1.3. Appendix: A duality argument in Cb(X × Y ) 39 u(t, x). One can imagine that the effort accomplished by this ant within an infinitesimal time dt will be proportional to the mass carried and to the modulus of the velocity of the ant so that effort can be modelled by the norm of the vector field σ = ρu. Summing over all ants, we obtain that the total effort should be |σ| (which does not depend on time). On the other hand, the divergence of σ coincides with the negative of the time-derivative of the heap (this may seem not intuitive to many readers, but should become so with the help of Chapters 5 and 8). From our assumptions, this is just μ − ν, so one should have ∇ · σ = μ − ν. 1.3. Appendix: A duality argument in Cb(X × Y ) In Section 1.1 we have used the Fenchel-Rockafellar duality theorem only in a particular case, when the underlying spaces X and Y were compact the rest of the proof of Theorem 1.3 was done by approximation. What happens if we try to directly apply the duality theorem in a non-compact case? This will not lead us to any improvement of Theorem 1.3, but will be rather instructive. To work out the method, we shall restrict the generality of the assumptions a little bit. Certainly variants of the proof would work in a more general setting (as suggested by some of the arguments in [211]), but since we already gave a general enough statement in Theorem 1.3, we shall not look for such extensions. Proposition 1.22 (Kantorovich duality again). Let X and Y be locally compact Polish spaces, let c be a lower semi-continuous nonnegative func- tion on X × Y , and let μ, ν be two Borel probability measures on X, Y respectively. Then (1.23) inf π∈Π(μ,ν) I[π] = sup (ϕ,ψ)∈Φc J(ϕ, ψ). Let us recall again that a topological space is locally compact if any of its elements admits a compact neighborhood. All the notions which we shall need about such spaces, in particular Urysohn’s extension lemma, can be found in Rudin [217]. The idea of the proof is to copy Step 1 in the proof of Theorem 1.3. We would now like to choose E = C0(X × Y ), because then E∗ would be M(X × Y ). But if we do so, then Ξ becomes trivial: as noticed in Exercise 1.11, it is identically 0... We therefore choose E = Cb(X × Y ). The tricky point is that if X × Y is non-compact, then E∗ is larger than M(X × Y ). Of course, if ∈ E∗, then acts continuously on the subset C0(X × Y ) of Cb(X × Y ), so there exists a unique π = π( ) ∈ M(X × Y ) such that ∀u ∈ C0(X × Y ), , u = X×Y u dπ.

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