28
GEORGIOS K. ALEXOPOULOS
LEMMA 3.4.1. Either
(l, l + r
2
) x ^ ( e ) G Q ,
or there is d\ #i(£) 0 (i.e. 9\ depends only on £ and the group G) such that
(3.4.4) \W\(l + Ol)\Ak\
and hence
(3.4.5) \W\ (l + 91)a\A\.
The proof of the above lemma will be given later.
If (1,1 + r2) x Br(e) £ Q then (3.3.2) follows from lemma 3.1.1 and corollary
3.1.2. So we shall assume that
(l, l + r
2
) x 5
r
( e ) ^ Q .
By using (3.4.3) repeatedly (or by using corollary 3.1.2), we can see that there
is 62 0 and m G N such that for all a 1,
(3.4.6)
* ( (
S
W
S
2
) x ^ ( e M r r V , ^
2
) x B ( e ) , ( 0 , ( T W ) x £?cs(e))
b2G~m.
Let us fix a 1 such that
(3.4.7) ( 1 + ^ ^ 1 + T -
If Q = (£, t + 52) x Bs(x) e Q then we set
Qi = (t,t + ^ V ) x B8(x),
Q1
= {Q\Qe Q}
and
W1=UQieQiQ1.
If
Q1
= (t,t +
a2r]2s2)
x Ba(x) e
Q1
then we set
Q2 = (t + s2, t + TVs 2 ) x Bs(x), Q2 = { Q \ Q G Q1}
and
W2 = UQ2eQ2Q2.
It follows from (3.4.2), (3.4.3) and (3.4.6) that if the constant c is chosen large
enough then there is £3 0 such that
(3.4.8)
^(W2,
A(0,c
2
r
2
) x Bcr{e)) 83.
Let
7
| ( l , l + r
2
) x B
r
( e ) |
and let u (0,1) to be determined later.
Case Ha : \W2 \ (1,1 + r2) x Br(e)\
LO\A\.
This assumption implies that
\W2
\ (1,1 +
r2)
x BP(e)| oryKl, 1 +
r2)
x Br{e)\.
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