26

GEORGIOS K. ALEXOPOULOS

Also by lemma 3.2.2 there is c$ 0 such that

PxKrt-to]6

for all t G

(a~2r2,a2r2).

So, if we take c 1 + a + c$ then by (3.2.3)

u(t,x) / pt-t0(x,y)dy - Px[r*r t-t0]

JAt0

25-5 = S

for all (t,x) G

(a~2r2,a2r2)

x Bar(e) and the lemma follows.

3.3. The second growth lemma.

LEMMA 3.3.1 (second growth lemma). For all b 0, there is (3 1, c 1,

S 0 and m G N such that such that

(3.3.1) ttf

(/?r2,(/?

+

fc2)r2)

x £

6r

(e), A (0, (/? +

&2)r2)

x Bcr(e)\

\A\

(l, l + r 2 ) x £

r

( e ) |

for all r 1 and eien/ measurable subset A C (1,1 +

r2)

x Br(e).

The above lemma will actually be a consequence of the following:

LEMMA 3.3.2. For all b 0, there is (3 1, c 1, ( 5 0, 5' 0, 0 0

and ra G N swc/i that such that for all r 1 and every measurable subset A C

(l,r

2

) x Br(e) either

(3.3.2) ^ ( / ? r 2 , ( / ? + 62)r2) x £

6r

(e), A, (0, (/? + a62)r2) x £cr(e)")

J(l,l + r2)xB

r

(e)|y '

or there is a measurable subset AQ C (1,1 -f

r2)

x Br(e) such that

(3.3.3) |4| (1 + 0)14

and

(3.3.4) ^ ( ^

0

, A ( 0 , l + r2) x£

c r

(e) ) J'.

3.4. Proof of the lemma 3.3.2. We shall use the notation

Q{s,t,x) = (t-

-s2,t+-s2)

x Bs(x).

Let us fix ro 3 and k r%.

If r2 6k, then (3.3.2) follows from (3.1.2) (or the local Harnack inequality

(2.2.1)).

So we shall assume from now on that

r2

6k.