FRAMES AND TIME-FREQUENCY ANALYSIS

Transcription

1 FRAMES AND TIME-FREQUENCY ANALYSIS LECTURE 2: GABOR FRAMES Christopher Heil Georgia Tech heil

2 READING For background on Hilbert spaces and operator theory: Chapters 2 in C. Heil, A Basis Theory Primer, Birkhäuser, Boston, 20. For background on the Fourier transform: Chapter 9 in C. Heil, A Basis Theory Primer, Birkhäuser, Boston, 20. Today s lecture is based upon: Chapter (Sections..5 and.9) in C. Heil, A Basis Theory Primer, Birkhäuser, Boston, 20. Also see: C. E. Heil and D. F. Walnut, Continuous and discrete wavelet transforms, SIAM Review, 3 (989), pp For further reading: K. Gröchenig, Foundations of Time-Frequency Analysis, Birkhäuser, Boston, 200. C. Heil, History and evolution of the Density Theorem for Gabor frames, J. Fourier Anal. Appl., 3 (2007), pp

3 { e 2πinxχ(x) } n Z { e 2πinxχ(x k) } n Z BAD EXAMPLE Let χ = χ [0,]. Then is an ONB for L 2 [0, ]. Likewise is an ONB for L 2 [k, k + ]. Since the intervals [k, k + ] are nonoverlapping, { e 2πinxχ(x k) } k,n Z is an ONB for L 2 (R). Disadvantage : χ is discontinuous (but it does decay quickly). πiξ sinπξ Disadvantage 2: χ(ξ) = e πξ decays on the order of /ξ (but it is smooth). Question: Can we somehow generate an ONB for L 2 (R) based on a single function g that is both smooth and decays quickly? Remark: Smoothness and decay are interchanged under the Fourier transform. 3

4 FUNDAMENTAL OPERATORS Translation: (T a f)(x) = f(x a), a R. Modulation: (M b f)(x) = e 2πibx f(x), b R. Dilation: D λ f(x) = λ /2 f(λx), λ > 0. Involution: f(x) = f( x). Time-frequency shifts are M b T a and T a M b (note T a M b = e 2πiab M b T a ) Figure. The Gaussian window φ(x) = e πx2 and the real part of the time-frequency shift M 3 T 5 φ. (Regular or Lattice) Gabor (Gah-bor) System: G(g, a, b) = {M bn T ak g} k,n Z = {e 2πibnx g(x ak)} k,n Z. g is the atom and az bz is the lattice. 4

5 WAVELETS Figure 2. The function ψ(x) = e πx2 cos 3x and a time-scale shift D 2 3T 36 ψ(x) = 8 /2 ψ(8x 36). (Regular) Wavelet (Wave-let) System: W(ψ, a, b) = {D a nt bk ψ} k,n Z = {a n/2 ψ(a n x bk)} n Z. Remark The Heisenberg group is H = { e 2πit T a M b }a,b,t R. The Affine group is A = {D a T b } a>0,b R. Both are nonabelian LCGs, but H is unimodular while A is not. H = T R R with group operation (z, a, b) (w, c, d) = (e 2πiad zw, a + c, b + d). The Schrödinger representation is (z, a, b) zm b T a. 5

6 TWO WAVELET ONBs Haar System (90): W(ψ, 2, ) = {2 n/2 ψ(2 n x k)} n Z with ψ = χ [0,/2] χ [/2,]. Daubechies W 4 System ( 988): W(W 4, 2, ) = {2 n/2 W 4 (2 n x k)} n Z Figure 3. Left: Haar wavelet ψ. Right: Daubechies wavelet W 4. 6

