Finite Frames and Sigma Delta Quantization

Transcription

1 Finite Frames and Sigma Delta Quantization with Matt Fickus Alex Powell and Özgür Yilmaz Aram Tangboondouangjit Finite Frames and Sigma Delta Quantization p.1/57

2 Frames Frames F = {e n } N n=1 for d-dimensional Hilbert space H, e.g., H = K d, where K = C or K = R. Any spanning set of vectors in K d is a frame for K d. Finite Frames and Sigma Delta Quantization p.2/57

3 Frames Frames F = {e n } N n=1 for d-dimensional Hilbert space H, e.g., H = K d, where K = C or K = R. Any spanning set of vectors in K d is a frame for K d. F K d is A-tight if x K d, A x 2 = N x, e n 2 n=1 Finite Frames and Sigma Delta Quantization p.2/57

4 Frames Frames F = {e n } N n=1 for d-dimensional Hilbert space H, e.g., H = K d, where K = C or K = R. Any spanning set of vectors in K d is a frame for K d. F K d is A-tight if x K d, A x 2 = N x, e n 2 n=1 If {e n } N n=1 is a finite unit norm tight frame (FUN-TF) for K d, with frame constant A, then A = N/d. Finite Frames and Sigma Delta Quantization p.2/57

5 Frames Frames F = {e n } N n=1 for d-dimensional Hilbert space H, e.g., H = K d, where K = C or K = R. Any spanning set of vectors in K d is a frame for K d. F K d is A-tight if x K d, A x 2 = N x, e n 2 n=1 If {e n } N n=1 is a finite unit norm tight frame (FUN-TF) for K d, with frame constant A, then A = N/d. Let {e n } be an A-unit norm TF for any separable Hilbert space H. A 1, and A = 1 {e n } is an ONB for H (Vitali). Finite Frames and Sigma Delta Quantization p.2/57

6 The geometry of finite tight frames The vertices of platonic solids are FUN-TFs. Finite Frames and Sigma Delta Quantization p.3/57

7 The geometry of finite tight frames The vertices of platonic solids are FUN-TFs. Points that constitute FUN-TFs do not have to be equidistributed, e.g., ONBs and Grassmanian frames. Finite Frames and Sigma Delta Quantization p.3/57

8 The geometry of finite tight frames The vertices of platonic solids are FUN-TFs. Points that constitute FUN-TFs do not have to be equidistributed, e.g., ONBs and Grassmanian frames. FUN-TFs can be characterized as minimizers of a frame potential function (with Fickus) analogous to Finite Frames and Sigma Delta Quantization p.3/57

9 The geometry of finite tight frames The vertices of platonic solids are FUN-TFs. Points that constitute FUN-TFs do not have to be equidistributed, e.g., ONBs and Grassmanian frames. FUN-TFs can be characterized as minimizers of a frame potential function (with Fickus) analogous to Coulomb s Law. Finite Frames and Sigma Delta Quantization p.3/57

10 Frame force and potential energy F : S d 1 S d 1 \ D R d P : S d 1 S d 1 \ D R, where P(a, b) = p( a b ), Coulomb force p (x) = xf(x) CF(a, b) = (a b)/ a b 3, f(x) = 1/x 3 Finite Frames and Sigma Delta Quantization p.4/57

11 Frame force and potential energy F : S d 1 S d 1 \ D R d P : S d 1 S d 1 \ D R, where P(a, b) = p( a b ), Coulomb force p (x) = xf(x) Frame force CF(a, b) = (a b)/ a b 3, f(x) = 1/x 3 FF(a, b) =< a, b > (a b), f(x) = 1 x 2 /2 Finite Frames and Sigma Delta Quantization p.4/57

12 Frame force and potential energy F : S d 1 S d 1 \ D R d P : S d 1 S d 1 \ D R, where P(a, b) = p( a b ), Coulomb force p (x) = xf(x) CF(a, b) = (a b)/ a b 3, f(x) = 1/x 3 Frame force FF(a, b) =< a, b > (a b), f(x) = 1 x 2 /2 Total potential energy for the frame force TFP({x n }) = Σ N m=1σ N n=1 < x m, x n > 2 Finite Frames and Sigma Delta Quantization p.4/57

13 Characterization of FUN-TFs For the Hilbert space H = R d and N, consider {x n } N 1 S d 1... S d 1 and TFP({x n }) = Σ N m=1σ N n=1 < x m, x n > 2. Theorem Let N d. The minimum value of TFP, for the frame force and N variables, is N; and the minimizers are precisely the orthonormal sets of N elements for R d. Finite Frames and Sigma Delta Quantization p.5/57

14 Characterization of FUN-TFs For the Hilbert space H = R d and N, consider {x n } N 1 S d 1... S d 1 and TFP({x n }) = Σ N m=1σ N n=1 < x m, x n > 2. Theorem Let N d. The minimum value of TFP, for the frame force and N variables, is N; and the minimizers are precisely the orthonormal sets of N elements for R d. Theorem Let N d. The minimum value of TFP, for the frame force and N variables, is N 2 /d; and the minimizers are precisely the FUN-TFs of N elements for R d. Finite Frames and Sigma Delta Quantization p.5/57

15 Characterization of FUN-TFs For the Hilbert space H = R d and N, consider {x n } N 1 S d 1... S d 1 and TFP({x n }) = Σ N m=1σ N n=1 < x m, x n > 2. Theorem Let N d. The minimum value of TFP, for the frame force and N variables, is N; and the minimizers are precisely the orthonormal sets of N elements for R d. Theorem Let N d. The minimum value of TFP, for the frame force and N variables, is N 2 /d; and the minimizers are precisely the FUN-TFs of N elements for R d. Problem Find FUN-TFs analytically, effectively, computationally. Finite Frames and Sigma Delta Quantization p.5/57

16 Given u 0 and {x n } n=1 u n = u n-1 + x n -q n q n = Q(u n-1 + x n ) u n = u n-1 + x n -q n x n D + Q qn First Order Σ Finite Frames and Sigma Delta Quantization p.6/57

17 A quantization problem Qualitative Problem Obtain digital representations for class X, suitable for storage, transmission, recovery. Quantitative Problem Find dictionary {e n } X: 1. Sampling [continuous range K is not digital] x X, x = x n e n, x n K (R or C). 2. Quantization. Construct finite alphabet A and Q : X { q n e n : q n A K} such that x n q n and/or x Qx small. Methods Fine quantization, e.g., PCM. Take q n A close to given x n. Reasonable in 16-bit (65,536 levels) digital audio. Coarse quantization, e.g., Σ. Use fewer bits to exploit redundancy. Finite Frames and Sigma Delta Quantization p.7/57

18 Quantization A δ K = {( K + 1/2)δ, ( K + 3/2)δ,...,( 1/2)δ, (1/2)δ,..., (K 1/2)δ} 6 (K 1/2)δ 4 2 δ δ { { q u 0 3δ/2 δ/2 δ { u u axis 2 4 f(u)=u ( K+1/2)δ Q(u) = arg min{ u q : q A δ K} = q u Finite Frames and Sigma Delta Quantization p.8/57

19 Setting Let x R d, x 1. Suppose F = {e n } N n=1 is a FUN-TF for R d. Thus, we have x = d N x n e n N n=1 with x n = x, e n. Note: A = N/d, and x n 1. Goal Find a good quantizer, given A δ K = {( K )δ, ( K )δ,...,(k 1 2 )δ}. Example Consider the alphabet A 2 1 = { 1, 1}, and E 7 = {e n } 7 n=1, with e n = (cos( 2nπ 7 ), sin(2nπ 7 )). Finite Frames and Sigma Delta Quantization p.9/57

20 A 2 1 = { 1, 1} and E Γ A 2 1 (E 7 ) = { n=1 q ne n : q n A 2 1} Finite Frames and Sigma Delta Quantization p.10/57

21 PCM Replace x n q n = arg{min x n q : q A δ K }. Then x = d N satisfies N q n e n n=1 x x d N N (x n q n )e n d N n=1 δ 2 N e n = d 2 δ. n=1 Not good! Bennett s white noise assumption Assume that (η n ) = (x n q n ) is a sequence of independent, identically distributed random variables with mean 0 and variance δ2 12. Then the mean square error (MSE) satisfies MSE = E x x 2 d 12A δ2 = (dδ)2 12N Finite Frames and Sigma Delta Quantization p.11/57

22 Remarks 1. Bennett s white noise assumption is not rigorous, and not true in certain cases. 2. The MSE behaves like C/A. In the case of Σ quantization of bandlimited functions, the MSE is O(A 3 ) (Gray, Güntürk and Thao, Bin Han and Chen). PCM does not utilize redundancy efficiently. 3. The MSE only tells us about the average performance of a quantizer. Finite Frames and Sigma Delta Quantization p.12/57

23 A 2 1 = { 1, 1} and E 7 Let x = ( 1 3, 1 2 ), E 7 = {(cos( 2nπ 7 A = { 1, 1}. ), sin(2nπ 7 ))}7 n=1. Consider quantizers with Finite Frames and Sigma Delta Quantization p.13/57

24 A 2 1 = { 1, 1} and E 7 Let x = ( 1 3, 1 2 ), E 7 = {(cos( 2nπ 7 A = { 1, 1}. 1.5 ), sin(2nπ 7 ))}7 n=1. Consider quantizers with Finite Frames and Sigma Delta Quantization p.14/57

25 A 2 1 = { 1, 1} and E 7 Let x = ( 1 3, 1 2 ), E 7 = {(cos( 2nπ 7 A = { 1, 1}. 1.5 ), sin(2nπ 7 ))}7 n=1. Consider quantizers with 1 x PCM Finite Frames and Sigma Delta Quantization p.15/57

