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for all h ∈ AC. Thus, integrating over all the h ∈ AC ∩ H and by continuity of the mapping
G(Y) = dh g(h, Y) (128)
AC∩H
1
− where g(h, Y) = e 2Yi·h 2 i , we can conclude that,
Now, we remark that
with
H
AC∩H
g(x∗ i )
1
−
dh e 2f ′′ (x∗ i )h2 i =
and
dh e
AC∩H
H
g(xi) e −f(xi)
AC∩H
AC∩H
1
−
dh e 2f ′′ (x∗ i )h2 i =
dh e− 1
2f ′′ (x ∗ i )h2 i
411
S→∞
−−−→ 1. (129)
1
−
dh e 2f ′′ (x∗ i )h2
i +
AC∩H
1
−
dh e 2f ′′ (x∗ i )h2 i , (130)
1
− 2f ′′ (x∗ i )h2 i ∼ O
(2π) N−1
2
N w
i=1
2 iN
j=1 f ′′
j (x∗ j )
f ′′
i (x∗ i )
, (131)
α
−
e f ′′ (x∗
) , α > 0, (132)
where x ∗ = max{x ∗ i } (note that 1/f ′′ (x) → ∞ with x by Proposition 1). Indeed, we clearly have
AC∩H
1
−
dh e 2f ′′ (x∗ i )h2 i ≤
=
AC
1
−
dh e 2f ′′ (x∗ i )h2 i , (133)
(2π) N/2
f ′′ (x ∗ i )
BC
du
i
1 u
− 2
e 2 i
f ′′ (x∗ i )
2π f ′′ (x∗ i )
, (134)
where we have performed the change of variable ui = f ′′ (x∗ i ) · hi and denoted by BC the set {h ∈
RN , u2 i ≤ C2 }. Now, let x∗ max = max{x∗ i } and x∗min = min{x∗i }. Expression (134) then gives
AC∩H
1
−
dh e 2f ′′ (x∗ i )h2 i ≤
SN−1
(2π) N/2
f ′′ (x∗ min )N/2
du
BC
e−
1
2 f ′′ (x∗ max )u 2 i
(2π f ′′ (x∗ min
f
= SN−1
′′ (x∗ max) N/2
f ′′ (x∗
N
Γ
min
)N 2 ,
C2 2 f ′′ (x∗ max)
N
2 −1
f ′′ (x ∗ max) N/2
f ′′ (x ∗ min )N
C 2
2 f ′′ (x ∗ max)
))N/2 , (135)
, (136)
· e −
C 2
2 f ′′ (x∗ max ) , (137)
which decays exponentially fast for large S (or large x ∗ max) as long as f ′′ goes to zero at infinity, i.e, for any
function f which goes to infinity not faster than x 2 . So, finally
AC∩H
1
−
dh e 2f ′′ (x∗ i )h2 i =
which concludes the proof of equation (113).
(2π) N−1
2
N w
i=1
2 iN
j=1 f ′′
j (x∗ j )
f ′′
i (x∗ i )
23
+ O e −
α
f ′′ (x∗ max )
, (138)