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Chapter 1: Preliminaries 177
p
The first one guarantees that ψ ∈ L (B R ) and the second one that ψ ∈
W 1,2 (B R ).The fact that ψ/∈ L ∞ (B R ) is obvious.
3) We have, denoting by δ ij the Kronecker symbol, that
2
∂u i δ ij |x| − x i x j 2 n − 1
= =⇒ |∇u| = .
3 2
∂x j |x| |x|
We therefore find
Z Z 1
p p/2 n−1−p
|∇u (x)| dx =(n − 1) σ n−1 r dr .
Ω 0
This quantity is finite if and only if p ∈ [1,n).
Exercise 1.4.2. The inclusion AC ([a, b]) ⊂ C ([a, b]) is easy. Indeed by defin-
ition any function in AC ([a, b]) is uniformly continuous in (a, b) and therefore
can be continuously extended to [a, b].
Let us now discuss the second inclusion, namely W 1,1 (a, b) ⊂ AC ([a, b]).
Let u ∈ W 1,1 (a, b). We know from Lemma 1.38 that
Z
b k
0
u (b k ) − u (a k )= u (t) dt .
a k
We therefore find
Z
X X b k
0
|u (b k ) − u (a k )| ≤ |u (t)| dt .
k k a k
Let E = ∪ k (a k ,b k ). A classical property of Lebesgue integral (see Exercise
1
1.3.7) asserts that if u ∈ L , then, for every > 0,there exists δ> 0 so that
0
Z
X
meas E = |b k − a k | <δ ⇒ |u | < .
0
E
k
The claim then follows.
Exercise 1.4.3. This follows from Hölder inequality, since
x ⎛ x ⎞ 1 ⎛ x ⎞ 1
Z Z p Z p 0
p 0
0 0 p
|u (x) − u (y)| ≤ |u (t)| dt ≤ ⎝ |u (t)| dt ⎠ ⎝ 1 dt ⎠
y y y
⎛ ⎞ 1
p
x Z
p 1
0
≤ ⎝ |u (t)| dt ⎠ |x − y| p 0
y