Page 91 - Complementarity and Variational Inequalities in Electronics

82 Complementarity and Variational Inequalities in Electronics

Theorem 4. If

(M, ) ∈ AC n ,
n
then, for each q ∈ R , problem VI(M,q, ) has at least one solution.
n
Proof. Let q ∈ R .From (3.5) and (4.2), Problem VI(M,q, ) is equivalent to
the fixed point problem

u = P (u − (Mu + q)).
n
n
Let us now define by H :[0,1]× R → R the continuous homotopy

H(t,u) = P (tu − t(Mu + q)) = P (u − t(Mu + q) + (1 − t)u ).
We claim that there exists R 0 > 0 such that, for all R ≥ R 0 and t ∈[0,1],
n
H(t,u)
= u, ∀u ∈ R ,||u|| = R. (4.37)

Indeed, if we suppose the contrary, then we may find sequences {t i } i∈N ⊂[0,1]
n
and {u i } i∈N ⊂ R satisfying ||u i ||→+∞ and u i = H(t i ,u i ). Then
n
t i (Mu i + q) + (1 − t i )u i ,v − u i + (v) − (u i ) ≥ 0,∀v ∈ R . (4.38)
It is clear from (4.38) that

u i ∈ D( ) (i ∈ N).

Moreover, ||u i ||
= 0for i large enough, and we may set
u i
z i = .
||u i ||
There exists subsequences, again denoted by {t i } and {z i }, such that lim t i =
i→+∞
t ∈[0,1] and lim z i = z with ||z|| = 1.
i→+∞
Let x 0 ∈ D( ) be any element in the domain of .Let λ> 0. For i large
enough, λ < 1, and thus
||u i ||
λ λ
u i + (1 − )x 0 ∈ D( )
||u i || ||u i ||
since the set D( ) is convex.
Recalling that the set D( ) is assumed to be closed and taking the limit as
i →+∞, we get λz + x 0 ∈ D( ). This result holds for any λ> 0, and thus
1

z ∈ (D( ) − x 0 ) = D( ) ∞ . (4.39)
λ
λ>0