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In-Depth Information
Z
(8.15)
is a Riesz basis of
V
0
. Now a basis is a Riesz basis, if it satisfies the Riesz
condition. If we have a function
φ
B
=
{
N
n
(
x
−
r
)
}
,
r
∈
∈
L
(
IR
) and two constants
A
and
B
with
0
<A
≤
B<
∞
, then we say that
{
φ
(
.
−
r
)
,
r
∈
Z
)
}
satisfies the Riesz
condition if
∞
2
2
2
,
l
2
,
A
{
c
r
}
≤
c
r
φ
(
.
−
r
)
≤
B
{
c
r
}
{
c
r
}∈
r
=
−∞
and the Fourier transform
φ
of
φ
satisfies
∞
|φ
(
x
+2
πr
)
2
A
≤
|
≤
B,
a.e.
r
=
−∞
In order to find the condition for the cardinal B-spline, we should detect the
lower and upper bounds
A
and
B
. From equation (8.14),
N
n
(
x
)=
N
n−
1
(
x
)
∗
N
1
(
x
)
=
N
n−
2
(
x
)
∗
N
1
(
x
)
∗
N
1
(
x
)
=
N
1
(
x
)
∗
N
1
(
x
)
∗
N
1
(
x
)
∗···∗
N
1
(
x
)
,
and hence taking the Fourier transform, we get
N
n
(
ω
)=(
N
1
)
n
(
ω
)
.
Since,
N
1
(
ω
)=
1
0
e
−iωx
dx
(8.16)
1
−
e
−
iω
iω
=
.
Therefore.
2
=
2
n
e
−iω
iω
1
−
| N
n
(
ω
)
|
.
Now,
e
−iω/
2
(
e
iω/
2
−e
−iω/
2
)
1
−
e
−
iω
iω
=
iω,
=
e
−iω/
2
ω
sin (
ω/
2)
,
=
e
−iω/
2
sin (
ω/
2)
(
ω/
2)
.
Therefore, considering 2
π
periodicity with replacement of
ω
by 2
x
and sum-
ming over
r
, the expression for
| N
n
(
ω
)
2
becomes
|
∞
∞
sin
2
n
(
x
+
πr
)
(2
x
+2
πr
)
2
n
,
| N
n
(2
x
+2
πr
)
2
=
e
−
4
inx/
2
2
2
n
|
r
=
−∞
r
=
−∞
∞
1
(
x
+
πr
)
2
n
,
=
e
−
2
inx
(sin
2
n
x
)
(8.17)
r
=
−∞
∞
1
(
x
+
πr
)
2
n
,
=(sin
2
n
x
)
r
=
−∞
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