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( x)(t) := α x(t τ ) e2πiνmt, t R, (2) H m − m ∀ ∈ m N X∈ where (τ ,ν ) R2 are delay-Doppler shift parameters and α are the corresponding complex m m ∈ m weights, for m N. Here, the (discrete) spreading function can be supported on unbounded subsets ∈ of the (τ,ν)-plane with the identifiability condition on the support area of the spreading function replaced by a density condition on the support set supp( ) := (τ ,ν ) : m N . Specifically, for H { m m ∈ } supported on rectangular lattices according to supp( ) = a 1Z b 1Z, Kozek and Pfander H H − × − established that is identifiable if and only if ab 6 1 [3]. In [13] a necessary condition for H identifiability of a set of Hilbert-Schmidt operators defined analogously to (2) is given; this condition is expressed in terms of the Beurling density of the support set, but the time-frequency pairs (τm,νm) are assumed to be confined to a lattice. Now, in practice the discrete spreading function will not be supported on a lattice as the parameters τm,νm correspond to time delays and frequency shifts induced, e.g. in wireless communication, by the propagation environment. It is hence of interest to understand the limits on identifiability in the absence of “geometry-discretizing” assumptions—such as a lattice constraint—on supp( ). Resolving this problem is the aim of the present paper. H
A. Fundamental limits on identifiability The purpose of this paper is twofold. First, we establish fundamental limits on the stable identi-
fiability of in (2) in terms of supp( ) and αm m N. Our approach is based on the following H H { } ∈ R2 insight. Defining the discrete complex measure µ := m N αm δτm,νm on , where δτm,νm denotes ∈ the Dirac point measure with mass at (τm,νm), the input-outputP relation (2) can be formally rewritten as 2πiνt ( µx)(t)= x(t τ)e dµ(τ,ν), t R, (3) H R2 − ∈ Z where we use throughout instead of for concreteness. Identifying the system thus amounts Hµ H Hµ to reconstructing the discrete measure µ from x. More specifically, we wish to find necessary and Hµ sufficient conditions on classes H of measures guaranteeing stable identifiability bounds of the form
dr(µ,µ′) 6 dm( x, x), for all µ,µ′ H , (4) Hµ Hµ′ ∈ for appropriate reconstruction and measurement metrics dr and dm, where µ is the ground truth measure to be recovered and µ′ is the estimated measure. The class H can be thought of as modelling the prior information available about the measure µ facilitating its identification by restricting the set 3
of potential estimated measures µ′. In particular, the smaller the class H , the “easier” it should be to satisfy (4). In addition to the class H of measures itself, the existence of a bound of the form (4) depends on the choice of the probing signal x, so we will later speak of identifiability by x. This formulation reveals an interesting connection to the super-resolution problem as studied by Donoho [14], where the goal is to recover a discrete complex measure on R, i.e., a weighted Dirac train, from low-pass measurements. The problem at hand, albeit formally similar, differs in several important aspects. First, we want to identify a measure µ on R2, i.e., a measure on a two-dimensional set, from observations in one parameter, namely ( x)(t), t R. Next, the low-pass observations Hµ ∈ in [14] are replaced by short-time Fourier transform-type observations, where the probing signal x appears as the window function. While super-resolution from STFT-measurements was considered in [15], the underlying measure to be identified in [15] is, as in [14], on R. Finally, [14] assumes that the support set of the measure under consideration is confined to an a priori fixed lattice. While such a strong structural assumption allows for the reconstruction metric dr to take a simple and intuitive form, it unfortunately bars taking into account the geometric properties of the support sets considered. By contrast, the general definition of stable identifiability (see Definition 1) analogous to [14] will pave the way for a theory of support recovery without a lattice assumption, as discussed in the next subsection. These differences make for very different technical challenges. Nevertheless, we can follow the spirit of Donoho’s work [14], who established necessary and sufficient conditions for stable iden- tifiability in the classical super-resolution problem. Donoho’s conditions are expressed in terms of the uniform Beurling density of the measure’s (one-dimensional) support set and are derived using density theorems for interpolation in the Bernstein and Paley-Wiener spaces [16] and for the balayage of Fourier-Stieltjes transforms [17]. We will, likewise, establish a sufficient condition guaranteeing stable identifiability for classes of measures whose supports have density less than 1/2 “uniformly over the class H ” (formally introduced in Definition 2). In addition, we show that this is also a necessary condition for classes of measures invariant under time-frequency shifts and closed under a natural topology on the support sets. We will see below that these requirements are not very restrictive as we present several examples of identifiable and non-identifiable classes of measures. The proofs of these results are based on the density theorem for interpolation in the Bargmann-Fock space [18]–[21], as well as several results about Riesz sequences from [22].
B. Robust recovery of the delay-Doppler support set The second goal of the paper is to address the implications of the identifiability condition on the recovery of the discrete measure µ. Concretely, suppose that we want to recover a fixed measure H µ := m N αm δτm,νm from a known class of measures assumed to be stably identifiable (in the ∈ H senseP of (4)) with respect to a probing signal x, and let µn n N be a sequence of “estimated (n) { } ∈ ⊂ candidate measures” µ := N α δ (n) (n) for the recovery of µ. We will show that, under a n m m τm ,νm ∈ H mild regularity condition onPx, the stable identifiability condition (4) on guarantees that
(n) µn x µx = supp( µn ) supp( µ) and αm m N αm m N, (5) H →H ⇒ H → H { } ∈ → { } ∈ 4 as n , where the topologies in which these limits take place will be specified in due course. → ∞ In words, this result says that the better the measurements match the true measurement , Hµn Hµ the closer the estimated measures µn are to the ground truth µ. This, in particular, shows that “measurement matching” is sufficient for recovery within stably identifiable classes H , i.e., any H algorithm that generates a sequence of measures µn n N satisfying µn x µx will { } ∈ ⊂ H → H succeed in recovering µ H . Crucially, we do not assume that the support sets supp( ) and ∈ Hµ supp( ), for n N, are confined to a lattice (or any other a priori fixed discrete set). To the best Hµn ∈ of our knowledge, this is the first known LTV system identification result on the robust recovery of the discrete support set of the measure, instead of its weights only.
Notation. We write BR(a) for the closed ball in C of radius R centered at a, and denote its boundary by ∂B (a). For a set S C, we let 1 : C R be the indicator function of S, taking on R ⊂ S → the value 1 on S and 0 elsewhere. We will identify C with R2 whenever appropriate and convenient. We say that a set Λ C is discrete if, for all λ Λ, one can find a δ > 0 such that λ λ > δ, ⊂ ∈ | − ′| for all λ Λ λ . Following the terminology employed in [22, §2.2], we say that a set Λ C is ′ ∈ \ { } ⊂ relatively separated if rel(Λ) := sup #(Λ B (x)) : x C < . { ∩ 1 ∈ } ∞ Further, we say that Λ is separated (usually referred to as uniformly discrete in the literature), if
sep(Λ) := inf λ λ′ : λ, λ′ Λ, λ = λ′ > 0. {| − | ∈ 6 } Finally, for two separated sets Λ , Λ R2, we define their mutual separation according to 1 2 ⊂
ms(Λ1, Λ2) := inf λ1 λ2 . (6) λ1 Λ1,λ2 Λ2 | − | ∈λ =λ ∈ 16 2 Note that points that are elements of both Λ1 and Λ2 are excluded from consideration in the expression for mutual separation.
For a Banach space , we write , ∗, and , to denote the norm, the topological dual of B k·kB B h· ·iB×B∗ , and the dual pairing on , respectively. Throughout the paper we use p and q to denote conjugate B B indices in [1, ] such that 1/p + 1/q = 1. We write M p for the vector space of all complex Radon ∞ C R2 measures on of the form µ = λ Λ αλδλ, where Λ is a relatively separated discrete subset of , ∈ αλ λ Λ is a sequence in C, and the norm { } ∈ P p 1/p λ Λ αλ , if p [1, ) µ := ∈ | | ∈ ∞ k kp sup Pλ Λ αλ , if p = ∈ | | ∞ is finite. For such measures we define supp(µ) := λ Λ : αλ = 0 . Furthermore, for s > 0, we { ∈ 6 } p p let Ms = µ M : sep(supp(µ)) > s . { ∈ } For a complex number λ = τ +iν (or the corresponding point (τ,ν) R2), we write ( x)(t) := ∈ Mν e2πiνtx(t) for the modulation operator, ( x)(t) := x(t τ) for the translation operator, and π(λ)= Tτ − for the combined time-frequency shift operator. Recall that, for a nonzero Schwartz test MνTτ function ϕ (R), the short-time Fourier transform (STFT) with respect to the window function ϕ ∈ S 5 is the map taking Schwartz distributions on R to complex-valued functions on C according to Vϕ
( ϕx)(λ)= x,π(λ)ϕ (R) (R), for x ′(R), λ C, V h iS′ ×S ∈ S ∈ 1 πt2 where (R) denotes the set of tempered distributions on R. We take ϕ(t) = 2 4 e to be the S′ − L2-normalized gaussian and, following [7], we write
1/p p R R p p p Mm( )= x ′( ) : x Mm(R) := ( ϕx)(λ) m(λ) dλ < , ∈ S k k C | V | ∞ ( Z ) for the weighted modulation space on R of index p and weight function m : C R> . When m 1, → 0 ≡ we write M p(R) for the unweighted modulation space. We remark that ϕ has the convenient property of being its own Fourier transform, i.e., ϕ = ϕ. According to [7, Thm. 11.3.5, Thm. 11.3.6], M p(R) is a Banach space, and, for p [1, ), its dual space can be identified with M q(R) via the dual ∈ ∞ pairing b p q f, g M p(R) M q (R) = ϕf, ϕg Lp(C) Lq (C), for f M (R), g M (R). (7) h i × hV V i × ∈ ∈
Finally, for real-valued functions f and g of several variables p1,...,pn (which may be real or complex numbers, or even functions), and a non-negative integer m 6 n, we write f . p1,...,pm g if there exists a non-negative function C = C(p ,...,p ) such that f 6 C g, as well as f g 1 m ≍ p1,...,pm if f . p1,...,pm g and g . p1,...,pm f. We use the notation f . g only if C is a universal constant, i.e., if it is independent of all of the p1,...,pn.
II. CONTRIBUTIONS A. Operators and identifiability In order to formalize our definition of identifiability (4), we first need to make sense of the integral (3). Concretely, we consider only probing signals x in the modulation space M 1(R) (also referred to in the literature as , the Feichtinger algebra) and, for a measure µ M p, we interpret (3) as a S0 ∈ linear operator : M 1(R) M p(R) given by Hµ → 2πiν µx = x( τ)e · dµ(τ,ν) := µ( λ ) π(λ)x. H R2 ·− { } Z λ supp(µ) ∈ X The convergence of this sum in the Banach space M p(R) is guaranteed by the following proposition whose proof can be found in the appendix.
