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Thus if jm ;1=2Zjj =2 foranyj =0:::3thenjSj ; ejj for large m, and by Taylor expansion
we get
3X @
Ym = fm(e0e1:::e3)+ fm(e0e1:::e3)(Sj ; ej)
@xj
In view of (5.8), (5.4),
(Sj ; ej) 2
+O(
3X
j=0
j=0
(Sj ; ej) 2 )+O((m=n) 2 )+m ;1 m:
2 (ESj ; ej) 2 + m ;1 Z 2
j = O((m=n) 2 )+2m ;1 Z 2
1
= O((m=n) 2 )+2m ;1 m (5.9)
since Z1 2X. This proves (5.5). (5.9) implies that jSj ; ejj = O((m=n) )+m ;1=2 m, and therefore
from (5.5) it follows that (5.6) holds. (5.6) implies (5.7).
Lemma 5.5 Let (4.4) and Assumptions f l m hold, and let
Then
for some >0 and m 2X.
m n 4
4 +1 ; some >0: (5.10)
m 1=2 R (1) ( 0)+UmR (2) ( 0)+
U 2 m
2m 1=2 R(3) ( 0) = 1+
m m (5.11)
Proof. Multiplying both sides of (4:3) by m1=2 , the left hand side of (5.11) can be written as
m with
1+
m
1=2 ;1;
m = O(m m jR(1) ;1 ;1;
(b)j + m m jUmj 3 jR (4) ( )j)
where j ; 0j jb ; 0j. It remains to show that m 2X. We show rst that
Now R (4) ( )isalinearcombination of terms
where
Now
F l0
0
( )F l1
1
( ) :::Fl4
4
F 4
0
( )
jR (4) ( )j C log 4 m: (5.12)
( )
l0 + :::+ l4 =4 0 l0:::l4 4 (5.13)
Fk( )= dk
d k F0( ) F0( )=
jFk( )j log k m
mX
j=1
mX
j=l
j I( `j) k 0:
j I( `j) = (log m) k F0( ):
Therefore the terms (5.13) are bounded bylog 4 m and (5.12) holds. By Lemma 5.7, Um = m 1=2 m e m
where e m 2X. Assumption (5.10) implies that
1=2 2
m m m some >0: (5.14)
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