Comlex methods in Gabor analysis Yurii Lyubarskii August, 14, 2013 - PowerPoint PPT Presentation

Comlex methods in Gabor analysis Yurii Lyubarskii August, 14, 2013 Yurii Lyubarskii Comlex methods in Gabor analysis

Main objects: ˜ π λ : f → e i πν t f ( t − µ ); Λ = { λ } ⊂ C , λ = µ + i ν , φ ( t ) = e − π t 2 ; G (˜ Λ) = { π λ φ } λ ∈ ˜ Λ Yurii Lyubarskii Comlex methods in Gabor analysis

Main objects: ˜ π λ : f → e i πν t f ( t − µ ); Λ = { λ } ⊂ C , λ = µ + i ν , φ ( t ) = e − π t 2 ; G (˜ Λ) = { π λ φ } λ ∈ ˜ Λ Low Beurling density: #(˜ D − (˜ Λ ∩ B ( z , R )) Λ) = lim R →∞ inf z ∈ C π R 2 Yurii Lyubarskii Comlex methods in Gabor analysis

Main objects: ˜ π λ : f → e i πν t f ( t − µ ); Λ = { λ } ⊂ C , λ = µ + i ν , φ ( t ) = e − π t 2 ; G (˜ Λ) = { π λ φ } λ ∈ ˜ Λ Low Beurling density: #(˜ D − (˜ Λ ∩ B ( z , R )) Λ) = lim R →∞ inf z ∈ C π R 2 We know (Seip ’92, L.’92): D − (˜ Λ) > 1 ⇔ G (˜ Λ) is a frame in L 2 ( R ) i.e. � |� f , π λ φ �| 2 ≤ B (˜ A (˜ Λ) � f � 2 Λ) � f � 2 , ∀ f ∈ L 2 ( R ) . 2 ≤ λ ∈ ˜ Λ Yurii Lyubarskii Comlex methods in Gabor analysis

Main objects: ˜ π λ : f → e i πν t f ( t − µ ); Λ = { λ } ⊂ C , λ = µ + i ν , φ ( t ) = e − π t 2 ; G (˜ Λ) = { π λ φ } λ ∈ ˜ Λ Low Beurling density: #(˜ D − (˜ Λ ∩ B ( z , R )) Λ) = lim R →∞ inf z ∈ C π R 2 We know (Seip ’92, L.’92): D − (˜ Λ) > 1 ⇔ G (˜ Λ) is a frame in L 2 ( R ) i.e. � |� f , π λ φ �| 2 ≤ B (˜ A (˜ Λ) � f � 2 Λ) � f � 2 , ∀ f ∈ L 2 ( R ) . 2 ≤ λ ∈ ˜ Λ Q: What happens with the frame constants A (˜ Λ) and B (˜ Λ) as density of ˜ Λ shrinks to 1 ? Yurii Lyubarskii Comlex methods in Gabor analysis

Setting of the problem Given Λ ⊂ C , D − (Λ) = 1. Let ˜ Λ = a Λ. What can be said about behavior of A (˜ Λ) and B (˜ Λ) as a → 1? Yurii Lyubarskii Comlex methods in Gabor analysis

Setting of the problem Given Λ ⊂ C , D − (Λ) = 1. Let ˜ Λ = a Λ. What can be said about behavior of A (˜ Λ) and B (˜ Λ) as a → 1? Another setting: Who is responsible for behavior of A (˜ Λ)? Yurii Lyubarskii Comlex methods in Gabor analysis

Setting of the problem Given Λ ⊂ C , D − (Λ) = 1. Let ˜ Λ = a Λ. What can be said about behavior of A (˜ Λ) and B (˜ Λ) as a → 1? Another setting: Who is responsible for behavior of A (˜ Λ)? Known: (Groechenig, Borichev, L.): Let Λ = α Z + i β Z , αβ = 1 then A (˜ Λ) ≍ (1 − a ) as a ր 1. (Numerically: Strohmer, Sondergaast) Yurii Lyubarskii Comlex methods in Gabor analysis

Setting of the problem Given Λ ⊂ C , D − (Λ) = 1. Let ˜ Λ = a Λ. What can be said about behavior of A (˜ Λ) and B (˜ Λ) as a → 1? Another setting: Who is responsible for behavior of A (˜ Λ)? Known: (Groechenig, Borichev, L.): Let Λ = α Z + i β Z , αβ = 1 then A (˜ Λ) ≍ (1 − a ) as a ր 1. (Numerically: Strohmer, Sondergaast) Tool: combination of Gabor techniques with estimates of entire functions. What about other configurations? Yurii Lyubarskii Comlex methods in Gabor analysis