Proposition 1. Let Λ be a relatively separated subset of C, and let p [1, ). Then ∈ ∞ (i) : ℓp(Λ) M 1(R) M p(R) given by HΛ × → (α, x) := α π(λ)x, for all α ℓp(Λ), x M 1(R), HΛ λ ∈ ∈ λ Λ X∈ is a well-defined continuous linear operator, in the sense of the sum converging unconditionally in the norm of M p(R). Moreover, this operator is bounded according to
(α, x) p R . rel(Λ) α p x 1 R , kHΛ kM ( ) k kℓ k kM ( ) for all α ℓp(Λ), x M 1(R). ∈ ∈ 6
(ii) For a fixed x M 1(R), the adjoint operator ( ,x) ∗ : M q(R) ℓq of the map ℓp α ∈ HΛ · → ∋ 7→ (α, x) is given by HΛ q Λ( ,x) ∗(y)= y,π(λ)x M q (R) M p(R) λ Λ, for y M (R). (8) H · {h i × } ∈ ∈ M p 1 R p R Next, for a measure µ = λ Λ αλδλ , define µ : M ( ) M ( ) by µ(x) = ∈ ∈ H → H (α, x). Then, Hsupp(µ) P (iii) for every µ M p, ∈ 1 p µ ℓ∞ . µ M (R) M (R). (9) k k kH k → As a consequence of item (iii) of Proposition 1, we have
1 p 1 p µ1 µ2 ℓ∞ . µ1 µ2 M (R) M (R) = µ1 µ2 M (R) M (R), k − k kH − k → kH −H k → and therefore µ = µ whenever = . In other words, the measures in M p are completely 1 2 Hµ1 Hµ2 characterized by their action on M 1(R), and thus there is a one-to-one correspondence between the measures in M p and the operators : µ M p . Note that this property is necessary for there to {Hµ ∈ } be any hope of recovering a measure µ from a measurement x with respect to a single probing Hµ signal x. We are now ready to state our definition of stable identifiability:
Definition 1 (Stable identifiability). Let p [1, ). We say that a class of measures H M p ∈ ∞ ⊂ is stably identifiable by a probing signal x M 1(R) if there exist constants C ,C > 0 (that may ∈ 1 2 depend on p and x) such that
C (ms(Λ , Λ ) 1) µ µ 6 x x p 6 C µ µ , (10) 1 1 2 ∧ k 1 − 2kp kHµ1 −Hµ2 kM (R) 2 k 1 − 2kp for all µ ,µ H , where Λ := supp(µ ), j 1, 2 . 1 2 ∈ j j ∈ { }
The significance of the term ms(Λ1, Λ2) in (10) becomes apparent when we consider classes H that contain measures with potentially arbitrarily close supports. For a concrete example, consider the class H = µ M p, #(supp(µ)) = 1 of single time-frequency shifts. This class contains the measures { ∈ } µ = δ and µ = δ , for all ǫ > 0. Let x M 1(R) be a probing signal satisfying the time- (0,0) ǫ (0,ǫ) ∈ localization constraint tx(t) L2(dt), but otherwise arbitrary. Then ms(supp(µ), supp(µ )) = ǫ, ∈ ǫ and 1 1 2πiǫ ǫ− ( x x)= ǫ− (1 e ·) x (2πi )x Hµ −Hµǫ − →− · in L2(R)= M 2(R) as ǫ 0, and hence →
x x 2 R /ms(supp(µ), supp(µ )) tx(t) 2 > 0, (11) kHµ −Hµǫ kM ( ) ǫ ≍ k kL (dt) as ǫ 0. On the other hand, µ µ = √2 is bounded away from 0 as ǫ 0. Thus, if → k − ǫk2 → the class H is to be identifiable, the lower bound in (4) needs to decay at least linearly with ms(supp(µ ), supp(µ )), for µ ,µ H . In contrast to (11), one could have another class K 1 2 1 2 ∈ containing measures µ′ and µ′ , for ǫ> 0, so that µ x µ x 2 R /ms(supp(µ′), supp(µ′ )) 0, ǫ kH ′ −H ǫ′ kM ( ) ǫ → as ǫ 0, i.e., µ x µ x 2 R decays superlinearly with ms(supp(µ′), supp(µ′ )). Classes such → kH ′ −H ǫ′ kM ( ) ǫ as K are not covered by our theory, and we hence exclude them from our definition of stable 7 identifiability. In summary, Definition 1 says that we consider a class of measures to be stably identifiable if the decay of x x 2 R as ms(supp(µ ), supp(µ )) 0 is not faster than kHµ1 −Hµ2 kM ( ) 1 2 → linear. This property will turn out to be crucial later when we discuss robust recovery (specifically, in the proofs of Theorems 3 and 4).
B. A necessary and sufficient condition for identifiability As already mentioned in the introduction, our necessary and sufficient condition for identifiability will be expressed in terms of the density of support sets measured uniformly over the class of measures under consideration. Concretely, we have the following definition:
Definition 2 (Upper Beurling class density). Let be a collection of relatively separated sets in R2, L and, for R > 0, define [0,R]2 = [0,R] [0,R] R2. For Λ , let n+(Λ, [0,R]2) be the largest × ⊂ ∈ L number of points of Λ contained in any translate of [0,R]2 in the plane. We then define the upper Beurling class density of according to L + 2 + n (Λ, [0,R] ) ( ) = limsup sup 2 . D L R Λ R →∞ ∈L We are now ready to state the first main result of the paper.
p Theorem 1 (A sufficient condition for identifiability). Let p (1, ) and s> 0, let H Ms be a ∈ ∞ ⊂ class of measures, and set = supp(µ) : µ H . Suppose that +( ) < 1 . Then the class H L { ∈ } D L 2 1 πt2 1 is identifiable by the standard gaussian ϕ(t) = 2 4 e , ϕ M (R). − ∈ Crucially, the support sets supp(µ) in Theorem 1 are not assumed to be subsets of a lattice or ∈ L 2 any other a priori fixed subset of R . In particular, one allows H to contain measures µ1 and µ2 with arbitrarily small ms(supp(µ1), supp(µ2)).
Note that a subclass H ′ of an identifiable class H is trivially identifiable, and accordingly the upper Beurling class density of the supports of measures in H ′ does not exceed that of the support sets corresponding to H . The sufficiency result in Theorem 1 is therefore “compatible” with the p inclusion relation on classes. By contrast, the “non-identifiability” of a class H Ms (i.e., the ⊂ nonexistence of a probing signal in M 1(R) by which the class would be identifiable) can only be meaningfully assessed in terms of the Beurling density +( supp(µ) : µ H ) for sufficiently D { ∈ } p rich classes of measures. For example, one can construct arbitrarily large finite subsets H of Ms with arbitrarily large +( supp(µ) : µ H ), and yet H will be identifiable (e.g. by the standard D { ∈ } gaussian, using the property that distinct time-frequency shifts of a gaussian are linearly independent). A converse statement to Theorem 1 can hence be meaningfully formulated only for classes H that are “sufficiently rich” in a suitable sense. In the present paper we will do this for classes of measures that are subspaces of M p, with support sets that are closed under limits with respect to weak convergence and invariant under time-frequency shifts. Before providing the precise definition of these classes of measures, we need to introduce the notion of weak convergence for subsets of C. Concretely, we say that a sequence of separated subsets w Λn n N converges weakly to Λ C, and write Λn Λ, if { } ∈ ⊂ −→ dist (Λ B (z)) ∂B (z), (Λ B (z)) ∂B (z) 0 as n , (12) n ∩ R ∪ R ∩ R ∪ R → →∞ 8 for all R> 0 and z C, where dist denotes the Hausdorff metric on the subsets of C. We are now ∈ ready to formalize the type of classes covered by our necessity result.
Definition 3 (Regular H ( )p classes). L Let p (1, ) and s> 0, and let be a collection of separated subsets of C. ∈ ∞ L (i) We say that is closed and shift-invariant (CSI) if it is closed under limits with respect to weak L convergence, and Λ+ z := λ + z : λ Λ , for all Λ and z C. { ∈ }∈L ∈L ∈ (ii) We define a class of measures H ( )p M p according to L ⊂
H ( )p = α δ : Λ , α ℓp(Λ) . L λ λ ∈L ∈ ( λ Λ ) X∈ We call H ( )p s-regular if is CSI and sep(Λ) > s, for all Λ . L L ∈L Even though the conditions in Definition 3 are rather technical, they are not overly restrictive, as evidenced by several examples of s-regular classes provided in §II-D. We are now ready to state our second main result, which is a necessary condition for identifiability of s-regular classes and as such constitutes a partial converse to Theorem 1.
Theorem 2 (A necessary condition for identifiability of s-regular classes). Let p (1, ) and s> 0, ∈ ∞ and let H ( )p M p be an s-regular class. If there exists an x M 1(R) such that H ( )p is L ⊂ ∈ L identifiable by x, then +( ) < 1 . D L 2
C. Identifiability and robust recovery In this subsection we formalize the claim (5) made in the introduction under the assumption that x M 1 (R), with the weight function m(z)=1+ z , for z C. Informally, this assumption imposes ∈ m | | ∈ faster-than-linear decay on x in both the time and frequency domains. Note that the L2-normalized gaussian ϕ is in M 1 (R), as its STFT decays exponentially (by virtue of ϕ (R) and [7, Thm. m ∈ S 11.2.5]). p p We begin by defining the weak-* topology on Ms , for p (1, ). Concretely, for µ Ms and p ∈ ∞ ∈ p a sequence µn n N Ms , we say that µn n N converges to µ in the weak-* topology of Ms , ∈ ∈ { w∗} ⊂ { } and write µ µ, if n −−→ lim fdµn = fdµ, (13) n C C →∞ Z Z C C for all continuous f : such that lim z f(z) = 0 and → | |→∞ sup f(z + y) < . y C, y 61 | | Lq (dz) ∞ ∈ | |
This definition corresponds to convergence in the weak-* topology on the Wiener amalgam space W ( ,Lp), which will be defined and treated systematically in §III. In order to formalize (5), it will M be helpful to first state the following weak-* recovery result for s-regular classes:
p p Theorem 3 (Weak-* Recovery Theorem). Let p (1, ) and s> 0, and let H ( ) Ms be an ∈ ∞ L ⊂ s-regular class. Assume furthermore that H ( )p is identifiable by a probing signal x M 1 (R), L ∈ m where m(z)=1+ z . Then | | p (i) if µ, µ H ( ) are such that e x = x, then µ = µ. ∈ L Hµ Hµ
e e 9
H p H p (ii) Let µ ( ) and let µn n N be a sequence in ( ) . Then µn x µx in the weak-* ∈ L { } ∈ L H →H p w∗ p topology of M (R) if and only if µ µ in Ms . n −−→ The proof of Theorem 3 relies crucially on the fact that the decay of the lower bound in (10) as a function of ms(supp(µ1), supp(µ2)) is not faster than linear. Note that item (i) of Theorem 3 guarantees perfect recovery of measures in H ( )p under perfect L measurement matching. However, this does not go a long way towards establishing (5) as item (ii) of the theorem deals with convergence in weak-* topologies “only”. To illustrate that a stronger form of convergence is needed, consider the 1/2-regular class H ( )2, where = Λ C : sep(Λ) > w∗ L L { ⊂ 1/2, #(Λ) 6 2 . In this class δ + δ δ as n (where δ is again the Dirac point } 0,0 n,0 −−→ 0,0 → ∞ τ,ν measure with mass at (τ,ν)), and so, if one were to rely on the weak-* convergence guarantee only, one could argue that δ0,0+δn,0 n N recovers δ0,0. This sequence does, indeed, capture the component { } ∈ δ0,0, but it also features the nonvanishing spurious component δn,0. Similarly, on the measurement side of (5), taking x = ϕ as the probing signal would yield ϕ+ϕ( n) ϕ in the weak-* topology ·− → of L2, but not in the norm topology. We can thus hope that upgrading from weak-* convergence to norm convergence on the measurement side of (5) might imply a stronger form of convergence of the sequence of candidate measures to the target measure. The following theorem establishes that this is, indeed, the case for s-regular classes H ( )p. Concretely, convergence of the measurements in L norm implies that the candidate measures µn approximate arbitrarily big finite sections of the target measure µ and do not have any spurious components.
