1.3. Appendix: A duality argument in Cb(X × Y ) 43 Proof of Proposition 1.22. Since μ and ν have σ-compact supports by Ulam’s lemma, we may assume without loss of generality that X and Y are σ-compact. We introduce again Θ and Ξ, defined as in Step 1 of the proof of Theo- rem 1.3, on Cb(X × Y ). If we want to apply Theorem 1.9, we immediately run into a problem: we would like to take z0 = 0, but Θ is plainly not continuous at 0 whenever inf c(x, y) = 0 (which is usually the case, since it is natural to assume c(x, x) = 0 if X = Y ). This difficulty is however easily remedied: it suffices to replace c by = c + ε, for some arbitrary ε 0. Indeed, with this modified cost, u ε Θ(u) = 0, so the assumptions of Theorem 1.9 are fulfilled with z0 = 0, and (1.9) holds. Since (as one readily checks) adding ε to the cost function results in adding ε to both sides of (1.4), the result for the modified cost c + ε will imply the result for the original cost c. In the sequel, we write c in place of cε. So we have the duality formula (1.30) sup u∈Cb(X×Y ) [−Θ(u) Ξ(u)] = inf ∈Cb(X×Y )∗ [Θ∗(− ) + Ξ∗( )]. There is no difficulty in evaluating the left-hand side, and we concentrate on the right-hand side. Let us first look at the Legendre transform of Θ. Let Cb(X × Y )∗ we have (1.31) Θ∗(− ) = sup u≥−c −, u = sup u≤c , u . Suppose that is not a nonnegative linear form then, by definition there exists some v Cb(X × Y ) such that v 0 but , v 0. Then, for any λ 0, we have λv c, and by using λv as a trial function, we see that the supremum in (1.31) is +∞, which will not contribute to the infimum in (1.30). We therefore only take into account the case when is nonnegative. Let us now have a look at Ξ∗. It is easy to check again that Ξ∗( ) = ⎪0 if ∀(ϕ, ψ) Cb(X) × Cb(Y ), , ϕ + ψ = X ϕ + Y ψ , +∞ else, with the shorthand + ψ)(x, y) = ϕ(x) + ψ(y). In particular, when is nonnegative, by using Lemma 1.25 we see that Ξ∗( ) is 0 if Π(μ, ν), and +∞ otherwise. One can then conclude as in Step 1 of the proof of Theorem 1.3.
Previous Page Next Page