Construction of Orthogonal Wavelets Using. Fractal Interpolation Functions. Abstract

Transcription

1 Construction of Orthogonal Wavelets Using Fractal Interpolation Functions George Donovan, Jerey S. Geronimo Douglas P. Hardin and Peter R. Massopust 3 Abstract Fractal interpolation functions are used to construct a compactly supported continuous, orthogonal wavelet basis spanning L (IR). The wavelets share many of the properties normally associated with spline wavelets in particular linear phase. Key Words: Wavelets, fractal interpolation functions, linear phase AMS Subject Classication: Primary 4A5 School of Mathematics, Georgia Institute of Technology, Atlanta, GA 333. Department of Mathematics, Vanderbilt University, Nashville, TN Department of Mathematics, Sam Houston State University, Huntsville, Tx, * G. Donovan is partially supported under a National Science Foundation Graduate Fellowship.

2 I. Introduction A wavelet basis of some function space, for example L (IR), is obtained by considering translates and dilates of one or several suitable functions [, 5, 6, 7, 9]. Much of the recent interest in these bases has stemmed from the fact that they can be built having various useful properties such as continuity, orthogonality, compact support, vanishing moments, etc. Most of the wavelets that have been investigated to date can be constructed using the notion of multiresolution analysis [6, 7]. Let be a function in L (IR) and set k;j (x) = ( k x? j). For each k ZZ, denote by V k the L {closure of the algebraic span of f k;j : j ZZg. The function is said to generate a multiresolution analysis if the following conditions are satised (i) V? V V ; (ii) clos L?S kzz V k = L ; (iii) T kzz V k = fg; and (iv) f ;j : j ZZg is an Riesz basis for V. Suppose generates a multiresolution analysis then is called a scaling function and it will satisfy the two scale dilation equation (x) = X c j (x? j): (:) If W is the orthogonal complement of V in V there exists a L (IR) such that spanf (? i); i ZZg = W, [6, 7]. Furthermore the function satises the equation (x) = X d j (x? j): (:) If f ;k g is an orthonormal basis for V, the formula d j = (?) j c?j ; (:3) gives a simple way of computing the coecients of equation (.). The celebrated work of Daubechies [5] gives explicit construction of nite sequences of coecients fc j g which give solutions of equation (.) that are orthogonal, compactly supported and have varying degrees of smoothness.

3 Recently Hardin, Kessler and Massopust [3] showed that certain classes of fractal interpolation functions (FIF) also generate a multiresolution analysis of L (IR). This multiresolution analysis has certain geometric features that are similar to the multiresolution analysis generated by splines [4]. These results have been generalized to several dimensions in [8] and [9]. In [] scaling functions were exhibited that form an orthonormal basis for the V given in []. Here we continue the investigation of the multiresolution analysis arising from FIF and construct orthogonal, compactly supported continuous wavelets. Wavelets with varying orders of dierentiability will be considered in a later paper [7]. These wavelets fall outside the class constructed in [5] and the multiresolution analysis from which they arise yields several scaling functions instead of just one. In this case (.) takes the form X (x) = C j (x? j); (:4) where each C j is a square matrix the size of which is determined by the number of scaling functions. Multiresolution analyses based upon several scaling functions have also appeared in the work of Micchelli [8], Goodman, Lee, and Wang [], Goodman and Lee [], Jia and Shen [4],and Loc [5]. We proceed as follows: in Section we review the relevant facts on FIF and the multiresolution analysis arising from these function spaces. Then in Section 3 we exhibit and solve the equations that give scaling functions whose dilates form an orthonormal basis for a certain V. We also examine the smoothness of these scaling functions and exhibit their Fourier transforms. In Section 4, we use the scaling functions constructed above to nd compactly supported, continuous, orthogonal wavelets. We investigate the support properties of these wavelets and discuss how to convert them into a wavelet basis for L [; ]. Finally in Section 5 we extend the methods developed in Sections 3 and 4 to integer scalings other than two. For these cases there are a number of parameters that are free to be specied. We examine this in the case of scaling by three and exhibit a one parameter family of symmetric scaling functions. II. Fractal Interpolation Functions Let I = [; ], and B(I) denote the Banach space of bounded real{valued functions

4 on I with the k k and C(I) B(I) the space of real-valued functions continuous on I. Let u i : [; )! [; ) and v i : [; ) IR! IR, i = ; ; : : : N? be as follows u i (x) = (x + i); (:) N v i (x; y) = i (x) + s i y; (:) where i (x) m, the set of polynomials with degree at most m. It will always be i assumed that s = max js i j <. Let I i = u i ([; )) = N ; i+ N for i = ; ; ; : : : ; N?, = ( ; ; : : : ; N? ), and dene : B(I)! B(I) by ( f)(x) = v i (u? i (x); f(u? i (x))); (:3) for x I i, i = ; ; : : : ; N?. Note that f () = ()=(? s ), and f (? ) = N? (? )=(? s N? ). s thus, Equations (.) and (.3) imply that is a contraction on B(I) with contractivity k f? gk skf? gk ; (:4) and so has a unique attractive xed point f B(I). In the event that satises the join-up conditions v i+ (; f ()) = v i (? ; f (? )); i = ; ; : : : ; N? ; (:5) then f is continuous and is called a fractal interpolation function (Barnsley []). general, G = graph f is typically a fractal set in IR made up of images of itself. To see this let w i : [; ) R! [; ) R be given by w i (x; y) = (u i (x); v i (x; y)); for i = ; ; : : : ; N?. Then (.3) implies that G = N? [ j= w j (G): Let = N N? j= m and then the following theorem gives the basic correspondence between elements in and functions in B(I). 3 In

5 Theorem.. [, ] The mapping 7! f is a linear isomorphism from to (). We will be interested in the case when m =. If f satises equation (.5) then \ C(I) = S is seen to be N + dimensional. Thus f C(I) and the space N N? j= each element g S is completely determined by g(i=n), i = ; ; : : : ; N. This allows us to view in a slightly dierent manner. Given y R N + let f y be the unique element of S passing through the points? i N ; y i, i = ; ; : : : ; N. We shall call functions fy S ane fractal interpolation functions (AFIF). Corollary.. The map : R N +! S is a linear isomorphism. The fact that S is isomorphic to R N + adds a geometric component to the multiresolution analysis associated with FIF and will play an important role in our construction of wavelets. Let C b (IR) be the space of bounded continuous functions on IR Fix s ; s ; : : : ; s N?, let ~ V = ff : fj [i;i+) is an AFIFg \ C b (IR) \ L (IR) and dene f ~ Vk, f(n?k ) ~ V. Then it was shown in [] and [3] that the sequence f ~ Vi g has the following properties, (a) : : : ~ V? ~ V ~ V : : : ; (b) T kr ~ V k = fg. Note that in contrast with [6] and [7], the spaces f ~ V i g given above are dened independently of any particular scaling functions which is one of the several properties these spaces share with the spline spaces S r d with integer knots (see [],[]). In fact ~ V is spanned by sets formed from the integer translates of several scaling functions. We say a multiresolution analysis is continuous and/or compactly supported if it is generated by a nite set of scaling functions f i (x)g N i= ; i (x) L (IR); i = : : : N such that each i is continuous and/or compactly supported on IR. If h i (); j (? k)i = i;j k; ; i = : : : N then the f i g generate an orthogonal multiresolution analysis. In order to nd orthogonal scaling functions f i g N i= that generate a continuous, compactly supported, orthogonal multiresolution analysis we develop some quadrature formulas for AFIF. Set I = R f ygdx where f y S and g L (I). Then from (.3) we 4

