36 1. The Kantorovich Duality that sup (ϕ,ψ)∈Φd J(ϕ, ψ) = sup X ϕ d(μ − ν) ϕ Lip ≤ 1 , where J(ϕ, ψ) = X ϕ dμ + Y ψ dν. From Remark 1.12 we know that sup (ϕ,ψ)∈Φd J(ϕ, ψ) = sup ϕ∈L1(dμ) J(ϕdd,ϕd), where ϕd(y) ≡ inf x∈X [d(x, y) − ϕ(x)], ϕdd(x) ≡ inf y∈X [d(x, y) − ϕd(y)]. Now, ϕd, being an infimum of 1-Lipschitz functions, bounded from below at some point x0, is 1-Lipschitz. So −ϕd(x) ≤ inf[d(x, y y) − ϕd(y)] ≤ −ϕd(x), where the right inequality follows by the choice x = y in the infimum, and the left inequality by the 1-Lipschitz property. This means that ϕdd = −ϕd, and sup Φc J(ϕ, ψ) ≤ sup ϕ∈L1(dμ) J(ϕdd,ϕd) = sup ϕ∈L1(dμ) J(−ϕd,ϕd) ≤ sup ϕ Lip≤1 J(ϕ, −ϕ) ≤ sup Φc J(ϕ, ψ). So there is equality everywhere, and the result follows. Exercise 1.17 (Total variation formula). Let X be a Polish space. Show that the assumptions of Theorem 1.14 are satisfied with the cost function c(x, y) = 1x=y. Use this to show that whenever μ and ν are Borel probability measures on X, inf π∈Π(μ,ν) π[{x = y}] = sup 0≤f≤1 X f d(μ − ν). Using the decomposition (μ − ν) = (μ − ν)+ − (μ − ν)−, where (μ − ν)± are singular to each other, show that sup 0≤f≤1 X f d(μ − ν) = (μ − ν)+[X] = (μ − ν)−[X] = 1 2 μ − ν TV . Using the regularity of (μ − ν)+, show that in fact (μ − ν)+[X] = sup K (μ − ν)+[K] = sup K (μ[K] − ν[K]), where the suprema are taken over all compact subsets K of X.

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