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 {xn}n∈N in a Hilbert space H is a frame if ∃ A, B > 0 such that ∀ x ∈ H, A x2 ≤

• n

|x, xn|2 ≤ B x2 In this case (DS, 1952):

• The frame operator Sx =

nx, xn xn is a topological

isormorphism S : H → H

• The canonical dual frame is {

xn}n∈N where xn = S−1xn

• x =
• n

x, xn xn =

• n

x, xn xn for every x ∈ H

• If x = cnxn, then
• n

|cn|2 =

• n
• x,

xn

• 2 +
• n
• cn − x,

xn

• 2
• ∗
• n=m

|xm, xn|2 = 1 − |xm, xm|2 − |1 − xm, xm|2 2

• If xm,

xm = 1, then xm, xn = 0 for n = m.

• The removal of a vector from a frame leaves either a frame or

an incomplete set: xm, xm = 1 = ⇒ {xn}n=m is a frame xm, xm = 1 = ⇒ {xn}n=m is incomplete

Multiplicative Completion (Boas and Pollard, 1948) Problem If {fk}k∈Z is an ONB for L2[0, 1] and F ⊆ Z is finite then {fk}k /

∈F is

incomplete. Does ∃ m ∈ L∞[0, 1] such that {fk · m}k /

∈F

is complete?

Multiplicative Completion (Boas and Pollard, 1948) Problem If {fk}k∈Z is an ONB for L2[0, 1] and F ⊆ Z be finite then {fk}k /

∈F is

incomplete. Does ∃ m ∈ L∞[0, 1] such that {fk · m}k /

∈F

is complete? Yes: Proof for F = {0} Chose m so f0/m / ∈ L2. If f, fk · m = 0 for k = 0 then f · m, fk = 0 for k = 0, so f · m = cf0 Hence f = cf0/m. Contradiction if c = 0. Therefore {fk · 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 |f0(x)| = ε > 0.

Choose ψ ∈ L∞\L2 with supp(ψ) ⊆ E. Set m(x) =      1, x / ∈ supp(ψ), 1/ψ(x), x ∈ supp(ψ). Then f0(x)/m(x) ≥ ε ψ(x), x ∈ supp(ψ), so f0/m / ∈ L2.

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 f1, . . . , fN ∈ A, then ∃ g ∈ L∞(µ) such that (a) g(x) = 0 for x ∈ X, and (b) f/g / ∈ A for all f ∈ span{f1, . . . , fN} \ {0}. Need “nonatomicness” Multiplicative completion fails in ℓ2(Z).

Surprising Equivalences (Talalyan, and Price and Zink, 1957–1975) If {fn}n∈N ⊆ L2(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 {fn χS} is complete in L2(S). (b) If f is finite a.e. and ε > 0 then ∃ S ⊆ X and g ∈ span{fn} such that µ(S) > 1 − ε and |f − g| < ε on S. (c) There exists a bounded, nonnegative function m such that {mfn} is complete in L2(µ). There are a wide variety of related results by Kazarian.

Weighted Exponentials: Fix g ∈ L2[0, 1], set Eg = {e2πinxg(x)}n∈Z (a) Eg is complete ⇐ ⇒ g = 0 a.e. (b) Eg is minimal ⇐ ⇒ 1/g ∈ L2[0, 1] (c) Eg is a Schauder basis ⇐ ⇒ |g|2 ∈ A2[0, 1] (d) Eg is a Bessel sequence ⇐ ⇒ g ∈ L∞[0, 1] (e) Eg is a frame sequence ⇐ ⇒ ∃ A, B > 0 such that A ≤ |g(x)|2 ≤ B for a.e. x with g(x) = 0 (f) Eg is a Riesz basis ⇐ ⇒ ∃ A, B > 0 such that A ≤ |g(x)|2 ≤ B a.e. (g) Eg is an orthonormal basis ⇐ ⇒ |g(x)| = 1 a.e. Attribution (c) Hunt, Muckenhaupt, Wheeden (1973) (e) Benedetto and Li (1998)

Multiplicative Completion for {e2πinx}n=0. Let en(x) = e2πinx.

• 1/x /

∈ L2[0, 1] so {xen}n=0 is complete

• Biorthogonal system is

en(x) = e2πinx − 1 x for n = 0

• {xen}n=0 is minimal (and complete, so exact).
• If {xen}n=0 were a Schauder basis then 1 ≤ xen2

en2 ≤ 2C

• xen2 = 3−1/2 and

en2 = 4πn πn

sin2 u u2 du → ∞

as n → ∞

• {xen}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) = {MnTkφ}k,n∈Z = {e2πinx e−(x−k)2}k,n∈Z Both systems have density 1. Why?

