INDUSTRIAL MATHEMATICS INSTITUTE 2000:15 Symmetric framelets A. Petukhov IMI Department of Mathematics Preprint Series University of South Carolina

Transcription

1 INDUSTRIAL MATHEMATICS INSTITUTE 2000:5 Symmetric framelets A. Petukhov IMI Preprint Series Department of Mathematics University of South Carolina

2 Symmetric framelets Alexander Petukhov June 27, 2000 Abstract We study tight wavelet frames associated with symmetric compactly supported re- nable functions, which are obtained with the unitary extension principle. We give a criterion for existence of two symmetric or antisymmetric compactly supported framelets. All renable masks of length up to 6, satisfying this criterion, are found. AMS subject classication: 42C5 Key Words: frames, tight frames, multiresolution analysis, wavelets Introduction The main goal of our paper is to present a criterion for existance of two symmetric or antisymmetric framelets generated by a symmetric renable function. We consider only functions of one variable in the space L 2 (R) with the inner product hf gi = Z ; f(x)g(x) dx: As usual we denote by ^f(!) Fourier transform of the function f(x) 2 L 2 (R), ^f(!) = Z ; f(x)e ;ix! dx: Suppose a real-valued function ' 2 L 2 (R) satises the following conditions: (a) ^'(2!) =m 0 (!)^'(!), where m 0 is an essentially bounded 2-periodic function (b) lim!!0 ^'(!) =(2) ;=2 then the function ' is called renable or scaling, m 0 is called asymbol of ', and the relation in item (a) is called a renement equation. This research was partially supported under Grants NSF KDI 578AO45, DoD-N , ONR/ARO-DEPSCoR-DAAG , and by Russian Foundation for Basic Research under Grant #

3 Every renable function generates multiresolution analysis (MRA) of the space L 2 (R), i.e., a nested sequence V ; V 0 V V j ::: of closed linear subspaces of L 2 (R) suchthat (a) \ j2zv j =? (b) [ j2zv j = L 2 (R) (c) f(x) 2 V j, f(2x) 2 V j+. To obtain the MRA we just have to take asabove V j the closure of the linear span of the functions f'(2 j x ; n)g n2z. Fulllment of item (a) and (b) for the obtained spaces V j was proved in []. Property (c) is evident. The most popular approach to the design of orthogonal and bi-orthogonal wavelets is based on construction of MRA of the space L 2 (R), generated with a given renable function. S.Mallat [6] showed that if the system f'(x;n)g n2zconstitutes a Riesz basis of the space V 0, then there exists a renable function 2 V 0 with a symbol m such that the functions f(x; n)g n2zform an orthonormal basis of V 0.Ifwe denote by W j the orthogonal complement of the space V j in the space V j+, then the function (which is called a wavelet), dened by the relation ^(2!) :=m (!)^(!) where m (!) =e i! m (! + ), generates orthonormal basis f (x ; n)g n2zof the space W 0. Thus, the system 2 k=2 (2 k x ; n) n k2z () constitutes an orthonormal basis of the space L 2 (R). We see that if wehave a renable function, generating a Riesz basis, then wehave explicit formulae for the wavelets, associated with this functions. It gives a simple method for constructing wavelets. Generally speaking, any orthonormal basis of L 2 (R) of the form () is called a wavelet system. However, wavelet construction based on a multiresolution structure has the advantage from the point of view eectiveness of computational algorithms, because it leads to the pyramidal scheme of wavelet decomposition and reconstruction (sf. [4]). It is well-known that the problem of nding orthonormal wavelet bases, generated by a scaling function, can be reduced to solving the matrix equation where M(!) = M(!)M (!) =I (2) m0 (!) m (!) m 0 (! + ) m (! + ) m 0 (!) m (!) are essentially bounded functions, and m 0 (;!) =m 0 (!), i.e., the Fourier series of these functions have real coecients. It is known (see [4]) that for any scaling function '(x) and associated wavelet (x), generating an orthogonal wavelet basis, the corresponding symbols m 0 (!) m (!) satisfy (2). Any renable function ', whose symbol m 0 is solution to (2), generates a tight frame (see [5] for the case when m 0 is polynomial, and [2] for the general case). 2

4 Let us recall that a frame in a Hilbert space H is a family of its elements ff k g k2zsuch that for any f 2H Akfk 2 X k2z jhf f kij 2 Bkfk 2 where optimal A and B are called frame constants. If A = B, the frame is called a tight frame. In the case when a tight frame has unit frame constants (for example, if it is an orthonormal basis) for any function f 2 L 2 (R) the expansion f = X n2z hf n fif n (3) is valid. The frame f l j kg j k2z n l=, where l j k (x) =2j=2 l (2 j x ; k), generated by translates and dilations of nite number of functions, is called an ane or wavelet frame. In the case when the symbol m 0 of a renable function ' does not satisfy the equation jm 0 (!)j 2 + jm 0 (! + )j 2 = we cannot construct an orthonormal bases of V of the form f'(x ; k) (x ; k)g. However, we can hope that there exists a collection of several framelets 2 ::: n 2 V, satisfying the following conditions: ) the functions f l j kg j k2z n l= form a tight frame of the space L2 (R) 2) for algorithms of decomposition and reconstruction the recurrent formulae h' j k fi X = c j l = k2z c j+ kh k;2l h g fi X j k = dq j l = k2z c j+ kg q k;2l q n (4) and where g q k c j+ l = X k2z c j kh l;k + are coecients of the expansions nx X q= k2z dq j k gq l;k (5) m 0 (!) =2 ;=2 X k2z h ke ;ik! m q (!) =2 ;=2 X k2z gq k e;ik! take place. Let ' be a renable function with a symbol m 0, ^k (!) =m k (!=2) ^'(!=2) 2 V, where each symbol m k is a 2-periodic and essentially bounded function for k = 2 ::: n. It is well-known that for constructing tight frames with property 2) the matrix M(!) = m0 (!) m (!) ::: m n (!) m 0 (! + ) m (! + ) ::: m n (! + ) plays an important role. It is easy to see that the equality is equivalent to (4) and (5). In our recent paper the following theorems were proved : M(!)M (!) =I (6) 3

5 Theorem A ([7]). If (6) holds, then the functions f j g k j= generate a tight frame of L2 (R). Remark. For n = this theorem was proved in [2]. For an arbitrary n it was proved in [8] under some additional decay assumption for ^'. In [3] this and the next theorem was proved for a special case when m 0 is a trigonometric polynomial. In [8] Theorem A was called the unitary extension principle. Theorem B ([7]). Equation (6) has a solution if and only if jm 0 (!)j 2 + jm 0 (! + )j 2 (a. e.): (7) Thus, the problem of constructing tight frames, generated by a renable function, can be reduced to nding m k, satisfying (6). It is clear that relation (6) can be re-written in the form where M (!) :=M (!)M (!) = M (!) = ;jm0 (!)j 2 ;m 0 (!)m 0 (! + ) (8) ;m 0 (!)m 0 (! + ) ;jm 0 (! + )j 2 m (!) m 2 (!) ::: m n (!) m (! + ) m 2 (! + ) ::: m n (! + ) Let us introduce the diagonal matrix (!) with eigenvalues of the matrix M (!) onthe diagonal and the matrix P (!) whose columns are the corresponding eigenvectors. Then 0 (!) = : For those! for which 0 ;jm 0 (!)j 2 ;jm 0 (! + )j 2 B(!)B(!) :=jm 0 (!)j 2 + jm 0 (! + )j 2 6=0 : we dene the matrix P (!) intheform 0 e i! m 0 (! + ) m 0 (!) B(!) B(!) P (!) = e ; i! m 0 (!) m0 (! + ) B(!) B(!) C A where B(!) is a -periodic function. At points, where B(!) = 0, we dene the matrix P (!) as the identity matrix. Thus, we have Now we can discribe all possible solutions to (6). M (!) =P (!)(!)P (!): (9) 4

6 Theorem C ([7]). Let a 2-periodic function m 0 (!) satisfy (7). Then there exists a pair of 2-periodic measurable functions m, m 2 which satisfy (6) for n =2.Any solution of (6) can be represented in the form of the rst row of the matrix M (!) =P (!)D(!)Q(!) (0) where D(!) is a diagonal matrix, D(!)D(!) =(!), Q(w) is an arbitrary unitary (a.e.) matrix with -periodic measurable components. Remark. To describe all possible solution to (6) for an arbitrary n we have to take an arbitrary n n unitary matrix Q with -periodic elements and a 2 n matrix D 0 which is extension of the matrix D(!) by mean of lling all new columns with zeros. Theorem D ([7]). Let a trigonometric polynomial m 0 (!) of degree n satisfy (7). Then there exists a pair of trigonometric polynomials m, m 2 of the degree at most n satisfying (6). Remark. In [3] this theorem was proved without the guaranteed degree of the polynomials m, m 2. Let now and in what follows a function m 0 (!) satisfy the condition m 0 (!) =e il! m 0 (!) l 2 Z () then the corresponding renable function '(x) are even after appropriate whole integer (l is even) or half integer (l is odd) shift of an argument. We shall call such funcions and their symbols symmetric. From here on without loss of generality we suppose l =0. We call antisymmetric those functions which after an appropriate whole or half shift are odd. We are interested in framelets k,which are either symmetric or antisymmetric. Symbols of antisymmetric framelets satisfy the relation m k (!) =;e il! m k (!) l 2 Z: (2) In applications such systems are of great practical importance. The numerical algorithms for them have low computational complexity. And the problem of signal edges is solved very easily by means of even or odd extension. The natural question about possibility to choose (anti)symmetric framelets for a given symmetric renable function ' arises. In the recent paper [3] positive answer this question was given. Theorem E ([3]). For any renable function ' with a polynomial symbol m 0 (anti)symmetric functions m, m 2, m 3,providing a solution to (6). there are3 However, many examples, when system (6) can be solved with 2 (anti)symmetric framelets, are known. For instance this is possible for the cases of piecewise-linear ([8]) and piecewisequadratic ([3]) B-splines. Our goal is to present a criterion for existence of 2 (anti)symmetric framelets. 5

7 2 Main result First, we introduce necessary denitions and notation. The degree of a trigonometric polynomial P k j=l a ke ijx,wherea l 6=0anda k 6= 0, is dened to be k ; l. We denote by L a set of all Laurent polynomials with real coecients, and by L n aset of Laurent polynomials with real coecients of degree at most n, i.e., L n := ( X k ) a j z j l k 2 Z aj 2 R 0 k ; l n j=l We denote by deg(f) the degree of the Laurent polynomial f. In what follows Laurent polynomials h k (z) are specied by the z-transform of the symbols m k (!), i.e., h k (e i! ):=m k (!). Theorem. Let h 0 (z) is a symmetric Laurent polynomial of degree n, satisfying (7). Then two (anti)symmetric solutions to (6) exist if and only if all roots of the Laurent polynomial h(z) :=; h 0 (z)h 0 (=z) ; h 0 (;z)h 0 (;=z) (3) have even multiplicity. Moreover, in this case polynomials m, m 2 of degree at most n can be chosen. This theorem has a simple consequence for B-spline multiresolutions. Let us recall that B-splines is dened to be : B 0 (x) := x 2 [0 ] 0 x 62 [0 ] B n+ := B n B 0 : Corollary. For the renable functions B n two (anty-)symmetric solutions exist for n = and do not exist for n = ::: 50. This corollary was obtained by direct computation of roots of the corresponding polynomials with Matlab. Remark. For n =0wehave Haar'swavelets. The solution for n =was found by A.Ron and Z.Shen [8] and the solution for n =2was found by C.Chui and W.He [3]. We consider the case n = 6 in Section 4. Asymbol h 0 (z) ==2+ P N n= a n(z ;2n+ + z 2n; ), P N n= a n ==4 iscalledinterpolatory. Corollary 2. An interpolatory symbol h 0 admit (anti)symmetric solutions to (6) if and only if h 0 (z) =(z ;2N +2+z 2N; )=4. Remark It is clear that only for N =wehave a real interpolatory renable function, satisfying the condition '(0) =, '(n) =0,n 2 Z, n 6= 0. 6

8 3 Proof of main result 3. Necessity We assume that the polynomials m 0 m m 2 provide a solution to (6), m 0 is symmetric and m, m 2 are either symmetric or antisymmetric. Let us introduce a matrix N(z) which is z-transform of the matrix M(!). Then equation (6) can be rewritten in the form N(z)N T (=z) =I: Obviuously we can extend the rectangular matrix up to a 3 3 para-unitary matrix N 0 (z) = h 0(z) h (z) h 2 (z) h 0 (;z) h (;z) h 2 (;z) a 0 (z 2 ) a (z 2 ) a 2 (z 2 ) A N 0 (z)n 0T (=z) =I: The new row can be obtained as a vector product of the known rows. It guarantees that a i (z) 2L. We note that any other possible polynomial choices of the last row dier from this one by the factor z k, k 2 Z. We used the factor z so that elements of the last row depend only on even powers. Thus, we have a 0 (z 2 )=z(h (=z)h 2 (;=z) ; h 2 (=z)h (;=z)): (4) We denote by S the set of all (anti)symmetric polynomials. We shall use the subscripts e, o, w, h to denote subsets of S, consisting of respectively even, odd, whole and half symmetric polynomials of S and integer superscripts to denote a value k in the relation f(=z) =z ;k f(z). For instance, Sw o 4 is the set of polynomial which are odd at a whole point and for any f 2 S 4 w o we have f(=z) =z ;4 f(z). Of course subscripts w and h is compatible correspondingly only with even and odd superscripts. Let us prove thata 0 2 S. This is enough to prove the necessity. Indeed, since we have the identity h 0 (z)h 0 (=z)+h 0 (;z)h 0 (;=z)+a 0 (z 2 )a 0 (=z 2 ) then the symmetry of a 0 implies that all non-zero roots of the polynomial ; h 0 (z)h 0 (=z) ; h 0 (;z)h 0 (;=z) =a 0 (z 2 )a 0 (=z 2 )=z 2k a 2 0 (z2 ) (5) have even multiplicity. First, we note that polynomials from S satisfy the followng obvious properties: (a) f g 2 S ) fg 2 S (a2) f g 2 S e ) fg 2 S e (a3) f g 2 S o ) fg 2 S e (a4) f 2 S o g2 S e ) fg 2 S o (a5) f g 2 S w ) fg 2 S w (a6) f g 2 S h ) fg 2 S w (a7) f 2 S w g2 S h ) fg 2 S h (b) f(x) 2 S w, f(;x) f(=x) 2 S w (b2) f(x) 2 S o, f(=x) 2 S o 7

