2 Infinite products and existence of compactly supported φ

Transcription

1 415 Wavelets 1 Infinite products and existence of compactly supported φ Infinite products.1 Infinite products a n need to be defined via limits. But we do not simply say that a n = lim a n N whenever the limit exists. (For series we do simply do that and define t n = lim N N t n.) This overly simple definition of products would allow (for example) (n 1) to converge. It represents and all the partial products N (n 1) = 0. But (n 1) = n= does not converge no finite limit for (n 1) = 1 (N 1) = (N 1)! n= Definition. An infinite product a n (where a n C n) is said to converge if there are at most finitely many n with a n = 0, say n 1, n,..., n k, and lim a n N n=n k +1 exists (as a finite limit in C) and is nonzero. (The same definition works for a n R). Example.3 has partial products and so this product is not convergent. 1 n 1 n = 1 N! 0 as N

2 415 Wavelets Proposition.4 If a n converges, then lim a n = 1 n Proof. Let us say that all the zero terms a n have n n 0. Then Also lim n n=n 0 +1 lim a n = p (say) with p 0 N+1 n n=n 0 +1 a n = p and dividing the latter by the former we have 1 = p N+1 p = lim n=n 0 +1 a n n N n=n 0 +1 a n Theorem.5 Assuming a n 1 for all n = lim N a N+1 a n converges (a n 1) < Proof. Note that a n = P N (say) is an increasing sequence of positive numbers and so it has a finite limit if and only if it is bounded above. Put S N = N (a n 1). S N is also increasing with N (as a n 1) and again lim N S N R if and only if S N is bounded above. Observe that 1 + x e x for all x 0 and so 1 + (a n 1) e an 1 a n e an 1 e an 1 = e N (an 1) a n P N e S N

3 415 Wavelets 3 But we also have P N = = a n (1 + (a n 1)) 1 + N (a n 1) = 1 + S N Hence 1 + S N P N e S N Thus (S N ) N=1 is bounded above if and only if (P N) N=1 and therefore lim N S N < lim N P N <. is bounded above Definition.6 An infinite product a n is called absolutely convergent if (1 + a n 1 ) is convergent. Lemma.7 An infinite product a n is convergent if and only if it obeys Cauchy s criterion for convergence of products: for each ε > 0 there exists N 0 so that N N N 1 N 0 a n 1 < ε n=n 1 Proof. Write Π l,m = m a n (m l) n=l Assume the product is convergent. We know that there exists n 0 1 and p 0 so that lim Π n 0,N = p N Given ε > 0, there exists N 0 > 0 so that ( ε p N N 0 Π n0,n p < min, p ) (1)

4 415 Wavelets 4 For N N 0, we have then ( 1 p Π n 0,N 1 ε < min, 1 ). Since z 1 < 1/ 1/ < z 1 <, we deduce that z 1 (1/p)Π n0 <,N and hence for N N 1 > N 0 we get Π N1,N 1 = Π n0,n 1 Π n0,n 1 1 = (1/p)Π n0,n 1 (1/p)Π n0,n 1 1 = (1/p)Π n0,n (1/p)Π n0,n 1 1 (1/p)Π n0,n 1 1 = (1/p)Π n0,n (1/p)Π n0,n 1 1 (1/p)Π n0,n 1 1 (1/p)Π n0,n (1/p)Π n0,n 1 1 ( (1/p)Π n0,n (1/p)Π n0,n 1 1 ) ( < ε ) = ε. Thus the Cauchy condition holds. Conversely, assume that the Cauchy condition holds. Then there is n 0 so that N N 1 n 0 Π N1,N 1 < 1/. In particular, for N > n 0 Π n0,n 1 < 1 () Note that z 1 < 1 z < 3/ <. Now, given any ε > 0 we can find n 1 (and we can assume that n 1 n 0 ) so that N N 1 n 1 Π N1,N 1 < ε Then Π n0,n Π n0,n 1 = Π n0,n 1 Π n0,n 1 Π n0,n 1 Π N1 +1,N 1 < ε = ε (3)

