5
$\begingroup$

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 $\mu$, $\nu\in\mathcal P(\Omega)$:

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

Any answer, comments and references are highly appreciated!

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.

$\endgroup$

1 Answer 1

2
$\begingroup$

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.

$\endgroup$
8
  • $\begingroup$ 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. $\endgroup$
    – alesia
    Oct 13, 2019 at 14:03
  • $\begingroup$ 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. $\endgroup$
    – user128095
    Oct 13, 2019 at 14:11
  • $\begingroup$ If there is a fixed compact set supporting all measures then the answer is positive by compactness. $\endgroup$ Oct 13, 2019 at 15:30
  • $\begingroup$ Thanks for the reply. Do you think there is a constructive proof? $\endgroup$
    – user128095
    Oct 13, 2019 at 17:02
  • $\begingroup$ 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. $\endgroup$ Oct 13, 2019 at 18:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy