Page 131 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

118 Regularity

references where this problem is discussed. We quote here only two of them,
namely Brézis [14] and John [63].
In this section we will treat only the problem of interior regularity, we refer
to the literature for the question of the regularity up to the boundary. We will,
however, give two different proofs. The first one is really specific to the present
problem but gives, in some sense, a sharper result than the second one, which
however applies to more general problems than the present context.

n
Theorem 4.7 (Weyl lemma). Let Ω ⊂ R be open and u ∈ L 1 (Ω) satisfy
loc
Z
u (x) ∆v (x) dx =0, ∀v ∈ C 0 ∞ (Ω) (4.7)
Ω
then u ∈ C ∞ (Ω) and ∆u =0 in Ω.

Remark 4.8 (i) The function u being defined only almost everywhere, we have
to interpret the result, as usual, up to a change of the function on a set of
measure zero.
(ii) Note that a solution of the weak form of Laplace equation
Z
1,2
(E w ) h∇u (x); ∇ϕ (x)i dx =0, ∀ϕ ∈ W 0 (Ω)
Ω
satisfies (4.7). The converse being true if, in addition, u ∈ W 1,2 (Ω).Therefore
(4.7) can be seen as a “very weak” form of Laplace equation and a solution of
this equation as a “very weak” solution of ∆u =0.

Proof. Let x ∈ Ω and R> 0 sufficiently small so that
n
B R (x)= {y ∈ R : |y − x| <R} ⊂ Ω .
Let σ n−1 =meas (∂B 1 (0)) (i.e. σ 1 =2π, σ 2 =4π,...). The idea is to show that
if Z
1
u (x)= udσ (4.8)
σ n−1 R n−1 ∂B R (x)
0
then u is independent of R, u ∈ C (Ω) and u = u a.e. in Ω. A classical result
(cf. Exercise 4.3.2) allows to conclude that in fact u ∈ C ∞ (Ω) and hence ∆u =0
in Ω, as claimed.
These statements will be proved in three steps.
Step 1. We start by making an appropriate choice of the function v in (4.7).
Let R be as above and choose ∈ (0,R) and ϕ ∈ C ∞ (R) with supp ϕ ⊂ ( , R).
Define then
v (y)= ϕ (|x − y|)