In my setting, $\mu$ and $\nu$ are probability measures on $\mathbb{R}^{2}$ with compact support. For any function $f\in{C^{2}\_{b}(\mathbb{R}^{2})}$, I have the estimate:
$$
|\mathbb{E}\_{\mu}(f)-\mathbb{E}\_{\nu}(f)|\leq{a\|Df\|\_{L^{\infty}}+b\|D^{2}f\|\_{L^{\infty}}}
$$
for some coefficients $a,b>0$. Does the estimate above imply an estimate for the Wasserstein distance of the form:
$$
W\_{2}(\mu,\nu)\leq{\eta(a,b)}
$$
where $\eta$ is a function of $a$ and $b$ such that $\eta(a,b)\rightarrow{0}$ as $(a,b)\rightarrow{(0,0)}$?



Getting Wasserstein closeness from a derivative estimate
Assuming the support of both measures is fixed, it is enough to bound $W\_1(\mu,\nu)$. That is, we want to bound $\vert \int f d\mu -d\nu \vert$ for any $1$-Lipschitz function $f$.


Take a triangulation of the plane with equilateral triangles of size $\sqrt b$ and let $f\_1$ be the linear interpolation of $f$. We have $\vert f - f\_1 \vert = O(\sqrt b)$. Also the derivative of $f\_1$ is $O(1)$. Take $f\_2$ to be the convolution of $f\_1$ with a Gaussian of width $\sqrt b$. It is not too hard to see that the second derivative of $f\_2$ is $O(1/\sqrt b)$, and its first derivative is $O(1)$, implying that $\vert \int f\_2 d\mu -d\nu \vert = O(a+\sqrt b)$. Now $\vert f\_2-f\_1\vert = O(\sqrt b)$ because the derivative of $f\_1$ is $O(1)$. Hence $\vert f-f\_2\vert $ is also $O(\sqrt b)$, meaning that $\vert \int f d\mu -d\nu \vert = O(a+\sqrt b) + O(\sqrt b)$.


