Page 183 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

170 Solutions to the Exercises

we get

kuk C 0,α = kuk C 0 +[u] C 0,α
≤ kuk C 0 +max {2 kuk C 0 , [u] C 0,β } ≤ 3 kuk C 0,β .

(iii) We now assume that Ω is bounded and convex and k =0 (for the sake
Ω ⊂ C
Ω .From the mean value
of simplicity) and let us show that C 1 ¡ ¢ 0,1 ¡ ¢
theorem, we have that for every x, y ∈ Ω we can find z ∈ [x, y] ⊂ Ω (i.e., z is an
element of the segment joining x to y)sothat
u(x) − u(y)= h∇u(z); x − yi .

We have, indeed, obtained that

|u(x) − u(y)| ≤ |∇u(z)||x − y| ≤ kuk C 1 |x − y| .

p
7.1.2 L spaces
Exercise 1.3.1. (i) Hölder inequality.Let a, b > 0 and 1/p +1/p =1,with
0
1 <p < ∞. Since the function f(x)= log x is concave, we have that
µ ¶
1 p 1 p 0 1 p 1 p 0
log a + b > log a + log b =log ab
p p 0 p p 0
and hence
1 p 1 p 0
a + b ≥ ab .
p p 0
Choose then
|u| |v|
a = ,b =
kuk kvk
L p L p 0
and integrate to get the inequality for 1 <p < ∞.The cases p =1 or p = ∞
are trivial.
Minkowski inequality.The cases p =1 and p = ∞ are obvious. We
thereforeassumethat 1 <p < ∞. Use Hölder inequality to get
Z Z Z
p p p−1 p−1
ku + vk = |u + v| ≤ |u||u + v| + |v||u + v|
L p
Ω Ω Ω
° ° ° °
° p−1° ° p−1°
≤ kuk L p °|u + v| ° + kvk L p °|u + v| ° .
L p 0 L p 0
The result then follows, since
° °
p−1
° p−1°
°|u + v| ° = ku + vk L p .
L p 0