Page 187 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

174 Solutions to the Exercises

We define
Z x Z x 1 Z νx
v ν (x)= u ν (t) dt = u (νt) dt = u (s) ds
ν
0 0 0
Z νx Z νx−[νx]
1 1
= u (s) ds = u (s) ds
ν [νx] ν 0
where [a] stands for the integer part of a ≥ 0 and where we have used the
R 1
periodicity of u and the fact that u = u =0.
0
We therefore find that
Z νx−[νx] Z 1
1 1 1 1
kv ν k L ∞ ≤ ν |u (s)| ds ≤ ν |u (s)| ds = ν kuk L 1 ≤ ν kuk L p .
0
0
(7.3)
Recall that we have to show that
Z 1
0
p
lim u ν (x) ϕ (x) dx =0, ∀ϕ ∈ L (0, 1) . (7.4)
ν→∞
0
p
0
Let > 0 be arbitrary. Since ϕ ∈ L (0, 1) and 1 <p ≤∞, which implies
1 ≤ p < ∞ (i.e., p 6= ∞), we have from Theorem 1.13 that there exists
0
0
ψ ∈ C 0 ∞ (0, 1) so that
kϕ − ψk L p 0 ≤ . (7.5)
We now compute
1 1 1
Z Z Z
u ν (x) ϕ (x) dx = u ν (x)[ϕ (x) − ψ (x)] dx + u ν (x) ψ (x) dx
0 0 0
Z 1 Z 1
0
= u ν (x)[ϕ (x) − ψ (x)] dx − v ν (x) ψ (x) dx
0 0
wherewehaveusedintegration by parts, thefact that v = u ν and ψ ∈ C ∞ (0, 1).
0
ν 0
Using Hölder inequality, (1.1), (7.3) and (7.5), we obtain that
1
¯Z ¯
¯ ¯ ° ° 1 ° °
0
0
¯ ¯ ° ° ° °
u ν (x) ϕ (x) dx ≤ kuk L p + kv ν k L ∞ ψ L 1 ≤ kuk L p + kuk L p ψ L 1 .
¯ ¯ ν
0
Let ν →∞, we hence obtain
1
¯Z ¯
¯ ¯
0 ≤ lim sup ¯ u ν ϕdx ≤ kuk L p .
¯
¯ ¯
ν→∞ 0
Since is arbitrary, we immediately have (7.4) and thus the result.