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

The Schwarz-Pick lemma states that if $D$ denotes the unit disk in the complex plane, and $f: D\rightarrow D$ is a holomorphic function, then it is a contraction with respect to the Poincare metric (which we shall denote as $\rho$) on the disk. A natural question to ask (for me at least) is are all functions $f: D\rightarrow D$ which are contractions with respect to $\rho$ holomorphic? This cannot be true exactly as stated since, if we denote the conjugation map by $\tau: z\mapsto \bar{z}$, we have, for an arbitrary holomorphic function $f$: \begin{align} \rho(f\circ\tau(z_1), f\circ\tau(z_2)) & \leq \rho(\tau(z_1),\tau(z_2))\\ &= \rho(z_1,z_2) \end{align} Since $\tau$ is an isometry for $\rho$; but $f\circ\tau$ is anti-analytic. So, my question is:

Is every function which is a contraction with respect to $\rho$ either analytic or anti-analytic? This seems too good to be true, so could anyone provide a simple counter-example?

Edit: contraction is the wrong word. It should be replaced with non-expansive or 1 Lispchitz as in 5P.M.'s answer.

share|cite|improve this question
up vote 1 down vote accepted

Let's say that $f$ is real differentiable. Its derivative matrix $Df$ has the operator norm $\|Df\|$. In order to be 1-Lipschitz with respect to $\rho$, it is necessary and sufficient that $$\|Df(z)\|\,\rho(f(z))\le \rho(z) \ \text{ for all } z\in D\tag{1}$$ or, explicitly, $$\|Df(z)\| \le \frac{1-|f(z)|^2}{1-|z|^2} \text{ for all } z\in D\tag{2}$$ Any Euclidean contraction that fixes the origin $0$ satisfies (2), because $\|Df(z)\|\le 1$ and $|f(z)|\le |z|$. For example, $f(z)=\operatorname{Re}z$ or $f(z)=z\,\min(1, \frac{1}{2|z|})$.

The above maps are not surjective, though. A surjective example can be obtained, for example, by moving every point $z$ toward the center by the same hyperbolic distance. Formally, $z$ goes to $\phi(|z|)\frac{z}{|z|}$ where $\phi:[0,\infty)\to [0,\infty)$ is defined by $$\log \frac{1+\phi(t)}{1-\phi(t)} = \max\left (0, \log \frac{1+t}{1-t}-1\right )$$

share|cite|improve this answer
Hi @5PM, thanks so much for your answer. I just wanted to clarify one thing; in my post I was denoting by $\rho$ the hyperbolic metric in the sense of metric spaces. Am I correct in saying that you mean $\rho(z) = \frac{dx^2+dy^2}{(1-|z|^2)^2}$ and that (1) comes from comparing $\rho(z)$ with $(f^{*}\rho)(z)$ (i.e. the pullback of $\rho$ with respect to $f$)? – Daniel Mckenzie Feb 7 '13 at 14:11
@DanielMckenzie It's the same thing. If the tangent vectors are shorter, then the curves are shorter, and two-point distance is the infimum of length of curves. You can see directly from (1) that for any smooth curve $\gamma$ the hyperbolic length of $f\circ \gamma$ is not greater than the hyperbolic length of $\gamma$. – user53153 Feb 7 '13 at 14:22
That makes sense. Great, thanks – Daniel Mckenzie Feb 7 '13 at 21:25

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.