Past exam of the optimal transport and partial differential equations course of the University of Oxford / 2025 / 2

Put $D=d_2(\mu ,\nu )$ and $z_t(x,y)=(1-t)x+ty$. The marginal distributions of the optimal transport plan $\Pi$ give ${\mu }_0=\mu$ and ${\mu }_1=\nu$. Alsoso ${\mu }_t$ has a finite second absolute moment. For $0\le s<t\le 1$, the pushforward measure $(z_s,z_t{)}_{\#}\Pi$ is a coupling of ${\mu }_s,{\mu }_t$. ThereforeApply the triangle inequality for the second Wasserstein distance along ${\mu }_0,{\mu }_s,{\mu }_t,{\mu }_1$:All inequalities are equalities, including when $D=0$. Thus this displacement interpolation is a constant speed metric geodesic:

Set $q=y-x$ and $z_t=x+tq$. The definition of pushforward measure gives the confinement energy $\mathcal{V}[{\mu }_t]=\int V(z_t)d\Pi$. The chain rule and differentiation under the integral sign, whenever justified, giveFor example, bounded Hessian matrix is sufficient: it gives $|\nabla V(z)|\le C(1+|z|)$, and the first two differentiated integrands are bounded by $C^{\prime }(1+|x{|}^2+|y{|}^2)$. The dominated convergence theorem applies because $\Pi$ has finite second absolute moments. Compactly supported endpoints also suffice because all interpolating points lie in one compact set.

The printed growth assumption alone does not ensure a classical second derivative. Here is a counterexample within the stated assumptions. On the line take the bounded nonnegative smooth functionand the probability measures $\mu ={\sum }_n{p}_n{\delta }_{n-1/2}$, $\nu ={\sum }_n{p}_n{\delta }_{n+1/2}$. Both have finite second absolute moments. Their translation transport plan is optimal: its cost is one, while every coupling has cost at least the square of its mean displacement, also one. At $t_0=1/2$, $z_{t_0}=n$, so $F(t)=\mathcal{V}[{\mu }_t]$ satisfies $F(t_0)=0$. Each summand is nonnegative, and the Fatou lemma givesThus the requested classical identity requires an additional integrability assumption on the Hessian matrix; quadratic growth controls the energy but not its second derivative.

For a confinement potential $V$ that is a $\lambda$-strongly convex function, the derivative formula, where valid, yieldsThe underlying displacement convexity conclusion is valid under the printed assumptions without that extra differentiation hypothesis. Apply convexity of $V(x)-\lambda |x{|}^2/2$ pointwise and integrate:The same argument on every subinterval makes $t\mapsto \mathcal{V}[{\mu }_t]-\lambda D^2{t}^2/2$ convex, so the second derivative lower bound also holds in the sense of distributions.

Use an independent copy $(x^{\prime },y^{\prime })$ of $(x,y)$ under the product measure $\Pi \otimes \Pi$. Write $q=y-x$, $q^{\prime }=y^{\prime }-x^{\prime }$, andSince the pushforward measure of $\Pi \otimes \Pi$ under $(z_t,z_t^{\prime })$ is ${\mu }_t\otimes {\mu }_t$, the interaction energy isTwo applications of the chain rule giveThe factor $1/2$ is retained. This classical identity requires differentiation under the integral sign to be justified, just as in part (b). Bounded Hessian matrix of $W$ is sufficient, since $|q-q^{\prime }{|}^2\le 2|q{|}^2+2|q^{\prime }{|}^2$ is integrable; compactly supported endpoints are another sufficient case. The printed quadratic-growth condition alone does not provide this domination.

The failure can occur for the interaction energy too. Take the even interaction potential $W(x)=1-\cos (2\pi x^3)$ and replace the endpoint probability measures in part (b) by $\frac{1}{2}{\delta }_0+\frac{1}{2}\mu$ and $\frac{1}{2}{\delta }_0+\frac{1}{2}\nu$. Keep the origin fixed and translate every other atom by one. This coupling is increasing on the line, hence is optimal: its assigned values extend to a nondecreasing transport map, which is the gradient of a convex function. Differences between translated atoms remain integers, so their mutual interaction vanishes. Only interactions with the origin contribute, giving $\mathcal{W}[{\mu }_t]=F(t)/4$, with $F$ from part (b). It therefore also lacks a finite classical second derivative at $t=1/2$.

Define the barycenter of a measure by $m_{\mu }=\int xd\mu (x)$. Under $\Pi$, the mean displacement is $\bar{q}=m_{\nu }-m_{\mu }$, and optimality gives $\int |q{|}^{2}d\Pi =D^2$. Expanding the square under the product measure yields the variance identity for independent copies:In the classical regime of part (c), the lower bound on the Hessian matrix of the interaction potential therefore givesFor equal barycenters, this becomes

For the full stated class of probability measures, prove the displacement convexity inequality directly. Pointwise strong convexity of $W$ givesIntegrating, including the $1/2$ in the interaction energy, and using the displayed variance identity for independent copies givesAll terms are finite by quadratic growth and finite second absolute moments. Equal barycenters give the claimed $\lambda$-displacement convexity. Also $m_{{\mu }_t}=(1-t)m_{\mu }+t{m}_{\nu }$, so this metric geodesic remains in the prescribed fixed-mean class. Applying the same argument to subintervals proves the second derivative bound in the sense of distributions even when a classical derivative has not been justified. The mean restriction is natural: translating all particles together leaves their pairwise differences and hence the interaction energy unchanged.

Choose a uniform lower bound $\kappa \in \mathbb{R}$ for the Hessian matrix of the confinement potential, so $D^{2}V(x)\ge \kappa I$ for every $x$. A sufficient condition isIndeed, convexity of $V(x)-\kappa |x{|}^2/2$ gives the confinement energy inequality from part (b), now with $\kappa$ possibly negative. Adding it to the equal-mean interaction energy inequality from part (d) givesThis proves positive displacement convexity on every class with fixed barycenter. In particular, if $D^{2}V\ge -KI$ with $K\ge 0$, the sufficient condition is $\lambda >K$ and one may take $\tilde{\lambda }=\lambda -K$. Wherever classical second derivatives exist, the equivalent differential calculation is ${\mathcal{E}}^{\prime \prime }[{\mu }_t]\ge (\kappa +\lambda )D^2$; the integrated proof also covers the printed growth assumptions.