5 415 Wavelets 5 as long as N > N 1 > n (and it is valid for N = N 1 also). This says that the sequence (Π n0,n) N=n 0 satisfies the usual Cauchy condition for sequences. Hence lim N Π n 0,N exists in C. By (), the limit is not zero. Hence the product converges. Theorem.8 Absolutely convergent infinite products a n are convergent. Proof. Let b n = a n 1, and Note that Π l,m = Q l,m = m a n = n=l m (1 + b n ) n=l m (1 + b n ) (m l). n=l Π l,l 1 = (1 + b l ) 1 = b l = Q l,l 1. By induction on m and for l fixed, we can verify that Here is the induction step Π m,l 1 Q m,l 1 (m l). Π m+1,l 1 = (1 + b m+1 )Π l,m 1 = (1 + b m+1 )(Π l,m 1) + (1 + b m+1 ) 1 = (1 + b m+1 )(Π l,m 1) + b m+1 Q m+1,l 1 = (1 + b m+1 )(Q l,m 1) + b m+1 Π m+1,l 1 (1 + b m+1 ) Π l,m 1 + b m+1 (1 + b m+1 ) (Q m,l 1) + b m+1 by the inductive hypothesis = Q m+1,l 1 Thus the fact that (1 + b n ) satisfies the Cauchy criterion for products implies the same of (1 + b n). The result follows from Lemma.7.

6 415 Wavelets 6 Example.9 (1 + z ) n converges no matter what value z C is given. To justify this example, it is sufficient (by Theorem.8) to show that the product is absolutely convergent, that is that ( 1 + z ) n converges. But the convergence of this follows from z n < and Theorem.5. To deal with infinite products where the terms are functions, we introduce a notion of uniform convergence. Definition.10 If f n : S C are complex-valued continuous functions defined on some compact set K, then the product (1 + f n (z)) is called uniformly convergent on K if there exists n 0 so that f n (z) 1 for n n 0 and there is a nowhere-vanishing function g: K C \ {0} so that given any ε > 0 there exists n 1 n 0 so that N n 1, z K (1 + f n (z)) g(z) < ε n=n 0 and inf g(z) > 0. z K Thus, we require not only that (after omitting a finite number of initial terms) the limits of the partial products be uniform but also that the limit be nonzero in a uniform way.

7 415 Wavelets 7 Proposition.11 If (1 + f n(z)) is a uniformly convergent product of continuous functions on a compact set K, then the sequence of partial products Π 1,N (z) = (1 + f n (z)) converges uniformly on K (as N ) to Π(z) = (1 + f n (z)) Proof. From the definition we know that there is n 0 and g: K C so that Π n0,n(z) = n=n 0 (1 + f n (z)) converges uniformly on K to g(z). By compactness of K and continuity of the f n sup Π 1,n0 1(z) = M < z K and then it follows easily that Π 1,N (z) = Π 1,n0 1(z)Π n0,n(z) Π 1,n0 1(z)g(z) = Π(z) uniformly on K. In fact, given ε > 0 we can choose N 1 > n 0 so that N N 1, z K Π n0,n(z) g(z) < ε M + 1 and then it follows that for N N 1, z K we have Π 1,N (z) Π(z) = Π 1,n0 1(z) Π n0,n(z) g(z) M ε M + 1 < ε The restriction to products of continuous functions on a compact set which we made in Definition.10 is used in the proof of the proposition. Without continuity and compactness, we could still have the result of the proposition by insisting only on uniform boundedness of each f n, but the applications are usually to the case in the definition. Exercise. Show that if a K is compact (in C, say) and f n : K C is a sequence of continuous functions so that (1 + f n (z))

8 415 Wavelets 8 is uniformly convergent on K, then Π(z) = (1 + f n(z)) is continuous on K and Π(z) = 0 n such that 1 + f n (z) = 0. As in the case of infinite products of constants, we can show that there is a Cauchy criterion for uniform convergence. Lemma.1 An infinite product (1+f n(z)), with continuous functions f n : K C on a compact set K, is uniformly convergent if and only if it satisfies the uniform Cauchy condition: given any ε > 0 there exist N 0 1 so that N N N 1 N 0, z K (1 + f n (z)) 1 < ɛ n=n 1 Proof. Copy the proof of Lemma.7 almost word for word, except that (1) has to be replaced by ( ε N N 0 Π n0,n g(z) < min, 1 ) inf g(w) w K Theorem.13 For continuous functions f n : K C on a compact set K, if (1 + f n (z) ) is uniformly convergent on K, then (1 + f n (z)) is uniformly convergent on K. Proof. Copy the proof of Theorem.8. Corollary.14 (M test) For continuous functions f n : K C on a compact set K, if M n = sup z K f n (z) < for all n and then is uniformly convergent on K. M n <, (1 + f n (z))