9 (b3) f(x) 2 S h e, f(;x) 2 S h o (b4) f(x) 2 S w e (f(x) 2 S w o ), f(;x) 2 S h e (f(x) 2 S w o ) (c) f 2 S w ) f(x) f(;x) 2 S. Now we consider the case h 0 2 S w.taking into account (a) (c), and (4), we obtain that the symmetry of a 0 may be violated only if h 2 (z)h (;z) 2 S h. In this case h and h 2 belong to dierent classes S w and S h.we suppose that h 2 S w, h 2 2 S h. Because of the orthogonality of the st and the 2nd rows of the matrix N 0 (z), we have h 0 (z)h 0 (;=z)+h (z)h (;=z)+h 2 (z)h 2 (;=z) 0: (6) We see that, according to properties (a2), (b), (c), the st and the 2nd summands in (6) belong to S e, whereas, by (a4) and (b2), the 3rd one belongs to S o. It means h 2 (z) 0. Hence, we have a 0 a 0, a 2 (z) =z k. Thus, we come to the case of one framelet. Of course this is impossible for h 0 2 S w (sf. [4], Chapter 8), though (4) gives us a permissible function a 0 (z) 0. We note that by the same reasons h h 2 2 S w. Actually, ifh h 2 2 S h, then the st term in (6) is even whereas the 2nd the 3rd ones are odd. It means that h 0 (z)h 0 (=z) 0. Moreover, it is easy to prove thath 2 S e, h 2 2 S o or vise versa. The point is that for the functions h h 2 with the same evenness, by (4), we get an even polynomial a 0.Thus, the polynomial a 0 (z 2 )a 0 (=z 2 ) has positive coetients of the lowest and the highest powers, whereas the corresponding coecients of the left-hand part of (5) are negative. h 2 Now ifwe suppose h 0 2 S h, in analogous way we obtain that the both functions h and belong to S h. Indeed, since the st term in (6) is odd then two other terms are odd. Hence, h h 2 2 S h. Easily to see that the case h h 2 2 S h e is not valid. It contradicts to the equality h 0 (z)h 0 (=z)+h (z)h (=z)+h 2 (z)h 2 (=z) = due to the positivity of coecients of the lowest and the highest powers of all terms. From(4) and (b3) we have a 0 2 S o for h h 2 2 S o and a 0 2 S e for h 2 S e, h 2 2 S o. Thus, we have 3 permissible cases: ) h 0 2 S w, h 2 S w e, h 2 2 S w o 2) h 0 2 S h, h 2 S h o, h 2 2 S h o 2) h 0 2 S h, h 2 S h e, h 2 2 S h o. Examples for all of them will be given in Section Suciency We assume that m 0 (!) is a trigonometric polynomial, then B(!) =jm 0 (!)j 2 + jm 0 (! + )j 2 and A(!) =;jm 0 (!)j 2 ;jm 0 (! + )j 2 are also non-negative trigonometric polynomials So according to Riesz lemma, we can take -periodic polynimials A(!) and B(!) such that ja(!)j 2 = A(!), jb(!)j 2 = B(!). Sinceh(z) is a symmetric polynomial of even power and its roots have even multiplicity, we can take an (anti)symmetric polynomial A(!). The choice of a 0 is not unique, in what follows we suppose for deniteness that either a 0 2 S 0 w or a 0 2 S h. We assume that the choice of matrices P (!) andd(!) in Theorem C corresponds the last assumptions. Our further proof repeats for the most part reasoning from [7]. 8

10 In fact, we cannot control the choice of the matrices P (!) and D(!) in (0). So we need to choose a unitary rational -periodic matrix Q(!) such that M (!) consists of trigonometric polynomials. Let us apply z-transform to (0). In what follows we consider the Laurent polynomials b(e 2i! )=B(!), a 0 (e 2i! )=A(!). After the same change of variable the matrix P (!) becomes H(z) = 0 z h 0 ; ; z b(=z 2 ) z h 0 z b(=z 2 ) h 0 (z) b(z 2 ) h 0 (;z) b(z 2 ) We put the last representation of the matrix H(z) through procedure of reduction. If polynomials h 0 (z), h 0 (;z), b(z 2 ) are divisible by z ; z 0, that in view of the symmetry of h(z) they are also divisible by z + z 0. Obviously, since h 0 () 6= 0, then z 0 6=. We cancel the fractions of H(z) by(=z 2 ; z 2 0)(z 2 ; z 2 0). After all possible cancellations we obtain the same matrix H 0 (z) =H(z) but its elements are expressed in terms of new functions h 0 0(z) and b 0 (z). It is clear that b 0 (z 2 )b 0 (=z 02 )=h 0 0(z)h 0 0(=z)+h 0 0(;z)h 0 0(;=z) andnumerators of the matrix H 0 (z) do not vanish simultaneously. Indeed, since the determinant ofh(z) is equal to =z, if for some z 0 wehave h 0 (z 0 )=h 0 (;z 0 )=h 0 (=z 0 )=h 0 (;=z 0 ) = 0, then either b(z 2 0)=0orb(=z 2 0) = 0. It means that the reduction of H(z) can be continued. We note that because the coecients of h 0 (z) andb(z) are real, the polynomials h 0 0(z) andb 0 (z) also have real coecients. Moreover, h 0 0(z) is symmetric. After taking z-transform the elements q (z 2 ), q 2 (z 2 ), q 2 (z 2 ), q 22 (z 2 ) of the matrix Q(!) satisfy the relations C A q 22 (z) =q (=z)z N q 2 (z) =;q 2 (=z)z N N 2 Z: Here, without loss of generality, we may suppose N = 0, because any other choice leads to the integer shift of one of the basic framelets. To reduce poles of the matrix H 0 (z) aftermultiplication by Q(!) we suppose that : q (z) = g (z) b 0 (z) q 2(z) = g 2(z) b 0 (=z) where g, g 2 are Laurent polynomials. Let R = fz z 2 ::: zn g be a set of all dierent roots of the polynomial b 0 (z 2 )b 0 (=z 2 ). We denote by k j the multiplicity of the root z j. It is clear that all four roots P z j have the same multiplicity. So the degree of polynomial b 0 (z 2 )b 0 (=z 2 ) is equal to 4 k j =4k, where k is the degree of polynomial b 0. To prove the theorem we need to nd polynomials g, g 2 which satisfy equations z h0 0 ; z g (z 2 )+a 0 (z 2 )h 0 0 (z)g 2(z 2 )=b 0 (z 2 )b 0 (=z 2 )f (z) (7) 9