26 A 2 1 = { 1, 1} and E 7 Let x = ( 1 3, 1 2 ), E 7 = {(cos( 2nπ 7 A = { 1, 1}. 1.5 ), sin(2nπ 7 ))}7 n=1. Consider quantizers with 1 x PCM 0.5 x Σ Finite Frames and Sigma Delta Quantization p.16/57

27 Σ quantizers for finite frames Let F = {e n } N n=1 be a frame for R d, x R d. Define x n = x, e n. Fix the ordering p, a permutation of {1, 2,...,N}. Quantizer alphabet A δ K Quantizer function Q(u) = arg{min u q : q A δ K} Define the first-order Σ quantizer with ordering p and with the quantizer alphabet A δ K by means of the following recursion. u n u n 1 = x p(n) q n q n = Q(u n 1 + x p(n) ) where u 0 = 0 and n = 1, 2,...,N. Finite Frames and Sigma Delta Quantization p.17/57

28 Stability The following stability result is used to prove error estimates. Proposition If the frame coefficients {x n } N n=1 satisfy x n (K 1/2)δ, n = 1,, N, then the state sequence {u n } N n=0 generated by the first-order Σ quantizer with alphabet A δ K satisfies u n δ/2, n = 1,, N. The first-order Σ scheme is equivalent to u n = n x p(j) j=1 n q j, n = 1,, N. j=1 Finite Frames and Sigma Delta Quantization p.18/57

29 Stability The following stability result is used to prove error estimates. Proposition If the frame coefficients {x n } N n=1 satisfy x n (K 1/2)δ, n = 1,, N, then the state sequence {u n } N n=0 generated by the first-order Σ quantizer with alphabet A δ K satisfies u n δ/2, n = 1,, N. The first-order Σ scheme is equivalent to u n = n x p(j) j=1 n q j, n = 1,, N. j=1 Stability results lead to tiling problems for higher order schemes. Finite Frames and Sigma Delta Quantization p.18/57

30 Error estimate Definition Let F = {e n } N n=1 be a frame for R d, and let p be a permutation of {1, 2,...,N}. The variation σ(f, p) is σ(f, p) = N 1 n=1 e p(n) e p(n+1). Finite Frames and Sigma Delta Quantization p.19/57

31 Error estimate Definition Let F = {e n } N n=1 be a frame for R d, and let p be a permutation of {1, 2,...,N}. The variation σ(f, p) is σ(f, p) = N 1 n=1 e p(n) e p(n+1). Theorem Let F = {e n } N n=1 be an A-FUN-TF for R d. The approximation x = d N q n e p(n) N n=1 generated by the first-order Σ quantizer with ordering p and with the quantizer alphabet A δ K satisfies x x (σ(f, p) + 1)d N δ 2. Finite Frames and Sigma Delta Quantization p.19/57

32 Order is important 0.2 Relative Frequency Relative Frequency Approximation Error Approximation Error Let E 7 be the FUN-TF for R 2 given by the 7th roots of unity. Randomly select 10,000 points in the unit ball of R 2. Quantize each point using the Σ scheme with alphabet A 1/4 4. The figures show histograms for x x when the frame coefficients are quantized in their natural order x 1, x 2, x 3, x 4, x 5, x 6, x 7 (left) and order x 1, x 4, x 7, x 3, x 6, x 2, x 5 (right). Finite Frames and Sigma Delta Quantization p.20/57

33 Even odd Approx. Error 5/N 5/N Frame size N E N = {e N n } N n=1, e N n = (cos(2πn/n), sin(2πn/n)). Let x = ( 1 π, 3 17 ). x = d N N x N n e N n, x N n = x, e N n. n=1 Let x N be the approximation given by the 1st order Σ quantizer with alphabet { 1, 1} and natural ordering. log-log plot of x x N. Finite Frames and Sigma Delta Quantization p.21/57

34 Improved estimates E N = {e N n } N n=1, Nth roots of unity FUN-TFs for R 2, x R d, x (K 1/2)δ. Quantize x = d N N x N n e N n, x N n = x, e N n n=1 using 1st order Σ scheme with alphabet A δ K. Theorem If N is even and large then x x δ log N. N 5/4 If N is odd and large then δ (2π+1)d δ N x x N 2. Remark The proof uses the analytic number theory approach developed by Sinan Güntürk, and the theorem is true more generally. Finite Frames and Sigma Delta Quantization p.22/57

35 Harmonic frames Zimmermann and Goyal, Kelner, Kovačević, Thao, Vetterli. H = C d. An harmonic frame {e n } N n=1 for H is defined by the rows of the Bessel map L which is the complex N-DFT N d matrix with N d columns removed. Finite Frames and Sigma Delta Quantization p.23/57

36 Harmonic frames Zimmermann and Goyal, Kelner, Kovačević, Thao, Vetterli. H = C d. An harmonic frame {e n } N n=1 for H is defined by the rows of the Bessel map L which is the complex N-DFT N d matrix with N d columns removed. H = R d, d even. The harmonic frame {e n } N n=1 is defined by the Bessel map L which is the N d matrix whose nth row is e N n = 2 d ( cos( 2πn N ), sin(2πn N ),...,cos(2π(d/2)n N ), sin( 2π(d/2)n ) ). N Finite Frames and Sigma Delta Quantization p.23/57