9 415 Wavelets 9 Proof. From M n < and Theorem.5 we deduce that (1 + M n) converges. The Cauchy condition for (1 + M n) implies the uniform Cauchy criterion for (1+ f n(z) ) and the result then follows from Theorem.13. Proposition.15 If m(ξ) is a trigonometric polynomial with m(0) = 1, then ( ) ξ m n converges uniformly on finite intervals [a, b] R. Proof. As trigonometric polynomials are differentiable and periodic with period 1 M = sup m (ξ) = sup m (ξ) < 0 ξ 1 ξ R It follows that m(ξ 1 ) m(ξ ) = and in particular that m (ξ) dξ M xi 1 ξ ξ=ξ ξ1 m(ξ) 1 = m(ξ) m(0) M ξ Now for any finite interval [a, b] we can find R > 0 so that [a, b] [ R, R] and then for ξ [a, b] we have ( ) ξ m 1 n M ξ MR n n As MR n < uniform convergence of the product follows from Corollary.14. Example.16 If then ( ) ξ m = n 1 0 m(ξ) = 1 + e πiξ e πixξ dx = F(χ [0,1) )(ξ) = e πiξ sin πξ πξ

10 415 Wavelets 10 Proof. By induction on N one can show that 1 + e πi(ξ/n ) = 1 N N 1 j=0 e πijξ/n which is a Riemann sum for the integral. It is easy to check that the integral is e πiξ 1 = e πiξ 1 e πiξ e πiξ πiξ sin πξ = e πiξ πξ i πξ Alternatively, one can sum the above sum as a geometric progression and arrive at the same limit. Theorem.17 If p(ξ) = k 1 k=k 0 c k e πikξ is a trigonometric polynomial with p(0) = and p(ξ) + p(ξ + 1/), then ˆφ(ξ) = p(ξ/ n ) defines a function ˆφ L (R). If in addition p(ξ) 0 for ξ < 1/4, then ˆφ = 1 and ˆφ(ξ + l) 1 for almost all ξ R. l Z Proof. Let m(ξ) = p(ξ)/. From Proposition.15, we know that the product for ˆφ converges uniformly on all finite intervals [a, b] R and thus ˆφ is a continuous function. Define a sequence of functions ˆφ r (r = 0, 1,,...) by and ˆφ 0 (ξ) = χ [ 1/,1/) (ξ) ˆφ r+1 (ξ) = m(ξ/) ˆφ r (ξ/) ( r+1 ( ) ) ξ = m χ j [ 1/,1/) j=1 r+1 ( ) ξ = χ [ r, r )(ξ) m j j=1 ( ) ξ r+1

11 415 Wavelets 11 We claim first that Φ r (ξ) = l Z ˆφ r (ξ + l) satisfies Φ r (ξ) 1 for all r = 0, 1,,.... This is clear for r = 0 and for r 0, separating even and odd terms, we get Φ r+1 (ξ) = l Z = l Z + l Z ˆφ r+1 (ξ + l) + l Z m(ξ/ + l) ˆφ r (ξ/ + l) = m(ξ/) l Z ˆφ r+1 (ξ l) m(ξ/ + 1/ + l) ˆφ r (ξ/ + 1/ + l) ˆφ r (ξ/ + l) + m(ξ/ + 1/) ˆφ r (ξ/ + 1/ + l) l Z since m is periodic with period 1 = m(ξ/) Φ r (ξ/) + m(ξ/ + 1/) Φ r (ξ/ + 1/) = m(ξ/) + m(ξ/ + 1/) = 1 (assuming Φ r 1 known). It then follows that ˆφ r = 1 for each r, because ˆφ r (ξ) dξ = χ [l,l+1) ˆφ r (ξ) dξ l Z = χ [l,l+1) ˆφ r (ξ) dξ l Z (by the monotone converegence theorem) = l Z l+1 l 1 ˆφ r (ξ) dξ = ˆφ r (η + l) dη l Z 0 1 = ˆφ r (η + l) dη = 0 l Z (by the monotone converegence theorem) 1 0 Φ r (η) dη = 1