7 GABOR FRAME OPERATOR Assuming G(g, a, b) = {M bn T ak g} k,n Z = {e 2πibnx g(x ak)} k,n Z is a frame, its frame operator is Sf = f, M bn T ak g M bn T ak g, f L 2 (R). n Z k Z Assignment. Prove that the Gabor frame operator S commutes with M bn and T ak, and this implies that S commutes with M bn and T ak. Therefore S (M bn T ak g) = M bn T ak (S g). Since the canonical dual is the image of G(g, a, b) under S, it has the form {S (M bn T ak g)} k,n Z = {M bn T ak g} k,n Z, = G( g, a, b), where g = S g. Assignment 2. Assuming G(g, a, b) is a frame, prove the following. (a) G(g, a, b) is a Riesz basis (i.e., exact) M bn T ak g, M bn T ak g = δ nn δ kk g, g =. (b) The canonical Parseval frame is G(g, a, b) where g = S /2 g. 7

8 PAINLESS NONORTHOGONAL EXPANSIONS Theorem (Daubechies, Grossmann, Meyer). Fix a, b > 0 and g L 2 (R). (a) If 0 < ab and supp(g) [0, b ], then G(g, a, b) is a frame for L 2 (R) if and only if there exist constants A, B > 0 such that Ab g(x ak) 2 Bb a.e. () k Z In this case, A, B are frame bounds for G(g, a, b). (b) If 0 < ab <, then there exist g supported in [0, b ] that satisfy () and are as smooth as we like (even infinitely differentiable). (c) If ab =, then any g that is supported in [0, b ] and satisfies () must be discontinuous. (d) If ab > and g is supported in [0, b ], then () is not satisfied and G(g, a, b) is incomplete in L 2 (R) Figure 4. Example of g(x) 2 and g(x a) 2 using a = 3 and b = /4. 8

9 Proof. (a) Set e bn (x) = e 2πibnx and I k = [ak, ak + b ]. Note {b /2 e bn } n Z is an ONB for L 2 (I k ). Taking f in the dense subspace C c (R), b f, M bn T ak g 2 = b 2 f(x) e 2πibnx g(x ak) dx n Z n Z = b ak+b f(x) g(x ak) e 2πibnx dx n Z ak = f T ak g, b /2 e bn 2 L 2 (I k ) n Z 2 = f Tak g L2 (I k ) (Plancherel) Hence, k,n Z = ak+b ak f(x) Tak g(x) 2 dx = f, M bn T ak g 2 = b k Z f(x) g(x ak) 2 dx f(x) g(x ak) 2 dx = b f(x) 2 g(x ak) 2 dx k Z (only translations!). f(x) 2 A dx = A f 2 L 2 (R). 9

10 Summary: If 0 < ab < then there exist nice Gabor frames for L 2 (R). If ab = then there exist Gabor frames G(g, a, b), but the ones we constructed are discontinuous. If ab > then we cannot construct a Gabor frame using this method. Assignment 3. Assume (with same conditions on g) that we have a frame, and set G 0 (x) = g(x ak) 2. k Z Prove the following. (a) The frame operator is Sf(x) = b G 0 (x) f(x), and S f(x) = bf(x)/g 0 (x). (b) If ab = then G(g, a, b) is a Riesz basis (= exact frame). (c) If ab < then G(g, a, b) is a redundant frame (overcomplete). Assignment 4. Given g C c (R), g 0, prove there exist a, b > 0 such that G(g, a, b) is a frame for L 2 (R). Question (extremely difficult): For which a, b > 0 is G(χ [0,], a, b) a frame? 0

11 THE NYQUIST DENSITY FOR GABOR FRAMES Theorem 2. Assume G(g, a, b) is a frame for L 2 (R) with frame bounds A, B > 0. Then In particular, g must be bounded. Ab k Z g(x ak) 2 Bb a.e. (2) Proof. Note that we are not assuming that g has compact support, but if f is supported in an interval I of length b, then f(x) 2 k Z g(x ak) 2 dx = b k,n Z f, M bn T ak g 2 (just as before) ba f 2 2 (since it s a frame) = ba f(x) 2 dx. Hence f(x) 2 ( k Z ) g(x ak) 2 ba dx 0, whenever supp(f) I. Therefore the lower inequality in (2) holds a.e. any interval I of length b.