6 nd I = N? X = N Z i= [i=n;(i+)=n ] N?Z X i= f y gdx = N? X i= Z v i (x; f y (x))g(u i (x))dx: [i=n;(i+)=n ] v i (u? i (x); f y (u? i (x))g(x)dx If we use equation (.) along with the assumption that i (x) = a i x+b i, i = ; ; : : : N?, in the above equation we nd that I = N N? X i= + N Z N? X i= (a i x + b i )g(u i (x))dx Z s i g(u i (x))f y (x)dx: (:6) If g = then (.6) yields [3] I = m = Z f y (x)dx = N P N?? ai + b i= i? N P N? i= s i ; (:7) while for g(x) = x we nd [3] I = m = Z xf y (x)dx = P N? N [a i= i(i= + =3) + b i (i + =) + is i m ] P? N N? s i= i (:8) With (.7) and (.8) integrals of two fractal functions may be computed. To this end let ^v i (x; y) = ^ i (x) + ^s i y with ^ i = ^a i x + ^b i, ^u i = u i, i = ; ; : : : ; N? and set g = ^f^y. Then [3] I = = Z N ^f^y (x)f y (x)dx P N? i= s i^a i m + ^s i a i ^m + s i^bi m + ^s i b i ^m + a i^a i 3 + (a i^b i +^a i b i ) + b i^bi? N P N? i= s i^s i ; (:9) where ^m and ^m are the zeroth and rst moments respectively of ^f^y. Consider the N + dimensional basis ff yi g N spanning i= S where y i = e i, < i < N, fe i g N being the standard basis in i= RN +, y = (; q ; : : : ; q N? ; ), and y N = (; p ; : : : ; p N? ; ). The sequences fp i g and fq i g are chosen so that hf y ; f yi i = = 5

7 hf yn ; f yi i for i = ; : : : ; N?. Extend f yi ; i = ; : : : ; N? to be functions in V ~ by dening each of them to be equal to zero for x 6 I. Let f i g N? i= be a sequence of orthogonal functions in ~ V obtained from ff yi g N? i= these functions are non-zero follows from Corollary.. Set and ^ i (x) = N? (x) = 8 < : i (x) k i (x)k, then we nd the following L by the Gram-Schmidt procedure. That f yn (x); x [; ) f y (x? ); x [; ) elsewhere, (:) Theorem.3. Let ~ V and i ; i = ; : : : ; N? be as above. Then ~ V = clos L spanf i (?l) : i = ; : : : ; N? ; l ZZg. Furthermore the set f ^ i g N? i= supported multiresolution analysis. generates a continuous, compactly Proof. From Corollary. and (.) we see that each i ; i = ; : : : ; N? is compactly supported and is an element of ~ V. Since every f ~ V is determined by its values at i N ; i ZZ; f has a unique expansion in terms of f y i ; i = ; : : : ; N? ; and N? and their integer translates. Thus every f ~ V has a unique expansion in terms of f ^ i g N? i= and their integer translates. In order to show that f ^ i g N? i= generates a multiresolution analysis we must show that, a) f ^ i g N? i= and its integer translates form a Riesz basis for V ~ and that b) clos L SkZ V ~ k = L. Since the set f ^ i g N? i= and its integer translates is an orthogonal set it follows that we need only show that there exits constants A; B; < A B < such that 8c = fc i g l ; Akck l k P c i ^N? (? i)k L Bkcjj l. It is easy to show that B = p 3 provides an upper bound while the lower is obtained by observing that k P c i ^N? (? i)k L = P i R i+ i (c i ^N? (x? i) + c i? ^N? (x? (i? ))) dx. Since f y and f yn are linearly independent the matrix, hfy ; f K = y i hf y ; f yn i hf y ; f yn i hf yn ; f yn i is positive denite. Let be the smallest eigenvalue of equal to p. ; K k N? k then A can be taken to be To show that S kz ~ V k is dense in L we note that for all x IR; = P i (P N? j= cj i f y j (x? i)) + N? (x? i) where c j i = cj =? p j? q j. Now b) follows from Proposition 3. in []. 6

8 III. Orthogonal Scaling Functions We begin by considering AFIF with scaling N =. By Theorem.3 f i g (note i= that in this case = f y ) and its integer translates span V ~ and we look for three mutually orthogonal functions f y ; f y ; f y. From (.), (.), (.3), and (.5) we nd that for f y, = x, =? x for f y, = (p? s )x, = (? s? p )x + p while for f y, = (q + s? )x +? s, = (s? q )x + q? s. Substituting these values into (.9) gives Z and Z Z f y f y dx = (4? 6s + 6p? s s? 4s? 4s + 4p s s + 3s 3 + 3s s? 4p s? 4p s ) 3(? s? s )(4? s? s )(? s? s ) ; f y f y dx = (4? 6s + 6q? s s? 4s? 4s + 4q s s + 3s 3 + 3s s? 4q s? 4q s ) 3(? s? s )(4? s? s )(? s? s ) ; f y f y dx = (3:) (3:) [4(p + q )(s + s + s s? ) + 8p q (s + s? s s? 4) + (s + s )? (s + s + )3? 4(s + s ) + s 3 (? + s? 6q ) + s 3 (?? 6p + s ) + 6q s (? s ) + 6p s (? s ) + 6s + 6s + s s ]=6(? + s + s )(?4 + s + s )(? + s + s ): Solving (3.) for p and (3.) for q yields (3:3) and p =?(4? 6s? s s? 4s? 4s + 3s3 + 3s s ) 6 + 4s s? 4s? ; (3:4) 4s q =?(4? 6s? s s? 4s? 4s + 3s 3 + 3s s ) 6 + 4s s? 4s? : (3:5) 4s If we substitute (3.4) and (3.5) into (3.3) we nd that Z f y f y dx = p(s ; s ) 6(6 + 4s s? 4s? 4s ) (? + s + s )(?4 + s + s )(? + s + s ); 7

