Effect of perturbing the atoms of a measure on the Wasserstein distance Let $(X,d)$ be a metric space, $x_1,\ldots,x_N\in X$ and $x_1',\ldots,x_N'\in X$ be atoms, and $G=\sum_{i=1}^Np_i\delta_{x_i}$, $G'=\sum_{i=1}^Np_i'\delta_{x_i}$, and $G''=\sum_{i=1}^Np_i'\delta_{x_i'}$ be mixing measures. In words: $G$ and $G'$ have the same atoms, but different weights. $G'$ and $G''$ have different atoms, but the same weights.
Assuming $G'\ne G''$ (i.e. there is at least one distinct atom), is it true that $W_p(G,G')\le W_p(G,G'')$? Here, $W_p$ is the usual $p$th Wasserstein distance between the measures $G$ and $G'$.
In other words, if two discrete measures have the same support, does "perturbing" the atoms in one of the measures always increase the Wasserstein distance? Or is it possible to move the atoms in one measure in such a way to decrease the Wasserstein distance?
 A: There is a nontrivial counterexample for $N=2$, $p=1$, and $X=\mathbb{R}$. Pick $x_1=-2$, $x_2=2$, $x'_1=-1$, $x'_2=1$ and $p_1=4/5$ and $p_1'=1/5$. Then $2.2=W_1(G,G'')<W_1(G,G')=2.4$. (I hope I did not mess up the calculation).
The intuition seems clear:
In the counterexample, you have to move $1/5$ of the total mass from $-2$ to $-1$ and $3/5$ of the total mass from $-2$ to $1$, while moving another $1/5$ of the total mass from $2$ to $1$. This is cheaper than moving $3/5$ of the mass all the way from $-2$ to $2$.
A: Without further constraints this is not true and easy to see if $X$ is Euclidean: Let $\Gamma$ be the support of an optimal coupling between $G$ and $G'$. If for fixed $y\in p_2(\Gamma)$ the set $\{x \in X ~|~(x,y)\in \Gamma \}$ lies in the interior of a half space whose boundary passes through $y$ then moving $y$ towards the interior of the half space decreases the optimal transport distance. For $N=2$ this works in any geodesic space, just let $x'_2$  be a point on a geodesic connecting $x_1$ and $x_2$.
