22 1. The Kantorovich Duality “inf sup” by a “sup inf”. For readers who are not familiar with this method, it is certainly more important to understand the formal proof, than the rigorous one. So let us write (1.7) inf π∈Π(μ,ν) I[π] = inf π∈M+(X×Y ) I[π] + 0 if π ∈ Π(μ, ν) +∞ else . Here M+(X ×Y ) denotes the space of nonnegative Borel measures on X ×Y . The function appearing inside brackets in the right-hand side of (1.7) is sometimes called the indicator function of Π. And since the constraints defining Π are linear, we can write this indicator function as the solution of a supremum problem involving only linear functionals: it is easy to check that 0 if π ∈ Π(μ, ν) +∞ else = sup (ϕ,ψ) ϕ dμ + ψ dν − [ϕ(x) + ψ(y)] dπ(x, y) , where the supremum runs over all (ϕ, ψ) ∈ Cb(X) × Cb(Y ), for instance. Thus the left-hand side of (1.4) is given by inf π∈M+(X×Y ) sup (ϕ,ψ) X×Y c(x, y) dπ(x, y) + X ϕ dμ + Y ψ dν − X×Y [ϕ(x) + ψ(y)] dπ(x, y) . Taking for granted that a minimax principle can be invoked, we rewrite this as sup (ϕ,ψ) inf π∈M+(X×Y ) X×Y c(x, y) dπ(x, y) + X ϕ dμ + Y ψ dν − X×Y [ϕ(x) + ψ(y)] dπ(x, y) (1.8) = sup (ϕ,ψ) X ϕ dμ + Y ψ dν − sup π∈M+(X×Y ) X×Y [ϕ(x) + ψ(y) − c(x, y)] dπ(x, y) . Let us compute the supremum inside the curly brackets. If the function ζ(x, y) ≡ ϕ(x) + ψ(y) − c(x, y) takes a positive value at some point (x0,y0), then by choosing π = λδ(x0,y0) and letting λ → +∞ (a Dirac mass at point (x0,y0) with very large mass), we see that the supremum is infinite. On

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