Page 184 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

Chapter 1: Preliminaries 171

(ii) Use Hölder inequality with α =(p + q) /q and hence α =(p + q) /p to
0
obtain
Z µZ ¶ 1/α µZ ¶ 1/α 0
0
pq/p+q pqα/(p+q) pqα /(p+q)
|uv| ≤ |u| |v|
Ω Ω Ω
µZ ¶ q/(p+q) µZ ¶ p/(p+q)
p q
≤ |u| |v| .
Ω Ω
p
(iii) The inclusion L ∞ (Ω) ⊂ L (Ω) is trivial. The other inclusions follow
from Hölder inequality. Indeed we have
Z Z µZ ¶ q/p µZ ¶ (p−q)/p
q q q·p/q p/(p−q)
|u| = (|u| · 1) ≤ |u| 1
Ω Ω Ω Ω
µZ ¶ q/p
(p−q)/p p
≤ (meas Ω) |u|
Ω
and hence
(p−q)/pq
kuk L q ≤ (meas Ω) kuk L p
which gives the desired inclusion.
If, however, the measure is not finite the result is not valid as the simple
1
2
example Ω =(1, ∞), u (x)= 1/x shows; indeed we have u ∈ L but u/∈ L .
Exercise 1.3.2. A direct computation leads to
1 1/ν
Z Z
p p αp αp−1
ku ν k L p = |u ν (x)| dx = ν dx = ν .
0 0
p
We therefore have that u ν → 0 in L provided αp − 1 < 0.If α =1/p,let us
p
p
0
show that u ν 0 in L . We have to prove that for every ϕ ∈ L (0, 1),the
following convergence holds
1
Z
lim u ν (x) ϕ (x) dx =0 .
ν→∞
0
By a density argument, and since ku ν k L p =1,itissufficient to prove the result
when ϕ is a step function, which means that there exist 0= a 0 <a 1 < ·· · <
a N =1 so that ϕ (x)= α i whenever x ∈ (a i ,a i+1 ), 0 ≤ i ≤ N − 1. We hence
find, for ν sufficiently large, that
Z Z
1 1/ν
ν 1/p dx = α 0 ν (1/p)−1 → 0 .
u ν (x) ϕ (x) dx = α 0
0 a 0
Exercise 1.3.3. (i) Wehavetoshowthatfor every ϕ ∈ L ,then
∞
Z
lim (u ν v ν − uv) ϕ =0 .
ν→∞
Ω