Page 63 - Complementarity and Variational Inequalities in Electronics

54 Complementarity and Variational Inequalities in Electronics

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Proposition 9. Let ∈ 0 (R ;R ∪{+∞}). Then
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(∀x ∈ D( ),e ∈ R ) : ∞ (e) ≥ (x + e) − (x).
Proof. We have
(x + λe) − (x)
∞ (e) = sup ≥ (x + e) − (x).
λ>0 λ
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Proposition 10. Let ∈ 0 (R ;R ∪{+∞}).If is bounded from below, then
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(∀x ∈ R ) : ∞ (x) ≥ 0.
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Proof. Let x 0 ∈ D( ).If (∀x ∈ R ) : (x) ≥ c, then
(tx + x 0 ) −1
(∀ t> 0) : ≥ ct ,
t
and thus

∞ (x) ≥ lim ct −1 = 0.
t→∞
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Proposition 11. If K is a nonempty closed convex subset of R , then
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( K ) ∞ = K ∞
Proof. Let x 0 ∈ K.If x ∈ K ∞ , then (see Proposition 6) for all λ> 0, x 0 +
1
λx ∈ K, and thus K (x 0 + λx) = 0. This results in
λ
(∀x ∈ K ∞ ) : ( K ) ∞ (x) = 0.

If x/∈ K ∞ , then there exists λ> 0 such that
x 0 + λx /∈ K,

and thus
1
( K (x 0 + λx) − K (x 0 )) =+∞.
λ
This results in
1
( K ) ∞ (x) = sup ( K (x 0 + λx) − K (x 0 )) =+∞.
λ>0 λ
Thus

(∀x/∈ K ∞ ) : ( K ) ∞ (x) =+∞.
We have proved that
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(∀x ∈ R ) : ( K ) ∞ ≡ K ∞ .