Bin Han Department of Mathematical Sciences University of Alberta Edmonton, Canada T6G 2G1

SYMMETRIC ORTHONORMAL SCALING FUNCTIONS AND WAVELETS WITH DILATION FACTOR d = Bin Han Department of Mathematical Sciences University of Alberta Edmonton, Canada T6G 2G1 email: bhan@math.ualberta.ca Abstract. It is well known that in the univariate case, up to an integer shift and possible sign change, there is no dyadic compactly supported symmetric orthonormal scaling function except for the Haar function. In this paper we are concerned with the construction of symmetric orthonormal scaling functions with dilation factor d =. Several examples of such scaling functions are provided in this paper. In particular, two examples of C 1 orthonormal scaling functions, which are symmetric about 0 and 1, respectively, are presented. We will then discuss how to construct symmetric 6 wavelets from these scaling functions. We explicitly construct the corresponding orthonormal symmetric wavelets for all the examples given in this paper. 1991 Mathematics Subject Classification. 2C05, 2C15, 3A30, 1A30. Key words and phrases. orthonormal scaling function, wavelets, symmetry, smoothness. Research supported in part by Killam Trust under Isaak Walton Killam Memorial Scholarship 1 Typeset by AMS-TEX

2 SYMMETRIC ORTHONORMAL SCALING FUNCTIONS AND WAVELETS 1. Introduction In her celebrated paper [6], Daubechies constructed a family of compactly supported orthonormal scaling functions and their corresponding orthonormal wavelets with dilation factor 2. Since then wavelets with compact support have been found to be very useful in applications. Also in that paper, Daubechies proved that up to an integer shift and possible sign change, except for the Haar scaling function (i.e., the characteristic function of the interval [0, 1]), there exists no dyadic orthonormal scaling function which is symmetric about some point on R. Therefore, with the dilation factor 2, there exists no symmetric or antisymmetric orthonormal wavelet except the Haar wavelet(also see [7]). On the other hand, symmetry is a much desired property in applications. In this paper, we consider how to achieve both symmetry and orthogonality of wavelets. It is very natural to consider using a different dilation factor other than 2. In fact it is of interest in its own right to consider other dilation factors in applications. In the engineering literature, decimation by an integer ratio other than d = 2 occurs in multi-rate digital processing(cf. [21]). The motivation for such signal splitting and coding before transmission is well understood in the literature where decompositions parallel the decompositions of quadrature mirror filtering. Such filtering is referred to as M channel filtering where the integer M corresponds to the integer d in this paper, see[20, 21, 23] for more details. The approach of using different dilation factor to construct wavelets was considered in [, 20, 23] and in other papers. A general method of constructing orthonormal scaling functions in L 2 (R) with general dilation factors was obtained in [20, 23]. As in the case of Daubechies orthonormal scaling functions, the orthonormal scaling functions constructed in [20, 23] lack symmetry. In [], Chui and Lian considered construction of symmetric and antisymmetric orthonormal wavelets with dilation factor 3. Several interesting examples of symmetric and antisymmetric wavelets with dilation factor 3 were reported there, but smoothness of the resulting wavelets constructed in [] was not considered. Let us now introduce some definitions and notations. The procedure for constructing orthonormal scaling functions with dilation factor d is very similar to that for constructing orthonormal dyadic scaling functions, which is discussed in details in [2, 7]. The reader is referred to [2, 7] for a complete picture of this construction. In the following, we are concerned with functional equations of the form (1.1) φ = k Z a(k)φ(d k), where φ is the unknown function defined on the real line R, a is a finitely supported sequence on Z, and here d is the dilation factor. We will use l 0 (Z) to denote the linear space of finitely supported sequences on Z and use δ to denote the sequence such that δ(0) = 1 and δ(k) = 0, for k Z\{0}. The equation (1.1) is called a refinement equation, and the sequence a is called the refinement mask. If a finitely supported sequence a satisfies (1.2) a(k) = d, k Z then it is known that(see [3]) there exists a unique compactly supported distribution φ satisfying the refinement equation (1.1) subject to the condition φ(0) = 1. This

SYMMETRIC ORTHONORMAL SCALING FUNCTIONS AND WAVELETS 3 distribution is said to be the normalized solution of the refinement equation (1.1). The Fourier transform of a function f L 1 (R) is defined to be ˆf(ξ) := R f(x)e ixξ dx, ξ R, and the domain of the Fourier transform can be naturally extended to include compactly supported distributions. Taking the Fourier transform on both sides of (1.1), we obtain (1.3) ˆφ(ξ) = d 1ã(e iξ/d ) ˆφ(ξ/d), ξ R, where ã(e iξ ) is given by its symbol ã(z) defined by (1.) ã(z) := k Z a(k)z k, z C\{0}. If the shifts of the normalized solution φ of (1.1) are orthonormal, that is, φ(x j)φ(x k) dx = δ jk j, k Z, then R (1.5) d 1 ã(e i(ξ+2πk/d) ) 2 = d 2, ξ R. k=0 In order to solve the refinement equation (1.1), we start with the hat function φ 0 defined by 1 + x for x [ 1, 0), φ 0 (x) := 1 x for x [0, 1], 0 for x R\[ 1, 1]. Then we study the iteration scheme Q n aφ 0, n = 0, 1, 2,, where Q a is the bounded linear operator on L p (R), 1 p given by Q a f := k Z a(k)f(d k), f L p (R). This iteration scheme is called a subdivision scheme(see [3]) or a cascade algorithm(see [9]) associated with a. We say that the subdivision scheme associated with a converges in the L p norm if there exists a function f L p (R) such that lim n Q n aφ 0 f p = 0. If this is the case, then it is necessary that f is equal to the normalized solution of (1.1) with the refinement mask a. The subdivision operator associated with a is defined by S a λ(i) := k Z a(i dk)λ(k), i Z,

