Frames, Erasures, and Excess in Infinite Dimensions Christopher Heil - PowerPoint PPT Presentation

Frames, Erasures, and Excess in Infinite Dimensions Christopher Heil Georgia Tech Codex Seminar, December 8, 2020

Frames (Duffin and Schaeffer, 1952) A sequence { x n } n ∈ N in a Hilbert space H is a frame if ∃ A, B > 0 such that � A � x � 2 ≤ |� x, x n �| 2 ≤ B � x � 2 ∀ x ∈ H, n In this case (DS, 1952): • The frame operator Sx = � n � x, x n � x n is a topological isormorphism S : H → H x n = S − 1 x n • The canonical dual frame is { � x n } n ∈ N where � � � • x = � x, � x n � x n = � x, x n � � x n for every x ∈ H n n

� � � � � � � • If x = � c n x n , then � 2 + | c n | 2 = � 2 � � x, � � c n − � x, � x n � x n � n n n x m �| 2 − | 1 − � x m , � � x m �| 2 x n �| 2 = 1 − |� x m , � • ∗ |� x m , � 2 n � = m • If � x m , � x m � = 1 , then � x m , � x n � = 0 for n � = m. • The removal of a vector from a frame leaves either a frame or an incomplete set: � x m , � x m � � = 1 = ⇒ { x n } n � = m is a frame � x m , � x m � = 1 = ⇒ { x n } n � = m is incomplete

Multiplicative Completion (Boas and Pollard, 1948) Problem If { f k } k ∈ Z is an ONB for L 2 [0 , 1] and F ⊆ Z is finite then { f k } k / ∈ F is incomplete. Does ∃ m ∈ L ∞ [0 , 1] such that { f k · m } k / is complete? ∈ F

Multiplicative Completion (Boas and Pollard, 1948) Problem If { f k } k ∈ Z is an ONB for L 2 [0 , 1] and F ⊆ Z be finite then { f k } k / ∈ F is incomplete. Does ∃ m ∈ L ∞ [0 , 1] such that { f k · m } k / is complete? ∈ F Yes: Proof for F = { 0 } ∈ L 2 . If � f, f k · m � = 0 for k � = 0 then � f · m, f k � = 0 for Chose m so f 0 /m / k � = 0 , so f · m = cf 0 Hence f = cf 0 /m. Contradiction if c � = 0 . Therefore { f k · m } k � =0 is com- plete. A (not entirely trivial) induction extends to finite F.

Constructing m Choose E ⊆ [0 , 1] with | E | > 0 and inf x ∈ E | f 0 ( x ) | = ε > 0 . Choose ψ ∈ L ∞ \ L 2 with supp ( ψ ) ⊆ E. Set    1 , x / ∈ supp ( ψ ) , m ( x ) =   1 /ψ ( x ) , x ∈ supp ( ψ ) . Then f 0 ( x ) /m ( x ) ≥ ε ψ ( x ) , x ∈ supp ( ψ ) , ∈ L 2 . so f 0 /m /

Generalizes to frames and less structured systems in function spaces. Lemma. Let A be a solid Banach space of measurable functions on a measure space ( X, Σ, µ ) . Assume for each measurable E ⊆ X there is a measurable function ψ on X such that { ψ � = 0 } ⊆ E and ψ / ∈ A . If f 1 , . . . , f N ∈ A , then ∃ g ∈ L ∞ ( µ ) such that (a) g ( x ) � = 0 for x ∈ X, and (b) f/g / ∈ A for all f ∈ span { f 1 , . . . , f N } \ { 0 } . Need “nonatomicness” Multiplicative completion fails in ℓ 2 ( Z ) .

Surprising Equivalences (Talalyan, and Price and Zink, 1957–1975) If { f n } n ∈ N ⊆ L 2 ( µ ) , where ( X, µ ) is a separable measure space with µ ( µ ) = 1 , then TFAE: (a) If ε > 0 then ∃ S ⊆ X such that µ ( S ) > 1 − ε and { f n χ S } is complete in L 2 ( S ) . (b) If f is finite a.e. and ε > 0 then ∃ S ⊆ X and g ∈ span { f n } such that µ ( S ) > 1 − ε and | f − g | < ε on S. (c) There exists a bounded, nonnegative function m such that { mf n } is complete in L 2 ( µ ) . There are a wide variety of related results by Kazarian.