11 ; z h0 0 ; z h0 0 z h0 0 ; z g 2 + a z 2 0 (z 2 )h 00(z)g z = b 0 (z 2 )b 0 (=z 2 )f 2 2 (z) (8) g (z 2 )+a 0 (z 2 )h 0 z 0(;z)g 2 (z 2 )=b 0 (z 2 )b 0 (=z 2 )f 3 (z) (9) z g 2 + a z 2 0 (z 2 )h 0 (;z)g 0 = b 0 (z 2 )b 0 (=z 2 )f z 2 4 (z): (20) where f f 2 f 3 f 4 2L. Moreover, we need to satisfy the condition of the unitarity ofthe matrix Q(!). Hence, g (z)g (=z)+g 2 (z)g 2 (=z) =b 0 (z)b 0 (=z): (2) Now weleave aside equation (2) and prove the existence of polynomials g g 2 2L k, satisfying (7) { (20). First we choose templates for the polynomials g and g 2. There are 2 dierent cases which depend on type of symmetry of the polynomials h 0, a 0, and on evenness of the number k. They can be classied into three groups with 4 cases in every group. ) h 0 2 S w, a 0 is odd. (a) a 0 2 Sh o, k is odd, g 2 Sh e, g 2 2 S ; h e,deg(g )=deg(g 2 )=k (b) a 0 2 Sw o, 0 k is odd, g 2 Sw o, 0 g 2 2 S ; h e, deg(g )=k ;, deg(g 2 )=k (c) a 0 2 Sh o, k is even, g 2 Sw e 2, g 2 2 Sw o 0,deg(g ) = deg(g 2 )=k (d) a 0 2 Sw o 0, k is even, g 2 Sh o, g 2 2 Sw e 0, deg(g )=deg(g 2 )=k 2) h 0 2 S h, a 0 is odd. (a) a 0 2 Sh o, k is odd, g 2 Sw e, 2 g 2 2 S ; h e, deg(g )=k ;, deg(g 2 )=k (b) a 0 2 Sw o 0, k is odd, g 2 Sh e, g 2 2 S ; h e, deg(g )=deg(g 2 )=k w e,deg(g )=k ;, deg(g 2 )=k (c) a 0 2 Sh o, k is even, g 2 Sh e, g 2 2 S ;2 (d) a 0 2 Sw o, 0 k is even, g 2 Sw e, 2 g 2 2 Sw e, 0 deg(g )=deg(g 2 )=k 3) h 0 2 S h, a 0 is even. (a) a 0 2 Sh e, k is odd, g 2 Sw e, 2 g 2 2 S ; h o, deg(g )=k ;, deg(g 2 )=k h o, deg(g ) = deg(g 2 )=k w o, deg(g )=k;, deg(g 2 )=k (b) a 0 2 Sw e, 0 k is odd, g 2 Sh e, g 2 2 S ; (c) a 0 2 Sh e, k is even, g 2 Sh e, g 2 2 S ;2 (d) a 0 2 Sw e, 0 k is even, g 2 Sw e, 2 g 2 2 Sw o, 0 deg(g )=deg(g 2 )=k For the cases 2d) and 3b) we have k + 2 unknown parameters and for others cases we have k + parameters. Obviously for all cases the left-hand parts of (7) { (20) are (anti)symmetric. First we show that there exist polynomials g and g 2, satisfying equations (7) { (20) at points of the set R. As it usually is in the case of a root ~z of multiplicity k,we ~ require that the left-hand parts of (7) { (20) are divisible by (z ; z 0 ) k. Equations (7) { (20) giveus6k homogeneous linear equations for k+ or k+2 unknown coetients of polynomials g and g 2. We shall prove that at most k of them are linearly independent. The proof of this fact we conduct in 4 steps. Three of these steps are based on the following lemma. 0

12 Lemma. Let a (z), a 2 (z), a 3 (z), a 4 (z), b (z), b 2 (z) c (z), c 2 (z) be Laurent polynomials, ja (z 0 )j 2 + ja 2 (z 0 )j 2 6=0, l is a positive integer. If a (z)b (z)+a 2 (z)b 2 (z) =(z ; z 0 ) l c (z) (22) then we have where c(z) 2L. a (z)a 4 (z) ; a 2 (z)a 3 (z) =(z ; z 0 ) l c 2 (z) (23) a 3 (z)b (z)+a 4 (z)b 2 (z) =(z ; z 0 ) l c(z) (24) Proof. Let us assume for deniteness that a (z 0 ) 6= 0. We express b from (22) and a 4 from (23). Using the obtained representations, we have a 3 (z)b (z)+a 4 (z)b 2 (z) = a 3 (z) (z ; z 0) l c (z) ; a 2 (z)b 2 (z) a (z) + b 2 (z) (z ; z 0) l c 2 (z)+a 2 (z)a 3 (z) a (z) (z ; z 0 ) l a 3(z)c (z)+b 2 (z)c 2 (z) a (z) = =: (z ; z 0 ) l c(z): In the rst step we prove that for every ~z 2Ronly one equation of the pairs f(7), (9)g and f(8), (20)g should be retained. Indeed, on the one hand det z h0 0 ; z ; z h0 0 z a 0 (z 2 )h 0 0(z) a 0 (z 2 )h 0 0(;z) = z a 0(z 2 )b 0 (z 2 )b 0 (=z 2 )=(z ; ~z) ~k c (z) c (z) 2L on the other hand, since a 0 (z)a 0 (=z) =;b(z)b(=z), a 0 (~z 2 ) 6= 0 for any ~z 2R. Hence, the last matrix at point ~z has at least one non-zero element. We assume for deniteness that the rst row contains non-zero element. Then by Lemma, if g and g 2 at the point ~z satisfy (7) with multiplicity ~ k, then they also satisfy (9) at least with the same multiplicity. So at the point ~z we can exclude equation (9) from consideration. In the same manner we eliminate one of equations (8) and (20). In the second step we reject equations, corresponding to the roots ~z and =~z. Now for two roots ~z and =~z we have 4 ~ k equations. It turns out that at most 2 ~ k of them are linearly independent. We show that we can keep only equations of the form (7) and (9). Indeed, let us assume that in the previous step we kept equation (7) for ~z 2Rand equation (8) for =~z. Nowwe prove that linear equations generated by (8) for =~z can be omitted. We apply the change of variable z 7! =z to (8). Then the left-hand part of (8) becomes a 0 (=z 2 )h 0 0(=z)g (z 2 ) ; zh 0 0(;z)g 2 (z 2 ): (25)

13 Since det z h0 0 ; z a 0 z 2 h 0 0 z a(z 2 )h 0 (z) ;zh 0 0(;z) = b 0 (z 2 )b 0 (=z 2 )(b 00 (z 2 )b 00 (=z 2 )h 0 0 (z)h0 0 (=z) ; ) where b 00 (z) =b(z)=b 0 (z), is divisible by (z ; ~z) ~k, expression (25) is also divisible by (z ; ~z) ~ k and the left-hand part of (8) is divisible by (z ; =~z) ~k. Dependence of the equations, generated by (20), is obtained by the same reasons. Indeed, after transform z 7! =z the left-hand part of (20) is equal to Since det z h0 0 ; z a 0 (=z 2 )h 0 0(;=z)g (z 2 )+zh 0 0(z)g 2 (z 2 ): a 0 (z 2 )h 0 0(z) a 0 z 2 h 0 0(;=z) zh 0 0(z) = b(z 2 )b(=z 2 )h 0 0(z)h 0 0(;=z) then the left-hand part of (20) is divisible by (z ; =~z) ~k. In the third step we prove that equations, corresponding ~z and ;~z, are linear dependent. Let us assume that wehavechosen equation (7) for the both roots ~z. After substitution z 7!;z the right-hand part of (7) is transformed to Since det ; z h0 0 z h0 0 ; z h0 0 ; z z z g (z 2 )+a 0 (z 2 )h 0 0 (;z)g 2(z 2 ): a 0 (z 2 )h 0 0(z) a 0 (z 2 )h 0 0(;z) = z a 0(z 2 )b 0 (z 2 )b 0 (=z 2 ) is divisible by (z ; ~z) ~k, then the equations for ;~z are linear dependent of the equations for ~z. In the case, when we take equation (7) for ~z and equation (9) for ;~z, the corresponding linear equations coincide. Finally, we note that because the left-hand parts of (7) { (20) belong to S, the linear equations for ~z =~z 2Rcoincide. Thus, we have proved the existence of a pair of polynomials g g 2 2L n, satisfying equations (7) { (20) on all of R. Although the polynomial b 0 (z 2 )b 0 (=z 2 )canhave complex roots, it is easy to check that wecanchoose polynomials g, g 2 with real coecients. Indeed, if z 0 isarootofb 0 (z 2 )b 0 (=z 2 ), then z 0 is also a root. Coecients of the equations, corresponding these roots, dier in complex conjugation. So instead of them we can consider real equations, corresponding to real and imagine parts of the initial equations. 2

