# Optimal transport and total variation distance

I have a question regarding the following concept equating total variation distance with a particular case of optimal transport.

I don't understand why equality (6.11) holds. We know by Kantorovich duality that the RHS is equal to $$2 \sup_{\phi \text{ Lipschitz} \\ |\phi|_{\text{Lip}} \leq 1} \int \phi d\mu - \int \phi d\nu \equiv f(\mu, \nu)$$ as a function is $$c-$$convex for a distance function $$c = 1(x \ne y)$$ if and only if it is $$1-$$Lipschitz.

As for the total variation, it is defined as $$T(\mu, \nu) \equiv \sup_{A \in \mathcal{F}} |\mu(A) - \nu(A)|$$ where $$\mathcal{F}$$ is our $$\sigma-$$algebra on whichever Polish space we're working with. It is obvious that for $$\phi(x) = 1_A(x)$$, we have that $$\phi$$ is $$1-$$Lipschitz and therefore $$T(\mu, \nu) \leq f(\mu, \nu)$$. I'm confused why we need the $$2$$ here, and how the other direction of the inequality would be shown?

Specifically, I need that for any $$1-$$Lipschitz function, there exists a set $$A \in \mathcal{F}$$ such that $$|\mu(A) - \nu(A)| \ge 2 \int \phi d\mu - \int \phi d \nu$$, but I have no idea how to get this right. Any help would be massively appreciated.

(The excerpt is from Villani (2009))

• Could you please clarify the bit "function is c−convex for a distance function c=1(x≠y) if and only if it is 1−Lipschitz"? This is my difficulty. Feb 11, 2023 at 15:39

I think there is some difference in definition. Look the lecture notes Probability in High Dimensions by Van-Handel. In example 4.14 the author writes:

$$||\mu - \nu||_{TV} = \inf_{M\in\mathcal C(\mu,\nu)}M(X\neq Y)$$

And he then goes on to prove this.

What might be happening is a different definition of the T.V metric.

Indeed, we can prove that using your definition of TV, the equality $$||\mu - \nu||_{TV} = \sup_A|\mu(A) - \nu(A)| = 2\inf P[X\neq Y]$$

Would be inconsistent. Note:

$$\mu(A) - \nu(A) = P[X \in A] - P[Y \in A] =$$ $$= P[X \in A, X=Y] - P[X \in A,X\neq Y]+ P[Y \in A,X=Y] - P[Y \in A,X\neq Y] =$$ $$= P[X \in A, X\neq Y] - P[Y \in A, X \neq Y] \leq P[X\neq Y]$$ Therefore, $$\sup_A|\mu(A) - \nu(A)| \leq P[X\neq Y]$$ Hence, $$\sup_A|\mu(A) - \nu(A)|>0 \implies 2P[X\neq Y]> \sup_A|\mu(A)-\nu(A)|$$

• Yes you're correct. I am making a mistake in my definition of total variation. Villani assumes the definition to be twice what I wrote down (but does not explicitly state this anywhere, as his way is probably standard/common). Thanks for this answer. Nov 12, 2020 at 21:22
• No problems. If this helped you, please consider upvoting. Cheers Nov 18, 2020 at 10:59