418 M. Bramanti - L. Brandolini
here ·
T
denotes transposition. Since k
γ
is locally integrable, the first integral is bounded, uniformly in γ by 59. As to the second term, by the vanishing property of the kernel P
h
k
γ
see Lemma 1, homogeneity and 59 we can write Z
kxkε
P
h
k
γ
x ux d x =
Z
ε kxk1
P
h
k
γ
x [ux − u0] dx
≤ ≤ sup
kyk≤1
|∇uy| Z
ε kxk1
c kxk
Q
· |x| dx ≤ c. The convergence of the last integral follows from 8.
5. Some properties of the Sobolev spaces S
k, p
We start pointing out the following interpolation inequality for Sobolev norms: P
ROPOSITION
2. Let X be a left invariant vector field, homogeneous of degree α 0. Then for every ε 0, u
∈ S
p
R
N
, p ∈ [1, ∞
60 kXuk
p
≤ ε X
2
u
p
+ 2
ε kuk
p
. Proof. The following argument is taken from the proof of Theorem 9.4 in [13]. Let γ t be the
integral curve of X with γ 0 = 0. Then, applying Taylor’s theorem to the function Ft =
u x ◦ γ t,
ux ◦ γ 1 = ux + Xux +
Z
1
1 − t X
2
ux ◦ γ tdt.
Using the translation invariance of k·k
p
and Minkowski’s inequality, we get kXuk
p
≤ X
2
u
p
+ 2 kuk
p
. Since X is homogeneous, by a dilation argument we get the result.
We will need a version of 60 for functions defined on a ball not necessarily vanishing at the boundary. For standard Sobolev norms, this result follows from the analog of 60 using an
extension theorem see for instance [14], pp. 169-173. However, it seems not easy to construct a continuous extension operator E : S
p
B
r
→ S
p
B
2r
. We are going to show how to bypass this difficulty.
First we construct a suitable family of cutoff functions. Given two balls B
r
1
, B
r
2
and a function ϕ
∈ C
∞
R
N
, let us write B
r
1
≺ ϕ ≺ B
r
2
to say that 0 ≤ ϕ ≤ 1, ϕ ≡ 1 on B
r
1
and sprtϕ
⊆ B
r
2
. L
EMMA
5 R
ADIAL CUTOFF FUNCTIONS
. For any σ ∈ 0, 1, r 0, k positive integer,
there exists ϕ ∈ C
∞
R
N
with the following properties: B
σ r
≺ ϕ ≺ B
σ
′
r
with σ
′
= 1 + σ 2;
L
p
estimates for hypoelliptic operators 419
P
j
ϕ ≤
cG, j σ
j −1
1 − σ
j
r
j
for 1 ≤ j ≤ k,
where P
j
is any left invariant differential monomial homogeneous of degree j . Proof. For simplicity, we prove the assertion for k
= 2. The general case is similar. Pick a function f : [0, r
→ [0, 1] such that: f
≡ 1 in [0, σr, f ≡ 0 in [σ
′
r, r , f ∈ C
∞
0, r , f
′
≤ c
1 − σ r
, f
′′
≤ c
1 − σ
2
r
2
. Setting ϕx
= f kxk, we can compute: X
i
ϕ x
= f
′
kxkX
i
kxk ; X
i
X
j
ϕ x
= f
′′
kxkX
i
kxk X
j
kxk + f
′
kxkX
i
X
j
kxk . Since X
i
kxk is homogeneous of degree zero for i = 1, . . . , q, X
i
X
j
kxk for i, j
= 1, . . . , q and X kxk are homogeneous of degree −1 and f
′
kxk 6= 0 for kxk σ
r , we get the result. Another tool we need in this context is an approximation result by suitable mollifiers. For a
fixed cutoff function ϕ, B
1
≺ ϕ ≺ B
2
0, set, for every ε 0, ϕ
ε
x = c · ε
−Q
ϕ D
1 ε
x with c
= R
R
N
ϕ x d x
−1
. Then L
EMMA
6. For u ∈ S
k, p
R
N
k nonnegative integer, 1 ≤ p ∞ and ϕ
ε
as above, define u
ε
= ϕ
ε
∗ u. Then u
ε
∈ C
∞
and u
ε
→ u in S
k, p
for ε → 0.
