I have been unable to solve the following problems from Kechris.  Any hints, advice, or solutions would be appreciated.

1. Exercise 4.9: Let X, Y be separable Banach spaces.  The weak topology on L(X,Y) is the one generated by the functions (from L(X,Y) into the scalar field) T \mapsto \langle Tx, y^* \rangle ; x \in X, y^* \in Y^*.  Show that if Y is reflexive, L_1(X,Y) = \{ T \in L(X,Y) : \| T \| \le 1 \} with the weak topology is compact metrizable.  Find a compatible metric.

2. Exercise 4.10:

A topological vector space is a vector space X (over \mathbb{K}, where \mathbb{K} = \mathbb{R} \text{ or } \mathbb{C}), equipped with a topology in which addition and scalar multiplication are continuous.  So Banach spaces and their duals with the weak*-topology are topological vector spaces. A point x in a convex set K is extreme (in K) if x = \lambda y + (1 - \lambda)z, with 0 < \lambda < 1, y,z \in K, implies y = z = x.  Denote by \partial_e K the extreme boundary of K, i.e., the set of extreme points of K.

Show that if K is a compact metrizable (in the relative topology) convex subset of a topological vector space, the the set \partial_e K is G_\delta in K.

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2 Responses to Questions

  1. Jindrich Zapletal says:

    I would suggest this: if L, M are disjoint compact subsets of K and e>0 is a number then the set A of all linear combinations of one point in L and another point in M, with the combination coefficients \geq e and \leq 1-e is compact too (continuous image of L times M times the interval). A is also disjoint from the extreme boundary, and the extreme boundary is just K setminus all sets A obtained in this way. Now observe that to get this equality it is enough to consider e rational and L,M of some fairly special form so that there are only countably many such pairs. I think any disjoint L,M can be increased to larger disjoint sets each of which is a complement of a finite union of basic open sets by compactness.

  2. alanmath says:

    Thank you, Dr. Zapletal. This solves the problem. To get countably many L,M, we could just take a countable dense subset P of K (which is Polish since it’s compact), and take all closed balls around each point of P of rational radii. Since the balls are closed, they are compact, and then to get the sets A, we could just consider disjoint pairs of these balls. Since if we have two distinct points, then there are two disjoint balls separating them, this should give countably many sets A which together equal the complement of \partial_e K. Thanks!

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