Fix g ∈ L 2 [0 , 1] , set E g = { e 2 πinx g ( x ) } n ∈ Z Weighted Exponentials: (a) E g is complete ⇐ ⇒ g � = 0 a.e. ⇒ 1 /g ∈ L 2 [0 , 1] (b) E g is minimal ⇐ ⇒ | g | 2 ∈ A 2 [0 , 1] (c) E g is a Schauder basis ⇐ ⇒ g ∈ L ∞ [0 , 1] (d) E g is a Bessel sequence ⇐ ⇒ ∃ A, B > 0 such that A ≤ | g ( x ) | 2 ≤ B (e) E g is a frame sequence ⇐ for a.e. x with g ( x ) � = 0 ⇒ ∃ A, B > 0 such that A ≤ | g ( x ) | 2 ≤ B a.e. (f) E g is a Riesz basis ⇐ (g) E g is an orthonormal basis ⇐ ⇒ | g ( x ) | = 1 a.e. Attribution (c) Hunt, Muckenhaupt, Wheeden (1973) (e) Benedetto and Li (1998)

Multiplicative Completion for { e 2 πinx } n � =0 . Let e n ( x ) = e 2 πinx . ∈ L 2 [0 , 1] so { xe n } n � =0 is complete • 1 /x / e n ( x ) = e 2 πinx − 1 • Biorthogonal system is � for n � = 0 x • { xe n } n � =0 is minimal (and complete, so exact). • If { xe n } n � =0 were a Schauder basis then 1 ≤ � xe n � 2 � � e n � 2 ≤ 2 C � πn sin 2 u • � xe n � 2 = 3 − 1 / 2 and � � e n � 2 = 4 πn u 2 du → ∞ as n → ∞ 0 • { xe n } n � =0 is not a Schauder basis • (Other results by Yoon/H., 2012) The original Gabor system, generated by the Gaussian at the critical density, has similar properties: G ( φ, 1 , 1) = { M n T k φ } k,n ∈ Z = { e 2 πinx e − ( x − k ) 2 } k,n ∈ Z Both systems have density 1 . Why?

Beurling Density for Λ ⊆ R 2 Let Q r ( z ) be the square centered at z with side lengths r . #(Λ ∩ Q r ( z )) Lower Beurling density: D − (Λ) = lim inf inf r 2 r →∞ z ∈ R 2 #(Λ ∩ Q r ( z )) Upper Beurling density: D + (Λ) = lim sup sup r 2 r →∞ z ∈ R 2 Examples: D ± ( α Z × β Z ) = 1 D + ( α Z × β Z + ) = 1 D − ( α Z × β Z + ) = 0 , αβ, αβ Conjecture ⇒ D − (Λ) ≥ 1 A Gabor system { M b T a g } ( a,b ) ∈ Λ is complete =

Beurling Density for Λ ⊆ R 2 Let Q r ( z ) be the square centered at z with side lengths r . #(Λ ∩ Q r ( z )) Lower Beurling density: D − (Λ) = lim inf inf r 2 r →∞ z ∈ R 2 #(Λ ∩ Q r ( z )) Upper Beurling density: D + (Λ) = lim sup sup r 2 r →∞ z ∈ R 2 Examples: D ± ( α Z × β Z ) = 1 D + ( α Z × β Z + ) = 1 D − ( α Z × β Z + ) = 0 , αβ, αβ Conjecture ⇒ D − (Λ) ≥ 1 A Gabor system { M b T a g } ( a,b ) ∈ Λ is complete = H./Walnut, 1995 (builds on H. Landau, 1964) False: ∃ complete Gabor system with D + (Λ) < ε

Translates of Gaussians (Zalik, 1978) φ ( x ) = e − x 2 ∃ Γ ⊆ R such that { T a φ } a ∈ Γ is complete Complete in L 2 ( R ) , not just a subspace like PW! For this example, D + (Γ) = ∞ as a subset of R R Complete Gabor system with no modulations { T a φ } a ∈ Γ is a Gabor system { M b T a φ } ( a,b ) ∈ Λ with Λ = Γ × { 0 } Complete, yet D − (Λ) = 0 and D + (Λ) = ∞ as a subset of R 2 R 2

Olevskii (1997) and Olevskii/Ulanovskii (2004) ∃ (nice!) g and Γ a (small!) perturbation of Z such that { T a φ } a ∈ Γ is complete in L 2 ( R ) . Density: D ± (Γ) = 1 as a subset of R : - 4 - 3 - 2 - 1 1 2 3 4 R As a Gabor System { T a φ } a ∈ Γ is a Gabor system with Λ = Γ × { 0 } . Complete, but D ± (Λ) = 0 as a subset of R 2 : 1 - 4 - 3 - 2 - 1 1 2 3 4 - 1 R 2 But how nice can these systems be? Frames? Bases?

