Undersampled windowed exponentials, spectra of Toeplitz operators ...

UNDERSAMPLED WINDOWED EXPONENTIALS, SPECTRA OF TOEPLITZ OPERATORS AND ITS APPLICATIONS

arXiv:1702.01887v1 [math.FA] 7 Feb 2017

CHUN-KIT LAI AND SUI TANG

Abstract. We characterize the completeness and frame/basis property of a union of under-sampled windowed exponentials of the form F (g) := {e2πinx : n ≥ 0} ∪ {g(x)e2πinx : n < 0} for L2 [−1/2, 1/2] by the spectra of the Toeplitz operators with symbol g. Using this characterization, we classify all real-valued functions g such that F (g) is complete or forms a frame/basis. Conversely, we use the classical Kadec-1/4-theorem in non-harmonic Fourier series to determine all ξ such that the Toeplitz operators with symbol e2πiξx is injective or invertible. These results demonstrate an elegant interaction between frame theory of windowed exponentials and Toeplitz operators. Finally, as an application, we use our results to answer some open questions in dynamical sampling, phase retrieval and derivative samplings on ℓ2 (Z) and Paley-Wiener spaces of bandlimited functions.

Contents 1. Introduction 2. Preliminaries on Toeplitz and Hankel operators 3. Frame properties of F (g) 4. Classification of admissible window functions 5. One-sided exponentials 6. Applications 7. Discussions and Open Problems References

1 6 9 12 15 17 21 22

1. Introduction Background. Let Ω be a measurable set of finite Lebesgue measure in Rd . Suppose that g ∈ L2 (Ω) \ {0} and Λ is a countable subset of Rd . We define the collection of windowed exponentials by E(g, Λ) := {g(x)e2πihλ,xi : λ ∈ Λ}. 2010 Mathematics Subject Classification. Primary 42C15, 42C30s. Key words and phrases. Completeness, Frames, Spectra, Toeplitz operators, windowed exponentials. 1

A natural question is to determine when a union of finitely many windowed expoSN nentials j=1 E(gj , Λj ) is complete or forms a frame/Riesz basis for L2 (Ω). Let us recall the definitions as below: S Definition 1.1. The collection of windowed exponentials N j=1 E(gj , Λj ) forms a 2 frame for L (Ω) if there exists 0 < A ≤ B < ∞ such that 2 N X Z X −2πihλ,xi 2 f (x)gj (x)e dx ≤ Bkf k2 . Akf k ≤ j=1 λ∈Λj

It is called a Riesz basis if the collection frame with the property that every P forms a2πihλ,xi 2 f ∈ L (Ω) is expanded uniquely as αj,λgj (x)e . The collection of windowed exponentials is called complete if Z f (x)gj (x)e−2πihλ,xi dx = 0, ∀λ ∈ Λj , j = 1, .., N, Ω

then f = 0 a.e. on Ω.

The question in determining the completeness and frame/basis properties of windowed exponentials is closely related to wavelet theory and Gabor analysis since problems about the frame of translate in the multi-resolution analysis and regular Gabor system can be reduced to those of windowed exponentials via respectively the Fourier transform and the Zak transform. For more details about the windowed exponentials, one can refer to [15] for an introduction and [14, 16, 18] for some recent advances. We also notice that if there is only one window, namely g = χΩ , then if E(χΩ , Λ) is called a Fourier frame if it forms a frame. Introduced by Duffin and Schaeffer [12] in the 1950s, Fourier frames have known to be a fundamental building block of sampling theory in applied harmonic analysis, and they have an elegant connection to complex analysis [24, 28]. It is well-known that every bounded set Ω admits a Fourier frame. Indeed, if we cover Ω by a cube, then the exponential orthonormal basis with frequency set denoted by Λ on the cube induces a tight Fourier frame for L2 (Ω). More generally, if 0 < m ≤ g ≤ M < ∞ on Ω, then E(g, Λ) forms a frame for L2 (Ω). We say S that {g1 , ..., gN } is admissible if there exists some Λj such that N j=1 E(gj , Λj ) forms 2 a frame of windowed exponentials for L (Ω). In [14] (See also [19]), the class of all admissible windows {g1 , ..., gN } were completely classified. The necessary and sufficient condition for {g1 , ..., gN } to be admissible for L2 (Ω) is that there exists c > 0 with the property that max

{j:kgj k∞ 0. a.e. on Ω

In particular, when there is only one window g, g is admissible if and only if 0 < m ≤ g ≤ M a.e. on Ω. If {g1 , ..., gN } is admissible, Λj are produced by over-sampling in the sense that the (Beurling) density of Λj are strictly greater than the measure of Ω. While oversampling is common in practice, there are real situations in which each Λj are indeed 2

undersampled or even of Beurling density zero, yet a union of windowed exponentials SN j=1 E(gj , Λj ) forms a frame and works perfectly.

This paper aims to study the possibility of forming a frame in the undersampling circumstances. In particular, we restrict our attention to a special case: let N0 = N+ ∪ {0} and N− be the set of all negative integers and define (1)

F (g) = E(χ[−1/2,1/2] , N0 ) ∪ E(g, N− ).

We ask for what g the set of functions F (g) is complete/ or forms a Riesz basis/frame for L2 [−1/2, 1/2]. It turns out the solution to the question above can be characterized completely by spectra of Toeplitz operators, which opens up a new connection between frame theory and spectral theory of Toeplitz operators. Furthermore, the consideration of the completeness properties F (g) has an immediate application to the dynamical sampling theory and phaseless reconstruction in applied harmonic analysis. Relation to Toeplitz operators. By expressing the analysis operator of F (g) in terms of the Toeplitz operators Tg , we have the following characterization regarding the invertibility of the Toeplitz operators Tg . Theorem 1.2. (1) F (g) is complete if and only if Tg is injective. (2) F (g) forms a frame if and only if Tg is both norm-bounded above and below. (3) F (g) forms a Riesz basis if and only if Tg is bounded and invertible. We will prove the theorem in Theorem 3.4. This theorem gives us in principle a complete characterization of the frame and basis property of F (g) regarding the invertibility of the Toeplitz operator and the triviality of the Toeplitz kernel. This leads us to look for classical results on Toeplitz operators. In particular, we need theorems characterizing the frame property by the function values of g. A complete classification for the invertible Toeplitz operators Tg was given analytically through the Widom-Devinatz theorem in the 1960s. We refer the reader to [6, 23] and a recent survey [17] for the detailed study. Makorov and Poltoratski also studied Toeplitz operators from Beurling-Malliavin theory and received considerable attention [22]. Unfortunately, as far as we know, there is no easily checkable geometric criterion to check if the Toeplitz operator Tg is invertible for a general complex-valued window g. We restrict our attention to two specific cases which demonstrate an elegant interaction between windowed exponentials and Toeplitz operators. First, in Theorem 4.2, using some classical results in the theory of Toeplitz operators, we will provide a complete characterization of the completeness and frame/bases property of F (g) when g is real-valued. Secondly, we will consider the complex windows gξ (x) = e2πiξx . Using the Kadec-1/4 theorem in non-harmonic Fourier series, we completely characterize all ξ for which F (g) is complete/ a frame or a Riesz basis. Without referencing to any results in Toeplitz operators, we classify the invertibility of all Tgξ . Theorem 1.3. Let gξ (x) = e2πiξx . Then 3

(1) Tgξ is not injective if ξ < −1/2. (2) Tgξ is invertible if −1/2 < ξ < 1/2. (3) Tgξ is injective but not normed-bounded below if ξ = −1/2 + n for n = 0, 1, 2, .... (4) Tgξ is injective and normed-bounded below but not surjective if ξ > 1/2 and ξ 6= −1/2 + n for n = 0, 1, 2, ... After establishing our results on F (g), we will also characterize the completeness and frame property for one-sided exponentials O(g) = E(χ[−1/2,1/2] , N− ) ∪ E(g, N−). This will help us prove our main results in the application section. Applications. Problems in determining the frame property of F (g) is also motivated by dynamical sampling/spatial-temporal sampling problems in the diffusion field. Spatial-temporal evolution of various physical phenomena can be modeled using diffusion fields such as temperature variation, and the distribution of pollution plumes in the atmosphere. In Wireless Sensor Networks (WSN), a lot of sensors are distributed to take spatial-temporal samples of the underlying physical field [1]. In practice, increasing the spatial sampling density is usually much more expensive than increasing the temporal sampling density [21], the hope is to use as fewer sensors as possible but to activate them sequentially at different times, which is economically more efficient. Formulated mathematically, Aldroubi and his collaborators [2, 3, 4, 5, 13] considered F ∈ ℓ2 (Z) and the spatial sampling sets {Ωi }, i = 1, ..., L, are collections of proper subsets in Z. They are usually assumed to be nested, i.e., Ω1 · · · ⊂ ΩL , and are well-spread. For example, Ω1 = · · · ΩL = mZ (m > 1) in [3]. Let also G1 , ..., GL ∈ ℓ2 (Z)1. We want to reconstruct F from the samples (G1 ∗ F (Ω1 ), ...., GL ∗ F (ΩL )) (Here G(Ωi ) = {G(n) : n ∈ Ωi }). Notice that this model is equivalent to sampling through the Paley-Wiener space of bandlimited functions supported on a bandwidth of length one as well as certain shift-invariant spaces [3]. As Ωi are proper subsets of Z, we may think they are below the Nyquist rate and hence samples on an individual Ωi is not able to recover F . We say that (Gi , Ωi )i=1,...,L allows a stable sampling if the linear map A : ℓ2 (Z) →

L M

ℓ2 (Ωi )

i=1

defined by (2)

AF = (G1 ∗ F (Ω1 ), ...., GL ∗ F (ΩL ))

1To

avoid confusion, we will use upper case letter to denote functions in ℓ2 (Z) and PW and lower case letter to denote functions in L2 [−1/2, 1/2] 4

admits a bounded left-inverse B such that BAF = F for all F ∈ ℓ2 (Z). We define the Fourier transform of G ∈ ℓ2 (Z) by b g(x) := G(x) =

∞ X

G(n)e−2πinx , x ∈ T,

n=−∞

where T is the circle group identified as [−1/2, 1/2]. We first notice that for A to be well-defined, one should naturally require Gi ∗ F ∈ ℓ2 (Z) (Lemma 6.1). In fact, this ci ∈ L∞ [−1/2, 1/2]. The following theorem connects dynamical is equivalent to G sampling problems with windowed exponentials we have been considering.

ci ∈ L∞ and let Theorem 1.4. Let G1 , ..., GL ∈ ℓ2 (Z) be sequences such that G ci (ξ). Then (Gi , Ωi )i=1,...,L allows a stable sampling if and only if SL E(gi , Ωi ) gi (ξ) = G i=1 forms a frame of windowed exponentials on L2 [−1/2, 1/2].

Because of this theorem, the frame property of F (g) and O(g) allows us to determine the set of all a such that {F (N− ), (G ∗ F )(N0 )} and {F (N− ), (G ∗ F )(N− )}

is stably recoverable (note that F (N− ) = (δ0 ∗F )(N− ) with δ0 is the Dirac function on ℓ2 (Z)). In contrast with Ωi = mZ in [3], N− and N0 has zero lower Beurling densities, which could not admit stable recovery individually on any subsets of [−1/2, 1/2]. Our study on F (g) also sheds some light in understanding the spatiotemporal sampling problem in the phaseless setting. Phaseless reconstruction or phase retrieval concerns about reconstructing f ∈ H by linear measurement up to a phase factor. We say that a collection of vectors {ϕn } in a Hilbert space H is (complex)-phase retireviable if (3)

|hf, ϕn i| = |hg, ϕn i| ∀n ⇒ f = eiθ g,

for some constant angle θ. It is called real-phase retrievable in H if H is a Hilbert space over the real numbers and f = ±g in the right-hand side of the implication of (3). Recently , there are tremendous works on phase retrieval due to its rapid application in engineering [7, 8, 9, 11, 25, and the references therein]. In our setting, we ask if there exist a window G that F ∈ ℓ2 (Z) can be uniquely determined up to a sign from the samples {|F (n)|, |G ∗ F (n)| : n ∈ Z}. This is a natural setting for complex-phase retrieval in Paley-Wiener space. Using our results on O(g), we will also give a partial answer to this question as well. Concerning sampling theory on Paley-Wiener space with bandwidth of length one, (4)

P W = {F ∈ L2 (R) : suppFb ⊂ [−1/2, 1/2]},

we notice that this space is isometric isomorphic to ℓ2 (Z) and thus our results on ℓ2 (Z) hold the same on Paley-Wiener space. As functions on PW space are differentiable (indeed entire on the complex plane), one may also think of differentiation as a linear evolution operator. Our study on F (g) shows unfortunately that the sam−1 ′ ples {F (n)}n≥0 ∪ 2πi F (n) n 0 such that k(A−λI)f k ≥ ckf k). (5) A − λI has a bounded left-inverse B such that B(A − λI) = I.

Proof. The equivalence of (1), (2) and (3) can be found in [10, Proposition 1.1 p.347]. The equivalence of (1) and (4) is obvious. (2) implies (5) follows from the inverse mapping theorem of bounded linear operators. Finally, (5) implies (4) follows from kxk = kBAxk ≤ kBkkAxk (we assume λ = 0 without loss of generality). So we can take c = kBk−1 . Our interest is to determine whether 0 is in the spectrum of a Toeplitz operator. The characterization depends closely on the range of Tφ . For φ ∈ L∞ , the essential range of φ is defined as essran(φ) := {y : m({x : |φ(x) − y| < ǫ}) > 0 for all ǫ > 0}. The essential infimum and essential supremum of a real-valued function φ are defined as essinf(φ) := inf essran(φ), esssup(φ) := sup essran(φ). The following theorem can be found in [23]. Theorem 2.4. Let Tφ be the Toeplitz operator associated with the L∞ function φ. Then (1) essran(φ) ⊂ σap (Tφ ). (2) (Coburn alternative) Tφ or Tφ is injective. In particular, if φ is real-valued, then Tφ is injective. (3) Suppose that φ is real-valued. Then σ(Tφ ) = [essinf(φ), esssup(φ)]. Moreover, for all λ in the open interval (essinf(φ), esssup(φ)), Tφ − λI is not surjective. The last part of the theorem follows from the proof in [23, Theorem 3.3.15]. Finally, we mention the general analytic invertibility theorem for Tφ due to WidomDevinatz (see e.g., [6, Theorem 2.23]). Theorem 2.5. [Widom-Devinatz] Tφ is invertible if and only if φ−1 ∈ L∞ [−1/2, 1/2] and (8)

φ = ei(eu+v+c) a.e. on [−1/2, 1/2] |φ|

where c ∈ R, u, v are bounded real-valued functions, kvk∞ < π/2 and u e is the harmonic conjugate of u. Condition in (8) is commonly referred as Helson-Szegö condition. 8

3. Frame properties of F (g) In this section, we will use the spectral properties of Toeplitz operators to characterize the frame properties of F (g). Recall that F (g) = E(χ[−1/2,1/2] , N0 ) ∪ E(g, N− ). F (g) forms a frame if and only if there exist 0 < A ≤ B < ∞ such that Akf k2 ≤

−1 X

|hf, gen i|2 +

n=−∞ 2

∞ X

|hf, en i|2 ≤ Bkf k2 ,

n=0

for all f ∈ L [−1/2, 1/2]. It forms a Bessel sequence if it only satisfies the upper bound. We now consider the analysis operator of F (g), Φg : L2 − 12 , 21 −→ ℓ2 (Z) defined by Φg f = (· · · , hf, ge−2i, hf, ge−1i, hf, e0 i, hf, e1 i, hf, ge2i · · · ) . The following theorem is well-known in the frame theory literature (See e.g [15, Theorem 8.29 and 8.32]). Theorem 3.1. (1) F (g) is complete if and only if Φg is injective. (2) F (g) is a frame if and only if Φg is both norm-bounded above and below. (3) F (g) is a Riesz basis if and only if Φg is invertible. Note that L2 [− 12 , 21 ] is unitarily equivalent to ℓ2 (Z) via the Fourier coefficients. 1 1 2 ←→ {fb(n)}n∈Z . f ∈L − , 2 2 P We denote en (x) = e2πinx . Suppose that g = +∞ n=−∞ bn en . Then g=

+∞ X

b−n en .

n=−∞

Hence,

hf, gen i = hf g, en i =

+∞ X

k=−∞

We also define the function g (x) = g(−x) = e −

∞ X

b−(n−k) fb(k). bn en (x).

n=−∞

Suppose that we identify Z = N ⊕ N0 and order N− = {−1, −2, ...} and N0 = {0, 1, 2, ...}. We can think of Φg as a mapping from ℓ2 (N− )⊕ℓ2 (N0 ) to ℓ2 (N− )⊕ℓ2 (N0 ), the map becomes Φg (a−1 , · · · , a0 , a1 , · · · ) = (

+∞ X

bk+1 ak ,

k=−∞

+∞ X

k=−∞

9

bk+2 ak , · · · , a0 , a1 , · · · )

Hence, Φg admits a matrix representation of the form   Teg | Hg  −− | −−  (9) O | I

where Teg : ℓ2 (N− ) −→ ℓ2 (N− ) and Hg : ℓ2 (N0 ) −→ ℓ2 (N− ) with the matrix representation     b0 b−1 b−2 · · · b1 b2 b3 · · ·   .  .   b2 b3 b4 . .   b1 b0 b−1 . .   Teg =  , Hg =  .. ..   b b ... ...  .  b . .   3 4  2 b1   .. . . .. . . . . .. . . . . . . With a slight abuse of notation, we identify ℓ2 (N− ) and ℓ2 (N0 ) by an obvious isometric isomorphism, so that Teg is a Toeplitz operator. We have a simple lemma concerning the spectra of Teg and Tg .

Lemma 3.2. σp (Teg ) = σp (Tg ), σap (Teg ) = σap (Tg ) and σc (Teg ) = σc (Tg ). Proof. This lemma follows from the a direct observation that for any f = (f1 , f2 , ...), Teg f = Tg f . We have the following proposition concerning the boundedness of the operators Tg . Proposition 3.3. F (g) forms a Bessel sequence for L2 ([−1/2, 1/2]) if and only if g ∈ L∞ [−1/2, 1/2] and Tg is a bounded operator on ℓ2 (N0 ). Proof. If g ∈ L∞ [−1/2, 1/2], then both Tg and Hg are bounded operators. The sufficiency part follows from a direct check. We now give the necessity part with two different proofs from two point of views. One is from the frame theory and the other one is from the Toeplitz operator theory. (Beurling density proof ). From [14, Theorem 3.2(i)], it was proved that if F (g) forms a Bessel sequence and g 6∈ L∞ , then the upper Beurling density of N− has to be zero. Recall that the upper Beurling density of Λ is #(Λ ∩ (x + [−R, R])) D + (Λ) = lim sup sup , 2R R→∞ x∈R From the definition, D + (N− ) = 1. This is a contradiction. Hence, g ∈ L∞ [−1/2, 1/2]. Thus, Tg is bounded by Remark 2.2. (From Toeplitz operator theory) From the definition of the Bessel sequence, we know that kΦg f k2 ≤ Bkf k2 . Let f = (f − , f + ) ∈ ℓ2 (N− )⊕ℓ2 (N0 ). Using the representation in (9), kTg˜f − + Hg f + k2 + kf + k2 ≤ B(kf − k2 + kf + k2 ). Let f + = 0, we obtain kTg˜f − k2 ≤ Bkf − k2 . Hence, Tg˜ is a bounded operator and g is bounded by Remark 2.2. 10

We are now ready to prove Theorem 1.2. From the facts about each spectra, we can formulate Theorem 1.2 in the following theorem. Theorem 3.4. (1) F (g) is complete if and only if 0 6∈ σp (Tg ). (2) F (g) forms a frame if and only if Tg is bounded and 0 6∈ σap (Tg ). (3) F (g) forms a Riesz basis if and only if Tg is bounded and 0 6∈ σ(Tg ). Proof. Because of Lemma 3.2, it suffices to prove all the statements with Tg replaced by Teg . Throughout the proof, we write f = (f − , f + ) ∈ ℓ2 (N− ) ⊕ ℓ2 (N0 ). (1) Suppose that 0 6∈ σp (Teg ). Then Teg is injective. We need to show that Φg is injective. To see this, let Φg f = 0 and we have Teg f − + Hg f + = 0, and f + = 0. Hence, Teg f − = 0. But Teg is injective, f − = 0. This shows Φg is injective and thus F (g) is complete. Conversely, that suppose Φg is injective and let Teg f = 0. Consider f Teg f (f, 0) and we have Φg = = 0. Hence, injectivity of Φg implies that 0 0 f = 0. Thus, f = 0. 0

(2). Suppose that Tg is bounded (and thus g is bounded) and 0 6∈ σap (Teg ). Then the proof of upper bound is trivial by the fact that g ∈ L∞ since now Tg˜ and Hg are bounded. We now show that it satisfies the lower bound. By Lemma 2.3, Teg is injective and has a closed range. Furthermore, it is norm bounded below. Let c1 be the constant such that kTeg f k ≥ c1 kf k, ∀ f ∈ ℓ2 (N− ) Let c2 = k − Hg k. As f = (f − , f + ), kf k2 = kf − k2 + kf + k2 . We need to prove that Φg is norm bounded below. kΦg f k2 =kTeg f − + Hg f + k2 + kf + k2 1 1 =kTeg f − + Hg f + k2 + kf + k2 + kf + k2 2 2 1 1 k − Hg f + k2 + kf + k2 ≥kTeg f − + Hg f + k2 + 2c2 2 1 1 ≥ min 1, · kTeg f − + Hg f + k2 + k − Hg f + k2 + kf + k2 2c2 2 2 1 1 1 ≥ min 1, · kTeg f − + Hg f + k + k − Hg f + k + kf + k2 2 2c2 2 1 1 1 ≥ min 1, kTeg f − k2 + kf + k2 2 2c2 2 c1 c1 1 ≥ min kf k2, , , 2 4c2 2

where we have used the inequality (a2 + b2 ) ≥ 21 (a + b)2 in the third last line and the triangle inequality in the second last line. This completes the proof. 11

Conversely, suppose that 0 ∈ σap (Teg ). Then there exists fn with kfn k = 1 such that kTeg fn k → 0 as n → ∞. Taking the vector (fn , 0). Then fn Tg fn → 0. Φg = 0 0 Hence, Φg is not bounded below and hence F (g) cannot be a frame. (3). Suppose that 0 6∈ σ(Teg ). Then T and hence Φg is injective by (1). g is injective e − y To see that Φg is surjective, we take and we try to solve y+ − − y f . = Φg + y+ f

Thus, f + = y + and Teg f − = y − −Hg f + . As Teg is surjective, Teg−1 is bounded by inverse mapping theorem of linear operators. Thus, we can find f − = Teg−1 (y − − Hg f + ) in ℓ2 (N). Hence, Φg is invertible and possesses a bounded inverse. 0 6∈ σ(Teg ) follows.

Conversely, Suppose that 0 ∈ σ(Teg ). Then 0 ∈ σap (Teg ) or 0 ∈ σc (Teg ). However, if 0 ∈ σap (Teg ), then (2) shows that F (g) cannot be a frame and hence cannot be a Riesz bases. On the other hand, if 0 ∈ σc (Teg ), then Teg cannot be surjective. Hence, Φg cannot be surjective either since it was, then Φg f = (y, 0) for any y and it makes Teg is surjective.

4. Classification of admissible window functions We will give some geometric characterizations of certain window functions for the frame/basis property of F (g). First of all, combining Theorem 3.4 and WidomDevinatz theorem (Theorem 2.5), we have an analytic characterization by HelsonSzegö. Theorem 4.1. F (g) forms a Riesz basis for L2 [−1/2, 1/2] if and only if g −1 ∈ L∞ and g satisfies the Helson-Szegö condition. As we have discussed in the introduction, Helson-Szegö condition does not give us a directly checkable criterion even for simple classes of functions. Furthermore, there is not even an analogous theorem for 0 to be outside the point/approximate point spectrum. We will turn our attention to certain classes of functions. We first give a complete characterization of the real-valued window functions. Theorem 4.2. Let g be real-valued. Then (1) F (g) is complete. (2) The following are equivalent. (a) F (g) forms a Riesz basis. (b) F (g) forms a frame. (c) 0 6∈ [essinf(g), esssup(g)]. 12

Proof. (1). By Theorem 2.4(2), we know that if g is real-valued, then Tg is injective. Hence, 0 6∈ σp (Tg ). Hence, by Theorem 3.4, F (g) is complete. (2) (a) implies (b) is from definition. (c) implies (a) follows from Theorem 2.4(iii) which states that σ(Tg ) = [essinf(g), esssup(g)]. As 0 6∈ σ(Tg ) by (c), Theorem 3.4(3) shows that F (g) is a Riesz basis. We now show (b) implies (c). We suppose that 0 ∈ [essinf(g), esssup(g)]. If 0 = essinf(g) or esssup(g), then 0 ∈ essran(g) ⊂ σap (Tg˜ ) by Theorem 2.4(1). By Theorem 3.4(2), F (g) does not form a frame. Suppose that 0 ∈ (essinf(g), esssup(g)). Then Theorem 2.4(3) shows that Tg is not surjective. As Tg is self-adjoint for realvalued g, ran(Tg∗ −0I) = ran(Tg )) is not the whole Hilbert space. By Lemma 2.3 (3), 0 ∈ σap (Tg ). By Theorem 3.4(2) again, F (g) does not form a frame. This completes the proof. Example 4.3. (1). Let g(x) = xχ[−1/2,1/2] . Then F (g) is complete as g is realvalued. However, it does not form a frame because 0 is in [essinf(g), esssup(g)] = [−1, 1]. (2). Let g(x) = −χ[−1/2,0) + χ[0,1/2] . Then F (g) is complete as g is real-valued. However, it does not form a frame because 0 is in the interval [essinf(g), esssup(g)] = [−1, 1] (note that essran(g) = {−1, 1}). On the other hand, if we consider the windowed exponentials {g(x)e2πinx : n ∈ Z}, then the system forms a frame for L2 [−1/2, 1/2] since |g| = 1. We now turn to study complex windows. In particular, we study the simple case when gξ (x) = e2πiξx . Then F (gξ ) = {e2πinx : n = 0, 1, 2, ...} ∪ {e2πi(ξ+n)x : n = −1, −2, ...}. In this special case, we define a map T : L2 [− 21 , 21 ] → L2 [− 12 , 12 ] by T f (x) = e−πiξx/2 f (x). It is easy to see that T is an invertible bounded linear operator. Hence F (gξ ) is complete/ a Riesz basis/ a frame if and only if T (F (gξ )) is complete/ a Riesz basis/ a frame. Note that ξ

