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Denote

$$K:=\Big\{ \rho :\mathbb{R}^d\to [0,\infty)~ \text{measurable}, : \int_{\mathbb{R}^d} \rho(x) dx=1, ~M(\rho)<\infty \Big\},$$

with

$$ M(\rho):=\int_{\mathbb{R}^d}|x|^2\rho(x)dx. $$

Think of $\rho$ as densities of probability distributions. We define the Wasserstein distance $d$ between two Borel probability measures on $\mathbb{R}^d$ as usual :

$$ d(\mu_1,\mu_2)^2=\inf_{\mu} \int_{\mathbb{R}^d\times \mathbb{R}^d} |x-y|^2 \mu(dx,dy), $$ with the infimum taken over all couplings with marginals $\mu_1,\mu_2$. Denote $d(\rho_1,\rho_2)=d(\mu_1,\mu_2)$ for distributions $\mu_1,\mu_2$ with densities $\rho_1,\rho_2$.


$\textbf{Question :}$

I am reading some text that says clearly : $$ M(\rho_1) \leq 2M(\rho_0)+2d(\rho_1,\rho_0)^2 $$

I don't see how this works? it seems like $|x|^2\leq 2|x|^2+2|x-y|^2$ could be used? Any help? My main difficulty is starting on the LHS how to introduce $\rho_0$, adding and subtracting in some way?

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1 Answer 1

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Let $\mu$ with marginals $\mu_1,\mu_0$, then $$ 2M(\rho_0)+2\int_{\mathbb{R}^d\times\mathbb{R}^d}|x-y|^2\mu(dx,dy)=2\int_{\mathbb{R}^d\times\mathbb{R}^d}\left(|y|^2+|x-y|^2\right)\mu(dx,dy) $$ because $$ \int_{\mathbb{R}^d\times\mathbb{R}^d}|y|^2\mu(dx,dy)=M(\rho_0)\int_{\mathbb{R}^d}\rho_1(x)dx=M(\rho_0) $$ Moreover $2|y|^2+2|x-y|^2=|x|^2+|x-2y|^2\geqslant |x|^2$ so that $$ \begin{aligned} 2M(\rho_0)+2\int_{\mathbb{R}^d\times\mathbb{R}^d}|x-y|^2\mu(dx,dy)&\geqslant\int_{\mathbb{R}^d\times\mathbb{R}^d}|x|^2\mu(dx,dy) \\ &=M(\rho_1)\int_{\mathbb{R}^d}\rho_0(y)dy\\ &=M(\rho_1) \end{aligned}$$ Taking the infimum on $\mu$ leads to $2M(\rho_0)+2d(\rho_1,\rho_0)\geqslant M(\rho_1)$.

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