Scaling behavior of Wasserstein distances

Question

Let $p>1$ and $\mu\neq \nu$ be two probability measures on $\Omega\subset \mathbb{R}^d$ a bounded set. For $\alpha \geq 0$, we let $$C_\alpha(\mu,\nu) = \inf_\sigma \frac{W_p(\mu+\sigma,\nu+\sigma)}{W_p(\mu,\nu)},$$ where $W_p$ is the $p$-Wasserstein distance and the infimum is taken over all nonnegative measures on $\Omega$ of mass $\alpha$. By using the compactness of such a set, it can easily be shown that $C_\alpha(\mu,\nu)>0$. Is it true that $C_\alpha=\inf_{\mu\neq \nu} C_\alpha(\mu,\nu) >0$ ? (here the infimum is taken over $\mu,\nu$ probability measures on $\Omega$). This result looks intuitive, but I could not find a proof of it in the litterature, nor could I prove it.

If $(\mu_n),(\nu_n)$ are sequences attaining the infimum, and if $C_\alpha=0$, then we may without loss of generality assume that $\mu_n$ and $\nu_n$ converge to the same limit measure $\mu$, and therefore I suspect that whether $C_\alpha=0$ or not is related to some fine properties of $W_p(\mu,\nu)$ for $\mu$, $\nu$ very close.

Note that this problem is related to Remark 2.4 in A description of transport cost for signed measures - Edoardo Mainini.

Answer 1

Actually, we have $C_\alpha =0$ for any $\alpha>0$. Indeed, let $\mu_t = (1-t)\delta_{x_0} + t\delta_0$ and $\nu_t = (1-t)\delta_{x_0} + t\delta_1$ for some $t\in (0,1)$ and $x_0$ far away enough from $0$ and $1$. Then $W_p^p(\mu_t,\nu_t)= t$, whereas if $\sigma = t\sum_{k=1}^{n} \delta_{k/(n+1)}$, then one can check that $W_p^p(\mu_t + \sigma,\nu_t + \sigma) \leq \frac{t }{(1+n)^{p-1}}$. By letting $t_n = \alpha/(n+1)$, we obtain that $C_\alpha(\mu_{t_n},\nu_{t_n})^p \leq (1+n)^{1-p}$. By letting $n$ goes to $\infty$, we see that $C_\alpha= \inf C_\alpha(\mu,\nu)=0$.