A survey on frame theory. Frames in general Hilbert spaces. Construction of dual Gabor frames. Construction of tight wavelet frames

Transcription

1 A survey on frame theory Frames in general Hilbert spaces Construction of dual Gabor frames Construction of tight wavelet frames Ole Christensen Technical University of Denmark Department of Mathematics Building 303, 2800 Lyngby Denmark 1

2 1 Goal and scope (H,, ): Hilbert space. Want: Expansions f = c k e k of signals f H in terms of convenient building blocks c k. Desirable properties could be: Easy to calculate the coefficients c k Only few large coefficients c k for the relevant signals f. 2

3 Part I: Expansions in general Hilbert spaces Bases; Shortcomings of bases; Frames; Central themes: tight frames, pairs of dual frames. Part II: Expansions in L 2 (R) of the (i) Gabor type: e k e 2πimbx g(x na), m, n Z, for some g L 2 (R), a, b > 0; (i) Wavelet type: e k 2 j/2 ψ(2 j x k), j, k Z, for some ψ L 2 (R). Central themes: Why are frames needed? Explicit construction of tight frames and dual frame pairs. 3

4 Operators on L 2 (R): Translation by a R: T a : L 2 (R) L 2 (R), (T a f)(x) = f(x a). Modulation by b R E b : L 2 (R) L 2 (R), (E b f)(x) = e 2πibx f(x). Dilation by a 0: D a : L 2 (R) L 2 (R), (D a f)(x) = 1 a f( x a ). With this notation, the Gabor system can be written {E mb T na g} m,n Z = {e 2πimbx g(x na)} m,n Z. The wavelet system has the form (D := D 1/2 ) {D j T k ψ} j,k Z = {2 j/2 ψ(2 j x k)} j,k Z. 4

5 1.1 The Fourier transform For f L 1 (R), the Fourier transform is defined by Ff(γ) = ˆf(γ) := f(x)e 2πixγ dx, γ R. The Fourier transform can be extended to a unitary operator on L 2 (R). Plancherel s equation: ˆf, ĝ = f, g, f, g L 2 (R), and ˆf = f. Important commutator relations: FT a = E a F, FE a = T a F, FD a = D 1/a F, FD = D 1 F. 5

6 2 Bessel sequences Let H be a separable Hilbert space. Definition 2.1 A sequence {f k } k=1 in H is called a Bessel sequence if there exists a constant B > 0 such that f, f k 2 B f 2, f H. k=1 Theorem 2.2 Let {f k } k=1 be a sequence in H, and B > 0 be given. Then {f k } k=1 is a Bessel sequence with Bessel bound B if and only if T : {c k } k=1 c k f k k=1 defines a bounded operator from l 2 (N) into H and T B. Corollary 2.3 If {f k } k=1 is a Bessel sequence in H, then k=1 c kf k converges unconditionally for all {c k } k=1 l2 (N). 6

7 Pre-frame operator associated to a Bessel sequence: T : l 2 (N) H, T {c k } k=1 = c k f k k=1 T is also called the synthesis operator. The adjoint operator - the analysis operator: T : H l 2 (N), T f = { f, f k } k=1. The frame operator: S : H H, Sf = T T f = f, f k f k. k=1 The series defining S converges unconditionally for all f H by Corollary

8 3 Various bases Definition 3.1 Consider a sequence {e k } k=1 of vectors in H. (i) The sequence {e k } k=1 is a basis for H if there for each f H exist unique scalar coefficients {c k (f)} k=1 such that f = c k (f)e k. k=1 (ii) A basis {e k } k=1 is an unconditional basis if the series in (i) converges unconditionally for each f H. (iii) A basis {e k } k=1 is an orthonormal basis if {e k } k=1 is an orthonormal system, i.e., if e k, e j = δ k,j. 8

9 3.1 Orthonormal bases Theorem 3.2 For an orthonormal system {e k } k=1, the following are equivalent: (i) {e k } k=1 is an orthonormal basis. (ii) f = k=1 f, e k e k, f H. (iii) f, g = k=1 f, e k e k, g, f, g H. (iv) k=1 f, e k 2 = f 2, f H. (v) span{e k } k=1 = H. (vi) If f, e k = 0, k N, then f = 0. Corollary 3.3 If {e k } k=1 is an orthonormal basis, then each f H has an unconditionally convergent expansion f = f, e k e k. k=1 9

10 Theorem 3.4 Let {e k } k=1 be an orthonormal basis for H. Then the orthonormal bases for H are precisely the sets {Ue k } k=1, where U : H H is a unitary operator. Proposition 3.5 Assume that {e k } k=1 is a sequence of normalized vectors in H and that f, e k 2 = f 2, f H. k=1 Then {e k } k=1 is an orthonormal basis for H. 10

11 3.2 Riesz bases Definition 3.6 A Riesz basis for H is a family of the form {Ue k } k=1, where {e k} k=1 is an orthonormal basis for H and U : H H is a bounded bijective operator. Theorem 3.7 If {f k } k=1 = {Ue k} k=1 is a Riesz basis for H, there exists a unique sequence {g k } k=1 in H such that f = f, g k f k, f H. k=1 The sequence {g k } k=1 is also a Riesz basis, and the series converges unconditionally for all f H. {g k } k=1 is called the dual Riesz basis. 11

12 From the proof: {g k } k=1 = {(U 1 ) e k } k=1. Dual of {g k } k=1 : {( ((U 1 ) ) 1 ) ek } k=1 = {Ue k} k=1 = {f k } k=1. So {f k } k=1 and {g k} k=1 are duals of each other: they are called a pair of dual Riesz bases. 12

13 Definition 3.8 Two sequences {f k } k=1 and {g k } k=1 in a Hilbert space are biorthogonal if f k, g j = δ k,j. Corollary 3.9 For a pair of dual Riesz bases {f k } k=1 and {g k} k=1 the following hold: (i) {f k } k=1 and {g k} k=1 are biorthogonal. (ii) For all f H, f = f, g k f k = k=1 f, f k g k. k=1 Proposition 3.10 If {f k } k=1 = {Ue k} k=1 is a Riesz basis for H, there exist constants A, B > 0 such that A f 2 f, f k 2 B f 2, f H. k=1 The largest possible value for the constant A is 1 U 1 2, and the smallest possible value for B is U 2. 13

14 Theorem 3.11 For a sequence {f k } k=1 in H, the following conditions are equivalent: (i) {f k } k=1 is a Riesz basis for H. (ii) {f k } k=1 is complete in H, and there exist constants A, B > 0 such that for every finite scalar sequence {c k } one has A c k 2 2 ck f k B ck 2. 14

15 4 The expansion property for a non-basis: Example 4.1 The family {e k } k Z = {e 2πikx } k Z forms an ONB for L 2 (0, 1). Consider an open subinterval I ]0, 1[ with I < 1. Identify L 2 (I) with the subspace of L 2 (0, 1) consisting of the functions which are zero on ]0, 1[\I. For a function f L 2 (I): f = k Z f, e k e k in L 2 (0, 1). (1) Since f f, e k e k k n 2 L 2 (I) 0 as n, 15

16 we also have f = k Z f, e k e k in L 2 (I). (2) That is, the (restrictions to I of the) functions {e k } k Z also have the expansion property in L 2 (I). However, they are not a basis for L 2 (I)! To see this, define the function f(x) = { f(x) if x I, 1 if x / I. Then f L 2 (0, 1) and we have the representation f = k Z f, e k e k in L 2 (0, 1). (3) By restricting to I, the expansion (3) is also valid in L 2 (I); since f = f on I, this shows that f = k Z f, e k e k in L 2 (I). (4) Thus, (2) and (4) are both expansions of f in L 2 (I), and they are non-identical; the argument 16

