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Denote v = v(x1x2). Then x1 =(;v + x2ES1)D ;1 where D =(2; ES1 ; ES2) ; m ;1=2 x2. Since
jf (x1x2)j C exp(;c(x 2 1 + x 2 2)) with some c>0, then for any >0,
P (Vm y) =
Z
v y:jx1jjx2j m
When jx1j jx2j m , and >0 is small, Lemma 7.1 implies
f ((;v + x2ES1)D ;1 x2)(;D ;1 )dvdx2 + o(e m):
D ;1 =1+hm(x2)+o(e m) hm(x2) =(e(1` )+e(2` ))(m=n) + m ;1=2 x2
x1 = ;v ; vhm(x2)+e(1` )x2(m=n) + o(e m)(jvj + jx2j) =;v + o(1): (6.31)
This and Taylor expansion imply
P (Vm y) =
Z
v y:jvjjx2j m
f (;v x2)+(vhm(x2) ; e(1` )x2(m=n) ) @
@v f (;v x2) dvdx2 + o(e m)
and integrating out x2, we arrive at the second order expansion:
F (Vm v) =
Z
v y
(v)Pm(v)dv + o(e m)
where Pm(y), is quadratic in y. Comparing this expansion with (6.18) we conclude that Pm(x)
1 ; m ;1=2 ep(x) where ep (6.19), as already shown, has coe cients (6.21), so that Pm(x) = 1+
m ;1=2 (2 + x 2 )=3 to prove (6.28).
To show (6.29) note that
F ((@By) )=
Z
(@By) :jxij m
f (x1x2)dx1dx2 + o(m ;1=2 )
for any >0. By (6.30), from y = v(x1x2) we can solve forx2 = h(y x1). Thus
Z
C
jx1j m
when = m ;1; and > :
Lemma 6.3 Let
Z
F ((@By) ) C
Z
(@By) :jxij m
dx1dx2 + o(m ;1=2 )
dx2dx1 + o(m
x22[h(yx1); h(yx1)+ ]
;1=2 ) C2 m + o(m ;1=2 )=o(m ;1=2 )
V 0 m = Vm + e 1+
m
where Vm is given by (4.8), m 2X and 0 <