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)

.