17 is that since f f in L 2 (0, 1), the expansions (1) and (3) show that { f, e k } k Z { f, e k } k Z. Conclusion: The restriction of the functions {e k } k Z to I is not a basis for L 2 (I), but the expansion property is preserved. It is natural to consider sequences which are not bases, but nevertheless have the expansion property. 17

18 5 Frames in Hilbert spaces Definition 5.1 A sequence {f k } k=1 of elements in H is a frame for H if there exist constants A, B > 0 such that A f 2 f, f k 2 B f 2, f H. k=1 The numbers A and B are called frame bounds. The optimal upper frame bound is the infimum over all upper frame bounds, and the optimal lower frame bound is the supremum over all lower frame bounds. Lemma 5.2 Suppose that {f k } k=1 is a sequence of elements in H and that there exist constants A, B > 0 such that A f 2 f, f k 2 B f 2 k=1 for all f in a dense subset V of H. Then {f k } k=1 is a frame for H with bounds A, B. 18

19 Definition 5.3 A sequence {f k } k=1 in H is a tight frame if there exist a number A > 0 such that f, f k 2 = A f 2, f H. k=1 The (exact) number A is called the frame bound. Recall: the pre-frame operator is T : H l 2 (N), T {c k } k=1 = c k f k. k=1 Frame operator associated to general frame {f k } k=1 : S : H H, Sf = T T f = f, f k f k. k=1 {f k } k=1 is a Bessel sequence, so the series defining S converges unconditionally for all f H by Corollary

20 Some important properties of S: Lemma 5.4 Let {f k } k=1 be a frame with frame operator S and frame bounds A, B. Then the following holds: (i) S is bounded, invertible, self-adjoint, and positive. (ii) {S 1 f k } k=1 is a frame with frame operator S 1 and frame bounds B 1, A 1. (iii) If A, B are the optimal frame bounds for {f k } k=1, then the bounds B 1, A 1 are optimal for {S 1 f k } k=1. {S 1 f k } k=1 of {f k } k=1. is called the canonical dual frame 20

21 Frame decomposition: Theorem 5.5 Let {f k } k=1 be a frame with frame operator S. Then f = f, S 1 f k f k, f H, and f = k=1 f, f k S 1 f k, f H. (5) k=1 Both series converge unconditionally for all f H. Proof. Let f H. Via Lemma 5.4, f = SS 1 f = S 1 f, f k f k = f, S 1 f k f k. k=1 k=1 Since {f k } k=1 is a Bessel sequence and { f, S 1 f k } k=1 l 2 (N), the series converges unconditionally by Corollary 2.3. The expansion (5) follows from f = S 1 Sf. 21

22 Corollary 5.6 If {f k } k=1 is a tight frame with frame bound A, then the canonical dual frame is {A 1 f k } k=1, and f = 1 f, f k f k, f H. (6) A Proof. k=1 For a tight frame, Sf, f = A f 2 = Af, f ; since S is self-adjoint, this implies that S = AI. By scaling of the vectors {f k } k=1 in a tight frame, we can always obtain that A = 1; in that case (6) has exactly the same form as the representation via an orthonormal basis. Tight frames can be used without any additional computational effort compared to the use of ONB s. 22

23 Other advantages of tight frames: If {f k } k=1 consists of functions with compact support or fast decay, the same is the case for the functions in the canonical dual frame. If {f k } k=1 consists of functions with a special structure (Gabor structure or wavelet structure) the same is the case for the functions in the canonical dual frame. The corresponding statements do not hold for a general frame {f k } k=1 and its canonical dual frame {S 1 f k } k=1! 23

24 To each frame {f k } k=1 one can associate a tight frame: Theorem 5.7 Let {f k } k=1 be a frame for H with frame operator S. Denote the positive square root of S 1 by S 1/2. Then {S 1/2 f k } k=1 is a tight frame with frame bound equal to 1, and f = f, S 1/2 f k S 1/2 f k, f H. k=1 Problems: Not easy to find {S 1/2 f k } k=1 ; Nice properties of {f k } k=1 not necessarily inherited by {S 1/2 f k } k=1. 24

25 5.1 Dual frames Another way to avoid the problem of inverting the frame operator S: A frame which is not a Riesz basis is said to be overcomplete. Theorem 5.8 Assume that {f k } k=1 is an overcomplete frame. Then there exist frames for which f = {g k } k=1 {S 1 f k } k=1 f, g k f k, f H. k=1 {g k } k=1 is called a dual frame of {f k} k=1. If the canonical dual frame is difficult to find, maybe there exist other duals which are easy to find??? 25

26 Lemma 5.9 Assume that {f k } k=1 and {g k} k=1 are Bessel sequences in H. Then the following are equivalent: (i) f = k=1 f, g k f k, f H. (ii) f = k=1 f, f k g k, f H. (iii) f, g = k=1 f, f k g k, g, f, g H. In case the conditions are satisfied, {f k } k=1 and {g k } k=1 are dual frames for H. Characterization of all dual frames: Theorem 5.10 Let {f k } k=1 be a frame for H. The dual frames of {f k } k=1 are precisely the families {g k } k=1 = S 1 f k + h k S 1 f k, f j h j j=1 where {h k } k=1 is a Bessel sequence in H. Does not (immediately) help! k=1, 26

27 5.2 Frames and Riesz bases Theorem 5.11 A Riesz basis {f k } k=1 for H is a frame for H, and the Riesz basis bounds coincide with the frame bounds. The dual Riesz basis equals the canonical dual frame {S 1 f k } k=1. Theorem 5.12 Let {f k } k=1 be a frame for H. Then the following are equivalent. (i) {f k } k=1 is a Riesz basis for H. (ii) If k=1 c kf k = 0 for some {c k } k=1 l2 (N), then c k = 0, k N. 27

28 5.3 Characterizations of frames Theorem 5.13 A sequence {f k } k=1 a frame for H if and only if T : {c k } k=1 c k f k k=1 in H is is a well-defined mapping of l 2 (N) onto H. Compare to the characterization of Bessel sequences in terms of T (T well-defined)! Theorem 5.14 Let {e k } k=1 be an arbitrary orthonormal basis for H. The frames for H are precisely the families {Ue k } k=1, where U : H H is a bounded and surjective operator. Compare to the characterization of ONB s (U unitary); Riesz bases (U bounded and bijective). 28

29 5.4 A frame where no subsequence is a basis Intuitively: A frame consists of a basis + some extra elements (redundance). Good as intuitive feeling - but wrong in the technical sense: be an orthonor- Example 5.15 Let {e k } k=1 mal basis for H. Let {f k } 1 1 k=1 := {e 1, e 2, e 2, e 3, 1 3 e 3, that is, each vector 1 l e l, l N, is repeated l times. Then {f k } k=1 is a tight frame for H with frame bound A = 1. No subfamily is a Riesz basis. 1 3 e 3, }; More complicated: In each separable Hilbert space, there exists a frame for which no subfamily is a basis! 29

30 Part II: Constructions in L 2 (R). Final goal: Obtain convenient expansions in L 2 (R), f = k I f, g k f k, f L 2 (R). Convenient means: f k, g k are given explicitly; f k, g k have compact support (no truncation necessary in numerical calculations); f k, g k decay fast in the Fourier domain; If f k has a certain structure (e.g., Gabor structure), g k has the same structure. In concrete applications, other requirements might be relevant (differentiability, zero-means, vanishing moments, approximation order etc.) 30

31 6 Special functions Typical generators: B-splines: The B-splines N n, n N, are given inductively by N 1 = χ [0,1], N n+1 (x) = N n N 1 (x) = 1 0 N n (x t) dt. The functions N n are called B-splines, and n is the order. Fundamental properties: Theorem 6.1 Given n N, the B-spline N n has the following properties: (i) supp N n = [0, n] and N n > 0 on ]0, n[. (ii) N n(x)dx = 1. (iii) k Z N n(x k) = 1 31

32 (iv) For any n N,. N n (γ) = ( ) 1 e 2πiγ n. 2πiγ For any function f : R R, let f(x) + = max{0, f(x)}. Theorem 6.2 For each n = 2, 3,..., the B- spline N n can be written 1 n ( ) n N n (x) = ( 1) j (x j) n 1 +. (n 1)! j j=0 32

33 Figure 1: The B-spline N 2. Figure 2: The B-spline N 3. 33

34 6.1 Symmetric B-splines: For n N, let B n (x) := T n 2 N n (x) = N n (x + n 2 ). The symmetric B-splines have similar properties: Corollary 6.3 For n N, the B-spline B n has the following properties: (i) k Z B n(x k) = 1. (ii) B ( n (γ) = e πiγ e πiγ 2πiγ ) n = ( sin(πγ) πγ ) n. 34

35 Figure 3: The B-spline B 2. Figure 4: The B-spline B 3. 35

36 7 Gabor systems Gabor systems, have the form {e 2πimbx g(x na)} m,n Z for some g L 2 (R), a, b > 0. Short notation: {E mb T na g} m,n Z = {e 2πimbx g(x na)} Gabor bases exist: Example 7.1 Let χ [0,1] denote the characteristic function for the interval [0, 1]. Then {e 2πikx χ [0,1] (x)} k Z is an ONB for L 2 (0, 1). By translation, for each n Z the space L 2 (n, n + 1) has the ONB {e 2πik(x n) χ [0,1] (x n)} k Z = {e 2πikx χ [0,1] (x n)} k Z. We conclude that L 2 (R) has the ONB { e 2πikx χ [0,1] (x n) } k,n Z. 36

37 Problem: The function χ [0,1] is discontinuous and has very slow decay in the Fourier domain: χ [0,1] (γ) = 1 0 e 2πixγ dx = e πiγsin πγ πγ. Thus, the function is not suitable for time-frequency analysis. Question: Can we obtain more suitable Gabor bases by replacing χ [0,1] by a smoother function g? No! A continuous function with compact support can not generate a Gabor Riesz basis. 37

38 A related short coming - the Balian Low Theorem: Theorem 7.2 Assume that {E mb T na g} m,n Z is ( a Riesz basis. Then ) ( ) xg(x) 2 dx γĝ(γ) 2 dγ =. R In words: A function g generating a Gabor Riesz basis can not be well localized in both time and frequency. R For example: impossible that the estimates C g(x) (1 + x 2 ) 1/2, C ĝ(γ) (1 + γ 2 ) 1/2 hold simultaneously. This motives the construction of Gabor frames! 38

39 Example 7.3 The Gaussian g(x) = e x2 generates a Gabor frame {E mb T na g} m,n Z for all a, b ]0, 1[. The Fourier transform ĝ(x) = πe π2 x 2 has exponential decay. Also, the dual generator S 1 g has exponential decay in time and frequency. The function h(x) = S 1/2 e x2 generates a tight Gabor frame {E mb T na h} m,n Z for all a, b ]0, 1[, and h as well as ĥ decay exponentially. Problem: S 1/2 e x2 and S 1 e x2 are not given explicitly. 39

40 Solution: Don t construct a nice frame and expect the canonical dual to be nice. Construct simultaneously dual pairs {E mb T na g},{e mb T na h} such that g and h have the required properties, and f = f, E mb T na h E mb T na g, f L 2 (R). m,n Z 40

41 7.1 Sufficient and necessary conditions Given a function g L 2 (R) and two numbers a, b > 0, consider the matrix-valued function M(x) := (g(x na m/b)) m,n Z, x R. Theorem 7.4 Let A, B > 0 and the Gabor system {E mb T na g} m,n Z be given. Then the following holds: (i) {E mb T na g} m,n Z is a Bessel sequence with bound B if and only if M(x) for a.e. x R defines a bounded operator on l 2 (Z) with norm at most bb. (ii) Assuming that {E mb T na g} m,n Z is a Bessel sequence, it is a frame for L 2 (R) with lower frame bound A if and only if bai M(x)M(x) a.e. x R, where I is the identity operator on l 2 (Z). 41

42 The following necessary condition for {E mb T na g} m,n Z to be a frame for L 2 (R) depends on the interplay between the function g and the translation parameter a, and is expressed in terms of the function G(x) := n Z g(x na) 2, x R. Proposition 7.5 Let g L 2 (R) and a, b > 0 be given, and assume that {E mb T na g} m,n Z is a frame with bounds A, B. Then ba n Z g(x na) 2 bb, a.e. x R. 42

43 Lemma 7.6 Suppose that f is a bounded measurable function with compact support and that the function G(x) = n Z g(x na) 2, x R is bounded. Then f, E mb T na g 2 m,n Z = 1 b + 1 b k 0 f(x) 2 n Z g(x na) 2 dx f(x)f(x k/b) n Z g(x na)g(x na k/b) dx. 43

44 Theorem 7.7 Let g L 2 (R), a, b > 0 and suppose that B := 1 b sup g(x na)g(x na k/b) x [0,a] <. k Z n Z Then {E mb T na g} m,n Z with bound B. If also is a Bessel sequence A := 1 b inf x [0,a] k 0 [ g(x na) 2 g(x na)g(x na k/b) > 0, n Z n Z then {E mb T na g} m,n Z with bounds A, B. is a frame for L 2 (R) 44

45 Corollary 7.8 Let g L 2 (R) be bounded and compactly supported. Then {E mb T na g} m,n Z is a Bessel sequence for any a, b > 0. Recall: G(x) = n Z g(x na) 2, x R. Corollary 7.9 Let a, b > 0 be given. Suppose that g L 2 (R) has support in an interval of length 1 b and that the function G is bounded above and below. Then {E mb T na g} m,n Z is a frame for L 2 (R) with bounds A, B. The frame operator S and its inverse S 1 are given by Sf = G b f, S 1 f = b G f, f L2 (R). 45

46 Corollary 7.10 Suppose that g L 2 (R) is a continuous function with support on an interval I with length I and that g(x) > 0 on the interior of I. Then {E mb T na g} m,n Z is a frame for all (a, b) ]0, I [ ]0, 1 I [. Corollary 7.11 For n N, the B-splines B n and N n generate Gabor frames for all (a, b) ]0, n[ ]0, 1/n[. Question: Characterization of (a, b) R 2 for which B n generates a Gabor frame? The exact answer is unknown, and is bound to be complicated: 1) {E mb T na B 2 } m,n Z can not be a frame for any b > 0 whenever a 2. 1) [Gröchenig, Janssen, Kaiblinger, Pfander]: For b = 2, 3,..., {E mb T na B 2 } m,n Z can not be a frame for any a > 0. 46

47 Example 7.12 Consider characteristic functions, g := χ [0,c[, c > 0. Surprisingly complicated to find the exact range of c > 0 and parameters a, b > 0 for which {E mb T na g} m,n Z is a frame! A complete answer has not been obtained yet. 47

48 Via a scaling, assume that b = 1. Janssen showed that (i) {E m T na g} m,n Z is not a frame if c < a or a > 1. (ii) {E m T na g} m,n Z is a frame if 1 c a. (iii) {E m T na g} m,n Z is not a frame if a = 1 and c > 1. Assuming now that a < 1, c > 1, we further have (iv) {E m T na g} m,n Z is a frame if a / Q and c ]1, 2[. (v) {E m T na g} m,n Z is not a frame if a = p/q Q, gcd(p, q) = 1, and 2 1 q < c < 2. (vi) {E m T na g} m,n Z is not a frame if a > 3 4 and c = L 1 + L(1 a) with L N, L 3. The graphical illustration of this result is known as Janssen s tie. 48

49 Theorem 7.13 Let g L 2 (R) and a, b > 0 be given. Then the following holds: (i) If ab > 1, then {E mb T na g} m,n Z can not be a frame for L 2 (R). (ii) If {E mb T na g} m,n Z is a frame, then ab = 1 {E mb T na g} m,n Z is a Riesz basis. 49

50 Duality principle: [Janssen], [Daubechies, Landau and Landau], [Ron and Shen]. Concerns the relationship between frame properties for a function g with respect to the lattice {(na, mb)} m,nz and the so-called dual lattice {(n/b, m/a)} m,nz. Theorem 7.14 Let g L 2 (R) and a, b > 0 be given. Then the Gabor system {E mb T na g} m,n Z is a frame for L 2 (R) with bounds A, B if and only if {E m/a T n/b g} m,n Z is a Riesz sequence with bounds aba, abb. 50

51 7.2 Time-frequency localization of Gabor expansions No function g 0 can have compact support simultaneously in the time-domain and the frequencydomain. But most signals appearing in practice are essentially localized in the time-frequency plane: The interesting part of the signal takes place on a finite time-interval, with frequencies belonging to a certain finite interval. How does this affects the Gabor frame expansion? 51

52 Given a number T > 0, define the operator Q T : L 2 (R) L 2 (R), (Q T f)(x) = χ [ T,T ] (x)f(x). Will use (I Q T )f as a measure for the content of the function f outside the interval [ T, T ]: that f essentially is localized on the interval [ T, T ] means that (I Q T )f is small compared to f. Similarly, for Ω > 0 define P Ω by P Ω : L 2 (R) L 2 (R), PΩ f(ν) = χ [ Ω,Ω] (ν) ˆf(ν). That ˆf essentially is localized on [ Ω, Ω] means that (I P Ω )f is small compared to f. Now assume that f is essentially localized in both domains, i.e., on [ T, T ] [ Ω, Ω] for some T, Ω > 0. Let B(T, Ω) = {(m, n) Z 2 : mb [ Ω, Ω], na [ T, T ]}. 52

53 Consider the frame expansion of f in terms of dual frames {E mb T na g} m,n Z and {E mb T na h} m,n Z, f = f, E mb T na h E mb T na g. m,n Z Do we obtain a reasonable approximation of f if we replace the sum over (m, n) Z 2 with a sum over (m, n) B(T, Ω)? Positive answer if we replace B(T, Ω) by a certain enlargement B(T + Λ, Ω + Γ) : 53

54 Theorem 7.15 Assume that the Gabor systems {E mb T na g} m,n Z and {E mb T na h} m,n Z form a pair of dual frames for L 2 (R) with upper frame bounds B and D respectively, and that for some constants C > 0, α > 1/2, the decay conditions h(x) C(1 + x 2 ) α, ĥ(ν) C(1 + ν2 ) α hold. Then, for any ɛ > 0 there exist numbers Λ, Γ > 0 such that for all T, Ω > 0, f f, E mb T na h E mb T na g {(m,n) B(T +Λ,Ω+Γ)} BD ( I Q T )f + (I P Ω )f + ɛ f ) for all f L 2 (R). 54

55 7.3 Tight Gabor frames Theorem 7.16 Let g L 2 (R) and a, b > 0 be given. Then the following are equivalent: (i) {E mb T na g} m,n Z is a tight frame for L 2 (R) with frame bound A = 1. (ii) For a.e. x R the following conditions hold: G(x) = n Z g(x na) 2 = b; n Z g(x na)g(x na k/b) = 0 for all k 0. Moreover, when the equivalent conditions hold, {E mb T na g} m,n Z is an orthonormal basis for L 2 (R) if and only if g = 1. 55

56 Corollary 7.17 Let a, b > 0 be given. Assume that ϕ L 2 (R) is a real-valued nonnegative function with support in an interval of length 1/b, and that ϕ(x + na) = 1, a.e. x R. n Z Then the function g(x) := bϕ(x) generates a tight Gabor frame {E mb T na g} m,n Z with frame bound A = 1. Example 7.18 For any n N, the B-spline ϕ = N n satisfies the requirements in Corollary 7.17 with a = 1 and any b ]0, 1/n]. Thus, for any b ]0, 1/n], the function g(x) = bn n (x) generates a tight Gabor frame {E mb T n g} m,n Z with frame bound A = 1. 56

57 7.4 The duals of a Gabor frame For a Gabor frame {E mb T na g} m,n Z with associated frame operator S, the frame decomposition shows that f = f, S 1 E mb T na g E mb T na g, f L 2 (R). m,n Z We need to be able to calculate the canonical dual frame {S 1 E mb T na g} m,n Z difficult! A simplification can be obtained via Lemma 7.19 Let g L 2 (R) and a, b > 0 be given, and assume that {E mb T na g} m,n Z is a Bessel sequence with frame operator S. Then the following holds: (i) SE mb T na = E mb T na S for all m, n Z. (ii) If {E mb T na g} m,n Z is a frame, then also S 1 E mb T na = E mb T na S 1, m, n Z. 57

58 Theorem 7.20 Let g L 2 (R) and a, b > 0 be given, and assume that {E mb T na g} m,n Z is a Gabor frame. Then the following holds: (i) The canonical dual frame also has the Gabor structure and is given by {E mb T na S 1 g} m,n Z. (ii) The canonical tight frame associated with {E mb T na g} m,n Z is given by {E mb T na S 1/2 g} m,n Z. Proposition 7.21 Let g L 2 (R), and assume that g as well as ĝ decay exponentially. Let a, b > 0 be given and assume that {E mb T na g} m,n Z is a frame. Then {E mb T na S 1/2 g} m,n Z is a tight frame, for which S 1/2 g as well as F(S 1/2 g) decay exponentially. Theoretically perfect! Can be applied to the Gaussian g(x) = e x2 /2 but the resulting generators are not given explicitly. 58

59 7.5 Characterizations of pairs of dual Gabor frames The dual Gabor pairs with Gabor structure are characterized in the Wexler Raz Theorem. Theorem 7.22 Let g, h L 2 (R) and a, b > 0 be given. Then, if the two Gabor systems {E mb T na g} m,n Z and {E mb T na h} m,n Z are Bessel sequences, they are dual frames if and only if { h, Em/a T n/b g = 0 for all (m, n) (0, 0) h, g = ab. 59

60 Janssen proved the following: Theorem 7.23 Two Bessel sequences {E mb T na g} m,n Z and {E mb T na h} m,n Z form dual frames if and only if g(x n/b ka)h(x ka) = bδ n,0, a.e. x [0, a]. k Z Consequence of Theorem 7.23: Lemma 7.24 Let g, h be real-valued, bounded, and compactly supported functions. Then for b > 0 sufficiently small the following two conditions are equivalent: (i) {E mb T n g} m,n Z and {E mb T n h} m,n Z form dual frames for L 2 (R); (ii) k Z g(x k)h(x k) = b, a.e. x [0, 1]. 60

61 7.6 Explicit construction of dual pairs of Gabor frames Dual Gabor frames with a = 1: Theorem 7.25 (C.) Let N N. Let g L 2 (R) be a real-valued bounded function with supp g [0, N], for which g(x n) = 1. n Z 1 Let b ]0, 2N 1 ]. Then the function g and the function h defined by h(x) = bg(x) + 2b N 1 n=1 g(x + n) generate dual frames {E mb T n g} m,n Z and {E mb T n h} m,n Z for L 2 (R). Proof. Note that supp g [0, N], supp h [ N + 1, N], so Lemma 7.24 applies for b 1/(2N 1). 61

62 Also, g and h have compact support, so {E mb T n g} m,n Z and {E mb T n h} m,n Z are Bessel sequences. We check the condition in Lemma For x [0, 1], 1 = ( N 1 k=0 g(x + k) ) 2 = (g(x) + g(x + 1) + + g(x + N 1)) = g(x) (g(x) + g(x + 1) + + g(x + N 1)) [g(x) + 2g(x + 1) + 2g(x + 2) + + 2g(x + N 1)] +g(x + 1) [g(x + 1) + 2g(x + 2) + 2g(x + 3) + + 2g(x + N 1)] +g(x + 2) [g(x + 2) + 2g(x + 3) + 2g(x + 4) + + 2g(x + N 1)] + + +g(x + N 2) [g(x + N 2) + 2g(x + N 1)] +g(x + N 1) [g(x + N 1)] = 1 N 1 g(x + k)h(x + k). b k=0 Thus the conclusion follows. 62

63 Corollary 7.26 (C.) For any N N and b ]0, 1 2N 1 ], the functions N N and h N (x) := bn N (x) + 2b N 1 n=1 N N (x + n) generate a pair of dual frames {E mb T n N N } m,n Z and {E mb T n h N } m,n Z. Important features of the dual pair of frame generators (N N, h N ) in Corollary 7.26: The functions N N and h N are splines; N N and h N have compact support, i.e., perfect time localization; By choosing N sufficiently large, polynomial decay of N N and ĥn of any desired order can be obtained. 63

64 Example 7.27 For the B-spline x x [0, 1[, N 2 (x) = 2 x x [1, 2[, 0 x / [0, 2[, we can use Corollary 7.26 for b ]0, 1/3]. For b = 1/3 we obtain the dual generator h(x) = 1 3 N 2(x) N 2(x + 1) 2 3 (x + 1) x [ 1, 0[, 1 = 3 (2 x) x [0, 2[, 0 x / [ 1, 2[. 64

65 Figure 5: The B-spline N 2 and the dual generator h for b = 1/3.! Figure 6: The B-spline N 3 and the dual generator h with b = 1/5. 65

66 Dual Gabor frames with general a > 0: Theorem 7.28 (C.) Let N N. Let g L 2 (R) be a real-valued bounded function with supp g [0, N], for which g(x n) = 1. n Z Let a, b > 0 and choose J N such that J ab(2n 1). Define the function h by h(x) = abg(x) + 2ab Then the functions N 1 n=1 g(x + n). g k = T a J kd a/j g, h k = T a J kd a/j h, k = 0,..., J 1 generate dual multi-gabor frames {E mb T na g k } m,n Z,k=0,...,J 1, {E mb T na h k } m,n Z,k=0,...,J 1 for L 2 (R). 66

67 Other choices? Yes! Theorem 7.29 (C., Kim, 2007) Let N N. Let g L 2 (R) be a real-valued bounded function with supp g [0, N], for which g(x n) = 1. n Z Let b ]0, 1 2N 1 ]. Define h L2 (R) by where h(x) = N 1 n= N +1 a n g(x + n), a 0 = b, a n + a n = 2b, n = 1, 2,, N 1. Then g and h generate dual frames {E mb T n g} m,n Z and {E mb T n h} m,n Z for L 2 (R). 67

68 Example ) Take a 0 = b, a n = 0 for n = N + 1,..., 1 a n = 2b, n = 1,... N 1 This is Theorem shortest support. This choice gives the 2) Take a N+1 = a N+2 = = a N 1 = b : if g is symmetric, this leads to a symmetric dual generator h(x) = b N 1 n= N +1 g(x + n). Note: h(x) = 1 on supp g. 68

69 x x Figure 7: The generators N 2 and N 3 and their dual generators via 2) in Example

70 8 Polynomial generators and duals [C., Kim] The restriction of any polynomial to a sufficiently large interval will generate a Gabor frame for small modulation parameters: Proposition 8.1 Let N N, and consider any bounded interval I R with I N. Then any (nontrivial) polynomial ( N 1 ) g(x) = c k x k χ I (x) k=0 generates a Gabor frame {E mb T n g} m,n Z for b ]0, 1/ I ]. 70

71 The condition on the relationship between the support size and the degree of the polynomial is necessary in Proposition 8.1: Example 8.2 Let g(x) = (x 1/2)(x 3/2)χ [0,2] (x) = (x 2 2x + 3/4)χ [0,2] (x). Then {E mb T n g} m,n Z is not a Gabor frame for any b > 0. In fact, for x [0, 1], g(x k) 2 = g(x) 2 + g(x + 1) 2, so k Z g(1/2 k) 2 = 0. k Z 71

72 In applications of frames: crucial that the frame generator as well as the dual frame generator have a convenient form. The polynomials are very convenient - but no pair of dual generators of this form exists: Theorem 8.3 Two compactly supported polynomials ( N1 ) g(x) = a k x k χ I1 (x) and h(x) = k=0 ( N2 k=0 b k x k ) χ I2 (x) of degree 1 can not generate dual Gabor frames {E mb T n g} m,n Z and {E mb T n h} m,n Z for any b > 0, no matter how the intervals I 1 and I 2 are chosen. 72

73 Remark: The proof shows that even finite linear combinations of the type g(x) = c k T k g, h(x) = dk T k h with g and h as in Theorem 8.3 can t generate dual Gabor frames {E mb T n g} m,n Z and {E mb T n h}m,n Z Conclusion: for polynomial generators we must search for other simple generators for the dual. Natural choices: B-splines! 73

74 Theorem 8.4 Let N N and g(x) := N 1 k=0 c kx k be a polynomial. Given an interval I R, let g(x) := g(x)χ I (x). Let h L 2 (R) be such that supp h I. Then the following are equivalent: (i) There exists β R such that g(x) and h(x) := βh(x) generate dual frames {E mb T n g} m,n Z and {E mb T n h}m,n Z for b > 0 sufficiently small; (ii) N 1 k=0 c k N 1 k=0 c k D k ĥ(n) = 0, n Z \ {0}; ( 2πi) k D k ĥ(0) ( 2πi) 0. k 74

75 Theorem 8.5 Let N N and g(x) := N 0 k=0 c kx k be a polynomial of degree N 0 < N. Given an interval I R, let g(x) := g(x)χ I (x). Let h L 2 (R) be such that supp h is an interval with supp h I. If h satisfy the Strang-Fix conditions, that is, ĥ(0) 0; D k ĥ(n) = 0, n N \ {0}, k = 0, 1,, N 1, then there always exist α, β R such that g(x) and h(x) := βh(x α) generate dual frames {E mb T n g} m,n Z and {E mb T n h}m,n Z for b > 0 sufficiently small. 75

76 Corollary 8.6 Let N N and g(x) := N 1 k=0 c kx k be a polynomial. Given an interval I containing [0, N], let g(x) := g(x)χ I (x). Then the following hold: (a) If I = [0, N] and κ := N 1 k=0 c k D k NN (0) ( 2πi) k 0, then g(x) and h(x) := b κ N N(x) generate dual frames {E mb T n g} m,n Z and {E mb T n h} m,n Z for b > 0 sufficiently small. (b) If I [0, N], then there exists α, β R such that g and h(x) := βn N (x α) generate dual frames {E mb T n g} m,n Z and {E mb T n h} m,n Z for b > 0 sufficiently small. 76