37 Harmonic frames Zimmermann and Goyal, Kelner, Kovačević, Thao, Vetterli. H = C d. An harmonic frame {e n } N n=1 for H is defined by the rows of the Bessel map L which is the complex N-DFT N d matrix with N d columns removed. H = R d, d even. The harmonic frame {e n } N n=1 is defined by the Bessel map L which is the N d matrix whose nth row is e N n = 2 d ( cos( 2πn N ), sin(2πn N ),...,cos(2π(d/2)n N ), sin( 2π(d/2)n ) ). N Harmonic frames are FUN-TFs. Finite Frames and Sigma Delta Quantization p.23/57

38 Harmonic frames Zimmermann and Goyal, Kelner, Kovačević, Thao, Vetterli. H = C d. An harmonic frame {e n } N n=1 for H is defined by the rows of the Bessel map L which is the complex N-DFT N d matrix with N d columns removed. H = R d, d even. The harmonic frame {e n } N n=1 is defined by the Bessel map L which is the N d matrix whose nth row is e N n = 2 d ( cos( 2πn N ), sin(2πn N ),...,cos(2π(d/2)n N ), sin( 2π(d/2)n ) ). N Harmonic frames are FUN-TFs. Let E N be the harmonic frame for R d and let p N be the identity permutation. Then N, σ(e N, p N ) πd(d + 1). Finite Frames and Sigma Delta Quantization p.23/57

39 Error estimate for harmonic frames Theorem Let E N be the harmonic frame for R d with frame bound N/d. Consider x R d, x 1, and suppose the approximation x of x is generated by a first-order Σ quantizer as before. Then x x d2 (d + 1) + d N δ 2. Hence, for harmonic frames (and all those with bounded variation), MSE Σ C d N 2 δ2. Finite Frames and Sigma Delta Quantization p.24/57

40 Error estimate for harmonic frames Theorem Let E N be the harmonic frame for R d with frame bound N/d. Consider x R d, x 1, and suppose the approximation x of x is generated by a first-order Σ quantizer as before. Then x x d2 (d + 1) + d N δ 2. Hence, for harmonic frames (and all those with bounded variation), MSE Σ C d N 2 δ2. This bound is clearly superior asymptotically to MSE PCM = (dδ)2 12N. Finite Frames and Sigma Delta Quantization p.24/57

41 Σ and optimal PCM The digital encoding MSE PCM = (dδ)2 12N in PCM format leaves open the possibility that decoding (reconstruction) could lead to Goyal, Vetterli, Thao (1998) proved MSE opt PCM O( 1 N ). MSE opt PCM C d N 2 δ2. Theorem The first order Σ scheme achieves the asymptotically optimal MSE PCM for harmonic frames. Finite Frames and Sigma Delta Quantization p.25/57

42 Uniform distribution and discrepancy A sequence (x n ) of real numbers is said to be uniformly distributed modulo 1 (u.d. mod 1) if 0 α < β 1, lim N 1 N N n=1 ½ [α,β) ({x n }) = β α, where {a} = a [a] denotes the fractional part of a real number a. Examples The following sequences are u.d. mod 1: 0 1, 0 2, 1 2, 0 3, 1 3, 2 3,..., 0 k, 1 k,..., k 1 k,... Finite Frames and Sigma Delta Quantization p.26/57

43 Uniform distribution and discrepancy A sequence (x n ) of real numbers is said to be uniformly distributed modulo 1 (u.d. mod 1) if 0 α < β 1, lim N 1 N N n=1 ½ [α,β) ({x n }) = β α, where {a} = a [a] denotes the fractional part of a real number a. Examples The following sequences are u.d. mod 1: 0 1, 0 2, 1 2, 0 3, 1 3, 2 3,..., 0 k, 1 k,..., k 1 k,... The sequence (nα) where α is an irrational number. Finite Frames and Sigma Delta Quantization p.26/57

44 Uniform distribution and discrepancy A sequence (x n ) of real numbers is said to be uniformly distributed modulo 1 (u.d. mod 1) if 0 α < β 1, lim N 1 N N n=1 ½ [α,β) ({x n }) = β α, where {a} = a [a] denotes the fractional part of a real number a. Examples The following sequences are u.d. mod 1: 0 1, 0 2, 1 2, 0 3, 1 3, 2 3,..., 0 k, 1 k,..., k 1 k,... The sequence (nα) where α is an irrational number. The sequence (log F n ) where F n denotes the Fibonacci sequence F 0 = 0, F 1 = 1, F n = F n 1 + F n 2 for n 2. Finite Frames and Sigma Delta Quantization p.26/57

45 Criteria for uniform distribution Weyl Criterion: The sequence (x n ) is u.d. mod 1 if and only if for all integers h 0 1 N lim e 2πihx n = 0. N N n=1 Finite Frames and Sigma Delta Quantization p.27/57

