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2
()
2
()
{}
∫
2
π
Λ⋅Λ⋅−
( ),
(
k
)
=Λ
ω
⋅
exp
ik
ω
d
ω
n
n
n
2
R
(
)
(
)
4
π
j
+
1
4
π
j
+
1
2
ω
1
2
∑
()
(
)
=
∫
∫
⋅
e
ik
ω
d
ω
Bzz
ν
,
⋅Λ
(
)
n
[]
12
2
8
2
jZ
∈
4
π
j
4
π
j
1
2
ω
2
4
π
4
π
∑
=
∫∫
()
(
)
j
ed
⋅
ik
ω
ω
Bz
ν
,
z
,
z
|
Λ+
(
2 )|
π
2
n
[]
123
0
0
2
8
2
jZ
∈
2
44
ππ
22
ππ
()
(
)
∫∫
ν
∫∫
=
B zzz ed
,,
⋅
ik
ω
ω
=
Π ⋅
ed
ik
ω
ω
=
δ
123
ok
,
00
00
Thus, we complete the proof of theorem 1.
Theorem 2.
For every
kZ
∈
2
mn Z
+
∈
and
, we have
Λ⋅Λ⋅−
(),
(
k
)
=
δδ
. (21)
m
n
m n
,
,
k
Proof.
For the case of
mn
=
, (20) follows from Theorem 1.As
mn
≠
and
mn
∈Ω
,
,
the result (20) can be established from Theorem 2, where
0
. In what follows, assuming that
m
is not equal to
n
and at least
one of
{,}
Ω=
{0,1, 2, 3}
0
mn
doesn't belong to
rewrite
m
,
n
as
mm
=
Ω
0
,
1
+
λ
,
nn
=
4
+
μ
,
mnZ
+
,
∈
,
λμ∈Ω
,
where
and
.
1
1
1
1
1
1
1
0
mn
=
1
,
λμ
≠
1
.
Case 1.
If
then
By (17), formulas (21) follows, since
1
1
2
(2
π
)
Λ⋅Λ⋅−
( ),
(
k
)
∫
=Λ
()
ω
Λ
()exp{
ω
⋅
ik
ω
}
d
ω
4
m
+
λ
4
n
+
μ
m
n
2
R
11
1 1
()
λ
(
)
( )
μ
(
)
∫
=
R
Bz
,
z
Λ
(
ω
2)
⋅ Λ
(
ω
2)
Bz
,
z
⋅
exp{
ik d
ω
}
ω
1
1
m
n
12
12
2
1
1
()
λ
(
)
∑
( )
μ
(
)
ik
ω
∫
=
Bz
,
z
Λ
( 22 )
ωπ
+
s
⋅ Λ
( 22 )
ωπ
+
s Bz
, ,
z
z
⋅
e d
ω
1
1
m
m
2
12
123
[0,4
π
]
1
1
2
sZ
∈
1
∫
=
Ξ
⋅
exp{
ik
ωω
}
d
=
O
λμ
,
2
2
(2
π
)
[0,2
π
]
11
mn
≠
mm
=+
n
4
λ
,
=
4
n
+
μ
2
,
mn Z
+
,
∈
,
Case 2.
If
we order
where
1
1
1
2
2
2
22
λ
2
,
μ ∈Ω
0
.
m
=
n
2
,
λμ
≠
2
.
and
If
then
Similar to Case 1, we have (21)
2
2
mn
≠
2
,
follows. That is to say, the proposition follows in such case. As
we order
2
mm
=+
3
,
λ
nn
=+
μ
mn Z
+
,
∈
,
, once more, where
and
2
3
2
3
3
33
Thus, after taking finite steps (denoted by
r
), we obtain
λμ∈Ω
,
.
mn
,
∈Ω
,
and
33
0
r
r
0
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