Page 134 - Textos de Matemática Vol. 34

124 JOSE´ MIGUEL URBANO
and since, due to (2.3),
24Λ222
|∇(vη)|≤ λ2+1 B1
wearriveat 2
|∇η|v B1
B1
2 22 (vη) ≤c(N,λ,Λ)|{vη̸=0}|N |∇η|v .
B1
Now for fixed 0 < r < R < 1, choose the cut-off function η ∈ C0∞(BR) such that
0≤η≤1,η≡1inBr and|∇η|≤ 2 . Putting,forρ>0,
R−r A(k,ρ)=x∈Bρ : u(x)>k,
we obtain (with C ≡ c(N, λ, Λ)) 2C22
(u−k) ≤ |A(k,R)|N A(k,r) (R − r)2
For h > k and 0 < ρ < 1,
(u−k) .
and
so we have
(u − k)2 (u−h) ≤ |A(h,R)|N
(R − r)2
(u−h)2 ≤ A(h,ρ)
(u−k)2 A(k,ρ)

(h − k)2|A(h, ρ)| ≤ 2C22
A(k,R)
A(h,r)
(u−h) A(h,R)
A(k,ρ)
1+ 2 C1N

or, equivalently, with ψ(s, ρ) = ∥(u − s)+∥L2(Bρ),
(2.4) ψ(h, r) ≤ C 1 R−r (h−k)N
≤ 2 4 (u−k)2 (R − r) (h − k)N A(k,R)
ψ(k, R)1+ 2 ,
We are now ready to use the iteration scheme devised by DeGiorgi. Define, for
N
1 11 kn = k 1 − 2n ; rn = 2 1 + 2n ,
where k is to be determined. Due to (2.4), we have, for n = 0,1,2,...,
2n+1+ 2n
2
for any h > k > 0 and 0 < r < R < 1. n = 0,1,2,...,
(2.5) ψ(kn, rn) ≤ C 2 kN
N 1+ 2 ψ(kn−1, rn−1) N