Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 7 4 Solution Created 2026-10-03 Updated 2026-10-07
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 Arzela-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.