12 Corollary 3 (Density and Frame Bounds). If G(g, a, b) is a frame for L 2 (R) with frame bounds A, B, then the following statements hold. (a) Aab g 2 2 Bab. (b) If G(g, a, b) is Parseval (A = B = ), then g 2 2 = ab. (c) 0 < ab. (d) g, g = ab, where g = S g. (e) G(g, a, b) is a Riesz basis if and only if ab = g, g =. Proof. (a) Recall that Ab k Z g(x ak) 2 Bb a.e. Integrating over [0, a]: Aab = a 0 Ab dx a 0 k Z g(x ak) 2 dx = g(x) 2 dx = g 2 2. (c) The canonical Parseval frame is G(g, a, b) where g = S /2 g. The elements of a Parseval frame have at most unit norm, so ab = g 2 2. (d) Since S /2 is self-adjoint, g, g = g, S /2 S /2 g = S /2 g, S /2 g = g 2 2 = ab. (e) G(g, a, b) is a Riesz basis if and only if g, g =. 2

13 Summary: If ab > then G(g, a, b) is not a frame. If G(g, a, b) is a frame and ab = then it is a Riesz basis. If G(g, a, b) is a frame and 0 < ab < then it is a redundant frame. The value /(ab) is called the density of the Gabor system G(g, a, b), because the number of points of az bz that lie in a given ball in R 2 is asymptotically /(ab) times the volume of the ball as the radius increases to infinity. We refer to the density /(ab) = as the critical density or the Nyquist density. Remark: 0 < ab is not sufficient for G(g, a, b) to be a frame. For example, G(χ [0, 2 ],, ) is not a frame because its elements are supported on intervals [k, k + 2 ] with k integer. 3

14 WIENER AMALGAM SPACES The following example shows some of the limitations of L p -norms. Example 4. If χ = χ [0,), then G(χ,, ) is an ONB for L 2 (R). Note that χ is discontinuous, but at least it is well-localized in time. Create a new function ( Walnut s function ) by dividing [0, ) into infinitely many pieces and then sending those pieces off to infinity : g = χ [0, 2 ) + T χ [ 2,3 4 ) + T 2 χ [ 3 4,7 8 ) + = χ [0, 2 ) + χ [+ 2,+3 4 ) + χ [2+ 3 4,2+7 8 ) +. Then g is not continuous and does not decay at infinity, but g 2 = χ 2. Further, because we translated the pieces by integers, G(g,, ) is an ONB for L 2 (R). We cannot distinguish a well-localized function from a poorly-localized function only by considering their L p -norms. 4

15 We create a space of functions which are locally L p and globally l q. Definition 5. Given p and q <, the Wiener amalgam space W(L p, l q ) consists of those functions f L p loc (R) for which the norm ( f W(Lp,l q ) = f χ [k,k+] q p k Z ) /q is finite. For q = we substitute the l -norm for the l q -norm above, i.e., We also define f W(Lp,l ) = sup f χ [k,k+] p. k Z W(C, l q ) = { f W(L, l q ) : f is continuous }, and we impose the norm W(L,l q ) on W(C, l q ). Example 6. We will be most interested in W(L, l ). χ = χ [0,] W(L, l ) L p (R) for every p. Walnut s function belongs to L p (R) for every p, but does not belong to W(L, l ). 5

16 We will be most interested in W(L, l ): f W(L,l ) = k Z f χ [k,k+]. Theorem 7. Let C a = max { + a, 2 }. (a) W(L, l ) L p (R) for p, and is dense for p <. (b) W(L, l ) is closed under translations, and T b f W(L,l ) 2 f W(L,l ). (c) W(L, l ) is an ideal in L (R), i.e., f L (R), g W(L, l ) = fg W(L, l ). (d) For each a > 0, f a = k Z f χ [ak,a(k+)] is an equivalent norm for W(L, l ), with C /a f a f W(L,l ) C a f a. 6

