1.1. General duality 29 From π0 we construct a π Π(μ, ν) in a natural way, by gluing together π0 with a little bit of π∗: π = π∗[X0 × Y0] π0 + 1(X0×Y0)c π∗ (check that π Π(μ, ν)!). So I[π] = π∗[X0 × Y0] I0[π0] + (X0×Y0)c c(x, y) dπ∗(x, y) I0[π0] + 2 c ∞δ = inf I0 + 2 c ∞δ. It follows that inf π∈Π(μ,ν) I[π] inf I0 + 2 c ∞δ. Now introduce the functional J0(ϕ0,ψ0) = X0 ϕ0 dμ0 + Y0 ψ0 dν0, defined on L1(dμ0)×L1(dν0). By Step 1 of the proof, we know that inf I0 = sup J0, where the supremum runs over all admissible couples (ϕ0,ψ0) L1(dμ0) × L1(dν0), i.e. those which satisfy ϕ0(x) + ψ0(y) c(x, y) for almost all x, y. In particular, there exist an admissible couple of functions ϕ0, ψ0 such that J0(ϕ0, ψ0) sup J0 δ. Our problem is to construct from (ϕ0, ψ0) a couple (ϕ, ψ) which would be very efficient in the problem of maximization of J(ϕ, ψ). It will be useful to ensure that the inequality ϕ0(x) + ψ0(y) c(x, y) is valid for all x and y, not just almost all. We can ensure this, provided that we allow ϕ0 and ψ0 to take values in R {−∞}. Indeed, we can introduce negligible sets Nx and Ny such that the inequality holds true for (x, y) Nx c × Ny c, and redefine the values of ϕ, ψ to be −∞ on Nx, Ny respectively. In a first step, we will control ϕ0, ψ0 from below at some point in X ×Y . Without loss of generality, we assume that δ 1. Since J0(0, 0) = 0, we know that sup J0 0, and hence J0(ϕ0, ψ0) −δ −1. By writing J0(ϕ0, ψ0) = X×Y [ϕ0(x) + ψ0(y)] dπ0(x, y), where π0 is any element of Π0(μ0,ν0), we deduce that there exists (x0,y0) X0 × Y0 such that (1.13) ϕ0(x0) + ψ0(y0) −1.
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