Page 89 - Complementarity and Variational Inequalities in Electronics

80 Complementarity and Variational Inequalities in Electronics

Indeed, if 0 <t ≤ 1, then the result follows from (4.13) since z ∈ D( ) ∞ ,z
= 0,
whereas if t = 0, then the result is trivial since z
= 0. We know that

(1 − t)z + tMz,h ≥ 0,∀h ∈ D( ∞ ).

j
j
Let j ∈{1,...,n}.Using (4.13),wemay set h = z,e e to get
2
(1 − t)z + t(Mz) j z j ≥ 0.
j
This last relation holds for all j ∈{1,...,n}, and since

(1 − t)z + tMz,z ≤ 0,

we get a contradiction:

2 2
0 ≥ (1 − t)z + t(Mz) k z k + (1 − t)z + t(Mz) j z j > 0.
k j
j
=k
c) We prove that PS n ⊂ AC n .Let (M, ) ∈ PS n , t ∈[0,1], and z ∈ B((1 −
t)I + tM, ). We claim that z = 0. Suppose on the contrary that z
= 0. Here
∗
(D( ∞ )) ={0}, and the second relation in (4.27) yields
(1 − t)z + tMz = 0.

If t = 0, then z = 0, a contradiction. If 0 <t ≤ 1, then
(1 − t)
Mz =− z,
t
∗
so that ν =− (1−t) ≤ 0 is a real eigenvalue of M, a contradiction to (4.17).
t
Let us now give some additional properties that will be used later.
n
Proposition 20. Let : R → R be a proper convex lower semicontinuous
function with closed domain, and let M ∈ R n×n .If

(M, ) ∈ PD0 n ∪ P0 n ∪ PS0 n ,

then
(∀λ> 0) : (λI + M, ) ∈ AC n .

Proof. Let λ> 0.
a) If (M, ) ∈ PD0 n , then condition (4.11) on M entails that λI + M satis-
fies condition (4.10), and the result is a consequence of Proposition 19.