Page 34 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

p
L spaces 21

see, for example, Theorem 2.1.5 in [31]. We will also assume, without loss of
generality, that
1
Z
u = u (x) dx =0 .
0
Step 1. Observe that if 1 ≤ p< ∞,then

Z 1 Z 1
p p p
ku ν k = |u ν (x)| dx = |u (νx)| dx
L p
0 0
Z ν Z 1
1 p p
= |u (y)| dy = |u (y)| dy .
ν 0 0
The last identity being a consequence of the 1-periodicity of u. We therefore
find that
ku ν k L p = kuk L p . (1.1)
The result is trivially true if p = ∞.
Step 2. (For a slightly different proof of this step see Exercise 1.3.5). We
p
thereforehavethat u ν ∈ L and, since u =0,wehavetoshowthat
Z 1
p
0
lim u ν (x) ϕ (x) dx =0, ∀ϕ ∈ L (0, 1) . (1.2)
ν→∞
0
0
p
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 h astep
0
0
function so that
kϕ − hk p 0 ≤ . (1.3)
L
Since h is a step function, we can find a 0 =0 <a 1 < ... < a I =1 and α i ∈ R
such that
h (x)= α i if x ∈ (a i−1 ; a i ) , 1 ≤ i ≤ I.
We now compute
Z Z Z
1 1 1
u ν (x) ϕ (x) dx = u ν (x)[ϕ (x) − h (x)] dx + u ν (x) h (x) dx
0 0 0
and get that

¯Z 1 ¯ Z 1 ¯Z 1 ¯
¯ ¯ ¯ ¯
¯ ¯ |u ν (x)||ϕ (x) − h (x)| dx + ¯ u ν (x) h (x) dx .
¯
u ν (x) ϕ (x) dx ≤
¯ ¯ ¯ ¯
0 0 0