Page 117 - Complementarity and Variational Inequalities in Electronics

108 Complementarity and Variational Inequalities in Electronics

u i
and we may set z i = . There exists a subsequence, again denoted by {z i },
||u i ||
such that lim z i = z with ||z|| = 1.
i→+∞
As in the proof of Theorem 5, we check that
z ∈ B(M,K).

Set
θ = max{ αα ;1 ≤ α ≤ n}.
j
1
Here, u i,j = e ,u i ∈ K j , (1 − jj ) ≥ 0 for all j ∈{1,...,n}, and thus
θ
1
(1 − jj )u i,j ∈ K j .
θ
Therefore
1
(I − )u i ∈ K.
θ
j
Moreover, it is also clear that 1 jj x 0,j ∈ K j (x 0,j = e ,x 0 ) for all j ∈
θ
{1,...,n}, and thus
1
x 0 ∈ K.
θ
1
1
Using now (4.79) with v = x 0 + (I − )u i ∈ K, we get
θ θ
1 1 1 1 1
( I + M)u i , u i − x 0 ≤ q, x 0 − u i .
i θ θ θ θ
Here
1 1
( I + M)u i , u i > 0,∀i ∈ N,i
= 0,
i θ
and thus
1
−Ð( I + M)u i , x 0 + q, u i − x 0 < 0.
i
Dividing this last relation by ||u i ||, we obtain

1 x 0
−Ð( I + M)z i , x 0 + q, z i − < 0.
i ||u i ||
Taking the limit inferior as i →+∞, we get
−ÐMz, x 0 + q, z ≤ 0,

that is,
T
q − M x 0 ,z ≤ 0.