Typical results ◮ Behavior of A (˜ Λ) can be arbitrarily bad: Statement: ∀ ω ( a ), ω ( a ) ց 0 as a ր 1 there exists Λ, D − (Λ) = 1 and { a n } , a n → 1 so that A ( a n Λ) < ω ( a n ); Yurii Lyubarskii Comlex methods in Gabor analysis

Typical results ◮ Behavior of A (˜ Λ) can be arbitrarily bad: Statement: ∀ ω ( a ), ω ( a ) ց 0 as a ր 1 there exists Λ, D − (Λ) = 1 and { a n } , a n → 1 so that A ( a n Λ) < ω ( a n ); ◮ If Λ is periodic (quasiperiodic) then A ( a Λ) ≍ (1 − a ); Yurii Lyubarskii Comlex methods in Gabor analysis

Typical results ◮ Behavior of A (˜ Λ) can be arbitrarily bad: Statement: ∀ ω ( a ), ω ( a ) ց 0 as a ր 1 there exists Λ, D − (Λ) = 1 and { a n } , a n → 1 so that A ( a n Λ) < ω ( a n ); ◮ If Λ is periodic (quasiperiodic) then A ( a Λ) ≍ (1 − a ); ◮ Let Λ = ( Z + i Z ) \ { 0 } . Then still A ( a Λ) ≍ (1 − a ); ◮ Let Λ = ( Z + i Z ) \ { 0 , 1 } . Then A ( a Λ) ≍ (1 − a ) 1+1 / 2 and so on. Yurii Lyubarskii Comlex methods in Gabor analysis

Typical results ◮ Behavior of A (˜ Λ) can be arbitrarily bad: Statement: ∀ ω ( a ), ω ( a ) ց 0 as a ր 1 there exists Λ, D − (Λ) = 1 and { a n } , a n → 1 so that A ( a n Λ) < ω ( a n ); ◮ If Λ is periodic (quasiperiodic) then A ( a Λ) ≍ (1 − a ); ◮ Let Λ = ( Z + i Z ) \ { 0 } . Then still A ( a Λ) ≍ (1 − a ); ◮ Let Λ = ( Z + i Z ) \ { 0 , 1 } . Then A ( a Λ) ≍ (1 − a ) 1+1 / 2 and so on. ◮ Adding finite number of points to Z + i Z does not improve decay of A ( a Λ) of course. Yurii Lyubarskii Comlex methods in Gabor analysis

Typical results ◮ Behavior of A (˜ Λ) can be arbitrarily bad: Statement: ∀ ω ( a ), ω ( a ) ց 0 as a ր 1 there exists Λ, D − (Λ) = 1 and { a n } , a n → 1 so that A ( a n Λ) < ω ( a n ); ◮ If Λ is periodic (quasiperiodic) then A ( a Λ) ≍ (1 − a ); ◮ Let Λ = ( Z + i Z ) \ { 0 } . Then still A ( a Λ) ≍ (1 − a ); ◮ Let Λ = ( Z + i Z ) \ { 0 , 1 } . Then A ( a Λ) ≍ (1 − a ) 1+1 / 2 and so on. ◮ Adding finite number of points to Z + i Z does not improve decay of A ( a Λ) of course. ◮ Conjecture: A ( a Λ) cannot decay slower than 1 − a independently of choice Λ with D − (Λ) = 1. Yurii Lyubarskii Comlex methods in Gabor analysis

The standard question Yurii Lyubarskii Comlex methods in Gabor analysis

The standard question Who is responsible ? Yurii Lyubarskii Comlex methods in Gabor analysis

The standard question Who is responsible ? The standard answer: Yurii Lyubarskii Comlex methods in Gabor analysis

The standard question Who is responsible ? The standard answer: Entire functions of course. Let S ( z ) be the generating entire function i.e. zero set of S is Λ. Example: Weierstarss σ -function: z � σ ( z ) = z (1 − m + in ) ( m , n ) � =(0 , 0) Yurii Lyubarskii Comlex methods in Gabor analysis

