46 1. The Kantorovich Duality u = 1 0 1u≥s ds (check!). In particular, whenever (ϕ, ψ) ∈ Φc, one can write (ϕ, ψ) = 1( 0 1ϕ≥s, −1ψ≤−s ) ds. Since ϕ is upper semi-continuous, the set 1ϕ≥s is closed, for all s ∈ [0, 1]. So we only have to check that for all s, 1ϕ≥s(x) − 1ψ≤−s(y) ≤ 1C(x,y). The only nontrivial case is when ϕ(x) ≥ s and ψ(y) −s in this situation (x, y) should belong to C. But this case implies ϕ(x)+ψ(y) 0, in particular c(x, y) ≥ ϕ(x) + ψ(y) 0. Since c takes values in {0, 1}, we deduce that c(x, y) = 1, which indeed means (x, y) ∈ C. The proof is complete.
Purchased from American Mathematical Society for the exclusive use of nofirst nolast (email unknown) Copyright 2017 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.
