Page 59 - Complementarity and Variational Inequalities in Electronics

50 Complementarity and Variational Inequalities in Electronics

1
= sup (x 0 + λx) − (x 0 ) .
λ>0 λ
The definition of ∞ does not depend on the choice of x 0 ∈ D( ).Wehave
indeed the following result.
n
Proposition 7. Let ∈ 0 (R ;R ∪{+∞}). We have
(tv)
∞ (x) = liminf
t→+∞ t
v→x
(t n x n ) n
= inf{liminf :{t n }⊂ R + \{0},{x n }⊂ R ,t n →+∞,x n → x}.
t n
Proof. Set
(tx)
∞ (x) = liminf .
t→+∞ t
v→x
n
Let x ∈ R and x 0 ∈ D( ). Consider a sequence {t n }⊂ R + \{0} such that t n →
+∞.Wehave
(t n z n )
∞ (x) ≤ liminf
n→∞ t n
x 0
with z n = x + → x. Therefore
t n
(t n x + x 0 )
∞ (x) ≤ liminf = ∞ (x). (4.3)
n→+∞ t n
n
Let {λ n }⊂ R + \{0}, {x n }⊂ R be sequences such that λ n →+∞ and x n → x.
For all λ> 0, we obtain (for n large enough) by the convexity and lower semi-
continuity of :
λ λ
(x 0 + λx) ≤ liminf ((1 − )x 0 + λ n x n )
n→∞ λ n λ n
λ λ
≤ liminf{(1 − ) (x 0 ) + (λ n x n )}
n→∞ λ n λ n
(λ n x n )
= (x 0 ) + λliminf .
n→∞ λ n
Thus
(x 0 + λx) − (x 0 ) (λ n x n )
≤ liminf .
λ n→+∞ λ n
n
This last inequality holds for any sequences {λ n }⊂ R + \{0} and {x n }⊂ R such
that λ n →+∞ and x n → x. Thus
(x 0 + λx) − (x 0 )
≤ ∞ (x),
λ