I am stuck on the following problem. I have a discrete distribution $\mu\_0$ (it is actually an empirical distribution). I have some $\mu\_i$ (again discrete, some empirical distribution). I have some bound on the Wasserstein distance $W\_2(\mu\_0, \mu\_i).$ I now want to consider a simple mixture of $\mu\_i,$ that is, $\nu=\sum\limits\_{i=1}^{m}\lambda\_i\mu\_i$ where $\sum\lambda\_i=1, \lambda\_i>0.$


My goal is to bound $W\_2^2(\mu\_0, \nu).$ I felt that it would be easy to get a bound on $W\_2^2(\mu\_0, \nu)$ in terms of $W\_2^2(\mu\_0, \mu\_i),$ but I am unable to prove anything. I want something like $$W\_2^2(\mu\_0, \nu)\le \sum \lambda\_i^2 W\_2^2(\mu\_0, \mu\_i).$$


This does not look terribly hard, but I am stuck. Can anyone please say if it is true or not? If anyone can give a simple demonstration of why this is true, it would be great.



$2$-Wasserstein distance between mixtures
$\newcommand\Ga{\Gamma}$ $\newcommand\ga{\gamma}$ $\newcommand\la{\lambda}$
Let $\Ga(\mu,\rho)$ denote the set of all measures with marginals $\mu$ and $\rho$. For each $i$, take any real $c\_i>W\_2(\mu\_0,\mu\_i)^2$, so that 
$$\int d(x,y)^2\ga\_i(dx\times dy)<c\_i$$ 
for some $\ga\_i\in\Ga(\mu\_0,\mu\_i)$. Let 
$$\ga:=\sum\_i\la\_i\ga\_i.$$
Then $\ga\in\Ga(\mu\_0,\nu)$ and hence 
$$W\_2(\mu\_0,\nu)^2\le\int d(x,y)^2\ga(dx\times dy)
=\sum\_i\la\_i \int d(x,y)^2\ga\_i(dx\times dy)<\sum\_i\la\_i c\_i.$$ 
Letting now $c\_i\downarrow W\_2(\mu\_0,\mu\_i)^2$ for each $i$ such that $W\_2(\mu\_0,\mu\_i)^2<\infty$, we get 
$$W\_2(\mu\_0,\nu)^2\le\sum\_i\la\_i W\_2(\mu\_0,\mu\_i)^2.$$ 




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The inequality you proposed, 
$$W\_2(\mu\_0,\nu)^2\le\sum\_{i=1}^k\la\_i^2 W\_2(\mu\_0,\mu\_i)^2,\tag{1}$$ 
cannot hold in general. Indeed, suppose that for some probability measure $\rho$ we have $0<W\_2(\mu\_0,\rho)<\infty$. Let $\mu\_i:=\rho$ and $\la\_i:=1/k$ for all $i=1,\dots,k$. Then $\nu=\rho$ and the left-hand side of (1) is a constant $>0$, whereas its right-hand side goes to $0$ as $k\to\infty$. 