SYMMETRIC ORTHONORMAL SCALING FUNCTIONS AND WAVELETS where λ l 0 (Z). By induction, we observe that (1.6) Q n aφ 0 = k Z S n a δ(k)φ 0 (d n k). Given a sequence λ l 0 (Z), the difference operator on l 0 (Z) is defined to be λ = λ λ( 1). It was proved by Han and Jia in [12] (also cf. Jia [13]) that the subdivision scheme associated with a converges in the L p norm if and only if (1.7) lim n Sn a δ 1/n p < d 1/p. If a satisfies (1.5) and the subdivision scheme associated with a converges in the L 2 norm, then the normalized solution φ of (1.1) with the refinement mask a lies in L 2 (R) and the shifts of φ are orthonormal(cf. [6, 13]). (The idea is to use the initial function χ [0,1), then we still have that lim n Q n aχ [0,1) φ 2 = 0. As the shifts of each Q n aχ [0,1) are orthonormal, the shifts of φ are also orthonormal.) Conversely, if the normalized solution φ of (1.1) with mask a lies in L 2 (R) and the shifts of φ are orthonormal, then a satisfies (1.5) and the subdivision scheme associated with a converges in the L 2 norm. The subdivision operator plays an important role in the study of convergence of subdivision schemes and the characterization of smoothness of refinable functions. The reader is referred to Goodman, Micchelli, Ward [11], Han and Jia [12] and Jia [13] for detailed discussion on subdivision operators. In this paper, to obtain symmetric orthonormal scaling functions, we will use the above equation such that the associated subdivision scheme converges in the L 2 norm, i.e., condition (1.7) is satisfied. The structure of this paper is as follows. In Section 2, we study the characterization of L p smoothness of a refinable function with dilation factor d. We derive two simple inequalities to estimate the L smoothness of a refinable function; then apply them in Section 3 to estimate the smoothness of several examples. Note that with dilation factor, up to a shift of an orthonormal scaling function φ, there are only two types of symmetry: φ is symmetric about either 0 or 1/6. In Section 3, for dilation factor, we present some examples of orthonormal scaling functions which are symmetric about 0 or 1/6, respectively. In particular, two C 1 orthonormal scaling functions, with one symmetric about 0 and the other symmetric about 1/6, are obtained. In Section, we discuss how to construct the corresponding wavelets with symmetry. With dilation factor, for any orthonormal scaling functions that are symmetric about 1/6, we can construct the corresponding three wavelets which are all either symmetric or antisymmetric about 1/6. However, on the contrary, with dilation factor, if an orthonormal scaling function is symmetric about 0, we shall prove that it is impossible to construct the corresponding three wavelets which are all either symmetric or antisymmetric about 0. This fact is true for any even dilation factor. But if other symmetry is allowed for wavelets, we can construct the corresponding wavelets with symmetry. For each of our examples such that the scaling function is symmetric about 0, we construct its three wavelets with one symmetric about 0, another antisymmetric about 0, and the third symmetric about 1/2.

SYMMETRIC ORTHONORMAL SCALING FUNCTIONS AND WAVELETS 5 2. L p Smoothness of Refinable Functions In this paper, we are concerned with construction of orthonormal scaling functions with symmetry. Our objective is to find smooth orthonormal scaling functions with symmetry for dilation factor. Before we do this, let us review some notations and results related to the characterization of smoothness of refinable functions. Following [1], we use the generalized Lipschitz space to measure smoothness of a given function. Let us recall the definition of the generalized Lipschitz space from [8]. For t R, the difference operator t is defined to be t f = f f( t) where f is a function from R to C. Let k be a positive integer. The k-th modulus of smoothness of f L p (R) is defined by ω k (f, h) p := sup k t f p, h 0. t h For ν > 0, let k be an integer greater than ν. The generalized Lipschitz space Lip (ν, L p (R)) consists of those functions f L p (R) for which (2.1) ω k (f, h) p Ch ν h > 0, where C is a constant independent of h. The optimal smoothness of a function f L p (R) in the L p norm is described by its critical exponent ν p (f) defined by ν p (f) := sup{ν : f Lip (ν, L p (R))}. The concept of stability plays an important role in the study of the smoothness of the normalized solution of (1.1). Let φ be a compactly supported distribution, we say that the shifts of φ are stable if for any ξ R, there exists k Z such that ˆφ(ξ + 2πk) 0. If φ lies in L p (R), where 1 p (when p =, we consider φ C 0 (R)), it is equivalent to saying that there exist two positive constants C 1 and C 2 such that (2.2) C 1 λ p k Z λ(k)φ( k) p C 2 λ p λ l 0 (Z), where λ p := ( k Z λ(k) p ) 1/p. Clearly, if the shifts of φ are orthonormal, then the shifts of φ are stable. See [15, 17, 2] for discussion of stability. Following Jia [13], we shall use the p-norm joint spectral radius of a finite collection of square matrices to characterize the critical exponent ν p (φ) of a refinable function φ. Let A be a finite collection of r r square matrices. For a positive integer n we denote by A n the Cartesian power of A: and for 1 p <, we define A n = {(A 1,, A n ) : A 1,, A n A}, A n p := A 1 A n p (A 1,,A n ) A n 1/p,

6 SYMMETRIC ORTHONORMAL SCALING FUNCTIONS AND WAVELETS where denotes a matrix norm such that AB A B for any r r matrices A and B. For p =, we define A n := max{ A 1 A n : (A 1,, A n ) A n }. For any 1 p, the l p -norm joint spectral radius of A is defined to be ρ p (A) := lim n An 1/n p = inf n 1 An 1/n p. For any finitely supported sequence a, we can write the symbol of a as (2.3) ã(z) = (1 + z + + z d 1 ) l b(z) and (1 + z + + z d 1 ) b(z). The characterization of L p smoothness of a refinable function is as follows: Theorem 2.1. Let φ L p (R) be the normalized solution of (1.1) with a finitely supported refinement mask a. Suppose that b l 0 (Z), given by (2.3), be supported on [0, N]. If the shifts of φ are stable, then (2.) ν p (φ) = 1 p log d ρ p ( {B0,, B d 1 } ), where B ε := ( b(di j ε) ) 1 i,j N d 1 integer which is no less than N d 1 (2.5) ν 2 (φ) = 1 2 1 2 log d ρ(t c ), N, 0 ε d 1 and d 1 is the smallest and 1. Moreover, when p = 2, where T c := (c di j ) N d 1 i,j N d 1 with c(z) := b(z) 2. Proof. From the definition of the subdivision operator, by induction, we have φ(x) = k Z S n a δ(k)φ(d n x k). Therefore for the integer l given in (2.3), we have, l d n φ(x) = k Z l S n a δ(k)φ(d n x k). Since the shifts of φ are stable, by (2.2), there exist positive constants C 1 and C 2 independent of n such that (2.6) C 1 l S n a δ p d n/p l d nφ p C 2 l S n a δ p. Hence if φ Lip (ν, L p (R)), by the definition of the generalized Lipschitz space, we see from (2.1) and (2.6) that (2.7) C 1 l S n a δ p Cd n/p d nν = Cd (1/p ν)n.

SYMMETRIC ORTHONORMAL SCALING FUNCTIONS AND WAVELETS 7 Let ρ p := lim n l Sa n δ 1/n p. It follows from (2.7) that ν 1 p log d ρ p. Hence ν p (φ) 1 p log d ρ p. To prove ν p (φ) 1 p log d ρ p, the proof is based on the following results from approximation theory: For a function f in L p (R) (f is continuous in the case p = ), f lies in Lip (ν, L p (R)) for ν > 0 if and only if, for some integer l > ν, there exists a constant C > 0 such that (2.8) l d nf p Cd nν n N. For these results, the reader is referred to the work of Boman [1] and Ditzian [10]. Since ρ p = lim n l Sa n δ 1/n p, for any η > 0, there exists a constant C η such that l Sa n δ p C η (ρ p + η) n. It follows from (2.6) that l d nφ p C 2 d n/p l S n a δ p C 2 C η d n/p (ρ p + η) n, η > 0. Therefore, by (2.8), φ Lip ( 1 p log d(ρ p + η), L p (R)) η > 0. Hence ν p (φ) 1 p log d ρ p. Now it suffices to prove that ρ p = ρ p ({B 0,, B d 1 }). Using the relation (2.3), it is easy to prove that, see [13, Theorem 3.3], ρ p = lim n l S n a δ 1/n lim n S n b δ 1/n p p =. By [12, Theorem 2.5], we see that ρ p := lim n S n b δ 1/n p = ρ p ({B 0,, B d 1 }). For the case p = 2, (2.5) is evident by (2.) and [12, Theorem.1]. The reader is referred to Han and Jia [12] and Jia [13] for more detailed discussion on joint spectral radius and its relation to subdivision operators. We mention that for the case p = 2, in the univariate case with dilation factor 2, Villemoes in [22] obtained a characterization of L 2 smoothness. In [1], Jia gave a characterization of L 2 smoothness for the multivariate refinement mask with an isotropic dilation matrix. It was illustrated in [1, 22] that ν 2 (φ) can be obtained by calculating the spectral radius of a finite matrix. (2.5) is a special case of [1]. In this paper, we are mainly interested in the case p = 2 and p =. Although it is difficult to apply Theorem 2.1 directly to obtain ν (φ), the following two inequalities provide an easy way to estimate ν (φ). Corollary 2.2. Under the same conditions as in Theorem 2.1 for p =. Then ( { }) (2.9) ν (φ) log d max b(dk + ε) : ε = 0, 1,, d 1. k Z If there exist 0 ε 0 d 1 and k 0 Z such that b(dk + ε 0 ) = 0 k Z\{k 0 }, then (2.10) ν (φ) log d ( b(dk0 + ε 0 ) ). Proof. For any r r matrix A = (a ij ) 1 i,j r, choose the following matrix norm: { r } A := max a ij : j = 1,, r. i=1

8 SYMMETRIC ORTHONORMAL SCALING FUNCTIONS AND WAVELETS Then by B ε := ( b(di j ε) ) 1 i,j N d 1, it is easily seen that for any ε Ω, B ε max{ k Z b(dk + ε) : ε = 0,, d 1}. Hence by the definition of l -norm joint spectral radius, we have ρ ({B 0,, B d 1 }) = inf {B 0,, B d 1 } n 1/n n 1 { } max b(dk + ε) : 0 ε d 1. k Z Hence by Theorem 2.1, (2.) holds true. If there exist 0 ε 0 d 1 and k 0 Z such that b(dk + ε 0 ) = 0 for any k Z\{k 0 }, then there exists ε 1 {0,, d 2} such that (d 1) (dk 0 +ε 0 +ε 1 ). If dk 0 +ε 0 = 0, we choose ε 1 = d 1. Set k 1 = dk 0+ε 0 +ε 1 d 1. Note that supp b [0, N] and N max{1, dk 0 + ε 0 }. It follows that 1 k 1 N d 1. Therefore the k 1-th column of B ε1 = ( b(di j ε 1 ) ) 1 i,j N d 1 is b(dk 0 + ε 0 )e k1 where e k1 has 1 at its k 1 th coordinate and 0 otherwise. In other words, B ε1 e k1 = b(dk 0 +ε 0 )e k1 ; therefore b(dk 0 + ε 0 ) is an eigenvalue of B ε1. It follows that ρ ({B 0,, B d 1 }) = lim {B 0,, B d 1 } n 1/n n Hence by Theorem 2.1, (2.5) holds. lim n Bn ε 1 1/n = ρ(b ε1 ) b(dk 0 + ε 0 ). We will apply the results in this section to several examples in Section 3. 3. Symmetric Orthonormal Scaling Functions With d = In this section, we are concerned with construction of symmetric orthonormal scaling functions with dilation factor. Throughout this section, we assume that d =. For any refinement mask a l 0 (Z), up to an integer shift, there are only two types of symmetry. The first type of symmetry is that the mask a is symmetric about 0, i.e., (3.1) a(k) = a( k) k Z. The second type of symmetry is that a is symmetric about 1 2, i.e., (3.2) a(k) = a(1 k) k Z. To find a mask of type (3.1) or (3.2) such that (1.5) holds true, we need the following result: Theorem 3.1. Let l be a positive integer and P l be the polynomial of degree l 1 16 determined by the first l terms of the Taylor series of expanded at (1 z) l (1 2z) 2l 16 z = 0, i.e., = P (1 z) l (1 2z) 2l l (z) + O(z l ), z 0. Let b(e iξ ) = 1 + e iξ + e i2ξ + e i3ξ 2l p(sin 2 ξ 2 )

SYMMETRIC ORTHONORMAL SCALING FUNCTIONS AND WAVELETS 9 for some polynomial p such that b satisfies the following equation (3.3) b(e iξ ) + b(e i(ξ+π/2) ) + b(e i(ξ+π) ) + b(e i(ξ+3π/2) ) = 16. Then the first l terms of p are given by P l. More precisely, p(z) = P l (z)+o(z l ), z 0. Proof. Since b(e iξ ) = ( sin(2ξ) ) 2lp(sin 2 ξ sin ξ 2 ), (3.3) can be rewritten as 2 ( sin(2ξ) For ξ near 0, we have ) ( 2l p(sin 2 ξ 2 ) (sin ξ + p(sin2 2ξ+π 2 )2l (sin 2ξ+π ( sin(2ξ) ) + p(sin ) 2l 2 ξ+π 2 ) (sin ξ+π + p(sin 2 )2l 2 2ξ+3π ) (sin 2ξ+3π ) 2l ) ( 2l p(sin 2 ξ 2 ) ) (sin ξ + O(1) = 16, ξ 0. 2 )2l Therefore ( ) 2l ( sin(2ξ) sin ξ p(sin 2 ξ 2 ) + ξ ) O(sin2l 2 ) = 16, ξ 0. 2 Let z = sin 2 ξ 2. Note that we have ( ) 2 sin(2ξ) = (1 sin 2 ξ sin(ξ/2) 2 )(1 2 ξ sin2 2 )2 = (1 z)(1 2z) 2, (1 z) l (1 2z) 2l (p(z) + O(z l )) = 16, z 0. Our assertion in this theorem follows from the above equality. ) = 16. In fact, if we take b(e iξ ) = 1+e iξ +e i2ξ +e i3ξ 2l P l (sin 2 ξ 2 ), then b satisfies (3.3). To find a mask a of type (3.1) such that (1.5) holds, when l = 2k, we are looking for ã(e iξ ) = 1 + e iξ + e i2ξ + e i3ξ 2k p(sin 2 ξ 2 ), or when l = 2k + 1, we are looking for ã(e iξ ) = e i3kξ ( 1 + e iξ + e i2ξ + e i3ξ ) 2k+1 ( e iξ + e i2ξ 2 ) p(sin 2 ξ 2 ), where p is a polynomial such that b(e iξ ) = ã(e iξ ) 2 satisfies (3.3). From Theorem 3.1, we see that when l = 2k, p 2 (z) = P l (z) + O(z l ) and when l = 2k + 1, (1 z)p 2 (z) = P l (z) + O(z l ). Hence the first l terms of p can be easily determined directly from P l while other terms of p can be found by using equation (3.3). Note that (3.3) is equivalent to (3.) b(0) = and b(k) = 0 k Z\{0}. A similar procedure can be used to find masks of type (3.2) such that (1.5) holds. By using the above procedure, we have the following examples.

10 SYMMETRIC ORTHONORMAL SCALING FUNCTIONS AND WAVELETS Example 3.1. Let l = 1. Then by Theorem 3.1 P 1 (z) = 16. We are looking for a mask given by ã(e iξ ) = 1 ( 1 + e iξ + e i2ξ + e i3ξ) ( e iξ + e i2ξ) p(sin 2 ξ 8 2 ), with a polynomial p(z) = + t z for some t R. By using (3.), we find that t satisfies the equation t 2 + 32 t = 0. Taking one root t = 8 6 16, we have the refinement mask a given by its symbol ã(z) = 2 6 z 3 + 1 2 z 2 + 2 + 6 z 1 + 1 + 2 + 6 z 1 + 1 2 z2 + 2 6 ) ( 2 6 6 6 = (1 + z + z 2 + z 3 z 2 + z 1 + + 2 6 z ). 1 Then a satisfies (1.5) and the normalized solution φ of (1.1) with the above refinement mask a lies in L 2 (R) with ν 2 (φ) = 1 2 log 8 2 6 2 0.5918 by Theorem 2.1. In fact, (2.) gives us ν p (φ) = 1 + 1 2p 1 p log ((2 6) p + 6 p ) for 1 p < and ν (φ) = log 6 0.3538. Therefore φ is a continuous orthonormal scaling function which is symmetric about the origin. Example 3.2. Let l = 2. Then by Theorem 3.1 P 1 (z) = 16 + 160z. We are looking for a mask given by ã(e iξ ) = 1 16 1 + e iξ + e i2ξ + e i3ξ 2 p(sin 2 ξ 2 ) with a polynomial p(z) = + 20 z + t z 2 for some t R. By using (3.), we find that t satisfies the equation 3t 2 + 22t + 3200 = 0. Taking one root t = 8 6 112 3, we have the refinement mask a given by its symbol 6 1 6 + 1 6 + 10 ã(z) = z 5 z + z 3 + 1 96 8 96 2 z 2 + 50 6 z 1 8 + 25 + 6 + 50 6 z 1 + 1 6 + 10 6 + 1 6 1 2 8 2 z2 + z 3 z + z 5 ) 96 ) 8 96 = (1 + z + z 2 + z 3 (1 + z 1 + z 2 + z 3 ( 6 1 z 2 + 13 2 6 6 z 1 + 96 8 16 + 13 2 6 6 1 z 1 + z ). 2 8 96 Then a satisfies (1.5) and the normalized solution φ of (1.1) with the above refinement mask a lies in L 2 (R) with ν 2 (φ) 1.0975 by Theorem 2.1. By (2.9) and (2.10), we have ν (φ) = log 6 16 0.6191. Therefore φ is a continuous orthonormal scaling function which is symmetric about the origin. Example 3.3. Let l = 3. Then By Theorem 3.1 P 3 (z) = 16 + 20z + 2016z 2. We are looking for a refinement mask given by ã(e iξ ) = 1 ei5ξ ( 1 + e iξ + e i2ξ + e i3ξ) 3 ( 1 + e iξ ) p(sin 2 ξ 2 ), with a polynomial p(z) = + 32 z + 156 z 2 + c 1 z 3 + c 2 z + c 3 z 5 + c z 6 for some parameters c 1, c 2, c 3, c R. Let c = 096. Let t 13.93708376395 satisfy the following equation: 36 t 8 + 992 t 7 + 30772 t 6 + 10993728 t 5 + 295090 t + 367729016 t 3 + 3322363672 t 2 + 18518676388 t + 136198081 = 0. z 3

SYMMETRIC ORTHONORMAL SCALING FUNCTIONS AND WAVELETS 11 From (3.), we have c 3 = 512 t, c 2 = 32(t 2 + 56 t + 72) and c 1 = ( 539136 t 7 + 65686176 t 6 + 39119110 t 5 + 105090578 t + 193133087616 t 3 + 216527600720 t 2 + 136750295366 t + 37190151300 ) /51790153. By computation, we have the refinement mask a given by its symbol where ã(z) = (1 + z 1 + z 2 + z 3 ) 3 q(z) q(z) =0.0078125 z 11.0319576659 z 10 + 0.03561721 z 9 + 0.01121382221 z 8 0.07626968211 z 7 0.02717616670 z 6 + 0.1037099788 z 5 + 0.1037099788 z 0.02717616670 z 3 0.07626968211 z 2 + 0.01121382221 z 1 + 0.03561721 0.0319576659 z + 0.0078125 z 2. Then a satisfies (1.5) and the normalized solution φ of (1.1) with the above refinement mask a lies in L 2 (R) with ν 2 (φ) 1.6688 by Theorem 2.1. By (2.10), we have ν (φ) log (0.158) > 1.329. Therefore φ is a C 1 orthonormal scaling function which is symmetric about the origin. Similarly, we have the following two examples of refinement masks of type (3.2) which satisfy (1.5). Example 3.. Let l = 2. Then by Theorem 3.1 P 2 (z) = 16+160z. We are looking for a mask given by ã(e iξ ) = 1 32 1 + e iξ + e i2ξ + e i3ξ 2 (1 + e iξ )p(sin 2 ξ/2) with a polynomial p(z) = + 22z + tz 2 for some t R. By using (3.), we find that t satisfies the equation t 2 + 152t + 1056 = 0. Taking one root t = 295 76, we have the refinement mask a given by its symbol 295 19 295 + 3 ã(z) = z 5 z + 13 295 295 + 29 z 3 + z 2 + 59 295 295 295 z 1 + + 6 6 6 z1 + 59 295 295 + 29 z 2 + z 3 6 + 13 295 295 + 3 295 19 z z 5 + z 6 ) ) ( 295 19 = (1 + z 1 + z 2 + z 3 (1 + z 1 + z 2 + z 3 z 2 + 35 3 295 295 295 z 1 + + 6 6 z1 + + 35 3 295 295 19 z 2 + z ). 3 Then a satisfies (1.5) and the normalized solution φ of (1.1) with the above refinement mask a lies in L 2 (R) with ν 2 (φ) 1.122 by Theorem 2.1. By (2.9) and (2.10), we have ν (φ) = log 295 6 0.989. Therefore φ is a continuous orthonormal scaling function which is symmetric about the point 1 6.

12 SYMMETRIC ORTHONORMAL SCALING FUNCTIONS AND WAVELETS Example 3.5. Let l = 2. Then by Theorem 3.1 P 2 (z) = 16+160z. We are looking for a mask given by ã(e iξ ) = 1 32 ei3ξ ( 1 + e iξ + e i2ξ + e i3ξ) 2 ( 1 + e iξ ) p(sin 2 ξ 2 ), with a polynomial p(z) = + 22z + c 1 z 2 + c 2 z 3 + c 3 z for some c 1, c 2, c 3 R. Let c 3 := 52. By solving (3.), we find that c 2 516.118718671358 satisfies the following equation: and c 2 + 1396 c 3 2 + 533080 c 2 2 28699 c 2 3386820220 = 0, c 1 = ( c 3 2 + 718 c 2 2 19666 c 2 115288928 ) /261256. By computation, we have the refinement mask a given by its symbol with ã(z) = ( 1 + z 1 + z 2 + z 3) (1 + z 1 + z 2 + z 3) q(z) q(z) = 0.05517578125 z.1321937569 z 3.03891892187 z 2 + 0.1980205855 z 1 + 0.0522310031 + 0.0522310031 z 1 + 0.1980205855 z 2 0.03891892187 z 3.1321937569 z + 0.05517578125 z 5. Then a satisfies (1.5) and the normalized solution φ of (1.1) with the above refinement mask a lies in L 2 (R) with ν 2 (φ) 1.205 by Theorem 2.1. By (2.10), we have ν (φ) log (.237) > 1.037. Therefore φ is a C 1 orthonormal scaling function which is symmetric about the point 1 6. The reader is referred to the appendix for the graphs of these scaling functions. Moreover, another C 1 orthonormal scaling function φ which is symmetric about 1/6 such that ν 2 (φ) 2.068 and ν (φ) > 1.583 is presented in the appendix. Since in the preceding examples we have symmetric orthonormal scaling functions, it is quite natural to ask whether or not we can construct the corresponding wavelets with symmetry. We shall answer this question in the next section.. Construction of Symmetric Wavelets With Dilation Factor For dilation factor 2, it is easy to construct orthonormal wavelets from orthonormal scaling functions (see [2, 6, 7]). For dilation factor d > 2, things are different. For any finitely supported sequence a on Z and an integer l Z, we define a new sequence on Z by (.1) a l (k) := a(l + dk) k Z. It is well known that to construct compactly supported orthonormal wavelets from compactly supported orthonormal scaling functions, we have to deal with the problem of extension of matrix whose entries are trigonometric polynomials. More precisely, for the general dilation factor d 2, given an orthonormal scaling function φ

SYMMETRIC ORTHONORMAL SCALING FUNCTIONS AND WAVELETS 13 with a finitely supported refinement mask a, constructing orthonormal wavelets with compact support from φ is equivalent to finding a unitary matrix whose entries are 2π periodic trigonometric polynomials such that the first column is (ãε0 (e iξ ), ã ε1 (e iξ ),, ã εd 1 (e iξ ) ) T (see [2, 7, 15, 16, 18]) where {ε0,, ε d 1 } is a complete set of the representatives of the distinct cosets of the quotient group Z/dZ. That is, the following matrix ã ε0 (e iξ ), m 1 ε 0 (e iξ ) m d 1 ε 0 (e iξ ) 1 ã ε1 (e iξ ), m 1 ε 1 (e iξ ) m d 1 ε 1 (e iξ ) (.2).. d......., ξ R ã εd 1 (e iξ ), m 1 ε d 1 (e iξ ) m d 1 ε d 1 (e iξ ) is a unitary matrix for some finitely supported sequences m i, i = 1,, d 1. The above problem of matrix extension was discussed by Jia and Shen in [16] and by Riemenschneider and Shen in [19]. A good constructive algorithm to practically work out matrix (.2) was reported by Lawton, Lee, and Shen in [18]. But if the refinement mask a is symmetric, the resulting wavelets by using the algorithm in [18] are not necessarily (anti)symmetric. In this section we shall discuss whether or not for dilation factor, we can construct symmetric wavelets from a refinement mask with symmetry that satisfies (1.5). The following two Lemmas will be needed later. Lemma.1. Let a be a finitely supported sequence on Z and φ be the normalized solution of (1.1) with the refinement mask a and dilation factor d 2. Then φ(x) = φ( c d 1 x) for some c R if and only if its associated refinement mask a satisfies a(k) = a(c k) k Z. Proof. Note that φ(x) = φ( c d 1 x) is equivalent to (.3) φ(ξ) = e i c d 1 ξ φ(ξ). From the fact that φ(ξ) = (ã(ξ/d j=1 j )/d ), we see that (.3) is equivalent to ã(ξ) = e icξ ã(ξ), which can be rewritten as a(k) = a(c k) k Z. Therefore, we complete the proof. Note that a(k) = a(c k) k Z is equivalent to saying that (.) ã l (ξ) = ã c l (ξ) l Z. Lemma.2. Let φ be the normalized solution of (1.1) with a refinement mask a l 0 (Z) and dilation factor d 2. Suppose φ(x) = φ( c d 1 x) for some c R. Let ψ(x) = b(k)φ(dx k), where b l 0 (Z). k Z Then ψ(x) = ψ(c 1 x), respectively ψ(x) = ψ(c 1 x) if and only if b(k) = b(dc 1 c/(d 1) k) k Z, respectively b(k) = b(dc 1 c/(d 1) k) k Z. Proof. The proof is similar to that of Lemma.1. In the rest of this section, without mention, we always assume that the dilation factor is d =. In the following, we first consider how to construct wavelets with symmetry from an orthonormal scaling function whose refinement mask a satisfies a(k) = a(1 k) k Z.

1 SYMMETRIC ORTHONORMAL SCALING FUNCTIONS AND WAVELETS Theorem.3. Let φ be an orthonormal scaling function and satisfies (1.1) with the refinement mask a. Suppose that φ(x) = φ( 1 3 x), x R. Let m 1 (e iξ ) = ã 2 (e iξ ) ã 0 (e iξ )e i2ξ + ã 2 (e iξ )e iξ ã 0 (e iξ )e iξ, m 2 (e iξ ) = ã 0 (e iξ ) + ã 2 (e iξ )e i2ξ ã 0 (e iξ )e iξ ã 2 (e iξ )e iξ, m 3 (e iξ ) = ã 2 (e iξ ) ã 0 (e iξ )e i2ξ ã 2 (e iξ )e iξ + ã 0 (e iξ )e iξ. Define ψ i (ξ) := 1 mi (e iξ/ ) φ(ξ/), i = 1, 2, 3. Then ψ 1, ψ 2, ψ 3 are orthonormal wavelets derived from φ. Moreover ψ 1 (x) = ψ 1 ( 1 3 x) and ψ i(x) = ψ i ( 1 3 x), i = 2, 3. Proof. By Lemma.1, we have ã 1 (e iξ ) = ã 0 (e iξ ) and ã 1 (e iξ ) = ã 2 (e iξ ). Hence from (1.5), we have ã 0 (e iξ ) 2 + ã 2 (e iξ ) 2 = 2. Now it is easy to check that the following matrix (.5) ã 0 (e iξ ) ã 2 (e iξ ) ã 0 (e iξ ) ã 2 (e iξ ) 1 ã 2 (e iξ ) ã 0 (e iξ ) ã 2 (e iξ ) ã 0 (e iξ ) 2 ã 0 (e iξ ) ã 2 (e iξ ) ã 0 (e iξ ) ã 2 (e iξ ), ã 2 (e iξ ) ã 0 (e iξ ) ã 2 (e iξ ) ã 0 (e iξ ) is a unitary matrix. In fact, set A(e iξ ) := ( ) ã0 (e iξ ) ã 2 (e iξ ) ã 2 (e iξ. ) ã 0 (e iξ ) ξ R Then A(e iξ )A(e iξ ) T = 2 which implies (.5) is a unitary matrix. Hence ψ i, i = 1, 2, 3 are the resulting orthonormal wavelets by the general theory of wavelets, for example see [2, 7, 15, 16, 18, 19]. It follows directly from Lemma.2 that the wavelet ψ 1 is symmetric about 1/6, and the wavelets ψ 2 and ψ 3 are antisymmetric about 1/6. Now we consider the first type of symmetry, that is, the orthonormal scaling function φ satisfies φ(x) = φ( x). It is quite natural to guess that we can obtain the corresponding wavelets which are all (anti)symmetric about the origin. In the following we shall prove that, on the contrary, this is impossible. Since in applications, we generally require that φ be a real function. Hence in the following, we assume that our orthonormal scaling functions and their associated wavelets are real functions. Theorem.. Let φ be a real orthonormal scaling function with refinement mask a. Suppose φ(x) = φ( x). Then there exist no wavelets ψ 1, ψ 2, ψ 3 derived from φ, which are compactly supported real functions, such that each of them is either symmetric or antisymmetric about the origin. Proof. Since φ is a real function, we have that a(k) R, k Z. From Lemma.1, we see (.6) ã 0 (e iξ ) = ã 0 (e iξ ), ã 2 (e iξ ) = ã 2 (e iξ ), and ã 1 (e iξ ) = ã 1 (e iξ ).

SYMMETRIC ORTHONORMAL SCALING FUNCTIONS AND WAVELETS 15 Suppose that ψ i, i = 1, 2, 3 are obtained from some sequences p, q, r l 0 (Z) respectively. Then ψ 1 (ξ) = 1 p(e iξ/ ) φ(ξ/), ψ2 (ξ) = 1 q(e iξ/ ) φ(ξ/), and ψ 3 (ξ) = 1 r(e iξ/ ) φ(ξ/). Moreover the following matrix (.7) M(e iξ ) := 1 2 ã 0 (e iξ ) p 0 (e iξ ) q 0 (e iξ ) r 0 (e iξ ) ã 2 (e iξ ) p 2 (e iξ ) q 2 (e iξ ) r 2 (e iξ ) ã 1 (e iξ ) p 1 (e iξ ) q 1 (e iξ ) r 1 (e iξ ) ã 1 (e iξ ) p 1 (e iξ ) q 1 (e iξ ) r 1 (e iξ ), ξ R is a unitary matrix. We first prove that among ψ 1, ψ 2, ψ 3, there is at most one wavelet which can be antisymmetric about the origin. Suppose not, without loss of generality, we assume that ψ 1 and ψ 2 are antisymmetric about the origin. From Lemma.2, we see that (.8) p 0 (e iξ ) = p 0 (e iξ ), p 2 (e iξ ) = p 2 (e iξ ) and p 1 (e iξ ) = p 1 (e iξ ) q 0 (e iξ ) = q 0 (e iξ ), q 2 (e iξ ) = q 2 (e iξ ) and q 1 (e iξ ) = q 1 (e iξ ). Consider the following matrix: M 1 (z) := 1 2 ã 0 (z) p 0 (z) q 0 (z) r 0 (z) ã 2 (z) p 2 (z) q 2 (z) r 2 (z) ã 1 (z) + ã 1 (z) p 1 (z) + p 1 (z) q 1 (z) + q 1 (z) r 1 (z) + r 1 (z) ã 1 (z) ã 1 (z) p 1 (z) p 1 (z) q 1 (z) q 1 (z) r 1 (z) r 1 (z) where z = e iξ, ξ R. Since M(e iξ ) is a unitary matrix, then for any ξ R, det M 1 (e iξ ) = 2 det M(e iξ ) 0. On the other hand, each component of p, q, r is a real number since ψ i, i = 1, 2, 3 are real functions. Note that for any sequence b l 0 (Z), we always have b 2 (1) = b 2 (1). Thus (.8) implies that p 0 (1) = 0, p 2 (1) = 0, and p 1 (1) + p 1 (1) = 0 q 0 (1) = 0, q 2 (1) = 0, and q 1 (1) q 1 (1) = 0. Hence det M 1 (1) = 0. A contradiction to det M 1 (e iξ ) 0 for any ξ R. Therefore there is at most one wavelet which can be antisymmetric about the origin. Next, we prove that among ψ 1, ψ 2, ψ 3, there is at most one wavelet which can be symmetric about the origin. Suppose not, we assume that ψ 1 and ψ 2 are symmetric about the origin. From Lemma.2, we see that (.9) p 0 (e iξ ) = p 0 (e iξ ), p 2 (e iξ ) = p 2 (e iξ ) and p 1 (e iξ ) = p 1 (e iξ ) q 0 (e iξ ) = q 0 (e iξ ), q 2 (e iξ ) = q 2 (e iξ ) and q 1 (e iξ ) = q 1 (e iξ ). Note that for any sequence b l 0 (Z), we always have b 2 (e iξ ) = e iξ b 2 (e iξ ). Hence (.6) and (.9) yield that ã 1 ( 1) ã 1 ( 1) = 0, p 1 ( 1) p 1 ( 1) = 0 and q 1 ( 1) q 1 ( 1) = 0 ã 2 ( 1) = 0, p 2 ( 1) = 0, and q 2 ( 1) = 0,

