Interpolating Spline Wavelet Packets
Jianzhong Wang
November 16, 2000
Abstract
Interpolating wavelets are used to construct the bases for smoothness
spaces and for which the coefficients are obtained by sampling rather
than integration. We construct the wavelet packets corresponding to the
interpolating spline wavelets and provide the interpolating schemes for
performing the wavelet transforms.
1
Introduction
Let Cu be the Banach space of the functions, which are bounded and uniformly
continuous on R, equipped with the uniform norm || · ||∞ . Chui and Li introduced the MRA’s and the corresponding wavelet structures in Cu (See [1]. The
similar discussion can also be found in [3]. ) Let Nm (x) be the m-th order
B-spline defined recursively by N1 (x) = χ[0,1) (x) and Nm (x) = Nm−1 ∗ N1 (x) =
R1
Nm−1 (x − t) dt. For convenience, we write ψ0 (x) = N2m (x). It is known that
0
ψ0 satisfies the following two-scaling equation:
¶
2m µ
X
2m
−(2m−1)
ψ0 (x) = 2
ψ0 (2x − k).
(1)
k
k=0
By setting
X
Vj = {
ck ψ0 (2j x − k);
(ck ) ∈ `∞ },
k∈Z
the authors of [1] point out that {Vj }j∈Z forms an MRA of Cu in the following
S
T
sense: Cu = Vj , Vj = R. Then the wavelet function corresponding to this
MRA can be defined by
X
ψ1 (x) =
(−1)k−1 ψ0 (k − 1)ψ0 (2x − k).
(2)
k∈Z
P
Letting P0 (z) =
Ψ0 (z) = 12 k∈Z ψ0 (k)z k , and Q0 (z) = zΨ0 (−z), we
can rewrite equations (1.1) and (1.2) to
ω
ψb0 (ω) = P0 (z)ψb0 ( ),
2
ω
z = e−iω/2 ,
ψb1 (ω) = Q0 (z)ψb0 ( ),
2
2m
( 1+z
,
2 )
1
R
where fb is the Fourier transform of f : fb(ω) := R f (x)e−ixω dx. Now we define
the wavelet subspace
X
Wj := {
dk ψ1 (2j x − k), (dk ) ∈ l∞ },
k∈Z
L
L
which is a complement of Vj in Vj+1 : W
Vj = Vj+1 , where
denotes
the
Tj
L
direct sum, i.e. Wj + Vj = Vj+1 and Vj Wj = ∅. Therefore Cu = R j∈Z Wj .
It can be verify that
g(2−j Z) = 0,
∀g ∈ Wj .
(3)
Noting that
Ψ0 (z) :
Ψ1 (z) :
1X
ψ0 (k)z k 6= 0,
2
k∈Z
1X
=
ψ1 (k + 1/2)z k 6= 0,
2
=
|z| = 1,
k∈Z
then ∀f ∈ Cu , we can define interpolating operators I0j : Cu → Vj such that
I0j f (2−j Z) = f (2−j Z), and I1j : Cu → Wj such that I1j f (2−j (Z + 1/2)) =
f (2−j (Z + 1/2)). By assertion (1.4), we have
M j
I1 ,
j ∈ Z.
(4)
I0j+1 = I0j
L
L
where
is theL“direct sum”
L of operators defined by P1L P2 = P1 +P2 −P2 P1 .
Generally, P1 P2 6= P2 P1 . Hence the notation α∈Λ Pα is only applied
for the ordered set Λ, meanwhile the “sum” takes the same order.
According to (1.5), decomposing a function into its wavelet components can
be performed by the interpolating scheme. Hence we call ψ1 an interpolating
wavelet.
The aim of this paper is to construct interpolating wavelet packets for this
structure.
2
Construction of Wavelet Packets
S
Let
the set of natural numbers and Z+ = N {0}. We define
Pn N denote
Pn αd =j
−j−1
d
2
for
an
integer
d
∈
N
whose
dyadic
expansion
is
d
=
j=0 j
j=0 dj 2
with dn 6= 0 and set α0 = 0. It is clear that α2d = αd /2 and α2d+1 = (αd + 1)/2.
In this paper, Id,j , (d, j) ∈ Z+ × Z, always denotes a dyadic interval [d2j , (d +
1)2j ). We also write I0,∞ = [0, ∞) . A set of dyadic intervals T = {Id,j }(d,j)∈ΓT
is called a dyadic decomposition
of an interval L ⊂ R if the intervals in T are
S
mutually disjoint and (d,j)∈ΓT Id,j = L. ( A dyadic interval may has (infinity)
many dyadic decompositions.) Since the intervals in a dyadic decomposition T
are mutually disjoint , we can order their indexes by setting (d, j) < (d0 , j 0 ) if
2
0
d2j < d0 2j . Later we always assume that the index set ΓT for T is well-ordered
in this way.
Now we are to construct the interpolating wavelet packet. Recall that functions ψ0 and ψ1 are already defined by (1.1) and (1.2) respectively. Now for any
positive integer d ∈ N, we inductively define
X
ψ2d (x) =
(−1)k ψd−1 (k + αd−1 )ψd (2x − k),
k∈Z
ψ2d+1 (x) =
X
(−1)k−1 ψd (k + αd − 1)ψd (2x − k).
k∈Z
Similarly, by setting
Ψd (z) =
1X
ψd (k + αd )z k ,
2
(5)
k∈Z
and
Pd (z) =
Qd (z) =
Ψd−1 (−z)
zΨd (−z)
we can rewrite (2.1) as following.
ψb2d (ω) =
ψb2d+1 (ω) =
ω
Pd (z)ψbd ( ),
2
ω
b
Qd (z)ψd ( ),
2
z = e−iω/2 .
It can be verify that
Ψ2d (z 2 ) =
Ψ2d+1 (z 2 ) =
Ψd (−z)Ψd−1 (z) + Ψd (z)Ψd−1 (−z),
2Ψd (z)Ψd (−z).
We shall establish some properties for Pd (z) and Qd (z). For this purpose we
introduce the notion of Pólya frequency polynomials (PF polynomials). A polynomial is called a PF polynomial if its coefficients is a Pólya frequency sequence
(cf. [4]). If the sequence is also symmetric, we call the corresponding polynomial an SPF polynomial. It is well-known that an SPF polynomial p(z) can be
factored as
n
Y
p(z) = z k
(z − rj ),
j=1
where each rj < 0 and rj rn−j = 1. If p(−1) 6= 0, 1 ≤ j ≤ n (which implies that
n is even), then p(z) 6= 0 on the unit circle in the complex plane.
Lemma 2.1 Let Gd (z) = Ψd ((−1)d z). Then Gd (z) is an SPF polynomial of
degree 2m − 1 and Gd (−1) 6= 0.
3
Proof. We use the mathematical induction to prove it. At first, G0 (z) =
2m (z)
Ψ0 (z) = zΠ
(2m−1)! , where Π2m (z) is the Euler-Frobenius Polynomial of degree
2m − 2 (See [2], [5]). Hence Ψ0 (z) is an SPF polynomial with Ψ0 (−1) 6= 0. The
statement is true for d = 0. Let r0,j , 1 ≤ j ≤ 2m − 2, be all roots of G0 (z). By
Q2m−2
2
identity (2.5), G1 (z) = 2z j=1 (z + r0,j
) which implies that the statement is
also true for d = 1. Now for any d ∈ N we assume that the statement is already
true for any l ∈ Z+ , l < d. We will prove it also true for d.
In case of d = 2n + 1, we have Ψd (z 2 ) = 2Ψn (−z)Ψn (z). Since n < d, by the
inductive assumption, Gn (z) in an SPF polynomial with Gn (−1) 6= 0. Similar
to the proof for d = 1, we can claim the statement is true for d.
Now consider the case of d = 2n. Setting Hn (z) = Gn (z)Gn−1 (z) :=
P2m−2
P2m−3
P4m−4
z 2 j=0 hj z j , Hne (z) = z j=0 h2j z j , and Hno (z) = z j=0 h2j+1 z j , we
have Hn (z) = Hne (z 2 ) + zHno (z 2 ) and Gd (z) = 2Hne (z). Since Hn (z) is an SFP
polynomial, so does Gd (z). Note that Hno (−1) = 0, then Gd (−1) = 2Hne (−1) =
2(Hn (i) − iHno (−1)) = 2Hn (i) 6= 0. Hence the statement is true for any d ∈ Z+ .
Lemma 2.2 For any integer d ∈ N, we have the following.
(1) supp ψd ⊂ [1, 2m];
(2) Ψd (z) 6= 0,
|z| = 1;
(3) ψd (Z + αn ) = 0,
n ∈ Z+ , n < d.
Proof. Since supp ψ0 ⊂ [0, 2m], by (1.2), supp ψ1 ⊂ [1, 2m]. Using the
mathematical induction we can verify that, for any d ∈ N, the degree of Ψd (z)
is 2m − 1 and z is a simple factor of Ψd (z). Therefore by (2.1), for any d ∈ N,
suppψd ⊂ [1, 2m].
Statement (2) is a direct consequence of Lemma 1.
