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ORTHOGONAL POLYNOMIALS AND THE CONSTRUCTIONOF PIECEWISE POLYNOMIAL SMOOTH WAVELETS∗

G. C. DONOVAN† , J. S. GERONIMO‡ , AND D. P. HARDIN§

SIAM J. MATH. ANAL. c© 1999 Society for Industrial and Applied MathematicsVol. 30, No. 5, pp. 1029–1056

Abstract. Orthogonal polynomials are used to construct families of C0 and C1 orthogonal,compactly supported spline multiwavelets. These families are indexed by an integer which representsthe order of approximation. We indicate how to obtain from these multiwavelet bases for L2[0, 1]and present a C2 example.

Key words. orthogonal polynomials, multiwavelets, splines

AMS subject classification. 41A15

PII. S0036141096313112

1. Introduction. Wavelet bases [2] for L2(R) have the nice property that onceone of the basis functions is known the rest may be obtained by dilation and integertranslation of this function. In this case the basis has one generator. Multiwavelets[1], [5], [9], [11], [13], [14], [19] are similar to wavelets except that the basis is obtainedby the dilation and integer translation of several functions instead of just one. Theconstruction of most wavelets and multiwavelets is based on multiresolution analysis(MRA) [17], [16]. Let φ0, . . . , φr be compactly supported L2-functions, and supposethat V0 = clL2 span{φi(· − j): i = 0, 1, . . . , r, j ∈ Z}. Then V0 is called a finitelygenerated shift invariant (FSI) space. Let (Vp)p∈Z be given by Vp = {φ(2p·): φ ∈V0}. Each space Vp may be thought of as approximating L2 at a different resolutiondepending on the value of p. The sequence (Vp) is called a multiresolution analysisgenerated by φ0, . . . , φr if (a) the spaces are nested, · · · ⊂ V−1 ⊂ V0 ⊂ V1 ⊂ · · ·, and(b) the generators φ0, . . . , φr and their integer translates form a Riesz basis for V0.Because of (a) and (b), we can write

(1.1) Vj+1 = Vj ⊕Wj ∀j ∈ Z.

The space W0 is called the wavelet space, and if ψ0, . . . , ψr generate a shift-invariantbasis for W0, then these functions are called wavelet functions. If, in addition,φ0, . . . , φr and their integer translates form an orthogonal basis for V0, then (Vp)is called an orthogonal MRA. It has been shown in Lemma 2.1 of [5] that we canalways assume that φj , j = 0, . . . , r can be chosen so that they are minimally sup-ported on [−1, 1]; i.e., each scaling function has support in [−1, 1], and the nonzerorestrictions of the scaling functions and their integer translates to [0, 1] are linearlyindependent. Set Φ = (φ0, . . . , φr)∗. Then (a) and (b) imply that Φ satisfies thematrix dilation equation

(1.2) Φ(t) =∑

CiΦ(2t− i).∗Received by the editors December 4, 1996; accepted for publication (in revised form) August 3,

1998; published electronically August 16, 1999. This research was supported in part by the Institutefor Mathematics and its Applications with funds provided by the NSF.

http://www.siam.org/journals/sima/30-5/31311.html†Department of Mathematics, Princeton University, Princeton, NJ 08544 (donovan@

math.princeton.edu). This author was supported by an NSF postdoctoral fellowship.‡School of Mathematics, Georgia Institute of Technology, Atlanta, GA 30332 (geronimo@

math.gatech.edu). This author was supported by an NSF grant.§Department of Mathematics, Vanderbilt University, Nashville, TN 37240 (hardin@math.

vanderbilt.edu). This author was supported by an NSF grant.

1029

1030 G. C. DONOVAN, J. S. GERONIMO, AND D. P. HARDIN

In order to obtain orthogonal MRAs it is useful to divide V0|[0,1] into the followingsubspaces:

• A0 = span{φ ∈ Φ: supp φ ⊂ [0, 1]} = {g ∈ V0: supp g ⊂ [0, 1]},• C0 = span{φ(·)χ[0,1]: φ ∈ Φ} ªA0,• C1 = span{φ(· − 1)χ[0,1]: φ ∈ Φ} ªA0.

Since the functions in A0 are orthogonal to their integer translates and by Gram–Schmidt may be made mutually orthogonal, it is not difficult to see [5] the following.

Theorem 1.1. (Vp) is orthogonal iff C0 ⊥ C1.In general it is not possible to obtain an orthogonal basis for V0 by taking finite

linear combinations of the original basis functions that generate it. However, wecan modify V0 by adding appropriate functions so that C0 ⊥ C1. The notion ofintertwining MRAs arises from the observation that some functions can be movedfrom one level of an MRA to another without destroying the defining properties of anMRA. Suppose that (Vp) is an MRA as defined above and that W is an FSI subspaceof V1 with Riesz basis {φr+1(·− j): j ∈ Z} for some compactly supported φr+1. Withthis we can generate a new space V0 = V0+W together with associated dilation spacesVp = {φ(2p·): φ ∈ V0}. Then (Vp) is also a multiresolution analysis. Indeed, it is clear

that V0 ⊂ V0. It is also the case that V0 ⊂ V1, since both V0 and W are subspaces ofV1. From these two inclusions and the definition of the dilated spaces, it follows that

(1.3) · · · ⊂ V−1 ⊂ V−1 ⊂ V0 ⊂ V0 ⊂ V1 ⊂ V1 · · ·

and hence

· · · ⊂ V−1 ⊂ V0 ⊂ V1 ⊂ · · · .

It is easy to see how the other conditions necessary for (Vp) to be an MRA also followfrom (1.3). Of course this process can be repeated if more functions are needed. LetA1 = {φ ∈ V1: suppφ ⊆ [0, 1]}. If W ⊂ A1ªA0 such that (I−PW )C0 ⊥ (I−PW )C1,where PW is the orthogonal projection onto W , then the intertwining MRA will bean orthogonal MRA. In Theorem 1 of [5] the following theorem was proved.

Theorem 1.2. If (Vp) is a multiresolution analysis generated by compactly sup-ported scaling functions, then there is some pair of integers (q, n) and some orthogonalmultiresolution analysis (Vp) such that

Vq ⊂ V0 ⊂ Vq+n.

We call (Vp) an intertwining MRA and the theorem implies in particular thatfor those MRAs generated by splines there exist orthogonal intertwining MRAs alsogenerated by splines. The main results of this paper give explicit constructions oforthogonal intertwining MRAs for C0 and C1 spline MRAs with various orders ofapproximation. In this paper we will use the notation

(1.4) fi,j = 2i/2f(2i· − j)

and P{f1,f2,...,fk}, fi ∈ L2(R), will denote the orthogonal projection onto the subspacespanned by f1, . . . , fk. In section 2 we make precise which MRAs we will be studyingand examine the spaces A0, C0, and C1 associated with these MRAs. It is here whereorthogonal polynomials play a role since the A0 spaces for the MRAs we will beconsidering will be spanned by subclasses of ultraspherical polynomials (see also [18],[15], and [10]). Furthermore, we introduce the functions that will span W , which we

ORTHOGONAL POLYNOMIALS 1031

will “borrow” in order to construct an orthogonal intertwining MRA. These functionsare chosen so as to be smooth as possible and symmetric or antisymmetric with respectto x = 1

2 . In section 3 we derive various properties of these functions needed to carryout the construction. In section 4 we give a description of how to construct waveletsfrom scaling functions. This explicit computation of the multiwavelets functions interms of the scaling functions allows us to preserve symmetry or antisymmetry. Alsoindicated in this section is how to construct multiwavelet bases for L2[0, 1]. In section 5we construct families of compactly supported continuous orthogonal scaling functionshaving an axis of symmetry. Each family is indexed by an integer which representsthe order of approximation. In section 6 families of C1 scaling functions and waveletsare constructed. Finally, in section 7, a C2 example is given.

2. Piecewise polynomial MRA. The MRAs we will study are those associatedwith piecewise polynomial splines. Let Snk be the space of polynomial splines of degree

n with Ck knots at the integers, and set V n,k0 = Snk ∩ L2(R). With V n,kp as above itis known that these spaces form a multiresolution analysis [1], [5], [11], [19]; however,with the exception of Sn−1, which were studied by Alpert, it is not possible to findcompactly supported, orthogonal generators for these spaces. We will consider Snkwhen n > 2k + 1.

In order to construct orthogonal intertwining MRAs we examine the spaces An,k0 ,

Cn,k0 , and Cn,k1 , associated with V n,k0 as described above, as well as An,k1 , defined by

analogy with An,k0 to be An,k1 = {g ∈ V1: supp g ⊂ [0, 1]}. Our goal is to find a space

W ⊂ An,k1 ªAn,k0 so that

(2.1) (I − PW )Cn,k0 ⊥ (I − PW )Cn,k1 .

We begin by describing a convenient basis for An,k0 . Note that from the description of

An,k0 , g ∈ An,k0 iff g(t) = tk+1(1 − t)k+1q(t) for some polynomial q of degree at most

n−2k−2. Thus the linear dimension of An,k0 is n−2k−1. If {tk+1(1−t)k+1qj−2k−2: j =

2k + 2, . . . , n} forms an orthogonal basis for An,k0 , then∫ 1

0

t2k+2(1− t)2k+2qi−2k−2(t)qj−2k−2(t) dt = ciδi,j ,

and we see that the qi’s are orthogonal with respect to the measure t2k+2(1− t)2k+2 dton [0, 1]. The monic ultraspherical polynomials pλi are monic polynomials of degree

i = 0, 1, . . . , which are orthogonal on [−1, 1] with respect to the measure (1−t2)λ−12 dt

(Szego [20]). From the above equation, we can choose qi(t) = p2k+ 5

2i (2t − 1), i =

0, . . . , n− 2k − 2. This leads to ([7], [8]) the following lemma.

Lemma 2.1. A basis for An,k0 is {tk+1(1−t)k+1p2k+ 5

2i (2t−1): i = 0, . . . , n−2k−2},

where each p2k+ 5

2i is a monic ultraspherical polynomial of degree i.

The ultraspherical polynomials already have been used in the construction ofwavelets from fractal interpolation functions (Donovan, Geronimo, and Hardin [6],Donovan [4]). For later computations we define the set {φki }, where φki (t) = (1 −t2)k+1p

2k+ 52

i−2k−2(t) for i ≥ 2k + 2 with φk2k+1 = 0.Some important properties of the ultraspherical polynomials that we will use later

[20] follow.(a) The Rodriguez formula:

(1− t2)mpm+ 1

2n (t) = (−1)n

(n+ 2m)!

(2n+ 2m)!Dn(1− t2)n+m.


(b) The recurrence formula:

pm+ 1

2n+1 (t) = tp

m+ 12

n (t)− anpm+ 12

n−1 (t), n = 1, 2, . . .

with an = (n+2m)n(2n+2m+1)(2n+2m−1) , p0(t) = 1 and p1(t) = t.

(c)

pm+ 1

2n (t) =

1

(n+ 2)(n+ 1)D2p

m−2+ 12

n+2 (t).

The polynomials also have the following useful representation in term of hyper-geometric functions [20]:

(2.2) pm+ 1

2n (t) =

2n(m+ 1)n(n+ 2m+ 1)n

2F1

(−n n+ 2m+ 1

m+ 1;

1− t2

),

where formally

pFq

(a1 . . . apb1 . . . bq

; t

)=∞∑i=0

(a1)i . . . (ap)i(b1)i . . . (bq)i(1)i

ti

with (a)i = (a)(a + 1) . . . (a + i − 1). Since one of the numerator parameters in thehypergeometric function in (2.2) is a negative integer, the series has finitely manynonzero terms and the result is a polynomial.

From the recurrence formula is it not difficult to see that

(2.3)

∫ 1

−1

pm+ 1

2n (t)2(1− t2)mdt =

2(n+ 2m)!n!

(2n+ 2m− 1)!!(2n+ 2m+ 1)!!,

where m!! = m(m− 2) . . .. Thus

(2.4) ‖φkn‖2L2 =2(n+ 2k + 2)!(n− 2k − 2)!

(2n− 1)!!(2n+ 1)!!.

We now consider the spaces Cn,ki , i = 0, 1. Since the dimension of V n,k0 |[0,1] is n + 1

and the dimension of An,k0 is n−2k−1, the dimension of Cn,k0 is k+1 and the same is

true for Cn,k1 . For computational compatibility with the ultraspherical polynomials,we scale these spaces so that they are defined on [−1, 1] instead of [0, 1] and denote

them as Cn,ki ( ·+12 ) and An,k0 ( ·+1

2 ).Let rki (t) = (1 − t)i(1 + t)k+1, lki (t) = rki (−t), and Pn,k denote the orthogonal

projection onto An,k0 . Then a basis for Cn,k0 ( ·+12 ) is rn,ki = (I − Pn,k)rki , i = 0, . . . , k,

while for Cn,k1 ( ·+12 ) the analogous basis is ln,ki = (I − Pn,k)lki . We shall refer to the

families of rki and lki , respectively, as right and left “ramps,” where right and leftdenote on which side of the interval [−1, 1] the unprojected functions do not vanishk + 1 times.

The action of the projection can be readily computed from the following innerproduct:

(2.5) 〈rki , φkn〉 =2k+i+n+2n!(i+ k + 1)!(n− k − i− 2)!(n+ 2k + 2)!

(k − i)!(k + i+ n+ 2)!(2n)!,


where k ≥ i and n ≥ 2k + 2 (Gradshteyn and Ryzhik [12, p. 826]). This formula canbe derived for instance by substituting (2.2) into the above inner product, integratingterm by term, then using Saalschutz’s formula for summing a 3F2 hypergeometricfunction (Bailey [3, p. 49]). The symmetry between rki and lki and the fact that theultraspherical polynomials are of definite parity imply that 〈lki , φkn〉 = (−1)n〈rki , φkn〉.

There are several simple useful relations among the ramp functions which we nowderive. The first is the obvious relation

(2.6) rn,ki = rn−1,ki − 〈r

ki , φ

kn〉

〈φkn, φkn〉φkn.

Note that for n ≥ 2k + 2, the polynomial rn,ki+1 is uniquely determined by the order

of its zeros at ±1, its orthogonality to An,k0 , its degree, and its leading coefficient.

The polynomial (1 − t)rn,ki vanishes at ±1 the same number of times as rn,ki+1, is of

degree n+ 1, and is orthogonal to An−1,k0 . Subtracting off φkn+1 times an appropriate

constant then projecting out φkn yields the following relation:

(2.7) rn,ki+1(t) = (1− t)rn,ki (t) +〈rki , φkn+1〉〈φkn, φkn〉

φkn(t)− 〈rki , φ

kn〉

〈φkn, φkn〉φkn+1(t).

Since (1− t2) ddtrki (t) = (k+1+ i)rki+1−2irki , an argument similar to that given above

leads to

(2.8)

(1− t2)d

dtrn,ki (t) = (k + 1 + i)rn,ki+1 − 2irn,ki + (n+ 2)

〈rki , φkn+1〉〈φkn, φkn〉

φkn(t)

+ n〈rki , φkn〉〈φkn, φkn〉

φkn+1(t).

Likewise,

(2.9)

d

dt

((1− t2)rn,ki (t)

)= (k + 3 + i)rn,ki+1 − 2(i+ 1)rn,ki + n

〈rki , φkn+1〉〈φkn, φkn〉

φkn(t)

+ (n+ 2)〈rki , φkn〉〈φkn, φkn〉

φkn+1(t).

Analogous formulas are easily obtained for ln,ki . In the multiresolution analysis given

above, V n,k1 is a spline space with knots at the half-integers. Thus, to do the inter-twining step, we will borrow from the space of splines supported on [0, 1] with a knotat 1

2 . Once again, for purposes of compatibility with the φki , we dilate these functionsso that they are supported on [−1, 1] and have a knot at 0. The dilated functionswill be denoted ukn. To construct these, we first define a sequence of spline functions{ukn} ∈ Ck(R), where

ukn(t) =

{1− |t| tn−1 +

∑ki=1

(−1)in!!2ii! (n−2i)!! (1− t2)i if t ∈ [−1, 1],

0 otherwise

for odd n ≥ 2k + 1, and

ukn(t) =

{t− |t| tn−1 +

∑ki=1

(−1)i(n−1)!!2ii! (n−2i−1)!! t(1− t2)i if t ∈ [−1, 1],

0 otherwise


for even n ≥ 2k+2. It should be noted that the parity of the function ukn is always theopposite of the parity of n. From this sequence, we obtain new sequences by applyingthe projection I − Pn,k:

ukn,m = (I − Pn,k)ukn−m.

The functions to be borrowed from V1( ·+12 ) to give an orthogonal multiresolution

analysis are obtained as appropriate linear combinations of this latter class of splines.To calculate the above projections, it is necessary to find the inner products

ck,n,j =

∫ 1

−1

ukj (t)φkn(t)dt

for n, j ≥ 2k + 1, which are given in the following lemma.Lemma 2.2. With n, j ≥ 2k+ 2, if (n− j) is even, then ck,n,j = 0, and if (n− j)

is odd and j > n, then

ck,n,j =−2

n−j+4k+32 (n− 2k − 2)!j!( j−2k−n

2 )k+1

(n+ 1)n(j − n)!!( j−2k+n−12 )!

× 3F2

(−k − 1 j+2

2j+1

2j−2k−n

2j−2k+n+1

2

; 1

),

=−2

n−j+4k+32 (n−2k

2 )k+1(−2k−n−12 )k+1(n− 2k − 2)!j!

(n+ 1)n(j − n)!!( j+n+12 )!

× 3F2

(−k − 1 j+1

2 k + 1−n2

n+12

; 1

),(2.10)

while if (n− j) is odd and n > j, then

ck,n,j =2 (−1)

n−j+12 +k+1(n− 2k − 2)!j!(n+ 2k − j + 1)!

(n+ 1)n(n−2k+j−12 )!(n−j+2k+1

2 )!

× 3F2

(−k − 1 j+2

2j+1

2j−2k−n

2j−2k+n+1

2

; 1

)

=(−1)

n−j+12 22k+3(n−2k

2 )k+1(−2k−n−12 )k+1 (n− 2k − 2)!j!(n− j − 1)!

(n+ 1)n(n+j+12 )!(n−j−1

2 )!

× 3F2

(−k − 1 j+1

2 k + 1−n2

n+12

; 1

).(2.11)

Remark. When n = 2k + 2 the 3F2 in the second part of (2.11) reduces to atruncated 2F1.

Proof. The fact that ck,n,j = 0 for (n− j) even follows from the parity of ukj and

φkn. From property (c) above we find the p2k+ 5

2n = n!

(n+2k+2)!D2k+2p

1/2n+2k+2, where

p1/2n is the monic Legendre polynomial of degree n [20]. Therefore from the definition

of φkn we find that φkn(t) = (1− t2)k+1 (n−2k−2)!n! D2k+2p

1/2n (t). Substitute this into the

integral for cm,n,j and integrate by parts 2k+2 times. Since j ≥ 2k+2, ukj has 2k+1


continuous derivatives in (−1, 1) and Di(1−t2)k+1ukj = 0 for t = ±1, i = 0, . . . , 2k+1.Therefore

ck,n,j =(n− 2k − 2)!

n!

∫ 1

−1

D2k+2[(1− t2)k+1ukj (t)]p1/2n (t)dt.

Observe that for i < k + 1, D2k+2(1 − t2)k+i+1 is a polynomial of degree less than

2k + 2. Consequently the orthogonality of p1/2n implies that only the term |t|tj−1 in

ukj will be nonzero in the above integral. Using the parity of ukj and φkn to integrate

on [0, 1], then expanding (1− t2)k+1 and differentiating yields

ck,n,j = −2(n− 2k − 2)!

n!

k+1∑i=0

(k + 1

i

)(−1)i

(j + 2i)!

(j + 2i− 2k − 2)!

∫ 1

0

tj+2i−2k−2p1/2n (t)dt.

With l = j + 2i− 2k − 2 we find, using (2.2) with m = 0, that∫ 1

0

tlp1/2n (t)dt =

2nn!l!

(n+ 1)n(l + 1)!2F1

(−n n+ 1

l + 2; 1

2

)=

2nn!l!Γ( l+32 )Γ( l+2

2 )

(n+ 1)n(l + 1)!Γ( l+3+n2 )Γ( l+2−n

2 ),

where Kummer’s Theorem has been used in summing the 2F1 [3, p. 11]. If n is even,l is odd and we find

Γ( l+3+n2 )

Γ( l+32 )

=( l+n+3−2i

2 )i(l+n−2i+1

2 )!

( l+12 )!

and

Γ( l+22 )

Γ( l+2−n2 )

=( l−2i+2−n

2 )k+1l!!2j−n−l

2

( l−2i+2−n2 )i(j − n)!!

.

The above formulas and the substitutions (j + 2i)! = 22ij!( j+12 )i(

j+22 )i and

(−1)i(k+1i

)= (−k−1)i

(1)iyield the first line of (2.10) for n even. To obtain the sec-

ond line of (2.10), use the transformation formula [3, p. 85]

3F2

(−(k + 1) a b

e f; 1

)=

(e− b)k+1(f − b)k+1

(e)k+1(f)k+13F2

(−(k + 1) b a+ b− k − e− fb− e− k b− f − k ; 1

).

To arrive at the first line of (2.11) for n odd perform manipulations similar to those

described above onΓ( l+3+n

2 )

Γ( l+22 )

andΓ( l+3

2 )

Γ( l+2−n2 )

. The second line of (2.11) is obtained in a

similar manner.The above formulas show that the hypergeometric functions have a fixed number

of terms for fixed smoothness. Furthermore, the hypergeometric functions given inthe second lines of (2.10) and (2.11) are a sum of positive terms. Any zero appearingin the denominator of the above formulas cancels with a zero in the numerator.


3. Recurrence formulas and inner products. As in the case of the rampfunctions there are useful difference and differential difference equations relating theukn,m’s. The spline ukn,m is uniquely defined in terms of the degree of its zeros at±1, the n − m − 1 continuous derivatives at zero, the values of the left and right(n − m)th derivatives at zero, its degree, and its orthogonality to An,k0 . Denotingby “rem” the remainder of integer division (i.e., p rem q ≡ q(pq − bpq c)), we find from

the definition of ukn−m that tukn−m−1 = ukn−m + ((n − m) rem 2)Kk,n,m(1 − t2)k+1.

Furthermore,∫ 1

−1tukn−1,m(t)φki (t)dt = 0 for i ≤ n− 2. The recurrence formula for φkn,

the fact that (1 − t2)k+1 ∈ An,k0 ( ·+12 ), and parity considerations now imply that for

n−m− 1 ≥ 2k + 1

(3.1) ukn,m = tukn−1,m − εk,n,mφkn−1+(m rem 2),

where

(3.2) εk,n,m =ck,n+(m rem 2),n−1−m

〈φkn−1+(m rem 2), φkn−1+(m rem 2)〉

.

Analogues of (2.8) and (2.9) can also be derived. To this end, note that (1 −t2)Dukn−m = −(n−m)ukn+1−m + (n−m)ukn−1−m and that

〈(1− t2)Dukn,m(t), φki (t)〉 = 〈ukn,m(t), D((1− t2)φki (t)

)〉= 0, i ≤ n− 1

since ukn,m is orthogonal to all functions of the form (1−t2)k+1πn−2k−2, where πn−2k−2

is an arbitrary polynomial of degree less than or equal to n − 2k − 2. The parity ofDukn,m now implies that

(3.3) (1−t2)Dukn,m(t) = −(n−m)ukn+1,m(t)+(n−m)ukn−1,m(t)+δk,n,mφkn+(m rem 2),

where

(3.4)

δk,n,m = −(n−m)ck,n+(m rem 2),n+1−m

〈φkn+(m rem 2), φkn+(m rem 2)〉

+ (n− 1 + (m rem 2))ck,n−1+(m rem 2),n−m


.

Another similar formula is(3.5)D((1− t2)ukn,m(t)

)= −(n−m+ 2)ukn+1,m(t) + (n−m)ukn−1,m(t) + γk,n,mφ

kn+(m rem 2)

with

(3.6)

γk,n,m = (n+ 1 + (m rem 2))ck,n−1+(m rem 2),n−m


+ (n+ 2−m)ck,n+(m rem 2),n+1−m

〈φkn+(m rem 2), φkn+(m rem 2)〉

.

One final useful formula is

(3.7) ukn,n−m = ukn−2,n−m−2 −ck,n−1,n−m+(m rem 2)


φkn−1+(m rem 2).


Note that the above formulas hold for k ≥ 0 and n −m ≥ 2k + 3. In order to findappropriate linear combinations of {ukn,m} that solve (2.1) we need to compute theinner products of the various functions defined above. We begin with the followinglemma.

Lemma 3.1. Given rn,ki and ln,kj , i, j ≤ k, as above,

〈rn,ki , ln,kj 〉 = (−1)n+122k+i+j+3(k+i+1)!(k+j+1)!(n−k−i−1)!(n−k−j−2)!(n+2k+3)!(k−i)!(k−j)!(k+n+j+3)!(k+n+i+2)!(n−2k−2)!

× 3F2

( −(k − j) 2k + i+ j + 4 1

−(n− k − j − 2) k + n+ j + 4; 1

).(3.8)

Proof. Multiply (2.8) by ln,kj and integrate by parts to find

−〈rn,ki , D((1− t2)ln,kj

)〉 = (k + 1 + i)〈rn,ki+1, ln,kj 〉 − 2i〈rn,ki , ln,kj 〉

+ (n+ 2)〈rki , φkn+1〉〈lkj , φkn(t)〉

〈φkn, φkn〉,

where the fact that 〈ln,kj , φkn+1〉 = 〈lkj , φkn+1〉 has been used. Now eliminate D((1 −

t2)ln,kj)

using the analogue of (2.9) for ln,kj and collect terms to get

0 = (k + i+ 1)〈rn,ki+1, ln,kj 〉+ 2(j + 1− i)〈rn,ki , ln,kj 〉 − (k + j + 3)〈rn,ki , ln,kj+1〉

+ n〈rki , φkn〉〈lkj , φkn+1〉

〈φkn, φkn〉+ (n+ 2)

〈rki , φkn+1〉〈lkj , φkn〉〈φkn, φkn〉

.

Multiply (2.7) by ln,kj and integrate to obtain

〈rn,ki+1, ln,kj 〉 = −〈rn,ki , ln,kj+1〉+ 2〈rn,ki , ln,kj 〉+

〈rki , φkn+1〉〈lkj , φkn〉〈φkn, φkn〉

+〈rki , φkn〉〈lkj , φkn+1〉

〈φkn, φkn〉.

Using the above two equations to eliminate 〈rn,ki+1, ln,kj 〉, solving for 〈rn,ki , ln,kj 〉, and

then using (2.5) gives

〈rn,ki , ln,kj 〉 = (2k+i+j+3)2(k+j+2) 〈rn,ki , ln,kj+1〉+ (−1)n+122k+i+j+3(n+2k+3)!(k+j+1)!(k+i+1)!(n−k−j−2)!(n−k−i−1)!

(k−j)!(k−i)!(k+j+n+3)!(k+i+n+2)!(n−2k−2)! .

Iterating this formula from j+ 1 to k and utilizing the fact that lkk+1 is in An,k0 yieldsthe formula

〈rn,ki , ln,kj 〉 =(−1)n+122k+i+j+3(k + i+ 1)!(k + j + 1)!(n− k − i− 1)!(n+ 2k + 3)!

(k − i)!(k + n+ i+ 2)!(n− 2k − 2)!

×k−j∑m=0

(n− k −m− j − 2)!(2k + i+ j + 4)m(k −m− j)!(n+ k +m+ j + 3)!

.

Finally, with the substitutions (n+ k+m+ j+ 3)! = (n+ k+ j+ 3)!(n+ k+ j+ 4)m,(a− j)! = (−1)j a!

(−a)j, and a ∈ {k −m,n− k −m− 2} we obtain (3.8).


An analogous argument gives

(3.9)

2(i+ j + 1)〈rn,ki , rn,kj 〉 = (k + i+ 1)〈rn,ki+1, rn,kj+1〉+ (k + j + 3)〈rn,ki , rn,kj+1〉

+ (n+ k + 1 + i)〈rn,ki , φkn〉〈rn,kj , φkn+1〉

〈φkn, φkn〉

+ (n− k − i+ 1)〈rn,ki , φkn+1〉〈rn,kj , φkn〉

〈φkn, φkn〉.

Since 〈rn,ki , ukn,m〉 = 〈rki , ukn,m〉 the following recurrence formula is easily obtained

from (3.1) using the relation rki (t) = (1− t)rki−1:

(3.10) 〈rn,ki , ukn,m〉 = 〈rn−1,ki , ukn−1,m〉 − 〈rn,ki+1, u

kn,m〉+ εk,n,m〈rki , φkn+(m rem 2)〉.

To obtain the next formula we will need the following lemma.Lemma 3.2. Formally

3F2

(a b c

d e; 1

)=c(e− a)

de3F2

(a b+ 1 c+ 1

d+ 1 e+ 1; 1

)+d− cd

3F2

(a b+ 1 c

d+ 1 e; 1

).

Proof. The proof will be based on the contiguous relations for 3F2 hypergeometricfunctions found in Wilson’s paper [21]. Although the proof will be formal in practicewe will apply the result when one of the numerator parameters is a negative integer,in which case the series has only a finite number of nonzero terms.

Term by term subtraction yields

3F2

(a b c

d e; 1

)=bc

de3F2

(a b+ 1 c+ 1

d+ 1 e+ 1; 1

)+ 3F2

(a− 1 b c

d e; 1

),

and equation 8 in [21] implies that

de 3F2

(c d a− 1

d e; 1

)= c(d+ e− a− b− c)3F2

(c+ 1 b+ 1 a

d+ 1 e+ 1; 1

)+ (d− c)(e− c)3F2

(c b+ 1 a

d+ 1 e+ 1; 1

).

Finally equation 7 in [21] gives

3F2

(c b+ 1 a

d+ 1 e+ 1; 1

)=

e

e− c 3F2

(c b+ 1 a

d+ 1 e; 1

)− c

e− c 3F2

(c+ 1 b+ 1 a

d+ 1 e+ 1; 1

).

Utilizing the above formulas and simplifying yields the result.Lemma 3.3. Let rn,ki and ukn,m be as above. Then for n− 2m ≥ 2k+ 1, k = 0, 1,

or 2,

(3.11)

〈rn,kk , ukn,2m〉 =(−1)m+k22k+2−n(n− 2k − 1)!(2k + 1)!(n− 2m)!(2m+ 2k)!

(n+ 2k + 1)!(n− k −m)!(m+ k)!

× 3F2

(−k n−2m+2

2n−2m+1

2−2m−2k+1

2 n−m− k + 1; 1

).


Proof. Although this result is true more generally we will use it only for the casesindicated, which simplifies the proof considerably. The above formula can be verifiedby hand for m = 0, k ∈ {0, 1, 2} and n ∈ {2k + 1, 2k + 2}. With m even set i = k in(3.10) to obtain

(3.12) 〈rn+1,kk , ukn+1,m〉 = 〈rn,kk , ukn,m〉 − εk,n+1,m〈rkk , φkn+1〉.

From (3.1) and (2.10) we find that

εk,n+1,m〈rkk , φkn〉 =(−1)

m2 +k+222k+1−n(n− 2k − 1)!(2k − 3)!(n−m)!(m+ 2k + 2)!

(n+ 2k + 2)!( 2n−2k−m2 )!(m+2k+2

2 )!

× 3F2

(−k − 1 n−m+2

2n−m+1

2−m−2k−1

22n−m−2k+2

2

; 1

).

Set a = n−m+22 , b = −k − 1, c = a − 1

2 , and d = −m−2k−12 . Then the hy-

pergeometric functions associated with 〈rn,kk , ukn,m〉, 〈rn+1,kk , ukn+1,m〉, and εk,n+1,m

are 3F2

(a b+1 cd+1 e ; 1

), 3F2

(a b+1 c+1d+1 e+1 ; 1

), and 3F2

(a b cd e ; 1

), respectively.

From Lemma 3.2 we find that (3.11) satisfies (3.12) for n ≥ 2k + 1 + 2m. Multiply(3.7) by rkk and integrate to obtain

(3.13) 〈rn,kk , ukn,2m〉 = 〈rn−2,kk , ukn−2,n−2m−2〉 −

ck,n−1,n−2m

〈φkn−1, φkn−1〉

〈rkk , φkn−1〉.

Since

ck,n−1,n−2m〈rkk , φkn−1〉〈φkn−1, φ

kn−1〉

=(−1)m+k22k−n+3(2n− 1)(2k + 1)!(n− 2k − 3)!(2k + 2m)!

(n− 2k + 1)!(n− k −m− 1)!(m+ k)!

× 3F2

(−k − 1 n−2m+2

2n−2m+1

2−2m−2k+1

2 n−m− k ; 1

),(3.14)

it is not difficult to show that (3.11) satisfies (3.13). To see this interchange d and e in

Lemma 3.2, eliminate 3F2

(a b+1 c+1d+1 c+1 ; 1

)and make the substitutions a = n−2m+2

2 ,

b = −k − 1, c = n−2m+12 , d = n−m− k, and e = −2m−2k+1

2 . Now multiply by

(−1)m+k22k−n+3(2n− 1)(2k + 1)!(n− 2k − 3)!(2k + 2m)!

(n− 2k + 1)!(n− k −m− 1)!(m+ k)!

and use (3.11) and (3.14). The result follows since (3.11) satisfies (3.12), (3.13), andthe initial conditions mentioned above.

The relation

(3.15) 〈rn,ki , ukn,2m+1〉 = 〈rn+1,ki , ukn+1,2m+2〉

shows that we need only compute the above inner products for m even. This relationcan be obtained by observing that ukn−(2m+1) = uk(n+1)−(2m+2) and that from the

parity of ukn+1−2m−2, 〈ukn+1−2m−2, φkn+1〉 = 0. Since lki (t) = rki (−t) the parity of ukn,m

implies that

(3.16) 〈ln,ki , ukn,m〉 = (−1)n+m+1〈rn,ki , ukn,m〉.


We finish this section by obtaining a recurrence formula for the inner products〈ukn,j , ukn,i〉. We will do this only for i and j even since the same reasoning as above

shows 〈ukn,2j+1, ukn,2i+1〉 = 〈ukn+1,2j+2, u

kn+1,2i+2〉. Set m = 2i and increment n by one

in (3.3). Then multiply by ukn,2j and integrate by parts to get

− 〈ukn+1,2i(t), D[(1− t2)ukn,2j(t)]〉= −(n− 2i)〈ukn+2,2i, u

kn,2j〉+ (n− 2i)〈ukn,2i, ukn,2j〉+ δk,n+1,2ick,n+1,n−2j .

From (3.5) we find that the term on the left-hand side of the above equation can beeliminated, giving

− (n− 2j)〈ukn+1,2i, ukn−1,2j〉+ (n− 2j + 2)〈ukn+1,2i, u

kn+1,2j〉

= −(n− 2i)〈ukn+2,2i, ukn,2j〉+ (n− 2i)〈ukn,2i, ukn,2j〉+ δk,n+1,2ick,n+1,n−2j .

Another useful equation is

〈ukn+2,2i, ukn,2j〉 = 〈ukn+1,2i, u

kn+1,2j〉+ εk,n+2,2ick,n+1,n−2j ,

which can be obtained from (3.1). The above two equations can be combined to give

(3.17) (2n−2j−2i+3)〈ukn+1,2i, ukn+1,2j〉 = (2n−2j−2i+1)〈ukn,2i, ukn,2j〉+κk,n,2i,2j ,

where(3.18)κk,n,2i,2j = (δk,n+1,2i− (n− 2i+ 1)εk,n+2,2i)ck,n+1,n−2j + (n− 2j)εk,n+1,2ick,n,n−1−2j .

4. Wavelets. Before we begin the actual construction of spline wavelets somegeneral results will be given that will help in the construction and also show howto modify the bases constructed to obtain wavelet bases for compact intervals. Letthe multiresolution analysis (Vp) be generated by n orthonormal scaling functionsφ0, . . . , φn−1 minimally supported on [−1, 1], with exactly k of these functions,φ0, . . . , φk−1 not having support in [0, 1]. Then from the theory of paraunitary oper-ators there are n orthonormal wavelet functions ψ0, . . . , ψn−1, which we may assumeare also minimally supported on [−1, 1]. Below, we give a method for constructingthe wavelet functions from the scaling functions. In the case of symmetric or an-tisymmetric scaling functions, this method allows the construction of symmetric orantisymmetric wavelet functions.

Using the notation given in (1.4), define Q0 = span{φ0, . . . , φk−1},Q1 = span{φ0

1,0, . . . , φk−11,0 }, Y = Q0 ∩Q1, and dimY = m. Also, let Y0 = Q0χ[−1,0] ∩

Q1χ[−1,0] and Y1 = Q0χ[0,1] ∩Q1χ[0,1] with dimensions m0 and m1, respectively.Lemma 4.1. The number of wavelet functions ψj such that PQ1

ψj 6= 0 is at leastk −m.

Proof. Suppose that there are k < k − m such wavelet functions ψ0, . . . , ψk−1.Suppose then that v ∈ Q1 ª Y is chosen to have unit length and to be orthogonal to

PQ1span{ψ0, . . . , ψk−1}. Such a choice is possible because Q1ªY has k − m dimen-

sions while PQ1span{ψ0, . . . , ψk−1} has at most k dimensions. Let y = (I − PQ0

)v ∈V1 ª V0. Since span{φk, . . . , φn−1} ⊥ Q1, (1.1) implies that y ∈W0.

On the other hand, we claim that y cannot be written as a linear combination ofthe wavelet functions. Indeed, suppose y can be expressed as a linear combination of


wavelet functions. Then in particular PQ1y must be in PQ1

span{ψ0, . . . , ψk−1} andhence orthogonal to v. That is,

0 = PvPQ1y

= PvPQ1(I − PQ0)v

= Pv(PQ1 − PQ1PQ0)v

= Pv(v − PQ1PQ0v)

= v − PvPQ1PQ0v

or

v = PvPQ1PQ0

v.

Because these projections are orthogonal, the above equality is possible only if v ∈ Q0,which is not the case since v ∈ Q1 ª Y .

Lemma 4.2. The number of wavelet functions not supported on [0, 1] is at leastj = 2k −m0 −m1.

Proof. Suppose that there are fewer than j such wavelet functions. By the abovelemma, there must be at least k−m of them, whose span will be denoted by Ψs, suchthat PQ1

ψ 6= 0 for ψ ∈ Ψs \ {0}. In addition, there are < k −m0 −m1 +m others,spanning Ψa. We assume that Ψs has exactly k −m dimensions and PQ1

Ψa = {0}.Since every wavelet function is orthogonal to the generators of V0,

PQ1(Ψs ⊕Ψa) = PQ1ªY (Ψs ⊕Ψa),

where Q1 ª Y has at most k −m dimensions. Thus, by taking linear combinations ifnecessary, one can always arrange for this assumption to be correct. Let Ψ1 be thespan of the remaining wavelets supported on [0, 1].

We begin by defining the following five spaces:

T = (I − PQ1)Q0,

T0 = {t ∈ T : supp t ⊂ [−1, 0]},T1 = {t ∈ T : supp t ⊂ [0, 1]},U = T ª T0 ª T1,

Z = (χ[0,1] − χ[−1,0])U

for which the dimensions are k − m, m0 − m, m1 − m, k − m0 − m1 + m, andk−m0−m1 +m, respectively. Note that since U ⊥ Q1, multiplying by characteristicfunctions to construct Z does not destroy the membership of these functions in V1.Also, since the supports of T0 and T1 are [−1, 0] and [0, 1], it follows that these spacesare orthogonal to U on each of said intervals and thus are orthogonal to Z.

Next, we observe that Q0 ⊕ Ψs = T ⊕ Q1. It is easy to show that both spaceshave dimension 2k −m. Furthermore, T ⊕Q1 = Q0 +Q1 ⊂ Q0 ⊕Ψs because amongthe wavelet functions and their translates, only those in Ψs are not orthogonal to Q1.

Now, by its construction, Z is orthogonal to any translates of Ψ1 as well asnonzero translates of Ψs, Ψa, and Q0. It is also orthogonal to arbitrary translates ofthe scaling functions supported on [0, 1]. However, Z ⊂ V1, so we have

Z ⊂ Q0 ⊕Ψs ⊕Ψa

= T ⊕Q1 ⊕Ψa

= U ⊕ T0 ⊕ T1 ⊕Q1 ⊕Ψa,


and since Z ⊥ T0 ⊕ T1 ⊕Q1, it follows that

Z ⊂ U ⊕Ψa.

But Z has more dimensions than Ψa, so there exists a nonzero z ∈ Z which is alsoin U . By the definition of Z, there is some u ∈ U such that (χ[0,1] − χ[−1,0])u = z.Finally, consider the function x = u + z. Since both u and z are in U , we knowthat x ∈ U ⊥ T1. On the other hand, x ∈ U ⊂ T , and clearly suppx ⊂ [0, 1], sox ∈ T1. The only way for both of these to be true is to have x = 0, which yields acontradiction.

For the scaling vectors constructed in the next section, we find that m0 = m1 =m = 0. In this event we have the following corollary.

Corollary 4.3. If m0 = m1 = m = 0, then the number of wavelet functionsnot supported on [0, 1] is at least 2k.

From the above lemmas we have the following theorem.Theorem 4.4. Let the multiresolution analysis (Vp) be generated by n orthonor-

mal scaling functions φ0, . . . , φn−1 minimally supported on [−1, 1]. Let k, m, m0,and m1 be defined as above. Then there exists n orthonormal wavelet functionsψ0, . . . , ψn−1 minimally supported on [−1, 1] such that exactly 2k −m0 −m1 are notsupported in [0, 1].

Proof. As indicated above, k − m wavelets Ψs not supported in [0, 1] may beconstructed as an orthonormal basis for (I − PQ0

)Q1 since this basis is in V1 and isorthogonal to V0. We construct k−m0−m1 +m more functions in the following way.As indicated in Lemma 4.2, Z is orthogonal to V0 ª

(Q0 ª (Q0 ∩Q1)

)as well as any

wavelet functions supported on [0, 1]. Thus the wavelet functions we seek are foundby taking an orthonormal basis for span(I − PQ0∪Ψs)Z.

We now show that the remaining n−2k+m0 +m1 may be chosen so that they aresupported in [0, 1]. Since the dimensions ofA0 andA1 are n−k and 2n−k, respectively,A1 ª A0 is n-dimensional. From this space we select a subspace K perpendicular to{φi|[0,1]}k−1

i=0 and {φi(· − 1)|[0,1]}k−1i=0 . The dimension of this space will determine the

maximum number of wavelets with support contained in [0, 1]. If after an appropriatechange of basis φ0|[0,1] is in Y0, then φ0 is perpendicular to A1ªA0. Thus K needs to

be chosen perpendicular only to a (k − 1)-dimensional subspace of span{φi|[0,1]}k−1i=1

and a k-dimensional subspace of span{φi(· − 1)|[0,1]}k−1i=0 . Proceeding in this way we

find that the dimension of K is n− 2k +m0 +m1, which completes the proof.Suppose Φ is a set of n scaling functions all supported on [−1, 1] and let Ψ be

a set of wavelet functions constructed as above. Since the wavelet functions are allsupported on [−1, 1] they satisfy the following equation:

(4.1) Ψ(t) =

1∑i=−2

DiΦ(2t− i),

where each Di is an n× n matrix.The above decomposition of the wavelet basis allows a description of how to obtain

a basis when the multiresolution analysis is restricted to [0, 1]. To this end let Ψs, Ψa,and Ψ1 be as above and set Ψn,j = Ψ(2n· − j).

Theorem 4.5. Let the multiresolution analysis (Vp) be generated by orthogonalscaling functions Φ = {φ0, . . . , φn−1}, minimally supported on [−1, 1], and let k,m, m0, and m1 be defined as above. If Φχ[0,1] is an orthogonal set, then for q ≥ 0{φiq,jχ[0,1]: i = 0, . . . , n−1, j = 0, . . . , 2q}\{0} is an orthogonal basis for Vq = Vqχ[0,1].


Furthermore, there exist orthogonal bases

(4.2) Ψq =

2q−1⋃j=0

Ψ1q,j ∪

2q⋃j=0

Ψ2q,j ∪

2q−1⋃j=0

Ψ3q,j ∪

2q⋃j=1

Ψ4q,j ∪

2q−1⋃j=1

Ψ5q,j

χ[0,1]

for the wavelet spaces Wq = Wqχ[0,1] such that V0 ⊕⊕

q≥0 Wq = L2[0, 1].Proof. First note that if Φχ[0,1] is orthogonal, then Φχ[−1,0] is as well. From this

it is easy to see that {φiq,jχ[0,1]: i = 0, . . . , n− 1, j = 0, . . . , 2q} \ {0} is an orthogonal

basis for Vq.Now, for the wavelet spaces, we know that Wq = Vq+1 ª Vq. Thus, if Ψ is any

orthogonal basis for Ψa ⊕Ψs, a basis for Wq may include

2q−1⋃j=0

Ψ1q,j ∪

2q−1⋃j=1

Ψq,j .

To complete the basis, however, it is necessary to see what happens near the ends ofthe interval [0, 1] and to choose a suitable Ψ.

Consider first the left endpoint. Since the remaining basis functions must beorthogonal to those given above as well as the generators for Vq, it is clear thatthey must be obtained from span Ψq,0 by truncation to the interval [0, 1]. In addi-tion, they must be orthogonal to Φq,0χ[0,1]. The number of functions needed canbe found by counting the dimensions associated with their support and subtract-ing the number of orthogonality restrictions. For the former, note that q ≥ 0Ψq,0χ[0,1] ⊂ span(Φq+1,0χ[0,1] ∪ Φq+1,1), a 2n-dimensional space. For the latter,note that these functions must be orthogonal to span(Φq,0χ[0,1] ∪ Ψ1

q,0), which has2n−2k+m0 +m1 dimensions, as well as a (k−m0)-dimensional subspace of span Φq,1.Thus 2n− (2n−2k+m0 +m1)− (k−m0) = k−m1 wavelet functions are required tocomplete the basis on the left. Similar arguments show that there are k −m0 on theright and that these two groups have k = max{k−m0 −m1, 0} functions in common(before truncation). We remark that while it is unusual for an MRA to have nonzerom, m0, or m1, it is even more unusual, but still possible, to have k −m0 −m1 < 0.

Finally, we choose Ψ so that it can be partitioned into four sets:• Ψ2, with k functions used on both ends,• Ψ3, with k −m1 − k functions used only on the left,• Ψ4, with k −m0 − k functions used only on the right, and• Ψ5, with k functions not used on either end.

With Ψ2, Ψ3, Ψ4, and Ψ5 defined in this fashion, (4.2) gives an orthogonal basis forWq.

5. Construction of continuous spline wavelets. We now set k = 0 so thatSn0 is the space of piecewise C0 polynomials with knots at the integers and V n,00 =Sn0 ∩ L2(R). By Lemma 2.1, {φ0

2, . . . , φ0n} forms an orthogonal basis for An,00 ( ·+1

2 ),

where φ0n(t) = (1 − t2)p

5/2n−2(t). In this case r0

0(t) = 1 + t while l00(t) = 1 − t, and

Cn,00 ( ·+12 ) and Cn,01 ( ·+1

2 ) are each one dimensional. Hence from Theorem 1.1 we seethat an orthogonal intertwining MRA can be constructed if we can find a functionw ∈ An,01 ( ·+1

2 )ªAn,00 ( ·+12 ) satisfying 〈(I − Pw)rn,00 , (I − Pw)ln,00 〉 = 0, which for non-

zero w is equivalent to

(5.1) 〈rn,00 , ln,00 〉〈w,w〉 = 〈rn,00 , w〉〈w, ln,00 〉.


As a member of An,00 ( ·+12 ), w must be a spline of degree n with knots at the integers.

Since we would like to construct wavelets that are symmetric or antisymmetric weshall choose w so that it is symmetric or antisymmetric and also so that the knot atzero is as smooth as possible. Since the dimension of the space of piecewise polynomialfunctions in An,01 ( ·+1

2 ) ª An,00 ( ·+12 ) that are Cn−2 at zero is two, there exists a basis

x1, x2 for this space where x1 is symmetric and x2 is antisymmetric (one of them beinga multiple of un,0). Thus if we wish to find a symmetric or antisymmetric w satisfying(5.1), it will be at most Cn−3 at zero. Naturally, if symmetry is not important, asmoother w may be constructed from the space spanned by x1 and x2. Because of(5.3) below we see that w must be chosen having the same parity as (−1)n+1. Thusfor symmetry and the greatest possible smoothness we are forced to choose

(5.2) w0n = α0(n)u0

n,0 + u0n,2

for n ≥ 3. From (3.8), (3.11), and (3.16) we have

(5.3) 〈rn,00 , ln,00 〉 =(−1)n+18

(n+ 2)(n+ 1)n,

(5.4) 〈rn,00 , u0n,2m〉 =

(−1)m(n− 1)!(n− 2m)!(2m)!

2n−2(n+ 1)!(n−m)!(m)!,

and

(5.5) 〈ln,00 , u0n,m〉 = (−1)m+n+1〈rn,00 , u0

n,m〉.Also from (3.9) we find

(5.6) 〈rn,00 , rn,00 〉 = (n+ 1)〈r0

0, φ0n〉〈r0

0, φ0n+1〉

〈φ0n, φ

0n〉

=8

n(n+ 2).

With this (5.1) becomes

0 =

∣∣∣∣ 〈u0n,0, u

0n,0〉〈rn,00 , u0

n,0〉〈rn,00 , u0

n,0〉 |〈rn,00 , ln,00 〉|∣∣∣∣α0(n)

2+ 2

∣∣∣∣ 〈u0n,0, u

0n,2〉〈rn,00 , u0

n,2〉〈rn,00 , u0

n,0〉 |〈rn,00 , ln,00 〉|∣∣∣∣α0(n)

+

∣∣∣∣ 〈u0n,2, u

0n,2〉〈rn,00 , u0

n,2〉〈rn,00 , u0

n,2〉 |〈rn,00 , ln,00 〉|∣∣∣∣ ,(5.7)

where we have used the sign structure of (5.3) and (5.5). In order to find solutions tothis equation we need the following formula, for n ≥ 2 max{i, j}+ 1 with i, j ∈ {0, 2}:(5.8)

〈u0n,2i, u

0n,2j〉 =

(−1)i+j(n− 2i)!(n− 2j)!(2i)!(2j)!(n− 1)!(n2 + 5n+ 2− 2i− 2j)

22n−1i!j!(n− i)!(n− j)!(n+ 1)!(2n+ 1− 2i− 2j).

For the cases n > 2 max{i, j}+ 1 the above formula follows by induction using (3.17)with

κ0,n,2i,2j =(−1)i+j+1(2i)!(2j)!(n− 2i)!(n− 2j)!(n− 1)!

2(2n− 1)i!j!(n+ 1− i)!(n+ 1− j)!(n+ 2)!Πn,

where

Πn = 3n5 + (27− i− j)n4 + (85− 14(i+ j))n3 + (117− 51(i+ j) + 2(i+ j)2)n2

+ (72− 62(i+ j) + 10(i2 + j2) + 24ij)n

+ 16− 24(i+ j)− 4ij(i+ j) + 20ij + 8(i2 + j2),


as well as the initial conditions 〈u02,0, u

02,0〉 = 1

15 , 〈u04,2, u

04,0〉 = − 17

13440 , and 〈u04,2, u

04,2〉 =

1960 . For n = 3, i = 0, j = 1 and n = 3, i = 1, j = 1 (5.8) can be verified by hand.

Substituting the above formulas into (5.7) and solving for α0(n) yields two solu-tions,

(5.9) α0(n) = 2

(n−2)(2n+1)(n−1) ± 2

√3√

(2n+1)(n+1)(2n−3)(n−1)

n(2n− 1),

either of which will suffice to give w as in (5.2). This w is then used to constructorthogonal the scaling functions. For j = 2, . . . , n, set

φj =

{φ0j (2· − 1) if t ∈ [0, 1),

0 otherwise,

φ1 =

{w(2· − 1) if t ∈ [0, 1),

0 otherwise,

and φ0 = (I − P{φ1,...,φn,φ1(·+1),...,φn(·+1)})h, where

h(t) =

{(1− |t|) if t ∈ [−1, 1),

0 otherwise,

where Px is the orthogonal projection onto the subspace spanned by x.Theorem 5.1 (see [7], [8]). For n ≥ 3 and α0(n) given by (5.9), Φ = {φ0, . . . , φn}>

generates an orthogonal multiresolution analysis {V n,0k }. Furthermore, the last n func-tions are symmetric or antisymmetric about 1

2 while the first function is symmetricabout 0.

Proof. Since V n,00 = clL2 span{h(· − i), φ2(· − i), . . . , φn(· − i) ∀i ∈ Z} and V n,00 =clL2 span{φ0(· − i), . . . , φn(· − i) ∀i ∈ Z} we find that V n,00 ⊂ V n,00 . The result nowfollows from the above construction.

Remark. As we have shown, the generators of Φ are the smoothest functionsderived from Φ having the support, symmetry, and orthogonality properties indicatedabove.

With the scaling functions above we may now construct the coefficients C0n,i,

i = −2,−1, 0, 1 in the matrix refinement equation. In light of Theorem 4.4, two ofthe wavelets will not be supported only in [0, 1] while the remaining n − 1 will besupported on [0, 1]. In particular, the following holds.

Corollary 5.2. Let

ψ0 =√

2(I − Pφ0)φ01,0,

ψ1 =√

2(χ[0,1] − χ[−1,0])(I − Pφ01,0

)φ0,

and {ψi: i = 2, . . . , n} be an orthonormal basis for Ψ1 consisting of functions thatare symmetric or antisymmetric with respect to 1

2 . Then {ψ0, . . . , ψn} generates a

shift-invariant orthonormal basis for W0. Furthermore, ψ0(0) = φ0(0).Proof. Up to the constants, the formulas for ψ0 and ψ1 follow directly from

Theorem 4.4. It is true in general that 〈φ0, φ00,1〉 = 1√

2, and hence the normalization

factors for both ψ0 and ψ1 are as shown. To see why this is the case, observe thatφ0

1,0 is the only generator of V1 that does not vanish at 0, and φ01,0(0) =

√2φ0(0). It

is also for this reason that ψ0(0) = φ0(0).


-1

-0.5

0

0.5

1

-1 -0.5 0.5 1

-4

-2

0

2

4

6

0.2 0.4 0.6 0.8 1

φ0 φ1

Fig. 1. Selected C0 scaling functions with approximation order 4 (n = 3).

Table 1Formulas for the scaling functions.

φ0(t) =

√30√

327+56√

14(1855−256√

14)25235210

(2002t3 − (645√

14 + 658)t2

−(1106− 285√

14)t+ 322− 16√

14) for 0 ≤ t < 12√

30√

327+56√

14(−6433+1040√

14)25235210

(20818t2 + (1835√

14− 17024)t

+2254− 1176√

14)(t− 1)for 1

2≤ t ≤ 1

φ0(−t) for − 1 ≤ t < 00 elsewhere

φ1(t) =

{3√

15(√

7+21√

2)1750

(280t2 − (75√

14 + 665)t+ 322 + 54√

14)t for 0 ≤ t ≤ 12

φ1(1− t) for 12< t ≤ 1

0 elsewhere

φ2(t) =√

30t(1− t)χ[0,1]

φ3(t) =√

210t(1− t)(2t− 1)χ[0,1]

In the examples given by Theorem 5.1, Ψ1 is symmetrical about 12 , and hence

it must have an orthonormal basis consisting of symmetrical and antisymmetricalfunctions.

An example is given in Figure 1, where the minus sign is chosen in (5.9), andthe analytic formulas associated with this example can be found in Table 1.1 Thewavelets with support [−1, 1] may be computed using Corollary 5.2 while those sup-ported in [0, 1] can be obtained by finding n− 1 orthogonal functions, symmetrical orantisymmetrical with respect to 1

2 , from the space (I−P{φ0,φ0(·−1)})An,01 ªAn,00 . The

matrix coefficients in the refinement equation for the scaling function may be easilycalculated using the orthogonality of these functions. This also holds for the matrixcoefficients in the expansion for the wavelets.

As a final remark to this section we note that because of the symmetry of thescaling functions and wavelets, these bases can easily be modified to bases for compactintervals. Using the notation φik,j(x) = φi(2kx−j), let φik,j = φik,j |[0,1], ψ

mk,j = ψmk,j |[0,1]

1Wavelets and the matrices in the refinement equation for this and other examples can be foundat www.math.gatech.edu/∼geronimo.


for m 6= 1, and

ψ1k,j =

{0 if supp ψ1

k,j ∩ [0, 1]c 6= Ø,

ψ1k,j otherwise;

we find the following theorem.Theorem 5.3. The set {φik,j : k ≥ 0, i = 1, . . . , n, 0 ≤ j ≤ 2k − 1 + δ0,i} is an

orthogonal basis for V n,0k = V n,0k ∩ L2[0, 1] while {ψik,j : k ≥ 0, i = 1, . . . , n, δ1,i ≤j ≤ 2k−1 + δ0,i} forms an orthogonal basis for Wn,0

k = Wn,0k ∩L2[0, 1]. Furthermore,

clL2 V n,00 ⊕⊕k≥0 Wn,0k = L2[0, 1].

6. Construction of differentiable spline wavelets. For differentiable splines,we take k = 1 and consider the space V n,01 = Sn1 ∩ L2(R). Lemma 2.1 says that

{φ14, . . . , φ

1n} forms an orthogonal basis for An,10 ( ·+1

2 ), where φ1n(t) = (1− t2)2p

9/2n−4(t).

In this case we have r1i (t) = (1 + t)2(1− t)i and l1i (t) = (1− t)2(1 + t)i, i = 0, 1. The

spaces Cn,10 ( ·+12 ) and Cn,11 ( ·+1

2 ) are each two dimensional. Hence from Theorem 1.1we see that an orthogonal intertwining MRA can be constructed if we can find twofunctions w1, w2 ∈ An,11 ( ·+1

2 )ªAn,10 ( ·+12 ) such that

(6.1) 〈(I − P{w1,w2})rn,1i , (I − P{w1,w2})l

n,1j 〉 = 0, i ≤ j = 0, 1.

In order to construct scaling functions with a symmetry axis one of these functionswill be constructed symmetric and the other antisymmetric. From (3.8) we find

(6.2) 〈rn,10 , ln,10 〉 =128(−1)n+1(n2 + 2n− 9)(n− 2)!

(n+ 3)!,

〈rn,10 , ln,11 〉 =768(−1)n+1(n− 2)!

(n+ 3)!,

and

(6.3) 〈rn,11 , ln,11 〉 =4608(−1)n+1(n− 3)!

(n+ 4)!.

With k = 1, (3.11) yields

〈rn,11 , u1n,2m〉 = 〈rn,1, u1

n,2m〉

=3(−1)m+1(2m)!(n− 2m)!(n2 + 5n+ 2− 8m)(n− 3)!

2(n−5)m!(n−m)!(n+ 3)!.(6.4)

Combining (6.4) with (3.7) and (3.10) and using initial condition 〈r2,10 , u1

2,2〉 = 103 , we

obtain

(6.5) 〈rn,10 , u1n,2m〉 =

(−1)m+1(2m)!(n− 2m)!(n2 + 7n+ 4− 12m)(n− 2)!

2(n−4)m!(n−m)!(n+ 2)!.

In order to simplify the computations somewhat we biorthogonalize the ramp

functions. Set rn,0 = rn,10 , ln,0 = ln,10 , rn,1 = rn,11 − 〈rn,11 ,ln,0〉〈rn,0,ln,0〉rn,0, and ln,1 = ln,11 −

〈ln,11 ,rn,0〉〈rn,01,ln,1〉 ln,1. With the help of the inner products given above, we find

(6.6) 〈rn,1, ln,1〉 =(−1)n4608(n− 3)!

(n+ 4)!(n2 + 2n− 9),


(6.7) 〈rn,1, u1n,2m〉 =

96(−1)m(2m)!(n− 2m)!(n− 3)!(n2 − 4mn+ 3n+ 6)

2nm!(n−m)!(n2 + 2n− 9)(n+ 3)!,

and

(6.8)

〈rn,1, u1n,2m+1〉 =

96(−1)m(2m+ 1)!(n− 2m− 1)!(n+ 3)(n− 2)!(3n2 − 4mn+ 9n− 8m)

2nm!(n−m)!(n2 + 2n− 9)(n+ 4)!.

As in the C0 case we can use (3.17) to compute the inner products 〈u1n,2i, u

1n,2j〉;

i, j = 0, 1, or 2; and n > 2 max{i, j} + 3. This computation was done using Mapleand yielded

(6.9) 〈u1n,2i, u

1n,2j〉 =

(−1)i+j(n− 2i)!(n− 2j)!(2i)!(2j)!(n− 3)!q(n, i, j)

22n−1(2n+ 1− i− j)(n+ 3)!i!j!(n− i)!(n− j)! ,

where

q(n, i, j) = n6 + 19n5 + (131− 8(i+ j))n4 + (368− 112(i+ j))n3

+ (372− 424(i+ j) + 24(i+ j)2)n2

+ (212− 320(i+ j) + 120(i2 + j2) + 432ij)n

+ 2(24− 48((i+ j)(1 + ij) + (i2 + j2)) + 81ij).

We now construct three orthogonal functions v0 = b0,0(n)u1n,0 + b0,2(n)u1

n,2 + u1n,4,

v2 = b2,0(n)u1n,0 + b2,2(n)u1

n,2 + u1n,4, and v4 = b4,0(n)u1

n,0 + b4,2(n)u1n,2 + u1

n,4, withthe additional constraints that v0 be orthogonal to rn,0 and rn,1, v2 be orthogo-nal to rn,1 and v0, and v4 be orthogonal to v0 and v2. Using the inner productformulas and a symbolic manipulation package such as Maple, we find b0,0(n) =

12(n−3)(n−2) , b0,2(n) = 12(n−1)

(n−3)(n−2) , b2,0(n) = 12(2n+1)(7n2−17n+18)(2n−7)(n−2)(n−3)(7n2−n+18) , b2,2(n) =

12(14n2−25n+54)(n−1)2

(2n−7)(n−2)(n−3)(7n2−n+18) ,

b4,0(n) =4(2n+ 1)(2n− 1)(9n5 + 175n4 + 285n3 − 5695n2 + 12786n− 8280)

(2n− 7)(2n− 5)(n− 3)(n− 2)(n2 + 11n− 6)(3n3 + 28n2 − 67n+ 28),

and

b4,2(n) =4(n− 1)(2n− 1)(9n5 + 179n4 + 477n3 − 4163n2 + 7458n− 4392)

(2n− 7)(n− 3)(n− 2)(n2 + 11n− 6)(3n3 + 28n2 − 67n+ 28).

These equations allow us to compute the following inner products between the newfunctions with the biorthogonal ramps:

〈v2, rn,0〉 =18432(n2 + 2n− 9)(n− 4)!

2n(n+ 1)!(2n− 7)(7n2 − n+ 18),

〈v4, rn,0〉 =−7680(n2 + 15n− 24)

2n(2n− 7)(2n− 5)(n+ 2)n(n− 3)(3n3 + 28n2 − 67n+ 28),

and

〈v4, rn,1〉 = 3072q(n)(n−1)(n−4)!2n(2n−7)(2n−5)(n−2)(n+3)!(n2+11n−6)(n2+2n−9)(3n3+28n2−67n+28) ,


where

q(n) = 107n7 + 1230n6 − 3580n5 − 9546n4 + 21437n3 + 30204n2 − 70956n+ 25920.

Also, we have

〈v0(n), v0(n)〉 =11052(2n− 9)!!

22n(2n+ 1)!!(n− 2)2(n− 3)2,

〈v2(n), v2(n)〉 = 11052(2n−9)!!q(n)22n(2n−1)!!(2n−7)(n−1)n(n+1)(7n2−n+18)2(n−3)2(n−2)2 ,

and

〈v4(n), v4(n)〉 = 2048(n2+13n−18)q(n)q1(n)(n−1)!(2n−3)((2n−9)!!)2

22n(n+3)!(n−2)3((n−3)(2n−3)!!(3n3+28n2−67n+28)(n2+11n−6))2

with

q1(n) = n6 + 39n5 + 445n4 + 585n3 − 7286n2 + 9816n− 2880.

From (6.1) we see that we will need to borrow two functions in order to make anorthogonal intertwining MRA, and in order for these to be symmetric or antisymmetricwe will set w1,n = α1,0(n)v0 + α1,2(n)v2 + v4 and w2,n = α2,1(n)u1

n,1 + u1n,3. Taking

note of the sign structure in (6.2), (5.5), and (6.6), (6.1) becomes the three equations

|〈rn,0, ln,0〉| = 〈rn,0, w1,n〉2 − 〈rn,0, w2,n〉2,0 = 〈rn,0, w1,n〉〈w1,n, rn,1〉 − 〈rn,0, w2,n〉〈w2,n, rn,1〉,

|〈rn,1, l1,n〉| = −〈rn,1, w1,n〉2 + 〈rn,1, w2,n〉2.These can be solved to give

〈rn,1, w1,n〉√〈w1,n, w1,n〉=

√|〈rn,1, l1,n〉| − 〈rn,1, w2,n〉2

〈w2,n, w2,n〉 ,

〈rn,0, w1,n〉√〈w1,n, w1,n〉=

〈rn,0, w2,n〉〈rn,1, w2,n〉√|〈rn,1, l1,n〉|〈w2,n, w2,n〉2 − 〈rn,1, w1,n〉2〈w2,n, w2,n〉

,

0 = |〈rn,1, l1,n〉| |〈rn,1, l1,n〉| 〈w2,n, w2,n〉+ |〈rn,1, l1,n〉| 〈rn,0, w2,n〉2− |〈rn,1, l1,n〉| 〈rn,1, w2,n〉2.(6.10)

With the definition of w2 the last of the above equations can be rewritten as

0 =

∣∣∣∣∣∣〈u1n,1, u

1n,1〉 −〈u1

n,1, rn,0〉〈u1n,1, rn,1〉

〈u1n,1, rn,0〉〈rn,0, rn,0〉 0〈u1n,1, rn,0〉 0 〈rn,1, rn,1〉

∣∣∣∣∣∣α22,1

+ 2

∣∣∣∣∣∣〈u1n,1, u

1n,3〉 −〈u1

n,3, rn,0〉〈u1n,3, rn,1〉


∣∣∣∣∣∣α2,1

+

∣∣∣∣∣∣〈u1n,3, u

1n,3〉 −〈u1

n,3, rn,0〉〈u1n,3, rn,1〉


∣∣∣∣∣∣ .


Given the values of the inner products appearing in this equation, we find that eitherof the two values

(6.11) α2,1(n) =2

(n− 1)(2n− 3)

{(2n− 1)(3n− 8)

(n− 2)± 2√

7

√(2n− 1)(n+ 2)

(2n− 5)(n− 2)

}

will suffice for α2,1. The first equation in (6.10) can be used to eliminate√〈w1,n, w1,n〉

in the middle equation, and we have

(6.12) α1,2(n) =(7n2−n−18)

{q3(n)(n−1)(n+1)(n+2)+q(n)

√7√

(2n−1)(2n−5)(n+2)(n−2)}

108(2n−5)(n+2)(n2+11n−6)(3n3+28n2−67n+28)(n4−6n3−31n2−22n+72)

with

q3(n) = 152n6 + 3893n5 + 19240n4 − 76625n3 − 105912n2 + 330372n− 129600.

Now we solve for α1,0(n) to find

(6.13) α1,0(n) =√

10108

q(n){

(n−4)(n−2)(2n+1)n(n+2)(2n−7)

(q4(n)+q5(n)

√7√

(2n−1)(2n−5)(n+2)(n−2))}1/2

(2n−5)(n2+11n−6)(3n3+28n2−67n+28)(n4−6n3−31n2−22n+27)

with

q4(n) = 62n6 + 2n5 + 1480n4 − 4526n3 − 2250n2 + 9774n− 5508

and

q5(n) = 8n4 + 95n3 − 126n2 + 297n− 162.

Knowing w1 and w2, we are now able to construct the orthogonal C1 scalingfunctions. Let

h0(t) =

{2 |t|3 − 3 |t|2 + 1 if t ∈ [−1, 1),

0 otherwise

and

h1(t) =

{(1− |t|)2t if t ∈ [−1, 1),

0 otherwise.

For j = 4, . . . , n, set

φj(·) =

{φ1j (2· − 1) if t ∈ [0, 1),

0 otherwise,

φ2(·) =

{w1(2· − 1) if t ∈ [0, 1),

0 otherwise,

φ3(·) =

{w2(2· − 1) if t ∈ [0, 1),

0 otherwise,

and φi = (I − P{φ2,...,φn,φ2(·+1),...,φn(·+1)})hi, i = 0, 1. Then the above computationsgive the following theorem.


Theorem 6.1. For n ≥ 6, α1,0(n), α1,2(n), and α2,1(n) given by (5.9), Φ =

{φ0, . . . , φn}∗ generates an orthogonal multiresolution analysis {V n,1k }. Furthermore,the last n − 1 functions are symmetric or antisymmetric about 1

2 . The first function

φ0 is symmetric about 0 while φ1 is antisymmetric about 0.Proof. Since h0 and h1 are linear combinations of similarly scaled versions of

r1j , j = 0, 1 and l1j , j = 0, 1 the result follows from the computations above.

With the scaling functions above we construct the coefficients C1n,i, i = −2,−1, 0,

1, in the matrix refinement equation. Theorem 4.4 implies that there will be fourwavelets not supported in [0, 1]. Using arguments similar to Corollary 5.2 leads to thefollowing.

Corollary 6.2. Suppose ψ0, . . . , ψ3 are chosen so that

ψ0 =√

2(I − Pφ0)φ01,0,

ψ1 ∝ (I − P{ψ0,φ0})(χ[0,1] − χ[−1,0])(I − Pφ11,0

)φ1,

ψ2 ∝ (I − P{ψ3,φ1})(χ[0,1] − χ[−1,0])(I − Pφ01,0

)φ0,

ψ3 =2√

2√7

(I − Pφ1)φ11,0

and ψ4, . . . , ψn form a basis for Ψ0 consisting only of functions symmetrical or an-tisymmetrical about 1

2 . Then {ψ0, . . . , ψn} generates a shift-invariant orthonormal

basis for W0. Furthermore, ψ0(0) = φ0(0), (ψ3)′(0) =√

7(φ1)′(0), ψ1(0) = 0, and(ψ2)′(0) = 0.

An example is given in Figure 2. In Figure 3 explicit formulas for these functions,accurate at least to 10 digits, are given. The coefficients in the refinement equationmay be calculated using inner products. The wavelets supported on [−1, 1] can becomputed using Corollary 6.2. The wavelets supported in [0, 1] can be obtained byfinding n−3 orthogonal set of functions, symmetrical or antisymmetrical with respectto 1

2 , from the space (I − P{φ4,...,φn−1})An,11 ªAn,10 .

Theorem 4.5 shows that in order to construct a wavelet basis for [0, 1] we needto rotate the scaling functions so that {χ[0,1]φ

i}i=0,1 is an orthogonal set as well as

{χ[−1,0]φi}i=0,1. Exploiting the symmetry of φ0 and φ1, we find that φ0 = − 1√

2φ0 −

1√2φ1 and φ1 = − 1√

2φ0 + 1√

2φ1 have the desired property. Let φik,j = φik,j |[0,1],

ψmk,j = ψmk,j |[0,1], m 6= 1, 3, and for m = 1, 3 set

ψmk,j =

{0 if supp ψmk,j ∩ [0, 1]c 6= Ø,

ψmk,j otherwise.

Then from Theorem 4.5 we have the following theorem.Theorem 6.3. The set {φik,j : k ≥ 0, i = 1, . . . , n, 0 ≤ j ≤ 2k−1+χ{0,1}(i)} is an

orthogonal basis for V n,1k = V n,1k ∩L2[0, 1] while {ψik,j : k ≥ 0, i = 1, . . . , n, χ{1,3}(i) ≤j ≤ 2k − 1 + χ{0,2}(i)} forms an orthogonal basis for Wn,1

k = Wn,1k ∩ L2[0, 1]. Fur-

thermore, clL2 V n,10 ⊕⊕k≥0 Wn,1k = L2[0, 1].

Examples of these functions for n = 6 can be found in Figure 4.2

2Wavelets as well as the matrices in the refinement equation for this and other examples may befound at the web site www.math.gatech.edu/∼geronimo.


-0.5

0

0.5

1

1.5

2

2.5

-1 -0.5 0 0.5 1

-2

-1

0

1

2

-1 -0.5 0 0.5 1

φ0 φ1

-1.5

-1

-0.5

0

0.5

1

1.5

0 0.2 0.4 0.6 0.8 1

-2

-1

0

1

0 0.2 0.4 0.6 0.8 1

φ2 φ3


7. A C2 example. The methods of the previous sections can be used to con-struct C2 multiwavelets as well, although the formulas become extremely complicated.We will therefore content ourselves with briefly describing the procedure and exhibit-ing an example which may be of use. We do not prove that the procedure worksfor arbitrary n; however, we have verified it for a number of cases. For the C2 casek = 2, r2

i = (1 + t)3(1 − t)i and l2i (t) = (1 − t)3(1 + t)i, i = 0, 1, 2. The spacesCn,20 ( ·+1

2 ) and Cn,21 ( ·+12 ) are each three dimensional. Hence we search for three or-

thonormal functions w1, w2, w3 ∈ An,21 ( ·+12 )ªAn,20 ( ·+1

2 ) such that

(7.1) 〈(I − P{w1,w2,w3})rn,2i , (I − P{w1,w2,w3})l

n,2j 〉 = 0, i ≤ j = 0, 1, 2.

The above equation yields six nonlinear equations and in order to ease the computationsomewhat we impose that 〈w3, r

n,20 〉 = 0. By examining (7.1) we find that w1 must

satisfy four equations, w3 five equations, and w2 two equations. Thus, we choosew1 = a1,0u

2n,0 + a1,2u

2n,2 + a1,4u

2n,4 + a1,6u

2n,6, w3 = a3,0u

2n,0 + a3,2u

2n,2 + a3,4u

2n,4 +

a3,6u2n,6 + a3,8u

2n,8, and w2 = a2,1u

2n,1 + a2,3u

2n,3 and observe that, as a consequence,

n must be at least 11 for these functions all to have C2 smoothness. For n = 11, weused Maple to obtain the following solution to 60 digits of accuracy:


φ0(t) =

4067.904397t6 − 3085.517213t5 − 739.5537604t4

+1113.129531t3 − 249.0932934t2 + 2.585173201 for 0 ≤ t ≤ 12

(−10946.18252t4 + 29698.96665t3 − 29673.11423t2

+12935.83259t− 2076.394058)(t− 1)2 for 12< t ≤ 1


φ1(t) =

t(−12435.14749t5 + 11555.74525t4 − 175.1583524t3

−2704.493962t2 + 885.9596767t− 79.47577362) for 0 ≤ t ≤ 12

(17496.63084t4 − 47730.65634t3 + 47977.37771t2

−21054.27094t+ 3403.826649)(t− 1)2 for 12< t ≤ 1

−φ1(−t) for − 1 ≤ t < 00 elsewhere

φ2(t) =

(2t− 1)t2(−12193.17741t3 + 8033.232335t2

−923.0732055t− 95.73716085) for 0 ≤ t ≤ 12

−φ2(1− t) for 12< t ≤ 1

0 elsewhere

φ3(t) =

(54655.48659t4 − 76071.45058t3 + 37177.81845t2

−7410.133638t+ 493.1155758)t2 for 0 ≤ t ≤ 12

φ3(1− t) for 12< t ≤ 1

0 elsewhere

φ4(t) = 3√

70t2(t− 1)2χ[0,1]

φ5(t) = 3√

770t2(2t− 1)(t− 1)2χ[0,1]

φ6(t) = 3√

182t2(22t2 − 22t+ 5)(t− 1)2χ[0,1]

Fig. 3. Formulas for the C1 scaling functions.

0 10.50-0.5-10

1

0

-2

-3

-1

0 10.50-0.5-10

1

0

-1

-2

-3

φ0 φ1

Fig. 4. Modified C1 scaling functions of degree 6 for truncation to [0, 1].


φ0(t) =

960.8390712t11 − 5925.739238t10 + 13283.80990t9

− 11118.85044t8 − 4083.423604t7 + 15463.88643t6

− 12421.39021t5 + 4470.423870t4 − 629.0433748 ∗ t3 − 1.5t2 + 1 for 0 ≤ t < 12

(1− t)3(−1188.938305t8 + 3170.302415t7 − 3222.784994t6

+ 1570.468582t5 − 377.8071111t4 + 37.83633397t3

+ 1.585938501t2 − .6720319852t+ .01241619153) for 12< t ≤ 1


φ1(t) =

t(−43.55084884t10 + 268.8737383t9 − 588.5150956t8

+ 426.0982641t7 + 364.8815486t6 − 923.7069266t5

+ 722.0921528t4 − 265.9177149t3 + 39.49465380t2 + .75t− .5) for 0 ≤ t < 12

(1− t)3(55.58117958t8 − 152.6750597t7 + 161.0387487t6

− 81.78617419t5 + 20.22050034t4 − 1.881383443t3

− .1197243402t2 + .03075948978t− .0002283505048) for 12≤ t ≤ 1

−φ1(−t) for − 1 ≤ t < 00 elsewhere

φ2(t) =

−4.290797794t11 + 25.45780265t10 − 49.23874622t9

+ 10.99949461t8 + 96.73006250t7 − 163.9201834t6 + 126.4483297t5

− 52.49395898t4 + 11.30900992t3 − 1.002944440t2 + .001962959931 for 0 ≤ t < 12

(1− t)3(5.245026448t8 − 14.95000551t7 + 16.41738822t6

− 8.639953559t5 + 2.137676365t4 − .1611170142t3

− .02104722553t2 + .002877121114t+ .00003155317905) for 12< t ≤ 1


φ3(t) =

t3(837.0653712t8 − 1385.345634t7 + 2498.141505t6

− 19046.31150t5 + 51025.24775t4 − 61978.36348t3

+ 38423.77644t2 − 11768.03455t+ 1393.219445) for 0 ≤ t ≤ 12

φ3(1− t) for 12< t ≤ 1

0 elsewhere

φ4(t) =

(13553.28638t7 − 72947.95448t6 + 155109.9841t5

− 159752.4874t4 + 73088.74021t3 − 2106.644683t2

− 8799.858303t+ 1848.967175)t3(t− 1/2) for 0 ≤ t ≤ 12

−φ4(1− t) for 12< t ≤ 1

0 elsewhere

φ5(t) =

t3(19651.08243t8 − 58017.27598t7 + 17880.79147t6

+ 126264.0351t5 − 207975.2506t4 + 145513.9138t3

− 51576.42468t2 + 8813.779535t− 552.7580636) for 0 ≤ t ≤ 12

φ5(1− t) for 12< t ≤ 1

0 elsewhere

φ6(t) =√

6006t3(1− t)3χ[0,1]

φ7(t) =√

30030t2(2t− 1)(1− t)3χ[0,1]

φ8(t) = 2√

7293t3(30t2 − 30t+ 7)(1− t)3χ[0,1]

φ9(t) = 2√

40755t3(2t− 1)(34t2 − 34t+ 7)(1− t)3χ[0,1]

Fig. 5. Formulas for the scaling functions.


-0.50

0.5

1

1.5

2

2.5

3

-1 -0.5 0 0.5 1 -2

-1

0

1

2

-1 -0.5 0 0.5 1

φ0 φ1

-2

-1

0

1

2

3

-1 -0.5 0 0.5 1

-1.5

-1

-0.5

0

0.5

1

1.5

0 0.2 0.4 0.6 0.8 1

φ2 φ3

-1.5

-1

-0.5

0

0.5

1

1.5

0 0.2 0.4 0.6 0.8 1

-2

-1

0

1

2

0 0.2 0.4 0.6 0.8 1

φ4 φ5


a1,0 = 837.065371210437626131768992156281384724545782634651643883901,

a1,2 = 34683.2805853131547547484051801591515443493456004150468616372,

a1,4 = 98577.9464165718483400146101646574256130573895257061737344447,

a1,6 = 23873.2858941591526073330349900845644544457875198167225175943,

a2,1 = 69361.5525108187609792638961429103692940232506514292283577945,

a2,3 = 135633.530246170747066239552110728177870041230865307938249067,

a3,0 = 19651.0824273352067272236935152142143793755548082840043700210,

a3,2 = 518517.565184033631919745869328732529406714729877896561460025,

a3,4 = 968629.606667579619572209561202671860306580349823867383091657,

a3,6 = 236239.020798546376511780424212516703801694760661827446183240,

a3,8 = 3210.54994458080183556006870726795783137235163979326960039700.


The corresponding functions are given with 10 digits of accuracy in Figure 5, andtheir graphs are shown in Figure 6. The orthonormal wavelets can be calculatedusing techniques similar to those of sections 5 and 6. One final Gram–Schmidt stepis necessary since two of the three functions in (I−PQ0)Q1 are symmetric. The sameis true of (I − PQ0∪Ψs)Z.

Acknowledgment. J. S. Geronimo would like to thank the members of theTheoretical Physics Division at Saclay and the Laboratoire d’Analyse Numerique atthe University of Pierre and Marie Curie for their hospitality and support during thetime this work was being completed.

REFERENCES

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[3] W.N. Bailey, Generalized Hypergeometric Series, Cambridge Tracts in Math. Math. Phys. 32,Cambridge University Press, Cambridge, UK, 1964.

[4] G.C. Donovan, Fractal Functions, Splines, and Wavelets, Ph.D. thesis, Georgia Institute ofTechnology, Atlanta, GA, 1995.

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