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120 Regularity
0
From (4.12) we deduce that u ∈ C (Ω). Indeed let x, y ∈ Ω and R be sufficiently
small so that B R (x) ∪ B R (y) ⊂ Ω. We then have that (denoting by ω n the
measure of the unit ball)
¯ ¯
1 ¯ Z Z ¯ 1 Z
¯
|u (x) − u (y)| = ¯ u (z) dz − u (z) dz¯ ≤ |u (z)| dz ,
n ¯ n
ω n R ¯ B R (x) B R (y) ¯ ω n R O
where O =(B R (x) ∪ B R (y)) r (B R (x) ∩ B R (y)). Appealingtothe fact that
1
u ∈ L (B R (x) ∪ B R (y)) andto Exercise1.3.7, wededucethat u is indeed
continuous.
It therefore remains to prove that u = u a.e. in Ω. This follows from Lebesgue
theorem and the fact that u ∈ L 1 (Ω).Indeed letting R tend to 0 in (4.12)
loc
we have that for almost every x ∈ Ω the right hand side of (4.12) is u (x).The
theorem has therefore been established.
We now present a second proof that uses the so called difference quotients,
introduced by Nirenberg.
n
Theorem 4.9 Let k ≥ 0 be an integer, Ω ⊂ R be a bounded open set with
C k+2 boundary, f ∈ W k,2 (Ω) and
½ Z ∙ ¸ ¾
1 2 1,2
0
(P ) inf I (u)= |∇u (x)| − f (x) u (x) dx : u ∈ W 0 (Ω) .
Ω 2
Then there exists a unique minimizer u ∈ W k+2,2 (Ω) of (P’). Furthermore there
exists a constant γ = γ (Ω,k) > 0 so that
kuk k+2,2 ≤ γ kfk k,2 . (4.13)
W W
¡ ¢
In particular if k = ∞,then u ∈ C ∞ Ω .
Remark 4.10 (i) Problem (P) and (P’) are equivalent. If in (P) the boundary
datum u 0 ∈ W k+2,2 (Ω), then choose f = ∆u 0 ∈ W k,2 (Ω).
(ii) A similar result as (4.13) can be obtained in Hölder spaces (these are
then known as Schauder estimates), under appropriate regularity hypotheses on
the boundary and when 0 <a< 1,namely
kuk k+2,a ≤ γ kfk k,a .
C C
If 1 <p < ∞, it can also be proved that
kuk k+2,p ≤ γ kfk k,p ;
W W
these are then known as Calderon-Zygmund estimates and are considerably harder
to obtain than those for p =2.