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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 $K>0$.

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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 $$d(\beta(\mu),\beta(\nu)) = |\beta(\mu) - \beta(\nu)|^a \le (d_{W_1([0,1])}(\mu,\nu))^a \le (d_{W_1([0,1]^a)}(\mu,\nu))^a.$$

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  • $\begingroup$ Perfect, so in principle any space (not only NPC spaces) could potentially admit such a map $\beta$? $\endgroup$ – MrMMS Jun 26 at 15:40

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