12
GEORGIOS K. ALEXOPOULOS
To this end, let us consider the function
u(x,
z)=
J2
&
(xi
+
^(*)) (* ^ * '
x M-
liq
Then, by (1.5.4), Lu = 0. Furthermore
(1.5.8)
Lu2
= 2uLu -
2[{Elu)2
+ ... +
{Epu)2]
= - 2
[(Em)2
+ ... +
(Epu)2}
0.
We want to prove that
(1.5.9) (Lu2) = -2 £
qij
?¥.
We have
^
2
= 5Z &&(* + ^ ) ( *; + ^ )
= Y2
&€J(X*X3
+
x*^j + XJ^ +
^V)-
Also
L [(Xi + ^ ) ( * i +
Vj)]
= - Cij - Oi + Ci^' + 0 ^ + L ( W )
- ] T [Zeibu^^ + baZ^ + Zzib^ + btjZ^].
l£m
Since (L(?/V)} - 0, we conclude that
(L((x
i
+ ^ i ) ( x
i
+ ^ ) ) ) = -2g
i j
which proves (1.5.9).
Now, if we had
then by (1.5.8), we would have that E{U 0, for all 1 i p, i.e. the func-
tion it would be constant along the integral curves of the vector fields Ei,...,Ep.
Since these vector fields satisfy Hormander's condition, u would then be a constant
function, and hence we would have that
^Xl + ... + €qxq = £ V 1 + - + W + const.
This is absurd because the right term is a bounded function and the left term is
not bounded.
6. A parabolic Harnack Inequality.
THEOREM
1.6.1. Let L be a centered left invariant sub-Laplacian on a con-
nected Lie group of polynomial volume growth G. Let also U a compact neighborhood
of the identity element e of G and let a, b G N. Then there are a,/3,0 G N, a (5
and c 0 such that for all n G N and all u 0 satisfying
(^+L)u = 0 in
(0,On2)
x
U9n,
we have
(1.6.1) sup
{u(an2,x)
:xe
Uan}
cinf
{u(f3n2,x)
: x G
Uhn)
.
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