'(\frac{1}{2})|$


Maybe it helps to know that the only inequality in the proof of the Schwarz-Pick lemma comes from Schwarz' lemma. So you should first understand equality there. Then look at the proof of the Schwarz-Pick lemma to find out what the condition for equality is there. All you need to know about Möbius transformations is that the map $\frac{z-z_0}{\overline{z_0}z-1}$ maps the unit disk into itself and the point $z_0$ into the origin.

By a sharp upper bound they mean some number $C>0$ such that $|f^\prime(1/2)|\le C$ for all $f$ as stated in the problem and such that for all $C^\prime<C$ there would exist some $f$ with $|f^\prime(1/2)|>C^\prime$. To do that, of course it suffices to find one $f$ such that $|f^\prime(1/2)|=C$.

What is meant is to plugin $z=1/2$ in the Schwarz-Pick Lemma and then use $(a)$ to prove that the resulting constant is sharp (find a counterexample for which equality holds).

  • $\begingroup$ Okay, I will try to understand the prove of Schwarz's Lemma. But, for the sharp upper bound, you mean that it is actually a upper bound of family of holomorphic function $f$ on unit disc satisfying $|f(z)|<1$ and $f(\frac{1}{2})$. So I precisely have to find a constant $M > 0$ fits for any functions, and $M$ somehow is similar to least upper bound of set, say ($\sup A$). So by Schwarz Pick, I should show that $M = \frac{1 - \frac{1}{9}}{1 - \frac{1}{2}}$ is roughly a supremum of family of functions. Is my understanding correct ? It sounds quite hard. $\endgroup$ – user117375 Mar 17 '15 at 22:22
  • $\begingroup$ You are correct. And all you have to do now to show that it is sharp is to exhibit some $f$ satisfying the constraints such that $|f^\prime(1/2)|=M$. $\endgroup$ – J.R. Mar 17 '15 at 22:24
  • $\begingroup$ So I need to find a holomorphic function $f$ on a disc satisfying $|f(z)| < 1$ on the disc and $|f'(1/2)| = \frac{16}{9}$. I will try construct such a function. But, if it is okay, could you give some hints ? I anyway suspect that the function have to be Mobius, which I am not familiar with. $\endgroup$ – user117375 Mar 17 '15 at 22:28
  • $\begingroup$ Yes, the function should be a Möbius transform. Try the one I gave you with suitable $z_0$.... $\endgroup$ – J.R. Mar 17 '15 at 22:30
  • $\begingroup$ Okay, I will try. Thank you very much for your help. $\endgroup$ – user117375 Mar 17 '15 at 22:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.