![]() 2)Įvery point of K is the barycente of a unique maximal law. The cone C is a lattice for its own order ≤. The two following properties are equivalent 1) ![]() Recall that each point of K is the barycentre of at least one maximal measure for the order ⊣ λ (41), and that the maximal measures are exactly those carried by the boundary when K is metrizable (43). Here is Choquet's uniqueness theorem (the proof says a little more than the statement). On the other hand, as E is locally convex and K compact, there exist coverings of the preceding type finer than any given covering, and the corresponding measures λ are arbitrarily close to μ for narrow convergence. The measure with finite support Σ it iε b(μ i) = λ thus admits the same barycentre as μ. To establish this consider a finite covering of K by convex open sets ω i it is easy to construct a representation of the law μ as a convex combination μ = σ it iμ i, where for each i the law μ i is carried by ω i. It thus suffices to be able to approximate each law μ, in the sense of vague convergence, by discrete measures with the same barycentre. We know moreover (40) thatį ˆ ( x ) = s u p μ ( f ), μ running over set of all the laws with barycentre x. Thanks to the relation between decompositions and barycentric representations, this formula can be writtenį ˆ ( x ) = s u p μ μ ( f ), μ running over the set of probability laws on K with barycentre x, and with finite support. (63.1) f ˆ ( x ) = sup ∑ i f ( y i ) the sup being taken over all the finite decompositions × = Σ iy i, y i ∈ C.
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |