Gabor orthonormal bases generated by the unit cubes

Transcription

1 Gabor orthonormal bases generated by the unit cubes Chun-Kit Lai, San Francisco State University (Joint work with J.-P Gabardo and Y. Wang) Jun, 2015

2 Background Background

3 Background Let 1 g 0 on L 2 (R d ). 2 Λ be a discrete countable set on R 2d, we write (t, λ) Λ where t, λ R d. The Gabor system G(g, Λ) = {E λ T t g : (t, λ) Λ}. (1) where E λ f (x) = e 2πi λ,x f (x), T t f (x) = f (x t).

4 Background Gabor frame: A f 2 (t,λ) Λ f, E λ T t g 2 B f 2, f L 2 (R d ). (2) Gabor orthonormal bases (GONB): 1 mutually orthonormal e 2πi λ 1 λ 2,x g(x t 1 )g(x t 2 )dx = 0. For distinct (t 1, λ 1 ), (t 2, λ 2 ) Λ and g 2 = 1. 2 Completeness f 2 = f, E λ T t g 2, f L 2 (R d ). (t,λ) Λ

5 Background Fundamental problem in Gabor analysis. Determine F full (g) = {Λ : G(g, Λ) is a frame}. F(g) = {(α, β) : G(g, αz d βz d ) is a frame}. F full (e πx2 ) = {Λ : D (Λ) > 1}. (Lyubarskii, and Seip and Wallsten) F(g) = {(α, β) : αβ < 1} for totally positive function of finite type. (Gröchenig and Stöckler, 2011) F(χ [0,1] ) is completely classified by Dai and Sun, 2014.

6 Background We are interested in the following problem: O full (g) = {Λ : G(g, Λ) is an GONB}. 1 Particular interest is O full (χ [0,1] d ). 2 On the other hand, if g belongs to the modulation space M 1 (R d ). O full (g) =. (Ascensi, Feichtinger and Kaiblinger, Partial result in the past: Balian-Low theorem for lattices, or symplectic lattice)

7 Background A direct observation: Proposition Let g = K 1/2 χ K, and K R d is measurable with finite Lebesgue measure. Suppose that Let K tiles R d by a discrete set J. For each t J, the set of exponentials {e 2πi λ,x : λ Λ t } is an orthonormal basis for L 2 (K). Λ = {t} Λ t. (3) t J Then G(g, Λ) is a Gabor orthonormal basis for L 2 (R d ).

8 Background 1 The first condition means that K is a translational tile (with J called tiling set) 2 The second condition means that L 2 (K) admits an orthonormal basis of exponentials. K is called a spectral set (and each Λ t is called spectrum). Conjecture Fuglede conjecture (1974): Spectral sets if and only if translational tile. 3 Disproved later by Tao in Lagarias, Reeds and Wang (2000) Tiling set and spectrum for [0, 1] d is equivalent. Qu: Calling the structure standard, is that all Λ in O(χ [0,1] d ) standard?

9 Main theory and results: R 1 Main theory and results: R 1

10 Main theory and results: R 1 Short time Fourier transform V g f (t, ν) = f (x)g(x t)e 2πiνx dx 0 = e 2πi λ 1 λ 2,x g(x t 1 )g(x t 2 )dx = e 2πi λ 1 λ 2,t 1 e 2πi λ 1 λ 2,x g( Proposition G(g, Λ) is a mutually orthonormal if and only if Λ Λ {(t, ν) : V g g(t, ν) = 0} {0}.

11 Main theory and results: R 1 Let g d = χ [0,1] d V g1 g 1 (t, ν) = 0, ( t 1; 1 2πiν ( e 2πiνt e 2πiν), 0 t 1; 1 e 2πiν(t+1) ), 1 t πiν (4) Z(V g1 g 1 ) = {(t, ν) : t 1} {(t, ν) : ν(1 t ) Z \ {0}}. Proposition If G(g 1, Λ) is mutually orthogonal, then Λ + [0, 1] 2 is a packing. χ [0,1] 2(x + λ) = χ [0,1] 2 δ Λ 1. λ Λ

12 Main theory and result: R 1 Completeness: f 2 = f, e 2πi λ, g( t) 2 f 2 = (t,λ) Λ f 2 = V g f (t, λ) 2 (t,λ) Λ (t,λ) Λ V g f (τ t, ξ λ) 2 (Replacing f by e 2πiξx f (x τ)) V g f 2 tiles R by Λ f 2 = δ Λ V g f 2

13 Main theory and results Theorem [Gabardo, Wang, and L.] Suppose that F, G L 1 (R n ) are two functions with F, G 0 and R n F (x) dx = R n G(x) dx = 1. Suppose that µ is a positive Borel measure on R n such that F µ 1 and G µ 1. Then, F µ = 1 if and only if G µ = 1. If G(g 1, Λ) is mutually orthogonal, then f 2 = δ Λ V g f 2 δ Λ χ [0,1] 2 = 1.

14 Main theory and results: R 1 Theorem G(χ [0,1], Λ) is a Gabor orthonormal basis if and only if Λ is standard. Proof There are only two types of tilings on R 2 Λ = (Z + a k ) {k} or Λ = (Z + a k ) (5) k Z k Z{k} The first type does not satisfy the mutually orthogonality. The second type is standard.

15 Main theory and results: R d Main theory and results: R d

16 Main theory and results: R d V gd g d (t, ν) = V g1 g 1 (t 1, ν 1 )... V g1 g 1 (t d, ν d ) where t = (t 1,..., t d ) and ν = (ν 1,..., ν d ). ( d ) Z(V gd g d ) = {(t, v) : t max 1} {(t, ν) : ν i (1 t i ) Z \ {0})} i=1 If G(g d, Λ) is mutually orthogonal, then f 2 = δ Λ V g f 2 δ Λ χ [0,1] d = 1. (6) Theorem G(χ [0,1] d, Λ) is a Gabor orthonormal basis if and only if the inclusion Λ Λ Z(V g g) {0} holds and Λ + [0, 1] 2d is a tiling.

17 Main theory and results: R d Remark: Tiling of cubes on R d can be very difficult. For example, Theorem (Lagarias and Shor, 1992) On R d, d 10, there exists a tiling of cubes whose adjacent cubes do not share (d 1)-dimensional face.

18 Main theory and results: R d Suppose that 1 G(χ [0,1] m, Λ 1 ) is a Gabor orthonormal basis for L 2 (R m ) 2 For each (s, λ) Λ 1, a discrete set Λ (s,λ) in R 2n such that G(χ [0,1] n, Λ (s,λ) ) is a Gabor orthonormal basis of L 2 (R n ). Λ = {(s, t, λ, ν), (t, ν) Λ (s,λ) }. (7) (s,λ) Λ 1 We say that a Gabor system G(χ [0,1] d, Λ) with Λ as in (7) is pseudo-standard.

19 Main theory and results: R d Proposition Λ = (s,λ) Λ 1 {(s, t, λ, ν), (t, ν) Λ (s,λ) }. Every pseudo-standard Gabor system G(χ [0,1] d, Λ) is a Gabor orthonormal basis of L 2 (R d ). It follows from V gd g d (s, t, λ, ν) = V gm g m (s, λ) V gn g n (t, ν), (s, λ) R 2m, (t, ν) R 2n. (8)

20 Main theory and results: R d Consider the two-dimensional case d = 2. Let Λ 1 = m Z{m} (Z + µ m ), µ m [0, 1). Associate with each γ = (m, j + µ m ) Λ 1, the set Λ γ = n Z{n + s m,j } (Z + ν n,m,j ), s m,j R, ν n,m,j [0, 1). Then, Λ := {(m, n + s m,j, j + µ m, k + ν n,m,j ) : m, n, j, k Z}

21 Main theory and results: R d

22 Main theory and results: R d Complete classification of G(χ [0,1] 2, Λ) on R 2 : Mixture of standard and pseudo-standard structure. Theorem Suppose that G(χ [0,1] 2, Λ) is a Gabor orthonormal basis for L 2 (R 2 ). Then we can partition Z into J and J such that either Λ = {(m + t n,k, n, j + µ k,m,n, k + ν n ) : m, j, k Z} or n J m Z n J {(m + t n, n)} Λ m,n

23 Main theory and results: R d Theorem Λ = {(m, n + t m,j, j + ν m, k + µ j,m,n ) : n, j, k Z} m J n Z m J {(m, n + t m )} Λ m,n. where Λ m,n + [0, 1] 2 tile R 2 and t n,k, µ k,m,n and ν n are real numbers in [0, 1) as a function of m, n or k.

24 Another problem Suppose that g L 2 (R d ) and suppg = Ω is a bounded set. Conjecture (Liu and Wang, 2003) Suppose that G(g, J Λ) is an GONB. Then (i) g = Ω 1/2 χ Ω. (ii) {e 2πiλx : λ Λ} is an orthonormal basis of L 2 (Ω) and (iii) J is a tiling set of Ω. Liu and Wang proved that the conjecture is true if Ω = [0, 1].

25 Another problem Theorem (Dutkay and L., 2014) Let dµ = g dx be an absolutely continuous measure on R d and it admits a Fourier frame A f 2 f (x)e 2πiλx g(x)dx 2 B f 2. λ Λ Then B A sup x Ω g(x) inf x Ω g(x). In particular, if B = A, then g = c a.e. on the support. Theorem (Dutkay and L., 2014) If the window function g 0 on the support Ω, then Liu-Wang conjecture holds.

26 Thank You!!