摘要
A "purely geometric" proof of the Lindenstrauss-Troyanski result ([2], [6]) on strongly exposed points of weakly compact sets in Banach spaces is given.I would like to thank the referee for his suggestions.We consider a Banach-space X, \\ ||.If £ G X and if e > 0, then B(£,e) = {x G X; \\x -¿|| < e}.The closed convex hull of a set A G X, is denoted c(A).Let C be a convex subset of X, then Ce is the set of the extremal points of C. If F is a convex, weakly compact, F a convex, closed, bounded set of X and J a closed subinterval of [0,1 ], (K, B,J) denotes the closed, convex setWe have to introduce a few geometrical definitions, referring to [4].Suppose that C is a nonempty, bounded, closed and convex subset of X.Let"strongly exposed" if there exists / G X* (\\f\\ = 1) such that Ve > 0, 3a > 0 with £ G S(fa, C) G B(íe).Let S be the set of all/ G X* such that 11/11 = 1 and/strongly exposes some point of C. Proposition I.If K is a nonempty, convex, weakly-compact subset of X and if B is convex, closed and bounded with K fl B = 0, then the set D = c(K U F) = (K,B, [0,1]) has the following property: Ve > 0, 3£ G K such that £ £ c(D\B(£,e)).First we observe that it is sufficient to prove the proposition for X separable.Indeed, suppose D does not have the required property.Then there exists e > 0 with Vx G K: x G c(D\B(x,e)) and hence, Vx G K, 3A(x) G D with the properties: x G c(A(x)), A(x) n B(x, e) = 0 and A(x) is countable.By induction we construct a sequence (Kn,Bn)n where Kn is countable in K and Bn is countable in F. Let (K0,B0) = ({x},0), where x is some element of K. Suppose we already found (Kn,Bn).Consider Vy G UxeKnA(x) an element ky in K and by in B, with y G c({ky,by)).Let