This is a test: I am recovering from a hand injury and I think that if I can work a Mathoverflow problem, then I am fit enough to resume work. Here is a non-constructive answer to the question. So, Roberto, let me know soon if I can go back to work.
I don't know of any reference where you can find the solution. I started thinking about the problem and found a simple solution (using standard results).
I read in another question about Wasserstein distance the suggestion of using a finite cover of the unit ball in $Lip(mathbb{R}^d) $ to reduce estimates of $ W_{mathbb{R}^d} $ to a max over a finite set. I see that idea working for Lipschitz functions on a bounded set, but I don't see how to apply that idea when working in the whole $mathbb{R}^d $.
Here is my solution.
Let
$quad Lip1(mathbb{R}^d) = {f:mathbb{R}^d longrightarrowmathbb{R}: forall x,y in mathbb{R} ^d text{ with } x neq y, frac{|f(x)-f(y)|}{|x-y|} leq 1) }$
be the set of 1-Lipschitz functions in $mathbb{R}^d$.
Let $mathcal{G}$ be the set of functions on $ mathbb{R}^d $ of the form:
$(1)quad f(x) = sum_{i in F} a_if_i(xcdot{v_i}) $
where
$quad F$ is a finite set,
$quad a_i in mathbb{R}, a_i geq 0, sum_{i in F} a_i = 1,$
$quad f_i in Lip1(mathbb{R}) ,$
$quad v_i in mathbb{R}^d , |v_i| = 1 text{ (Euclidean norm)}$
i.e.: $mathcal{G}$ is the convex hull of $Lip1(mathbb{R})$ functions composed with one dimensional projections. Clearly
$quad mathcal{G} subseteq overline{mathcal{G}} subseteq Lip1(mathbb{R}^d)$
where
$overline{mathcal{G}}$ is the closure of
$mathcal{G}$ in any topology in which $Lip1(mathbb{R}^d)$ is closed.
We will work with a weak* topology, namely the one in which the neighborhoods of 0 are generated by the sets
$(2)quad mathcal{N} = {f in Lip(mathbb{R}^d):
arrowvert int_{mathbb{R}^d} f(x) u(x) dx +
int_{mathbb{R}^d} nabla f(x) cdot{w(x)} dxarrowvert < epsilon } $
for some $ u in L^1(mathbb{R}^d, (1+|x|)dx)$ with $int u(x) dx = 0$,
some $ w in L^1(mathbb{R}^d)^d $, and some $ epsilon > 0 $.
Here are some known facts:
1) If $f in Lip(mathbb{R}^d)$, then $ f$ is differentiable a.e and $ sup{ frac{|f(x)-f(y)|}{|x-y|}: x neq y} = |nabla f |_infty $ is finite, denoted $|f|_{Lip}$.
2) If $f in Lip(mathbb{R}^d)$, then $ |f(x) - f(0)| leq |f|_{Lip} |x| $, so the integral $int_{mathbb{R}^d} f(x) u(x) dx $ is well defined for any $ u in L^1(mathbb{R}^d ,(1+|x|)dx)$.
3) With this topology $ Lip(mathbb{R}^d) $ is not separable. We will be actually working in $ Lip(mathbb{R}^d)/constants $, but we will not need be very explicit about it.
4) With this topology, any linear continuous function is of the form
$quad int_{mathbb{R}^d} f(x) u(x) dx +
int_{mathbb{R}^d} nabla f(x) cdot{w(x)} dx $
for some $ u in L^1(mathbb{R}^d, (1+|x|)dx)$ with $int u(x) dx = 0$,
and some $ w in L^1(mathbb{R}^d)^d $. The representation is not unique.
Claim 1: Let $mathcal{S}$ be the space of functions like (1) with arbitrary $a_i$, i.e.: $ mathcal{S}=mathbb{R}mathcal{G} $. The closure of $ mathcal{S}$ in the topology defined in (2) is $Lip(mathbb{R}^d)$.
Proof: If $ overline{mathcal{S}} neq Lip(mathbb{R}^d)$ then, by Hahn-Banach, there would by a non-zero linear function $ L(f)=int_{mathbb{R}^d} f(x) u(x) dx +
int_{mathbb{R}^d} nabla f(x) cdot{w(x)} dx$ such that $L(f) = 0$ for all $f in mathcal{S}$.
Let $rho$ be a smooth function with compact support. Since $mathcal{S}$ is invariant under translations it follows that
$(3) quad 0 = int uast{rho}(x) f(x) + wast{rho}(x)cdot{nabla f(x)} dx = $
$quad int (uast{rho}(x) - div(wast{rho})(x))f(x) dx $ for all $f in mathcal{S}$.
The real and imaginary parts of functions of the form $ f(x)=e^{-2pi xcdot{xi}}$ are in $mathcal{S}$, therefore the Fourier transform of $(uast{rho}(x) - div(wast{rho}))$ is zero, so (3) holds for any $f in Lip(mathbb{R}^d)$. Letting $rho longrightarrow delta $, we get that $L(f)=0$ for any $f in Lip(mathbb{R}^d)$, contradicting the fact that $L$ is not null.
Claim 2: $ Lip(mathbb{R}^d) = bigcup_{n in mathbb{N}} noverline{mathcal{G}}$.