p p Theorem 4 (Robust Recovery Theorem). Let p (1, ) and s > 0, and let H ( ) Ms be an ∈ ∞ L ⊂ s-regular class. Assume furthermore that H ( )p is identifiable by a probing signal x M 1 (R), L ∈ m where m(z)=1+ z , and let C and C be the corresponding constants such that (10) is fulfilled. | | 1 2 p p Fix a µ H ( ) , write Λ = supp(µ), and let µn n N be a sequence in H ( ) such that ∈ L { } ∈ L x x p R 0 as n . kHµn −Hµ kM ( ) → →∞ Then, for every ǫ> 0 and every finite subset Λ of Λ such that µ µ1e <ǫ, there is an N N k − Λkp ∈ so that, for all n > N, the measures µn take the form e (n) ε µn = αλ δλ+ n(λ) + ρn, λ Λ˜ X∈ (n) ε 4C2 where n(λ) 6 ǫ and αλ αλ 6 ǫ, for all λ Λ, and ρn p 6 C (s 1) ǫ . | | | − | ∈ k k 1 ∧ (n) One can view ε as the “successfully recovered finite section” of , which approx- λ Λ˜ αλ δλ+ n(λ) e µ ∈ imates both the time-frequencyP shifts and their weights within ǫ error, whereas ρn is the “spurious” component, whose norm is also proportional to ǫ. The constant of proportionality (C /C ) (s 1) 1 2 1 · ∧ − in the bound on ρ can be interpreted as a “condition number”, indicating that the spurious k nkp component is more difficult to suppress when the ratio of identifiability constants C2/C1 is large, or when the separation s of the measures under consideration is excessively small, which agrees with our intuition on the behavior of the “difficult cases”. 10
D. Examples of identifiable and non-identifiable s-regular classes Finally, we present several explicit families of s-regular classes and discuss their identifiability in view of Theorems 1 and 2. Let p (1, ), s> 0, N N, θ> 0, and R> 0, and define the sets ∈ ∞ ∈ sep = Λ R2 : sep(Λ) > s , Ls { ⊂ } fin = Λ R2 : sep(Λ) > s, #(Λ) 6 N , and Ls,N { ⊂ } Ray = Λ R2 : sep(Λ) > s, n+ Λ, (0,R)2 6 θR2 . Ls,θ,R { ⊂ } sep p fin p Ray p 2 We call the corresponding sets H ( s ) , H ( ) , and H ( ) , the ℓ -separated, finite, and L Ls,N Ls,θ,R Rayleigh classes, respectively. The following proposition shows that these classes are s-regular.
sep fin Ray Proposition 2. Let s > 0, N N, θ > 0, and R > 0. Then the collections s , , and ∈ L Ls,N Ls,θ,R are CSI and so the corresponding ℓ2-separated, finite, and Rayleigh classes are s-regular.
Theorems 1 and 2 can be used to obtain the following identifiability results for these classes.
Corollary 5 (Finite class). Let p (1, ), s> 0, and N N. Then the class H( fin )p is identifiable ∈ ∞ ∈ Ls,N 1 πt2 by the gaussian ϕ(t) = 2 4 e− .
Corollary 6 (ℓ2-separated class). Let p (1, ) and s> 0. Then, ∈ ∞ 1 sep p (i) if s> 2 3− 4 , H ( s ) is stably identifiable by ϕ, and · L 1 sep p (ii) if s 6 2 3− 4 , H ( s ) is not stably identifiable by any probing signal. · L Corollary 7 (Rayleigh class). Let p (1, ) and s (0,θ 1/2). Then, ∈ ∞ ∈ − (i) if θ < 1 , H ( Ray )p is stably identifiable by ϕ, for all R> 0, and 2 Ls,θ,R (ii) if θ > 1 , there exists an R > 0 such that H ( Ray )p is not stably identifiable by any probing 2 0 Ls,θ,R signal, for all R > R0.
One could also consider the class H ( Ξ )p for a fixed lattice Ξ= A(Z Z)+b, where A R2 2 { } × ∈ × and b R2, in which case H ( Ξ )p is sep(Ξ)-separated, stably identifiable by ϕ if det(A) > 1, ∈ { } and not stably identifiable by any probing signal if det(A) 6 1.
III. LATTICES, BEURLINGDENSITIES, AND WIENER AMALGAM SPACES In this section we introduce various technical tools used throughout the paper. We begin with square lattices in C and write Ωγ = ωm,n = γ(m + in) m,n Z for the square lattice in C of mesh { } ∈ size γ > 0. Whenever we identify C with R2, Ω is equivalently given by (γm,γn) : m,n Z . γ { ∈ } Next, we define the (standard) upper Beurling density, which is analogous to our Definition 2, but is defined for individual subsets of R2, instead of classes of subsets.
Definition 4 (Upper Beurling density, [17, p. 346] [23, p. 47]). Let Λ be a relatively separated set in R2, and, for R > 0, let [0,R]2 R2. Let n+(Λ, [0,R]2) be the largest number of points of Λ ⊂ contained in any translate of [0,R]2. We then define
+ 2 + n (Λ, [0,R] ) D (Λ) := limsup 2 , R R →∞ and we call this quantity the upper (standard) Beurling density of Λ. 11
The following three lemmas, whose proofs can be found in the appendix, relate the lattices Ωγ, the upper Beurling class density, and the standard Beurling density.
Lemma 8. Let be a collection of relatively separated sets in R2, and suppose that +( ) < . L D L ∞ Then, (i) for every θ > +( ), there exists an R > 0 such that D L 0 n+(Λ, (0,R)2) 6 θR2,
for all Λ and R > R , and ∈L 0 + + (ii) ( ) > supΛ D (Λ). D L ∈L Definition 5. Let Λ be a non-empty relatively separated subset of C, and let γ > 0 and R> 0. We say that Λ is R-uniformly close to Ωγ = ωm,n = γ(m + in) m,n Z if there exists an enumeration { } ∈ Z Z λm,n (m,n) of Λ (with index set ) such that λm,n ωm,n 6 R, for all (m,n) . { } ∈I I ⊂ × | − | ∈I 2 Lemma 9. Let Λ be a non-empty discrete set in C, and let θ > 0, γ > 0, and R > 0. If γ− > θ + 2 2 and n (Λ, (0,R) ) 6 θR , then there exists an R′ = R′(θ, γ, R) > 0 such that Λ is R′-uniformly close to Ωγ.
