Page 194 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

Chapter 1: Preliminaries 181

∗
Replacing this value in the definition of f we have obtained that
∗ p 0
|x |
f (x )= .
∗
∗
p 0
(ii) We do not compute f , but instead use Theorem 1.55. We let
∗
⎧ ¡ 2 ¢ 2
⎨ x − 1 if |x| ≥ 1
g (x)=
⎩
0 if |x| < 1
and we wish to show that f ∗∗ = g. We start by observing that g is convex,
0 ≤ g ≤ f, and therefore according to Theorem 1.54 (ii) we must have

g ≤ f ∗∗ ≤ f.
First consider the case where |x| ≥ 1; the functions g and f coincide there and
hence f ∗∗ (x)= g (x),for such x. We next consider the case |x| < 1.Choose in
Theorem 1.55
1+ x 1 − x
x 1 =1,x 2 = −1,λ 1 = ,λ 2 =
2 2
to get immediately that f ∗∗ (x)= g (x)= 0. Wehavethereforeproved the claim.
(iv) This is straightforward since clearly

∗ ∗ ∗
f (ξ )= sup {hξ; ξ i − det ξ} ≡ +∞
ξ∈R 2×2
and therefore
∗
∗
f ∗∗ (ξ)= sup {hξ; ξ i − f (ξ )} ≡−∞ .
∗
ξ ∈R 2×2
∗
n
∗
Exercise 1.5.5. (i) Let x ,y ∈ R and λ ∈ [0, 1]. It follows from the definition
∗
that
∗
∗
∗
∗
f (λx +(1 − λ) y )= sup {hx; λx +(1 − λ) y i − f (x)}
∗
x∈R n
∗ ∗
=sup {λ (hx; x i − f (x)) + (1 − λ)(hx; y i − f (x))}
x
∗ ∗
≤ λ sup {hx; x i − f (x)} +(1 − λ)sup {hx; y i − f (x)}
x x
∗
∗
∗
∗
≤ λf (x )+(1 − λ) f (y ) .
(ii) For this part we can refer to Theorem I.10 in Brézis [14], Theorem 2.2.5 in
[31] or Theorem 12.2 (coupled with Corollaries 10.1.1 and 12.1.1) in Rockafellar
[87].