77 Conclusion: polynomial generators on sufficiently large intervals almost always generate Gabor frames with B-spline duals for sufficiently small modulation parameters. 77

78 Example 8.7 Let g(x) = (x 1)χ I (x). First, let I = [0, 2]. A direct calculation shows that Thus N 2 (0) = 1, κ = 1 k=0 D N 2 (0) = 2πi. c k D k N2 (0) ( 2πi) k = 0. One can prove that g(x k)n 2 (x k) = 0. k Z Hence N 2 is not a dual of g for any b > 0 by Lemma Now, let I = [ 1, 3]. Then βn 2 (x α 0 ) is dual generator of g for b > 0 sufficiently small for some constant β (which can be explicitly calculated). 78

79 Example 8.8 Let g(x) = x 2 (5 x) 2 χ [0,5] (x) = ( 25x 2 10x 3 + x 4) χ [0,5] (x) ( 4 ) =: c k x k χ [0,5] (x). k=0 For the B-spline N 5, Thus κ := D k N5 (0) ( 2πi) k = 4 k=0 20/3, k = 2; 75/4, k = 3; 331/6, k = 4. D k N5 (0) c k = 103/3 0. ( 2πi) k Hence h(x) := b κ N 5 is a dual of g(x) for b > 0 sufficiently small by Corollary 8.6 (a). See Figure 8. 79

80 x x 5 6 Figure 8: The generators g and h for b = 1/5 in Example

81 Example 8.9 Let g(x) = x 3 (6 x) 2 χ [0,6] (x) = ( 36x 3 12x 4 + x 5) χ [0,6] (x) ( 5 ) = c k x k χ [0,6] (x). For N 6, we have Thus κ := k=0 D k N6 (0) ( 2πi) k = 5 k=0 63/2, k = 3; 1087/10, k = 4; 777/2, k = 5. D k N6 (0) c k = 2181/10 0. ( 2πi) k Hence h(x) := b κ N 6 is a dual of g(x) for b > 0 sufficiently small by Corollary 8.6 (a). See Figure 9. 81

82 x x 6 Figure 9: The generators g and h for b = 1/6 in Example

83 Extensions: 1) Higher dimensions: (C., R. Y. Kim) For f L 2 (R d ) and d d matrices B, C, and m, n Z d, let E Bm f(x) = e 2πiBm x f(x), x R d T Cn f(x) = f(cn x), x R d. 83

84 Theorem 8.10 Let N N and assume that g L 2 (R d ) has supp g [0, N] d, and that n Z d g(x n) = 1. Assume that the d d matrix B is invertible and B 1 d(2n 1). For i = 1,..., d, let F i be the set of lattice points {k j } d j=1 Zd for which the coordinates k j, j = 1,..., d, satisfy if j = 1,..., i 1, then k j N 1; if j = i, then 1 k j N 1; if j = i + 1,..., d, then k j = 0. Define h L 2 (R d ) by h(x) := det B g(x) + 2 d g(x + k). k F i i=1 Then {E Bm T n g} m,n Z d and {E Bm T n h} m,n Z d are dual frames for L 2 (R d ). 84

85 2) Tight frames (C., R.Y. Kim): Theorem 8.11 Let N N. Let g L 2 (R d ) be a non-negative function with supp g [0, N] d, for which g(x n) > 0, a.e. x R d. n Z d Assume that the d d matrix B is invertible and B 1 d N. Define h L 2 (R d ) by h(x) := det B g(x) n Z d g(x n). Then the function h generates a Parseval frame {E Bm T n h} m,n Z d for L 2 (R d ). The generator in Theorem 8.11 has fast decay in the Fourier domain if it is smooth: Lemma 8.12 Let k N and let f C dk (R d ) be compactly supported. Then ˆf(γ) A(1 + γ 2 ) k/2. 85

86 3) Irregular B-splines (C., Wenchang Sun): Theorem 8.13 Let {x n : n Z} R be a sequence such that and lim x n = ±, x n 1 x n, n ± x n+2n 1 x n M, n Z, for some constants N N and M > 0. Let g n be the N-th order normalized B-spline with knots (x n, x n+1,, x n+n ) and h n (x) = bg n (x) + 2b N 1 k=1 g n k (x). Then {E mb g n } m,n Z and {E mb h n } m,n Z are a pair of dual frames for L 2 (R) provided that 0 < b 1/M. 86

87 9 Wavelet frames 9.1 Wavelet bases Given a function ψ L 2 (R) and j, k Z, let ψ j,k (x) := 2 j/2 ψ(2 j x k), x R. In terms of T k and Df(x) = 2 1/2 f(2x), ψ j,k = D j T k ψ, j, k Z. If {ψ j,k } j,k Z is an orthonormal basis for L 2 (R), the function ψ is called a wavelet. Definition 9.1 A multiresolution analysis for L 2 (R) consists of a sequence of closed subspaces {V j } j Z of L 2 (R) and a function φ V 0, such that the following conditions hold: (i) V 1 V 0 V 1. (ii) j V j = L 2 (R) and j V j = {0}. (iii) f V j [x f(2x)] V j+1. (iv) f V 0 T k f V 0, k Z. (v) {T k φ} k Z is an orthonormal basis for V 0. 87

88 An MRA can be used to construct an ONB for L 2 (R): For j Z, let W j denote the orthogonal complement of V j in V j+1. Then L 2 (R) = j Z W j. The spaces W j satisfy the same dilation relationship as V j, i.e., ψ W 0 [x ψ(2 j x)] W j. In order to obtain an orthonormal basis {ψ j,k } j,k Z for L 2 (R) it is now enough to find ψ W 0 such that {ψ( k)} k Z is an orthonormal basis for W 0. 88

89 We will now explain how to find a suitable function ψ. The condition φ V 0 V 1 implies by (iii) that 1 2 D 1 φ V 0. Since {T k φ} k Z is an orthonormal basis for V 0, there exist coefficients {c k } k Z l 2 (Z) such that 1 D 1 φ = c k T k φ. (7) 2 k Z We will now rewrite (7) in terms of the Fourier transform of φ. Let E k (x) = e 2πikx, x R. Now, applying the Fourier transform on both sides of (7), 1 D ˆφ = c k E k ˆφ. 2 k Z Defining the 1-periodic function H 0 := k Z c ke k, this can be written as ˆφ(2γ) = H 0 (γ) ˆφ(γ), a.e. γ R. (8) 89

90 The equation (8) is called a scaling equation or refinement equation. Now, a certain choice of a 1-periodic function H 1, implies that the function ψ defined via ˆψ(2γ) = H 1 (γ) ˆφ(γ) (9) generates a wavelet orthonormal basis {D j T k ψ} j,k Z. One choice of H 1 is to take H 1 (γ) = H 0 (γ )e 2πiγ. Note that (9) leads to an explicit expression of the function ψ in terms of the given function φ: 90

91 Lemma 9.2 Assume that ˆψ(2γ) = H 1 (γ) ˆφ(γ) holds for a 1-periodic and bounded function H 1 with Fourier expansion H 1 = k Z c ke k. Then ψ(x) = 2 k Z c k DT k φ(x) = 2 k Z c k φ(2x + k), a.e. x R. In particular, if H 1 is a trigonometric polynomial, H 1 (x) = N 2 k=n 1 c k e 2πikx, then ψ(x) = 2 = 2 N 2 k=n 1 c k DT k φ(x) N 2 k=n 1 c k φ(2x + k), x R. 91

