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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 $X =Y=Z \subset \mathbb R^d$, $p \in [1, +\infty)$, $\mathcal P (X)$ be 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: $W_p$ satisfies triangle inequality.

Akira
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  • I'm not the one who has downvoted, but I think that who did has a point. The question is not self-contained: though being clear to someone who already know the subject what $\Pi(\mu,\nu)$ stands for, I think that for all the others it would be worthwhile to define it directly in the question. – Bob Jul 10 '22 at 08:02
  • @Bob thank you for your suggestion. I'm going to add related definitions. – Akira Jul 10 '22 at 08:09

1 Answers1

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Let $\mu, \nu, \omega \in \mathcal P_p (X)$. Let $\pi_1 \in \Pi (\mu, \nu)$ and $\pi_2 \in \Pi (\mu, \nu)$ be optimal, i.e., $$ W^p_p (\mu, \nu) := \int_{X \times Y} |x-y|^p \mathrm d \pi_1 (x, y) \quad \text{and} \quad W^p_p (\nu, \omega) := \int_{Y \times Z} |y-z|^p \mathrm d \pi_2 (y, z). $$

Let $P^{X \times Y}$ and $P^{Y \times Z}$ be the projection maps from $X \times Y \times Z$ to $X \times Y$ and $Y \times Z$ respectively. Let $\gamma \in \mathcal P(X \times Y \times Z)$ such that $P^{X \times Y}_\sharp \gamma = \pi_1$ and $P^{Y \times Z}_\sharp \gamma = \pi_2$. Here $P^{X \times Y}_\sharp \gamma$ is the push-forward of $\gamma$ by $P^{X \times Y}$. Such $\gamma$ does exists by gluing lemma. Let $\pi = P^{X \times Z}_\sharp \gamma$. It's easy to prove that $\pi \in \Pi(\mu, \omega)$. Finally, $$ \begin{align} W_p(\mu, \omega) &\le \left [ \int_{X \times Z} |x-z|^p \mathrm d \pi (x, z) \right ]^{1/p} \\ &= \left [ \int_{X \times Y \times Z} |(x-y) - (y-z)|^p \mathrm d \gamma(x,y,z) \right ]^{1/p} \\ &\le \left [ \int_{X \times Y} |x-y|^p \mathrm d \pi_1 (x, y) \right ]^{1/p} + \left [ \int_{Y \times Z} |y-z|^p \mathrm d \pi_2 (y, z) \right ]^{1/p} \\ &= W_p(\mu, \nu) + W_p(\nu, \omega). \end{align} $$

This completes the proof.

Akira
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