2. TH E CONTROL DISTANCE AND TH E LOCAL HARNACK INEQUALITY 21
THEOREM
1.15.3. Let us assume that L is centered. Then, for allYeg, the
Riesz transform operators
R^ =
YL-1^00,
and R*^ = L'^^Y
are bounded on
Lp,
for 1 p oo and from
L1
to
weak-L1.
Note that in general the second order Riesz transforms YiY^
- 1
'
0 0
and
i-i,ooy
1
y
2 j
Y1,Y2 G Q may be unbounded even on L? (cf. [A2]).
If G is nilpotent then there is no homogenisaton phenomena and so we can also
consider higher order Riesz transforms:
THEOREM
1.15.4. Let us assume that L is centered and that G is nilpotent.
Then, for all Yi,..., Yn G Q the Riesz transform operators
Rn,oc = yi...y
n
£"
n / 2
'°° and K,oo =
L-nl2^Yx..Yn
are bounded on LP, for 1 p oo and from L1 to weak-L1.
Combining the theorems 1.15.2 and 1.15.4 we have the following:
COROLLARY
1.15.5. Let L and
EQ
be as in theorem 1.15.1 and let us assume
that G is nilpotent. Then, for allY\,...,Yn E t\ the Riesz transform operators
Rn = Yx..YnL-nl2 and R*n = L-n'2Yx..Yn
are bounded on LP, for 1 p oo and from L1 to weak-L1.
2. The control distance and the local Harnack inequality
2.1. The control distance associated to a sub-Laplacian. To every
left invariant sub-Laplacian L {E\ + ... + E2) + EQ (or to every choice of left
invariant vector fields Ei,...,Ep satisfying Hormander's condition) on a connected
Lie group G we can associate a so called control distance d^.,.). This distance is
left invariant, i.e d^x^y) = diJ(e,x~1y), x,y G G and it is defined as follows:
Let us denote by C the set of all absolutely continuous paths c : [0,1] G, satisfying
£(£) EiK p bi(t)Ei (c(£)), for almost all t G [0,1] and set
|c|= / (b1(t)2 + ... + bp(t)2)1/2dt.
Jo
We define
dL(x, y) = inf {|c| : c G C, c(0) = x, c(l) = y} .
We denote by B^(x) = {y G T V : d(x,y) r} the associated balls. We shall drop
the index L when there is no risk of confusion.
The behavior of the balls Br(x) as r oo is actually a group invariant. More
precisely, let us fix a compact neighborhood U of the identity element e of G and
let \.\G be defined as in (1.1). Let us also consider a constant c 0 such that
B1/c(e)CUCBc(e).
Then
(2.1.1)
Bn/C(e) C
Un
C Bcn(e), n G N.
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