17 Corollary 8. If f W(L, l ), then k Z T ak f χ [0,a] = k Z f T ak χ [0,a] = k Z f χ [ak,a(k+)] C f W(L,l ). Corollary 9. If f W(L, l ), then its a-periodization ϕ(x) = n Z T ak f(x) = n Z f(x ak) is bounded, and T ak f = k Z T ak f χ [0,a] C f W(L,l ). k Z 7

18 THE WALNUT REPRESENTATION The Painless Nonorthogonal Expansions require supp(g) [0, b ]. For a generic g L 2 (R) we can break g into pieces supported on intervals of length b, analyze each piece, and past the pieces back together (to do this we will need to assume that g belongs to g W(L, l )). Definition 0. Given g W(L, l ) and a, b > 0, we define associated correlation functions G n (x) = k Z g(x ak) g(x ak n b ), n Z. In particular, G 0 (x) = k Z g(x ak) 2. The usual lattice az bz and the adjoint lattice b Z az implicitly appear here. Equivalent forms: G n = T ak g T ak+ n bḡ = ( T bḡ) ak g T n. k Z k Z Thus G n is the a-periodization of g T n bḡ. Since g W(L, l ), we have G n L (R). Lemma. If g W(L, l ) then G n C g 2 W(L,l ). n Z 8

19 Proof. We compute that Therefore G n = k Z ( T bḡ) ak g T n C bḡ g T n W(L,l ). G n C g T n g b W(L,l ) n Z n Z = C g χ [k,k+] T ng χ b [k,k+] n Z k Z C k Z g χ [k,k+] ( n Z T n b g χ [k,k+] ). The series in parentheses on the last line resembles the W(L, l ) norm of T n g, but it is not b since the summation is over n instead of k. But for an appropriate constant C we have T ng χ b [k,k+] = g χ [ n b +k, n b +k+] n Z n Z C m Z g χ [m,m+] = C g W(L,l ). Hence G n C k Z g χ [k,k+] C g W(L,l ) C C g 2 W(L,l ). n Z 9

20 Theorem 2 (Walnut Representation). If g W(L, l ) and a, b > 0, then G(g, a, b) is a Bessel sequence, with Bessel bound B = C g 2 W(L,l ) and frame operator Sf = f, M bn T ak g M bn T ak g = b T nf G b n, f L 2 (R). (3) n Z k Z n Z Proof. Because g W(L, l ), the series in (3) converges absolutely in L 2 (R), and Sf 2 b T nf b 2 G n B f 2, where B = C g 2 W(L,l ). n Z Fix f C c (R) and k Z. Then f T ak ḡ is bounded and compactly supported, so its b - periodization F k (x) = j Z f ( x j b ) ( ) g x ak j b belongs to L [0, b ]. We compute that f, M bn T ak g 2 = 2 f(x) e 2πibnx g(x ak) dx n Z n Z = b f ( x j ) ( b e 2πibn x j ) b g ( x ak j ) 2 b dx n Z 0 j Z = b f ( x j ) ( 2 b g x ak j b) e 2πibnx dx n Z 0 j Z = F k, e bn 2 L 2 [0,b ] n Z 20

21 Using Fubini, f, M bn T ak g 2 = b k Z k Z n Z = Fk 2 L 2 [0,b ] = b b = b k Z = b k Z = b k Z = b j Z b 0 b 0 0 F k (x) F k (x)dx j Z F k (x) 2 dx. f ( x j ) ( ) ( b g x ak j b Fk x j b) dx f(x) g(x ak) F k (x)dx f(x) g(x ak) j Z f ( x j b ) g ( x ak j b) dx f(x) f ( x j ) b g(x ak) g ( x ak j b) dx k Z = b f(x) f ( x j b) Gj (x)dx j Z = f, b T jf G j (x) b j Z L 2 (R) = f, Sf f 2 Sf 2 B f 2. 2

