# How to control Wasserstein distance in terms of characteristic function

Let $$\mathcal P(\Omega)$$ be the set of probability measures supported on some compact subset $$\Omega\subset\mathbb R^d$$. For $$\mu\in\mathcal P(\Omega)$$, denote by $$F_{\mu}$$ its characteristic function, i.e.

$$F_{\mu}(x)~:=~\int_{\mathbb R^d}e^{i\langle x,y\rangle}\mu(dy)~=~\int_{\Omega}e^{i\langle x,y\rangle}\mu(dy),\quad \forall x\in\mathbb R^d.$$

Let $$W(\cdot,\cdot)$$ be the Wasserstein distance of order $$1$$. My question is whether there exists a continuous function $$c:\mathbb R_+\to\mathbb R_+$$ with $$c(0)=0$$ s.t. it holds for all $$\mu$$, $$\nu\in\mathcal P(\Omega)$$:

$$W(\mu,\nu)~\le~c\big(\|F_{\mu}-F_{\nu}\|\big),\quad \mbox{with } \|F_{\mu}-F_{\nu}\|:=\max_{x\in\mathbb R^d}|F_{\mu}(x)-F_{\nu}(x)|.$$

Personal thought: We use Kantorovich's duality, i.e.

$$W(\mu,\nu)~=~ \sup_{f\in \rm{Lip}_1(\mathbb R^d)} \int f d\mu -\int fd\nu ~=~\sup_{f\in \rm{Lip}_1(\Omega)} \int f d\mu -\int fd\nu,$$

where $$\rm{Lip}_1(\mathbb R^d)$$ and $$\rm{Lip}_1(\Omega)$$ denote respectively the collection of $$1-$$Lipschitz functions on $$\mathbb R^d$$ and $$\Omega$$. It remains to write the integral $$\int f d\mu$$ in terms of $$F_{\mu}$$. Denote by $$\hat f$$ the Fourier transform of $$f$$, i.e.

$$\hat{f}(x)~:=~ \int_{\mathbb R^d}e^{i\langle x,y\rangle}f(y)dy.$$

We notice $$\hat{f}$$ is well defined for each $$f\in \rm{Lip}_1(\Omega)$$. Then we may apply Parseval's equality under "suitable conditions"

$$\int_{\mathbb R^d}f(x)\mu(dx)~=~\int_{\mathbb R^d}\hat{f}(x) F_{\mu}(x)dx.~~~~~~~~~~~~ (\ast)$$

I am not familiar with the regularity analysis of $$\hat f$$. If someone knows how to proceed based on $$(\ast)$$, I am happy to know.

For measures on the real line, There is no such function $$c$$. Given $$n$$, let $$\mu_n$$ assign mass $$1/n$$ to each of $$-n,n$$ and remaining mass to zero. Let $$\nu_n$$ be Dirac measure at zero. As $$n$$ grows, the Fourier transforms get uniformly close, with sup norm distance of order $$1/n$$, yet the Wasserstein distance remains 2.

For measures on a fixed compact set there is such a function $$c$$ since both metrics metrize the weak * topology and the space of probability measures on a compact set is itself weak * compact.

• On the positive side, for distributions with a fixed compact support, and for any epsilon>0, there is a function c such that the inequality holds up to epsilon. Maybe the epsilon slack can be removed, I don't know. Oct 13, 2019 at 14:03
• Thanks for the kind reply. Actually, for the case of unbounded support, this example is not very surprising. As in your example, $\mu_n$ converges weakly to $\nu\equiv \nu_n$, but does not converge to $\nu$ under Wasserstein metric.
– user128095
Oct 13, 2019 at 14:11
• If there is a fixed compact set supporting all measures then the answer is positive by compactness. Oct 13, 2019 at 15:30
• Thanks for the reply. Do you think there is a constructive proof?
– user128095
Oct 13, 2019 at 17:02
• I am sure there is a constructive proof but it would require more work. Is the compactness argument I sketched clear? If not I can add details. Oct 13, 2019 at 18:05