16 SYMMETRIC ORTHONORMAL SCALING FUNCTIONS AND WAVELETS which implies det M 1 ( 1) = 0. A contradiction to det M 1 (e iξ ) 0 for any ξ R. Thus there is at most one wavelet which can be symmetric about the origin. Therefore it follows that there exist no wavelets ψ 1, ψ 2, ψ 3 which are all (anti)symmetric about the origin. We are done. In a similar way, we can prove that if the dilation factor d = 2k and φ is an orthonormal scaling function with dilation d such that φ is symmetric about the origin, then we can obtain at most k 1 wavelets which are symmetric about the origin and at most k 1 wavelets which are antisymmetric about the origin. If other symmetry is allowed, for Example 3.1, 3.2 and 3.3, we can construct their associated wavelets ψ 1, ψ 2, ψ 3 such that ψ 1 is symmetric about the origin, ψ 2 is antisymmetric about the origin, and ψ 3 is symmetric about 1/2. Example.1. Let φ be the orthonormal scaling function given in Example 3.1. By direct computation, it is easy to verify that the following matrix 1 2 2 6 + 2+ 6 e iξ 1 2 + 1 2 e iξ 2+ 6 + 2 6 e iξ 1 2 3 3 2 12 + 2 3+3 2 12 e iξ 3 6 + 3 6 e iξ 2 3+3 2 12 + 2 3 3 2 12 e iξ 3 2 3 3 2 6 2 3+3 2 6 e iξ 3 3 3 3 e iξ 2 3+3 2 6 2 3 3 2 6 3 6 e iξ 0 2 6 3 6 3 0 is a unitary matrix. Let sequences p, q, r be defined by their symbols as p(z) = 2 3 3 2 ( z 3 + z 3) 3 ( + z 2 + z 2) + 2 3 + 3 2 ( z 1 + z 1) 3 12 6 12 q(z) = 2 3 3 2 ( z 3 z 3) 3 ( + z 2 z 2) + 2 3 + 3 2 ( z 1 z 1) 6 3 6 6 r(z) = 3 z1 + 2 6 6 3 z2 3 z3. Then the three wavelets are given by ψ 1 (ξ) := 1 p(e iξ/ ) φ( ξ ), ψ2 (ξ) := 1 q(e iξ/ ) φ( ξ ), and ψ 3 (ξ) := 1 r(e iξ/ ) φ( ξ ). By Lemma.2, it is clear that ψ 1 is symmetric about the origin, ψ 2 is antisymmetric about the origin and ψ 3 is symmetric about 1 2. Example.2. Let φ be the orthonormal scaling function given in Example 3.2. Let sequences p, q, r be defined by their symbols as T p(z) = 1 23 23 2 (z 5 + z 5 ) + 23 2 + 23 288 1 23 6 (z 2 + z 2 ) + 1 23 + 55 2 1 23 7 2 2 + 2 23 q(z) = (z 5 z 5 ) 36 36 + 2 2 3 (z 2 z 2 ) 2 2 3 (z 1 z 1 ), 6 1 r(z) = (z 1 + z 5 ) 1 + 6 (1 + z ) + 18 9 (z + z ) 23 2 + 10 23 (z 3 + z 3 ) 288 (z 1 + z) + 7 23 55 2, 72 (z z ) + 5 2 + 23 (z 3 z 3 ) 36 6 + 10 (z 1 + z 3 ) + 2 18 3 z2.

SYMMETRIC ORTHONORMAL SCALING FUNCTIONS AND WAVELETS 17 Then the three wavelets are given by ψ 1 (ξ) := 1 p(e iξ/ ) φ( ξ ), ψ2 (ξ) := 1 q(e iξ/ ) φ( ξ ), and ψ 3 (ξ) := 1 r(e iξ/ ) φ( ξ ). By Lemma.2, it is clear that ψ 1 is symmetric about the origin, ψ 2 is antisymmetric about the origin and ψ 3 is symmetric about 1 2. The construction of symmetric wavelets from Example 3.3 is presented in the appendix. Finally the graphs of all the above examples of orthonormal scaling functions and their associated wavelets with symmetry are presented in the appendix. Acknowledgement: The author is grateful to Professor Rong-Qing Jia for discussion on the material of this paper and for his advice on presentation of this paper. References [1] J. Boman, On a problem concerning moduli of smoothness, in Colloquia Mathematica Soc, János Bolyai, Fourier Analysis and Approximation, Proceedings of Conference in Budapest, Hungary, 1976, 273 179. [2] C. K. Chui, An Introduction to Wavelets, Academic Press, Boston, 1992. [3] A. S. Cavaretta, W. Dahmen, and C. A. Micchelli, Stationary Subdivision, Memoirs of Amer. Math. Soc. 93 (1991). [] C. K. Chui and J. A. Lian, Construction of Compactly Supported Symmetric and Antisymmetric Orthonormal Wavelets with Scale = 3., Appl. Comp. Harmonic Anal. 2 (1995), 21 51. [5] A. Cohen and I. Daubechies, Orthonormal bases of compactly supported wavelets: III. Better frequency resolution, SIAM J. Math. Anal. 2 (1993), 520 527. [6] I. Daubechies, Orthonormal bases of compactly supported wavelets, Comm. Pure Appl. Math. 1 (1988), 909 996. [7] I. Daubechies, Ten Lectures in Wavelets, SIAM, Philadelphia, 1992. [8] R. A. DeVore and G. G. Lorentz, Constructive Approximation, Springer-Varlag, Berlin, 1993. [9] I. Daubechies and J. C. Lagarias, Two-scale difference equations:ii. Local regularity, infinite products of matrices and fractals, SIAM J. Math. Anal. 23 (1992), 1031 1079. [10] Z. Ditzian, Moduli of smoothness using discrete data, J. Approx. Theory 9 (1987), 115 129. [11] T. N. T. Goodman, C. A. Micchelli and J. D. Ward, in Recent Advances in Wavelet Analysis, L.L. Schumaker and G. Webb (eds.), Academic Press, 199, pp. 335 360. [12] B. Han and R. Q. Jia, Multivariate refinement equations and subdivision schemes, manuscript. [13] R.Q. Jia, Subdivision schemes in L p spaces, Advances in Comp. Math. 3 (1995), 309 31. [1] R.Q. Jia, Characterization of smoothness of multivariate refinable functions in Sobolev spaces,, Trans. Amer. Math. Soc. (to appear). [15] R. Q. Jia and C. A. Micchelli, Using the refinement equation for the construction of prewavelets V: extensibility of trigonometric polynomials, Computing 8 (1992), 61 72. [16] R. Q. Jia and Z. W. Shen, Multiresolution and wavelets, Proc. Edinburgh Math. Soc. 37 (199), 271-300. [17] R. Q. Jia and J. Z. Wang, Stability and linear independence associated with wavelet decomposition, Proc. Amer. Math. Soc. 117 (1993), 1115 112. [18] W. Lawton, S. L. Lee and Z. W. Shen, An algorithm for matrix extension and wavelet construction, Math. Comp. 21 (1996), 723 737. [19] S. D. Riemenschneider and Z. W. Shen, Wavelets and pre-wavelets in low dimensions, J. Approx. Theory 71 (1992), 18 38. [20] P. Steffen, P. N. Heller, R. A. Gopinath and C. S. Burrus, Theory of regular M-band wavelet bases, IEEE Transaction on Signal Processing 1 (1993), 397 3510. [21] A. K. Soman, P. P. Vaidyanathan, and T. Q. Nguyen, Linear phase paraunitary filter banks: theory, factorizations, and applications,, IEEE Trans. Signal Processing 1 (1993), 380 396.

