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

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

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Comlex methods in Gabor analysis Yurii Lyubarskii August, 14, 2013 Yurii Lyubarskii Comlex methods in Gabor analysis Main objects: : f e i t f ( t ); = { } C , = + i , ( t ) = e t 2 ; G (

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Comlex methods in Gabor analysis

Yurii Lyubarskii August, 14, 2013

Yurii Lyubarskii Comlex methods in Gabor analysis

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Main objects:

˜ Λ = {λ} ⊂ C, λ = µ + iν, πλ : f → eiπνtf (t − µ); φ(t) = e−πt2; G(˜ Λ) = {πλφ}λ∈˜

Λ

Yurii Lyubarskii Comlex methods in Gabor analysis

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SLIDE 3

Main objects:

˜ Λ = {λ} ⊂ C, λ = µ + iν, πλ : f → eiπνtf (t − µ); φ(t) = e−πt2; G(˜ Λ) = {πλφ}λ∈˜

Λ

Low Beurling density: D−(˜ Λ) = limR→∞ infz∈C

#(˜ Λ∩B(z,R)) πR2

Yurii Lyubarskii Comlex methods in Gabor analysis

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Main objects:

˜ Λ = {λ} ⊂ C, λ = µ + iν, πλ : f → eiπνtf (t − µ); φ(t) = e−πt2; G(˜ Λ) = {πλφ}λ∈˜

Λ

Low Beurling density: D−(˜ Λ) = limR→∞ infz∈C

#(˜ Λ∩B(z,R)) πR2

We know (Seip ’92, L.’92): D−(˜ Λ) > 1 ⇔ G(˜ Λ) is a frame in L2(R) i.e. A(˜ Λ)f 2

2 ≤

  • λ∈˜

Λ

|f , πλφ|2 ≤ B(˜ Λ)f 2, ∀f ∈ L2(R).

Yurii Lyubarskii Comlex methods in Gabor analysis

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SLIDE 5

Main objects:

˜ Λ = {λ} ⊂ C, λ = µ + iν, πλ : f → eiπνtf (t − µ); φ(t) = e−πt2; G(˜ Λ) = {πλφ}λ∈˜

Λ

Low Beurling density: D−(˜ Λ) = limR→∞ infz∈C

#(˜ Λ∩B(z,R)) πR2

We know (Seip ’92, L.’92): D−(˜ Λ) > 1 ⇔ G(˜ Λ) is a frame in L2(R) i.e. A(˜ Λ)f 2

2 ≤

  • λ∈˜

Λ

|f , πλφ|2 ≤ B(˜ Λ)f 2, ∀f ∈ L2(R). 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

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Setting of the problem

Given Λ ⊂ C, D−(Λ) = 1. Let ˜ Λ = aΛ. What can be said about behavior of A(˜ Λ) and B(˜ Λ) as a → 1?

Yurii Lyubarskii Comlex methods in Gabor analysis

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Setting of the problem

Given Λ ⊂ C, D−(Λ) = 1. Let ˜ Λ = aΛ. 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

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Setting of the problem

Given Λ ⊂ C, D−(Λ) = 1. Let ˜ Λ = aΛ. 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 = αZ + iβZ, αβ = 1 then A(˜ Λ) ≍ (1 − a) as a ր 1. (Numerically: Strohmer, Sondergaast)

Yurii Lyubarskii Comlex methods in Gabor analysis

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Setting of the problem

Given Λ ⊂ C, D−(Λ) = 1. Let ˜ Λ = aΛ. 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 = αZ + iβZ, 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

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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 {an}, an → 1 so that A(anΛ) < ω(an);

Yurii Lyubarskii Comlex methods in Gabor analysis

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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 {an}, an → 1 so that A(anΛ) < ω(an);

◮ If Λ is periodic (quasiperiodic) then A(aΛ) ≍ (1 − a);

Yurii Lyubarskii Comlex methods in Gabor analysis

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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 {an}, an → 1 so that A(anΛ) < ω(an);

◮ If Λ is periodic (quasiperiodic) then A(aΛ) ≍ (1 − a); ◮ Let Λ = (Z + iZ) \ {0}. Then still A(aΛ) ≍ (1 − a); ◮ Let Λ = (Z + iZ) \ {0, 1}. Then A(aΛ) ≍ (1 − a)1+1/2 and so

  • n.

Yurii Lyubarskii Comlex methods in Gabor analysis

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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 {an}, an → 1 so that A(anΛ) < ω(an);

◮ If Λ is periodic (quasiperiodic) then A(aΛ) ≍ (1 − a); ◮ Let Λ = (Z + iZ) \ {0}. Then still A(aΛ) ≍ (1 − a); ◮ Let Λ = (Z + iZ) \ {0, 1}. Then A(aΛ) ≍ (1 − a)1+1/2 and so

  • n.

◮ Adding finite number of points to Z + iZ does not improve

decay of A(aΛ) of course.

Yurii Lyubarskii Comlex methods in Gabor analysis

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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 {an}, an → 1 so that A(anΛ) < ω(an);

◮ If Λ is periodic (quasiperiodic) then A(aΛ) ≍ (1 − a); ◮ Let Λ = (Z + iZ) \ {0}. Then still A(aΛ) ≍ (1 − a); ◮ Let Λ = (Z + iZ) \ {0, 1}. Then A(aΛ) ≍ (1 − a)1+1/2 and so

  • n.

◮ Adding finite number of points to Z + iZ does not improve

decay of A(aΛ) of course.

◮ Conjecture: A(aΛ) cannot decay slower than 1 − a

independently of choice Λ with D−(Λ) = 1.

Yurii Lyubarskii Comlex methods in Gabor analysis

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The standard question

Yurii Lyubarskii Comlex methods in Gabor analysis

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The standard question

Who is responsible ?

Yurii Lyubarskii Comlex methods in Gabor analysis

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The standard question

Who is responsible ? The standard answer:

Yurii Lyubarskii Comlex methods in Gabor analysis

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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: σ(z) = z

  • (m,n)=(0,0)

(1 − z m + in)

Yurii Lyubarskii Comlex methods in Gabor analysis

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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: σ(z) = z

  • (m,n)=(0,0)

(1 − z m + in) σ(z) = z

  • (m,n)=(0,0)

(1 − z m + in)e

z m+in + 1 2( z m+in) 2 Yurii Lyubarskii Comlex methods in Gabor analysis

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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: σ(z) = z

  • (m,n)=(0,0)

(1 − z m + in)

Yurii Lyubarskii Comlex methods in Gabor analysis

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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: σ(z) = z

  • (m,n)=(0,0)

(1 − z m + in) |σ(z)| ≍ e

π 2 |z|2, dist(z, Z + iZ) > ǫ. Yurii Lyubarskii Comlex methods in Gabor analysis

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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: σ(z) = z

  • (m,n)=(0,0)

(1 − z m + in) |σ(z)| ≍ e

π 2 |z|2, dist(z, Z + iZ) > ǫ.

Statement: Let S(z) ≍ e

π 2 |z|2, dist(z, Λ) > ǫ. Then

A(aΛ) ≍ 1 − a, as a ր 1.

Yurii Lyubarskii Comlex methods in Gabor analysis

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SLIDE 23

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: σ(z) = z

  • (m,n)=(0,0)

(1 − z m + in) |σ(z)| ≍ e

π 2 |z|2, dist(z, Z + iZ) > ǫ.

