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)
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.
Previous Page Next Page