Lemma 10. Let be a set of relatively separated subsets of C, and let γ > 0 and R > 0. If Λ is L R-uniformly close to Ω , for all Λ , then +( ) 6 γ 2. γ ∈L D L − We conclude this section by formalizing Wiener amalgam spaces [24], [25] on C and relating them p to weak-* convergence on Ms defined in (13). We adopt most of our terminology from [22]. Let (C) D be the test space of smooth compactly supported functions on C, with its usual inductive limit topology and the corresponding topological dual (C), called the space of distributions. Let be a Banach D′ B space that admits a continuous embedding into (C). Furthermore, fix a non-negative compactly D′ C supported continuous function ψ ( ) forming a partition of unity, i.e., Z2 ψ( z) = 1, and ∈ D z ·− C R r ∈ let m : >0 be a weight function of the form m(z) =(1+ z ) , forP some r > 0. Then, for → p | | p [1, ], the Wiener amalgam space W ( ,Lm) is defined as ∈ ∞ B
p C p W ( ,Lm)= f ′( ) : f W ( ,Lm) := f ψ( z) m(z) < . B ∈ D k k B k ·− kB Lp(dz) ∞ p The definition of W ( ,Lm) is independent of the choice of ψ, and different ψ define equivalent p B p norms on W ( ,Lm). Informally, W ( ,Lm) is the space of distributions (i.e., generalized functions) B B p on C that are “locally in ” and “globally in Lm”. B p p Next, we claim that Ms W ( ,L ), for p (1, ], where is the space of regular complex- ⊂ M ∈ ∞ M valued Borel measures on C with the total variation norm. To see this, let r > 0 be such that M p p p supp(ψ) Br(0). Now, for a measure µ = λ Λ αλδλ s denote µ := λ Λ αλ δλ. Then ⊂ ∈ ∈ | | ∈ | | H¨older’s inequality yields P P 1/q 1/p q p µ ψ( z) 6 αλ 6 1 αλ k ·− kM | | | | λ Λ Br (z) λ Λ Br (z) ! λ Λ Br (z) ! ∈ X∩ ∈ X∩ ∈ X∩ 1/p 2 1/q 1 p C . ψ sep(Λ)− y z 6r d µ (y) , for z , C {| − | } | | ∈ Z ! 12
2 2 where the last inequality follows since one can pack at most r /(sep(Λ)/2)− spheres of radius sep(Λ)/2 in Br(z). Therefore, as sep(Λ) > s Tonelli’s theorem yields
p 1 p µ ψ( z) . ψ,s z y 6r d µ dz k ·− kM Lp(dz) C " C {| − | } | | # Z Z
1 p 2 p = z y 6r dz d µ = πr µ p < , C C {| − | } | | k k ∞ Z ´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹Z ¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶ =πr2 p p and so µ W ( ,L ). As µ Ms was arbitrary, we have therefore shown that ∈ M ∈ M p µ W ( ,Lp) . ψ,p,s µ p, for all µ s , (14) k k M k k ∈ p p which establishes Ms W ( ,L ). ⊂ M Now, by the Riesz-Markov-Kakutani representation theorem [26, Thm. 6.19], can be identified M with the topological dual C0∗ of
C0 = f L∞(C) : f continuous, lim f(z) = 0 , { ∈ z | | } | |→∞ via the pairing µ,f = fdµ. Therefore, by [25, Thm. 2.8], we have that f(y) 1 (y) is h i C | | Br (z) integrable w.r.t. the product measure d µ (y) dz on C C, for µ W ( ,Lp) and f W (C ,Lq), R | | × × ∈ M ∈ 0 and W ( ,Lp) can be identified with the topological dual of W (C ,Lq) via the dual pairing M 0 1 µ,f := f Br(z)dµ dz. h i C C Z Z An application of Fubini’s theorem hence yields
1 1 2 µ,f = f(y) Br (z)(y)dµ(y) dz = f(y) z y 6r dz dµ(y)= πr fdµ. h i C C C C {| − | } Z Z Z Z Z Thus, as πr2 is a constant depending only on the choice of ψ through supp(ψ) B (0), one can ⊂ r instead use the following simpler dual pairing to effect the correspondence between W ( ,Lp) and M q W (C0,L )∗: µ,f = fdµ = f(λ)µ( λ ), h i { } Z λ supp(µ) ∈ X p q p for µ W ( ,L ) and f W (C ,L ). Therefore, definition (13) of weak-* convergence in Ms ∈ M ∈ 0 corresponds precisely to convergence in the weak-* topology on W ( ,Lp) (i.e., the weak topology M q generated by W (C0,L )). Finally, in the special case of weak convergence of subsets of R2 defined in (12), following [22, p. w 398], we have that, if infn N sep(Λn) > 0, then weak convergence of subsets Λn Λ is equivalent ∈ −→ to δ δ in the weak-* topology W (C ,L1). λ Λn λ λ Λ λ 0 ∈ → ∈ P P IV. PROOFOF THEOREM 1 As already mentioned in the introduction, the proof of Theorem 1 relies on the theory of interpo- lation of entire functions. The idea for the proof is based on [27, Thm. 1], where the lower bound (analogous to the left-hand side of (10)), however, depends in a non-explicit manner on the supports of the individual measures in the identifiability condition. As our goal is to obtain an explicit lower 13 bound, namely, a constant multiple of the minimum separation of the supports, our theorem needs to be stated in terms of the class density (according to Definition 2) instead of simply considering the standard Beurling density (according to Definition 4) of the supports of the individual measures in the class. This difference will also require us to delve deeper into the interpolation theory underlying the proof of [27, Thm. 1].