92 The Haar basis can be constructed via the multiresolution analysis defined by φ = χ [0,1[ and V j = {f : f is const. on [2 j k, 2 j (k + 1)[, k Z}. In terms of the function φ, the Haar function is ψ = 1 φ 1,0 1 φ 1, The Haar function is a special case of a spline wavelet. One can consider higher order splines N n and define associated multiresolution analyses, which leads to wavelets of the type ψ(x) = k Z c k N n (2x k). These wavelets are called Battle Lemarié wavelets. Except for the case n = 1, all coefficients c k are non-zero, which implies that the wavelet ψ has support equal to R. However, the wavelets have exponential decay. 92

93 For φ L 2 (R), let V j = span{d j T k φ} k Z. Then the MRA-conditions (iii) and (iv) are satisfied. When does V j generate an MRA? Lemma 9.3 Let φ L 2 (R). Then the following holds: (i) j V j = {0}. (ii) Assume that the spaces V j are nested. If ˆφ > 0 on a neighborhood of 0, then j V j is dense in L 2 (R). Thus, if the spaces V j are nested and the condition in (ii) is satisfied, then the MRA-conditions (i) (iv) are satisfied. We thus need a condition ensuring that V j are nested: 93

94 Lemma 9.4 Assume that φ L 2 (R) and that {T k φ} k Z is a Bessel sequence. Define the spaces V j by V j = span{d j T k φ} k Z. Then the following holds: (i) If ψ L 2 (R) and there exists a bounded 1-periodic function H 1 such that then ψ V 1. ˆψ(2γ) = H 1 (γ) ˆφ(γ), (ii) If there exists a bounded 1-periodic function H 0 such that ˆφ(2γ) = H 0 (γ) ˆφ(γ), then V j V j+1 for all j Z. 94

95 Via Lemma 9.3 and Lemma 9.4 we obtain the following: Theorem 9.5 Let φ L 2 (R), and assume that ˆφ > 0 on a neighborhood of 0. Assume further that ˆφ(2γ) = H 0 (γ) ˆφ(γ), is satisfied for a bounded 1-periodic function H 0. Define the spaces V j by V j = span{d j T k φ} k Z. Then the following holds: (i) If {T k φ} k Z is an orthonormal system, then φ and the spaces V j form a multiresolution analysis. (i) If {T k φ} k Z is a Bessel sequence, then the spaces V j satisfy the conditions (i) (iv) in Definition

96 Important properties for wavelet bases {ψ j,k } j,k Z : that ψ has a computationally convenient form, for example that ψ is a piecewise polynomial (a spline); regularity of ψ; symmetry (or anti-symmetry) of ψ, i.e., that ψ(x) = ψ( x) or ψ(x) = ψ( x), x R; compact support of ψ, or at least fast decay; that ψ has vanishing moments, i.e., that for a certain m N, x l ψ(x) dx = 0 for l = 0, 1,..., m. 96

97 Short description of the role of these conditions: Vanishing moments: needed if we want smooth wavelets. If ψ is an m times differentiable function with bounded derivatives with reasonable decay and {ψ j,k } j,k Z is an orthonormal system, then ψ has m 1 vanishing moments. Vanishing moments are essential in the context of compression. Assuming that {ψ j,k } j,k Z is an orthonormal basis for L 2 (R), every f L 2 (R) has the representation f = f, ψ j,k ψ j,k. (10) j,k Z All information about f is stored in the coefficients { f, ψ j,k } j,k Z, and (10) tells us how to reconstruct f based on the coefficients. In practice one can not store an infinite sequence of non-zero numbers, so one has to select a finite number of the coefficients to keep. Done by 97

98 Thresholding: Chooses a certain ɛ > 0 and keep only the coefficients f, ψ j,k for which f, ψ j,k ɛ. If ψ has a large number of vanishing moments, then only relatively few coefficients f, ψ j,k will be large: Theorem 9.6 Assume that the function ψ L 2 (R) is compactly supported and has N 1 vanishing moments. Then, for any N times differentiable function f L 2 (R) for which f (N) is bounded, there exists a constant C > 0 such that f, ψ j,k C 2 jn 2 j/2, j, k Z. 98

99 Further relevant properties: Compact support (or at least fast decay) of ψ is essential for the use of computer-based methods, where a function with unbounded support always has to be truncated. For the same reason we often want the support to be small. The condition of ψ being symmetric is helpful in image processing, where a non-symmetric wavelet will generate non-symmetric errors, which are more disturbing to the human eye than symmetric errors. 99

100 One can not combine the classical multiresolution analysis with the desire of having a symmetric wavelet ψ: Proposition 9.7 Assume that φ L 2 (R) is real-valued and compactly supported, and let V j = span{d j T k φ} k Z, j Z. Assume that (φ, {V j }) constitute a multiresolution analysis. Then, if the associated wavelet ψ is real-valued and compactly supported and has either a symmetry axis or an antisymmetry axis, then ψ is necessarily the Haar wavelet. Thus, under the above assumptions we are back at the function we want to avoid! 100

101 Definition 9.8 Let ψ L 2 (R). A frame for L 2 (R) of the form {D j T k ψ} j,k Z is called a dyadic wavelet frame. The associated frame operator: S : L 2 (R) L 2 (R), Sf = j,k Z f, D j T k ψ D j T k ψ. The frame decomposition: f = f, S 1 D j T k ψ D j T k ψ, f L 2 (R). j,k Z Inconvenient - one needs to calculate f, S 1 D j T k ψ, j, k Z. 101

102 Improvement: can show that S 1 D j T k ψ = D j S 1 T k ψ. Unfortunately, in general D j S 1 T k ψ D j T k S 1 ψ. We can not expect the canonical dual frame of a wavelet frame to have wavelet structure. Bownik and Weber: example of a wavelet system {ψ j,k } j,k Z for which The canonical dual does not have the wavelet structure; There exist infinitely many functions ψ for which { ψ j,k } j,k Z is a dual frame. 102

103 Example 9.9 Let {ψ j,k } j,k Z be a wavelet ONB for L 2 (R). Given ɛ ]0, 1[, let θ = ψ + ɛdψ. Then {θ j,k } j,k Z is a Riesz basis and the canonical dual frame is given by S 1 θ j,2k+1 = ψ j,2k+1 for all j, k Z. Also, for any k 0, S 1 θ j,2k = ψ j,2k ɛψ j 1,k + + 0, j Z, k 0. For k = 0, S 1 θ j,0 n=0 ( ɛ) n ψ j n,0, j Z. In particular, the canonical dual frame of {θ j,k } j,k Z does not have the wavelet structure. For a Riesz basis, the dual is unique: so no dual with wavelet structure exists! 103

104 Other properties which are not inherited by the canonical dual frame: If ψ has compact support, then θ also has compact support, and all the functions {θ j,k } j,k Z have compact support. For the canonical dual frame {S 1 θ j,k } j,k Z, the functions have compact support when k 0. However, the functions S 1 θ j,0 do not have compact support. 104

105 Fom general frame theory: Two ways to awoid inconvenient frame expansions: Look for tight frames; Look for convenient dual frame pairs. Some more aspects are relevant in the wavelet case: 105

106 The popular wavelet bases are based on multiresolution analysis, which leads to a very convenient algorithmic structure. This implies a special form for the function ψ generating the wavelet basis: ψ = k Z c k DT k φ for a certain function φ satisfying a scaling equation, ˆφ(2γ) = H 0 (γ) ˆφ(γ) for some 1-periodic function H 0. The algorithmic structure offered by a multiresolution analysis is a great advantage compared to the use of general wavelet orthonormal bases. Thus: While constructing wavelet frames, it is desirable to maintain the important aspects of the multiresolution analysis. 106

