For Lebesgue-absolutely continuous probability measures $\rho\ll \mathcal{L}^d$ in the whole space $\mathbb{R}^d$ with finite second moments (i-e $\rho\in \mathcal{P}^2_{ac}(\mathbb{R}^d)$), let $$ \mathcal{H}(\rho)=\int_{\mathbb{R}^d}\rho\log \rho \,dx $$ be the Boltzmann entropy.

Question: Is $\mathcal{H}$ continuous with respect to the quadratic Wasserstein distance $\mathcal{W}_2$? i-e is it true that $$ \mathcal{W}_2(\rho_n,\rho)\to 0 \quad\Rightarrow\quad \mathcal{H}(\rho_n)\to\mathcal{H}(\rho)\qquad ??? $$

I guess this should be known and written down somewhere. I looked at the classical references (Villani's books [topics...] and [old and new...], also in [Ambrosio-Gigli-Savaré]) but surprisingly enough I couldn't find the precise statement.

It is well known that $\mathcal{H}$ is $0$-displacement convex (in the sense of McCann). By analogy with the Euclidean setting, where any convex function $\phi:\mathbb{R}^d\to \mathbb{R}$ is continuous, I suspect that any displacement convex functional $\mathcal{F}:\mathcal{P}_2(\mathbb{R}^d)\to\mathbb{R}$ should be continuous with respect to the wasserstein distance $\mathcal{W}_2$. But maybe not...

This question arised in the context of my research: I am trying to construct weak solutions to some system of PDEs by means of the Jordan-Kinderlehrer-Otto/DeGiorgi's minimizing scheme, and I need the above statement in some technical step. Unfortunately the mere lower semi-continuity would not be enough for my purpose, I really need full continuity.


1 Answer 1


Let $A_n=\bigcup_{i=0}^{n-1}[2i/(2n),(2i+1)/2n)$, $\rho_n=2\chi_{A_n}$, $A=[0,1]$ and $\rho=\chi_{[0,1]}$. Then it seems pretty clear that $\mathcal W_2(\rho_n,\rho)\to 0$. On the other hand $H(\rho_n)\not\to H(\rho)$, right?

  • $\begingroup$ yes you're perfectly right, very nice counterexample. Yet another naive belief goes to the bin... $\endgroup$ Commented Jun 26, 2015 at 9:04
  • $\begingroup$ what about lower semi-continuity then? after thinking on it for a while this should be enough for my purpose, contrarily to what I said in my post $\endgroup$ Commented Jun 26, 2015 at 10:00
  • $\begingroup$ Yes, it is l.s.c; You have that $\rho_n\to \rho$ weakly (as probability measures). Since $\rho_n\to H(\rho_n|\rho)$ is l.s.c., you are done. $\endgroup$ Commented Jun 26, 2015 at 15:48
  • $\begingroup$ Well, sure, if the relative H is lsc then of course the original one is lsc too. But do you have a reference? $\endgroup$ Commented Jun 26, 2015 at 20:02
  • $\begingroup$ The relative entropy is the rate function in Sanov's theorem of large deviations, and thus is by definition l.s.c. The proof is not hard - H can be written as a Legendre transform, that is a supremum over continuous functions. Any book on large deviations (in particular, Deuschel-Stroock, Varadhan's lecture notes, Dembo-Zeitouni) contains a proof of these facts. $\endgroup$ Commented Jun 26, 2015 at 21:05

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