Past exam of the mathematics course of the University of Cambridge 2007 iii Paper 9 5 iv Solution 2026-10-14
Every function with Lipschitz constant at most yields a feasible pair , and replacing by reverses the objective. HenceConversely, the Lipschitz regularization of transport potentials in part iii dominates every feasible pair by with , so . Together with the preceding parts this proves the Kantorovich–Rubinstein theorem here in all four requested forms:The seminorm convention is essential. If instead meant , the equality would fail: on a two-point space at distance , with point masses at the two different points, whereas the latter unit-ball supremum is . Indeed, an oscillation needs supremum norm at least and Lipschitz constant , so the sum norm bounds by , attained by values . This distinguishes the Lipschitz seminorm used above from a full bounded-Lipschitz norm.
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 7 4 Solution Created 2026-10-03 Updated 2026-10-14
We prove Kantorovich duality theorem here by a positive-functional extension argument; it produces the minimizing transport plan at the same time as equality of the values. Let and . This is a bounded continuous function because is a compact metric space. Write for the real space of continuous functions on a compact space, and letOn this vector subspace define . It is well-defined: two representations differ by a constant in the first variable and the opposite constant in the second, and both measures have mass one. It is positive, since implies and hence . Also .
Define the majorant envelope, which is a sublinear functional:Constants majorize every , and positivity gives , so this quantity is finite. Taking approximate minimizing majorants proves subadditivity; rescaling majorants proves positive homogeneity. Moreover for , and subtracting from a majorant provesThe dual value in the question is exactlySubadditivity at gives .
On , assign the value to . If the representation is unique. If , then , so the assignment is still well-defined. It is dominated by : for , use ; for , positive homogeneity gives . Together with the translation identity, these verify domination in both cases.
The real Hahn-Banach theorem extends this functional to with . If , then , so . Thus is a positive linear functional, with and . Positivity also gives , so is continuous. The Riesz-Markov-Kakutani representation theorem supplies a Borel probability measure on the compact space , satisfying .
For every , its pullback belongs to , so . Likewise . Uniqueness in the Riesz-Markov-Kakutani representation theorem shows that these are exactly the two marginal distributions. Finally, every feasible pair gives a lower bound for the cost of every transport plan, by integration. Our constructed plan achieves that bound:This Kantorovich duality by positive extension establishes the requested existence and equality without assuming an optimal plan in advance.
The metric structure additionally gives the Kantorovich–Rubinstein theorem formulation. For a feasible pair define . The triangle inequality for makes one-Lipschitz continuous, , and . Hence the dual objective is at most . Conversely is feasible for every one-Lipschitz . ThereforeNormalizing for a fixed gives a uniformly bounded equicontinuous family; it is closed and compact in the uniform topology by the Arzelà-Ascoli theorem. The objective is uniformly continuous on this family, so the supremum is attained as well. In particular, is the first Wasserstein distance between the two measures.
Past exam of the mathematics course of the University of Cambridge 2019 iii Paper 348 3 c Solution Created 2026-10-03 Updated 2026-10-05
The Kantorovich–Rubinstein theorem concerns a Polish space and probability measures with finite first moments, meaning for one, hence every, . It statesHere means for every . The Wasserstein distance with ground cost therefore equals a supremum over functions with Lipschitz constant at most one. One may normalize , since adding a constant does not change the difference of integrals. This normalization bounds by , so finite first moments ensure integrability. The supremum is unchanged if an absolute value is placed around the difference, because is admissible whenever is.