Bounding probability densities on a Wasserstein-2 geodesic

Question

Consider two probability measures which are supported on a bounded domain $\Omega$ with density functions $p_0$ and $p_1$. It is well-known that for the Wasserstein-2 distance, there exists uniquely a geodesic $\mathcal{L}_t, 0\le t\le 1,$ connecting $p_0$ and $p_1$. That is to say, $Law(\mathcal{L}_0)=p_0,\ Law(\mathcal{L}_1)=p_1$,and for any $0 \le s \le t\le1$, $$ W_2(\mathcal{L}_s,\mathcal{L}_t)=(t-s)W_2(\mathcal{L}_0,\mathcal{L}_1). $$
We use $p_t$ to denote the density function of $\mathcal{L}_t $. My question is that, under what conditions on $p_0$ and $p_1$, there exists a constant $\lambda>0$ such that for all $0\le t\le1$, inequality $ 1/\lambda< p_t < \lambda $ holds universally on $\Omega$?

Answer 1

It turns out that for most convex domains $\Omega$, one can find smooth probability measures $\rho_0$ and $\rho_1$ which are supported on $\Omega$, have strictly positive density everywhere on $\Omega$, but whose barycenter (in the $2$-Wasserstein sense) has non-convex support (so has support strictly smaller than $\Omega$). This phenomena was recently discovered by Santambrogio and Wang [1]. As such, I'm not sure of sufficient conditions to ensure that a lower bound on the density holds throughout the displacement interpolation. To the best of my knowledge, this question is still open.

Given this, the natural question is whether it is possible to establish lower bounds for $d \rho_t$ on its support. The densities of $\rho_0$ and $\rho_1$ do not go to zero at the boundary $\partial \Omega$, so one might imagine that this continues to be the case for $\rho_t$ (although the support is evolving). By contrast, it might also be the case that $d \rho_t$ vanishes continuously at some points within the domain. Presumably both are possible.

 Later edit:  After thinking about it further, I realized that Caffarelli's $C^2$ estimate can be used to obtain a lower bound for $d \rho_t$ on its support. As such, the following can be considered as a lower bound of the densities, with the caveat that the support along displacement interpolation can be smaller than the initial or target measures. 

Proposition:

Suppose that $\Omega$ is a smooth and strongly convex domain and that $\rho_0$ and $\rho_1$ are smooth probability measures supported in $\Omega$. Suppose further that both measures are absolutely continuous with respect to the Lebesgue measure and satisfy $1/C \leq d \rho_i \leq C$ for some $C \geq 1.$ Let $\rho_t$ be 2-Wasserstein geodesic between $\rho_0$ and $\rho_1$. Then for $0 \leq > t \leq 1$, there is a constant $C^\prime>0$ so that $$ d\rho_t > > C^\prime$$ on the support of $\rho_t$ (which might be strictly smaller than $\Omega$).

Proof:

By Brenier's theorem [2], optimal transport from $\rho_0$ to $\rho_1$ is induced by the sub-differential of some convex function $u$. Furthermore, the 2-Wasserstein geodesic between $d \rho_0$ and $d \rho_1$ is induced by displacement interpolation, which transports mass along straight lines. As such, we have that$$ \rho_t = (\nabla u_t )_\sharp \rho_0.$$ In this formula, $u_t =(1-t) \frac{x^2}{2}+ t \cdot u$ and the sharp notation indicates pushforward of measures. It may seem odd, but $ \nabla \frac{x^2}{2}$ serves as a convenient way to write the identity map.

The goal is to derive a lower bound for $d \rho_t $ on its support. Using the change of variables formula, we have the following:  $$ d \rho_t = \frac{d \rho_0}{\det(\nabla^2 u_t)} $$

 From this, we see that we need to bound the Jacobian of $\nabla u_t$, which is simply the determinant of the Hessian matrix $$\det \left(\frac{\partial^2 u_t}{\partial x^i \partial x^j} \right). $$ For convenience, we will use the notation $ H(u)$ for the Hessian matrix  $\frac{\partial^2 u}{\partial x^i \partial x^j} $.

