26
Since B is an open set, from separation theorems for convex sets we can deduce the
existence of I-' E W' and A E lR such that
Now we show that A must be nonnegative. If A be negative, by taking "'(1 = 0, WI = G(u),
W2 E int(C) and '12 = -k, with k a positive integer number, in (3.18) we get
Letting k go to 00 we obtain a contradiction. Therefore A ~ O.
Notice that the inequality (3.18) is strict and hence it is not possible to have
A =11 1-' 11= 0, from which (3.8) follows.
On the other hand, since B = C x (-00,0], we get
(3.19)
Hence, in order to prove (3.9) it suffices to take WI = G(u), W2 = wE C and '11 = '12 = O.
Finally, equality (3.10) can be obtained by taking WI = G(u) + DG(u)(v - u), '11 =
f(u)(v - u), with u E Uad, '12 = 0 and W2 = G(u).
Corollary 3.1 In addition to the assumptions of theorem 2, suppose the following Slater
condition holds:
3va E Uad such that G(u) + DG(u)(va - u) E int(C)
(3.21 )
then A can be taken equal to 1 in (3.8)-{3.10).
Proof- Let us suppose that (3.21) holds but). = O. Then we can deduce
(I-',W - G(u)) < 0, Vw E int(C).
(3.22)
Indeed, if this is not true, by using (3.9) we obtain the existence of Wa E int(C) such that
(3.23)
and then, by using again (3.9) we get
(1-', W + Wa - G(u)) ~ 0, Vw E B.(O),
(3.24)
for f small enough in order to have Wa + B.(O) C int(C). Then we have (1-' , w) ~ 0, Vw E
B.(O), which implies I-' = 0 in contradiction with (3.8).
Therefore (3.22) holds. Now, by taking W = G(u) + DG(u)(va - u) E int(C) in (3.22)
we get
([DG(u)]* 1-' , Va - u} = (1-' , DG(u)( Va - u)} < 0
(3.25)