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By G. Chilov

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If T[E Ao (Ce. ), let (k) = fx GI* x if k0. -(1-(k),01 { o3 if k=0. S ince X * is positively generated, I(k)4 0. It is clear that '(k) is closed, and convex. W e show that 1? 7 willlensure that there is an affine selection /0 of f. It is clear that this tp satisfies 40o/IT,O, thus completing the proof. Suppose x 1§(k),x 1 E Cie), then :e;,;1T(k),0 and xn;TT(k'),O. It will immediately follow that x + )0x' Tr k +( l-A )kl If k (1 X) k' #0, this shows that 52 AI(k) (1 --)n )(k') C= 1(X k (1 -4)1e).

Because all of our special spaces are positively generated, there can be few cases in which L(X,Y) is one of these. The converse of this result is also known. 1. (ELLIS). If L(X,Y) is order unit normed, then X , is base—normed and Y is order unit normed. Proof. Let B be the closed set fxe X t II x II =3 • If f E X * , yE. Y let (f wy)(x)= f(x)y, so that II f ey = It f II II Y Let E be the order unit defining the norm in L(X,Y), so that u n til y inf inf f N : —XE fOy f : \E(x) f(x)y NE(x),\/x e si , 39 If be B, and y II =1 0 there is f e X * with II f f(b)=1.

E. x that Y. is closed, we ( for some and noting y+ fes Letting A( see that y>,x. Hence E y is an upper bound for C. On the other hand, if z is any other upper bound for C, y = sup (C). supremum of then z yk . Hence z )/ sup Note C on fykl 3r, so that here that from the proof, y is the point-wise 'be/C. < t and the class of all finite subsets of choose k(1),•••,k(r)1 e A f 1, . If A e , C2 so that: x eA,]xk(j),(14j‘r), with xeS(xk(j),1/n). n A S(xk(j)11/n) for 1‘,jr. It follows that for 1‘. j. r, x x -t- (1/n)e k(j) v i ) with x(J)/ for some x 6A.

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