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 dμ + Y ψ0 dν = X×Y ϕ0(x) + ψ0(y) dπ∗(x, y)

Purchased from American Mathematical Society for the exclusive use of nofirst nolast (email unknown) Copyright 2003 American Mathematical Society. Duplication prohibited. Please report unauthorized use to cust-serv@ams.org. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.