28 1. The Kantorovich Duality Exercise 1.11. Let us try to extend this proof to the non-compact case. Why would we like to replace Cb(X × Y ) by C0(X × Y )? Show that if we do so in the definition of Ξ, then the latter turns out to be trivial: Ξ ≡ 0. This shows that the proof as such does not work in a non-compact setting. Still, a variation of it can be used, as we shall see later in Appendix 1.3. 2. We shall now relax the assumption of compactness this step is the most technical. For the moment we keep the assumption that c is bounded and uniformly continuous. We define c ∞ = sup X×Y c(x, y). We will reduce to the compact case by a careful truncation procedure. First of all, let π∗ be an optimal transference plan in the Kantorovich prob- lem, in the sense that I[π∗] = inf π∈Π(μ,ν) I[π]. Since c is bounded, the infimum is obviously finite. The existence of π∗ follows by the compactness of Π(μ, ν), by an argument which will be detailed in Step 3. We do not give the argument here because in fact it is not essential at this point of the proof: one could work with approximate minimizers as well. Let δ 0 be arbitrarily small. Since X and Y are Polish, so is X × Y (easy exercise). In particular, π∗ is tight, and there exist compact sets X0 ⊂ X, Y0 ⊂ Y such that (1.12) μ[X \ X0] ≤ δ, ν[Y \ Y0] ≤ δ. Note that, as a consequence of (1.12) (exercise), π∗[(X × Y ) \ (X0 × Y0)] ≤ 2δ. Define π∗0 = 1X0×Y0 π∗[X0 × Y0] π∗ note that it is a probability measure on X0 × Y0, and let μ0, ν0 be the marginals of π∗0 onto X0, Y0 respectively. We naturally define Π0(μ0,ν0) as the set of probability measures π0 on X0 × Y0 with marginals μ0,ν0, and we define I0 on Π0(μ0,ν0) by I0[π0] = X0×Y0 c(x, y) dπ0(x, y). Let π0 ∈ Π0(μ0,ν0) be such that I0[π0] = inf π0∈Π0(μ0,ν0) I0[π0].

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