Page 218 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

Chapter 4: Regularity 205

(ii) The preceding result does not apply to f (ξ)= e −ξ 2 and α = β =0.
Indeed we have f ∗∗ (ξ) ≡ 0 and we therefore cannot find λ ∈ [0, 1], a, b ∈ R so
that
½
λa +(1 − λ) b =0
λe −a 2 +(1 − λ) e −b 2 =0 .
In fact we should need a = −∞ and b =+∞. Recall that in Section 2.2 we
already saw that (P) has no solution.
¡ 2 ¢ 2
(iii) If f (ξ)= ξ − 1 ,wethen find
⎧ ¡ 2 ¢ 2
⎨ ξ − 1 if |ξ| ≥ 1,
f ∗∗ (ξ)=
⎩
0 if |ξ| < 1 .
Therefore if |β − α| ≥ 1 choose in (i) λ =1/2 and a = b = β − α. However
if |β − α| < 1,choose a =1, b = −1 and λ =(1 + β − α) /2. In conclusion, in
both cases, we find that problem (P) has u as a minimizer.
Exercise 3.6.2. If we set ξ =(ξ ,ξ ), we easily have that
2
1
⎧
⎨ f (ξ) if |ξ | ≥ 1,
1
f ∗∗ (ξ)=
⎩ 4
ξ 2 if |ξ | < 1 .
1
From the Relaxation theorem (cf. Theorem 3.28) we find m =0 (since u ≡ 0 is
1,4
such that I (u)=0). However no function u ∈ W (Ω) can satisfy I (u)= 0,
0
hence (P) has no solution.
Exercise 3.6.3. It is easy to see that
⎧
∗
0 if ξ =0
n o ⎨
2
∗
∗
∗
f (ξ )= sup hξ; ξ i − (det ξ) =
ξ∈R 2×2 ⎩ +∞ if ξ 6=0
∗
and therefore
∗
f ∗∗ (ξ)= sup {hξ; ξ i − f (ξ )} ≡ 0 .
∗
∗
ξ ∈R 2×2
∗
7.4 Chapter 4: Regularity
7.4.1 The one dimensional case
Exercise 4.2.1. (i) We first show that u ∈ W 2,∞ (a, b), by proving (iii) of
Theorem 1.36. Observe firstthatfrom (H1’) andthe fact that u ∈ W 1,∞ (a, b),
0
we can find a constant γ > 0 such that, for every z ∈ R with |z| ≤ ku k L ∞,
1
f ξξ (x, u (x) ,z) ≥ γ > 0, ∀x ∈ [a, b] . (7.17)
1