Page 185 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

172 Solutions to the Exercises

Rewriting the integral we have
Z Z Z
(u ν v ν − uv) ϕ = u ν (v ν − v) ϕ + (u ν − u) vϕ .
Ω Ω Ω
Since (v ν − v) → 0 in L p 0 and ku ν k L p ≤ K, wededucefromHölderinequality
that the first integral tends to 0. The second one also tends to 0 since u ν −u 0
p
in L and vϕ ∈ L p 0 (this follows from the hypotheses v ∈ L p 0 and ϕ ∈ L ).
∞
Let us show that the result is, in general, false if v ν v in L p 0 (instead of
p
0
v ν → v in L ). Choose p = p =2 and u ν (x)= v ν (x)= sin νx, Ω =(0, 2π).We
0
2
have that u ν ,v ν 0 in L but the product u ν v ν does not tend to 0 weakly in
2
2
2
L (since u ν (x)v ν (x)= sin νx 1/2 6=0 in L ).
(ii) We want to prove that ku ν − uk L 2 → 0.We write
Z Z Z Z
2 2 2
|u ν − u| = u − 2 uu ν + u .
ν
Ω Ω Ω Ω
R 2 2 2 1
The first integral tends to u since u u in L (choosing ϕ(x) ≡ 1 ∈ L ∞
ν
R
2
in the definition of weak convergence). The second one tends to −2 u since
2
2
u ν u in L and u ∈ L . The claim then follows.
Exercise 1.3.4. (i) The case p = ∞ is trivial. So assume that 1 ≤ p< ∞.We
next compute
Z Z
+∞ +∞
u ν (x)= ϕ (x − y) u (y) dy = ν ϕ (ν (x − y)) u (y) dy
ν
−∞ −∞
Z
+∞ ³ z ´
= ϕ (z) u x − dz .
ν
−∞
We therefore find
Z
+∞ ¯ ³ ´¯
¯ z ¯
|u ν (x)| ≤ ϕ(z) ¯u x − ¯ dz
ν
−∞
Z
+∞ h ¯ ³ ´¯i
1/p 0 1/p ¯ z ¯
= |ϕ(z)| |ϕ(z)| ¯u x − ¯ dz .
ν
−∞
Hölder inequality leads to
0
µZ ¶ 1/p µZ ¶ 1/p
+∞ +∞ ¯ ³ ´¯ p
¯
|u ν (x)| ≤ ϕ(z) dz ϕ(z) ¯u x − z ¯ ¯ dz .
ν
−∞ −∞
R
Since ϕ =1, we have, after interchanging the order of integration,
Z Z Z
+∞ +∞ +∞ n ¯ ³ ´¯ p o
p p ¯ z ¯
ku ν k = |u ν (x)| dx ≤ ϕ (z) ¯u x − ¯ dz dx
L p ν
−∞ −∞ −∞
Z ½ Z ¾
+∞ +∞ ¯ ³ ´¯
p p
z ¯
≤ ϕ (z) ¯ ¯u x − ¯ dx dz ≤ kuk L p .
ν
−∞ −∞