Statement: Let S(z) ≍ e

π 2 |z|2, dist(z, Λ) > ǫ. Then

A(aΛ) ≍ 1 − a, as a ր 1. X X X

Yurii Lyubarskii Comlex methods in Gabor analysis

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Fock space F

F = {F ∈ Hol(C); F2 =

  • C

|F(z)|2e−π|z|2dmz < ∞} F ∈ F ⇒ |F(z)| ≤ Const e

π 2 |z|2. Yurii Lyubarskii Comlex methods in Gabor analysis

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Fock space F

F = {F ∈ Hol(C); F2 =

  • C

|F(z)|2e−π|z|2dmz < ∞} F ∈ F ⇒ |F(z)| ≤ Const e

π 2 |z|2.

Properties

◮ eπz ¯ w – reproducing kernel: F(w) = F(·), eπ ¯ w·. ◮ Fock shift: λ = µ + iν ∈ C, ⇒ unitary mapping βλ : F → F

βλ : F → βλF(z) = eiπµνe− π

2 |λ|2eπ¯

λzF(z − λ)

Yurii Lyubarskii Comlex methods in Gabor analysis

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Bargmann transform B : L2(R) → F

B : f → F(z) = 2

1 4 e− π 2 z2

R

f (t)e−πt2e2πtzdt Equivalent definition: B : Hn(t) → πn n! 1

2

zn, n = 0, 1, . . . Hn(t) = cne−2πt2 dn dtn eπt2

Yurii Lyubarskii Comlex methods in Gabor analysis

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Bargmann transform B : L2(R) → F

B : f → F(z) = 2

1 4 e− π 2 z2

R

f (t)e−πt2e2πtzdt Equivalent definition: B : Hn(t) → πn n! 1

2

zn, n = 0, 1, . . . Hn(t) = cne−2πt2 dn dtn eπt2 Properties

◮ B – unitary mapping

Yurii Lyubarskii Comlex methods in Gabor analysis

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Bargmann transform B : L2(R) → F

B : f → F(z) = 2

1 4 e− π 2 z2

R

f (t)e−πt2e2πtzdt Equivalent definition: B : Hn(t) → πn n! 1

2

zn, n = 0, 1, . . . Hn(t) = cne−2πt2 dn dtn eπt2 Properties

◮ B – unitary mapping ◮ B : f → F(z) ⇒ B : ˆ

f → F(iz)

Yurii Lyubarskii Comlex methods in Gabor analysis

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Bargmann transform B : L2(R) → F

B : f → F(z) = 2

1 4 e− π 2 z2

R

f (t)e−πt2e2πtzdt Equivalent definition: B : Hn(t) → πn n! 1

2

zn, n = 0, 1, . . . Hn(t) = cne−2πt2 dn dtn eπt2 Properties

◮ B – unitary mapping ◮ B : f → F(z) ⇒ B : ˆ

f → F(iz)

◮ Bπλf = βλBf

Yurii Lyubarskii Comlex methods in Gabor analysis

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SLIDE 30

Bargmann transform B : L2(R) → F

B : f → F(z) = 2

1 4 e− π 2 z2

R

f (t)e−πt2e2πtzdt Equivalent definition: B : Hn(t) → πn n! 1

2

zn, n = 0, 1, . . . Hn(t) = cne−2πt2 dn dtn eπt2 Properties

◮ B – unitary mapping ◮ B : f → F(z) ⇒ B : ˆ

f → F(iz)

◮ Bπλf = βλBf

In particular B : φ → 1; B : πλφ → βλ1 = ce− π

2 |λ|2ezλ, |c| = 1. Yurii Lyubarskii Comlex methods in Gabor analysis

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Frame inequality

A(˜ Λ)f 2

2 ≤

  • λ∈˜

Λ

|f , πλφ|2 ≤ B(˜ Λ)f 2, ∀f ∈ L2(R) ⇓ A(˜ Λ)F2

F ≤

  • λ

e−π|λ|2|F, eπλF|2 ≤ B(˜ Λ)F2

F, F ∈ F

Yurii Lyubarskii Comlex methods in Gabor analysis

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Frame inequality

A(˜ Λ)f 2

2 ≤

  • λ∈˜

Λ

|f , πλφ|2 ≤ B(˜ Λ)f 2, ∀f ∈ L2(R) ⇓ A(˜ Λ)F2

F ≤

  • λ

e−π|λ|2|F, eπλF|2 ≤ B(˜ Λ)F2

F, F ∈ F

⇓ A(˜ Λ)F2

F ≤

  • λ

e−π|λ|2|F(¯ λ)|2

  • Fa

≤ B(˜ Λ)F2

F,

F ∈ F We need to estimate sampling constants in Fock space.