We begin our exposition of the required technical tools by defining the Weierstrass σγ -function associated with Ωγ = ωm,n = γ(m + in) m,n Z : { } ∈ 2 z z 1 z C σγ(z)= z 1 exp + 2 , z . − ωm,n ωm,n 2 ω ∈ (m,n) Z2 (0,0) m,n ∈Y\{ } We will need several basic facts about this function, which can be found in [28] along with a more detailed account of its properties. Concretely, we note that the infinite product in the definition of σγ converges absolutely uniformly on compact subsets of C, and therefore defines an entire function.
Moreover, σγ satisfies the following growth estimate:
π 2 2 γ− z Lemma 11 ([28, Cor. 1.21]). We have σ (z) e− 2 | | d(z, Ω ), where d(z, Ω ) = min z ω : | γ | ≍ γ γ γ {| − | ω Ω denotes the Euclidean distance from z to the lattice Ω . ∈ γ} γ In order to enable working with measures µ whose supports are not subsets of lattices, we will need to perturb the zeros of the Weierstrass σγ -function. We will do so following [28] and [20, p. 109].
Concretely, let Z Z be an index set with (0, 0) , and let Λ= λm,n (m,n) be a discrete I ⊂ × ∈I { } ∈I subset of C with λ = 0 for (m,n) (0, 0) . We now define the modified Weierstrass function m,n 6 ∈ I\{ } associated with Λ by
2 z z 1 z C gΛ(z) = (z λ0,0) 1 exp + 2 , z . (15) − − λm,n λm,n 2 ω ∈ (m,n) (0,0) m,n ∈I\{Y } According to [28, Lem. 4.21], provided there exist γ > 0 and R > 0 such that Λ is R-uniformly close to Ωγ, expression (15) converges uniformly on compact subsets of C to an entire function with zero set Λ. The proof of Theorem 1 relies on constructing and controlling the growth of an entire function interpolating a sequence of values βλ λ Λ at the points of Λ = supp(µ1) supp(µ2), ∈ p { } ∪ where µ ,µ Ms are the measures for which (10) is to be established. This will be accomplished 1 2 ∈ 1 by means of “basis functions” that interpolate the one-hot sequences λ=λ λ Λ, for λ′ Λ. The { { ′}} ∈ ∈ following lemma furnishes a prototype for these basis functions, obtained by “dividing out” a zero of the modified Weierstrass function associated with Λ, as well as a growth bound reminiscent of [28] and [20, §2.2], with the crucial difference that our bound makes the dependence on the mutual separation of supp(µ1) and supp(µ2) explicit. The proof of the lemma largely follows [28], the only difference being that we need to take the specific form Λ = supp(µ ) supp(µ ) of Λ into account, 1 ∪ 2 carrying out the calculations more explicitly to extract the dependence on the mutual separation of supp(µ1) and supp(µ2).
Lemma 12. Let Λ= λm,n (m,n) be a relatively separated subset of C with λ0,0 = 0. Furthermore, { } ∈I let ρ, s, θ, γ, and R be positive real numbers, and set Ωγ = ωm,n = γ(m + in) m,n Z . Define { } ∈ = (m,n) : λ 6 s and suppose that Is { ∈I | m,n| 2 } (i) #( ) 6 2, and, if = (0, 0), (m ,n ) , then λ > ρ , Is Is { ′ ′ } | m′,n′ | 14
+ 2 2 (ii) n (Λ, (0,R′) ) 6 θR′ , for all R′ > R, and (iii) λ ω 6 R, for all (m,n) . | m,n − m,n| ∈I Now, let g be given by (15) and define g : C C according to Λ Λ → 2 gΛ(z) z z z z 1 z gΛ(z)= exp e 1 exp + . (16) z ω λ ω ω 2 ω2 m,n − m,n 2 − m,n m,n m,n (m,n) s (m,n) Z Y∈I Y∈ \I Thene (a) g (0) = 1 and g (λ ) = 0, for (m,n) (0, 0) , and Λ Λ m,n ∈I\{ } (b) there exist constants C > 0 and c> 0 depending only on s, θ, γ, and R such that e e π 2 2 γ− z 1 c z log z g (z) e− 2 | | 6 C(ρ 1)− e | | | |, for all z C. (17) | Λ | ∧ ∈
The proof of Lemma 12e can be found in the appendix. The next preparatory step towards the proof of Theorem 1 is to relate Gabor systems generated 1 πt2 by ϕ(t) = 2 4 e− with entire functions of suitably bounded growth by means of the Bargmann transform. Concretely, we will work with a definition of the Bargmann transform consistent with [29] in order to facilitate arguments involving the isometry property between modulation spaces and Bargmann-Fock spaces introduced next. For conjugate indices p,q [1, ], the Bargmann-Fock ∈ ∞ p C space ( ) is defined as the set of all entire functions F for which F p(C) < , where F k kF ∞ 1/p p pπ z 2/2 F p(C) := F (z) e− | | dz , for p [1, ), k k C | | ∈ ∞ F Z and π z 2/2 F := sup F (z) e− | | . ∞(C) k kF z C | | ∈ The Bargmann transform is now defined as the linear map B : M p(R) p(C) given by → F 1 πz2/2 2πtz πt2 (Bf)(z) = 2 4 e− e − f(t)dt, z C. R ∈ Z According to [29, §1.4], the Bargmann transform is an isometric isomorphism between the Banach spaces M p(R) and p(C), i.e., it is bijective and F B p R f p(C) = f M p(R), for all f M ( ). (18) k kF k k ∈ Following [30], when p [1, ), the topological dual of p(C) can be identified with q(C) via ∈ ∞ F F the pairing π z 2 F, G p(C) q (C) = F (z)G(z)e− | | dz, h i C F ×F Z for F p(C) and G q(C). ∈ F ∈ F The following lemma is a generalization (from L2(R) to M q(R)) of the standard identity [7, Prop. 1 πt2 3.4.1] relating the Bargmann transform with time-frequency shifts of the gaussian ϕ(t) = 2 4 e− .
