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Fix the closed unit disk (for simplicity) $\mathbb{D}^n\subset \mathbb{R}^n$. Given two probability measures $\mu,\nu$ on $\mathbb{D}^n$ we define the Kantorovich-Rubinstein distance between them to be $$ F(\mu,\nu)=\sup\{ \int f d\mu-\int fd\nu: \Vert f\Vert_L \leq 1\} $$ where here $$ \Vert f \Vert_L=\sup_{x\in \mathbb{D}^n} |f(x)|+\sup_{x,y\in \mathbb{D}^n, x\neq y} \frac{|f(x)-f(y)|}{|x-y|}. $$

It seems to be the case that $\mu_i \to \mu$ in the weak* topology if and only if $F(\mu_i, \mu)\to 0$.

My question, is what happens if one restricts the space of test functions used to define $F$. E.g., let $$ F_k(\mu,\nu)=\sup \{ \int f d\mu-\int f d\nu: \Vert f\Vert_k \leq 1\} $$ where $$ \Vert f\Vert_k =\sum_{i=0}^k \sup_{x\in \mathbb{D}} |D^i f(x)|. $$ is the $C^k$ norm.

Is it still true that $\mu_i\to \mu$ in the weak* topology if and only if $F_k(\mu_i, \mu)\to 0$ for some $k\geq 1$?

Clearly, for $k'\geq k\geq'$ one has $$ F_{k'}\leq F_k \leq F. $$ so one direction is trivial. My understanding is the reverse direction just uses some equicontinuity of families with bounded Lipschitz norm so should also work for the stronger norms, but am not sure if I am missing something.

If this is true, are there any subtle differences between $F_k$ and $F$?

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  • $\begingroup$ usually in Kantorovich -- Rubinstein distance you do not include supremum norm of $f$ to $\|f\|_L$ $\endgroup$ – Fedor Petrov Feb 22 at 16:54
  • $\begingroup$ @FedorPetrov Sure, but either definition gives (strongly) equivalent $F$ so is not really that big a deal. $\endgroup$ – RBega2 Feb 22 at 16:57
  • $\begingroup$ The span of functions with finite $k$-norm is still dense in $C(\mathbb{D}^n)$, thus the convergence of the values of measures against such functions implies weak star convergence, right? $\endgroup$ – Fedor Petrov Feb 22 at 18:02
  • $\begingroup$ @FedorPetrov Yes you are correct I confused myself. Density means $F_k$ convergence implies weak* convergence. The harder direction (for $F$) is the other way and this uses equicontinutity. However, as $F$ dominates $F_k$, one gets it for free for $F_k$ once one knows it for $F$. $\endgroup$ – RBega2 Feb 22 at 19:41

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