Wasserstein distance in R^d from one dimensional marginals

This question occurred to me while I was reading Klartag's papers on central limit theorems for convex bodies.

Given probability measures $\mu$, $\nu$ on (the Borel $\sigma$-field of) $R^d$ with finite first moments, their Wasserstein distance is given by: $$W_{R^d}(\mu,\nu) = \sup \mbox{ of }\int_{R^d}f d\mu-\int_{R^d}f d\nu\mbox{ over all 1-Lipschitz }f:R^d\to R.$$

(NB: there was an error in this formula -- an inf in the place of the sup.)

Given a vector $v\in R^d$, let $\mu_v$ be the distribution of $X.v$, where $X$ has distribution $\mu$. Define $\nu_v$ analogously. Note that $\mu_v$ and $\nu_v$ are distributions over $R$.

Question: is there a constant $C_d>0$ depending on $d$ only such that: $$W_{R^d}(\mu,\nu)\leq C_d\sup_{v\in R^d, |v|=1}W_{R}(\mu_v,\nu_v)?$$ If so, how does $C_d$ grow with $d$?

An illustrative example: Assume $Z$ is uniform over a $D-1$ dimensional sphere $S^{D-1}$ in $R^D$. Any one-dimensional marginal of $\sqrt{D-1}Z$ is approximatelly Gaussian. Now let $\mu$ be the law of the first $d$ coordinates of $\sqrt{D-1}Z$, and $\nu$ be the standard Gaussian distribution on $R^d$. Can one deduce from the previous statement alone that $\mu$ and $\nu$ are close?

Another example: Let $Z$ be a random vector in $R^D$ with mean $0$ and covariance matrix $I_D$, $D\gg 1$. Old results of Sudakov (quoted here) show that "most" one-dimensional marginals of $Z$ are close to $|Z|N$ where $N$ is standard normal and independent from $Z$. A positive answer to the above question would lead to typical results for $d$-dimensional projections of $Z$.

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I don't have time to think about your question at the moment, but you may be interested in the results for typical $d$-dimensional projections here: arxiv.org/abs/0912.2044 –  Mark Meckes Sep 4 '10 at 11:40
There is an error in your definition of the Wasserstein metric: you have to replace inf by sup. –  Florian Oct 1 '10 at 8:51
Thanks! I guess I had the dual definition in mind when writing this. –  Roberto Imbuzeiro Oliveira Oct 1 '10 at 10:04
That a probability measure $\mu$ on $\mathbb{R}^d$ is entirely defined by its projections on lines is more or less Radon's theorem. What you want is a quantitative version of this, you could therefore try to use the inversion formula for Radon transform: it gives you, at least when $\mu$ has a density, an expression of $\mu$ in terms of the $\mu_v$. However it seems difficult to relate this formula with the Wasserstein distance. –  Benoît Kloeckner Oct 1 '10 at 14:04

There is a result which contains an answer to your question in a somewhat different form. Instead of the transportation metric it uses another metric which metrizes the weak topology in the space of measures on $\mathbb R^d$: $$\lambda(\mu,\nu) \le \delta \iff \exists\; T\ge 1/\delta : \langle \exp(i(t,\cdot)),\mu-\nu\rangle \le \delta \quad\forall\; |t|\le T \;,$$ which might still be OK for your purposes. This article by Klebanov and Rachev actually contains a stronger result (Theorem 4): it gives an explicit upper estimate for $\lambda(\mu,\nu)$ in terms of the maximal distance $\lambda(\mu_v,\nu_v)$ between the projections of $\mu$ and $\nu$ onto a finite (growing) number of directions $v$ in $\mathbb R^d$.