46 Criteria for uniform distribution Weyl Criterion: The sequence (x n ) is u.d. mod 1 if and only if for all integers h 0 1 N lim e 2πihx n = 0. N N n=1 The sequence (x n ) is u.d. mod 1 if and only if for every Riemann-integrable function f on [0, 1], the following equation holds: lim N 1 N N f({x n }) = n=1 1 0 f(x)dx. Finite Frames and Sigma Delta Quantization p.27/57

47 Discrepancy The discrepancy D N of a finite sequence x 1,...,x N of real numbers is defined by D N = D N (x 1,...,x N ) = sup 0 α<β 1 1 N N ½ [α,β) ({x n }) (β α). n=1 We define D N by D N = D N(x 1,...,x N ) = sup 0<α 1 1 N N ½ [0,α) ({x n }) α. n=1 The discrepancy of a sequence measures how well the sequence is distributed over a unit interval. Finite Frames and Sigma Delta Quantization p.28/57

48 Properties of discrepancy The sequence ω is u.d. mod 1 if and only if lim N D N (ω) = 0. Finite Frames and Sigma Delta Quantization p.29/57

49 Properties of discrepancy The sequence ω is u.d. mod 1 if and only if lim N D N (ω) = 0. For any sequence of N numbers we have 1/N D N 1. Finite Frames and Sigma Delta Quantization p.29/57

50 Properties of discrepancy The sequence ω is u.d. mod 1 if and only if lim N D N (ω) = 0. For any sequence of N numbers we have 1/N D N 1. The discrepancy D N and DN are related by the inequality DN D N 2DN. Finite Frames and Sigma Delta Quantization p.29/57

51 Properties of discrepancy The sequence ω is u.d. mod 1 if and only if lim N D N (ω) = 0. For any sequence of N numbers we have 1/N D N 1. The discrepancy D N and DN are related by the inequality DN D N 2DN. Explicit formula for D N : If x 1 x 2 x N are N numbers in [0, 1), x 0 = 0, and x N+1 = 1, then D N = max j=0,...,n max( x j j/n, x j+1 j/n ). Finite Frames and Sigma Delta Quantization p.29/57

52 Estimation of discrepancy LeVeque inequality Let x 1,...,x N be a finite sequence of real numbers. Then D N ( 6 π 2 h=1 1 h 2 1 N N n=1 e 2πihx n 1/3 2). Finite Frames and Sigma Delta Quantization p.30/57

53 Estimation of discrepancy LeVeque inequality Let x 1,...,x N be a finite sequence of real numbers. Then D N ( 6 π 2 h=1 1 h 2 1 N N n=1 e 2πihx n 1/3 2). Erdös-Turán inequality Let m be a positive integer and x 1,...,x N a finite sequence of real numbers, then D N 6 m π m k=1 ( 1 k 1 ) 1 m + 1 N N n=1 e 2πikx n. Finite Frames and Sigma Delta Quantization p.30/57

54 Estimation of discrepancy LeVeque inequality Let x 1,...,x N be a finite sequence of real numbers. Then D N ( 6 π 2 h=1 1 h 2 1 N N n=1 e 2πihx n 1/3 2). Erdös-Turán inequality Let m be a positive integer and x 1,...,x N a finite sequence of real numbers, then D N 6 m π m k=1 ( 1 k 1 ) 1 m + 1 N N n=1 e 2πikx n. Corollary of Erdös-Turán inequality There exists C such that for all m > 0 and for any x 1,...,x N R, ( 1 m D N C m + h=1 1 1 h N N n=1 e 2πihx n ). Finite Frames and Sigma Delta Quantization p.30/57

55 Proof of Erdös-Turán inequality Assume WLOG that x 1 x 2 x N are numbers in [0, 1). Set N (x) = 1 N N n=1 ½ [0,x) (x n ) x for 0 x 1 and extend this function with period 1 to R. Note a typical graph of N (x), with x 1 = 1/8, x 2 = 2.3/8, x 3 = 5.4/8, x 4 = 7.2/8. Finite Frames and Sigma Delta Quantization p.31/57

56 Proof of E-T inequality (continued) Assume that 1 0 N (x) = 0. Finite Frames and Sigma Delta Quantization p.32/57

57 Proof of E-T inequality (continued) Assume that 1 0 N (x) = 0. Set S h = 1 N N n=1 e2πihx n for h Z. Then by computing the Fourier coefficients of N (x) we have for integer h 0 S h 2πih = 1 0 N (x)e 2πihx dx. Finite Frames and Sigma Delta Quantization p.32/57

58 Proof of E-T inequality (continued) Let m be a positive integer and a a real number. Then m (m + 1 h )e 2πiha S h 2πih h= m = 1 a a ( m N (x + a) h= m (m + 1 h )e 2πihx )dx, where means 0 is not included in the range of the sum. Finite Frames and Sigma Delta Quantization p.33/57