9 where p(s ; s ) = s 4 + 6s3? 7s3 s + 8s s? 8s? 7s s 3 + 8s s? 4s s + s + s 4 + 6s 3? 8s + s + 8: (3:6) Consequently, we have Lemma 3.. The AFIF f y, f y and f y with y = (; q ; ), y = (; ; ), and y = (; p ; ) constitute an orthogonal basis for S only for pairs (s ; s ) such that js j <, js j < and p(s ; s ) =. Corollary 3.. The only pairs (s ; s ) such that the basis ff x ; f x ; f x3 g with x = (; ; a), x = (; b; ) and x 3 = (; c; ) can be made an orthogonal basis for S are those pairs for which p(s ; s ) =. The same is true for bases of the form ff z ; f z ; f z3 g with z = (a; ; ), z = (; b; ) and z 3 = (; c; ). Proof. Let (s ; s ) be such that ff x ; f x ; f x3 g is an orthogonal basis. Suppose a 6= otherwise the result follows from Lemma 3. and let f ^f x ; ^f x ; ^f x3 g be the corresponding orthonormal basis. If ^fx () = a and ^f x () = b set w = b ^fx? a ^fx and w = a ^fx + b ^fx where a + b =. Then fw ; w ; ^f x3 g is an orthonormal basis of S with w () = w () = and w () =. But by Lemma 3. this can only happen for values (s ; s ) such that p(s ; s ) =. An analogous argument can be applied to the basis ff z ; f z ; f z3 g. From Theorem.3 and Lemma 3. we have the following Theorem 3.3. Suppose that the pair (s ; s ) is such that p(s ; s ) = with js j < and js j <. Then ^ i, i = ;, generate a continuous, compactly supported, orthogonal multiresolution analysis of L (IR). It follows from Theorem.3 that for general pairs (s ; s ) with js j < and js j <, (x) = = 3X i= C i (x? i): (3:7) 8

10 The matrices C i, i = ; ; ; 3 may be computed by evaluating (3.7) at x = i=4, i = ; ; : : : ; 8. From the values of i, i = ; for f y, f y and f y computed earlier we nd s C = ; C = + =? q?s?p ; C = s + =? p p?s + p(s? p) p?s?q + q(s? p) q For later use we dene the inner product ; C 3 = + p(s? q) q?s : + q(s? q) (3:8) Z Z h; i = IR (x) dx = (x) (x) (x) IR (x) (x) (x) dx = E (x) ; where E k k =. k k We will now show that even if longer supports are considered compactly supported, continuous orthogonal scaling functions can only be constructed for those values (s ; s ) for which p(s ; s ) = with js j < and js j <. Lemma 3.4. Suppose for a given three dimensional subspace V of C(I), there is no orthonormal basis with two functions vanishing at one endpoint and the remaining function vanishing at the other endpoint. Then any continuous, compactly supported pair of functions and composed of linear combinations of the basis elements of V and their integer translates constructed so that h i (x); j (x? k)i = i;j k;, must have the property that the leftmost nonzero components of, and are linearly dependent as are their rightmost nonzero components. Proof. Suppose the support of and are [; N] and [; M] respectively with N +M 3. Because of continuity (); (); (x + N? )j x=, and (x + M? )j x= all vanish, furthermore (x)j [;] and (x)j [;] are orthogonal to (x + N? )j [;] and (x + M? )j [;]. The result now follows since F spans V. We note that by rotation it can always be arranged so that M 6= N. Lemma 3.5. If the hypotheses of the above lemma are satised, then no such pair of functions ; exists. 9

11 Proof. Suppose that and exist, that M < N, and set y = jj jj, and z = Then yand z are orthonormal and for a suitable basis can be represented as y = (y ; y ; : : : ; y M ; ; ; : : : ; ) = ((y ; ; ; ); (y ; ; y ; ; y ;3 ); : : : ; (; ; y M;3 ); (; ; ); : : : ; (; ; )) jj jj. and z = (z ; z ; : : : ; z N ) = ((z ; ; ; ); (z ; ; z ; ; z ;3 ); : : : ; (; ; z N;3 )) where M < N and z ; >. Consider the rotation r dened on pairs of vectors of the above form r(y; z) = q (z ; y? y ; z; y ; y + z ; z) = (~y; ~z); y; + z ; where ~y = ((; ; ); (~y ; ; ~y ; ; ~y ;3 ); : : : ; (~y N; ; ~y N; ; ~y N;3 )) ~z = (( q z ; + y ; ; ; ); (~z ;; ~z ; ; ~z ;3 ); ); : : : ; (~z N; ; ~z N; ; ~z N;3 )); and the shift map s dened on the range of r by s(~y; ~z) = ((~y ; ~y 3 ; : : : ; ~y N ; ); (~z ; : : : ; ~z N )) = (^y; ^z): Both s and r are continuous, and k^zk = k^yk =, and ^y and ^z satisfy the necessary orthogonality relations. The operations also preserve continuity of the components, so the functions corresponding to ^y and ^z are continuous. By Lemma 3.4, we know that ^y and ^z are linearly dependent, so ^y ; = = ^y ;3. It follows that (^y; ^z) is back in the domain of r, so we can iterate the map s r on (y; z) to produce a sequence (s r) j (y; z) = (y (j) ; z (j) ) for j = ; ; ; : : :. Since this sequence is contained in a compact set we can extract a convergent subsequence with limit say (Y; Z). By continuity of the inner product, we have ky k = kzk =, Y and Z satisfy the orthogonality relations, and the functions corresponding to Y and Z are continuous. Observe that z (j) ; is monotone increasing in j. Since it is a component of a unit vector, it is bounded above by, and hence converges to some number, Z ;. We also have that (z (j) ; ) converges (to Z ;), so that the increments of this sequence, (z (j+) ; )?

12 (z (j) ; ) = (y (j) ; ), must converge to zero. We claim that for each i f; ; : : : ; N? g, lim j! y (j) i; =. We have just shown that this is true for i =, so suppose it is true for some i. Then for each j, we have jy (j+) i; j = z (j) ; y(j)? i+; y(j) ; z(j) i+; q (y (j) ; ) + (z (j) Since y (j) ; and z(j) ; are both components of unit vectors, each is no larger than. Thus the ; ) denominator (above) is no larger than p <, and we have : jy (j+) i; j jz (j) ; y(j) i+; j? jy (j) ; z(j) i+; j; or jz (j) ; y(j) j i+; jy(j+) i; j + jy (j) ; z(j) j i+; jy (j+) i; j + jy (j) ; j: By our induction hypothesis, this last expression converges to zero. Furthermore, z (j) ; z ; >, so lim k! y (j) i+; =, and the claim is established by induction. Now, our limit point (Y; Z) must have the property that Y i; =, for i = ; ; : : : ; N. But by the lemma, we also know that the leftmost nonzero 3-tuple in Y, say Y p, is linearly dependent on Z = (Z ; ; ; ). So, Y p has the form (Y p; ; ; ), and we have just shown that Y p; =. Therefore Y must be zero, which contradicts the fact that ky k =. With the above lemmas we are now able to prove, Theorem 3.6. If s and s do not satisfy the relation p(s ; s ) = (equation (3.6)) then there are no continuous, compactly supported, orthogonal scaling functions formed from the AFIF generated by s and s such that the L closure of the linear span these functions and their integer translates is ~ V. Proof. It is easy to show that V ~ cannot be spanned by the integer translates of one continuous compactly supported scaling function. This is because if the scaling function is supported on [; M]; M > and we consider an interval of length N there will be at most N + M? translates of supported on this interval. However the same interval will

13 zero-set of p zero-set of q y x - -3 Figure contain N + interpolation points. Consequently for large enough N there will not be enough translates of to match all the necessary conditions. By Lemma 3.5 we need now only consider pairs s and s that allow an orthogonal basis for S in which at least two basis vectors vanish at either zero or one. But Lemma 3. and Corollary 3. show that this can occur only in the case when p(s ; s ) =, which proves the result. Since compactly supported continuous scaling functions constructed from AFIF occur only for the pairs (s ; s ), js j <, js j <, such that p(s ; s ) =, we examine the zeroset of this polynomial. The next lemma shows that the particular set we are interested in is convex. This conrms the contour plot given in Figure.

14 Lemma 3.7. The zero-set of the polynomial p(s ; s ) = s 4 + 6s3? 7s3 s + 8s s? 8s? 7s s 3 + 8s s? 4s s + s + s 4 + 6s 3? 8s + s + 8; has two connected components, one a convex closed curve and the other a pair of asymptotically linear curves that cross at a point. Proof. First, we transform the polynomial using the change of variables to get a new polynomial, s = x? y s = x + y; q(x; y) = 8y 4 + (?4 + 4x )y? x x 3? 7x + 4x + 8: Note that q is symmetric with respect to y for all values of x. The transformation is just a rotation by 45 and a dilation by p, so it preserves all relevant properties of the zero-set. Now, we want to nd out where the transformed polynomial, q, takes the value zero. To do this we consider the equation q(x; y) = and solve for y as a function of x. This gives y = q p 7? 6 3 x + 6p 3 x4? 3x 3 + 8x? 6x + ; or y = q p 7? 6 3 x? 6p 3 x4? 3x 3 + 8x? 6x + : The rst solution gives a pair of asymptotically linear curves that cross at x =, y =. The second solution, which is real-valued only for x [? ; ], gives two halves of a symmetric 5 closed curve,, and is the one that we are primarily interested in. Let f denote the positive branch of. If f is a concave down function, it follows by symmetry that is a convex curve. Furthermore, since the square root function is monotone concave down, it suces to show that f (= f f) is concave down. Thus, we wish to demonstrate that the second derivative of f is nonpositive on [? ; ]. That is, 5 p p? 4 + 3(48x3? 96x + 56x? 6) 3(44x? 9x + 56)? ; 3 4p 3= p = 3

15 where p = x 4? 3x 3 + 8x? 6x +. Note that p is positive for all values of x, so the above expression is well-dened. Also, since the denominators are powers of p, they are positive as well, and we can multiply them out to get an equivalent inequality,?4p 3=? p 3(?44x x 5? 888x x 3? 78x + 58x? ): To show this, we approximate p p with a linear polynomial, q = p 3( 7? 3 x). Observe that 4 4 q p p on [? 5 ; ], since p? q = 6 (x? )(9x3? 3x + x? 9) is nonnegative on that interval. Thus, by substituting pq for p 3= in the above and expanding, we have? 4p 3=? p 3(?44x x 5? 888x x 3? 78x + 58x? ) p 3(?44x 6 + 6x 5? 68x 4 + 4x 3? 4x + 673x? 99) p 3(?44x 6 + 6x 5? 68x 4 + 4x 3? 4x + 673x? 99) + p 3 (37683? 9989x) 35 =? 4p 3 35 (5x? 4) (45x 4? 95x x? 979x + 98) : The last inequality is justied, since both factors involving x are nonnegative for all values of x. Let (a; b) be the appropriate pair that solves the equations dp=ds = and p =, which yields a?:468. Then by examining p(s ; s ) and using Lemma 3.7 we see that for every s [a; ) there is at least one and at most two values of s, js j < with p(s ; s ) =. If s = s then x = in the polynomial q(x; y) and we nd q(; y) =?y y 3? 7y + 4y + 8 =?(5y + )(y? )(? + y). Thus a solution of this equation is y =?=5 = s = s and equations (3.4) and (3.5) give p = q =?3=. Consequently for s = s =?=5 the functions f y ; f y, and f y g with y = [;?3=; ], y = [; ; ], and y = [;?3=; ] are mutually orthogonal. The scaling functions and given as in Theorem 3.3 are shown in Figure. Note that is symmetric about / while is symmetric about. Consequently individually both exhibit linear phase [4]. If we normalize ^ so that h^; ^i = I then the coecients in (3.7) can be re-expressed as ^(x) = P 3 p i= b C i p ^(x? i) where b Ci = E? C i E bc = 3 p = 4=5?=?3 p = ; b C = 4. The f Ci b g for s = s =?=5 are p 3 = p ; 9= =

16 bc = 9=?3 p = Figure ; b C3 = :?= The values of p and q for other acceptable pairs s, s are given in Table. We now consider the smoothness and approximation order of the function i, i = ;. Recall that the Holder exponent of f C(I) at x is x = lim! infflog jf(x)? f(y)j= log jx? yj : y B(x; )g and = inff x ; x Ig is called the Holder exponent of f. Let S be a closed subspace of L (IR), E(f; S) = minfkf?sk : s Sg; ^f be the Fourier transform of f, and w k(ir) = ff L (IR) : kfk w k = k( + j j k ) ^fk < g. Following de Boor, DeVore, and Ron [6] we say that S provides approximation order k, if for every 5

17 f w k (IR), E(f; S h ) C S h k kfk w k (IR) where S h = fs( h ) : s Sg. Lemma 3.8. Given any pair (s ; s ), js j <, js j <, let f S. If js j < = and js j < = then f is Lipschitz continuous, i.e., there exists an M < such that 8 x; y [; ]; jf(x)? f(y)j Mjx? yj. If max js i j > = and f is not a line then f has Holder exponent =? log max js i j= log. We note that the second part of the lemma has already been proved by Bedford [3] and we include the proof for the convenience of the reader. Proof. Let s = max i js i j, i n = fi ; i ; : : : ; i n g, i j f; g, j = ; : : : ; n, and a(i n ) = i = + i =4 + + i n = n. Then for x [a(i n ); a(i n ) + = n ] we nd from (.3) that f(x) = nx k= + kx Y! k? k? i m x + b ik? s m? ij m= j=! nx s ij A f n x? n?m i m n Y j= m= (3:9) Suppose = n+ jx? yj = n and x; y [a(i n ); a(i n ) + = n ], then the above formula shows us that jf(x)? f(y)j nx k= where C = kfk. Suppose rst that s < = then jf(x)? f(y)j nx k= (s) k? jx? yj + s n C; (3:) (s) k? jx? yj + (s) n Cjx? yj; where the fact that n+ jx? yj has been used to obtain the last term in the above expression. Since s < we nd that jf(x)? f(y)j + C jx? yj: (3:)? s 6

18 If = n+ jx? yj = n but x and y are not in the same dyadic interval let x be the boundary point between the respective intervals x and y are located in. jf(x)? f(y)j jf(x)? f(x )j + jf(x )? f(y)j and applying (3.) gives jf(x)? f(y)j ( + C)? s jx? yj for all x; y [; ]. This gives the Lipschitz continuity of f. Then If s = max js i j > = assume without loss of generality that f() = = f(). For if that is not the case we can modify f by adding linear functions L (x) and L (x) so that ^f = f + L + L is still in S, ^f() = ^f() =. If = n+ jx? yj = n, a(i n ), x; y [a(i n ); a(i n ) + = n ] (3.) implies jf(x)? f(y)j s n C + nx k= (s) k?n?! = s n C + s? s Since = n+ jx? yj = n we nd that jx? yj?(n+) =??n =? s n with = ln s. Hence ln = jf(x)? f(y)j C + s jx? yj (3:)? s for x, y in the same dyadic interval of length = n. If x, y are not in the same dyadic interval then using the same argument as in the case when s < we nd jf(x)? f(y)j + C + s jx? yj? s for all x; y [; ]. We need only show that is the largest possible exponent. Suppose without loss of generality that s = js j. Since f vanishes at and and f( ) 6= ( since f is not a line), there exist distinct points x and y [; ] such that f(x ) 6= f(y ) and x?y f (x )?f (y ) >. With x n = x = n and y n = y = n we nd from (3.9) that Therefore f(x n )? f(y n ) = n? X k= (s ) x k?? y n + s n (f(x )? f(y )): jf(x n )? f(y n )j = s n jf(x )? f(y )j + x? y f(x )? f(y ) which proves the result. We can now prove s n C kjx n? y n j 7 :? ( s ) n? s? s

19 Theorem 3.9. Suppose the pair (s ; s ) is such that js j < and js j < with p and q satisfying (3.4) and (3.5) respectively. Then V provides approximation order. If s and s are both in magnitude less than = then i, i = ; are both Lipschitz. If s = max js i j > = then i, i = ; have Holder exponent =? log s= log. Proof. In order to show that V provides approximation order. We need only prove that the hat function g(x) = n x x,? x x is in V []. It is easy to see from (.3) that for any pair s, s with js j < and js j <, f [;=;] (x) = x. Consequently, g(x) = (=? p ) (x) + (x) + (=? q ) (x? ): The fact that and are Lipschitz when s < = follows from Lemma 3.7. Furthermore Lemma 3.7 also shows that if s > = then the Holder exponent of is log jsj= log =. This will also be true of once it is shown that is not piecewise linear. For s > = this can happen only if p = q = =, however from (3.) we nd with p = =, Z f y f y dx = which is a contradiction. Therefore f y cannot be a line and the result now follows.? + s (? + s + s )(?4 + s + s ) 6= is not a line. A similar argument shows that f y We complete this section by computing the Fourier transform of and. To this end set g(x) = e ikx, N = in (.6) to nd ^f y (k) = Z e ikx f y (x)dx = + = Z Z (a x + b )e ikx= dx e ik(x+)= (a x + b )dx + =(s + e ik= s ) ^f y (k=): For, a =, b =, a =? and b =. Consequently, ^ (k) = 8e ik= sin k=4 k + (s + e ik= s ) ^ (k=): 8

20 With C(k) = (s + e ik= s ) and h (k) = 8e ik= sin k=4 k ^ (k) = X n n= j= we nd that C(k= j? ) A h (k= n ): The above series converges uniformly for k IR since jc(k= j )j js j+js j = r <, j = ; : : : and since j sin x=xj for all x IR. To compute the Fourier transform of we use the fact that can be written in terms of the hat function g and as was shown above. Thus ^ (k) = ^g(k)? (=? p? (=? q )e ik ) ^ (k): Below the Fourier transforms of and are plotted when s = s =?= ^ ^ Figure 3 IV. Construction of Compactly Supported Wavelets Having constructed a multiresolution analysis based on the vector =? and its dilates and translates we now construct compactly supported, continuous wavelets. If ~ W is the orthogonal complement of ~ V in ~ V then any ~ W can be written as (x) =? (x) (x) where i (x)j [j=;(j+)=), i = ; is an AFIF with interpolation points at the quarter integers. Since (x) =? (x) (x) is supported on [; ], we shall look for (x) to be supported 9

21 on the same interval. Therefore set i (x) = 8 >< >: g y i (x); x =, g y i (x); = x, g y i 3 (x); x 3=, g y i 4 (x); 3= x, i = ; : (4:) Here y j i are the values i (x) taken at the quarter integer points in the intervals indicated. In order to preserve continuity and keep the support of (x) on [; ], y i = [; ai ; ai ], y i = [a i ; a i 3; a i 4], y i 3 = [a i 4; a i 5; a i 6] and y i 4 = [a i 6; a i 7; ], i = ;. The coecients a i j, j = ; ; : : : ; 7, i = ; are adjusted so that (x) is orthogonal to (x) and its translates and h (x); (x)i =. In the case when s =?=5 = s and p =?3= = q it follows from (.7), (.8), and (.9) that for i = or, Z i (x) (x)dx = 5 3 ai ai ai ai 4; (4:) Z Z i (x) (x + )dx = 3 64 ai ai ai ai 7 ; (4:3) i (x) (x)dx =? 48 ai? ai ai ai ai? 5 ai? 6 48 ai 7 ; (4:4) Z Z i (x) (x + )dx = 3 6 ai? ai? 48 ai 3? 6 ai 4 ; (4:5) i (x) (x? )dx =? 6 ai 4? 48 ai 5? ai ai 7 ; (4:6) and Z (x) (x)dx = 5 96 (a a + a 3a 3 + a 5a 5 + a 7a 7) (a a + a 6a 6 + a 4a 4 + a 6a 6) (a 4a + a 4a + a 6a 4 + a 6a 4) (a a + a a + a 3 a + a a 3 + a 4a 3 + a 3a 4 + a 5a 4 + a 4a 5 + a 6a 5 (4:7) + a 5a 6 + a 7a 6 + a 6a 7): The above equations once set equal to zero will x all but three of the unknowns. Two of these are used to normalize the integrals of and. One unknown remains because of

22 the one parameter family of rotations taking into other mother wavelets. A remarkable fact to be shown below is that once (4.3), (4.5) and (4.6) are satised then h ; (+i)i = 8 i ZZ, i 6=. A solution to the above equations that give h ; i = I with having the smallest possible support is " y = ;?3p ; 9 p # ; " p y 3 = ; 5 p ;?3 p Since ~ V we have # y = [; ; ] ; y = 3?; 48 5 ;?3 ; 5 (x) = ; 3X i= y = " 9 p y 4 = ;?73 p p ; p # ; ; "?3 p ; y = ; 3 ;? ; y 4 =?3 5 ; 5 ; ; # ; (4:8) (4:9) D i (x? i); (4:) where for the particular given by (4.8) and (4.9) p p p p = 3 6=?9 3= = 6 D = ; D =?= p ; 3 p p p 3 3=? 6= D = 3 p 6=? p? 3=6 ; D 3=5 3 =? p : 6=3 In order to see why the orthogonality of to the translates of implies the orthogonality of (4:) to its non zero integer translates we examine the multiresolution analysis arising from and consider the slightly more general case where IR N which we will use in section V. In this case each C i in (3.7) is an N N matrix and satises the N-scale dilation equation Likewise the mother wavelet (x) = (x) = N? X i= satises N? X i= C i (Nx? i): D i (Nx? i);

