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.



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