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](https://link.springer.com/article/10.1007/s10958-012-0718-2).



Scaling behavior of Wasserstein distances
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 $C\_\alpha= \inf C\_\alpha(\mu,\nu)=0$.