14 Thus, we have k homogeneous linear equation for k + or k + 2 unknown values. Let us take any non-degenerate solution of the system. Now we prove that the corresponding pair of polynomials g and g 2 of degree at most k, satisfying (7) { (20) and the relation satises also the equation g 2 () + g2 2 () = b02 () (26) g (z)g (=z)+g 2 (z)g 2 (=z) =b 0 (z)b 0 (=z): (27) Indeed, let us assume for deniteness that ~z 2Rand jh 0 (;=~z)j 2 + ja(z 2 )h 0 (z)j 2 6=0. By (8), we have det z h0 0 ; z g z 2 a 0 (z 2 )h 0 0 (z) g 2 z 2 = z h0 0 ; z g 2 ; a z 2 0 (z 2 )h 00(z)g z = 2 ; b 0 (z 2 )b 0 (=z 2 )f 2 (z): Thus, by Lemma and from (7), the expression g (z)g (=z)+g 2 (z)g 2 (=z) is divisible by (z ; ~z) ~k. It means that polynomials in the left-hand and right-hand parts of (27) have 2k common zeros. It remains to normalize the left-hand polynomial according to (26). The normalization is impossible only in the case when g () = g 2 () = 0. However, it implies that the left-hand part of (27) has 2k + zeros. It follows from this that g (z)g (=z) + g 2 (z)g 2 (=z) 0. Hence, g (z) g 2 (z) 0. It contradicts to the assumption that at least one of the polynomials g and g 2 is non-degenerate. 3.3 Proof of Corollary 2 Indeed, for interpolatory symbols we have h 0 (z) ==2+ NX n= a n (z ;2n+ + z 2n; )=:=2+ ~ h 0 (z) a 0 (z)a 0 (=z) =;(=2+ ~ h 0 (z)) 2 ;(=2; ~ h 0 (z)) 2 =2(=2+ ~ h 0 (z))(=2; ~ h 0 (z))=2h 0 (z)h 0 (;z): The functions h 0 (z) andh 0 (;z) have distinct roots. Hence, by Theorem, we have to nd those interpolatory symbols which have roots of even multiplicity. Such polynomials have to be non-negative. Otherwise, they have roots of odd multiplicity on the circle jzj =. Thus, by Riesz lemma, h 0 (z) =p(z)p(=z), where p 2 S. On the other hand, h 0 (z)+h 0 (;z) =. It implies the equality p(z)p(=z) + p(;z)p(;=z) =, which is valid (see [4], Chapter 8) only for p(z) =z M ( + z 2N; )=2, M N 2 Z. 4 Examples In this section we discribe all possible renable functions of the class S with degrees of their symbols up to 5, satisfying criterion for the existence of 2 (anti)symmetric framelets and 3

15 give examples of their symbols and graphs. Besides, we construct 2 framelets corresponding to the B-spline B 6. The degree of its symbol is equal to 7. The case deg(h 0 ) = corresponds to the Haar wavelets. It gives an example of an orthonormal basis with symmetry. 4. The case deg(h 0 )=2 This case is also trivial. A.Ron and Z.Shen [8] presented the construction of 2 framelets associated with the piecewise linear B-spline B with the symbol h 0 (z) =z ; =4+=2+z=4 (see Fig. ). The symbols of framelets can be represented in the form h (z) =;z ; =4+ =2 ; z=4 and h (z) = (z ; ; z)=2 p 2. Easily to see that this is a unique example of a renable function for which the unitary extension principle is applicable. Indeed, let h 0 (z) =z ; + + z. On the one hand, 2 + =. On the other hand, by Theorem B, we have h 2 0() + h 2 0(;). Hence, h 0 (;) = 0 that implies ; 2 = 0. Hence, ==2, ==4. Fig.. deg(h 0 )=2 4.2 The case deg(h 0 )=3 As we know at the moment only one example of a renable function, admiting 2 framelets is known for this case. C.Chui and W.He [3] have done it for the B-spline B 2, h 0 (z) = z ; =8+3(+z)=8+z 2 =8. Then the framelets are dened by the symbols h (z) = p 3(;z)=4, h 2 (z) =(z ; +3; 3z ; z 2 )=8 (see Fig.3). An arbitrary symbol h 0 (z) 2 S h of degree 3, satisfying the condition h 0 () =, can be represented in the form h 0 (z) =z ; + + z + z 2 + ==2: (28) We prove that such symbols, admit constructing 2 antisymmetric framelets if and only if 0, 0. It follows from (28) that h(z) =; h 0 (z)h 0 (=z) ; h 0 (;z)h 0 (;=z) =;4z ;2 +(; 4( ) ; 4z 2 = ; 4z ;2 +8 ; 4z 2 = ;4z ;2 (z 2 ; ) 2 =4(z ; =z)(=z ; z): 4

16 Hence, h(z) ispositive for jzj =, if and and only if ab 0. If =0, ==2 we the Haar wavelets. If ==2, =0wehave a tight frame with a unique framelet. All other choice of positive parameters leads to a framelet system with two antisymmetric framelets with h (z) =;z ; ; + z + z 2 h 2 (z) =2 p (;+z) Several examples of graphs of such renable functions and framelets are illustrated by Figures 2{6. Fig. 2. deg(h 0 )=3, =7=6 Fig. 3. deg(h 0 )=3, =3=8 (piecewise quadratic splines) Fig. 4. deg(h 0 )=3, =5=6 5

17 Fig. 5. deg(h 0 )=3, ==4 (piecewise linear splines) Fig. 6. deg(h 0 )=3, =3=6 4.3 The case deg(h 0 )=4 For this case an arbitrary admissible symbol h 0 has the form Thus, h 0 (z) =;z ;2 +0:25z ; +(0:5+2)+0:25z ; ; z 2 6= 0: h(z) = ; 2 2 z ;4 +( ; =8)z ;2 ; ( ; =4) + ( ; =8)z 2 ; 2 2 z 4 = (z ; =z)(=z ; z)(2 2 z ;2 ; ( ; =8) z 2 ) (29) On the one hand, the last factor in (29) has to have roots of even degree. On the other hand, if z 0 is its root, then ;z 0 and z ; are also its roots. It means that only, i are permissible roots. It follows from this that ; =8 =4 2.However, the positive sign in this relatoin leads to the negative function h(z). So we have just two solutions with the correspondig framelet symbols 2 = ; p 2 4 h (z) =;0:25z ; +0:5 ; 0:25z h 2 (z) = i z ;2 ; 0:25z ; +0:25z ; i z ;2 : Figures 7 and 8 show plots of basic framelets. 6