In order to get the desired estimate, we need to use the structure of the optimal transport. In particular, Brenier's theorem shows that the potential $u$ is a weak solution to the Monge-Ampere equation $$ \det \left( H(u) \right) = \frac{d \rho_0(x)}{d \rho_1( \nabla u (x))} \hspace{2in} (\ast)$$ with appropriate boundary conditions (which are not important here). From our assumptions, we can estimate the right hand side as $$     \frac{1}{ C^2 }     \leq  \frac{d \rho_0(x)}{d \rho_1( \nabla u (x))} \leq C^2, $$ so we have uniform bounds on $\det H(u)$.

However, this is not enough. What we actually want is upper bounds on $\det H(u_t)$. Using the fact that the Hessian of $\frac{x^2}{2}$ is the identity matrix, we can expand $\det(H(u_t))$ to obtain the following: $$\det( H(u_t))= \det((1-t) I + t H(u) ) = \Pi_{i=1}^n \left( (1-t)+t \lambda_i \right ).$$ Here, $ \lambda_i$s are the eigenvalues of $H(u)$. From this, we can see that in order to bound $\det(H(u_t))$, we need to bound the eigenvalues of $H(u)$, not simply the determinant. To do this, we use a deep result of Caffarelli.

Theorem (Caffarelli [3]):

Suppose that $\Omega$ is strongly convex and $\rho_0$ and $\rho_1$ are smooth probability measures supported in $\Omega$ satisfying $1/C \leq > d \rho_i \leq C$ for some $C \geq 1$. Then there exists a $C^\prime$ so that the convex function $u$ solving $(\ast)$ satisfies$$\| u > \|_{C^2} \leq C^\prime.$$

This establishes an a priori $C^2$ estimate for solutions to the Monge-Ampere equation $(\ast)$. Even more strikingly, it implies that $u$ is smooth, uniformly strongly convex, and in fact a classical solution to $(\ast)$.

For our purposes, with a uniform $C^2$ estimate on $u$, we can bound all the eigenvalues of $H(u)$, which gives a bound on the Jacobian of $\nabla u_t$. Together with the assumption that $d \rho_0 > 1/C$, this implies a lower bound on the density of $d  \rho_t$, wherever it is nonzero. $ \hspace{2in} \square $

To give an intuitive explanation for this result, we can imagine optimal transport moving a large body of water. Santambroggio and Wang's result shows that dry spots might emerge during this transport, even if there are no dry spots for the initial and final configurations. However, the argument above shows that under additional assumptions, dry spots cannot emerge from the water smoothly receding. Instead, it's more akin to "Moses parting the Red Sea," where there are tall walls of water around the dry areas. For those not familiar with the imagery, I've included a painting by Cornelis de Wael.

 

[1] Santambrogio, Filippo; Wang, Xu-Jia, Convexity of the support of the displacement interpolation: counterexamples, Appl. Math. Lett. 58, 152-158 (2016). ZBL1345.49056.

[2] Brenier, Yann, Décomposition polaire et réarrangement monotone des champs de vecteurs. (Polar decomposition and increasing rearrangement of vector fields), C. R. Acad. Sci., Paris, Sér. I 305, 805-808 (1987). ZBL0652.26017.

[3]Caffarelli, Luis A., The regularity of mappings with a convex potential, J. Am. Math. Soc. 5, No. 1, 99-104 (1992). ZBL0753.35031.

Answer 2

Only half an answer: If $(M,g)$ is a complete Riemannian manifold with nonnegative Ricci curvature (or more generally a $\mathrm{RCD}(0,\infty)$ space), then $\{\rho m\in P_2(M)\mid \lVert\rho\rVert_\infty\leq c\}$ is geodesically convex in $(P_2(M),W_2)$. In particular, this result applies to measures on flat Euclidean space.

One can also extend this theorem to manifolds with a negative lower Ricci curvature bound if one additionally assumes that the endpoints of the geodesic have bounded support. All this can be found in Section 3 of Metric measure spaces with Riemannian Ricci curvature bounded from below by Ambrosio, Gigli, and Savaré.

I don't know an answer for lower bounds, but checking out the other papers by these authors might be a good start.