Is unit ball in 2-Wassestein metric weakly compact?

Question

This might be a trivial question, but I am trying to prove equi-coerciveness of some family of functions on the space of Probability measures on some space. I could reduce the problem to showing that $$\{\nu:\mathcal{W}_2^2(\mu, \nu)\le t\}$$ is compact (or is at least contained in some compact subset of $\mathcal{P}(\mathbb{R})$.)

The space of probability measures on $\mathbb{R}$ is equipped with weak topology. Now, I know from Banach Alaglou theorem unit ball in weak topology is compact. But here I am taking the ball in $\mathcal{W}_2$ metric. Can anyone tell me if this is correct or I am just wasting time?

Answer 1

Yes, it is true. It follows from Prokhorov's theorem that in order to prove (pre-)compactness, it suffices to prove tightness. However, if we define $K$ to be the compact set such that $\mu(\mathbb{R}\setminus K)<\varepsilon$, and $K_T:=\{x\in \mathbb{R}:\mathrm{dist}(K,x)\leq T\}$, then $\nu(\mathbb{R}\setminus K_T)>2\varepsilon$ implies that $\mathcal{W}_2^2(\mu,\nu)>\varepsilon T^2$ (since you have to move a mass $\geq\varepsilon$ over a distance $\geq T$), and so it is impossible for $\nu$ in your set provided that $\varepsilon T^2>t$. This proves tightness.

Answer 2

Since we care about probability measures, a sufficient condition for compactness is that the supports of the respective probability measures are compact. Here is the reasoning:

1. Let $X$ be a compact subset of $\mathbb{R}^d$. Denote by $C_0^\ast(\mathbb{R}^d)$ the space of all continuous functions on $\mathbb{R}^d$ that vanish at infinity. Then the Banach-Alaoglu theorem implies that the unit ball of $C^\ast_0(\mathbb{R}^d)$ is compact in the weak$^\ast$-topology. This implies that the set of probability measures is compact in the weak$^\ast$ topology.

2. In general, the weak$^\ast$ and the weak topology do not coincide for measures on $\mathbb{R}^d$, but they do for measures on the compact $X$, as then $C_0^\ast(X)=C_b^\ast(X)$, where $C_b(X)$ is the space of all bounded and continuous functions on $X$. This means that the set $\mathcal{M}(X)$ of all probability measures on $X$ equipped with the total variation norm is weak$^*$- and therefore weakly compact.

3. On compact sets the $p$-Wasserstein distance (for $p\in[1,\infty)$) metrizes weak convergence (Theorem 5.10 in Santambrogio 2015), from which it follows that the set you defined is compact since it is closed.

In my opinion, a great reference for these sorts of questions is:

Santambrogio, F (2015): "Optimal Transport for Applied Mathematicians", Birkhäuser

Answer 3

It is not true since $R^d$ is not compact. See page 34 of Ambrosio and Gigli's User's Guide (https://www.math.umd.edu/~yanir/OT/AmbrosioGigliDec2011.pdf). To quote:

...if X is unbounded, then P2(X) is not locally compact. Actually, for any measure $\mu\in P_2(X)$ and any $r > 0$, the closed ball of radius $r$ around µ is not compact. To see this, fix $\bar{x}$ ∈ X and find a sequence $x_n$ ⊂ X such that $d(x_n, \bar{x})$ → ∞. Now define the measures $\mu_n := (1 − \epsilon_n)\mu + \epsilon_n \delta_{x_n}$, where $\epsilon_n$ is chosen such that $\epsilon_n d^2(\bar{x},x_n)=r^2.$ To bound from above $W_2^2(\mu,\mu_n)$, leave fixed $(1-\epsilon_n)\mu$, move $\epsilon_n \mu$ to $\overline{x}$ and then move $\epsilon \delta_{\overline{x}}$ into $\epsilon_n \delta_{x_n},$ this gives $$W_2^2(\mu,\mu_n)\leq \epsilon_n \left(\int d^2(x,\overline{x})d\mu(x) + d^2(x_n,\bar{x})\right),$$

so that $\limsup W_2(\mu, \mu_n) ≤ r$. Conclude observing that $$\liminf_{n\rightarrow\infty}\int d^2(x,\bar{x})d\mu_n =\liminf_{n\rightarrow\infty}(1-\epsilon_n)\int d^2(x,\bar{x})d\mu+\epsilon_n d^2(x_n,\bar{x})=\int d^2(x,\bar{x})d\mu+r^2,$$ thus the second moments do not converge. Since clearly $\mu_n$ weakly converges to $\mu$, we proved that there is no local compactness.