59 Proof of E-T inequality (continued) Let m be a positive integer and a a real number. Then m (m + 1 h )e 2πiha S h 2πih h= m = 1 a a ( m N (x + a) h= m (m + 1 h )e 2πihx )dx, where means 0 is not included in the range of the sum. Because of the periodicity of the integrand, the last integral can be taken over [ 1/2, 1/2]. Finite Frames and Sigma Delta Quantization p.33/57

60 Proof of E-T inequality (continued) Note the following discrete Fejér kernel: m h= m (m + 1 h )e 2πihx = sin2 (m + 1)πx sin 2. πx Finite Frames and Sigma Delta Quantization p.34/57

61 Proof of E-T inequality (continued) Note the following discrete Fejér kernel: m h= m (m + 1 h )e 2πihx = sin2 (m + 1)πx sin 2. πx We have 1/2 1/2 N (x + a) sin2 (m + 1)πx sin 2 dx πx 1 m (m + 1 h ) S h 2π h = 1 π h= m m (m + 1 h) S h h. h=1 Finite Frames and Sigma Delta Quantization p.34/57

62 Proof of E-T inequality (continued) From the nature of the graph of N (x) we either have N (b) = DN or N(b + 0) = lim x b + N (x) = DN for some b [0, 1]. We will deal with the second case. Finite Frames and Sigma Delta Quantization p.35/57

63 Proof of E-T inequality (continued) From the nature of the graph of N (x) we either have N (b) = DN or N(b + 0) = lim x b + N (x) = DN for some b [0, 1]. We will deal with the second case. For b < t b + DN, we have N (t) = DN + N (t) N (b + 0) DN + b t. Finite Frames and Sigma Delta Quantization p.35/57

64 Proof of E-T inequality (continued) From the nature of the graph of N (x) we either have N (b) = DN or N(b + 0) = lim x b + N (x) = DN for some b [0, 1]. We will deal with the second case. For b < t b + DN, we have N (t) = DN + N (t) N (b + 0) DN + b t. Choose a = b D N. Then for x < 1 2 D N, N (x + a) D N + b x a = 1 2 D N x. Finite Frames and Sigma Delta Quantization p.35/57

65 Proof of E-T inequality (continued) 4 (x) D4 slope = 4(t) 4 (b+0) t b 1 b 0 t 1 x slope = 1 Finite Frames and Sigma Delta Quantization p.36/57

66 Proof of E-T inequality (continued) Breaking the integral into [ 1 2, D N 2 ], [ D N 2, D N 2 ], [ D N 2, 1 2 ] and using the symmetry argument and the inequality for x [0, π/2], we obtain sinx 2 π x, Finite Frames and Sigma Delta Quantization p.37/57

67 Proof of E-T inequality (continued) ( D N /2 1/2 + D N /2 1/2 + DN /2 DN /2 ) N (x + a) sin2 (m + 1)πx sin 2 πx dx D N /2 D N /2 ( D N D N = D N = D N 1/2 D N D N /2 0 1/2 0 m (m + 1)πx x)sin2 sin 2 dx DN πx sin 2 (m + 1)πx sin 2 dx πx sin 2 (m + 1)πx sin 2 dx 2DN πx sin 2 (m + 1)πx sin 2 dx 3DN πx D N 3D N 1/2 D N /2 dx 1/2 1/2 D N /2 1/2 sin 2 (m + 1)πx sin 2 dx πx sin 2 (m + 1)πx DN /2 sin 2 dx πx sin 2 (m + 1)πx DN /2 sin 2 dx πx 4x > m + 1 D 2 N Finite Frames and Sigma Delta Quantization p.38/57

68 Proof of E-T inequality (continued) D N 3 m π Since D N 2DN, we have D N 6 m π m ( 1 h 1 m + 1 ) S h. h=1 m ( 1 h 1 m + 1 ) S h. h=1 Finite Frames and Sigma Delta Quantization p.39/57

69 Proof of E-T inequality (continued) D N 3 m π Since D N 2DN, we have D N 6 m π m ( 1 h 1 m + 1 ) S h. h=1 m ( 1 h 1 m + 1 ) S h. h=1 Show that for any finite sequence x 1,...,x N in [0, 1), there exists c [0, 1) such that {x 1 + c},..., {x N + c} satisfies 1 0 N (x)dx = 0. This completes the proof of the Erdös-Turán inequality. Finite Frames and Sigma Delta Quantization p.39/57

70 Koksma inequality Theorem Let f be a function of bounded variation on [0, 1] with total variation Var(f) and let x 1,...,x N be given points in [0, 1) with discrepancy D N. Then 1 N N f(x n ) n=1 1 0 f(t)dt Var(f) D N. Proof Assume that x 1 x 2 x N are points in [0, 1). Set x 0 = 0 and x N+1 = 1. Riemann-Stieltjes integration by parts gives 1 N N f(x n ) n=1 1 0 f(t)dt = N n=0 xn+1 x n (t n N )df(t). Finite Frames and Sigma Delta Quantization p.40/57

