Page 92 - Complementarity and Variational Inequalities in Electronics

A Variational Inequality Theory Chapter | 4 83

Let e ∈ D( ∞ ). Then from (4.29), e ∈ D( ) ∞ , and thus u i + e ∈ D( ).We
may set v = u i + e in (4.38) to get

(1 − t i )u i + t i (Mu i + q),e + (u i + e) − (u i ) ≥ 0,

and thus
(1 − t i )u i ,e + t i (Mu i + q),e + ∞ (e) ≥ 0.

Note that ∞ (e) < +∞ since e ∈ D( ∞ ), and we may therefore divide this
last relation by ||u i || to get

q 1
(1 − t i )z i ,e + t i Mz i + t i ,e + ∞ (e) ≥ 0.
||u i || ||u i ||
Taking the limit as i →+∞, we get (1 − t)z + tMz,e ≥ 0. This holds for any
e ∈ D( ∞ ), and thus
∗
(1 − t)z + tMz ∈ (D( ∞ )) . (4.40)
Setting now v = x 0 in (4.38), we obtain

2
(1 − t i )||u i || + t i Mu i ,u i ≤ (1 − t i )u i ,x 0 + t i Mu i ,x 0
+ t i q,x 0 − u i + (x 0 ) − (u i ).

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

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