Proving that integral is larger than Wasserstein metric I’m trying to understand a proof that appeared in a paper that I’m reading. The proof is about an inequality regarding the Wasserstein metric. My question is only related to showing that a particular integral is larger than the Wasserstein metric of two probability measures.
First, the p-th Wasserstein metric is given by:
$$
W_p(\nu,\mu):=\inf \left \{
\left(\int_{X^2}\mid x-y\mid^pd\pi(x,y)\right)^{\frac{1}{p}};  \pi\in P(X^2);\pi_o = \nu; \pi_1=\mu 
\right \}
$$
Where $\nu,\mu$ are probability measures and $\pi$ is a coupling of $\nu$ and $\mu$.
Now, let $T:X \rightarrow X$, and assume that
$$
\nu((-\infty,T(x)]) = \mu((-\infty,x]), \ \forall x \in \mathbb R
$$
Also, assume that $\mu$ is the standard Gaussian measure on $\mathbb R$. How does one then shows that:
$$
\int_\mathbb R \frac{(T(x) - x)^2}{2} d\mu(x)\geq \frac{W_2^2(\nu,\mu)}{2}
$$
I believe that is not necessary for $\mu$ to be a Gaussian, because this was used in the a different part of the proof. But I included this information, since this is how it was presented.
 A: First observe that T is a transport map, because $\nu=T_\#\mu$ is the pushforward measure of $\mu$ along $T$. This follows from the fact that intervals of the form $(-\infty,x]$ are Borel sets.
Since $\mu$ has a density w.r.t. the Lebesgue measure, the Monge-Kantorovich equivalence gives us that the optimal coupling $\pi$ is unique and $\pi=(Id, \tilde{T})_\#\mu$, for some transport map $\tilde{T}: X\to X$.
This implies that
$$\int_{X^2}|x-y|^2d\pi(x,y)=\int_X|x-\tilde{T}(x)|d\mu(x).$$
Your inequality therefore holds with equality iff $T=\tilde{T}$.
For a proof of the Monge-Kantorovich equivalence refer to Computational Optimal Transport Theorem 2.1 for a sketch. You will find references to more detailed proofs there.
A: Although I think the answer by @dirich1337 is correct, there might be a simpler answer without using the Monge-Kantorovich equivalence.
What I came up with is the following. Let $\pi$ be the product measure, hence $\pi = \mu \times \nu$. This is indeed a coupling, so we immediately have
$$
\int_{X^2} (x-y)^2d\pi\geq W_2^2(\mu,\nu)
$$
The only thing left to show would be that using such coupling, we end up with the desired integral. Which is true, as shown below:
$$
\int y d\nu(y) = \int T(x) d\mu(x)
$$
$$
\int_{\mathbb R^2}(x-y)^2d(\mu\times\nu)=
\int_{\mathbb R}\int_{\mathbb R}
(x-y)^2d\nu d\mu=
\\ =
\int_{\mathbb R}\int_{\mathbb R}
x^2d\nu d\mu
-
2\int_{\mathbb R}x d\mu
\int_{\mathbb R}
y d\nu+
\int_{\mathbb R}\int_{\mathbb R}
y^2d\nu d\mu =
\\
=
\int_{\mathbb R}
x^2d\mu
-
2\int_{\mathbb R}x d\mu
\int_{\mathbb R}
T(x) d\mu+
\int_{\mathbb R}
T(x)^2d\mu =
\int_{\mathbb R}(x-T(x))^2d\mu
$$