Beurling Density for Λ ⊆ R2 Let Qr(z) be the square centered at z with side lengths r. Lower Beurling density: D−(Λ) = lim inf

r→∞

inf

z∈R2

#(Λ ∩ Qr(z)) r2 Upper Beurling density: D+(Λ) = lim sup

r→∞

sup

z∈R2

#(Λ ∩ Qr(z)) r2 Examples: D±(αZ×βZ) = 1 αβ, D−(αZ×βZ+) = 0, D+(αZ×βZ+) = 1 αβ Conjecture A Gabor system {MbTag}(a,b)∈Λ is complete = ⇒ D−(Λ) ≥ 1

Beurling Density for Λ ⊆ R2 Let Qr(z) be the square centered at z with side lengths r. Lower Beurling density: D−(Λ) = lim inf

r→∞

inf

z∈R2

#(Λ ∩ Qr(z)) r2 Upper Beurling density: D+(Λ) = lim sup

r→∞

sup

z∈R2

#(Λ ∩ Qr(z)) r2 Examples: D±(αZ×βZ) = 1 αβ, D−(αZ×βZ+) = 0, D+(αZ×βZ+) = 1 αβ Conjecture A Gabor system {MbTag}(a,b)∈Λ is complete = ⇒ D−(Λ) ≥ 1 H./Walnut, 1995 (builds on H. Landau, 1964) False: ∃ complete Gabor system with D+(Λ) < ε

Translates of Gaussians (Zalik, 1978) φ(x) = e−x2 ∃ Γ ⊆ R such that {Taφ}a∈Γ is complete Complete in L2(R), not just a subspace like PW! For this example, D+(Γ) = ∞ as a subset of R R Complete Gabor system with no modulations {Taφ}a∈Γ is a Gabor system {MbTaφ}(a,b)∈Λ with Λ = Γ × {0} Complete, yet D−(Λ) = 0 and D+(Λ) = ∞ as a subset of R2 R2

Olevskii (1997) and Olevskii/Ulanovskii (2004) ∃ (nice!) g and Γ a (small!) perturbation of Z such that {Taφ}a∈Γ is complete in L2(R). Density: D±(Γ) = 1 as a subset of R:

• 4
• 3
• 2
• 1

1 2 3 4

R As a Gabor System {Taφ}a∈Γ is a Gabor system with Λ = Γ × {0}. Complete, but D±(Λ) = 0 as a subset of R2:

• 4
• 3
• 2
• 1

1 2 3 4

• 1

1

R2 But how nice can these systems be? Frames? Bases?

Olson–Zalik Conjecture (1992) No set of translates {Tag}a∈Γ can be a Schauder basis for L2(R) Results in this direction:

• Olson/Zalik: {Tag}a∈Γ cannot be a Riesz basis for L2(R)
• Christensen/Deng/H. (1999):

{MbTag}(a,b)∈Λ is a frame for L2(R) = ⇒ D−(Λ) ≥ 1 Corollary: {Tag}a∈Γ cannot be a frame for L2(R) Definition: {Tag}a∈Γ is a Schauder basis if there is some enumeration Γ = {ak}k∈N such that f =

∞

• k=1

ck(f) Takg uniquely, for all f ∈ L2(R)

Olson–Zalik Conjecture (1992) No set of translates {Tag}a∈Γ can be a Schauder basis for L2(R) Results in this direction:

• Olson/Zalik: {Tag}a∈Γ cannot be a Riesz basis for L2(R)
• Christensen/Deng/H. (1999):

{MbTag}(a,b)∈Λ is a frame for L2(R) = ⇒ D−(Λ) ≥ 1 Corollary: {Tag}a∈Γ cannot be a frame for L2(R)

• Deng/H. (2000):

{MbTag}(a,b)∈Λ is a Schauder basis for L2(R) = ⇒ D+(Λ) ≤ 1

• Deng/H. (2000):

g ∈ L2\L1 = ⇒ {Tag}(a,b)∈Γ not a Schauder basis

Olson–Zalik Conjecture (1992) No set of translates {Tag}a∈Γ can be a Schauder basis for L2(R) The full Olson–Zalik conjecture is still open! Conjecture: Density of Gabor Schauder bases (Deng/H.) If {MbTag}(a,b)∈Λ is a Schauder basis for L2(R), then D±(Λ) = 1 Olson–Zalik is a Corollary if true: {Tag}a∈Γ can never be a Schauder basis for L2(R), because Λ = Γ × {0} ⊆ R2 has density D−(Λ) = 0

Gabor Schauder bases do exist: If g(x) = |x|−1/4 χ[−1

2,1 2] then {MnTkg}k,n∈Z is a Gabor Schauder basis that

is not a Riesz basis H./Powell (2006) {MnTkg}k,n∈Z is Schauder basis for L2(R) ⇐ ⇒ |Zg|2 ∈ A2(T × T) Product A2 weights w ∈ A2(T × T) if ∀ intervals I, J ⊆ R,

• 1

|I| |J|

• J
• I

w(x, y) dx dy 1 |I| |J|

• I
• J

1 w(x, y) dx dy

• ≤ C.

