Define the Wasserstein metric for two probability measures $\mu$ and $\nu$ as follows: $$ d_W(\mu,\nu)=sup_h\Big[\Big|\int h(x)\mu (x)-\int h(x)\nu (x): h(\cdot) \mathrm{\;is\;}1\mathrm{-Lipschitz\;continuous}\Big|\Big] .$$
Suppose $g(x)$ is $\epsilon$-Lipschitz continuous, do we have $$\Big|\int g(x)\mu (x)-\int g(x)\nu (x)\Big|\leq \epsilon \cdot d_W(\mu,\nu)$$
Any hint?
For a $\epsilon$-Lipshitz continuous $g$, we have $|g(x)-g(y)|\leq \epsilon|x-y|$ so if you define $g'=g/\epsilon$ then $g'$ is $1-$Lipshitz continuous.
Therefore, $|\int g'd\mu-\int g'd\nu|\leq \sup_{h:\text{h is 1-Lipshitz}}|\int hd\mu-hd\nu|=d_W(\mu,\nu)$
Therefore, $|\int gd\mu-\int gd\nu|\leq \epsilon d_W(\mu,\nu)$
hint: consider new function $g'(x) = g(x)/\epsilon$, which is $1$-Lipschitz continuous.