Page 195 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

182 Solutions to the Exercises

∗
(iii) Since f ∗∗ ≤ f,we find that f ∗∗∗ ≥ f .Furthermore, by definition of
n
n
∗∗
∗
f ,we find, for every x ∈ R , x ∈ R ,
∗
hx; x i − f ∗∗ (x) ≤ f (x ) .
∗
∗
Taking the supremum over all x in the left hand side of the inequality, we get
f ∗∗∗ ≤ f , and hence the claim.
∗
(iv) By definition of f ,wehave
∗
f (∇f (x)) = sup {hy; ∇f (x)i − f (y)} ≥ hx; ∇f (x)i − f (x)
∗
y
and hence
∗
f (x)+ f (∇f (x)) ≥ hx; ∇f (x)i .
We next show the opposite inequality. Since f is convex, we have
f (y) ≥ f (x)+ hy − x; ∇f (x)i ,
which means that

hx; ∇f (x)i − f (x) ≥ hy; ∇f (x)i − f (y) .
Taking the supremum over all y, we have indeed obtained the opposite inequality
and thus the proof is complete.
(v) We refer to the bibliography, in particular to Mawhin-Willem [72], page
35, Rockafellar [87] (Theorems 23.5, 26.3 and 26.5 as well as Corollary 25.5.1)
and for the second part to [31] or Theorem 23.5 in Rockafellar [87]; see also the
exercise below.
Exercise 1.5.6. We divide the proof into three steps.
Step 1.We know that
∗
f (v)= sup {ξv − f (ξ)}
ξ
1
and since f ∈ C and satisfies
f (ξ)
lim =+∞ (7.11)
|ξ|→∞ |ξ|
we deduce that there exists ξ = ξ (v) such that
0
f (v)= ξv − f (ξ) and v = f (ξ) . (7.12)
∗
Step 2.Since f satisfies (7.11), we have

Image [f (R)] = R .
0