Getting Wasserstein closeness from a derivative estimate

Question

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 $(a,b)\rightarrow{(0,0)}$?

Answer 1

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 $\vert \int f d\mu -d\nu \vert = O(a+\sqrt b) + O(\sqrt b)$.

Answer 2

$\newcommand\de\delta\newcommand\R{\mathbb R}$No. (This is an answer assuming, according to the letter of the original post, that the support sets of the measures $\mu$ and $\nu$ are compact but not necessarily fixed. I did not notice the latter comment by the OP before posting this answer).

Indeed, let $\mu:=\de_0$, the Dirac measure supported on $\{0\}$. Let $X$ be any random vector in $\R^2$ such that $a:=E|X|<\infty$ but $E|X|^2=\infty$, where $|\cdot|$ is the Euclidean norm. For any natural $n$, let $\nu_n$ be the distribution of the random vector $X_n:=X\,1(|X|<n)$.

Then for any $f\in C^{2}_{b}(\R^{2})$, $g:=f-f(0)$, and all $x\in\R^2$ we have $|g(x)|\le \|Df\|_{L^\infty}\,|x|$, and hence $$|E_\mu(f)-E_{\nu_n}(f)|=|E_\mu(g)-E_{\nu_n}(g)|=|E_{\nu_n}(g)| \le E_{\nu_n}(|g|) \\ \le\int_{\R^2}\|Df\|_{L^\infty}\,|x|\,\nu_n(dx) \le a\|Df\|_{L^\infty}\le a\|Df\|_{L^\infty}+b\|D^2f\|_{L^\infty}$$ for any $b>0$.

However, $$W_2(\mu,\nu_n)=\int_{\R^2}|x|^2\,\nu_n(dx)=E|X_n|^2\to E|X|^2=\infty.$$ $\Box$