18 Fig. 7. deg(h 0 )=4, =0: Fig. 8. deg(h 0 )=4, = ;0: The case deg(h 0 )=5 Here we consider symbols, satisfying the relation where + + ==2, 6= 0. We have h 0 (z) =z ;2 + z ; + + z + z ; + z 3 h(z) = ; 4z ;4 ; 4( + )z ;2 +(; 4( )) ; 4( + )z 2 ; 4z 4 = Thus, the parameters have to satisfy either or (z ; =z)(=z ; z)(4z ;2 + 4(2 + + )+4z 2 ): (30) 8 + 4( + ) = 8 (3) 8 + 4( + ) =;8: (32) First consider equation (3). It is reduced to the relation ( + ) = 0 that implies two possibilities: (a) + =0, ==2 (b) + ==2, =0. Solution (a) is not valid because in this case and have dierent signs that leads to the negative polinomial h(z) 7

19 (for jzj = ). Solution (b) is valid when >0, 0. Every such a pair of the parameters gives two odd framelets. Their symbols can be written in the form h (z) =2 p ( ; z) h 2 (z) =(z ;3 ; z 4 )+(z ;2 + z 3 ): Figures 9{ shows several examples of their plots. Fig. 9. deg(h 0 )=5, =0:2 Fig. 0. deg(h 0 )=5, =0:25 Fig.. deg(h 0 )=5, =0:3 Now we consider equation (32). Let x, then ==2 ; ;. We substitute to (32) and solve it with respect to. Then 2 = ; 2 p ; = ; 2 p ; 2 : 4 8

20 Hence, =2. We consider four cases (a) ==2 (b) 0 <<=2 (c) = 0 (d) <0. The case (a) corresponds to the Haar wavelets. In the case (b) we have two framelets of dierent evenness. The case (c) is well-known. It gives one framelet. The case (d) is invalid, because it leads to the negative h. We give examples for 3 sets of parameters. ) =0:0680, = ;0:00680, =0:499, (see Fig. 2). h (z) =;0:26099(z ;2 + z 3 )+0:23858(z ; + z 2 )+0:022338( + z) h 2 (z) =0:2680(z ;2 ; z 3 ) ; 0:23889(z ; ; z 2 ) Fig. 2. deg(h 0 )=5, =0:0680, = ;0:00680, =0:499 2) = ;0:043402, =0:068402, =0:475, (see Fig. 3). h (z) =;0:8984(z ;2 + z 3 )+0:29856(z ; + z 2 ) ; 0:08972( + z) h 2 (z) =0:94098(z ;2 ; z 3 ) ; 0:30590(z ; ; z 2 ) Fig. 3. deg(h 0 )=5, = ;0:043402, =0:068402, =0:475 3) = ;0:05777, =0:30777, =0:25, (see Fig. 4). h (z) =0:05777(z ;2 + z 3 ) ; 0:30777(z ; + z 2 ) ; 0:25( + z) h 2 (z) =0:073223(z ;2 ; z 3 ) ; 0:426776(z ; ; z 2 ) 9

21 Fig. 4. deg(h 0 )=5, = ;0:05777, =0:30777, =0: B-spline B 6 We give here symbols of framelets generated by theb-spline B 6 with the symbol 7 +z h 0 (z) =z ;3 : 2 Symbols of framelets are dened by the relations h (z) =;0:00783(z ;3 ; z 4 ) ; 0:054687(z ;2 + z ; + z 2 ; z 3 )+0:49287( ; z) h 2 (z) =0:04340(;z ;2 + z 3 )+0:289379(;z ; + z 2 )+0:248039(; +z): Their plots are presented by Figure 5. Fig. 5. deg(h 0 ) = 7, The framelets associated with the spline B 6 Acknowlegements The author thanks Professor R.DeVore for the presented opportunity to conduct this research in creative atmosphere of Industrial Mathematics Institute of the University of South Carolina. The author is also grateful to Dr.M.Nielsen for the careful reading the manuscript and for his useful remarks. 20

22 References [] C. de Boor, R.A. DeVore and A. Ron, On the construction of multivariate (pre)wavelets, Constr. Approx. 9 (993), 23 { 66. [2] M.Bownik, Tight frames of multidimensional wavelets, J. Fourier Anal. Appl. 3 (997), 525 { 542. [3] C.K.Chui, W.He, Compactly supported tight frames associated with renable functions, Appl. and Comp. Harm. Anal 8 (2000),#3, 293 { 39. [4] I. Daubechies, Ten lectures on wavelets, CBMF conference series in Applied Mathematics, Vol 6, SIAM, Philadelphia, 992. [5] W.Lawton, Tight frames of compactly supported ane wavelets, J. Math. Phys. 3 (990), 898 { 90. [6] S. Mallat, Multiresolution approximation and wavelet orthonormal bases of L 2 (R), Trans. Amer. Math. Soc. 35 (989), 69 { 87. [7] A.Petukhov, Explicit construction of framelets, Preprint IMI of University of South Carolina. 2000: # 3. [8] A. Ron and Z. Shen, Ane systems in L 2 (R d ) II: the analysis of the analysis operator, J. Func. Anal. Appl.48 (997), 408 {