Olson–Zalik Conjecture (1992) No set of translates { T a g } a ∈ Γ can be a Schauder basis for L 2 ( R ) Results in this direction: • Olson/Zalik: { T a g } a ∈ Γ cannot be a Riesz basis for L 2 ( R ) • Christensen/Deng/H. (1999): { M b T a g } ( a,b ) ∈ Λ is a frame for L 2 ( R ) = ⇒ D − (Λ) ≥ 1 Corollary: { T a g } a ∈ Γ cannot be a frame for L 2 ( R ) Definition: { T a g } a ∈ Γ is a Schauder basis if there is some enumeration Γ = { a k } k ∈ N such that ∞ � for all f ∈ L 2 ( R ) f = c k ( f ) T a k g uniquely , k =1

Olson–Zalik Conjecture (1992) No set of translates { T a g } a ∈ Γ can be a Schauder basis for L 2 ( R ) Results in this direction: • Olson/Zalik: { T a g } a ∈ Γ cannot be a Riesz basis for L 2 ( R ) • Christensen/Deng/H. (1999): { M b T a g } ( a,b ) ∈ Λ is a frame for L 2 ( R ) = ⇒ D − (Λ) ≥ 1 Corollary: { T a g } a ∈ Γ cannot be a frame for L 2 ( R ) • Deng/H. (2000): { M b T a g } ( a,b ) ∈ Λ is a Schauder basis for L 2 ( R ) = ⇒ D + (Λ) ≤ 1 • Deng/H. (2000): g ∈ L 2 \ L 1 = ⇒ { T a g } ( a,b ) ∈ Γ not a Schauder basis

Olson–Zalik Conjecture (1992) No set of translates { T a g } a ∈ Γ can be a Schauder basis for L 2 ( R ) The full Olson–Zalik conjecture is still open! Conjecture: Density of Gabor Schauder bases (Deng/H.) If { M b T a g } ( a,b ) ∈ Λ is a Schauder basis for L 2 ( R ) , then D ± (Λ) = 1 Olson–Zalik is a Corollary if true: { T a g } a ∈ Γ can never be a Schauder basis for L 2 ( R ) , because Λ = Γ × { 0 } ⊆ R 2 has density D − (Λ) = 0

Gabor Schauder bases do exist: If g ( x ) = | x | − 1 / 4 χ [ − 1 2 ] then { M n T k g } k,n ∈ Z is a Gabor Schauder basis that 2 , 1 is not a Riesz basis H./Powell (2006) ⇒ | Zg | 2 ∈ A 2 ( T × T ) { M n T k g } k,n ∈ Z is Schauder basis for L 2 ( R ) ⇐ Product A 2 weights w ∈ A 2 ( T × T ) if ∀ intervals I, J ⊆ R , � � � � � � � � 1 1 1 w ( x, y ) dx dy w ( x, y ) dx dy ≤ C. | I | | J | | I | | J | J I I J Conditionality The Schauder basis condition is with respect to a particular class of enumerations of Z 2 .

What about exact Gabor systems? Exact = minimal and complete Exact is “slightly less” than a Schauder basis Example from before with φ ( x ) = e − x 2 { M n T k φ } ( k,n ) � =(0 , 0) is exact; Λ = Z 2 \ { (0 , 0) } has density D ± (Λ) = 1 Ascensi/Lyubarskii/Seip (2008) with φ ( x ) = e − x 2 � � √ √ If Λ = ( − 1 , 0) , (1 , 0) , (0 , ± 2 n ) , ( ± 2 n, 0) then { M b T a φ } ( a,b ) ∈ Λ is exact Density: D − (Λ) = 0 , D + (Λ) = ∞ . Used in the proof: Λ is the zero set of the entire function s ( z ) = z 2 − 1 sin πz 2 2 . z 2 Question Does there exist a set of translations { T a g } a ∈ Γ that is exact in L 2 ( R ) ?