Bounded set in $(P_p(\mathbb{R}^d),W_p(\cdot,\cdot))$ is precompact?

Question

Does the Wasserstein space $(P_p(\mathbb{R}^d),W_p(\cdot,\cdot))$ posesses the property that any bounded set is precompact? Or at least, if $\mu_0\in P_p(\mathbb{R}^d)$ and given $r>0$ a real number, is the Wasserstein ball $B(\mu_0,r):=\{\mu\in P(\mathbb{R}^d) : W_p(\mu_0,\mu)<r\}$ a tight set?

Answer 1

No, the Wasserstein space does not have the property that bounded sets are precompact. A counterexample is given in Remark 2.8 here: https://www.math.umd.edu/~yanir/OT/AmbrosioGigliDec2011.pdf. The preceding theorem also gives a characterization of precompactness in $W_2$ ($W_p$ is very similar).

PS. I do not agree that $W_p (0, \mu)$ is a norm, at least in the obvious way. After all, given a constant $c\neq 0$, we would require that $\Vert c \mu \Vert_p = |c| \Vert \mu \Vert_p$. But what is $c \mu$ supposed to be? Usually in measure theory we would say that $c \mu$ is defined so that $ c\mu(A) = c (\mu(A))$ for every measurable set. But if $c\neq1$ then $c\mu$ is not even a probability measure, hence not even in $W_p$, so $W_p (0, c\mu)$ is ill-defined.

However, $W_p (\mathbb(R)^d)$ is a complete separable metric space which is in some sense infinite-dimensional. It is just not a Banach space. It is better to think of the space of probability measures as a surface inside the space of all signed measures (which is actually a Banach space), but with $W_p$ we have equipped that surface with an exotic metric that is not induced from the ambient space.