# Barycenter Map on Wasserstein Space

Let $$(X,d)$$ be a complete separable metric space, $$P_1(X,d)$$ be the set of Radon probability measures on $$X$$ satisfying $$P_1(X,d)\triangleq \left\{ \nu:\,(\exists x_0\in X)\, \int_{x\in X} d(x,x_0)d\nu(x) <\infty \right\},$$ which will be equipped with the 1-Wasserstein metric $$d_{W_1}$$.

In Section 3.1 of this text, it is remarked that the existence of a $$1$$-Lipschitz map $$\beta:P_1(X,d)\rightarrow (X,d)$$ satisfying $$\beta(\delta_x)=x,$$ implies that $$(X,d)$$ is a Busemann space; hence does not exist in general.

Question: If the requirement the $$\beta$$ is $$1$$-Lipschitz is relaxed to the existence of a lsc function $$\rho:(0,\infty)\rightarrow (0,\infty)$$ satisfying $$\lim\limits_{x \downarrow \infty} \rho(0)=0$$ and $$\rho(d_{W_1}(\beta(x),\beta(y)))\leq d(x,y),$$ then does this still imply the existence of geodesics on $$(X,d)$$. In particular, do such maps exist in general for some such $$\rho$$? For example, $$\rho$$ may be $$\rho(x)\triangleq L x^{\alpha},$$ for some $$\alpha \in (0,1)$$ and $$K>0$$.

## 1 Answer

No. It does not imply the existence of geodesics. Fix $$0 < a < 1$$ and consider $$a$$-snowflake of $$[0,1]$$. I.e. a metric on $$[0,1]$$ given by $$d(x,y) = |x-y|^a,$$ the resulting metric space is denoted by $$[0,1]^a$$. Note that this metric is greater then the original one and does not allow curves of finite length.

Take $$\beta$$ to be a standard barycenter map for measures on $$\mathbb{R}$$. Now for two probability measures $$\mu$$ and $$\nu$$ on $$[0,1]$$ we have $$d(\beta(\mu),\beta(\nu)) = |\beta(\mu) - \beta(\nu)|^a \le (d_{W_1([0,1])}(\mu,\nu))^a \le (d_{W_1([0,1]^a)}(\mu,\nu))^a.$$