Contractions in convex metric spaces

Question

Let $M \subset \mathbb{R}^n$ be open, bounded and convex and equip $M$ with an unbounded metric $d$ that induces the Euclidean topology. Is there always a map $f : M \to M$ and two constants $C_1 > 0$ and $C_2 < 1$ such that $$C_1d(x,y) \leq d(f(x),f(y)) \leq C_2d(x,y) \quad$$ for all $x,y \in M$?

I tried fixing a point and contracting along straight lines, but I somehow fail to see why this yields such a map...

What if we take an unbounded (simply connected) Riemannian manifold $(M,g)$ with induced metric $d$ instead?

Edit: Since it turned out to be wrong: Is it still wrong if we require the map to be proper instead? I.e. is there always a contractive proper map?

Answer 1

The answer is negative. Consider a round $2$-sphere, but with a cusp smoothly glued at the antipode of a fixed point $x_0$. Assume a circular symmetry, and that the cusp has a radius decaying as $2^{-d}$ with the distance $d$ to $x_0$, where $r$ is some function to be decided later.

Assume a $C_2$-Lipschitz, $C_1$ co-Lipschitz map $f$ exists with $0<C_1<c_2<1$. Note that $f$ is a homeomorphism: it is obviously injective, and if it where not onto there would be some boundary point $p$ to its image. There would exist a sequence $(x_n)_n$ going out of any compact such that $f(x_n)\to p$. Then $(f(x_n))_n$ would be a bounded sequence, contradicting the co-Lipschitz property.

The cusp circular curves which are far enough ($d\gg 1$) must then be mapped to curves circling the cusp, but which stay $C_2$ times closer (up to an additive constant) from the pole $x_0$ (but still behind the equator). The image curve must then have length of the order of $2^{-C_2d}=(2^{-d})^C_2$, which is $2^{(1-C_2)d}$ times larger than the original curve. Taking $d$ large enough, this ratio can be made arbitrarily large, thus contradicting the Lipschitz property.

I guess that the hyperbolic plan should be a counter example too (a large ball must be mapped into a smaller ball, which has much smaller volume because of the exponential growth; but since the map is co-Lipschitz, it can only decrease volume by a factor).