Page 191 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

178 Solutions to the Exercises

and by the properties of Lebesgue integrals (see Exercise 1.3.7) the quantity
x p
µZ ¶ 1
p
|u (t)| dt tends to 0 as |x − y| tends to 0.
0
y
Exercise 1.4.4. Observe firstthatif v ∈ W 1,p (a, b), p> 1 and y> x,then
y
¯Z ¯
¯ ¯
|v (x) − v (y)| = ¯ v (z) dz ¯
0
¯ ¯
x
(7.7)
µZ y ¶ 1/p µZ y ¶ 1/p 0
p 1/p 0
0
0
≤ |v (z)| dz dz ≤ kv k L p |x − y| .
x x
∞
Let us now show that if u ν u in W 1,p ,then u ν → u in L . Without loss
of generality, we can take u ≡ 0. Assume, for the sake of contradiction, that
∞
u ν 9 0 in L . We can therefore find > 0, {ν i } so that
ku ν i L ∞ ≥ , ν i →∞. (7.8)
k
} is equicontinuous (note also that
From (7.7) we have that the subsequence {u ν i
0
k
k
by Theorem 1.42 and Theorem 1.20 (iii) we have ku ν i L ∞ ≤ c ku ν i W 1,p ≤ c)
and hence from Ascoli-Arzela theorem, we find, up to a subsequence,
→ v in L . (7.9)
∞
u ν i j
p
v in L and by
We, however, must have v =0 since (7.9) implies u ν i j
p
u =0 in L ) we deduce
uniqueness of the limits (we already know that u ν i j
that v =0 a.e., which contradicts (7.8).
Exercise 1.4.5. Follows immediately from Theorem 1.20.
Exercise 1.4.6. It is clear that u ν → 0 in L .We also find
∞
∂u ν √ ν ∂u ν √ ν−1
= ν (1 − y) cos (νx) , = − ν (1 − y) sin (νx)
∂x ∂y
which implies that, there exists a constant K> 0 independent of ν, such that
ZZ
2
|∇u ν (x, y)| dxdy ≤ K.
Ω
Apply Exercise 1.4.5 to get the result.
Exercise 1.4.7. Since u ∈ W 1,p (Ω),we have that it is weakly differentiable
and therefore Z
£ ¤
ψ + uψ dx =0, ∀ψ ∈ C ∞ (Ω) .
u x i 0
x i
Ω
1,p 0
Let ϕ ∈ W (Ω) and > 0 be arbitrary. We can then find ψ ∈ C ∞ (Ω) so that
0 0
kψ − ϕk L p 0 + k∇ψ −∇ϕk L p 0 ≤ .