For two probability distributions $\mu$ and $\nu$ defined on $X$, the $p$-th Wasserstein distance between the two of them is defined as $$W_p(\mu,\nu) = \left(\inf_{\pi\in\Pi(\mu,\nu)}\int_{X\times X}c(x,x')^{p}\,\mathrm{d}\pi(x,x')\right)^{1/p}.
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We also have $\infty$-Wasserstein distance defined as the limit of p-th Wasserstein distance, i.e., $$W_\infty(\mu,\nu) = lim_{p \rightarrow \infty} W_p(\mu,\nu).$$ However, I am having a hard time understanding this $\infty$-Wasserstein distance. We know $\infty$-norm is the element with the largest absolute value, for $\infty$-Wasserstein distance do we have such a simplification as well?
Best Answer
You can see the discussion 5.5 in the book of Filippo Santambrogio "Optimal Transport for Applied Mathematician". We can prove that the infinite wasserstein distance is given by $$ W_{\infty}(\mu;\nu)=\inf \{ \lvert\lvert x-y\rvert\rvert_{L^{\infty}}(\gamma): \gamma \in \Pi(\mu,\nu) \} $$ where we have defined $\lvert\lvert x-y\rvert\rvert_{L^{\infty}}(\gamma)=\inf\{ m\in \mathbb{R} : \lvert x-y\rvert\leq m \quad \text{ a.e-} \gamma \text{ for } (x,y) \}$.
It means that the infinite Wasserstein distance is the infimum of the infinite distance over transport plans from $\mu$ to $\nu$. I hope it answers your question.