# Is unit ball in 2-Wassestein metric weakly compact?

This might be a trivial question, but I am trying to prove equipment-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?

• Do you mean "contained in some compact subset of the space of probability measures", or do you mean "contained in some compact subset of the space of probability measures for which $\cal W_2$ is defined"? – user95282 Jan 8 '20 at 14:59

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.

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