23 where the matrices D i are N(N?)N matrices. The orthogonality of (x) to its integer translates can be re-expressed as, Z N? X (x) (x? i)dx = i; I N = C k Ck?Ni; 8 i IR; (4:) where I N is the N N identity matrix. Likewise the orthogonality of against its translates gives N? X k= That W is orthogonal to V in V means k= D k D k?ni = i; I N (N?) ; 8 i IR: (4:3) N? X k= while V = V W can be expressed as [5] X k C k D k?ni = ; 8 i IR; (4:4) C m?nkc n?nk + D m?nkd n?nk = m;n I N 8 n; m ZZ: (4:5) If we set H = (C ; C ; : : : ; C N? ), H = (C N ; C N + ; : : : ; C N? ), G = (D ; D ; : : : ; D N? ), and G = (D N ; D N + ; : : : ; D N? ). Then (4.) can be recast as H H = (4:6) and Likewise (4.3) and (4.4) become H H + H H = I N (4:7) G G = (4:8) G G + G G = I N (N?) (4:9) H G = (4:) H G = (4:) and H G + H G = (4:)

24 Equation (4.5) can be recast as H H + H H + G G + G G = I N (4:3) and H H + G G = : (4:4) The general solution to (4.6) is H = P Y where P : IR N! IR N is an orthogonal projection onto the null space of H and Y is any N N matrix. Likewise from (4.) it is easy to see that G = P X where X is any N N(N? ) matrix. If we set H = (I N? P )Y then (4.6) and (4.7) are satised and Y = H + H. Observe that Y Y = I N. Equations (4.), (4.8) and (4.9) now suggest that G = (I N? P )X with X X = I N (N?), and (4.) says that Y X =. Since Y is an N N matrix and X is an N N(N? ) matrix it is an idea of Gil Strang that X can be obtained by letting its columns be the remaining orthonormal basis vectors for the IR N the rst N basis vectors being the columns of Y. That is we choose X so that the matrix (Y X) is an orthogonal matrix. If this is done (4.3) and (4.4) will also be satised since Y Y + XX = I N. Thus equations (4.6) { (4.4) and the assumptions on G ; G ; H ; H can be summarized as Y Y X (Y X) = I N = (Y X) X : The lters arising from the above equations are closely related to those found in Vetterli [] (also see Strang & Strela []). Thus we have shown Theorem 4.. Let fc i g N? i= be N N matrices satisfying (4.) then there exist N(N? )N matrices fd i g N? i= constructed as above so that equations (4.3), (4.4) and (4.5) are satised. We now show that for any wavelet that is constructed using the above scaling function the minimum length of the support of any of its components is 3/ and at least one component must have a support greater than or equal to two. Before proving the next lemma we note that for js j < and js j <, and f ~ V, f(x) = X i (c i (x? i) + c i (x? i)); 4:5a 3

25 where c i = f(i + ); 4:5b and c i = f(i + )? c i ( )? c i? ( 3 ): 4:5c Lemma 4.. For any pair (s ; s ) such that p(s ; s ) = with js j < and js j < there is no wavelet function supported on [; ] or on [ ; 3 ]. Proof. Denote by U the restriction of ~ V to the interval [; ]. That is, U = ffj [;] : f ~ V g. We see that U is a 5-dimensional vector space. For notational simplicity, let x = j [;] x = ( + )j [;] x 3 = ( +)j [;] x 4 = (?)j [;] : Note that rankf ; x ; x ; x 3 ; x 4 g = 5. Suppose that is a wavelet supported on [; ]. Then is orthogonal to x, x, and because of the orthogonality between wavelets and scaling functions. Also, from (4.5) applied to functions in ~ V we see that is orthogonal to ( +) and (?) (and hence to x 3 and x 4 ) since vanishes at both and. Thus f ; x ; x ; x 3 ; x 4 g? = fg. This cannot be, so there is no such wavelet,. Now, suppose that is supported on [ ; 3 ], and let y = j [; ] y = j [; ] y 3 = ( + )j [; ] y 4 = ( + 3)j [; ] z = ( + )j [; ] z = ( + )j [; ]: Note that the above are AFIF's on [; ]. Since f y () is orthogonal to f y () it follows from (3.8) that if s 6= then y and y are linearly independent. A similar argument shows that if s 6= then y 3 and y 4 are linearly independent. Orthogonality between and i (? j) dictates that z must be orthogonal to y 3 and y 4. This implies that z is orthogonal to all functions in ~ V j [; ] vanishing at since this space is three dimensional with basis ff yi ()g 3 i=. Similarly, z is orthogonal to all functions in ~ V j [; ] vanishing at. Thus it follows from (4.5) that is a multiple of (?). Now, does not vanish at, but we require that h ; i =. This cannot be, since among the functions 4

26 i (?j), of which is a linear combination, the only one that does not vanish at is (?). For the case of s = (or similarly for s = ), we can apply the quadrature formulas (.9). Solving p(; s ) =, yields s = p 7? 3 which implies that y 3 and y 4 linearly independent. Thus z is still a multiple of the left half of (). Rescale so that z ( ) =, hence z must have the form f [;r;] (), where r is yet to be determined. Since s = ; (x)j [; ] = f [; ;] which is a line. From (.9), h ( + ); i = hz ; y i = + r? 6s + s 6(s? )(s? 4) : In order for this to be equal to zero r = s? s? 6 = p 7? 3. We must also have h ; i =. From the above remarks we nd that, j [ ;] = f [;p;] (?), while j [ ;] = f?s [p; +p+s p (?), and ;] j [; 3 ] = f [; +q ;q](?), where from (3.5) q = 3s +s? 4s = p 7? 3, and from (3.4) p =?s +8 s = p 7?4. From (.9) we nd after change of?4 6 variables, h ; i = Z f [;p;] (x)f [p;?s +p+s p ;] (x)dx + = (7 + 5p 7)(6r? 4 p 7 + 5) : 756 Z f [;r;] (x)f [; +q ;q](x)dx Thus in order for the above integral to be equal to zero r = 4p 7?5 6 which cannot be. This leads to Theorem 4.3. For any pair (s ; s ) with p(s ; s ) = ; js j <, and js j < let be a wavelet. Then the support of one component of must be of length 3= while the support of the other component must be of length. Proof. Lemma 4. shows that both components of must have supports with lengths at least 3/. It is easy to see that not both may be supported on [; 3=] since in that case the pieces supported on [; 3=] or [; =] must be linearly dependent. A rotation of the components would then allow us to nd a wavelet whose support length is which would contradict Lemma 4.. A similar argument shows that both components of may not have support [=; ]. 5

