30 1. The Kantorovich Duality If we replace (ϕ0, ψ0) by (ϕ0 + s, ψ0 s) for some real number s, we do not change the value of J0(ϕ0, ψ0) and the resulting couple is still admissible. By a proper choice of s, we can ensure ϕ0(x0) 1 2 , ψ0(y0) 1 2 . This implies that, for all (x, y) X0 × Y0, ϕ0(x) c(x, y0) ψ0(y0) c(x, y0) + 1 2 , ψ0(y) c(x0,y) ϕ0(x0) c(x0,y) + 1 2 . To go further we shall use a key trick for “improving” admissible pairs of functions, which was popularized by R¨uschendorf. It will be encountered again several times in this book. Define, for x X, ϕ0(x) = inf y∈Y0 [c(x, y) ψ0(y)]. From the inequality ϕ0(x) c(x, y) ψ0(y) we see that ϕ0 ϕ0 on X0. This implies J0(ϕ0, ψ0) J0(ϕ0, ψ0). Moreover, for all x X we have a control of ϕ0(x) from above and below, in terms of the cost function: ϕ0(x) inf y∈Y0 [c(x, y) c(x0,y)] 1 2 , ϕ0(x) c(x, y0) ψ0(y0) c(x, y0) + 1 2 . Finally we define, for y Y , ψ0(y) = inf x∈X [c(x, y) ϕ0(x)], and we still have (ϕ0, ψ0) Φc. Then, it is straightforward to check that J0(ϕ0, ψ0) J0(ϕ0, ψ0) J0(ϕ0, ψ0). Moreover, for all y Y , ψ0(y) inf x∈X [c(x, y) c(x, y0)] 1 2 , ψ0(y) c(x0,y) ϕ0(x0) c(x0,y) ϕ0(x0) c(x0,y) + 1 2 . In particular, ϕ0(x) −c 1 2 , ψ0(y) −c 1 2 . Once we have these bounds, we are almost done! Indeed, J(ϕ0, ψ0) = X ϕ0 + Y ψ0 = X×Y ϕ0(x) + ψ0(y) dπ∗(x, y)
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