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y Pm j=1 log j = m log m ; m + O(log m), from (5.24) it follows that
Case a). Let 2 F1. Then (5.25), (5.27) imply
em( )=m ;1
mX
j=l
dj( )=(ej=m) ; 0 (1 + o(m ;1=2 )): (5.27)
dj( ) ; 1+m ;1
= m ;1
mX
j=l
mX
j=l
(m=j) 1; O((m=n) + j ;1 log j)
0 ;
dj( ) ; 1+O(m )
for some 0 > 0 when >0ischosen small enough. Hence for large m
= e ; 0
em( )=m ;1
mX
j=l
( m
ej ) 0; ; 1+O(m
0
; 1+O(m; ) cj ; 0j
; 0 +1 2 0 ; + O(m ) c(log n) ;2s =2
using the inequality e y ; (1 + y) cy 2 for y ;1+ , c>0, and j 0 ; j (log n) ;s . This proves
(5.20) in case i =1.
Case b). Let 2 F2. Then 1 ; 0 ; 1+ Then
m ;1
mX
j=l
dj( ) C ;1 m ;1
C ;1 m ;1
mX
j=l
; 0
(j=m) ; 0 = C ;1 m ;1
mX
j=l
(m=j) 1;
(C ) ;1 :
Choosing >0 small, (5.25) and assumption l n imply (5.20) for i =2.
Case c). Let 2 F3. Then 0 ; 1+ , and thus
m ;1
mX
j=l
dj( ) C ;1 m ;1
mX
j=l
)
mX
j=l
(m=j) 0;
(j=m) ; 0 C ;1 (m=l) 0; ;1
which together with (5.25) and assumption l n imply (5.20) for i =3.
To prove (5.22), set
By Lemma 7.3,
i( )=p ;1
i sm( )=m ;1
mX
j=l
Ej i( ) ; i( 0 )j 2k CDm(
p ;1
i dj( ) 0
`j (I( `j) ; EI( `j)) i =1 2 3:
0 ) k Ej i( )j 2k CDm( ) k (5.28)