Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$f:D\rightarrow D$ holomorphic with $f(0)=\frac{1}{2}$ and $f(1/2)=0$ where $D$ is closed unit disk. which of the following are correct?

1.$|f'(0)|\le 3/4$

2.$|f'(1/2)|\le 4/3$

3.both $1$ and $2$

4.$f(z)=z$ ,$z\in D$

Not able to guess which theorem I apply? Swartz lemma? well $4$ is wrong.

share|cite|improve this question
What's $x$ here? – Dylan Moreland Jun 13 '12 at 16:25
What is the point $x$? – copper.hat Jun 13 '12 at 16:25
Dear sir, thats my question also, I am copying from an exam paper.It was $x$ there. – Un Chien Andalou Jun 13 '12 at 16:26
up vote 4 down vote accepted

I strongly suspect there is a misprint in the exam paper, and that the holomorphic map is supposed to satisfy $f(0)={1\over 2}$ and $f({1\over 2})=0$, as everything makes a lot more sense then.

You cannot use the Schwarz lemma, since you do not have $f(0)=0$. But there is a generalization of this result called the Schwarz-Pick theorem, which is valid for any holomorphic map from $D$ to $D$. The Schwarz-Pick theorem supplies you with the inequality ${|f'(z)|\over 1-|f(z)|^2}\leq {1\over 1-|z|^2}$, and when you use this on the two points in the problem you get that (3) is true.

In fact, even more can be said about $f(z)$! Since the two points $0$ and $1\over 2$ are interchanged, the hyperbolic distance between them is preserved, and $f(z)$ has to be a Möbius transformation. A little computation shows that the only possible solution is $f(z)={2z-1 \over z-2}$, and that the inequalities on $|f'(0)|$ and $|f'({1\over 2})|$ are in fact equalities.

It is quite easy to get the Schwarz-Pick theorem from the Schwarz lemma. If the point you are interested in is mapped to some other point, you just compose your function with suitable Möbius transformations to get a map which sends $0$ to $0$, and use the Schwarz lemma on this composition. Unraveling the result gives you the Schwarz-Pick theorem.

One nice way to state the Schwarz-Pick theorem is that no holomorphic map from $D$ to $D$ can increase hyperbolic distance between any two points. You can get to a lot of very beautiful mathematics if you study this topic.

share|cite|improve this answer
" Since the two points 0 and 12 are interchanged, the hyperbolic distance between them is preserved, and f(z) has to be a Möbius transformation." Can you please elaborate more. How did you get to the conclusion that it has to be mobius? – UserB95 Mar 11 '15 at 11:07

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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