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)\.