# Compactness of the unit ball in the space of Radon measures w.r.t. the Kantorovich-Rubinstein norm

This question was posted previously but has not attracted any responses so I am repharising it in a slightly different language hoping to reach a wider community

Let $$(X,d)$$ be a pointed metric space with base point $$x_0$$. Denote by $$Lip_0(X)$$ the set of all Lipschitz functions on $$X$$ vanishing at $$x_0$$. The norm in $$Lip_0$$ is defined as $$\|f\|_{Lip_0} := Lip(f)$$, where $$Lip(f)$$ denotes the Lipschitz constant.

Denote by $$\mathcal M_0^1(X)$$ the space of balanced Radon measures on $$X$$ with a finite first moment, i.e. such that $$\mu(X)=0$$ and $$\int_X d(x,x_0) d|\mu|(x) < +\infty.$$ The total variation norm on $$\mathcal M_0^1(X)$$ is defined by $$\|\mu\|_{\mathcal M} := |\mu|(X)$$.

Another norm on $$\mathcal M_0^1(X)$$ can be defined using pairings with functions from $$Lip_0(X)$$ $$\|\mu\|_{KR} := \sup\left\{ \int f \, d\mu \colon \|f\|_{Lip_0} \leq 1 \right\},$$ where KR stands for Kantorovich-Rubinstein because of a connection to optimal transport.

If $$X$$ is compact then the closed unit ball $$\{\mu \in \mathcal M_0(X) \colon \|\mu\|_{\mathcal M} \leq 1\}$$ is compact w.r.t. the Kantorovich-Rubinstein norm, see Theorem VIII.4.3 in Kantorovich and Akilov. Functional Analysis.

Does a similar result hold in more general cases? It is likely to require a uniform bound on the first moments, so more presicely:

Question. Let $$M := \left\{\mu \in \mathcal M_0^1(X) \colon \|\mu\|_{\mathcal M} \leq 1, \, \int_X d(x,x_0) d|\mu|(x) \leq 1 \right\}.$$ Is $$M$$ compact w.r.t. the Kantorovich-Rubinstein norm?

I am interested in the following three scenarios:

(i) $$X$$ is locally compact;
(ii) $$X$$ is the unit ball in a Banach space;
(iii) $$X$$ is a Banach space.

Any help will be much appreciated.