Yurii Lyubarskii Comlex methods in Gabor analysis

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Two things to do

The standard mean value theorem ⇒ B(˜ Λ) stays bounded. Let, for definiteness, |S(z)| ≍ exp(π|z|2/2) off the zero set and we want to prove A(˜ Λ) ≍ (1 − a) as a ր 1,

Yurii Lyubarskii Comlex methods in Gabor analysis

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Two things to do

The standard mean value theorem ⇒ B(˜ Λ) stays bounded. Let, for definiteness, |S(z)| ≍ exp(π|z|2/2) off the zero set and we want to prove A(˜ Λ) ≍ (1 − a) as a ր 1, We need

◮ F2 F ≤ Const 1−a F2 a,F

Yurii Lyubarskii Comlex methods in Gabor analysis

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SLIDE 35

Two things to do

The standard mean value theorem ⇒ B(˜ Λ) stays bounded. Let, for definiteness, |S(z)| ≍ exp(π|z|2/2) off the zero set and we want to prove A(˜ Λ) ≍ (1 − a) as a ր 1, We need

◮ F2 F ≤ Const 1−a F2 a,F ⇒

A(˜ Λ) ≤ Const(1 − a)

Yurii Lyubarskii Comlex methods in Gabor analysis

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SLIDE 36

Two things to do

The standard mean value theorem ⇒ B(˜ Λ) stays bounded. Let, for definiteness, |S(z)| ≍ exp(π|z|2/2) off the zero set and we want to prove A(˜ Λ) ≍ (1 − a) as a ր 1, We need

◮ F2 F ≤ Const 1−a F2 a,F ⇒

A(˜ Λ) ≤ Const(1 − a)

◮ ∃F ∈ F; F2 F ≥ Const 1−a F2 a,F

Yurii Lyubarskii Comlex methods in Gabor analysis

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SLIDE 37

Two things to do

The standard mean value theorem ⇒ B(˜ Λ) stays bounded. Let, for definiteness, |S(z)| ≍ exp(π|z|2/2) off the zero set and we want to prove A(˜ Λ) ≍ (1 − a) as a ր 1, We need

◮ F2 F ≤ Const 1−a F2 a,F ⇒

A(˜ Λ) ≤ Const(1 − a)

◮ ∃F ∈ F; F2 F ≥ Const 1−a F2 a,F ⇒

A(˜ Λ) ≥ Const(1 − a)

Yurii Lyubarskii Comlex methods in Gabor analysis

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SLIDE 38

Two things to do

The standard mean value theorem ⇒ B(˜ Λ) stays bounded. Let, for definiteness, |S(z)| ≍ exp(π|z|2/2) off the zero set and we want to prove A(˜ Λ) ≍ (1 − a) as a ր 1, We need

◮ F2 F ≤ Const 1−a F2 a,F ⇒

A(˜ Λ) ≤ Const(1 − a) - Atomization

◮ ∃F ∈ F; F2 F ≥ Const 1−a F2 a,F ⇒

A(˜ Λ) ≥ Const(1 − a) - Interpolation

Yurii Lyubarskii Comlex methods in Gabor analysis

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SLIDE 39

Interpolation

Λ is zero set of S(z) ⇒ ˜ Λ is zero set of ˜ S(z) = S(z/a).