ξ

T (F (gξ )) = {e2πi(n− 2 )x : n ∈ N0 } ∪ {e2πi( 2 +n)x : n ∈ N− }. Now define the real sequence ξ ξ − Γ ξ = n − , n ∈ N0 ∪ n + : n ∈ N 2 2 and denote by E(Γξ ) = {e2πiλn ξ : λn ∈ Γξ } the complex exponential system associated with Γξ . Then we have T (F (gξ )) = E(Γξ ) and it is sufficient to determine whether E(Γξ ) is complete/ a Riesz basis/ a frame. To this end, results from Young and Levinson gave us essentially an important insight. 13

Theorem 4.4. (1) (Kadec-1/4 theorem) (see e.g. [31]): If |λn − n| ≤ L < 41 for all n ∈ Z, then {e2πiλn x }n∈Z is a Riesz basis for L2 [−1/2, 1/2].   n + 14 + ǫ, n > 0; 0, n = 0; , (2) (Levinson, [20, Theorem V] or [31, Theorem 4]) If λn =  1 n − 4 − ǫ, n < 0. then {e2πiλn x } is incomplete in L2 [−1/2, 1/2] for any ǫ > 0.   n + 14 , n > 0; 0, n = 0; , (3) (Redheffer-Young, [27, Theorem 4 and Section 8]) If λn =  1 n − 4 , n < 0. 2πiλn x 2 then {e } is exact in L [−1/2, 1/2] but it is not a Riesz basis.

Proposition 4.5 ([31], Theorem 7). The completeness of the system {e2πiλn x } in L2 [−1/2, 1/2] is unaffected if finitely many λn is replaced by another numbers. Now we are ready to state our results for this special case and summarize them in the following theorem

Theorem 4.6. (1) If ξ < −1/2, then F (gξ ) is incomplete. (2) If −1/2 < ξ < 1/2, then F (gξ ) is a Riesz basis. (3) If −1/2 + n < ξ < 1/2 + n, n ∈ N+ , then F (gξ ) is a frame but not a Riesz basis. (4) If ξ = −1/2 + n, n ∈ N0 , then F (gξ ) is complete but not a Riesz basis nor a frame. Proof.

(1) Suppose that ξ < −1/2. Then −ξ/2 > 1/4. By Theorem 4.4(2), and Proposition 4.5 with 0 being replaced by 0 − ξ/2, E(Γξ ) is not complete for ξ < −1/2.

(2) Using the Kadec-1/4 theorem, if −1/2 < ξ < 1/2, then Γξ satisfies the assumption with L = |ξ|/2, so E(Γξ ) is a Riesz basis. This gives us our conclusion. (3) Suppose that ξ = ξ0 + n for ξ0 ∈ (−1/2, 1/2) and n is a positive integer. Then the original system F (gξ0 ) ( F (gξ ). Note that we have proved that F (gξ0 ) is a Riesz basis by (2), hence F (gξ ) is a frame, but not a Riesz basis since there are excessive elements. (4) Using Theorem 4.4 (3), we know that E(Γ−1/2 ) is exact and thus complete and it is not a Riesz basis. It is not a frame either. Otherwise, E(Γ−1/2 ) will be an exact frame, which means it is a Riesz basis, which is a contradiction. Hence, we conclude that F (g−1/2 )) is complete but not a Riesz basis nor a frame. In general, note that we have F (g−1/2 ) ⊂ F (g1/2 ) ⊂ F (g3/2 ) ⊂ · · · . Suppose that there exists a set in this chain forms a frame. Then we can remove finitely many elements from this set to obtain F (g−1/2 ). Note that the removal of finitely many elements from a frame leaves either a frame or an incomplete set. Now, F (g−1/2 ) is complete, so if one of the sets forms a 14

frame, then F (g−1/2 ) is a frame also. This is a contradiction to what we just proved. Hence, every set in the above chain is complete and is not a Riesz basis nor frame. Proof of Theorem 1.3. It follows directly from Theorem 4.6 and Theorem 3.4.

5. One-sided exponentials In this section, we are interested in the following collection of windowed exponentials O(g) = E(χ[−1/2,1/2] , N− ) ∪ E(g, N− ). We ask for what g does the above collection is complete/ a Riesz basis/ a frame for L2 ([−1/2, 1/2]).This is related to phase retrieval problems. Similar we consider the analysis operator of O(g) defined by Ψg : 1 1 to F (g), 2 L − 2 , 2 −→ ℓ (Z) 2

Ψg f = (· · · , hf, ge−2i, hf, ge−1i, hf, e−1 i, hf, e−2 i, hf, e−3 i · · · ) . P Recall that en (x) = e2πinx and let g = +∞ n=−∞ bn en . Then g=

+∞ X

b−n en ,

n=−∞

g (x) = g(−x) = e

∞ X

bn en (x).

