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 $F$?