Page 61 - Complementarity and Variational Inequalities in Electronics

52 Complementarity and Variational Inequalities in Electronics

λ( (x 0 + tx 1 ) − (x 0 )) (1 − λ)( (x 0 + tx 2 ) − (x 0 ))
≤ lim + lim
t→+∞ t t→+∞ t
= λ ∞ (x 1 ) + (1 − λ) ∞ (x 2 ).

Let us now check that ∞ is positively homogeneous of degree 1. We have

(x 0 + λαx)
∞ (αx) = lim ,
λ→+∞ λ
and setting t = αλ, we get
(x 0 + tx)
∞ (αx) = lim α = α ∞ (x).
t→+∞ t
n
It remains to prove that ∞ is lower semicontinuous. Let {x n }⊂ R be a se-
quence such that x n → x and

∞ (x n ) ≤ c

with c ∈ R. Then
(x 0 + λx n ) − (x 0 )
liminfsup ≤ c,
n→∞ λ
λ>0
where x 0 is chosen in D( ). We know that, for each λ> 0,
(x 0 + λx) − (x 0 ) (x 0 + λx n ) − (x 0 )
≤ liminf .
λ n→+∞ λ
Thus
(x 0 + λx n ) − (x 0 )
∞ (x) ≤ sup liminf
λ>0 n→+∞ λ
(x 0 + λx n ) − (x 0 )
≤ liminf sup ≤ c.
n→+∞ λ
λ>0
Therefore the set
n
{x ∈ R : ∞ (x) ≤ c}
is closed for any c ∈ R ∪{+∞} (the case c =+∞ is trivial), and we may con-
clude that ∞ is lower semicontinuous.

Example 18. Let f : R → R be defined by

x 2 if x< 0
f(x) =
3x if x ≥ 0.