Yurii Lyubarskii Comlex methods in Gabor analysis

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SLIDE 40

Interpolation

Λ is zero set of S(z) ⇒ ˜ Λ is zero set of ˜ S(z) = S(z/a). |S(z)| ≍ e

π 2 |z|2 ⇒ |˜

S(z)| ≍ e

π 2a2 |z|2, it grows faster than any

functions F ∈ F. Standard interpolation formula: F(z) = ˜ S(z)

  • λ∈˜

Λ

F(λ) ˜ S′(λ)(z − λ)

Yurii Lyubarskii Comlex methods in Gabor analysis

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SLIDE 41

Interpolation

Λ is zero set of S(z) ⇒ ˜ Λ is zero set of ˜ S(z) = S(z/a). |S(z)| ≍ e

π 2 |z|2 ⇒ |˜

S(z)| ≍ e

π 2a2 |z|2, it grows faster than any

functions F ∈ F. Standard interpolation formula: F(z) = ˜ S(z)

  • λ∈˜

Λ

F(λ) ˜ S′(λ)(z − λ) is of no use: we need to estimate sum with the weight e− π

2 |z|2 but

˜ S grows too fast! There is NO interpolation formula which gives estimates in · F-norm for all z ∈ C.

Yurii Lyubarskii Comlex methods in Gabor analysis

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SLIDE 42

Recipe

Use various interpolation formulas for various rays! Let for example z = x > 0. Take ˜ S0(z) = e

π 2 (1− 1 a2 )|z|2 ˜

S(z) : |˜ S0(x)| ≍ e

π 2 x2. Yurii Lyubarskii Comlex methods in Gabor analysis

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SLIDE 43

Recipe

Use various interpolation formulas for various rays! Let for example z = x > 0. Take ˜ S0(z) = e

π 2 (1− 1 a2 )|z|2 ˜

S(z) : |˜ S0(x)| ≍ e

π 2 x2.

Still it is possible to prove F(z) = ˜ S0(z)

  • λ∈˜

Λ

F(λ) ˜ S′

0(λ)(z − λ)

One can obtain needed estimates for |F(x)| !

Yurii Lyubarskii Comlex methods in Gabor analysis

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SLIDE 44

Recipe

Use various interpolation formulas for various rays! Let for example z = x > 0. Take ˜ S0(z) = e

π 2 (1− 1 a2 )|z|2 ˜

S(z) : |˜ S0(x)| ≍ e

π 2 x2.

Still it is possible to prove F(z) = ˜ S0(z)

  • λ∈˜

Λ

F(λ) ˜ S′

0(λ)(z − λ)

One can obtain needed estimates for |F(x)| ! CHEATING: Generally no way to guess the behavior of the constant in advance. One has to START with interpolation formula which hints the answer.

Yurii Lyubarskii Comlex methods in Gabor analysis

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SLIDE 45

Atomization

Relation for the σ-function

log |σ(z)| = log |z|+

  • (m,n)=0

log

  • 1 −

z m + in

  • = π

2 |z|2+O(1), dist(z, Z+iZ) > ǫ

Yurii Lyubarskii Comlex methods in Gabor analysis

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SLIDE 46

Atomization

Relation for the σ-function

log |σ(z)| = log |z|+

  • (m,n)=0

log

  • 1 −

z m + in

  • = π

2 |z|2+O(1), dist(z, Z+iZ) > ǫ Integral relation:

  • C

log

  • 1 − z

ζ

  • dmζ = π

2 |z|2 + atomization procedure

Yurii Lyubarskii Comlex methods in Gabor analysis

slide-47
SLIDE 47

Atomization

Relation for the σ-function

log |σ(z)| = log |z|+

  • (m,n)=0

log

  • 1 −

z m + in

  • = π

2 |z|2+O(1), dist(z, Z+iZ) > ǫ Integral relation:

  • C

log

  • 1 − z

ζ

  • dmζ = π

2 |z|2 + atomization procedure

We get a function with zero discrete norm !

