Page 193 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

180 Solutions to the Exercises

Combining the two inequalities we have immediately (7.10).
Let us now show the converse and assume that (7.10) holds. Let λ ∈ (0, 1),
x, y ∈ R and define

y − x
z = + x ⇔ y = x + λ (z − x)
λ
ϕ (λ)= f (x + λ (z − x)) .
Observe that

ϕ (λ) − ϕ (0) = [f (x + λ (z − x)) − f (x)] (z − x)
0
0
0
0
1
= [f (x + λ (z − x)) − f (x)] (x + λ (z − x) − x) ≥ 0
0
0
λ
since (7.10) holds. Therefore, integrating the inequality, we find
ϕ (λ) ≥ ϕ (0) + λϕ (0)
0
and thus, returning to the definition of y,we find
f (y) ≥ f (x)+ f (x)(y − x)
0
which is equivalent by (i) to the convexity of f.
Exercise 1.5.2. Since f is convex, we have, for every α, β ∈ R,
f (α) ≥ f (β)+ f (β)(α − β) .
0
R
Choose then α = u (x) and β =(1/ meas Ω) u (x) dx, and integrate to get the
Ω
inequality.
Exercise 1.5.3. We easily find that
√
½ ∗2
− 1 − x if |x | ≤ 1
∗
∗
∗
f (x )=
+∞ otherwise.
√
2
Note, in passing, that f (x)= 1+ x is strictly convex over R.
Exercise 1.5.4. (i) Wehavethat
½ ¾
p
|x|
∗ ∗ ∗
f (x )= sup xx − .
x∈R p
The supremum is, in fact, attained at a point y where
0
∗ p−2 ∗ p −2 ∗
x = |y| y ⇔ y = |x | x .