Page 169 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

156 Isoperimetric inequality

R 1
Note that there is no constant term since u (x) dx =0. We know at the
−1
same time that
∞
X
0
u (x)= π [−na n sin nπx + nb n cos nπx] .
n=1
We can now invoke Parseval formula to get

Z
1 ∞
X ¡ 2 2 ¢
2
u dx = a + b n
n
−1 n=1
1 ∞
Z
X ¡ 2 2 ¢ 2
02
2
u dx = π a + b n n .
n
−1 n=1
The desired inequality follows then at once
1 1
Z Z
2
2
02
u dx ≥ π 2 u dx, ∀u ∈ X ∩ C .
−1 −1
Moreover equality holds if and only if a n = b n =0,for every n ≥ 2. This implies
that equality holds if and only if u (x)= α cos πx + β sin πx, for any α, β ∈ R,
as claimed.
2
Step 2. We now show that we can remove the restriction u ∈ X ∩ C [−1, 1].
By the usual density argument we can find for every u ∈ X a sequence u ν ∈
2
X ∩ C [−1, 1] so that
u ν → u in W 1,2 (−1, 1) .
Therefore, for every > 0,wecan find ν sufficiently large so that
Z Z Z Z
1 1 1 1
02 02 2 2
u dx ≥ u dx − and u dx ≥ u dx − .
ν
ν
−1 −1 −1 −1
Combining these inequalities with Step 1 we find
1 1 ¡ ¢
Z Z
2
2
02
u dx ≥ π 2 u dx − π +1 .
−1 −1
Letting → 0 we have indeed obtained the inequality.
We still need to see that equality in X holds if and only if u (x)= α cos πx +
β sin πx, for any α, β ∈ R. This has been proved in Step 1 only if u ∈ X ∩
2
C [−1, 1]. This property is established in Exercise 6.2.2.
We get as a direct consequence of the theorem