12 415 Wavelets 1 Now we can argue that R R ˆφ(ξ) R dξ = lim ˆφ r (ξ) dξ r R by uniform convergence on [ R, R] and the limit is at most 1 because each term is at most 1 (the r th term in the limit is less than ˆφ r = 1). As for each R > 0, it follows that R R ˆφ = ˆφ(ξ) dξ 1 ˆφ(ξ) dξ 1 Assuming now that p(ξ) 0 for ξ < 1/4, we have ˆφ(ξ) 0 for ξ < 1/. Since ˆφ is continuous, c = inf ξ 1/ ˆφ(ξ) > 0. Then for ξ < r 1, we have p(ξ/ ˆφ(ξ) n ) r p(ξ/ n ( ) r ) p(ξ/ n ) p(ξ/ n ) = = = ˆφ(ξ/ r ) and so ˆφ r r (ξ) = n=r+1 p(ξ/ n ) ˆφ(ξ) / ˆφ(ξ/ r ) 1 c ˆφ(ξ) for ξ r 1 and also for ξ > r 1 (when ˆφ r (ξ) = 0). As we now have (1/c) ˆφ(ξ) dξ (1/c) < and ˆφ r (ξ) (1/c) ˆφ(ξ), we can apply the dominated convergence theorem to show which tells us Finally, we have N l= N lim ˆφ r (ξ) dξ = lim r ˆφ(ξ + l) = lim r ˆφ(ξ) dξ = 1 N r l= N lim sup r ˆφ r (ξ) dξ = 1 ˆφ r (ξ + l) l= = lim sup Φ r (ξ) = 1 r ˆφ r (ξ + l)

13 415 Wavelets 13 and hence But then 1 = l= ˆφ(ξ) dξ = ˆφ(ξ + l) l= ˆφ(ξ + l) dξ = 1 and so l= ˆφ(ξ + l) = 1 for almost all ξ [0, 1]. As the sum is periodic with period 1, and a countable union of sets of measure zero still has measure zero, the sum is 1 for all ξ R except possibly for a set of measure zero. Corollary.18 If (c k ) k Z is a finitely nonzero sequence with { 1 for l = 0 c k c k l = 0 for l Z, l 0. k Z k c k = and k c ke πikξ 0 for ξ < 1/4, then there is a function φ L (R) with φ = 1, φ = k Z c k D T k φ and orthonormal translates {T k φ : k Z}. Proof. We know from Lemma 1.13 that p(ξ) = k c ke πikξ satisfies p(ξ) + p(ξ + 1/) and also we have p(0) = k c k =. From Theorem.17, we have p(ξ) ˆφ(ξ) = L (R) with ˆφ = 1. If we define φ to be the inverse Fourier transform of ˆφ, then we have φ = ˆφ = 1 and Fφ(ξ) = ˆφ(ξ) = p(ξ/) 1 ˆφ(ξ/) = p(ξ/) 1 Fφ(ξ/) ( ) = F c k D T k φ (ξ) k Z

14 415 Wavelets 14 Since F is bijective on L (R), we can conclude that φ satisfies the dilation equation. Finally, for j k T j φ, T k φ = T j T j φ, T j T k φ since T j is an isometry = φ, T k j φ = Fφ, F(T k j φ) = = since the Fourier transform is an isometry on L (R) = l Z ˆφ(ξ)e πi(k j)ξ ˆφ(ξ) dξ ˆφ(ξ) e πi(k j)ξ dξ l+1 l 1 ˆφ(ξ) e πi(k j)ξ dξ = ˆφ(ξ + l) e πi(k j)ξ dξ l Z 0 1 = ˆφ(ξ + l) e πi(k j)ξ dξ = = 0 0 l Z by the dominated convergence theorem, since ˆφ(ξ + l) = 1 ae l Z 1 0 e πi(k j)ξ dξ Our next aim is to show that the solution φ of the dilation equation that we have constructed is compactly supported (and hence in L 1 (R) as well as L (R)). There are a number of ways to do this directly, but we will take an approach that relies on a classical result in Fourier analysis that we will not prove. By a compactly supported function in L (R) we mean one that is almost everywhere zero outside some bounded closed interval [a, b]. Recall that an entire function g is a complex-valued function defined and holomorphic (or analytic) on the whole complex plane C. That means that g (w 0 ) = lim w w 0 g(w) g(w 0 ) w w 0

