Page 96 - Complementarity and Variational Inequalities in Electronics

A Variational Inequality Theory Chapter | 4 87

Theorem 5. If (M, ) ∈ Q0 n and B(M, ) ={0}, then

n
R(M, ) = R .
Proof. There exists λ 0 > 0 such that
n
(∀ 0 <λ ≤ λ 0 ) : R(λI + M, ) = R . (4.41)
1 n
Therefore, for all i ∈ N,i ≥ , there exists u i ∈ R such that
λ 0
1 n
( I + M)u i + q,v − u i + (v) − (u i ) ≥ 0, ∀v ∈ R . (4.42)
i
We claim that the sequence {u i }≡{u i ;i ∈ N\{0}} is bounded. Suppose on the
contrary that ||u i || → +∞ as i →+∞. Then, for i large enough, ||u i ||
= 0,
and we may set
u i
z i = .
||u i ||
There exists a subsequence, again denoted by {z i }, such that lim z i = z with
i→+∞
||z|| = 1.
It is clear from (4.42) that

u i ∈ D( ) (i ∈ N,i
= 0).

Let x 0 ∈ D( ) be any element in the domain of .Let α> 0. For i large
enough, α < 1, and thus
||u i ||
α α
u i + (1 − )x 0 ∈ D( )
||u i || ||u i ||
since D( ) is convex and x 0 ∈ D( ). Taking the limit as i →+∞, we get
αz + x 0 ∈ D( ) since D( ) is assumed to be closed. This result holds for any
α> 0, and thus
1
z ∈ (D( ) − x 0 ) = D( ) ∞ . (4.43)
α
α>0
Let e ∈ D( ∞ ). Then e ∈ D( ) ∞ and u i + e ∈ D( ).Wemay set v = u i + e
in (4.42) to get
1
( I + M)u i + q,e + (u i + e) − (u i ) ≥ 0,
i
and thus we obtain
1
u i ,e + Mu i + q,e + ∞ (e) ≥ 0.
i