22 Remark: The Bessel bound can be given explicitly as B = 2 b C /a C b g 2 W(L,l ), where C a = max { + a, 2 }. Assignment 5 (Frame Perturbations). Let {x n } n N be a frame for a Hilbert space H with frame bounds A, B. Given {y n } n N H, prove that if {x n y n } n N is Bessel with Bessel bound K < A, then {y n } n N is a frame. (Typos in printouts.) Assignment 6 (Perturbation in amalgam norm). (a) Assume G(g, a, b) is a frame for L 2 (R). Show that there exists a δ > 0 such that if h L 2 (R) and g h W(L,l ) < δ, then G(h, a, b) is a frame for L 2 (R). (b) Does this remain true if we perturb in L 2 -norm instead? Here is another consequence of the Walnut representation (but this takes some work to prove, e.g., Theorem 4..8 in H/Walnut survey Continuous and discrete wavelet transforms ). Corollary 3. If g W(L, l ), then G(g, a, b) is a frame for all small enough a, b. 22

23 BONUS: THE HRT CONJECTURE We have studied Gabor systems G(g, a, b) that are complete, incomplete, a frame, exact, inexact, a Riesz basis, an ONB, etc. But what about the most basic linear algebra property of all? Question 4 (HRT, 996). Are Gabor systems finitely independent? (Reference: C. Heil, J. Ramanathan, and P. Topiwala, Linear independence of time-frequency translates, Proc. Amer. Math. Soc., 24 (996), pp ) Independence here is linear independence in the most ordinary linear algebra sense: If there is some finite linear combination of elements of G(g, a, b) that equals zero, then the system is dependent, otherwise it is independent. If G(g, a, b) is minimal (has a biorthogonal system), then certainly it is independent. In particular, every exact system, every Riesz basis, and every ONB is independent. But what about redundant Gabor frames? Are they redundant because they are linearly dependent? For lattice Gabor systems the answer is known (but the proof is far from trivial!). Theorem 5 (Linnell, 999). If g L 2 (R)\{0} and a, b > 0, then G(g, a, b) is independent. In particular, if Λ = {(a k, b k )} N k= az bz then G(g, Λ) = {M b k T ak g} N k= 23 is independent.

24 For general Gabor systems the answer is unknown. Conjecture 6 (HRT Conjecture). If g L 2 (R) is not the zero function and Λ = {(a k, b k )} N k= is any set of finitely many distinct points in R 2, then G(g, Λ) = { M bk T ak g } N k= is a linearly independent set of functions in L 2 (R). That is, N c k e 2πibkx g(x a k ) = 0 = c = = c N = 0. k= Assignment 7. Either prove that the HRT Conjecture is valid, or find a counterexample. Or, for simplicity, just prove the special case given in the next conjecture. Conjecture 7 (HRT Subconjecture). If g L 2 (R)\{0} then { g(x), g(x ), e 2πix g(x), e 2πi 2x g(x 2) } is linearly independent ( here, Λ = {(0, 0), (, 0), (0, ), ( 2, 2)} ). Assignment 8. Let Λ = {(0, 0), (, 0), (0, b)}. Linnell s theorem implies that HRT holds for G(g, Λ) for any nonzero g L 2 (R). Find a simple proof that HRT holds for this case! 24

25 Evidence for HRT If we use only time-shifts g(x a k ) or frequency shifts e 2πib kx g(x) then it is true: N c k g(x a k ) = 0 a.e. = k= ( N c k g(x a k ) k= ) (ξ) = 0 a.e. N = c k e 2πia kξ ĝ(ξ) = 0 a.e. k= ( N ) = ĝ(ξ) c k e 2πia kξ = 0 a.e. k= = ĝ = 0 a.e. = g = 0 a.e. (Interesting side question: What if g L p with p > 2? ĝ is a distribution!) Assignment 9. Use the fact that a nontrivial trigonometric polynomial N k= c k e 2πia kξ cannot vanish on any set of positive measure to prove that the HRT holds for all compactly supported atoms g (using any finite set Λ). 25