15 415 Wavelets 15 exists in C for each w 0 C. An entire function is said to be of exponential type if there are constants A, B > 0 so that g(w) Ae B w holds for all w C. Examples of such functions are g(w) = cos w = (e iw e iw )/ and polynomial functions g(w) = n j=0 a jw j (which satisfy g(w) a(1 + w ) n Ae B w for a suitable a > 0 and any B > 0 with a suitable A). Theorem.19 (Paley-Wiener) A function f L (R) is compactly supported if and only if its Fourier transform Ff(ξ) is almost everywhere equal for ξ R to g(ξ) for an entire function g: C C of exponential type. Proof. Omitted. Theorem.0 If (c k ) k Z is a finitely nonzero sequence with { 1 for l = 0 c k c k l = 0 for l Z, l 0. k Z k c k = and k c ke πikξ 0 for ξ < 1/4, then there is a compactly supported function φ L (R) with φ = 1 and φ = k Z c k D T k φ Proof. What we do is show that the Fourier transform of the φ constructed in the proof of Corollary.18 above satisfies the conditions in Theorem.19. Let m(w) = (1/ ) k c ke πikw (with w C now replacing ξ R) and g(w) = ( w ) m n A look at the proof of Proposition.15 will show that we did not really make use of the fact that ξ was real in any essential way. The same argument shows that the product for g(w) is uniformly convergent for w R, any R > 0. Thus g: C C is well defined and continuous. In fact it is also entire because of a standard result in complex analysis that if a sequence of entire functions on C converges uniformly on compact subsets of C, then the limit is also entire. (In this case the partial products N m(w/n ) are clearly entire.) We have Fφ(ξ) = g(ξ) for ξ R and what remains is to establish the exponential type of g.

16 415 Wavelets 16 For w < 1 we know there is a constant M 0 so that m(w) 1 M w and so ( w ) g(w) = m n (1 + m(w/ n ) 1 ) (1 + M/ n ) = M < holds for w <. Now for w, choose N so that N w < N+1. Then ( N ( w ) ) ( w ) g(w) = m g n N and we need a bound on this finite product in terms of a power of e w. Suppose c k is zero outside the (finite) range k 0 k k 1. Consider for a moment the polynomial so that P (z) = (1/ ) k 1 k=k 0 c k z k k0 m(w) = e πik 0w P (e πiw ) Now, being a polynomial of degree d = k 1 k 0 we can find C > 0 so that and P (z) C for z < 1. Thus P (z) C z d for z 1 m(w) e π k 0 w C(e π w ) d = Ce π(d+ k 0 ) w holds for all w C (as e w = e Rw e w and e w 1 always). It follows that for N w < N+1 ( N ) g ( g(w) Ce π(d+ k 0 ) w / w ) n M C N e π(d+ k 0 ) w N+1 As N < log w / log we have C N = e N log C < e (log C/ log ) log w (log C/ log ) w e (since w 1). Thus for we have g(w) M e (log C/ log +π(d+ k 0 )) w

17 415 Wavelets 17 no matter what N is. Taking A = M and B = max(log C/ log + π(d + k 0 ), 0) we have g(w) Ae B w for all w C. Theorem.1 Suppose (c k ) k Z is a finitely nonzero sequence with c k = 0 for k outside the range k 0 k k 1, { 1 for l = 0 c k c k l = 0 for l Z, l 0. k Z k c k =, k even kc k = k odd kc k, and k c ke πikξ 0 for ξ < 1/4. Factor P 1 (z) = k 1 k=k 0 c k z k k0 = ( ) a 1 + z Q(z) for a polynomial Q(z) with Q( 1) 0 and a an integer. Assume sup{ Q(z) : z C, z = 1} = B < a 1 Then there is a compactly supported continuous function φ L (R) with φ = 1 and φ = c k D T k φ k Z Proof. From Lemma 1.13, we know that p(ξ) = e πik0ξ P 1 (e πiξ ) satisfies p(ξ) + p(ξ + 1/) =, p(0) = P 1 (1) = and consequently p(1/) = P 1 ( 1) = 0. The condition k even kc k = k odd kc k guarantees that the polynomial P 1 (z) has derivative 0 at 1. We conclude that P 1 (z) is divisible by (1 + z). Let a be the highest power of 1 + z which divides P 1 (z) and then we can factor P 1 (z) as in the statement (with a ). Our goal is to show that ˆφ(ξ) as constructed in Theorem.17 must be in L 1 (R). It follows that the inverse Fourier transform of ˆφ is a continuous function φ(x) = ˆφ(ξ)e πixξ dx and we already know from the proof of Corollary.18 that this satisfies the dilation equation. Take m(ξ) = p(ξ)/ ( ) = e πik 0ξ 1 + e πiξ a Q(e πiξ )

