By Johann Schroder

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1t for all t =p 0 in R*. 9. Suppose that A c R is a segmentally closed convex set with A =p 0 and denote by 2 the set ofallfunctionals t =p 0 in R* which support A at some v E oA. Then 2 describes A completely. 1t for all t E 2, leaving the remaining proof to the reader. Suppose the above inequalities hold for all t E 2, but u ¢ A. Due to our assumptions there exist a z E AC and atE (0, 1) such that v := (1 - t)u + tz E oA. 1t (cf. 8a). 1t, which contradicts the definition of t. 2. The set A = {u: U1UZ ~ 1, U1 > O} C [Rz is completely described by the set 2 of all functionals tu = (11U1 + (1zUz with (11 > 0, (1z > O.

5 Suppose that for M E IR n , " there exists a vector Z >- 0 with M Z >- o. Then M is inverse-positive if and only if Ilull z ~ IIMullMz for all u E IR". 6 If A, B, C E IR n , " such that A and C are inverse-positive and A ~ B ~ C, then B is inverse-positive. 7 If A, BE IR"'" are inverse-positive and A ~ B, then the matrices A + B, B- 1 A, AB- 1 are inverse-positive, and B- 1 A ~ f. 8 Suppose that for a given M E IR n ," with n > 1 there exists a vector Z > 0 with M Z ~ 0, but that for any matrix 1\1 which is obtained from M by omitting at least one column no vector' > 0 with 1\1, ~ 0 exists.

11. (i) u E K if and only if u + AW ~ 0 for some H' ~ 0 and all A> O. (ii) K is a wedge. ) (iv) Suppose (R, ~) contains an element z definite if and only if K is a cone. = u = o. 4) >- o. 3(i). (ii) Suppose that u, v E K. Then there exist x ~ 0, y ~ 0 such that u + AX ~ 0 and v + Ay ~ 0 for all A > 0; consequently, (Xu + [3v + A«(XX + [3y) ~ 0 for all A > 0, (X ~ 0, [3 ~ O. Since (Xx + [3y ~ 0, we conclude that (Xu + [3v E K for all (X ~ 0, [3 ~ O. Therefore, K is a wedge. 4) holds for some u, v E R.