n=−∞

We can identify L2 [−1/2, 1/2] with ℓ2 (Z) via Fourier coefficient. Identify Z = N × N0 and order N− = {−1, −2, ...} and N0 = {0, 1, 2, ...} in the same way as Case ˜ g as a mapping from ℓ2 (N− ) ⊕ ℓ2 (N0 ) to ℓ2 (N− ) ⊕ ℓ2 (N0 ), I. We can also think of Φ then it becomes +∞ +∞ X X ˜ Φg (a−1 , · · · , a0 , a1 , · · · ) = ( bk+1 ak , bk+2 ak , · · · , a−1 , a−2 , · · · ) −

k=−∞

k=−∞

˜ g admits a matrix representation of the form Hence, Φ   Teg | Hg  −− | −−  I | O

where Teg : ℓ2 (N− ) −→ ℓ2 (N− ) and Heg : ℓ2 (N0 ) −→ ℓ2 (N− ) with the matrix representation     b0 b−1 b−2 · · · b1 b2 b3 · · ·   .  .   b1 b0 b−1 . .   b2 b3 b4 . .     Teg =  . . . .  , Hg =  .. ..  .  b . . . . b b b 2 1 3 2     .. . . . . .. .. . . . . . ··· 15

We first have the following lemma. Lemma 5.1. Let H be a bounded Hankel operator with matrix representation (6), 0 ∈ σap (H). Proof. Suppose that 0 6∈ σap (H), then we know that H is bounded below by Lemma 2.3. Thus for any c ∈ ℓ2 (N0 ), there exists a positive number m > 0, kHck ≥ mkck. On ℓ2 (N0 ), we define the unilateral shift operator U by U(b0 , b1 , b2 , b3 , · · · ) = (0, b0 , b1 , b2 , b3 , · · · ) for (b0 , b1 , b2 , b3 , · · · ) ∈ ℓ2 (N0 ). The adjoint operator of U is defined by U ∗ (b0 , b1 , b2 , b3 , · · · ) = (b1 , b2 , b3 , · · · ). Observe that (U ∗ )n H = HU n and U is an isometry. For a fixed c ∈ ℓ2 (N0 ), 0 = lim k(U ∗ )n Hck = lim kHU n ck ≥ mkU n ck = mkck. n→∞

n→∞

Hence m = 0. We obtain a contradiction. So 0 ∈ σap (H).

Theorem 5.2. (1) O(g) is complete if and only if 0 6∈ σp (Hg ). (2) O(g) never forms a frame nor a Riesz basis. Proof. The proof of (1) is similar to Theorem 3.4(1), so we omit it. For (2), we just need to show O(g) never forms a frame. We first note by Lemma 5.1 that 2 0 6∈ σap (H g ). Thus, there exists fn∈ ℓ(N0 ) with norm 1 such that kHg (fn )k → 0. 0 0 Hg fn Consider . Then Ψg = . Hence, fn fn 0

0

Ψg

= kHg fn k → 0.

fn

Thus, the analysis operator of O is not normed bounded below and hence, it never forms a frame.

˜ ∞ is an inner function if |h| ≡ 1 on the unit circle. We say that a function h ∈ H As a consequence of Theorem 5.2 and Theorem 2.3 in [26], we have the following analytic characterization for completeness of O(g) when g is bounded. Theorem 5.3. Let g ∈ L∞ . The following are equivalent: (1) O(g) is not complete in L2 [−1/2, 1/2] (2) Hg has nontrivial kernel; (3) Range Hg is not dense in ℓ2 (N) ˜ ∞ and v is an inner function. (4) g = vψ, where ψ ∈ H An explicit example is given in the following proposition. P Proposition 5.4. Suppose that g = +∞ n=−∞ bn en , where (bn ) has only finite many nonzero entries. Then 0 ∈ σp (Hg ) and O(g) is not complete. 16

Proof. Suppose that g has all Fourier coefficients bn = 0 for all |n| ≥ N. Then the Hankel operator representation (7) will be given by   H | O  −− | −−  , O | O

where H is a non-zero N by N matrix and all other corners are zero matrices. Hence, it is clear that the image of ℓ2 (N) under U N U N [ℓ2 (N)], is in the kernel of Hg . ˜ ∞ and v In [26, Theorem 2.5], we know that the set {vψ, where ψ ∈ H is an inner function } is dense in L∞ . Thus, the set of g for which O(g) is not complete in L2 [−1/2, 1/2] is dense in L∞ . 6. Applications We will apply our theory to several applications discussed in the introduction. Spatial-Temporal Samplings. Consider F ∈ ℓ2 (Z) and recall if we have G1 , ..., GL ∈ ℓ2 (Z) and Ω1 , ..., ΩL ⊂ Z. With all notation and definitions in the introduction, our problem is to characterize {(Gi , Ωi )}i=1,..,L by windowed exponentials so that it stably recovers F by the samples (F (Ω0 ), G1 ∗ F (Ω1 ), ...., GL ∗ F (ΩL )). We first have the following lemma stating a sufficient condition for Gi ∗ F ∈ ℓ2 (Z), guaranteeing the well-definedness of A in (2). b ∈ L2 [−1/2, 1/2]. G ∗ F ∈ ℓ2 (Z) for all Lemma 6.1. Let G ∈ ℓ2 (Z) and let g = G F ∈ ℓ2 (Z) if and only if g ∈ L∞ [−1/2, 1/2]. Consequently, if gi ∈ L∞ [−1/2, 1/2] for all i = 1, ..., L, then A in (2) is well-defined . Proof. For any F ∈ ℓ2 (Z), note (G ∗RF )b(x) = g(x)f (x) where f = Fb. Clearly, R that ∞ 2 if g ∈ L [−1/2, | ≤ M |f |2 and Rthus gf ∈ L2 [−1/2, 1/2]. This P 1/2], then 2 |gfP means that n∈Z |G ∗ F (n)| = n∈Z |(gf )b(n)|2 = |gf |2 by the Parseval identity, so G ∗ F ∈ ℓ2 (Z) for all F ∈ ℓ2 (Z).

