Regularity near the characteristic boundary for sub-laplacian operators

374 DIMITER VASSILEV
From the strict subadditivity (3-19) of I λ , we conclude that exactly one of the
numbers α and d j is nonzero. We claim that α = 1 (and thus that all d j ’s are zero).
Suppose that there is a d j with d j = 1 and dν = δ gj . From the normalization,
Q n (1) = 1 / 2 , and hence
(3-24)
∫
1
2
∫B ≥ |v n | p∗ d H → dν = 1,
1 (g j )
B 1 (g j )
which is a contradiction. Thus, we proved that ‖v‖ L p ∗ (G) = α = 1 and v n → v in
L p∗ (G), which shows that v is a solution of the variational problem; see (3-9).
Suppose we have dichotomy. We have to show that this leads to a contradiction.
A simple process of taking a diagonal subsequence (see the remark after Lemma
3.3) shows that we can find a sequence R n > 0 such that
(3-25)
(3-26)
supp νn 1 ⊂ B R n
(g n ), supp νn 2 ⊂ G \ B 2R n
(g n ),
∫
∫
∣
∣
lim ∣λ − ∣ + ∣(1 − λ) − ∣ = 0.
n→∞
νn
1
G
νn
2
G
We fix a number ε such that
(3-27) 0 < ε < λ p/p∗ + (1 − λ) p/p∗ − 1.
Such a choice of ε is possible, as for 0 < λ < 1 and p/p ∗ < 1 we have λ p/p∗ +
(1 − λ) p/p∗ − 1 > 0.
Let ϕ be a cut-off function 0 ≤ ϕ ∈ C0
∞ (
B2 (e) ) with ϕ ≡ 1 on B 1 (e), and set
ϕ n = δ R
−1 τ g n n
ϕ. We have
∫
∫
∫
(3-28) |Xv n | p d H = |X (ϕ n v n )| p d H + |X ( ) p
(1 − ϕ n )v n | d H + εn .
G
G
G
Note that the remainder term ε n is expressed by an integral over an annulus
(3-29) A n = B 2Rn (g n ) \ B Rn (g n ).
Furthermore, we claim that
∫
(3-30) ε n ≥ o(1) − ε |Xv n | p d H, where o(1) → 0 as n → ∞.
G
Indeed, using the inequality
(
|a| + |b|
) p ≤ (1 + ε)|a| p + C ε,p |b| p ,