Proof. The proof follows the same line as in the Euclidean case. We just point out the following facts:
i convergence in L
p
is established first for u ∈ 3
β
G, R
N
, u with bounded support. The density of this space in L
p
can be proved in a general space of homogeneous type see for instance [4];
ii since X
i
is right invariant, X
i
u
ε
= X
i
u
ε
: from this remark and convergence in L
p
we get convergence in S
k, p
; iii to see that u
ε
∈ C
∞
, one has to consider right invariant vector fields X
R i
, write X
R i
ϕ
ε
∗ u = X
R i
ϕ
ε
∗u, and iterate. The possibility of representing any Euclidean derivative in terms of right invariant vector fields see §2.5 provides the conclusion.
The above Lemma is useful for us mainly in view of the following C
OROLLARY
1. If u ∈ S
k, p
1
≤ p ∞, k ≥ 1 and ϕ ∈ C
∞
, then uϕ
∈ S
k, p
.
420 M. Bramanti - L. Brandolini
Proof. The function uϕ is compactly supported in and, extended to zero outside , belongs to S
k, p
R
N
. Then uϕ
ε
converges to uϕ in S
k, p
. Since, for ε small enough, uϕ
ε
is compactly supported in , uϕ
ε
∈ C
∞
and uϕ
∈∈ S
k, p
.
T
HEOREM
21 I
NTERPOLATION INEQUALITY IN
C
ASE
A. Assume we are in Case A. For any u
∈ S
H, p
B
r
, p ∈ [1, ∞, H ≥ 2, r 0, define the following quantities:
8
k
= sup
1 2
σ 1
1 − σ
k
r
k
D
k
u
L
p
B
rσ
for k = 0, 1, 2, . . . , H.
Then for every integer j , 1 ≤ j ≤ H − 1, there exist positive constants c, δ
depending on G, j, H such that for every δ
∈ 0, δ we have
61 8
j
≤ δ 8
H
+ c
δ
jH − j
8 .
Proof. We proceed by induction on H . Let H = 2.
Let u ∈ S
p
B
r
and ϕ a cutoff function as in Lemma 5. By Corollary 1, uϕ ∈ S
p
B
r
, hence by density we can apply Proposition 2, writing
kX
i
uϕ k
p
≤ ε n
kϕ X
i
X
i
u k
p
+ k2 X
i
u X
i
ϕ k
p
+ ku X
i
X
i
ϕ k
p
o +
2 ε
kϕuk
p
for any ε 0. Hence kDuk
L
p
B
σ r
≤ ε n
D
2
u
L
p
B
σ ′ r
+ cG
1 − σ r
kDuk
L
p
B
σ ′ r
+ +
cG σ
1 − σ
2
r
2
kuk
L
p
B
σ ′ r
o +
2 ε
kuk
L
p
B
σ ′ r
. Multiplying both sides for 1
− σ r and choosing ε = δσ 1 − σ r we find 1
− σ r kDuk
L
p
B
σ r
≤ δσ 1 − σ
2
r
2
D
2
u
L
p
B
σ ′ r
+ +cGδσ 1 − σ r kDuk
L
p
B
σ ′ r
+ cG δ
+ 1
δσ kuk
L
p
B
σ ′ r
≤ noting that 1
− σ
′
= 1 − σ 2 ≤ 4δ8
2
+ CGδ8
1
+ cG δ
+ 2
δ 8
. Therefore
8
1
≤ 4δ
1 − cGδ
8
2
+ cG
δ +
2 δ
1 − cGδ
8 which, for δ small enough, is equivalent to
62 8
1
≤ δ8
2
+ c
δ 8
. Assume now that 61 holds for H
− 1. An argument similar to that used to obtain 62 applied to D
H −2
u yields 8
H −1
≤ δ8
H
+ c
δ 8
H −2
L
p
estimates for hypoelliptic operators 421
while, by induction, 8
H −2
≤ η8
H −1
+ c
η
H −2
8 .
Therefore 8
H −1
≤ δ8
H
+ c
δ η8
H −1
+ c
η
H −2
8 and, choosing η
=
δ 2c
we get 63
8
H −1
≤ 2δ8
H
+ c
δ
H −1
8 .
