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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 $X := \mathbb R^d$, $p \in [1, +\infty)$, and $$ \mathcal P_p(X) := \left \{\mu \in \mathcal P(X) \,\middle\vert\, \int_X |x|^p \mathrm d \mu < +\infty \right \}. $$

Theorem: Let $\mu, \mu_m \in \mathcal P_p(X)$. Then $W_p (\mu_m, \mu) \to 0$ if and only if $\mu_m \overset{\ast}{\rightharpoonup} \mu$ and $\int_X |x|^p \mathrm d \mu_m \to \int_X |x|^p \mathrm d \mu$. Here $W_p$ is the $p$-th Wasserstein metric.

Akira
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1 Answers1

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

We have $W_1 (\mu_m, \mu) \le W_p (\mu_m, \mu)$, so $W_1 (\mu_m, \mu) \to 0$. Then just as the first part of this answer, we can show that $\mu_m \overset{\ast}{\rightharpoonup} \mu$. Notice that $$ \int_X |x|^p \mathrm d \mu_m = W^p_p(\mu_m, \delta_0) \quad \text{and} \quad \int_X |x|^p \mathrm d \mu = W^p_p(\mu, \delta_0). $$

We have $W_p (\mu_m, \mu) \ge |W_p (\mu_m, \delta_0) - W_p (\mu, \delta_0)|$ with $\delta_0$ the Dirac measure at $0\in X$, so $$ W_p (\mu_m, \delta_0) \to W_p (\mu, \delta_0). $$

  1. Assume $\mu_m \overset{\ast}{\rightharpoonup} \mu$ and $\int_X |x|^p \mathrm d \mu_m \to \int_X |x|^p \mathrm d \mu$.

Fix $\varepsilon > 0$. For $R>0$, we define $c_R:X \to \mathbb R$ by $c_R (x) := \min \{R, |x|\}$. Then $c_R$ is a bounded metric on $X$ such that $c_R (x) \le |x|$. Notice that $c_R$ and $|\cdot|$ induce the same topology on $X$. By weak convergence of measures and convergence of $p$-th moment, we get $$ \int_X (|x|^p-c_R^p (x)) \mathrm d \mu_m \to \int_X (|x|^p-c_R^p (x)) \mathrm d \mu = \int_{|x|>R} (|x|^p-R) \mathrm d \mu \quad \forall R>0. $$

Notice that $c_R \nearrow |\cdot|$ as $R \nearrow +\infty$. By monotone convergence theorem, we pick $R>0$ such that $$ 0 \le \int_{|x|>R} (|x|^p-R) \mathrm d \mu < \frac{\varepsilon}{2}. $$

Then for sufficiently large $m$, we have $$ \int_{|x|>R} (|x|^p-R) \mathrm d \mu_m < \varepsilon. $$

For $x\in X$ such that $|x|>R$, we have $$ (|x|-R)^p \le |x|^p - R^p. $$

So for sufficiently large $m$, we have $$ \int_{|x|>R} (|x| - R)^p \mathrm d \mu_m < \varepsilon \quad \text{and} \quad \int_{|x|>R} (|x| - R)^p \mathrm d \mu < \varepsilon. $$

Let $P^R: X \to \overline B(0, R)$ be the projection onto the closed ball centered at $0$ with radius $R$. Then $P^R$ is continuous and coincides with the identity map on $\overline B(0, R)$. Moreover, for $x \notin \overline B(0, R)$, $|x-P^R (x)| = |x|-R$. Hence $$ \begin{align} W^p_p (\mu, P^R_\sharp \mu) &\le \int_X |x-P^R(x)|^p \mathrm d \mu (x) \\ &\le\int_{|x|>R} (|x|-R)^p \mathrm d \mu (x) \\ &< \varepsilon. \end{align} $$

Similarly, $W^p_p (\mu_m, P^R_\sharp \mu_m) \le \varepsilon$. Notice that $P_R$ is continuous and bounded, so $P^R_\sharp \mu_m \overset{\ast}{\rightharpoonup} P^R_\sharp \mu$. Also, $$ \mathrm{supp} (P^R_\sharp \mu_m) \subset \overline B(0, R) \quad \text{and} \quad \mathrm{supp} (P^R_\sharp \mu) \subset \overline B(0, R). $$

Then by this result, $W_p (P^R_\sharp \mu_m, P^R_\sharp \mu) \to 0$ as $m \to \infty$. Finally, $$ \begin{align} \limsup_m W_p(\mu_m, \mu) &\le \limsup_m W_p (\mu_m, P^R_\sharp \mu_m) + \limsup_m W_p (P^R_\sharp \mu_m, P^R_\sharp \mu) + \limsup_m W_p (P^R_\sharp \mu, \mu) \\ &\le \varepsilon^{1/p} + 0 + \varepsilon^{1/p}\\ &= 2 \varepsilon^{1/p}. \end{align} $$

The result then follows by taking the limit $\varepsilon \to 0^+$.

Akira
  • 17,367