Yurii Lyubarskii Comlex methods in Gabor analysis

slide-48
SLIDE 48

Atomization

Relation for the σ-function

log |σ(z)| = log |z|+

  • (m,n)=0

log

  • 1 −

z m + in

  • = π

2 |z|2+O(1), dist(z, Z+iZ) > ǫ Integral relation:

  • C

log

  • 1 − z

ζ

  • dmζ = π

2 |z|2 + atomization procedure

We get a function with zero discrete norm ! But not in our space !

Yurii Lyubarskii Comlex methods in Gabor analysis

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SLIDE 49

Idea # 1: Make a function which belongs to our space

Let for definiteness Λ = Z + iZ. Take uR(z) =

  • |ζ|<R

log

  • 1−z

ζ

  • dmζ =
  • π

2 |z|2,

|z| < R, πR2 log |z| − πR2 log R + π

2 R2,

|z| > R. and atomize it at integers.

Yurii Lyubarskii Comlex methods in Gabor analysis

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SLIDE 50

Idea # 1: Make a function which belongs to our space

Let for definiteness Λ = Z + iZ. Take uR(z) =

  • |ζ|<R

log

  • 1−z

ζ

  • dmζ =
  • π

2 |z|2,

|z| < R, πR2 log |z| − πR2 log R + π

2 R2,

|z| > R. and atomize it at integers. If R << a−1 then (Z + iZ) ∩ RD is close to a(Z + iZ) ∩ RD and we will have something small at a(Z + iZ) ∩ RD and in F.

Yurii Lyubarskii Comlex methods in Gabor analysis

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SLIDE 51

Idea # 1: Make a function which belongs to our space

Let for definiteness Λ = Z + iZ. Take uR(z) =

  • |ζ|<R

log

  • 1−z

ζ

  • dmζ =
  • π

2 |z|2,

|z| < R, πR2 log |z| − πR2 log R + π

2 R2,

|z| > R. and atomize it at integers. If R << a−1 then (Z + iZ) ∩ RD is close to a(Z + iZ) ∩ RD and we will have something small at a(Z + iZ) ∩ RD and in F. Trouble: Boundary effects of truncation.

Yurii Lyubarskii Comlex methods in Gabor analysis

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SLIDE 52

Idea # 1: Make a function which belongs to our space

Let for definiteness Λ = Z + iZ. Take uR(z) =

  • |ζ|<R

log

  • 1−z

ζ

  • dmζ =
  • π

2 |z|2,

|z| < R, πR2 log |z| − πR2 log R + π

2 R2,

|z| > R. and atomize it at integers. If R << a−1 then (Z + iZ) ∩ RD is close to a(Z + iZ) ∩ RD and we will have something small at a(Z + iZ) ∩ RD and in F. Trouble: Boundary effects of truncation. We can obtain just A(aΛ) > const(1 − a)4

Yurii Lyubarskii Comlex methods in Gabor analysis

slide-53
SLIDE 53

Idea # 2: Reducing the boundary effects

Take instead uR((1 − ǫ)z)

Yurii Lyubarskii Comlex methods in Gabor analysis

slide-54
SLIDE 54

Idea # 2: Reducing the boundary effects

Take instead uR((1 − ǫ)z) Trouble: Boundary effects of atomization

Yurii Lyubarskii Comlex methods in Gabor analysis

slide-55
SLIDE 55

Idea # 3: Modify the set of atomization points near the boundary !

Yurii Lyubarskii Comlex methods in Gabor analysis

slide-56
SLIDE 56

Idea # 3: Modify the set of atomization points near the boundary ! Comment: all this is not that simple as it looks like: all choices R = R(a), ǫ = ǫ(a), etc are unique possible. But

Yurii Lyubarskii Comlex methods in Gabor analysis

slide-57
SLIDE 57

Idea # 3: Modify the set of atomization points near the boundary ! Comment: all this is not that simple as it looks like: all choices R = R(a), ǫ = ǫ(a), etc are unique possible. But Finally we get A(aΛ) > const(1 − a)

Yurii Lyubarskii Comlex methods in Gabor analysis