If j = H − 1, this is exactly what we have to prove; if j H − 1, by induction
8
j
≤ ε8
H −1
+ c
ε
jH −1− j
8 ≤ by 63
≤ ε 2δ8
H
+ c
δ
H −1
8 +
c ε
jH −1− j
8 .
Choosing 2δ = η
1H − j
and ε = η
1 −1H− j
we get the result. R
EMARK
5. Note that the second part of the above proof does not hold in Case B since in that case D
k
u cannot be obtained as D D
k −1
u . However, the proof for H
= 2 holds also in Case B, since the field X
does not play any role in the definition of Du. We are going to prove an analogous interpolation inequality, in Case B, which will hold for H even. This proof will be
achieved in several steps. Let
L ≡
q
X
i =1
X
2 i
+ X ,
and let Ŵ be the fundamental solution of L homogeneous of degree two; recall that the transpose of L is just
L
T
≡
q
X
i =1
X
2 i
− X .
L
EMMA
7. Let Q 4. For every integer k ≥ 2 and any couple of left invariant differential
monomials P
2k −1
and P
2k −2
, homogeneous of degrees 2k − 1 and 2k − 2, respectively, we
can determine two kernels K
1
, K
2
depending only on these monomials which are smooth outside the origin and homogeneous of degrees 1
− Q , 2 − Q, respectively, such that for any test function u
P
2k −1
ux =
L L . . . Lu
k times
∗ K
1
x ;
64 P
2k −2
ux =
L L . . . Lu
k times
∗ K
2
x. Proof. By induction on k. Let k
= 2. By Lemma 1, we can write u
= Lu ∗ Ŵ = L Lu ∗ Ŵ ∗ Ŵ = L Lu ∗ K ,
422 M. Bramanti - L. Brandolini
where K = Ŵ ∗ Ŵ is homogeneous of degree 4 − Q. Hence
P
3
u = L Lu ∗ P
3
K = L Lu ∗ K
1
; P
2
u = L Lu ∗ P
2
K = L Lu ∗ K
2
, with K
1
, K
2
homogeneous of degree 1 − Q, 2 − Q, respectively.
Now, assume 64 holds for k − 1, that is
P
2k −3
ux =
L L . . . Lu
k −1 times
∗ K
1
x ;
P
2k −4
ux =
L L . . . Lu
k −1 times
∗ K
2
x. Any differential monomial P
2k −2
can be written either as X
i
P
2k −3
for some i = 1, . . . , q or
as X P
2k −4
. In the first case, we can write P
2k −2
ux = X
i
L L . . . Lu
k −1 times
∗ K
1
x =
= X
i
L L . . . Lu
k times
∗ Ŵ
∗ K
1
x =
= L L . . . Lu
k times
∗ X
i
Ŵ ∗ K
1
x =
L L . . . Lu
k times
∗ e K
2
x, with e
K
2
homogeneous of degree 2 − Q + 1 − Q + Q − 1 = 2 − Q we have applied
Lemma 1. In the second case, P
2k −2
ux = X
L L . . . Lu
k −1 times
∗ K
2
x =
= L L . . . Lu
k times
∗ X Ŵ
∗ K
2
x =
L L . . . Lu
k times
∗ e e
K
2
x, with, again, e
e K
2
homogeneous of degree 2 − Q.
Similarly, any differential monomial P
2k −1
can be written either as P
2
P
2k −3
with P
2
= X
or P
2
= X
i
X
j
for some i, j = 1, . . . , q or as P
3
P
2k −4
with P
3
= X
i
X for some
i = 1, . . . , q. Reasoning as above we get the result for this case, too.
L
EMMA
8. For every integer k ≥ 2 there exists a constant cG, k such that for every ε 0
and every test function u, D
2k − j
u
p
≤ ε D
2k
u
p
+ cG, k
ε
2k − j
kuk
p
for j = 1, 2, k ≥ 2.
L
p
estimates for hypoelliptic operators 423
Proof. Let K
1
, K
2
be as in Lemma 7. We split the kernel K
1
as K
1
= ϕ K
1
+ 1 − ϕ K
1
≡ K
1
+ K
1 ∞
, where ϕ is a cutoff function, B
1
≺ ϕ ≺ B
2
0. Therefore K
1
is homogeneous of degree 1
− Q near the origin and has compact support, hence it is integrable, while K
1 ∞
is homoge- neous of degree 1
− Q near infinity and vanishes near the origin. Writing e f x
= f x
−1
, we can compute
Z K
1 ∞
y
−1
◦ x L L . . . Luy d y
= Z
e K
1 ∞
x
−1
◦ y L L . . . Luy d y
= =
Z L
T
e K
1 ∞
x
−1
◦ y L L . . . L
k −1 times
uy d y = . . .
