Page 123 - Complementarity and Variational Inequalities in Electronics

114 Complementarity and Variational Inequalities in Electronics

since the set D( ) is convex and 0 ∈ D( ). Recalling that the set D( ) is
assumed to be closed and taking the limit as i →+∞, we get λz ∈ D( ).This
result holds for any λ> 0, and thus

1
z ∈ D( ) = D( ) ∞ . (4.85)
λ
λ>0
Setting now v = 0in (4.84), we obtain

1 2
||u i || + F(u i ),u i + q,u i ≤ (0) − (u i ).
i
Therefore
F(u i ),u i + q,u i − (0) + (u i ) ≤ 0.
Dividing this last relation by ||u i ||, we get

(0) (||u i ||z i )
F(||u i ||z i ),z i + q,z i − + ≤ 0.
||u i || ||u i ||
Taking the limit inferior as i →+∞, we get

(||u i ||z i )
liminf F(||u i ||z i ),z i + q,z + liminf ≤ 0.
i→+∞ i→+∞ ||u i ||
Therefore
r (z) + q,z + ∞ (z) ≤ 0
F
for some z ∈ D( ) ∞ ,z
= 0. This is a contradiction to our assumption.
The sequence {u i } is thus bounded. There exists a converging subsequence,
again denoted {u i }. Let us set

u = lim u i .
i→+∞
n
Let v ∈ R .Wehave
1
( u i + F(u i ) + q,u i − v − (v) + (u i ) ≤ 0.
i
Taking the limit inferior as i →+∞ and using the lower semicontinuity of ,
we obtain

F(u) + q,u − v − (v) + (u) ≤ 0. (4.86)
n
The vector v has been chosen arbitrarily in R , and thus the result in (4.86)
n
holds for all v ∈ R . The existence result follows.