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Proof.
As we have
c
t
=MSB(Tr(
f
(
ξ
t
))), and by (1), we obtain that (
1)
c
t
is
−
equal to:
2
l
2
l
−
1
−
1
μ
(Tr(
f
(
ξ
t
))) =
μ
j
ψ
j
(Tr(
f
(
ξ
t
))) =
μ
j
Ψ
j
(
f
(
ξ
t
)))
.
j
=0
j
=0
Changing the order of summation, we obtain that:
2
l
2
l
−
1
−
1
k
+
H−
1
Ψ
(
g
(
ξ
t
))
,
Θ
(
τ
1
,...,τ
r
)=
...
μ
j
1
...μ
j
r
j
1
=0
j
r
=0
t
=
k
where
g
(
X
)=
j
1
f
(
Xξ
τ
1
)+
j
2
f
(
Xξ
τ
2
)+
...
+
j
r
f
(
Xξ
τ
r
)
.
By Lemma 4.3
g
(
X
)
=0,andmoreover
g
(
X
)
∈
S
D
. Applying Lemma 2, we
have:
<D
2
m/
2
2
+1
.
k
+
H−
1
π
ln
4(2
m
−
1)
Ψ
(
f
(
ξ
j
))
π
j
=
k
Applying Corollary 7.4 of [5] (for
l
≥
4), we have:
⎛
⎞
r
2
l
π
ln(2) + 1
r
2
l
2
l
2
l
−
1
j
r
=0
|
−
1
j
=0
|
−
1
⎝
⎠
...
μ
j
1
...μ
j
r
|
=
μ
j
|
≤
.
j
1
=0
The Lemma follows.
The main result is the following estimate:
Theorem 7.2.
With notation as above, we have the bound:
2
l
π
ln(2) + 1
r
2
+1
D
√
2
m
.
2
r
≤
π
ln
4(2
m
H
1
2
r
−
1)
N
(
v
)
−
π
r
2
.Let
u
=
∈
Z
Proof.
For any
t
∈
[
k, k
+
H
−
1], let
c
t
=(
c
t
+
τ
1
, ..., c
t
+
τ
r
)
r
2
.Then
∈
Z
(
u
1
,...,u
r
)
r
u
·
c
t
=
u
i
c
t
+
τ
i
i
=1
and by definition of
N
(
v
)wehave
k
+
H
−
1
1)
u·
c
t
=
1)
u·v
N
(
v
)
.
S
(
u
)=
(
−
(
−
t
=
k
∈
Z
2
v
Thus we have
N
(
v
)=
1
2
r
1)
u·v
S
(
u
)
.
(
−
2
u∈
Z
