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](https://en.wikipedia.org/wiki/Wasserstein_metric#Definition) $d\_{W\_1}$. 


In [Section 3.1 of this text](http://www.numdam.org/article/RSMUP_2010__124__91_0.pdf), 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;](https://en.wikipedia.org/wiki/Busemann_G-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 $K>0$. 



Barycenter Map on Wasserstein Space
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 
$$d(\beta(\mu),\beta(\nu)) = |\beta(\mu) - \beta(\nu)|^a \le (d\_{W\_1([0,1])}(\mu,\nu))^a \le (d\_{W\_1([0,1]^a)}(\mu,\nu))^a.$$


