Page 22 - Complementarity and Variational Inequalities in Electronics

12 Complementarity and Variational Inequalities in Electronics

FIGURE 2.6 For any w ∈ R, there exists v ∈ R such that i (v) < wv, and thus ∂ i (0) =∅
(i = 1,2). However, for any given w ∈ R, we see that 1 (v) + 2 (v) ≥ wv,∀v ∈ R. Thus ∂( 1 +
2 )(0) = R.

and

√
− −x if x ≤ 0
(∀x ∈ R) : 2 (x) =
+∞ if x> 0.
Then

0 if x = 0
(∀x ∈ R) : 1 (x) + 2 (x) =
+∞ if x = 0.
We have
∂ 1 (0) =∅,∂ 2 (0) =∅,∂( 1 + 2 )(0) = R,
and thus

∂ 1 (0) + ∂ 2 (0) ⊂ ∂( 1 + 2 )(0)
and

∂ 1 (0) + ∂ 2 (0) = ∂( 1 + 2 )(0).
m
Proposition 3. Let ∈ 0 (R ;R ∪{+∞}) and A ∈ R m×n be given. Suppose
n
that there exists y 0 = Ax 0 (x 0 ∈ R ) at which is finite and continuous. Then
T
n
(∀x ∈ R ) : ∂( (A))(x) = A ∂ (Ax).
2
Example 7. Let : R → R be defined by
2
(∀x ∈ R ) : (x) = ϕ 1 (x 1 ) + ϕ 2 (x 2 )