The converse to Schwarz Pick lemma?

Question

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.

Answer 1

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 $$\log \frac{1+\phi(t)}{1-\phi(t)} = \max\left (0, \log \frac{1+t}{1-t}-1\right )$$