34 GEORGIOS K. ALEXOPOULOS
Let us observe that
- ( m i + Mi) - m1 = - ( M i - mi) .
So the function
v =(u - mi ) f - ( M i - mi ) j = 2(u - m1)(M1 - m i ) - 1
satisfies v(t,x) 1 for (t,x) G A.
So, by (3.3.1) there is S 0 such that
mf{v; {(3r\ (/? + b2)r2) x £
6 r
(e) } 8.
This implies that
m,2 m i -f - ^ ( M i mi) .
and hence
M
2
- m
2
M
2
- m i 8(M± - mi)
M i - m i - -5(M1 - mi ) = (1 - - 5 ) ( M i - mi) .
which proves (3.7.1).
Case II : \A\ ±|(1,1 + r 2 ) x J3r(e)|.
Then we consider the function v = Mi u and we are in the previous case.
3.8. En d of th e proof of theore m 3.1. We shall use again the notation
Dr(t,x) = (t-r2,t) x Br{x).
Let us fix b 0. Then by corollary 3.1.2 there is (3 1, c 1, 8 0 and m G N
such that such that
1. for all r 1 and all (t,x) G ( l , r 2 ) x S
r
(e) ,
(3.8.1) *((/?r 2 ,(/ ? + 6 2 )r 2 ) x 5
6 r
(e) , (t,x), (0, (/^ + 6 2 )r 2 ) x £
c r
(e) ) 5r~™.
2. for all r 1 and every ball D
£ r
C ( l , r 2 ) x J3r(e),
(3.8.2) *((/?r 2 ,(/ ? + fc2)r2) x B
6 r
(e), Z
er
(t,x), (0, (/? + 6 2 )r 2 ) x £
c r
(e) ) fem.
We shall prove that there is a constant A 0 such that
s u p L ; ( - r 2 , r 2 ) xBr/2(e)\ A
for all r 4 all functions u 0 satisfying
+ L)u = 0 in (0,(/3 + 6 2 )r 2 ) x £
c r
( e )
inf {u;
(/3r2,
(p + 6
2
)r
2
) x Bhr(e)} = 1.
This of course will prove the theorem.
Let us observe that by (3.7.1)
(3.8.3) sup{w;(l,r 2 ) x Br(e)\ -rm
o
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