18 SYMMETRIC ORTHONORMAL SCALING FUNCTIONS AND WAVELETS [22] L. F. Villemoes, Wavelet analysis of refinement equations, SIAM J. Math. Anal. 25 (199), 133-160. [23] G. V. Wellend and M. Lundberg, Construction of compact p-wavelets, Constr. Approx. 9 (1993), 37 370. [2] D. X. Zhou, Stability of refinable functions, multiresolution analysis and Haar bases, SIAM J. Math. Anal. 27 (1996), 891 90.

SYMMETRIC ORTHONORMAL SCALING FUNCTIONS AND WAVELETS 19 Appendix In this appendix, we give another example of an orthonormal scaling function φ such that φ is symmetric about 1/6. Example. Let l =. Then by Theorem 3.1 P (z) = 16+320 z+388 z 2 +2756 z 3. We are looking for a mask given by with a polynomial ã(e iξ ) = 1 1 + e iξ + e i2ξ + e i3ξ (1 + e iξ ) p(sin 2 ξ 512 2 ), p(z) = + 2 z + 515/2 z 2 + 825/z 3 + c 1 z + c 2 z 5 + c 3 z 6 + c z 7, for some parameters c 1, c 2, c 3, c R. To avoid complicated equations, we just present the numerical results here. Set c = 2600. By solving (3.), we find the following solution: c 1 = 953.913983167, c 2 = 31571.83383567 and c 3 = 7597.7033816055. The refinement mask a is given by its symbol as following: where ã(z) = z 6 (1 + z 1 + z 2 + z 3 ) q(z) q(z) = 0.00293258522992 (z 7 + z 8 ) + 0.01526777180859782 (z 6 + z 6 ) 0.0363685872066781(z 5 + z 6 ) + 0.0821621991858122 (z + z 5 ) 0.0123612317507572 (z 3 + z ) 0.01668829920195519(z 2 + z 3 ) 0.0086982565532505(z 1 + z 2 ) + 0.0567808667865175 (1 + z). Then a satisfies (1.5) and the normalized solution φ of (1.1) with the above refinement mask a lies in L 2 (R) with ν 2 (φ) 2.068 by Theorem 2.1. By (2.10), we have ν (φ) log (0.116897) > 1.583. Therefore φ is a C 1 orthonormal scaling function which is symmetric about 1/6. See the last page of this paper for the graphs of this orthonormal scaling function and its associated wavelets. In fact, in the above example, c can be an arbitrary number. c 3 is one of the real roots of a polynomial of degree 8. c 1 and c 2 can be expressed as rational functions of c 3 and c. From our experience, with the help of Theorem 3.1, it seems quite possible to find a C 2 symmetric orthonormal scaling function. But the computation and presentation of such refinement masks will be quite messy. Hence in this paper we don t go further to search for a C 2 orthonormal scaling function.

20 SYMMETRIC ORTHONORMAL SCALING FUNCTIONS AND WAVELETS Finally, the construction of symmetric wavelets from Example 3.3 is presented in the following. Let φ be the orthonormal scaling function given in Example 3.3. Let sequences p, q, r be defined by their symbols as p(z) = 0.5862062 0.07163869 (z 1 + z) + 0.9695267 (z 2 + z 2 ) 0.76735 (z 3 + z 3 ) + 0.181115 (z + z ) 0.231055 (z 5 + z 5 ) 0.213970 (z 6 + z 6 ) + 0.180303 (z 7 + z 7 ) + 0.051173 (z 8 + z 8 ) 0.00598797 (z 9 + z 9 ) 0.011671 (z 10 + z 10 ) + 0.01138051 (z 11 + z 11 ) q(z) =1.068180 (z 1 z 1 ) 0.21953 (z 2 z 2 ) + 0.231199 (z 3 z 3 ) 0.702186 (z z ) + 0.3212620 (z 5 z 5 ) + 0.320030 (z 6 z 6 ) 0.2331660 (z 7 z 7 ) 0.05855950 (z 8 z 8 ) + 0.007916712 (z 9 z 9 ) + 0.01535182 (z 10 z 10 ) 0.018887 (z 11 z 11 ) r(z) = 0.2879602 z 2 + 0.8739313 (z 1 + z 3 ) 0.07717986 (1 + z ) 0.512722 (z 1 + z 5 ) 0.757800 (z 2 + z 6 ) + 0.5337008 (z 3 + z 7 ) 0.2016173 (z + z 8 ) + 0.177715 (z 5 + z 9 ) + 0.217600 (z 6 + z 10 ) 0.1211 (z 7 + z 11 ) Then the three wavelets are given by ψ 1 (ξ) := 1 p(e iξ/ ) φ( ξ ), ψ2 (ξ) := 1 q(e iξ/ ) φ( ξ ), and ψ 3 (ξ) := 1 r(e iξ/ ) φ( ξ ). By Lemma.2, it is clear that ψ 1 is symmetric about the origin, ψ 2 is antisymmetric about the origin and ψ 3 is symmetric about 1 2. We should mention that in this paper we do not provide an explicit way of constructing symmetric wavelets from an orthonormal scaling function which is symmetric about 0. It would be nice if a constructive way(like the algorithm given in [18]) of deriving symmetric wavelets from such scaling functions can be found. If an orthonormal scaling function is symmetric about 1/6, the reader is referred to Theorem.3 in this paper for the construction of symmetric wavelets.