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

Each analytic function mapping the right half complex plane into itself must satisfy $$ \left|\frac{f(z)-f(1)}{f(z)+f(1)}\right| \leqslant \left|\frac {z-1}{z+1} \right|$$ for $\text{Re}\; z > 0.$

I have a hunch that this is an application of Schwarz's Lemma. I don't know how to proceed though. Thanks in advance.

share|cite|improve this question
Great question! I'm asking a similar question here, which might be a more general result:… – EthanAlvaree Apr 22 '15 at 12:47
up vote 1 down vote accepted

The map $h$ from $\{z,\Re z>0\}$ to the open unit disk $D$ given by $h(z)=\frac{z-1}{z+1}$ is one-to-one, then use Schwarz lemma with $g(z):=\dfrac{f\left(\frac{1+z}{1-z}\right)-f(1)}{f\left(\frac{1+z}{1-z}\right)+f(1)}$.

share|cite|improve this answer
this is actually a past qual question. I am thinking $f$o$h^{-1}$ is actually a mapping from unit disk to right half plane, but don't we need the mapping from unit disk to unit disk in Swartz's Lemma? I am confused. Could you please elaborate? – Deepak Dec 29 '12 at 1:11
,But, I still have difficulty seeing ${h\circ}f\circ h^{-1}(0)=0$. I can see after that Swartz lemma gives the result. – Deepak Dec 30 '12 at 3:43
But, in what ground can we assume that? @Davide Giraudo – Deepak Dec 30 '12 at 17:23
I don't know if you can access this because this is the departmental website, you can find it here if the link works – Deepak Dec 30 '12 at 19:13
I'm asking a similar question here:… – EthanAlvaree Apr 22 '15 at 12:48

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.