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This thread is meant to record a question that I feel interesting during my self-study. I'm very happy to receive your suggestion and comments.

Let $p \in [1, +\infty)$, $X=Y$ be compact subsets of $\mathbb R^d$, $\mathcal P (X)$ the set of all Borel probability measures on $X$, and $$ \mathcal P_p (X) := \left \{\mu \in \mathcal P(X) \,\middle\vert\, \int_X |x|^p \mathrm d \mu < +\infty \right \}. $$

We define the $p$-th Wasserstein metric $W_p$ by $$ W_p (\mu, \nu) := \inf_{\gamma \in \Pi(\mu, \nu)} \left [ \int_{X \times Y} |x-y|^p \mathrm d \gamma (x, y) \right ]^{1/p} \quad \forall \mu, \nu \in \mathcal P_p (X). $$

Here $\Pi(\mu, \nu)$ is the set of all Borel probability measures on $X\times Y$ whose marginals are $\mu, \nu$ respectively.

Theorem: Let $\mu, \mu_m \in \mathcal P_p(X)$. Then $\mu_m \overset{\ast}{\rightharpoonup} \mu$ if and only if $W_p (\mu_m, \mu) \to 0$.

Akira
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1 Answers1

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We have $$ W^p_1 (\mu, \nu) \le W^p_p (\mu, \nu) \le (\mathrm{diam} X)^p W_1 (\mu, \nu) \quad \forall p \in [1, +\infty). $$

So it suffices to prove for $p=1$. Let $$ \mathrm{Lib} (f) := \sup_{x\neq y} \frac{|f(x) - f(y)|}{|x-y|}. $$

  1. Assume $W_1 (\mu_m, \mu) \to 0$.

By Kantorovich-Rubinstein theorem, we have $$ W_1 (\mu_m, \nu) = \sup \left \{\int_X f \mathrm d (\mu_m-\nu) \,\middle\vert\, f \in L_1 (|\mu_m-\nu|) , \mathrm{Lib} (f) \le 1\right \}. $$

Let $f$ be bounded Lipschitz-continuous such that $\mathrm{Lib} (f) > 0$. Then $\frac{f}{\mathrm{Lib} (f)}$ is $1$-Lipschitz. Then $$ \int_X \frac{f}{\mathrm{Lib} (f)} \mathrm d (\mu_m-\nu) \to 0. $$

So $$ \int_X f \mathrm d (\mu_m-\nu) \to 0, $$

The result then follows from Portmanteau theorem.

  1. Assume $\mu_m \overset{\ast}{\rightharpoonup} \mu$.

Let $m_k$ be a subsequence such that $$ \lim_k W_1 (\mu_{m_k}, \mu) = \limsup_m W_1 (\mu_m, \mu). $$

Fix $a \in X$. Let $g_k$ be a Lipschitz funtion such that $\mathrm{Lib} (g_k) \le 1$ and $$ W_1 (\mu_{m_k}, \mu) \le \int_X g_k \mathrm d (\mu_{m_k} - \mu) + \frac{1}{k} = \int_X (g_k - g_k (a)) \mathrm d (\mu_{m_k} - \mu) + \frac{1}{k} . $$

It follows from $(\mu_{m_k} - \mu) (X) = 0$ that $$ \int_X g_k (a) \mathrm d (\mu_{m_k} - \mu) = 0 \quad \forall k. $$

Let $g'_k := g_k - g_k (a)$. Then $\mathrm{Lib} (g'_k) \le 1$ and $(g'_k)_k$ is bounded by $\mathrm{diam} X$. By Arzelà–Ascoli theorem, there is $g$ and a subsequence (re-labelled) such that $g'_k \to g$ in $\|\cdot\|_\infty$. Clearly, $\mathrm{Lib} (g) \le 1$. Finally, $$ \begin{align} \int_X g'_k \mathrm d (\mu_{m_k} - \mu) &= \int_X (g'_k-g) \mathrm d (\mu_{m_k} - \mu) + \int_X g \mathrm d (\mu_{m_k} - \mu) \\ &= \int_X (g'_k-g) \mathrm d (\mu_{m_k} - \mu)^+ - \int_X (g'_k-g) \mathrm d (\mu_{m_k} - \mu)^- + \int_X g \mathrm d (\mu_{m_k} - \mu) \\ &\le 2 \|g'_k-g\|_\infty (\mu_{m_k} - \mu)^+ (X) + \int_X g \mathrm d (\mu_{m_k} - \mu) \\ &\le 4 \|g'_k-g\|_\infty + \int_X g \mathrm d (\mu_{m_k} - \mu). \end{align} $$

The result then follows by taking the limit $k \to \infty$.

Akira
  • 17,367