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Chapter 1: Preliminaries 175
R
Exercise 1.3.6. (i) Let f ∈ C 0 ∞ (Ω) with Ω f (x) dx =1,bea fixed function.
Let w ∈ C 0 ∞ (Ω) be arbitrary and
∙Z ¸
ψ (x)= w (x) − w (y) dy f (x) .
Ω
R
We therefore have ψ ∈ C ∞ (Ω) and ψ =0 which leads to
0
Z Z Z Z
0= u (x) ψ (x) dx = u (x) w (x) dx − f (x) u (x) dx · w (y) dy
Ω Ω Ω Ω
Z ∙ Z ¸
= u (x) − u (y) f (y) dy w (x) dx .
Ω Ω
R
Appealing to Theorem 1.24, we deduce that u (x)= u (y) f (y) dy = constant
a.e.
R b
(ii) Let ψ ∈ C 0 ∞ (a, b),with a ψ =0, be arbitrary and define
Z x
ϕ (x)= ψ (t) dt .
a
Note that ψ = ϕ and ϕ ∈ C ∞ (a, b). We may thus apply (i) and get the result.
0
0
Exercise 1.3.7. Let, for ν ∈ N,
u ν (x)= min {|u (x)| ,ν} .
The monotone convergence theorem implies that, for every > 0,wecan find ν
sufficiently large so that
Z Z
|u (x)| dx ≤ u ν (x) dx + .
Ω Ω 2
Choose then δ = /2ν. We therefore deduce that, if meas E ≤ δ,then
Z Z Z
|u (x)| dx = u ν (x) dx + [|u (x)| − u ν (x)] dx ≤ ν meas E + ≤ .
E E E 2
For a more general setting see, for example, Theorem 5.18 in De Barra [37].
7.1.3 Sobolev spaces
Exercise 1.4.1. Let σ n−1 =meas (∂B 1 (0)) (i.e. σ 1 =2π, σ 2 =4π,...).
(i) The result follows from the following observation
Z Z R
p p n−1 p
kuk L p = |u (x)| dx = σ n−1 r |f (r)| dr.
0
B R