The standard question Who is responsible ? The standard answer: Entire functions of course. Let S ( z ) be the generating entire function i.e. zero set of S is Λ. Example: Weierstarss σ -function: z � σ ( z ) = z (1 − m + in ) ( m , n ) � =(0 , 0) z 2 � m + in + 1 z 2 ( m + in ) z σ ( z ) = z (1 − m + in ) e ( m , n ) � =(0 , 0) Yurii Lyubarskii Comlex methods in Gabor analysis

The standard question Who is responsible ? The standard answer: Entire functions of course. Let S ( z ) be the generating entire function i.e. zero set of S is Λ. Example: Weierstarss σ -function: z � σ ( z ) = z (1 − m + in ) ( m , n ) � =(0 , 0) Yurii Lyubarskii Comlex methods in Gabor analysis

The standard question Who is responsible ? The standard answer: Entire functions of course. Let S ( z ) be the generating entire function i.e. zero set of S is Λ. Example: Weierstarss σ -function: z � σ ( z ) = z (1 − m + in ) ( m , n ) � =(0 , 0) 2 | z | 2 , dist( z , Z + i Z ) > ǫ . π | σ ( z ) | ≍ e Yurii Lyubarskii Comlex methods in Gabor analysis

The standard question Who is responsible ? The standard answer: Entire functions of course. Let S ( z ) be the generating entire function i.e. zero set of S is Λ. Example: Weierstarss σ -function: z � σ ( z ) = z (1 − m + in ) ( m , n ) � =(0 , 0) 2 | z | 2 , dist( z , Z + i Z ) > ǫ . π | σ ( z ) | ≍ e 2 | z | 2 , dist( z , Λ) > ǫ . Then π Statement: Let S ( z ) ≍ e A ( a Λ) ≍ 1 − a , as a ր 1 . Yurii Lyubarskii Comlex methods in Gabor analysis

The standard question Who is responsible ? The standard answer: Entire functions of course. Let S ( z ) be the generating entire function i.e. zero set of S is Λ. Example: Weierstarss σ -function: z � σ ( z ) = z (1 − m + in ) ( m , n ) � =(0 , 0) 2 | z | 2 , dist( z , Z + i Z ) > ǫ . π | σ ( z ) | ≍ e 2 | z | 2 , dist( z , Λ) > ǫ . Then π Statement: Let S ( z ) ≍ e A ( a Λ) ≍ 1 − a , as a ր 1 . X X X Yurii Lyubarskii Comlex methods in Gabor analysis

Fock space F � F = { F ∈ Hol( C ); � F � 2 = | F ( z ) | 2 e − π | z | 2 dm z < ∞} C 2 | z | 2 . π F ∈ F ⇒ | F ( z ) | ≤ Const e Yurii Lyubarskii Comlex methods in Gabor analysis

Fock space F � F = { F ∈ Hol( C ); � F � 2 = | F ( z ) | 2 e − π | z | 2 dm z < ∞} C 2 | z | 2 . π F ∈ F ⇒ | F ( z ) | ≤ Const e Properties w – reproducing kernel: F ( w ) = � F ( · ) , e π ¯ ◮ e π z ¯ w · � . ◮ Fock shift: λ = µ + i ν ∈ C , ⇒ unitary mapping β λ : F → F 2 | λ | 2 e π ¯ β λ : F �→ β λ F ( z ) = e i πµν e − π λ z F ( z − λ ) Yurii Lyubarskii Comlex methods in Gabor analysis

Bargmann transform B : L 2 ( R ) → F 2 z 2 � 1 f ( t ) e − π t 2 e 2 π tz dt 4 e − π B : f �→ F ( z ) = 2 R Equivalent definition: � 1 � π n H n ( t ) = c n e − 2 π t 2 d n 2 dt n e π t 2 z n , n = 0 , 1 , . . . B : H n ( t ) �→ n ! Yurii Lyubarskii Comlex methods in Gabor analysis

Bargmann transform B : L 2 ( R ) → F 2 z 2 � 1 f ( t ) e − π t 2 e 2 π tz dt 4 e − π B : f �→ F ( z ) = 2 R Equivalent definition: � 1 � π n H n ( t ) = c n e − 2 π t 2 d n 2 dt n e π t 2 z n , n = 0 , 1 , . . . B : H n ( t ) �→ n ! Properties ◮ B – unitary mapping Yurii Lyubarskii Comlex methods in Gabor analysis