30 GEORGIOS K. ALEXOPOULOS
and
u e ( 0 , | ] ,
then we shall have
\A0\(l+d±)\A\
which proves (3.3.3).
3.5. Proof of lemma 3.4.1. Let us assume that (1,1 -f r2) ^ Q and let us
fix an e 0 such that
(3.5.1) £ (l + e ) £ l .
We set
= {Q e Q : £|Q| \QnASl\(l + s)\Q\}.
We have the following :
LEMMA 3.5.1.
(3.5.2)
UQO
GQ
OQ
0
= UQGQQ = W.
PROOF.
We shall use the notation
Ds(t,x) = (t-
s2,t)
x Bs(x).
We shall prove that for all D G Q and all (r, y) G D there is G such
that (r,y) G L°. This of course will imply the lemma.
So let us fix a ball Ds(t,x) = (t s2,t)x Bs(x) G Q and a point (r, y) G Ds(t,x).
If
£|£,(t,x)| |^
5
(t,x) n A5l| (l + e)£\Da{t,x)\,
then there is nothing to prove. So let us assume that
(3.5.3) \Ds(t, x) n \ | ( l + e)S\Da(t, x)\.
Since d(x,y) s, by chosing a slightly smaller value for s if necessary, we can
assume that s + d(e, x) r.
If there is a G [s -f- d(e, x), r) such that
f |(1, 1 + r2) x B„(e)| |(1,1 + (J2) x B„(e) n 4 | ( l + e)£|(l, 1 + a2) x J3ff(e)|
then we just take = (1,1 +
a2)
x Ba(e).
So, let us assume that this is not the case, i.e. that
(3.5.4) |(1,1 + a2) x Ba(e) D A6l\ f |(1,1 + a2) x 5
a
(e)|
for all a G [s + d(e,x),r).
We set
C = r d(e, x) s and T = d(e, x) + -
and we consider an admissible curve (cf. section 2.1) c : [0,T] T V such that
c(0) = e and c(T) = x.
For every a G [0, T], we set v(a) s + T a.
If
i;(0)2
t - 1, then we set
= (t- v(a)2,t) x £.(j) (c(r)), a G [0,T].
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