Page 97 - Complementarity and Variational Inequalities in Electronics

88 Complementarity and Variational Inequalities in Electronics

Noting that ∞ (e) < +∞ since e ∈ D( ∞ ), we may divide this last relation
by ||u i || to get
1 q 1
z i ,e + Mz i + ,e + ∞ (e) ≥ 0.
i ||u i || ||u i ||
Taking the limit as i →+∞, we get Mz,e ≥ 0. This holds for any e ∈
D( ∞ ), and thus

z ∈ K(M, ). (4.44)
Setting now v = x 0 in (4.42), we obtain
1 2 1
||u i || + Mu i ,u i ≤ ( I + M)u i ,x 0 + q,x 0 − u i + (x 0 ) − (u i ).
i i
(4.45)

The function is proper, convex, and lower semicontinuous, and thus (see e.g.
Theorem 1.1.11 in [46]) there exist a ≥ 0 and b ∈ R such that
n
(∀x ∈ R ) : (x) ≥−a||x|| + b.
Thus

1
Mu i ,u i ≤ a||u i || − b + ( I + M)u i ,x 0 + q,x 0 − u i + (x 0 ).
i
2
Dividing this last relation by ||u i || , we get
a b 1 x 0
Mz i ,z i ≤ − + ( I + M)z i ,
||u i || ||u i || 2 i ||u i ||
q x 0 (x 0 )
+ , − z i + .
||u i || ||u i || ||u i || 2
Taking the limit as i →+∞, we get Mz,z ≤ 0. Thus

z ∈ N − (M). (4.46)
Using (4.43), (4.44), and (4.46), we obtain that

z ∈ B(M, ).

We get a contradiction since we have proved that z
= 0 and z ∈ B(M, ).
The sequence {u i } is thus bounded. There exists a converging subsequence,
again denoted {u i }.Let

u = lim u i .
i→+∞