Page 103 - Complementarity and Variational Inequalities in Electronics

94 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, ).

Using now (4.61) with v = x 0 ,wealsoget

1
( I + M)u i ,u i − x 0 ≤ q,x 0 − u i + (x 0 ) − (u i ).
i
Here
1
( I + M)u i ,u i > 0,∀i ∈ N,i
= 0,
i
and thus
1
−Ð( I + M)u i ,x 0 + q,u i − x 0 − (x 0 ) + (u i )< 0.
i

Dividing this last relation by ||u i ||, we get
1 x 0 (x 0 ) (||u i ||z i )
−Ð( I + M)z i ,x 0 + q,z i − − + < 0.
i ||u i || ||u i || ||u i ||

Taking the limit inferior as i →+∞, we get
(||u i ||z i )
T
q − M x 0 ,z + liminf ≤ 0,
i→+∞ ||u i ||
and we obtain

T
q − M x 0 ,z + ∞ (z) ≤ 0.
This is a contradiction to condition (4.60) since we have already proved that
z ∈ B(M, ) and z
= 0.
The sequence {u i ;i ∈ N\{0}} is thus bounded, and we may conclude as in
the proof of Theorem 5.

Remark 25. Recall that, for a positive semidefinite matrix M ∈ R n×n ,
T
T
T
ker(M) = ker(M ) ⊂ ker(M + M ) = N 0 (M + M ).
Remark 26. i) Let us set
n
R + (M, ∞ ) ={z ∈ R : z,v + ∞ (v) > 0, ∀v ∈ B(M, ),v
= 0}