26 Evidence against HRT It s false if we use time-scale instead of time-frequency. Let χ = χ [0,]. Then χ(x) = χ(2x) + χ(2x ). = + The Haar wavelet, which generates a wavelet ONB for L 2 (R), is ψ(x) = χ(2x) χ(2x ). The box function χ is the scaling function for the Haar ONB Figure 5. Left: Haar wavelet ψ. Right: Daubechies wavelet W 4. 26

27 The Daubechies D 4 function: Time-scale shifts of D 4 are dependent: D 4 (x) = D 4 (2x) D 4 (2x ) D 4 (2x 2) D 4 (2x 3) The function D 4 is a scaling function that generates a multiresolution analysis. This MRA gives rise to the Daubechies wavelet W 4, which generates a wavelet ONB for L 2 (R). Dependency is critical for the construction of wavelet ONBs! 27

28 METAPLECTIC TRANSFORMATIONS Recall the unitary dilation operator D r f(x) = r /2 f(rx). We have D r (M bn T ak g)(x) = r /2 M bn T ak g(rx) = r /2 e 2πibnrx g(rx ak) = e 2πi(bnrx) r /2 g(r(x ak/r)) = M brn T ak/r D r g(x). Therefore ( ) { D r G(g, a, b) = Dr (M bn T ak g) } k,n Z = { M brn T ak/r D r g } k,n Z = G(D rg, a/r, br). Since D r is unitary, G(g, a, b) is independent G(D r g, a/r, br) is independent. The dilation D r is one example of a metaplectic transform. We can replace the lattice az bz with the rescaled lattice (a/r)z (br)z, at the cost of replacing g with its image D r g under a unitary map. More generally, if Λ is an arbitrary set of points in R 2, then the image of G(g, Λ) under the unitary map D r is G(D r g, r (Λ)), where r = [ /r 0 0 r ]. Note that r is area preserving (has determinant ). 28

29 The Fourier transform is another metaplectic transform. We have (M b T a g) = T b M a ĝ = e 2πiab M a T b. Scalars of unit modulus do not affect properties like completeness, independence, or being a frame: G(g, Λ) is a frame { M b T a g } (a,b) Λ is a frame { e 2πiab M a T b ĝ } k,n Z is a frame G(ĝ, R(Λ)) is a frame, where R = [ 0 0 ] is rotation by π/2 (an area-preserving transformation!). Now, every 2 2 matrix with det(a) = can be written as a product of rotations R, dilations r, and shears S r = [ 0 r ]. Assignment 0. Given g L 2 (R) and Λ R 2, show that G(g, Λ) is independent G(h, S r (Λ)) is independent, where h(x) = e πirx2 g(x). Remark: If d >, then it is not true that every 2d 2d matrix with det(a) = can be factored this way. Only the symplectic matrices can be so factored. 29

30 Combining the previous results gives us the following corollary. Corollary 8. Let A be a 2 2 matrix with det(a) =. Then there exists a unitary transform U on L 2 (R) such that G(g, Λ) is independent G(Ug, A(Λ)) is independent. We can also translate Λ by z = (c, d), at the cost of replacing g by M d T c g. Combining these results with Linnell s Theorem gives the following corollary. Corollary 9. If g L 2 (R) is nonzero and Λ is any finite subset of A(aZ bz) + z where det(a) = and z R 2, then G(g, Λ) is linearly independent. In other words, HRT holds whenever g is nonzero and Λ is a finite subset of a translate of a full-rank lattice. Any 3 points in R 2 lie on a translate of a full-rank lattice, so HRT holds when Λ 3. Corollary 20. If g L 2 (R) is nonzero and Λ contains, 2, or 3 distinct points of R 2, then G(g, Λ) is linearly independent. The 4-point set Λ = { (0, 0), (, 0), (0, ), ( 2, 2) } from the HRT Subconjecture is not a subset of a translate of a full-rank lattice! HRT is open for 4-point sets (though some special cases are known). 30