71 Proof of K inequality (continued) For a fixed n with 0 n N and for x n t x n+1, one can prove t n ( N max x n n N, x n+1 n ) N DN. Since D N D N, the Koksma inequality follows by elementary properties of the Riemann-Stieltjes integral. Finite Frames and Sigma Delta Quantization p.41/57

72 Additional results and definitions Exponential sum theorem Let a and b be integers with a < b, and let f be twice differentiable on [a, b] with f (x) ρ > 0 for x [a, b]. Then b n=a e 2πif(n) ( f (b) f (a) + 2)( 4 ρ + 3). The class of bandlimited functions B Ω consists of all real valued functions f L (R) whose f (as a distribution) is a finite Borel measure supported on [ Ω, Ω]. By the Paley-Wiener theorem, each element of B Ω is the restriction of an entire function to the real line. A function h B Ω is said to belong to the class M Ω if h L (R) and all zeros of h on [0, 1] are simple. Finite Frames and Sigma Delta Quantization p.42/57

73 Bernstein inequality If f B Ω then for all integers r 0, f (r) Ω r f. Finite Frames and Sigma Delta Quantization p.43/57

74 Bernstein inequality If f B Ω then for all integers r 0, f (r) Ω r f. Theorem (Güntürk) Let f B Ω. Then for each λ > 1, there exists an analytic function X λ B Ω such that for all t R, X λ (t) X λ (t 1) = f( t λ ) and there exists a constant C λ (f) > 0 such that X λ f( λ ) C λ (f) 1 λ. Finite Frames and Sigma Delta Quantization p.43/57

75 Theorem Let {F N } N=d be a family of unit norm tight frames for Rd, with F N = {e N n } N n=1. Suppose x R d satisfies x (K 1/2)δ for some positive integer K and δ > 0 in the Σ scheme. Let {x N n } N n=1 be the sequence of frame coefficients of x with respect to F N. If, for some Ω > 0, there exists h M Ω such that and if N is sufficiently large, then N and 1 n N, x N n = h(n/n), v N n C(x)δN 3/4 log N, where v N n = n j=1 un j and C(x) is a constant that depends on x. Finite Frames and Sigma Delta Quantization p.44/57

76 Proof of theorem For each N d, let u N n, for n = 1,...,N, be the state variables obtained from the Σ scheme and define Set u N 0 = 0. ũ N n = un n δ. Finite Frames and Sigma Delta Quantization p.45/57

77 Proof of theorem For each N d, let u N n, for n = 1,...,N, be the state variables obtained from the Σ scheme and define Set u N 0 = 0. Then, by the stability result, ũ N n = un n δ. ũ N n 1 2. Finite Frames and Sigma Delta Quantization p.45/57

78 Proof of theorem For each N d, let u N n, for n = 1,...,N, be the state variables obtained from the Σ scheme and define Set u N 0 = 0. Then, by the stability result, ũ N n = un n δ. ũ N n 1 2. Identify the interval [ 1/2, 1/2] with [0, 1] and apply Koksma s inequality: Finite Frames and Sigma Delta Quantization p.45/57

79 Proof of theorem (continued) vn N = δ n j=1 ũ N j = nδ 1 n n ũ N j j=1 1/2 1/2 y dy nδvar(x)d n (ũ N 1,...,ũ N n ) = nδd n (ũ N 1,...,ũ N n ) Next apply the Erdös-Turán inequality: There exists a constant C > 0 such that for all K N, ( 1 D n (ũ N 1,...,ũ N n ) C K + 1 n K k=1 1 k n j=1 e 2πikeuN j ). Finite Frames and Sigma Delta Quantization p.46/57

80 Proof of theorem (continued) Using the Güntürk s theorem, we have that for each integer N d there exists an analytic function X N B Ω such that for all 1 n N, X N (n) = u N n + c n δ/2, for some c n Z. Each c n has the same parity as n. Moreover, for all t R, X N(t) h( t N ) C N(x) N, for some constant C N (x) > 0. The Bernstein inequality yields for some constant C N (x) > 0. X N(t) 1 N h ( t N ) C N(x) N 2, Finite Frames and Sigma Delta Quantization p.47/57

81 Proof of theorem (continued) Let z 1 < z 2 < < z n be zeros of h in [0, 1]. Let 0 < α < 1 be a fixed constant to be determined later. Define intervals I j and J j by j = 1,...,n, I j = [Nz j N α, Nz j + N α ] j = 1,...,n 1, J j = [Nz j + N α, Nz j+1 N α ] J 0 = [1, Nz 1 N α ] and J n = [Nz n + N α, N]. Adjust intervals appropriately when z 1 = 0 or z n = 1. Finite Frames and Sigma Delta Quantization p.48/57

