Appendix ETopological vector spaces1. Locally convex topological vector spacesA vector space X equipped with a topology is said to be a topological vectorspace (TVS) if(i) (x 1 , x 2 ) ↦→ x 1 + x + 2 is continuous as a map from X ⊗ X into X(ii) (λ, x) ↦→ λx is continuous as a map from R ⊗ X into XA TVS is said to be locally convex if it is Hausdorff and if every neighborhoodof 0 contains a convex neighborhood of 0. Abbreviate “locally convextopological vector space” to lcTVS. Example. A TVS whose topology is defined by a norm is a lcTVS. The□ balls centered at the origin are convex. Example. Let X be a vector space and Ɣ be a collection of linear functionsfrom X into R that separate points (that is, if x and y are distinct points of Xthen f (x) ≠ f (y) for at least one f in Ɣ). Then the weakest topology on Xmaking all functions in Ɣ continuous makes X a lcTVS. The sets{x ∈ X : | f i (x)| sup λxx∈FIn particular, X ∗ separates points of X.If f is a convex, lower semi-continuous map from X into R ∪ {∞}, theepigraphepi( f ) ={(x, t) ∈ X ⊗ R : t ≥ f (x)}is a closed, convex subset of the lcTVS X ⊗ R. The hyperplanes that separateepi( f ) from points (x, s) with s < f (x) define a collection of continuous linearfunctionals on X whose pointwise supremum equals f . That is, f (x) = sup{x ∗ (x) : f ≥ x ∗ on X}Asymptopia: 4Jan99 c○David Pollard 1
Section E.1Locally convex topological vector spacesBarycenterLet K be a compact, convex subset of a lcTVS X. Let λ be a probabilitymeasure on the Borel sigma-field of K . The barycenter of λ is the point x 0in K for whichλx ∗ = x ∗ (x 0 ) for all x ∗ in X ∗If X = R the barycenter is just the mean, λx.Uniqueness of the barycenter in general follows from the fact that X ∗separates points of X. Existence follows from the separating hyerplane propertyfor Euclidean space. For each finite subset D of X ∗ defineK D ={x ∈ K : x ∗ (x) = λ(x ∗ ) for all x ∗ ∈ D}Each K D is closed, and hence compact. If each K D were nonempty then theintersection of all such K D would be nonempty. A point in the intersectionwould be a barycenter. (So the intersection can contain at most one point.)Suppose D = x1 ∗,...,x k ∗. To prove that K D ≠∅,define a continuouslinear map ψ from X into R k by x ↦→ (x1 ∗(x),...,x k ∗ (x)). The image ψ(K ) iscompact and convex. For K D to be nonempty we need ψ(K ) to contain the pointL = (λx1 ∗,...,λx k ∗). If it didn’t, there would exist a vector α = (a 1,...,α k )such that α ′ L < inf x∈K α ′ ψ(x). The linear functional x ∗ = ∑ i α i xi∗ wouldthen have the property λ(x ∗ ) = α ′ L < inf x ∗ (x),x∈Kan inequality impossible for a probability measure. (Proof borrowed fromAlfsen 1971, page 11).Notice that we didn’t really need to identify λ as a probability measure.It would suffice if λ were an increasing linear functional on the space C(K ) ofall continuous real functions on K , with λ1 = 1. We could then interpret λx ∗as the result of applying λ to the continuous function obtained by restricting x ∗to K . Of course the Riesz Representation Theorem ensures that λ correspondsto a Borel probability measure on K , but one doesn’t need to invoke thatrepresentation in order to prove existence of the barycenter. It was enough thatλ be an increasing, continuous linear functional on cc(K ) with λ(1) = 1.2. Minimax theoremNotation: C(K ) denotes the set of all bounded, continuous real functions on atopological space K , equipped with the norm for uniform convergence. It is alcTVS. Write C + (K ) for the set of nonnegative functions in C(K ). Theorem. Let f be a map from K ⊗C into R∪{∞}, where K is a compact,convex subset of a lcTVS X and C is a convex subset of a vector space Y.Suppose:(i) the map x ↦→ f (x, y) is convex and lower semi-continuous on K ;(ii) the map y ↦→ f (x, y) is concave on C.Then infx∈K supy∈Cf (x, y) = sup infy∈C x∈Kf (x, y)Proof. Clearly the left-hand side of equality is no smaller than theright-hand side. Also, if the right-hand side is infinite the equality holds fortrivial reasons.For convenience of notation define f (x, y) =∞for all x in K c .2 Asymptopia: 4Jan99 c○David Pollard