Conversely, if G ∗ F ∈ ℓ2 (Z) for all F ∈ ℓ2 (Z), then gf ∈ L2 [−1/2, 1/2] for all f ∈ L2 [−1/2, 1/2]. Now, define En = {x ∈ [−1/2, 1/2] : n ≤ |g(x)| < n + 1} and F = {n : |En | = 6 0}, where | · | denotes the Lebesgue measure. We consider ∞ X 1 f= χE (n + 1) · |En |1/2 n n=0 R P 1 Then |f |2 = ∞ n=1 (n+1)2 < ∞ and Z X n 2 2 |gf | dx ≥ , n + 1 n∈F

which is finite if and only if F has only finitely many elements. This is equivalent to say g ∈ L∞ . 17

b ∈ L∞ gives us a simple sufficient condition for A to be wellRemark 6.2. g = G defined. However, it is not necessary as the boundedness depends on the choice to Ωi . For example, if Ωi is a finite set, then G ∗ F (Ωi ) is clearly in ℓ2 (Ωi ) and is thus well-defined. Proof of Theorem 1.4. This is actually a standard proof. Note that the map A in (2) has a bounded left inverse if and only if A is normed bounded below by Lemma 2.3. i.e. L X X X (10) |Gi ∗ F (λ)|2 ≥ c |F (n)|2. i=1 λ∈Ωi

n∈Z

ci and f = Fb. Note that Denote by gi = G b

Gi ∗ F (λ) = (gf ) (λ) =

Z

f (x)g(x)e−2πiλx dx.

Therefore, (10) is equivalent to Z L X Z X −2πiλx 2 | f (x)g(x)e dx| ≥ c |f (x)|2 dx. i=1 λ∈Ωi

As all L2 [−1/2, 1/2] can be written as Fb for unique F ∈ ℓ2 (Z), we have shown a S bounded left inverse of A exists if and only if the window exponentials Li=1 E(gi , Ωi ) possess a positive lower frame bound. As gi are bounded, the upper bound is automatic. This completes the proof. Because of the above theorem, the set of all G for which the samples {F (N− ), (G ∗ F )(N0 )} can stably recover F is completely characterized. Theorem 6.3. {(δ0 , N− ), (G, N0 )} admits a stable sampling on ℓ2 (Z) if and only if b the Toeplitz operator Tg is norm-bounded below, i.e. 0 ∈ σap (Tg ), where g = G. Some special windows can be found in Section 4. Similarly, for {F (N− ), (G ∗ F )(N− )}, we notice however that none of the G forms a stable sampling in this case by Theorem 5.2.

Phaseless reconstruction. In phaseless setting, one wish to reconstruct F ∈ ℓ2 (Z) by the samples {|F |(Z), |G ∗ F |(Z)} up to a constant phase angle. In [30], Thakur proved that a real-valued f in the Paley-Wiener space (see (4) for definition) can be recovered up to a sign from the samples 1 (11) |F ( n)| : n ∈ Z = {|F (n)| : n ∈ Z} ∪ {|G ∗ F (n)| : n ∈ Z}, 2

where G(x) = sinc(π(x − 1/2)) and sinc(x) = sin πx/πx. Indeed, 21 Z = Z ∪ (Z + 1/2) and for those elements in Z + 1/2, we have Z 1/2 Z 1/2 −2πi(n+1/2)x b F (n + 1/2) = F (x)e dx = Fb(x)e2πi1/2x e−2πinx dξ = G ∗ F (n). −1/2

−1/2

Therefore the G(x) = sinc(π(x − 1/2)) allows real-phase retrieval. 18

Notice however that (11) is not complex phase-retrievable because for complexvalued F , |F (x)| = |F (x)| but F and F in general does not differ from a constant phase angle. Nonetheless, considering G ∗ F provides us a natural generalization. In order to be complex phase retrievable, G has to be complex-valued. We now ask for which G are real/complex phase retrievable. Let us recall a general criterion for phase retrievability: Lemma 6.4. [9] If a set of vectors {ϕi }i∈I on a Hilbert space H is complex-phase retrievable, then {ϕi }i∈I satisfies the complement property: if I = J ∪J c , then either {ϕi }i∈J or {ϕi }i∈J c is complete in H. Conversely, the complement property is sufficient for {ϕi }i∈I to be real-phase retrievable We have the following partial answers. Proposition 6.5. (1). Suppose that {|F |(Z), |G ∗ F |(Z)} is phase-retrievable over ˜ ∞ and v is an inner function. the complex field. Then g 6= vψ, where ψ ∈ H P (2). Suppose that G(x) = N n=−N an sinc(x − n). Then {|F (Z)|, |G ∗ F (Z)|} is not real phase-retrievable. Proof. (1) Suppose that G is of the form stated in the conclusion. Then consider the partition {F (N− ), G ∗ F (N− )} ∪ {F (N0 ), G ∗ F (N0 )}. By Theorem 5.2, we know {F (N− ), G ∗ F (N− )} is not complete. In fact, by a symmetry argument, {F (N0 ), G ∗ F (N0 )} is not complete either. This shows the system has no complement property. Thus it is not complex-phase retrievable. b has a (2). Note that from Proposition 5.4, the Hankel operator Hg , g = G, non-trivial kernel with non-zero real vectors. Using the same proof as in (1), our conclusion follows. Derivative Sampling. the sampling on Paley-Wiener space P W by the −1Consider ′ samples {F (n)}n≥0 ∪ 2πi F (n) n