107 Therefore, it is natural to require the generator ψ for a tight frame {D j T k ψ} j,k Z to have the form ψ = k Z c k DT k φ for some function φ; and, if we want to construct two dual wavelet frames {D j T k ψ} j,k Z and {D j T k ψ}j,k Z, it is natural to require that both ψ and ψ have that form. We want the coefficients in these formulas to be finite sequences. The B-splines B m are obvious candidates for the function φ. However, we can not obtain all of these properties simultaneously: Theorem 9.10 Consider B m for m > 1. Then there does not exist pairs of dual wavelet frames {D j T k ψ} j,k Z and {D j T k ψ}j,k Z for which ψ and ψ are finite linear combinations of functions DT k B m, j, k Z. 107

108 Thus, neither the approach of looking at tight frames, nor the idea of considering wavelet frame pairs work if we want the generator (respectively, generators) to have the form ψ = k Z c k DT k B m for a finite sequence {c k }. Solution: we will consider systems of the wavelet-type, but generated by more than one function. 108

109 Definition 9.11 Consider two sequences of functions ψ 1,..., ψ n L 2 (R) and ψ 1,..., ψ n L 2 (R). We say that {D j T k ψ l } j,k Z,l=1,...,n and {D j T k ψl } j,k Z,l=1,...,n are a pair of dual multiwavelet frames if both are Bessel sequences and n f = f, D j T k ψ l D j T k ψl, f L 2 (R). l=1 j,k Z A pair of dual multiwavelet frames is also called sibling frames or bi-frames. The frame {D j T k ψ l } j,k Z,l=1,...,n itself is called a multiwavelet frame. 109

110 Characterization of all dual wavelet frame pairs: Theorem 9.12 Assume that {D j T k ψ l } j,k Z,l=1,...,n and {D j T k ψl } j,k Z,l=1,...,n are Bessel sequences. Then {D j T k ψ l } j,k Z,l=1,...,n and {D j T k ψl } j,k Z,l=1,...,n are a pair of dual wavelet frames if and only if the two equations n ψ l (2 j γ) ψl (2 j γ) = 1, n l=1 l=1 j=0 j Z hold for a.e. γ R. ψ l (2 j γ) ψl (2 j (γ + q)) = 0, q 2N

111 Characterization of tight wavelet frames generated by one generator ψ: Theorem 9.13 A function ψ L 2 (R) generates a tight wavelet frame {ψ j,k } j,k Z with frame bound A if and only if the equations ˆψ(2 j γ) 2 = A, j Z ˆψ(2 j γ) ˆψ (2 j (γ + q)) = 0, q 2N + 1 j=0 hold for a.e. γ R. 111

112 10 The unitary extension principle We will state the unitary extension principle of Ron and Shen, which enables us to construct tight frames for L 2 (R) of the form {D j T k ψ l } j,k Z,l=1,...,n. Terminology: The interval ] 1 2, 1 2 [ is identified with the torus T The class of 1-periodic functions on R whose restriction to ] 1 2, 1 2 [ belongs to Lp ( 1 2, 1 2 ) is denoted by L p (T). The functions ψ 1,..., ψ n will be constructed on the basis of a function satisfying a refinement equation. It is convenient to denote the refinable function by ψ 0 instead of φ. 112

113 Standing assumptions and conventions: Let ψ 0 L 2 (R) and assume that (i) There exists a function H 0 L (T) such that ψ 0 (2γ) = H 0 (γ) ψ 0 (γ). (ii) lim γ 0 ψ0 (γ) = 1. Further, let H 1,..., H n L (T), and define ψ 1,..., ψ n L 2 (R) by ψ l (2γ) = H l (γ) ψ 0 (γ), l = 1,..., n. Finally, let H denote the (n + 1) 2 matrixvalued function defined by H 0 (γ) T 1/2 H 0 (γ) H 1 (γ) T 1/2 H 1 (γ) H(γ) =, γ R. H n (γ) T 1/2 H n (γ) 113

114 We want to find conditions on the functions H 1,..., H n such that ψ 1,..., ψ n generate a multiwavelet frame for L 2 (R). Explicit expression for ψ l : expanding H l in a Fourier series, H l (γ) = k Z c k,l e 2πikγ, we have ψ l (x) = 2 k Z c k,l DT k ψ 0 (x) = 2 k Z c k,l ψ 0 (2x + k). Note: If H l are trigonometric polynomials, the sums are finite. Therefore the functions ψ l have compact support if ψ 0 has compact support. 114

115 The general setup presented here preserves the algorithmic structure of a multiresolution analysis: by Theorem 9.5, the spaces V j := span{d j T k ψ 0 } k Z, j Z, satisfy the conditions for a multiresolution analysis in Definition 9.1, except (v). Also, ψ 1,..., ψ n V 1. The unitary extension principle: Theorem 10.1 Let {ψ l, H l } n l=0 be as in the general setup on page 113, and assume that H(γ) H(γ) = I for a.e. γ T. Then the multiwavelet system {D j T k ψ l } j,k Z,l=1,...,n constitutes a tight frame for L 2 (R) with frame bound equal to

116 The matrix H(γ) H(γ) has four entries, but it is enough to verify two sets of equations: Corollary 10.2 Let {ψ l, H l } n l=0 be as in the general setup on page 113, and assume that n H l (γ) 2 = 1, l=0 n H l (γ)t 1/2 H l (γ) = 0, l=0 for a.e. γ T. Then {D j T k ψ l } j,k Z,l=1,...,n constitutes a tight frame for L 2 (R) with frame bound equal to

117 Example 10.3 For any m = 1, 2,..., we consider the B-spline ψ 0 := B 2m of order 2m. Then ψ 0 (γ) = ( ) 2m sin(πγ). πγ It is clear that lim γ 0 ψ0 (γ) = 1, and by direct calculation, ( ) 2m sin(2πγ) ψ 0 (2γ) = 2πγ ( ) 2m 2 sin(πγ) cos(πγ) = 2πγ = cos 2m (πγ) ψ 0 (γ). Thus ψ 0 satisfies a refinement equation with two-scale symbol H 0 (γ) = cos 2m (πγ). 117

118 Now, consider the binomial coefficient ( ) 2m (2m)! := l (2m l)!l!, and define the functions H 1,..., H 2m L (T) by ( ) 2m H l (γ) = sin l (πγ) cos 2m l (πγ). l Using that cos(π(γ 1/2)) = sin(πγ) and sin(π(γ 1/2)) = cos(πγ), ( ) 2m T 1/2 H l (γ) = ( 1) l cos l (πγ) sin 2m l (πγ). l 118

119 Thus, the matrix H is given by H(γ) = H 0 (γ) T 1/2 H 0 (γ) H 1 (γ) T 1/2 H 1 (γ) H 2m (γ) T 1/2 H 2m (γ) = cos 2m (πγ) sin 2m (πγ) ( ) ( ) 2m 2m sin(πγ) cos 1 2m 1 (πγ) cos(πγ) sin 1 2m 1 (πγ) ( ) ( ) 2m 2m sin 2 2 (πγ) cos 2m 2 (πγ) cos 2 2 (πγ) sin 2m 2 (πγ). ( ) ( ) 2m 2m sin 2m 2m (πγ) cos 2m 2m (πγ) 119

120 We now verify the conditions in Corollary Using the binomial formula 2m ( ) 2m (x + y) 2m = x l y 2m l, l 2m l=0 H l (γ) 2 = l=0 2m l=0 ( 2m l ) sin 2l (πγ) cos 2(2m l) (πγ) = ( sin 2 (πγ) + cos 2 (πγ) ) 2m = 1, γ T. Using the binomial formula with x = 1, y = 1, 2m l=0 H l (γ)t 1/2 H l (γ) = sin 2m (πγ) cos 2m (πγ) 2m l=0 ( 1) l ( 2m l = sin 2m (πγ) cos 2m (πγ)(1 1) 2m = 0. ) 120