Page 21 - Complementarity and Variational Inequalities in Electronics

The Convex Subdifferential Relation Chapter | 2 11

(∀i ∈{1,...,n}) : ϕ i (v i ) − ϕ i (x i ) ≥ w i (v i − x i ),∀v i ∈ R.

Thus
⎛ ⎞
∂ϕ 1 (x 1 )
⎜ ⎟
⎜ ∂ϕ 2 (x 2 ) ⎟
∂ (x) = ⎜ . ⎟ .
.
⎜ ⎟
⎝ . ⎠
∂ϕ n (x n )
Remark 3. We have (see e.g. [15])
D(∂ ) ⊂ D( ) ⊂ D(∂ ).

Let us here recall some basic calculus rules (see e.g. [72], [79]).
n
Proposition 2. Let ∈ 0 (R ;R∪{+∞}). Then for all x ∈ D(∂ ) and λ> 0,
we have

∂(λ )(x) = λ∂ (x).
n
Let 1 , 2 ∈ 0 (R ;R ∪{+∞}). Then for every x ∈ D(∂ 1 ) ∩ D(∂ 2 ),we
have
∂ 1 (x) + ∂ 2 (x) ⊂ ∂( 1 + 2 )(x).

Remark 4. If there exists a point x 0 ∈ D( 1 ) ∩ D( 2 ) at which 1 is contin-
uous, then for every x ∈ D(∂ 1 ) ∩ D(∂ 2 ),wehave(see [12])

∂ 1 (x) + ∂ 2 (x) = ∂( 1 + 2 )(x).

Example 5. Let : R → R be defined by

(∀x ∈ R) : (x) = σ|x|
with σ> 0. Then

⎧
−σ if x< 0
⎪
⎪
⎨
∂ (x) = [−σ,+σ] if x = 0
⎪
⎪
+σ if x> 0.
⎩
Example 6. Let 1 , 2 : R → R be defined as in Fig. 2.6 by

+∞ if x< 0
(∀x ∈ R) : 1 (x) = √
− x if x ≥ 0