Proof: Let $f in Lip(mathbb{R}^d)$. From Claim 1, there is a net ${f_lambda}_lambda$ in $mathcal{S}$ such that $f_lambda rightarrow f$. So the functionals defined as $L_lambda(w)=int_{mathbb{R}^d} w(x)cdot{nabla f_lambda(x)} dx $ for $w in L^1(mathbb{R}^d)^d$ are pointwise bounded. By the uniform boundness theorem, they are uniformly bounded. It follows that
$ sup_lambda | nabla f_lambda |_infty $
is finite, since $ | T_lambda | = | nabla f_lambda |_infty $.
Taking $ n in mathbb{N} $ sufficiently large we have $ sup_lambda | nabla f_lambda |_infty leq n $, and so $ f in noverline{mathcal{G}} $.
Claim 3: $ overline{mathcal{G}} $ is close in the strong topology (i.e.: the topology defined by the (semi) norm $|f|=_{def} | nabla f |_infty $).
Proof: Since $ L^1(mathbb{R}^d) $ is separable, the weak* topology of $ Lip(mathbb{R}^d) $ restricted to $Lip1(mathbb{R}^d) $ is metrizable; let $d$ be a metric on $Lip1(mathbb{R}^d) $ that defines the weak* topology. Let $ f $ be in the closure of $ overline{mathcal{G}} $ with the strong topology. Let $ {f_n}_n$ converge to $f $ in the strong topology, $f_n in overline{mathcal{G}} $, i.e.: $ |nabla f_n - nabla f |_infty rightarrow 0 $; in particular, $ d(f_n,f) rightarrow 0 $.
For each $n$, since $f_n in overline{mathcal{G}} $, there is $ h_n in mathcal{G} $ such that $ d(h_n,f_n) < frac{1}{n} $. Then $ d(h_n,f) leq d(h_n, f_n) + d(f_n,f) rightarrow 0$, so $ f in overline{mathcal{G}} $.
Claim 4: There is a constant $C_d$ such that
$ Lip1(mathbb{R}^d) subseteq C_doverline{mathcal{G}} $.
Proof: From Claim 2, $ Lip(mathbb{R}^d) = bigcup_{n in mathbb{N}} noverline{mathcal{G}}$. From Claim 3, for each $ n in mathbb{N}, noverline{mathcal{G}}$ is closed in the strong topology. From Baire's theorem, at least one of the sets $noverline{mathcal{G}} text{ } (n in mathbb{N}) $ has non-empty interior, i.e.: there is $n in mathbb{N}, epsilon > 0, f in Lip(mathbb{R}^d)$ such that $f + epsilon Lip1(mathbb{R}^d) subseteq noverline{mathcal{G}}$. From Claim 2, there is $a in mathbb{N}$ such that $ f in aoverline{mathcal{G}}$. Therefore,
$ Lip1(mathbb{R}^d) subseteq Coverline{mathcal{G}}$ with $ C = frac{n+a}{epsilon}$.
Finally, we can give a non-constructively answer the question in the affirmative. Since the definition of Wasserstein distance requires integration against functions in $Lip1(mathbb{R}^d)$, we assume the measures involved have finite first moment.
Claim 5: If $mu, nu $ are measures in $mathbb{R}^d$ with finite first moment, then
$ quad W_{R^d}(mu,nu)leq C_dsup_{vin R^d, |v|=1}W_{R}(mu_v,nu_v) $
where $C_d$ is the constant in Claim 4.
Proof: Let's assume first that $ mu, nu $ have densities $ u, w $ with respect to Lebesgue measure in $mathbb{R}^d $. Then
$ quad W_{R^d}(mu,nu) = sup_{f in Lip1(mathbb{R}^d)} int_{mathbb{R}^d} f(x) (u(x)-w(x)) dx leq $
$quad sup_{f in C_doverline{mathcal{G}}} int_{mathbb{R}^d} f(x) (u(x)-w(x)) dx $, by Claim 4 .
But $sup_{f in C_doverline{mathcal{G}}} int_{mathbb{R}^d} f(x) (u(x)-w(x)) dx = sup_{f in overline{mathcal{G}}} int_{mathbb{R}^d} C_d f(x) (u(x)-w(x)) dx =$
$ C_d sup_{f in overline{mathcal{G}}} int_{mathbb{R}^d} f(x) (u(x)-w(x)) dx $.
In turn,
$ sup_{f in overline{mathcal{G}}} int_{mathbb{R}^d} f(x) (u(x)-w(x)) dx =
sup_{f in mathcal{G}} int_{mathbb{R}^d} f(x) (u(x)-w(x)) dx$, since
$L(f)=int_{mathbb{R}^d} f(x) (u(x)-w(x)) dx$ is continuous in the weak* topology (2).
By definition, $mathcal{G}$ is the convex hull of functions on $mathbb{R}^d$ of the form
$f(xcdot{v}) $ with $|v|=1$ and $f in Lip1(mathbb{R})$, so
$quad sup_{f in mathcal{G}} int_{mathbb{R}^d} f(x) (u(x)-w(x)) dx = $
$quad sup_{f in Lip1(mathbb{R}), |v|=1} int_{mathbb{R}^d} f(xcdot{v}) (u(x)-w(x)) dx = sup_{|v|=1} W_{mathbb{R}}(mu_v,nu_v) $.
By continuity of the Wasserstein distance, we can pass from probability distributions with density to arbitrary distributions with finite first moments.$square$
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