27 Suppose =? where supp = [; 3=] and supp = [=; ]. In this case the matrices in (4.) would have the form d d D = ; d7 D = ; d 9 d 8 d3 d D = 4 ; d 5 d 6 D 3 = : d d Set A = C D +C D 3, B = C D +C 3 D and C = C D +C D +C D +C 3 D 3, then (3.8) implies that A = =(s +?p)d 7, A = =(?p)(p?s )d 7, B = (?q)(q?s )d 5 and C = =(s +? q)d 5. If d 7 is not to be zero then the above equations imply that p = = and s =. However this would make (3.4) become = = 4+4s 6?4s which has no solution for js j <. Likewise if d 5 is not equal to zero then q = and s =?= which is outside the region of admissible pairs. Therefore d 5 = d 7 =. But in order for h ; i = either d 4 or d 6 must be equal to zero. Hence one of them must have a support of length which is impossible by Lemma 4.. Although it has been shown by Daubechies [5] that with one scaling function compactly supported continuous wavelets cannot be symmetric, this is not the case with two scaling functions. In order to obtain wavelets supported on [,] that are symmetric or antisymmetric with respect to one it must be that s = s. To see this suppose that is such a wavelet with support [,]. Since ~ V it follows using (.) and (.3) that for x [; ]; ( x ) = a x + s (x), and (? x ) =?a 8x + s (? x). The symmetry of implies that ( n ) = (? n ) where the plus sign is used if is symmetric with respect to one and the minus sign if it is antisymmetric with respect one. Comparing these equations we see that s = s which implies that both s values are equal to?=5. Suppose that one wavelet s is symmetric while the other a is antisymmetric with respect to one and set, y = [; a ; a ], y = [a ; a 3; a 4], y 3 = [a 4; a 3; a ], y 4 = [a ; a ; ], and y = [; a ; a ], y = [a ; a 3; ], y 3 = [;?a 3;?a ], y 4 = [?a ;?a ; ], then equations similar to (4.){(4.5), ((4.6) is automatically satised) give a =, a =?3, a 3 =, a 4 =?, and a =, a =?3, a 3 = 8. Here k s k L 6 = p k a k L. If s and a

28 =?3 p =?9 p = 3= Figure 4 normalize to then the corresponding matrices in (4.) are p p?=?3 = D =? p 9=?= ; D =?3= = 9 p = D = ; D 3 = p?= : = These wavelets are plotted in Figure 5. Consequently s and a individually exhibit linear phase. A similar computation where both wavelets are assumed to be symmetric with respect to one and supported on [,] yields no solution. Finally we note that the multiresolution analysis arising from AFIF is well suited for compact intervals. In fact, if we let V k = ~ V k \ L [; ] and ~ i k;j = i k;j j [;] then fv k g 7 ;

29 s a Figure 5 provides a multiresolution analysis for [; ] and ~ i k;j, j ZZ, i = ; is an orthogonal basis for Vk. To see what to do with the wavelets write j i = i j [j?;j], and i j = i j [j?;j], i = ;, j = ;. We now rotate the wavelets i.e., +; = a + b ; +; =?b + a (4:6) with jaj + jbj = so that h +; j ; ji = ; j = ; : (4:7) To see that this can be done note that (4.7) is equivalent to ah ; i + bh ; i = ; (4:8) 8

30 and ah ; i + bh ; i = : (4:9) However the fact that h ; i = h ; i+h ; i = and h ; i = h ; i+h ; i = shows us that (4.8) and (4.9) are not independent which allows us to construct the desired rotation. With +; and +; constructed as in (4.6) set ~ k;j = ;+ k;j j [;] and ~ k;j = ( if +; k;j \ [; ]c 6= ; otherwise +; k;j (4:3) then the non-zero components of f ~ k;j g jzz form an orthogonal basis for W k ; k where V = V k+ k W k. This leads to Theorem 4.4. f ~ i k;j = i k;j j [;],k,i = ; ;?i j k? g is an orthogonal basis for V k = V k \ L [; ] while f ~ i k;j ; k ; i = ; ; i? j k? (i + )g forms an orthogonal basis for W k. Furthermore L k W k = L [; ]. For the case where s = s =?=5 it is easy to see that ~ k;j are just the symmetric wavelets restricted to [; ] i.e., ~ k;j = s k;j j [;] while ~ k;j = a k;j if the support of a k;j [; ] and otherwise. V. Scaling by Other Integers Many of the results of the previous sections can be used to produce scaling functions and wavelets satisfying dilation equations of the form (x) = X C i (Nx? i); and X (x) = D i (Nx? i); with N >. Note that in this case the matrices D i will not in general be square matrices. We begin by considering the N + dimensional basis ff yi g N spanning i= S where y i = e i, < i N, fe i g N being the standard basis in i= RN + and last vector y is given by y = [; q ; q ; : : : ; q N? ; ]. What is needed is to adjust q i, i N? and s j, j N so that f y is nonzero and orthogonal to f yi, i N. Once this has 9

31 been accomplished, orthogonal scaling functions i, i N? can be obtained by applying the Gram-Schmidt procedure to the set ff yi g N i=. The functions i, i N? obtained from the functions ff yi g N? i= will be continuous and supported on [; ] since each f yi, i N? vanishes at zero and one. The last function N? can be obtained by subtracting from f yn its projection onto the subspace spanned by f i g N? i= then piecing it together continuously with f y as was done in the case N = in Section 3. That it is sucient to consider only a basis for S of this type follows from Theorem 5.3. Since there are N orthogonality relations and N? unknowns (q i, i N?, and s j, j N? ), it may be possible to impose other desirable conditions besides orthogonality and still obtain the required basis f i g N? i=. If (x) = ( ; ; : : : ; N? ) then will satisfy (x) = P N? i= C i (Nx? i). (x) may now be obtained from (x) using Theorem (4.) or the orthogonality equations (.9) and solving as in Section 4. In order to compute the orthogonality relations hf yi ; f y i =, i N, i j (x) = a i j x + bi j, i N, j N? need to be computed. From (.), (.), (.3) and (.5) we nd that a i j = i?;j? i;j? s j N;i and b i j = i;j for i N and j N? with q = and q N =. Also a j = q j+? q j + s j, b j = q j? s j for j N?. If S = P N? i= s i and S = P N? i= is i it follows from (.7) and (.8) that m i = N?S, m i = (N?S )i+s (N?S )(N?S ) for i N?, mn =?S (N?S ), m N = N(3N? ) + (? 5N)S + 3(? N)S + S 6(N? S )(N? S ) ; P m =? S N? + (N? S ) i= q i ; and P m = S N? (S? (N + ))? 3(N? )S + N + 6 q i= i((n? S )i + S ) : 6(N? S )(N? S ) 3

