Page 56 - Complementarity and Variational Inequalities in Electronics

A Variational Inequality Theory Chapter | 4 47

have
α α
λ n d n + (1 − )x 0 ∈ K,
λ n λ n
provided that n is large enough to have α ≤ 1. We obtain
λ n
α
αd n + (1 − )x 0 → αd + x 0 ∈ K as n →∞
λ n
since K is closed. The parameter α> 0 has been chosen arbitrarily, and thus

(∀ α> 0) : αd + x 0 ∈ K,

which means that d ∈ K ∞ .
Let us now show that K ∞ is a closed convex cone.

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Proposition 5. Let K ⊂ R be a nonempty closed convex set. Then K ∞ is a
closed convex cone.
Proof. 1) The set K ∞ is closed as the intersection of a family of closed sets.
It is also convex as the intersection of a family of convex sets. 2) Let λ> 0. If
1 1
x 0 ∈ K, then 0 ∈ (K − x 0 ), which results in 0 ∈ (K − x 0 ).3)If z ∈
λ λ>0 λ
αK ∞ (α > 0), then z = αd for some d ∈ K ∞ , and thus there exists λ n →+∞,
z λ n
d n → such that λ n d n ∈ K.Ifweset μ n = and z n = αd n then we see that
α α
μ n z n ∈ K,μ n →+∞, and z n → z. This yields z ∈ K ∞ .
Remark 13. If K is a nonempty closed convex cone, then K ∞ = K.
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Proposition 6. Let K be a nonempty closed convex subset of R . Then (i) If
x ∈ K and e ∈ K ∞ , then x + e ∈ K; (ii) If K is bounded, then K ∞ ={0}; and
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(iii) (∀z 0 ∈ R ) : (K − z 0 ) ∞ = K ∞ .
Proof. i) This property is a consequence of the relation

1
K ∞ = (K − x) ⊂ K − x.
λ
λ>0
ii) Suppose by contradiction that d
= 0 ∈ K ∞ . Then there exist sequences
λ n →+∞ and d n → d such that λ n d n ∈ K. However, λ n d n →+∞, which
is a contradiction since K is bounded.
iii) If z ∈ (K − z 0 ) ∞ , then there exist λ n →+∞ and z n → z such that
z 0
λ n z n ∈ K −z 0 . Set w n = z n + .Wehave λ n w n = λ n z n +z 0 ∈ K and w n → z.
λ n
This means that z ∈ K ∞ . Conversely, if z ∈ K ∞ , then there exist λ n →+∞ and
z 0
z n → z such that λ n z n ∈ K.Ifweset w n = z n − , then we have λ n w n =
λ n
λ n z n − z 0 ∈ K − z 0 and w n → z, so that z ∈ (K − z 0 ) ∞ .