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4
uZ
∈
mn Z
+
∈
Theorem 3.
For every
and
, we have
Γ⋅Γ⋅−
(),
(
k
)
=
δδ
. (21)
m
n
m n
,
,
k
Proof.
For the case of
mn
=
, (20) follows from Theorem 1. As
mn
≠
and
mn
,
∈Ω
,
equality (20) can be established from Theorem 2, where
0
Λ =
{0,1, 2,
⋅⋅⋅
,15}
. In what follows, assuming that
m
is not equal to
n
and at least
{,}
mn
Γ
0
,
m
n
one
of
doesn't
belong
to
rewrite
,
as
mm
=
16
+
λ
,
nn
=
16
+
μ
,
mn
∈
Z
+
,
λμ∈Γ
,
.
where
and
1
1
1
1
1
1
11
0
mn
=
1
,
λμ
≠
1
.
Case 1.
If
then
By (14), (16) and (18), (20) holds, since
1
1
4
∫
(2
π
)
Γ⋅Γ⋅−
( ),
(
k
)
=Γ
(
ω
)
Γ
(
ω
) exp{
⋅
ik
ω
}
d
ω
16
m
+
λ
16
n
+
μ
m
n
4
11
1 1
R
ik
ω
()
λ
∑
( )
μ
∫
(
)
⋅
ed
ω
=
Dz
(,
z
,
z
,
z
)
Γ
( 22 )
ω
+
v
π
⋅ Γ
( 22 )
ω
+
v
π
D z
,
z
,
z
,
z
1
1
m
m
4
1234
1234
[0,4
π
]
1
1
4
vZ
∈
∫
=
Ξ
⋅
exp{
ik
ωω
}
d
=
O
4
λμ
,
[0,2
π
]
11
Λ⋅Λ⋅−
(),
(
kO
)
=
.
Therefore,
m
n
and
()
φφ
(), (), ()
xx x
ι
,
x
ι ∈
J
()
Theorem 4
[7]
.
Let
LR
de-
fined by (28), (29), (33) and (34), respectively. Assume that conditions in Theorem 1
are satisfied. Then, for any function
be functions in
2
ι
fx
()
LR
∈
()
, and any integer n,
2
7
n
−
1
∑
∑ ∑ ∑
f
,
φφ
( )
x
=
f
,
. (27)
( )
x
ι
:,
su
nu
,
nu
,
ι
: ,
su
3
3
uZ
∈
ι
=
1
s
∞
uZ
∈
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