= Z
L
T
L
T
. . . L
T k times
e K
1 ∞
x
−1
◦ y uy d y
= = u ∗
L
T
L
T
. . . L
T k times
e K
1 ∞
∼
x ≡
u ∗ K
1 1
x. Therefore
P
2k −1
ux =
L L . . . L
k times
u ∗
h K
1
+ K
1 ∞
i x
= = L L . . . Lu ∗ K
1
+ u ∗ K
1 1
, where K
1 1
is homogeneous of degree 1 − Q − 2k near infinity and vanishes near the origin,
hence it is integrable. Integrability of K
1
, K
1 1
gives D
2k −1
u
p
≤ cG, k D
2k
u
p
+ kuk
p
. The same reasoning applied to K
2
gives D
2k −2
u
p
≤ cG, k D
2k
u
p
+ kuk
p
. The conclusions follows from the last two inequalities and a dilation argument.
T
HEOREM
22. [Interpolation inequality in Case B] For any function u
∈ S
2k, p
B
r
k ≥ 1, p ∈ [1, ∞, r 0, let 8
h
h = 0, 1, . . . , 2k be the seminorms defined
in Theorem 21. Then 8
j
≤ ε8
2k
+ cε, k8 for every integer j with 1
≤ j ≤ 2k − 1 and every ε 0. Proof. Let ϕ be a cutoff function as in Lemma 5. By Corollary 1 uϕ
∈ S
2k, p
B
r
, hence, by density, we can apply Lemma 8 to uϕ. By standard arguments see the first part of the proof of
Theorem 21 we get, for every δ 0, 8
2k − j
≤ δ
2k
X
h =0
8
h
+ c
δ
2k − j j
8 j
= 1, 2.
424 M. Bramanti - L. Brandolini
The previous inequality clearly holds if k is replaced by any integer i with 2 ≤ i ≤ k. Adding
up these inequalities for 2 ≤ i ≤ k, j = 1, 2, we get
k
X
i =2
8
2i −1
+ 8
2i −2
=
2k −1
X
i =2
8
i
≤ 2δ
k
X
i =2
2i
X
h =0
8
h
+ cδ, k · 8
≤ ≤ 2δk
2k
X
h =0
8
h
+ cδ, k · 8 .
Adding also 62 which holds in Case B, too, as noted in Remark 5: 8
1
≤ δ8
2
+ c
δ 8
we can write
2k −1
X
h =0
8
h
≤ 2δk
2k
X
h =0
8
h
+ cδ, k · 8 and, finally, for every ε 0,
2k −1
X
h =0
8
h
≤ ε · 8
2k
+ cε, k · 8 which proves the result.
Next, we need the following Sobolev-type embedding theorem: T
HEOREM
23. Let u ∈ S
p
B
r
for some r 0. Then: a if 1 p
Q 2
and
1 p
∗
=
1 p
−
2 Q
, then kuk
p
∗
≤ c p, G D
2
u
p
; b if
Q 2
p Q and β = 2 −
Q p
, then kuk
3
β
B
r
≤ c p, G, r D
2
u
L
p
B
r
. Proof. Let L , Ŵ be as in the proof of Lemma 7. Then for any u
∈ C
∞
B
r
we can write u
= Lu ∗ Ŵ. Theorem 18 then gives the assertion.
The results contained in Theorem 23 have been proved by Folland in [11], where a more complete theory of Sobolev and H¨older spaces defined by the vector fields X
i
is developed see §§4, 5 in [11], in particular Theorems 4.17 and 5.15. However, Folland’s theory relies on a deep
analysis of the sub-Laplacian on stratified groups, and therefore does not cover completely the cases we are considering here: remember that under our assumptions, the group G is graded but
not necessarily stratified.
L
p
estimates for hypoelliptic operators 425
6. Local estimates for solutions to the equation Lu = f in a domain