Now we use the mathematical induction to prove statement (3). It is clear
that statement (3) is true for d = 1. We will prove it also true for any d ∈ N
under the assumption that it is true for any l, l < d. For this purpose, at first we
point out the following fact. If for integers i ∈ Z+ and n ∈ N, ψn (Z + αi ) = 0,
then for any k ∈ Z, ψn (2(Z + α2i ) − k) = 0 and ψn (2(Z + α2i+1 ) − k) = 0. Now
we consider the case of d = 2n. By assertion (2.5), the inductive assumption,
and the fact we just mentioned, we can obtain ψd (Z + αl ) = 0, 0 ≤ l < d. The
statement is proven. In the case of d = 2n + 1, we can obtain ψd (Z + αl ) =
0, 0 ≤ l < d − 1 in the same way. Then we only need to verify ψd (Z + αd−1 ) = 0.
It can be derived from the calculation of
X
ψ(l + α2n ) =
(−1)k−1 ψn (k − l) + αn )ψ(2l + 2α2n − k)
k∈Z
=
X
(−1)k−1 ψn (k − 1 + αn )ψ(2l + αn − k) = 0, ∀l ∈ Z.
k∈Z
From Lemma 1 and Lemma 2 we obtain the following.
P
Lemma 2.3 Let Wjd = { k∈Z ck ψd (2j x−k); (ck ) ∈ `∞ } and Idj be the interpolating operator from Cu to Wjd such that Idj f (2−j (Z + αd )) = f (2−j (Z + αd )).
Then for any d ∈ Z+ and j ∈ Z, we have
4
0
(1) If Id,j ⊂ Id0 ,j 0 , then Wjd ⊂ Wjd0 ;
L 2d+1
2d
Wj−1 ;
(2) Wjd = Wj−1
j−1 L j−1
(3) Idj = I2d
I2d+1 .
From Lemma 3, we can prove the following theorem.
Theorem 2.4 Let T c be a dyadic decomposition of [0, ∞), T o a dyadic composition of (0, ∞), and T a dyadic decomposition of ID,J . Then the following
statements hold.L
(1) Cu = R (d,j)∈Γ 0 Wjd ;
T
L
(2) Cu = (d,j)∈ΓT c Wjd ;
L
(3) WJD = (d,j)∈ΓT Wjd ;
L
J
(4) ID
= (d,j)∈ΓT Idj , provided that set ΓT is finite.
Since ψd has the properties established in Theorem 1, we call W := {ψd }d∈N
an interpolating wavelet packet. We do not add ψ0 into the packet, because ψ0
is not a wavelet, but a scaling function.
3
Dual Wavelet Packet
In order to form the dual wavelet packet of W, we introduce the inner product
space
L2(−m) = {f ; f (i) ∈ L2 , i = 0, 1, · · · , m},
R
which is equipped with the inner product < f, g >:= f (m) (x)g (m) (x) dx. We
write f ⊥ g, if < f, g >= 0. It is obvious that W ⊂ L2(−m) . Let
Wjd,2 = {
X
ck ψ(2j x − k),
(ck ) ∈ `2 }.
k∈Z
We have the following.
T
Lemma 3.1 If Id,j
0
Id0 ,j 0 = ∅, then Wjd,2 ⊥ Wjd0 ,2 .
Proof. Write f [2m−1] (x) = f (2m−1) (x+) − f (2m−1) (x−). By using the
mathematical induction we can prove that
ψd∗ (l + αd ) = 22+(2m−1) ψd−1 (l + αd−1 ),
where [x] is the integer part of x. By assertion (3.1), we obtain
∗
ψ2d
(l + α2d+1 )
X
2m−1
(−1)k ψd−1 (k + αd−1 )ψd∗ (2l + αd + 1 − k)
= 2
k∈Z
= 2
d+3
2 (2m−1)
X
(−1)k ψd−1 (k + αd−1 )ψd−1 (2l + 1 − k + αd−1 ) = 0.
k∈Z
5
(6)
It follows that Sd := {x;
from (3.1) and Lemma 1,
ψd∗ (x) 6= 0} ⊂
< ψd (x − k), ψd0 (x − k 0 ) >= (−1)m
Sd
n=0 (Z
X
+ αn ). Note that if d < d0 ,
ψd∗ (x)ψd0 (x + (k − k 0 )) = 0,
x∈Sd
0
then Wjd,2 ⊥ WjdT,2 . The lemma is true in the case of j = j 0 . In the case of
j < j 0 , since Id,j Id0 ,j 0 = ∅. there is a d1 , d1 6= d0 , such that Id,j ⊂ Id1 ,j 0 , and
0
0
therefore Wjd,2 ⊂ Wjd01 ,2 . Since Wjd01 ,2 ⊥ Wjd0 ,2 , Wjd,2 ⊥ Wjd0 ,2 . In the case of
j > j 0 , the proof is similar.
Definition 3.1 W̃ := {ψ̃d }d∈N ⊂ L2(−m) is called a dual wavelet packet of
W(with respect to L2(−m) ) if
0
< ψd (2j x − k), ψ̃d0 (2j x − k 0 ) >= 0
for Id,j
T
Id0 ,j 0 = ∅, and < ψ̃d , ψd >= 1.
By lemma 4, if ψ̃d is chosen from W0d , then the first condition in Definition
1 is automatically satisfied for j 6= j0. In this case, W̃ := {ψ̃d }d∈N will be
a dual wavelet packet of W provided that {ψ̃d (x − k)}k∈Z is a dual basis of
{ψd (x − k)}k∈Z in space W0d for each d ∈ N.
P
Lemma 3.2 Let Ψd (z) = k∈Z ψd (k + αd )z k . then Ψd (z) 6= 0 on |z| = 1.
Proof. From (3.1), we have
Ψd (z) = 2[
d+1
2 ](2m−1)
Ψd−1 (z).
(7)
Hence the lemma follows Lemma 1.
Theorem 3.3 Let Gd (z) = (−1)m [Ψd (z)Ψd (z)]−1 =
X
ψ̃d (x) =
gk ψd (x − k).
P
k∈Z gk z
k
and set
(8)
k∈Z
Then W̃ = {ψ̃d }d∈N is a dual wavelet packet of W (with respect to L2(−m) ).
Proof. Since Ψd (z)Ψd (z) 6= 0 on the unit circle |z| = 1, the sequence
{gk } is exponential decay as |k| → ∞, and therefore ψ̃d ∈ L2(−m) . Defining
P
Ψ̃d (z) = k∈Z ψ̃d (k + αd )z k , we have Ψ̃d (z) = Gd (z)Ψd (z). Then
X
< ψd (x), ψ̃d (x − l) > z l
= (−1)m Ψ(z)Ψ̃d (z)
l∈Z
= Ψd (z)[Ψd (z)Ψd (z)]−1 Ψd (z) = 1.
The theorem is proven.
6
Now we turn back to the space Cu . Let the dual space of Cu be denoted
by Cu∗ , which contains all linear bounded functionals on Cu . For f ∗ ∈ Cu∗ and
f ∈ Cu , the functional value of f ∗ at f is denoted by (f, f ∗ ).
It is in Cu∗ that the classical Dirac delta function (distribution) δ defined by
2m
(f, δ) = f (0), f ∈ Cu . It can be verified that dd2m x ψ̃d (x) is also a functional in
Cu∗ . In fact, it is a linear combination of the Dirac delta function and its integer
translates with exponential decay coefficients.
Definition 3.2 W ∗ := {ψd∗ }d∈N ⊂ Cu∗ is called a dual wavelet packet of W
0
(with respect to Cu ) if (ψd (2j x − k), ψd∗0 (2j x − k 0 )) = 0 for Id,j ∩ Id0 ,j 0 = ∅, and
(ψd , ψd∗ ) = 1.
Now we give a dual wavelet packet of W with respect of Cu .
P
k
Theorem 3.4 Let (−1)m Ψ−1
k∈Z βk z , and set
d (z) :=
X
ψd∗ (x) :=
βk δ(x − k).
(9)
k∈Z
Then W ∗ = {ψd∗ }d∈N is a dual wavelet packet of W (with respect to Cu ).
Proof. We have
j
j0
m 2mj 0
< ψd (2 x), ψ̃d (2 (x − l)) >= (−1) 2
and
µ
¶
d2m
j
j0
ψd (2 x), 2m ψ̃d (2 (x − l))
d x
µ
¶
0
0
d2m
(ψd (2j x), ψd∗ (2j (x − l))) = (−1)m ψd (2j x), 2m ψ̃d (2j (x − l)) .
d x
It follows that
0
0
0
(ψd (2j x), ψd∗ (2j (x − l)) = 2−2mj < ψd (2j x), ψ̃d (2j (x − l)) > .
(10)
Then the theorem can be derived from (3.5) and Theorem 2.
References
[1] Chui, C. K. and C. Li, Dyadic affine Decompositions and functional wavelet
transforms, SJMA, 1994, to appear.
[2] Chui, C., K. and J. Z. Wang, On compactly supported wavelet and a duality
principle, TAMS 330 (1992), 903–915.
[3] Donoho, D., L., Interpolating wavelet transform, 1992, preprint.
[4] Karlin, S., Total Positivity, Stanford University Press, Stanford, California,
1968.
[5] Scheoberg, I., J., Cardinal Spline Interpolation, CBMS-NSF Series in Applied Math. #12, SIAM Pub., Philadephia, 1973.
7