Let $(X, \| \cdot \|)$ be a reflexive Banach space and $\mathbb{P}_n$, $\mathbb{P}_r$ be measures on $X$. Let the support of $\mathbb{P}_r$, $M := \text{supp}(\mathbb{P}_r)$ be a weakly compact set and $$P_M \colon D \to M, \qquad x \mapsto \text{argmin}_{y \in M} \| x - y \|$$ be the projection onto $M$ and $D \subset X$ the set of points for which the projection is unique (and thus the map $P_M$ is well-defined). Assume further that $\mathbb{P}_n(D) = 1$ and $(P_M)_{\#} \mathbb{P}_n = \mathbb{P}_r$.
(As $M$ is weakly compact, the projection always exists, but it need not be unique, as $M$ need not be convex.)
The Wasserstein-1 distance for measures is $$ W_1(\mathbb{P}_r, \mathbb{P}_n) = \sup_{f \in 1\text{-Lip}} \int_{X} f(x) \; \text{d}(\mathbb{P}_n - \mathbb{P}_r)(x), $$ where $1$-Lip denotes the set of 1-Lipschitz-continuous functions $f \colon X \to \mathbb{R}$.
Remark 2 in the paper Adversarial Regularizers in Inverse Problems by Lunz et al. states that the maximiser $f$ in the above formula is not unique: it can be changed to an arbitrary 1-Lipschitz function outside of $\text{conv}(\text{supp}(\mathbb{P}_r) \cap \text{supp}(\mathbb{P}_n))$.
But the statement is wrong (see below) but does somebody know what could be meant instead? If we replace the $\cap$ by a $\cup$ do we even need the convex hull anymore?
Counterexample (thanks to @MaoWao)
But taking $X = \mathbb{R}$ and $\mathbb{P}_r := \delta_0$ we have $M = \text{supp}(\mathbb{P}_r) = \{ 0 \}$, which is weakly compact and the projection is $$ P_M \colon \mathbb{R} \to \{ 0 \}, \qquad x \mapsto 0. $$ Choosing $\mathbb{P}_n := \delta_1$, we have $\text{supp}(\mathbb{P}_n) \cap \text{supp}(\mathbb{P}_r) = \emptyset$. Furthermore, for any measurable set $A \subset \mathbb{R}$ we have $$ \mathbb{P}_n(P_M^{-1}(A)) = \begin{cases} \mathbb{P}_n(\emptyset) = 0, & \text{if } 0 \notin A, \\ \mathbb{P}_n(\mathbb{R}) = 1,& \text{if } 0 \in A. \end{cases} = \mathbb{P}_r(A), $$ so $(P_M)_{\#} \mathbb{P}_n = \mathbb{P}_r$. But $$ W_1(\mathbb{P}_r, \mathbb{P}_n) = \sup_{f \in 1\text{-Lip}} f(1) - f(0) = 1 $$ for e.g. $f(x) = x$. But if we are allowed to change $f$ to on the complement of $\emptyset$, that is, anywhere, (such that it remains 1-Lipschitz) we can instead consider $\tilde{f}(x) = - x$, for which the objective function takes the value $- 1 \ne 1$.