31 JUST THREE POINTS Let s sketch the proof that { g(x), g(x a), g(x)e 2πix} is independent. Suppose c g(x) + c 2 g(x a) + c 3 g(x)e 2πix = 0, all x R. Rewriting, g(x a) = m(x) g(x), where m(x) = c c 2 c 3 c 2 e 2πix is -periodic. Iterating: Hence g(x 2a) = g((x a) a) = m(x a) g(x a) = m(x a) m(x) g(x). g(x ka) = Taking absolute values and logarithms: ln g(x ka) = ( k ) m(x ja) g(x). j=0 k ln m(x ja) + ln g(x) = j=0 where p is -periodic. Therefore k p(x ja) + ln g(x), j=0 k ln g(x ka) = k k p(x ja) + k j=0 ln g(x). 3

32 k ln g(x ka) = k k p(x ja) + k j=0 ln g(x). The term p(x ja) is the area of the box with base centered at x ja (mod ) k with height p(x ja) and width /k. The sum is the sum of the areas of the boxes: It s like a Riemann sum approximation to p(x) dx, except the boxes are randomly distributed at least, if a is irrational. 0 Ergodic Theorem: k k p(x ja) j=0 0 p(x) dx as k. 32

33 Consequently: k k p(x ja) j=0 0 p(x) dx = C k ln m(x ja) Ck j=0 k m(x ja) e Ck j=0 g(x ka) = ( k j=0 ) m(x ja) g(x) e Ck g(x) If C > 0 then g is growing as x, contradicting g(x) 2 dx <. Symmetric argument if C < 0. And C = 0 takes more work. 33

34 THE SUBTLETIES OF REDUNDANCY Why care? (a) It s an amazingly simple-to-state question. (b) The HRT Conjecture can be reformulated in terms of the the Heisenberg group. In this form, it is somewhat related to the following deep open conjecture. Conjecture 2 (Zero Divisor Conjecture; Higgins, 940). The group algebra F G of a torsionfree group G over a field F is a domain. (c) The nature of redundancy in infinite-dimensional settings is surprisingly subtle. Here is an example. Conjecture 22 (Feichtinger Conjecture, 990). Every frame can be written as a finite union of nonredundant (= Riesz) subsequences. 34

35 The Feichtinger conjecture has been shown to be equivalent to another famous problem. Conjecture 23 (Kadison Singer (Paving) Conjecture, 959). ε > 0, M such that n, n n matrices S having zero diagonal, partition {σ j } M j= of {,..., n} such that P σj SP σj ε S, j =,..., M, where P I is the orthogonal projection onto span{e i } i I. KADISON SINGER HAS RECENTLY BEEN SETTLED!! Adam Marcus, Dan Spielman, and Nikhil Srivastava, 203 Interlacing Families II: Mixed Characteristic Polynomials and the Kadison-Singer Problem K-S is true; so Feichtinger is true as well but how many subsequences is still unknown. 35

36 SOME TIME-SCALE DEPENDENCIES JUST FOR FUN The Devil s Staircase (Cantor Lebesgue Function) 2 ϕ(x) = 2 ϕ(3x) + 2 ϕ(3x ) + ϕ(3x 2) + 2 ϕ(3x 3) + 2ϕ(3x 4). Refinable functions like these play important roles in wavelet theory and in subdivision schemes in computer-aided graphics. 36

37 de Rham s Nowhere Differentiable Function ϕ(x) = 2 3 ϕ(3x) + 3 ϕ(3x ) + ϕ(3x 2) + 3 ϕ(3x 3) + 2 3ϕ(3x 4). 37