18 415 Wavelets 18 and, as in the proof of Theorem.17, ˆφ(ξ) = m(ξ/n ). From Proposition.15, we know that each of the three products ( ) e πik 0ξ/ 1 + e πiξ/ n n, and Q(e πiξ/n ) converges. The first product has constant modulus 1, the second has been πiξ sin πξ computed in Example.16 to be e. It is easy to see that the infinite product of a product is the product of the individual infinite products πξ provided each of the individual products converges. Thus ( ) ( ) a ˆφ(ξ) = e πik 0ξ/ n πiξ sin πξ e Q(e πiξ/n ) πξ To achieve our goal of showing ˆφ(ξ) dξ < we need an estimate on the last product. For convenience, let q(ξ) = Q(e πiξ ), a trigonometric polynomial. For 1 ξ 1 We have an estimate based on the largest value of the derivative M = sup ξ q (ξ) that q(ξ) 1 = q(ξ) q(0) M ξ and so for ξ 1 we have q(ξ/ n ) (1 + q(ξ/ n ) 1 ) (1 + M/ n ) < e M For ξ > 1 we argue in a somewhat similar way to the previous proof. Fix ξ. Choose N 0 so that N 1 < ξ N. Then ( q(ξ/ n ) = q(ξ/ n N ) ) q(ξ/ n ) q(ξ/ n ) e M n=n+1 But each q(ξ/ n ) = Q(e πiξ/n ) B and so q(ξ/ n ) BN e M But N < ξ and so B N e M < e M N log B < e M ( ξ ) log B. We conclude that, for ξ > 1 ˆφ(ξ) a 1 πξ e M ( ξ ) log B 1 = C ξ a log B

19 415 Wavelets 19 for some C. As we have the assumption a log B > 1, it follows that ˆφ(ξ) dξ < ξ >1 and 1 1 ˆφ(ξ) dξ is also finite (bounded by 1 Thus ˆφ L 1 (R). 1 sin πξ πξ a e M dξ < ). Remark. Theorem.1 can be used to show that there exist compactly supported continuous solutions (with orthonormal translates) to certain dilation equations. Daubechies produced a family of examples for a =, 3,.... The corresponding p a (ξ) = a 1 k=0 c ke πikξ is of the form p a (ξ) = ( ) 1 + e πiξ a q a (ξ) with q a (ξ) a trigonometric polynomial that is chosen in such a way that Now p a (ξ) + p a (ξ + 1/) =. p a (ξ) = p a (ξ)p a (ξ) ( ) 1 + e πiξ a ( ) 1 + e πiξ a = q a (ξ) ( ) + e πiξ + e πiξ a = q a (ξ) 4 ( ) a 1 + cos(πξ) = q a (ξ) ( ) sin a (πξ) = q a (ξ) = (1 sin (πξ)) a q a (ξ) and Daubechies chose q a so that Using a 1 ( a + k 1 q a (ξ) = k k=0 ) (sin (πξ)) k sin (π(ξ + 1/)) = sin (πξ + π/) = cos (πξ) = 1 sin (πξ)

20 415 Wavelets 0 we can show p a (ξ) + p a (ξ + 1/) a 1 ( a + k 1 = (1 sin (πξ)) a k k=0 a 1 ( a + k 1 +(sin (πξ)) a k =. k=0 ) (sin (πξ)) k ) (1 sin (πξ)) k For example in the case a =, this is true because (with s = sin (πξ)) it comes down to (1 s) (1 + s) + s (1 + (1 s)) = ((1 s + s )(1 + s) + s (3 s)) = (1 4s + s + s 3 + 3s s 3 ) =. For the Daubechies 4 coefficient example, writing z = e πiξ, we have p(ξ) = z z z3 = ( ) 1 (1 + z + z )((1 3)z + (1 3)) 8 using long division = ( ) ( 1 + z (1 3)z + (1 + ) 3) and so If we compute q(ξ) = q(ξ)q(ξ) q(ξ) = Q(z) = (1 3)z + (1 + 3) = 1 4 ((1 3) zz + (1 + 3) + (1 3)z + (1 3)z) = 1 4 ((4 3) + (4 + ) (z + z)) using z = e πiξ = 1 = cos(πξ) = (1 sin (πξ)) = 1 + sin (πξ) = 1 + s

21 415 Wavelets 1 (where s = sin (πξ) still). We can see then that B = sup{ q a (ξ) : ξ R} = (1 + s) = 3 < a 1 = 1 = sup 0 s 1 and by Theorem.1, the corresponding φ is continuous. So is the mother wavelet ψ constructed from φ. We should check all the hypotheses of Theorem.1. April 1, 001