Page 39 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

26 Preliminaries

Example 1.28 Let Ω=R and the function u (x)= |x|. Its weak derivative is
then given by
⎧
⎨ +1 if x> 0
u (x)=
0
−1 if x< 0 .
⎩
Example 1.29 (Dirac mass). Let
⎧
⎨ +1 if x> 0
H (x)=
0 if x ≤ 0 .
⎩
We will show that H has no weak derivative. Let Ω =(−1, 1). Assume, for the
sake of contradiction, that H = δ ∈ L 1 (−1, 1) and let us prove that we reach a
0
loc
contradiction. Let ϕ ∈ C 0 ∞ (0, 1) be arbitrary and extend it to (−1, 0) by ϕ ≡ 0.
Wethereforehavebydefinition that
1 1 1
Z Z Z
0 0
δ (x) ϕ (x) dx = − H (x) ϕ (x) dx = − ϕ (x) dx
−1 −1 0
= ϕ (0) − ϕ (1) = 0 .
We therefore find
1
Z
δ (x) ϕ (x) dx =0, ∀ϕ ∈ C 0 ∞ (0, 1)
0
which combined with Theorem 1.24, leads to δ =0 a.e. in (0, 1).With an
analogous reasoning we would get that δ =0 a.e. in (−1, 0) and consequently δ =
0 a.e. in (−1, 1). Let us show that we already reached the desired contradiction.
Indeed if this were thecasewewould have,for every ϕ ∈ C 0 ∞ (−1, 1),

Z 1 Z 1
0= δ (x) ϕ (x) dx = − H (x) ϕ (x) dx
0
−1 −1
Z
1
0
= − ϕ (x) dx = ϕ (0) − ϕ (1) = ϕ (0) .
0
This would imply that ϕ (0) = 0, for every ϕ ∈ C ∞ (−1, 1),which is clearly
0
absurd. Thus H is not weakly differentiable.
Remark 1.30 By weakening even more the notion of derivative (for example,
by not requiring anymore that v is in L 1 loc ), the theory of distributions can give a
meaning at H = δ, it is then called the Dirac mass. We will however not need
0
this theory in the sequel, except, but only marginally, in the exercises of Section
3.5.