82 Proof of theorem (continued) Let z 1 < z 2 < < z n be zeros of h in [0, 1]. Let 0 < α < 1 be a fixed constant to be determined later. Define intervals I j and J j by j = 1,...,n, I j = [Nz j N α, Nz j + N α ] j = 1,...,n 1, J j = [Nz j + N α, Nz j+1 N α ] J 0 = [1, Nz 1 N α ] and J n = [Nz n + N α, N]. Adjust intervals appropriately when z 1 = 0 or z n = 1. When N is sufficiently large then J 0 I 1 J 1 I n J n = [1, N]. Finite Frames and Sigma Delta Quantization p.48/57

83 Proof of theorem (continued) J 0 I 1 J 1... I n J n 1 Nz 1 N α Nz 1 + N α Nz 2 N α Nz n N α Nz n + N α N Finite Frames and Sigma Delta Quantization p.49/57

84 Proof of theorem (continued) It follows from the properties of h M Ω that there exist constants C 1 (x), C 2 (x) and C 3 (x) such that for sufficiently large N and for all t J 0 J n, h ( t N ) C 1 1(x) N 1 α, X n(t) 1 C 2 (x) N 2 α, X N(t) C 3 (x). By the exponential sum theorem, we have that there exists a constant C(x) > 0 and an index N 0 such that for all N N 0 and k 1 n j=0 n N J j e 2πikeu N n k C(x)( N 1 α/2 + k δ δ ). Finite Frames and Sigma Delta Quantization p.50/57

85 Proof of theorem (continued) For all N d and k 1, n j=1 n N I j e 2πikeu N n 2n N α. There exists a constant C(x) > 0 such that for all k 1 and N N 0, n j=1 e 2πikeuN j C(x)(Nα + kn 1 α/2 δ + k δ ). Now choose α = 3/4 and K, an integer larger than N 1/4 0 and note that n k=1 1 k 1 + log n, n k=1 1 k 2 n. Finite Frames and Sigma Delta Quantization p.51/57

86 Proof of theorem (continued) Then there exist constants C 1 (x), C 2 (x), C 3 (x) > 0 such that for sufficiently large N ( 1 vn N nδc 1 (x) K + 1 n nδc 2 (x)( K k=1 1 N3/4 + N1/4 n ( N3/4 k + N1 3/8 δ ) k δ ) log N + 2N3/4 n δ + N1/4 nδ ) C 2 (x)( δn N 1/4 + δn3/4 log N + δn 3/4 + N 1/4 ) C 3 (x)δn 3/4 log N. This completes the proof of the theorem. Finite Frames and Sigma Delta Quantization p.52/57

87 Corollary hypotheses {F N } N=d be a family of unit norm tight frames for Rd, for which each F N = {e N n } N n=1 satisfies the zero sum condition. Suppose x R d satisfies x (K 1/2)δ for some positive integer K and δ > 0 in the Σ scheme. Let {x N n } N n=1 be the sequence of frame coefficients of x with respect to F N, and suppose there exists h M Ω, Ω > 0, such that N and 1 n N, x N n = h(n/n). Additionally, suppose that fn N = e N n e N n+1, n = 1,...,N, satisfies fn N c 1 N for some constant c 1 > 0 and fn N fn+1 N c 2 N, for some 2 constant c 2 > 0, and set u N 0 = 0 in the Σ scheme. Then Finite Frames and Sigma Delta Quantization p.53/57

88 Corollary conclusion If N is even and sufficiently large, then x x N C 1 (x) δ log N N 5/4, for some constant C 1 (x) > 0. If N is odd and sufficiently large, then C 2 (x) δ N x x N δd 2N (σ(f N, p N ) + 1), for some constant C 2 (x) > 0. Finite Frames and Sigma Delta Quantization p.54/57

89 Proof of corollary x x N = d N = d N ( N 1 n=1 ( N 2 n=1 u N n (e N n e N n+1) + u N Ne N N v N n (f N n f N n+1) + v N N 1f N N 1 + u N Ne N N ) ), where f N n = e N n e N n+1, v N n = n j=1 en j, vn 0 = 0. Finite Frames and Sigma Delta Quantization p.55/57

90 Proof of corollary x x N = d N = d N ( N 1 n=1 ( N 2 n=1 u N n (e N n e N n+1) + u N Ne N N v N n (f N n f N n+1) + v N N 1f N N 1 + u N Ne N N ) ), where f N n = e N n e N n+1, v N n = n j=1 en j, vn 0 = 0. By the Theorem, we have d N ( N 2 n=1 v N n (f N n f N n+1) + v N N 1f N N 1 for some constant C 1 (x) > 0. ) C 1 (x) δ log N N 5/4, Finite Frames and Sigma Delta Quantization p.55/57

91 Proof of corollary (continued) If N is even then u N N = 0 and so x x N C 1 (x) δ log N N 5/4, for some constant C 1 (x) > 0 and for sufficiently large N. If N is odd then u N N = δ/2 and we have, combining with previous results, C 2 (x) δ N x x N δd 2N (σ(f N, p N ) + 1), for some constant C 2 (x) > 0 and for sufficiently large N. This completes the proof of the Corollary. Finite Frames and Sigma Delta Quantization p.56/57

92 Finite Frames and Sigma Delta Quantization p.57/57