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(
)
Lemma 2. Assuming that
φ
( )
x
is an orthogonal scaling function.
Pz z z z is
,,,
1234
the symbol of the sequence {()}
du defined in (3).Then we have
2
2
2
Ω=
Pzzzz
(,,,)
+
P z z z z
( ,
−
−
, ,
−
−
)
+
P zzzz
( ,,,)
−
1234
1
2
3
4
1234
2
2
2
+
Pz
(,
−
z z z
, , )
+
Pz z
(, ,
−
z z
, )
+
Pz z z
(, , ,
−
z
)
1
2
3
4
1
2
3
4
1
2
3
4
2
2
2
+−−
Pz zzz
( ,
,,)
+−
Pzz zz
( ,, ,)
−
+−
Pzzz z
( ,,,
−
)
1
234
12
34
123
4
2
2
2
+
Pz
(
,
−
z
,
−
z z
,
)
+
Pz
(
,
−
z z
,
,
−
z
)
+
Pz z
(
,
,
−
z
,
−
z
)
1
2
3
4
1
2
3
4
1
2
3
4
2
2
=
1
2
− +−
2
Pzz z z
(
,
,
−−
,
)
+
Pz z z z
(
,
−−−
,
,
)
+−−−
Pz z zz
(
,
,
,
)
+−−
Pz zz z
(
,
,
,
)
12
3
4
1
2
3
4
1
2
3
4
1
2
3
4
fx
ν
()
ν =
0,1,
⋅⋅⋅
,15
Lemma 3. If
(
) are orthogonal wavelet functions associated
φ
()
x
with
. Then we have
{
Ξ= ∑
1
D
()
λ
(( 1)
−
j
z
,( 1)
−
j
z
,( 1)
−
j
z
,( 1)
−
j
z
).
D
()
ν
(( 1)
−
j
z
,( 1)
−
j
z
,( 1)
−
j
z
,( 1)
−
j
z
)
λμ
,
j
=
0
1
2
3
4
1
2
3
4
+
D
()
λ
((
−
1)
j
+
1
z
, (
−
1)
j
z
, (
−
1)
j
z
, (
−
1)
j
z
)
⋅
D
()
ν
((
−
1)
j
+
1
z
, (
−
1)
j
z
, (
−
1)
j
z
, (
−
1)
j
z
)
1
2
3
4
1
2
3
4
+
D
()
λ
((
−
1)
j
z
, (
−
1)
j
+
1
z
, (
−
1)
j
z
, (
−
1)
j
z
).
D
()
ν
((
−
1)
j
z
, (
−
1)
j
+
1
z
, (
−
1)
j
z
, (
−
1)
j
z
)
1
2
3
4
1
2
3
4
D
()
ν
((
−
1)
j
z
, (
−
1)
j
z
, (
−
1)
j
+
1
z
, (
−
1)
j
z
)}
+
D
()
λ
((
−
1)
j
z
, (
−
1)
j
z
, (
−
1)
j
+
1
z
, (
−
1)
j
z
)}
1
2
3
4
1
2
3
4
+
D
()
λ
((
−
1)
j
z
, (
−
1)
j
z
, (
−
1)
j
z
, (
−
1)
j
+
1
z
)
⋅
D
()
ν
((
−
1)
j
z
, (
−
1)
j
z
, (
−
1)
j
z
, (
−
1)
j
+
1
z
)
1
2
3
4
1
2
3
4
(
λ
)
j
+
1
j
j
+
1
j
(
ν
)
j
+
1
j
j
+
1
j
+
D
((
−
1)
z
, (
−
1)
z
, (
−
1)
z
, (
−
1)
z
).
D
((
−
1)
z
, (
−
1)
z
, (
−
1)
z
, (
−
1)
z
)
1
2
3
4
1
2
3
4
()
λ
j
+
1
j
j
j
+
1
()
ν
j
+
1
j
j
j
+
1
+
D
((
−
1)
z
, (
−
1)
z
, (
−
1)
z
, (
−
1)
z
).
D
((
−
1)
z
, (
−
1)
z
, (
−
1)
z
, (
−
1)
z
)
1
2
3
4
1
2
3
4
=
δ
λν∈
,
, , ,
⋅⋅⋅
, .
,
(18)
λν
4
Theorem 1 [6] . For nZ +
∈
vZ
∈
,
, we have
Γ⋅ Γ⋅−
(),
(
v
)
=
δ
(19)
n
n
0,
v
uZ
∈
4
and nZ +
∈
ι ∈
{0,1,
2,
⋅⋅⋅
,14,15}
Theorem 2 [6] . For
,
, we have
Γ⋅
(),
Γ
(
⋅ − =
u
)
δδ
. (20)
16
n
16
n
+
ν
0,
0,
u
{
}
ν ∈
0,1,
⋅⋅⋅
,15
Proof. By Lemma 1 and 3, and formulas (14) and
, we get that
4
(2
π
)
Γ⋅
( ),
Γ
(
⋅ −
u
)
16
n
16
n
+
ν
2
()
(
)
()
(
)
∫
0
ν
=
DzzzzDzzzz
,,,
,,,
⋅
Γ
(
ω
2)
⋅
exp{
iu
ω
}
d
ω
n
1234
1234
4
R
()
(
)
()
(
)
∫
0
ν
=
DzzzzDzzzz
,,,
,,,
1234
1234
4
[0,4
π
]
 
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