32 With the above moments (.9) yields and I n; = hf yn ; f y i X N? = (s n?? s n )m + (m n? m n ) s i (q i? s i ) + m n N? X i= i= s i? q i + s n m? s n? + s n (q n? + 4q n + q n+ ) I N; = m s N?? X N? + i= N? X i= s i!! = N? N? X i= s i! ; n N? ; X N? + (m N? m N? =6) s i (q i? s i ) s i? q i (m N? =3) + q N?? s N? 6 i=!. N? N? X i= s i! : (5:) (5:) For N = (5.) and (5.) yield (3.) and (3.3) with p =. For N = 3 (5.) and (5.) become more complicated and we shall restrict ourselves to considering s values that give symmetric or antisymmetric wavelets. Just as when N =, we require that the s-values be arranged symmetrically, so s must be the same as s. In this case a one parameter family of continuous, compactly supported, orthogonal scaling functions each with linear phase will be produced. We proceed as before, examining what conditions S must satisfy for compactly supported scaling functions to exist. Let be the space of vectors vanishing on both sides, and be those vanishing on the left. Then is a -dimensional space, and has a basis of the form x = [; ; ; ] and x = [; ;?; ]. Let x 3 = [; p ; p ; ] be a vector orthogonal to in, so that fx ; x ; x 3 g forms an orthogonal basis for. Now, if we can nd x 4 = [; q ; q ; ] orthogonal to, then we easily generate the scaling functions fx on [; ] = elsewhere = fx on [; ] elsewhere and = 8 < : f x3 on [; ] f x4 (? ) on (; ] elsewhere. If there exists no such x 4, then there are no compactly supported scaling functions as we can see from the following general result 3

33 Theorem 5.. Let V be an N-dimensional subspace of C [; ] such that there does not exist an orthonormal basis with N? of the basis functions vanishing at zero and the remaining function vanishing at one, or vice versa. Then there does not exist compactly supported C (IR) functions f j g N? j= composed of linear combinations of basis elements of V and their integer translates constructed so that h j (x); i (x? l)i = j;i ;l. Proof. The proof is by induction on N. The case N = 3 was established in Lemma 3.5. We only consider the step from N = 3 to N = 4 as the general argument is similar. For ease of notation set x = ; y =, and z = 3 and suppose that each is supported on a subset of [; M]. Suppose V satises the hypotheses above then there are three cases we need to consider: Case : One of the functions (say, x) is supported on [; ], or x, y, and z can be transformed by a nite number of rotations and shifts so that this is the case. First, perform this transformation, if necessary. We see that the components of y and z are restricted to a 3- dimensional space, which has no orthonormal basis of vectors with two functions vanishing on the left and one vanishing on the right or vice versa. Thus by Lemmas 3.4 and 3.5, the scaling functions y and z cannot exist. Case : All three functions have support [; ] or longer, and the leftmost components, fx ; y ; z g, have rank. In this case, we may begin to apply the rotate and shift strategy of Lemma 3.5 on y and z. If for some nite j, y (j) and z (j) are not linearly dependent, then we go to Case 3, below. Otherwise, the proof of Lemma 3.5 applies, and we are done. Case 3: All three functions have support [; ] or longer, and the leftmost components, fx ; y ; z g, have rank. Note that fx ; y ; z g cannot have rank 3 because then the rightmost nonzero component of any of them would be orthogonal to all functions vanishing on the left, which we assumed was impossible. Here, we apply an argument similar to the one found in Lemma 3.5 using two rotation maps instead of one. Many of the details in this proof are analogous to those in the proof of Lemma 3.5, so they are omitted here. Two among x, y, and z must be linearly independent, so assume y and z are they. We can rotate y and z so that y and z are orthogonal, and for an appropriate basis, we 3

34 have x = ((x ; ; x ; ; ; ); x ; x 3 ; : : : ; x M ) y = ((y ; ; ; ; ); y ; y 3 ; : : : ; y M ) z = ((; z ; ; ; ); z ; z 3 ; : : : ; z M ) where y ; > and z ; >. As before, we have a rotation r which transforms x and y into x = ((; x ; ; ; ); x ; : : : ; x M ) y = ((y ;; y ;; ; ); y ; : : : ; y M) leaving z unchanged and r which transforms x and z into x = (; x ; : : : ; x M ) z = ((; z ; ; ); z ; : : : ; z M ) leaving y unchanged. We also have the shift map s, which takes x to (x ; x 3; : : : ; x M ; ) = x leaving y and z unchanged. The resulting vectors x, y, and z are themselves orthogonal scaling functions, and thus the leftmost components cannot have rank 3. Clearly, y and z are linearly independent, so their rank is. Now, we iterate the map s r r to obtain the sequence (x (j) ; y (j) ; z (j) ) = (s r r ) j (x; y; z), which must have some limit point, (X; Y; Z). Analogously to Lemma 3.5, both fy (j) ;g and fz(j) ;g are monotone increasing from which it follows that X k; = X k; = for k = ; ; : : : ; M and hence that X =, which we know is not possible. To proceed to higher N we need only consider the analog of Case 3 above since the other cases are eliminated by the induction hypothesis. The analog of Case 3 is when all N functions have support [; ] or longer and the dimension of the space spanned by the leftmost components of these functions is N?. In this case N? rotations are needed to reduce by iteration one of the functions to zero thereby forcing a contradiction. Next, we investigate what conditions must be imposed on s and s in order that such a basis should exist. The vectors [; ; ; ], [; ; ; ], and [; ; ; ] form a basis (not orthonormal) for, so it suces to check for orthogonality between x 3 and each of these vectors. From (5.) and (5.) we nd (where y i ; i = ; ; ; 3 are used), after some 33

35 manipulation, I ; = 8s s q? 8q? 7q + s + 4s + 6s q? 48s q? 8s s? 4s q? 4s q + 76s + 3s + 8s s q? 4s 3? 6s 3? 7 + s q? s s? s s? s q + 8s q? 6q s = ; I ; = 8s s q? 7q? 8q + s + 4s? 48s q + 6s q? 8s s? 4s q? 4s q + 8s + 6s + 8s s q? 6s q + 8s q? s q + q s = ; I 3; =?8s s q? 7q + 48s + s q + s q? 4s s + 4s q + 4s q? 5s? s? 8s s q? s s q? 3s 3? 8s s q? 6s s? 6s s + 3sq + 6s q + 8q s? 4s 3 q + 8s 3 s + 6s s + 4s 3 s? 6s 3 q? s s q + 8s 4 + s4 = : Solving the rst two for q and q gives q = 3s3? 76s? 3s s + 4s s + 6s s? 4s + 36? 56s? 6s 3 ; (6s + s + 45) (s? s? 3) q = 4s3? s? 4s s + s s + s s? 36s? 3s? 9s? 9? s 3 : (6s + s + 45) (s? s? 3) If we substitute these expressions into the third equation and simplify, we get the desired relationship, = 48s 4? 56s 3 + 6s s + 9s s + 4s + 6s s 3? 96s s? 7s s? 9s + s 4 + 6s3 + 7s + 48s + 9: Let the polynomial on the right be denoted by ^p(s ; s ). The zero set of ^p is shown in Figure 6. From the above, it is now clear that Theorem 5.. Continuous, compactly supported, orthogonal scaling functions f j g j= can be constructed so that this set and its integer translates span V ~ (N = 3; s = s ) if and only if js j <, js j <, and ^p(s ; s ) =. 34