Conditionality The Schauder basis condition is with respect to a particular class of enumerations of Z2.

What about exact Gabor systems? Exact = minimal and complete Exact is “slightly less” than a Schauder basis Example from before with φ(x) = e−x2 {MnTkφ}(k,n)=(0,0) is exact; Λ = Z2 \ {(0, 0)} has density D±(Λ) = 1 Ascensi/Lyubarskii/Seip (2008) with φ(x) = e−x2 If Λ =

• (−1, 0), (1, 0), (0, ±

√ 2n), (± √ 2n, 0)

• then {MbTaφ}(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) = z2 − 1 z2 sin πz2 2 . Question Does there exist a set of translations {Tag}a∈Γ that is exact in L2(R)?

Translations Again Translations {Tag}a∈Γ are finitely linearly independent in L2(R) Same is true in Lp(R) if 1 ≤ p ≤ 2 via the Fourier transform:

N

• k=1

ckg(x − ak) = 0 = ⇒ N

• k=1

cke2πiξak

• g(ξ) = 0

= ⇒

• g = 0.

But if p > 2 then g is a distribution:

N

• k=1

ckg(x − ak) = 0 = ⇒ ϕ g = 0 = ⇒ /

• g = 0

Distributions can be supported at a single point, functions cannot.

• Q. ∃ g ∈ Lp(R)

s.t. N

• k=1

cke2πiξak

• g(ξ) = 0?

Theorem (Edgar/Rosenblatt, 1979) Assume S ⊆ Rn is closed and has Hausdorff dimension ≤ n − 1. If g ∈ Lp(Rn) and supp( g) ⊆ S, then p ≥ 2n/(n − 1). Corollary If g ∈ Lp(Rn) and p < 2n/(n − 1), then translates of g are linearly independent. Proof sketch The zero set of a trigonometric polynomial ϕ on Rn is the intersection

• f an analytic variety in Cn with Rn. The Hausdorff dimension of this

zero set is ≤ n − 1.

Corollary If g ∈ Lp(Rn) and p < 2n/(n − 1), then translates of g are linearly

• independent. (Rosenblatt, 1995, added p = 2n/(n − 1))

Theorem (Edgar/Rosenblatt, 1979) If 2n/(n − 1) < p < ∞, then there exists a function g ∈ Lp(Rn) that has linearly dependent translates. Example Translates are independent in L1(R) for 1 ≤ p < ∞, but translates are independent in L2(R2) for 1 ≤ p ≤ 4.

The HRT Conjecture (1996) Time-frequency translates are finitely linearly independent. That is, if: (a) 0 < ∞

−∞

|g(x)|2 dx < ∞, (b) {(ak, bk)}N

k=1 are finitely many distinct points in R2,

then:

N

• k=1

ck e2πibkx g(x − ak) = 0 = ⇒ c1 = · · · = cN = 0. This conjecture is open in the generality stated, even if we require g to be continuous (or smoother).

HRT Subconjecture If g: R → R is continuous and 0 < ∞

−∞

|g(x)|2 dx < ∞, then c1 g(x) + c2 g(x − 1) + c3 g(x)e2πix + c4 g(x − √ 2)e2πi

√ 3x = 0

implies c1 = c2 = c3 = c4 = 0. That is,

• g(x), g(x − 1), g(x)e2πix, g(x −

√ 2) e2πi

√ 3x

is linearly independent. This is open even if we assume that g is Schwartz-class.

Frames (Duffin and Schaeffer, 1952) A sequence {xn}n∈N in a Hilbert space H is a frame if ∃ A, B > 0 such that ∀ x ∈ H, A x2 ≤

• n

|x, xn|2 ≤ B x2 Quote from H., “What is a Frame?” Sadly, Duffin and Schaeffer both passed away before anyone thought to ask why they called such a system a “frame”. Is it because A f2 and B f2 “frame” the sum between them? We will never know.

Frames (Duffin and Schaeffer, 1952) A sequence {xn}n∈N in a Hilbert space H is a frame if ∃ A, B > 0 such that ∀ x ∈ H, A x2 ≤

• n

|x, xn|2 ≤ B x2 Quote from H., “What is a Frame?” Sadly, Duffin and Schaeffer both passed away before anyone thought to ask why they called such a system a “frame”. Is it because A f2 and B f2 “frame” the sum between them? We will never know. But there’s more to the story . . .

THANK YOU.