Past exam of the optimal transport and partial differential equations course of the University of Oxford / 2025 / 1

Compact support of a measure gives finite second absolute moments for both probability measures. Since the source satisfies absolute continuity of measures with respect to Lebesgue measure, Brenier's theorem identifies the quadratic Kantorovich optimal transport problem with the Monge optimal transport problem:Here $S_{\#}\mu$ is the pushforward measure. The optimal transport plan is unique and equals $(\mathrm{Id},T{)}_{\#}\mu$. The optimal transport map is unique up to a $\mu$-null set, withThe gradient of this convex function is c-cyclically monotone for $c(x,y)=|x-y{|}^2$ on a full-measure set. In particular, summing the two supporting inequalities for the convex function gives $(T(x)-T(x^{\prime }))\cdot (x-x^{\prime })\ge 0$ wherever both gradients exist. No regularity of the target beyond the stated probability measure assumptions is needed.

Write $P=\omega {\omega }^T$. Since $P^T=P$ and $P^2=P$, the orthogonal reflection $R_{\omega }=I-2P$ satisfies $R_{\omega }^T{R}_{\omega }=I$ and $R_{\omega }^{-1}=R_{\omega }$. Thus, for every Borel set $E$, invariance under the orthogonal group gives

The spherical average must use normalized surface measure without an additional area factor. Let $d\sigma$ be the normalized surface measure on a sphere, and setThe correct radial disintegration of an orthogonally invariant measure isIndeed, take normalized Haar measure $H$ on the compact orthogonal group $O(d)$. Invariance and Fubini's theorem giveRight invariance of Haar measure and transitivity of the orthogonal group on each sphere show that the inner average depends only on $r=|x|$. Averaging it once more over $x=r\theta$, using invariance of surface measure on a sphere and Fubini's theorem, identifies its value as $A\eta (r)$. At $x=0$ the same identity reads $\eta (0)=\eta (0)$. The definition of pushforward measure now proves the boxed formula. This argument also works for invariant probability measures with an atom of a measure at the origin.

For the printed normalization, choose a smooth cutoff function $\eta$ equal to one on a ball containing the support of a measure $\mu$. Its left integral is one, whereas the printed right integral is ${\sigma }_d$. Thus the extra ${\sigma }_d$ cannot be retained. Equivalently, with unnormalized area measure $dS$, the average is ${\sigma }_d^{-1}\int \eta (r\theta )dS(\theta )$.

The reflection-averaging hint also fails in general dimension. For $x=e_1$,The first equality uses $\int {\omega }_i{\omega }_{j}d\sigma ={\delta }_{ij}/d$, which follows from coordinate reflections, coordinate permutations, and ${\sum }_i{\omega }_i^2=1$. Taking a test function equal to the first coordinate near the unit sphere disproves the hint when $d\ne 2$. Averaging over the full orthogonal group, as above, proves the required corrected result without that hint.

Let $\alpha ={\Phi }_{\#}{\mu }_0={\Phi }_{\#}{\mu }_1$. The corrected radial disintegration of an orthogonally invariant measure, whose proof did not require absolute continuity of measures, gives for every test function $\eta$Finite Borel measures on Euclidean space are determined by their integrals against test functions: the difference defines the zero distribution, or one can approximate continuous compactly supported functions uniformly by smooth functions and use regularity of the measures. Therefore

Write $d\mu =f(x)dx$, with $f\ge 0$. Applying polar coordinates and Fubini's theorem gives, for bounded nonnegative measurable $h$,Also $\tilde{\mu }(\{0\})=\mu (\{0\})=0$ by absolute continuity of measures. Extending $g$ by zero on the negative half-line proves

Both radial pushforward measures have compact support of a measure, so Brenier's theorem on the line supplies a unique optimal transport plan and a transport map $\tau$, unique $\tilde{\mu }$-almost everywhere, withIn one dimension, the gradient of a convex function is a nondecreasing function. Nonnegativity holds $\tilde{\mu }$-almost everywhere because ${\tau }_{\#}\tilde{\mu }=\tilde{\nu }$ is carried by $[0,\infty )$. To choose a nonnegative, nondecreasing representative everywhere, let $F,G$ be the radial cumulative distribution functions and use the monotone rearrangementAt $F(r)=0$ set $\tau (r)=0$, and at $F(r)=1$ use the largest point of the support of a measure $\tilde{\nu }$. These choices preserve monotonicity and boundedness. Since $F$ is continuous, $F(R)$ has the uniform distribution on $(0,1)$ when $R$ has law $\tilde{\mu }$, and the quantile function $G^{-1}$ sends this uniform distribution to $\tilde{\nu }$. Thus this representative has the required pushforward measure, even if the target has atoms.

Choose the bounded, nonnegative nondecreasing function $\tau$ from part (d), and defineThe function $h$ is convex and nondecreasing. For $0\le a\le 1$, the triangle inequality, monotonicity, and convexity giveHence $U$ is a convex function. By the gradient of a radial function formula and the fundamental theorem of calculus,for $x\ne 0$ whenever $h^{\prime }(|x|)=\tau (|x|)$. A monotone function has at most countably many discontinuities, so part (d) makes the exceptional radii $\tilde{\mu }$-null. The origin is also $\mu$-null. Define $\mathcal{T}(0)=0$; thus $\mathcal{T}=\nabla U$ $\mu$-almost everywhere.

For every $Q$ in the orthogonal group, $\mathcal{T}(Qx)=Q\mathcal{T}(x)$, including at zero. Consequently ${\mathcal{T}}_{\#}\mu$ is invariant under the orthogonal group. Moreover $|\mathcal{T}(x)|=\tau (|x|)$ away from the null origin, and composition of pushforward measures givesPart (c) therefore gives ${\mathcal{T}}_{\#}\mu =\nu$. The Knott–Smith optimality criterion makes a transport map given by a convex function's gradient optimal for the quadratic transport cost. Uniqueness from Brenier's theorem yieldsSince directions agree, this also identifies the optimal value: