Quintic deficient spline wavelets

Quintic deficient spline wavelets F. Bastin and P. Laubin January 19, 4 Abstract We show explicitely how to construct scaling functions and wavelets which are quintic deficient splines with compact support and symmetry properties. 1 Introduction For m IN, it is well known that the functions N m+1. k k ZZ, where N m+1 = χ [,1]... χ [,1] m + 1 factors, constitute a Riesz basis of the set of smoothest splines of degree m, V = {f L IR : f [k,k+1] = P m k, k ZZ and f C m 1 IR} where P m k is a polynomial of degree at most m; for m =, it is simply the set of functions in L IR which are constant on every interval [k, k + 1], k ZZ. Moreover, if we define V j = {f L IR : f j. V }, j ZZ then the sets V j j ZZ constitute a multiresolution analysis of L IR and the function N m+1 is a scaling function for it; hence one gets bases of wavelets from standard constructions [6]; Chui-Wang biorthogonal wavelets,[]; Battle-Lemarié orthonormal wavelets, [7]. P. Laubin died on February 1, 1. Bastin and Laubin started working on this paper together. That s why both are quoted as authors. Key words: Wavelets, quintic deficient splines MSC Classification: 4C4,41A15 1

For numerical analysis purposes, splines of odd degree are of special interest; moreover, it is also useful to consider the set of deficient splines of degree m + 1 m IN, that is to say V := {f L IR : f [k,k+1] = P m+1 k, k ZZ and f C m+1 IR} see [3], [8]. As for the space V, a standard argument shows that the space V is a closed subspace of L IR. For m = 1, this is the set of smoothest cubic splines; for m =, we denote this set as the set of deficient quintic splines. In what follows, we want to show explicitely how to construct scaling functions and wavelets which are quintic deficient splines with compact support and symmetry properties. We go straithforward to the heart of the problem of the construction of the multiresolution analysis, with all direct computations and without referring or using other results. The construction of the wavelets is also a direct computation adapted to the problem. The idea of the proof that they are a Riesz basis comes from [4], [5]. For the sake of completeness, we give here all the justifications. Definitions and notations We say that a sequence of functions f k k ZZ in a Hilbert space H,. satisfies the Riesz condition if they are A, B >, A B such that A c k c k f k B c k RC for every finite sequence c k of complex numbers. If we denote by L the closed linear hull of the f k k ZZ then the map T : l L c k k ZZ k= is then a topological isomorphism. We say that the functions f k k ZZ constitute a Riesz basis for L. We use the notation fξ for the Fourier transform IR e ixξ fx dx of f. c k f k

In case H = L IR and f k x = fx k k ZZ, taking Fourier transforms, the inequality RC of the Riesz condition can be written as follows A c k p w L [,π] B c k RCF with wξ = fξ + lπ L 1 loc, pξ = c k e ikξ. Finally, using a classical argument based on Fejer kernel for example, one shows that RFC is satisfied for every finite sequence c k if and only if A wξ B a.e. see for example [1],[7]. For the sake of completeness, we also recall the standard definition of multiresolution analysis. We say that a sequence of closed linear subspaces V j j ZZ of L IR constitutes a multiresolution analysis of L IR if the following properties hold: i V j V j+1 j ZZ, j ZZ V j = L L IR, j ZZ V j = {} ii f V f. k V k ZZ iii j Z, f V j f j. V iv there is ϕ L IR such that the functions ϕ. k, k ZZ, are a Riesz basis for V. From a mutiresolution analysis, one constructs a Riesz basis of L IR from a standard procedure see for example [6], [7], using the spaces W j, orthogonal complement of V j in V j+1 j ZZ. Here we use this procedure but with two functions instead of one for property iv. 3 Construction of a multiresolution analysis Let us denote by V the following set of quintic splines V := {f L IR : f [k,k+1] = P 5 k, k ZZ and f C 3IR}. Looking for f V with support [, 3] smaller interval does not give anything, we are lead to a homogenous linear system of 18 unknowns and 16 equations; this let us think that two scaling functions will be needed to generate V. 3

Proposition 3.1 A function f with support [, 3] belongs to V if and only if nx 4 + ax 5 if x [, 1] b 5 4 3 x 3 + c x 3 + d x 3 fx = +e x 3 + f x 3 + g if x [1, ] h3 x 4 + j3 x 5 if x [, 3] if x [, 3] with b = 19a + 57 5 n c = 15 a + 11 n d = 45 a 7 n e = 45 4 a 33 4 n f = 135 16 a + 81 16 n g = 117 3 a + 67 16 n h = 15a + 1n j = 1a 33 5 n Proof. The particular form in which we write the polynomials are due to the fact that we have in mind to construct functions with symmetry. Moreover, the polynomial on [, 1] resp. [, 3] can immediately be written in this form because we want C 3 regularity at the point resp. 3 and support in [, 3]. The coefficients are obtained using the definition of the quintic splines; we get an homogenous system of 8 linear equations with 1 unknowns. Among the functions described above, there exists symmetric and antisymmetric ones the symmetry is naturally considered relatively to 3. We are also going to show that they generate V. Theorem 3. The following functions ϕ a and ϕ s x 4 11 15 x5 if x [, 1] 9 ϕ a x = x 3 + 3x 3 8 3 38x 3 15 5 if x [1, ] 3 x 4 + 11 3 15 x5 if x [, 3] if x < or x > 3 ϕ s x = x 4 3 5 x5 if x [, 1] 57 8 3 x 3 + x 3 4 if x [1, ] 3 x 4 3 5 3 x5 if x [, 3] if x < or x > 3 are respectively antisymmetric and symmetric with respect to 3 constitutes a Riesz basis of V. {ϕ a. k, k ZZ } {ϕ s. k, k ZZ } 4 and the family

Here is a picture of ϕ s, ϕ a. 1.8.6.4..4. -. -.4.5 1 1.5.5 3.5 1 1.5.5 3 Proof. Construction of ϕ a, ϕ s. We use the notations and the result of Proposition 3.1. We look for a, n such that n = h a = j b = d = f = resp. This system is equivalent to the single equation With n = 1, a = 3 5 Riesz condition. For every k ZZ, we define n = h a = j c = d = g =. 5a + 3n = resp. 15a + 11n =. resp. n = 1, a = 11 15, we get ϕ s resp. ϕ a. ϕ a,k x = ϕ a x k and ϕ s,k x = ϕ s x k. We first prove that the functions ϕ a,k k ZZ resp. ϕ s,k k ZZ form a Riesz family. Indeed, since we have c k ϕ a,k L IR = 1 π c k e ikξ ϕ a ξ L IR = 1 π c k e ikξ ω a ξ L [,π] for every finite sequence c k of complex numbers and where ω a ξ = ϕ a ξ + lπ, 5

it suffices to show that there are constants c, C > such that c ω a ξ C, ξ [, π]. Using the definition of ϕ a, we get ϕ a ξ = 16i e 3iξ/ 3ξcos 3ξ ξ 6 + 9 cosξ 11 sin3ξ 7 sinξ = 16i e 6ξ 3iξ/ cos ξ ξ 4 + cos ξ sinξ 19 + 11 cos ξ. 6 Using some computations lead to ω a ξ = 1 1r = ξ + k r+ r + 1! Dr ξ ϕ a ξ + lπ = π sin, r IN, ξ IR \ ZZ, πξ 347 136 cos ξ 385 cosξ 31185 hence to the conclusion. The same can be done for ϕ s. We get and ω s ξ = ϕ s ξ = 96 ξ 6 e 3iξ/ sin ξ ξ + cos ξ 3 sin ξ ϕ s ξ + lπ = 14445 + 7678 cos ξ + 53 cosξ. 3465 Now, let us consider both families ϕ a,k k ZZ and ϕ s,k k ZZ together. For every finite sequence c k and d k of complex numbers, we have c k ϕ a,k + d k ϕ s,k L IR = c k ϕ a,k j + d k ϕ s,k j L [,1]. j= On [, 1], only ϕ a,l, ϕ s,l with l =, 1, are not identically ; moreover, these functions are linearly independant see appendix for a proof. As on a finite dimensional space, all norms are equivalent, we get that there are r, R > such that r c k ϕ a,k j L [,1] + d k ϕ s,k j L [,1] c k ϕ a,k j + d k ϕ s,k j L [,1] R c k ϕ a,k j L [,1] + d k ϕ s,k j L [,1]. 6

Now, writing again j= j= c k ϕ a,k j L [,1] = d k ϕ s,k j L [,1] = c k ϕ a,k L IR, d k ϕ s,k L IR and using what has been done on each family separately, we conclude. Riesz basis for V. Let us show that V is the closed linear hull of the ϕ a,k, ϕ s,k k ZZ. On one hand, as the set V is a closed subspace of L IR containing each ϕ a,k and ϕ s,k, it contains the closed linear hull of these functions. On the other hand, using Fourier transforms, we see that it suffices to show that for every f V, there are p, q L loc and π periodic such that fξ = pξ ϕ s ξ + qξ ϕ a ξ Let f V. Because of the definition of V, there are c k k ZZ, d k k ZZ l such that m D 6 f = lim c k δ k + d k δ k m + k= m in the distribution sense, where δ k and δ k are respectively the Dirac and the derivative of the Dirac distribution at k see appendix for proof. Taking Fourier transforms, we get also iξ 6 fξ = lim m m + k= m a.e. c k e ikξ + id k ξe ikξ ; it follows that there are mξ, nξ L loc and π periodic such that iξ 6 fξ = mξ + ξnξ a.e. Hence the problem is to find p, q L loc and π periodic such that mξ + ξnξ iξ 6 = pξ ϕ s ξ + qξ ϕ a ξ. Using the explicit expression of the Fourier transform of ϕ s and ϕ a, we are lead to look for p, q such that mξ = e 3iξ/ 3 96 sin ξ sin ξ 3ξ pξ + 16i11 sin + 7 sin ξ qξ nξ = e 3iξ/ 96 + cos ξ sin ξ 3ξ pξ 48icos + 9 cos ξ qξ. 7

For fixed ξ, this is a linear system of two equations and two unknowns; as 3. 96 sin ξ sin ξ 3ξ 16i11 sin det + 7 sin ξ 96 + cos ξ sin ξ 3ξ 48icos + 9 cos ξ = C sin 6 ξ with C = 3 1 i, we get pξ = 16ie3iξ/ C sin 6 ξ 3mξcos 3ξ + 9 cosξ + nξ11 sin3ξ + 7 sinξ and qξ = 3nξ 96e3iξ/ C sin 6 ξ sin ξ sin ξ + mξ + cos ξ sinξ. These functions are π periodic; it remains to prove that they are L loc. Indeed, using mξ = ξnξ ξ 6 fξ we get and 3mξcos 3ξ + 9 cosξ + nξ11 sin3ξ + 7 sinξ = 3ξ 6 fξ cos 3ξ [ ] 3 + 9 cosξ + nξ 8 ξ7 + Oξ 9 3nξ sin ξ sin ξ + mξ + cos ξ sinξ = sin ξ ξ 6 + cos ξ fξ [ + nξ 1 ] 6 ξ5 + Oξ 7 and we conclude. Remark 3.3 1 The previous proof also shows that a function f of L IR belongs to V if and only if there exist m, n L loc, π periodic such that and Since we get 3ξ cos 3ξ + 9 cosξ iξ 6 fξ = mξ + ξnξ a.e. 11 sin 3ξ 7 sinξ = 3 8 ξ7 + Oξ 9 ξ + cos ξ 3 sin ξ = 1 6 ξ5 + Oξ 7 ϕ a =, ϕ s = 4 5 8

For every j ZZ we define V j = {f L IR : f j. V }. Proposition 3.4 The sequence V j j ZZ is an increasing sequence of closed sets of L IR and V j = {}, V j = L IR. j ZZ j ZZ Moreover, the functions ϕ a, ϕ s satisfy the following scaling relation ϕ s ξ ϕ s ξ = M ξ ϕ a ξ ϕ a ξ where M ξ is the matrix called filter matrix M ξ = e 3iξ/ = e 3iξ/ 64 1 cos ξ 19 + 13 cos ξ 9i 3 16 cos ξ sin ξ i sin ξ 16 + 11 cos ξ 1 cos ξ 8 7 cos ξ 3 3 51 cos ξ 3ξ + 13 cos 9isin ξ 3ξ + sin i11 sin 3ξ + 1 sin ξ 3ξ 7 cos + 9 cos ξ Expressed in terms of the functions instead of the Fourier transform, the scaling relation can be written as follows ϕ s ξ = 1 64 13ϕ sx + 51ϕ s x 1 + 51ϕ s x + 13ϕ s x 3 9ϕ a x 9ϕ a x 1 + 9ϕ a x + 9ϕ a x 3 ϕ a x = 1 64 11ϕ sx + 1ϕ s x 1 1ϕ s x 11ϕ s x 3 7ϕ a x + 9ϕ a x 1 + 9ϕ a x 7ϕ a x 3 Proof. Using the definition of V and of the V j j ZZ, it is clear that V j V j+1 for every j ZZ. The density of the union is due to the facts that a smoothest spline is also a deficient spline V j V j and that the union j ZZ V j is dense in L IR. Now, let f j ZZ V j. For every j, there is then a polynomial P j resp. Q j such that f = P j on [, j ] resp. f = Q j on [ j, ]. It follows that P j = P j resp. Q j = Q j for every j, j hence f is a polynomial on [, + [ resp. ], ]. Since f L IR, this implies f = on [, + [ resp. ], ]. 9.

Let us show how to obtain the scaling relation. We have We define ϕ s ξ = 3e 3iξ sin ξ + cosξ ξ 5 9e 3iξ sin ξ sinξ ξ 6 ϕ a ξ = 3ie 3iξ cos3ξ + 9 cos ξ + ie 3iξ 11 sin3ξ + 7 sin ξ ξ 5 4ξ 6 m s ξ = 9 e 3iξ sin ξ sinξ, n s ξ = 3e 3iξ sin ξ + cosξ m a ξ = i 4 e 3iξ 11 sin3ξ + 7 sin ξ, n a ξ = 3i e 3iξ cos3ξ + 9 cos ξ and use the resolution of the linear system * occuring in the proof of Theorem 3. to get p s ξ = 8 e 3iξ/ 3 sin 6 ξ 3m s ξcos 3ξ + 9 cosξ + n sξ11 sin 3ξ + 7 sinξ = e 3iξ/ cos ξ 19 + 13 cos ξ 3 q s ξ = 7 ie 3n 3iξ/ sin 6 ξ s ξ sin ξ sin ξ + m sξ + cos ξ sin ξ. = 9e 3iξ/ cos ξ 16 sinξ p a ξ = 8 e 3iξ/ 3 sin 6 ξ 3m a ξcos 3ξ + 9 cosξ + n aξ11 sin 3ξ + 7 sinξ = ie 3iξ/ sin ξ 16 + 11 cos ξ 3 q a ξ = 7 ie 3n 3iξ/ sin 6 ξ a ξ sin ξ sin ξ + m aξ + cos ξ sin ξ. such that = e 3iξ/ 64 9 cosξ 7 cos3ξ ϕ s ξ = p s ξ ϕ s ξ + q s ξ ϕ a ξ ϕ a ξ = p a ξ ϕ s ξ + q a ξ ϕ a ξ 1

The scaling relation leads to the following formula 1 Property 3.5 We have W ξ = M ξw ξm ξ + M ξ + πw ξ + πm ξ + π R1 where with ωs ξ ω W ξ = m ξ ω m ξ ω a ξ ω a ξ = ω s ξ = ω m ξ = ϕ a ξ + lπ = ϕ s ξ + lπ = 347 136 cos ξ 385 cosξ 31185 14445 + 7678 cos ξ + 53 cosξ 3465 ϕ s ξ + lπ ϕ a ξ + lπ = i sin ξ 691 + 193 cos ξ. 51975 Proof. Define Using the scaling relation, we have φξ = ϕs ξ ϕ a ξ. φξ φ ξ = M ξφξ φ ξm ξ. As we also have φξ φ ξ = ϕs ξ ϕ s ξ ϕ a ξ ϕ a ξ ϕ s ξ ϕ a ξ 1 In case V is generated by one single function ϕ, we recall that we have m ξ ωξ + m ξ + π ωξ + π = ωξ where m is the filter and where ωξ = + k= ϕξ + kπ. 11

hence we finally get from ** φξ + lπ φ ξ + lπ = W ξ W ξ = M ξw ξm ξ + M ξ + πw ξ + πm ξ + π. From the previous results, we obtain that the closed subspaces V j j ZZ form a multiresolution analysis of L IR with the difference that V is generated using two functions. A next step is then to define W as the orthogonal complement of V in V 1 and to construct mother wavelets in that context, that is to say functions which will genererate W and which will be compactly supported deficient splines with symmetry properties. 4 Construction of wavelets Proposition 4.1 A function f belongs to W if and only if there exists p, q L loc, π periodic such that and M ξ W ξ pξ qξ fξ = pξ ϕ s ξ + qξ ϕ a ξ + M ξ + π W ξ + π pξ + π qξ + π = a.e. where M is the filter matrix obtained in Proposition 3.4 and W ξ is the matrix defined in Property 3.5. Proof. We have f W f V 1 and f V p, q L loc, π per. : fξ = pξ ϕ s ξ + qξ ϕ a ξand f V. Let us develop the orthogonality condition, assuming the decomposition of f in terms of p, q. We have f V f, ϕ s,k = and f, ϕ a,k = k ZZ dξ e ikξ pξ ϕ s ξ + qξ ϕ a ξ φξ = IR k ZZ 1

where φξ = ϕs ξ ϕ s ξ using the scaling relation φξ = M ξφξ we get f V We finally obtain f V π IR π dξ e ikξ M ξpξ ϕ s ξ + qξ ϕ a ξ φξ = dξ e ikξ pξωs ξ + qξω M ξ m ξ pξω m ξ + qξω a ξ dξ e ikξ M ξw ξ M ξ W ξ pξ qξ pξ qξ. = k ZZ + M ξ + π W ξ + π k ZZ = k ZZ pξ + π qξ + π = a.e. Property 4. Define pξ = 8 p k e ikξ, 8 qξ = q k e ikξ. k= k= Then M ξ W ξ pξ qξ + M ξ + π W ξ + π pξ + π qξ + π = if and only if p = 388966976749167 q 6 59941398618818 + 131897133485383 q 7 199846687966 p 1 = 3946599711643653 q 6 59941398618818 p = 91616639337153 q 6 11988796557636 + 31475411718145575 q 7 59941398618818 + 984633993868151 q 7 399693174191 p 3 = 631169434995 q 6 11988796557636 + 7587715798335339 q 7 11988796557636 p 4 = 11876645619117 q 6 399693174191 13 35446674749693691 q 7 11988796557636

p 5 = 5864774773555 q 6 18749164977478479 q 7 11988796557636 399693174191 p 6 = 8697 q 6 97 815185 q 7 9166 q 6 p 7 = 9166 + 7671 q 7 97 p 8 = q = 8179838954188 q 6 99933435483 83134695564643 q 7 99933435483 q 1 = 83134695564643 q 6 99933435483 q = 3764966693617437 q 6 399693174191 q 3 = 66516461318335 q 6 399693174191 q 4 = 39373357761159 q 6 399693174191 q 5 = 31899113463613 q 6 399693174191 q 8 = 84648115943933 q 7 99933435483 3158164744637865999 q 7 399693174191 6769716351838981 q 7 399693174191 37145958857943389 q 7 399693174191 869366331131 q 7 399693174191 It follows that there exists deficient spline wavelets with support in [, 5], i.e. functions ψ such that or 7 ψξ = p k e ikξ ϕ s ξ + k= 1 ψx = 7 k= p k ϕ s x k + 7 q k e ikξ ϕ a ξ k= 7 q k ϕ a x k. k= Proof. The degree of the polynomials p, q are due to a look to the system that has to be solved. The resolution of the linear system is a Mathematica computation. Property 4.3 For every q 6, q 7, the function ψ has at least one vanishing moment. Proof. We have with ψξ = pξ ϕ s ξ + qξ ϕ a ξ 7 pξ = p k e ikξ, 7 qξ = q k e ikξ. k= k= 14

As ϕ a =, ϕ s it suffices to check that p =. To obtain this property, we just use the relation *** with ξ = the relation is in fact an equality everywhere since p, q are polynomials in that case. Indeed, since and W = M = 1 1 3 ωs ω a, M π =, W π = 5 3 ωs π ω a π from *** we obtain ω p = hence the conclusion. Moreover, symmetric compactly supported wavelets can be constructed: take q 6, q 7 such that p = p 7 ; then p 1 = p 6, p = p 5, p 3 = p 4, q = q 7, q 1 = q 6, q = q 5 ; q 3 = q 4 we denote these coefficients with an s and we get after some normalisation 1 ψ s x = 17951959ϕ s x + ϕ s x 7 16355665 ϕ s x 1 + ϕ s x 6 9 16989967 3 ϕ s x + ϕ s x 5 + 61668897 ϕ s x 3 + ϕ s x 4 9 + 67958549 ϕ a x ϕ a x 7 + 7686815ϕ a x 1 ϕ a x 6 3 + 577361163 ϕ a x ϕ a x 5 + 15594499ϕ a x 3 ϕ a x 4 3 In the same way, antisymmetric compactly supported wavelets can be constructed: take q 6, q 7 such that p = p 7 ; then p 1 = p 6, p = p 5, p 3 = p 4, q = q 7, q 1 = q 6, q = q 5 ; q 3 = q 4 we denote these coefficients with an a and we get after some normalisation 1 ψ a x = 8619155ϕ s x ϕ s x 7 31634977ϕ s x 1 ϕ s x 6 579681 ϕ s x ϕ s x 5 56567 ϕ s x 3 ϕ s x 4 +3619536ϕ a x + ϕ a x 7 + 3717576ϕ a x 1 + ϕ a x 6 + 7494639675 ϕ a x + ϕ a x 5 + 57769199 ϕ a x 3 + ϕ a x 4 15,,

Here are ψ s, ψ a up to a multiplicative constant 4 1 1 3 4 5 1 3 4 5-1 - - -4 The preceeding definitions can also be written using Fourier transforms. define 7 7 p s ξ = p s ke ikξ, q s ξ = qke s ikξ k= k= 7 7 p a ξ = p a ke ikξ, q a ξ = qke a ikξ. k= k= With ps ξ q M 1 ξ = s ξ p a ξ q a ξ we get from *** We M 1 ξw ξm ξ + M 1ξ + πw ξ + πm ξ + π = R and ψ s ξ ψ a ξ = M 1 ξ ϕs ξ ϕ a ξ. Now, we want to show that the family {ψ s,k : k ZZ } {ψ a,k : k ZZ } is a Riesz basis for W. First, we give a lemma which will be of great help to get the Riesz condition. We note here that this way of proving the Riesz condition is different from the one used for the scaling functions. We could have used the same method but computations became much more complicated; that s why we tried to get the result through another way. 16

Lemma 4.4 [5] Let f, g L IR. We define f k x = fx k, g k x = gx k, k ZZ and ωf,f ξ ω f,g ξ Hξ = ω f,g ξ ω g,g ξ where ω f,f ξ = ω g,g ξ = ω f,g ξ = k= k= k= fξ + kπ ĝξ + kπ fξ + kπ ĝξ + kπ. The following properties are equivalent: i the family {f k : k ZZ } {g k : k ZZ } satisfies the Riesz condition ii there exists A, B > such that π A c k l + d k l Hξ pξ qξ for every finite sequences c k, d k and where pξ =, pξ qξ c k e ikξ, qξ = d k e ikξ dξ B c k l + d k l iii there exists A, B > such that A λ i ξ B i = 1, where λ 1 ξ, λ ξ are the eigenvalues of Hξ. Proof. We have c k f k + d k g k L IR = 1 π c k e ikξ fξ + d k e ikξ ĝξ L IR = 1 π = 1 π π π pξ ω ff ξ + qξ ω aa ξ + pξqξω fg ξ + pξqξω fg ξ dξ pξ pξ Hξ, dξ, qξ qξ 17

which shows that i and ii are equivalent. Now, for every ξ, the matrix Hξ is hermitian. Therefore, for every ξ, there is a unitary matrix Uξ such that U ξhξuξ = diagλ 1 ξ, λ ξ. As we have p π pξ pξ U = Uξ, Uξ dξ q L [,π] qξ qξ π pξ pξ =, dξ qξ qξ p = q we obtain that ii is equivalent to L [,π] = c k l + d k l π A c k l + d k l λ 1 ξpξ + λ ξqξ dξ B c k l + d k l for every finite sequences c k, d k. Now, it is clear that iii implies ii. To get that ii implies iii, it suffices for example to use the Fejer kernel as p, q same proof as for the Riesz condition. where Now we want to use this lemma to obtain the desired result about the wavelets. Let us give some notations: define the matrix ωψs ξ ω ψs,ψ W ψ ξ = a ξ ω ψs,ψ a ξ ω ψa ξ ω ψa ξ = ω ψs ξ = ω ψs,ψ a ξ = ψ a ξ + lπ ψ s ξ + lπ ψ s ξ + lπ ψ a ξ + lπ. Theorem 4.5 The family {ψ s,k : k ZZ } {ψ a,k : k ZZ } constitutes a Riesz basis for W. The functions with index s resp. a are symmetric resp. antisymmetric. The support of ψ s, and ψ a, is included in [, 5]. 18

It follows that the functions j/ ψ s j x k, j/ ψ a j x k j, k ZZ form a Riesz basis of compactly supported deficient splines of L IR with symmetry properties. Proof. Using the expression of ψ a, ψ s in terms of ϕ a, ϕ s, i.e. ψ s ξ ψ a ξ = M 1 ξ ϕs ξ ϕ a ξ and by a computation similar to the one leading to R1, we get. W ψ ξ = M 1 ξw ξm 1 ξ + M 1 ξ + πw ξ + πm 1 ξ + π. R3 Then, since W ξ is hermitian positive definite for every ξ, the matrix W ψ has the same property if and only if the matrices M 1 ξ and M 1 ξ+π are not simultaneously singular. This is the case since we have up to an exponential function and a multiplicative constant det M 1 ξ = sin ξ/ 6494444315995 +148314169491171 cos ξ + 1193535397974745 cosξ +6516381148114 cos3ξ + 49964973943594 cos4ξ +554976754968 cos5ξ + 53917177111 cos6ξ which gives the graph for 1 37 det M 1 ξ + det M 1 ξ + π 1.5 1.5 -.5 1 3 4 5 6 Finally, since the elements of W ψ are trigonometric π-periodic polynomials, the eigenvalues are also periodic and continuous. Since they are strictly positive, condition iii of Lemma 4.4 follows. Hence the family of wavelets satisfies the Riesz condition. 19

To prove that the closure of the linear hull of the functions ψ s,k, ψ a,k k ZZ is W, it remains to show that f W, { f, ψs,k = f, ψ a,k = f =. For f W, we have see Proposition 4.1 p, q L loc, π periodic such that fξ = pξ ϕ s ξ + qξ ϕ a ξ and M ξ W ξ pξ qξ + M ξ + π W ξ + π pξ + π qξ + π = a.e. 1 The same computation as the one leading to the equality above in Proposition 4.1, but using orthogonality to ψ s,k, ψ a,k instead of to ϕ s,k, ϕ a,k, leads to M 1 ξ W ξ pξ qξ + M 1 ξ + π W ξ + π pξ + π qξ + π = a.e. Then 1 and are equivalent to M ξ W ξ M ξ + π W ξ + π M 1 ξ W ξ M 1 ξ + π W ξ + π pξ qξ pξ + π qξ + π = a.e. 3 We have M ξ W ξ M ξ + π W ξ + π M 1 ξ W ξ M 1 ξ + π W ξ + π M ξ M = ξ + π W ξ M 1 ξ M 1 ξ + π W ξ + π Using the relations R1, R, R3, we get M ξ M ξ + π M 1 ξ M 1 ξ + π W ξ W ξ + π = W ξ W ψ ξ M ξ M ξ + π M 1 ξ M 1 ξ + π..

For every ξ, the matrices W ξ, W ψ ξ are not singular. matrix M ξ M ξ + π M 1 ξ M 1 ξ + π Hence, for every ξ the is not singular. The conclusion follows: from 3 we obtain pξ = qξ = a.e. 5 Appendix with l =, 1, are linearly inde- Property 5.1 The functions ϕ a,l [,1], ϕ s,l [,1] pendant. Proof. For x [, 1], we have P a, x := ϕ a, x = ϕ a x = x 4 11 15 x5 P a, 1 x := ϕ a, 1 x = ϕ a x + 1 = 9 8 x 1 + 3x 1 3 38 15 x 1 5 P a, x := ϕ a, x = ϕ a x + = 1 x 4 + 11 1 x5 15 P s, := ϕ s, x = ϕ s x = x 4 3 5 x5 P s, 1 := ϕ s, 1 x = ϕ s x + 1 = 57 8 3 x 1 + x 1 4 P s, := ϕ s, x = ϕ s x + = 1 x 4 3 5 3 x5. If r j j = 1,..., 6 are such that r 1 P a, + r P a, 1 + r 3 P a, + r 4 P s, + r 5 P s, 1 + r 6 P s, = then the coefficients of x j j =,..., 5 are equal to. We get the system 3r + 3r 3 + r 5 r 6 = 3r 3r 3 + r 5 + r 6 = r 5 r 6 = 3r + 4r 3 5r 5 5r 6 = 3r 1 + 3r 6r 3 + 4r 4 + 19r 5 + 8r 6 = 9r 1 + 9r 3 11r 4 38r 5 11r 6 = which is easy to solve; the unique solution is r 1 = r = r 3 = r 4 = r 5 = r 6 =. 1

Property 5. For every f V, there are c k k ZZ, d k k ZZ l such that D 6 f = lim m m + k= m c k δ k + d k δ k in the distribution sense, where δ k and δ k are respectively the Dirac and the derivative of the Dirac distribution at k. Proof. Let f V and, for every k ZZ, let f [k,k+1] = P 5 = polynomial of degree at most 5. If a, a 1 are respectively the coefficients of x 4, x 5 in P 5, then D 4 P 5 x = 5!a 1 x + 4!a and, for h C IR with compact support, fxd 6 hx dx IR = 5! k ZZ a 1 a k 1 1 h + 4!a a k 1 + 5!ka 1 a k 1 1 Dh. For every k ZZ, we define c k = 5!a 1 a k 1 1, d k = 4!a a k 1 5!ka 1 a k 1 1 = 4! a + 5ka 1 a k 1 + 5ka k 1 1 hence to conclude, it suffices to prove that a 1 k ZZ l, a + 5ka 1 k ZZ l. Do obtain this, we first remark that, on the linear space of polynomials of degree at most 5, all norms are equivalent. Hence, there are r, R > such that 5 1 5 r A j P x dx R A j j= j= for every polynomial P x = 5 j= A j x j. Next, for f V, using the same notations as just above, we have f L IR = = k+1 k= k 1 k= P 5 x dx P 5 x + k dx.

Moreover, in P 5 x + k, the coefficient of x 5 is a 1 and the coefficient of x 4 is a + 5ka 1. It follows that a k= Hence the conclusion. 1 + a + 5ka 1 1 r k= 1 P 5 x + k dx 1 r f L IR. References [1] F. Bastin, Ondelettes, notes de séminaire, ULg, 199. [] C.K. Chui and J.Z. Wang, On compactly supported spline wavelets and a duality principle, Transactions of the AMS, 33,, 199, 93-915. [3] O. Gilson, O. Faure, P. Laubin, Quasi-optimal convergence using interpolation by non-uniform deficient splines, preprint January 1 [4] T.N.T. Goodman, S.L. Lee, Wavelest of multiplicity r, Trans. Amer. Math. Soc. 34, 1, 1994, 37-34 [5] T.N.T. Goodman, S.L. Lee, W.S. Tang, Wavelets in wandering subspaces, Trans. Amer. Math. Soc., 338,, 1993, 639-654 [6] S. Mallat, Multiresolution approximations and wavelet orthonormal bases of L IR, Trans. Amer. Math. Soc., 3151, 1989, 69-88. [7] Y. Meyer, Ondelettes et opérateurs, I,II,III, Hermann, Paris, 199. [8] S.S. Rana, Dubey Y.P., Best error bounds of deficient quintic quintic splines interpolation, Indian J. Pure Appl. Math., 8, 1997, 1337-1344 Address Françoise BASTIN University of Liège, Institute of Mathematics, B37 4 LIEGE-BELGIUM email: F.Bastin@ulg.ac.be Fax: xx3 4 366 95 47 3