Lemma 13. Let q [1, ). Then, for every y M q(R) and λ = τ + iν C, we have ∈ ∞ ∈ ∈ πiτν π λ 2/2 B ¯ y,π(λ)ϕ M q (R) M p(R) = e− e− | | ( y)(λ). h i × Before finally embarking on the proof of Theorem 1, we state the following two lemmas about 15 abstract Banach spaces and Wiener amalgam spaces that will facilitate the application of the more specialized theory of Bargmann-Fock spaces. Their proofs can be found in the appendix.
Lemma 14. Let : X Y be a continuous linear operator between Banach spaces X and Y . We A → then have the following: (i) If is bounded below (i.e. there exists a c > 0 such that x > c x , for all x X), then A kA k k k ∈ the adjoint : Y X is surjective. A∗ ∗ → ∗ (ii) Suppose that there exists a constant a> 0 such that, for every f X , there is a g Y with ∈ ∗ ∈ ∗ g = f and a g 6 f . Then is bounded from below by a. A∗ k kY ∗ k kX∗ A Lemma 15. Let p [1, ) and Λ R2 a separated subset and set s = sep(Λ) > 0. Then ∈ ∞ ⊂
αλ f( λ) . p,s f W (L ,L1) α ℓp(Λ), ·− Lp(R2) k k ∞ k k λ Λ X∈ p 1 for all αλ λ Λ ℓ (Λ) and f W (L∞,L ). { } ∈ ⊂ ∈ Proof of Theorem 1. Fix µ ,µ H and let Λ = supp(µ ), for j 1, 2 , and Λ = Λ Λ . 1 2 ∈ j j ∈ { } 1 ∪ 2 p We can then write µ1 µ2 = λ Λ αλδλ, where α ℓ (Λ), so that µ1 µ2 p = α ℓp and − ∈ ∈ k − k k k ϕ ϕ = (α, ϕ), for ( , ϕ) : ℓp(Λ) M p(R) as defined in the statement of Proposition Hµ1 −Hµ2 HΛ HPΛ · → 1. With this, (10) is equivalent to
C ms(Λ , Λ ) 1 α p 6 (α, ϕ) p R 6 C α p , (19) 1 1 2 ∧ k kℓ kHΛ kM ( ) 2k kℓ and hence it suffices to find constants C1 = C1(p, H , ϕ) > 0 and C2 = C2(p, H , ϕ) > 0 such that (19) holds. To this end, first note that by item (i) of Proposition 1 we have
(α, ϕ) p R . rel(Λ) α . kHΛ kM ( ) ϕ k kp p Furthermore, as µ ,µ Ms , we get 1 2 ∈ 2 rel(Λ Λ ) 6 rel(Λ ) + rel(Λ ) . s− , 1 ∪ 2 1 2 and so the upper bound in (19) holds for some C2 > 0 depending on ϕ and s, as desired.
We proceed to establish the lower bound in (19). Note that this bound holds trivially if ms(Λ1, Λ2)= 0 or Λ = Λ = ∅, so suppose w.l.o.g. that ms(Λ , Λ ) > 0 and Λ = ∅. Then, in particular, Λ = ∅. 1 2 1 2 1 6 6 Now, as ( , ϕ) : ℓp(Λ) M p(R) is a continuous linear operator between Banach spaces, Lemma HΛ · → 14 implies that it suffices to find a C1 = C1(p, H , ϕ) > 0 such that the following statement holds: q q (P1) For every β ℓ (Λ), there exists a y M (R) such that ( ( , ϕ))∗ (y)= β and ∈ ∈ HΛ ·
C ms(Λ , Λ ) 1 y q R 6 β q . 1 1 2 ∧ k kM ( ) k kℓ By item (ii) of Proposition 1 and Lemma 13, we have the following expression for ( ( , ϕ))∗ in HΛ · terms of the Bargmann transform:
πiτν π λ 2/2 q ( Λ( , ϕ))∗ (y)= e− e− | | (By)(λ) λ=τ+iν Λ, y M (R). (20) H · { } ∈ ∈ Thus, as the Bargmann transform is an isometric isomorphism between M p(R) and p(C), and the F πiτν q map βλ λ=τ+iν Λ βλe− λ=τ+iν Λ is an isometric isomorphism on ℓ (Λ), the statement { } ∈ 7→ { } ∈ 16
(P1) is equivalent to the following statement about interpolation: (P2) For every β ℓq(Λ), there exists an F q(C) such that e π λ 2/2F (λ) = β , for all ∈ ∈ F − | | λ λ Λ, and ∈ C1 ms(Λ1, Λ2) 1 F q (C) 6 β ℓq . (21) ∧ k kF k k To prove (P2), we will make use of the interpolation basis functions provided by Lemma 12. To this end, fix θ > 0 and γ > 0 such that 2 +( ) < 2θ < γ 2 < 1, and let β ℓq(Λ) be arbitrary. Then, D L − ∈ + 2 2 by Lemma 8, there exists an R0 > 0 (depending only on H ) such that n (Λj, (0,R) ) 6 θR , for j 1, 2 and R > R . Now, for each λ Λ, define the set Λ = λ λ : λ Λ . We will seek ∈ { } 0 ∈ λ { ′ − ′ ∈ } to apply Lemma 12 to each of the sets Λλ as λ ranges over Λ. To this end, first note that e + 2 + 2 + 2 + 2 2 n (Λλ, (0,R) )= n (Λ, (0,Re) ) 6 n (Λ1, (0,R) )+ n (Λ2, (0,R) ) 6 2θR ,
1/2 for all R > Re 0. Therefore, as γ < (2θ)− , it follows by Lemma 9 that there exists an R′ = R (θ, γ, R ) such that Λ is R -uniformly close to Ω = ω = γ(m + in) : m,n Z . In ′ 0 λ ′ γ { m,n ∈ } e particular, there exists an enumeration Λλ = λm,n (m,n) such that λm,n ωm,n 6 R′, for e { } ∈I | − | all (m,n) . Note that 0 Λλ by definition of Λλ. In order to apply Lemma 12 we need to ∈ I ∈ e e e additionally ensure that we work with an enumeration of Λλ = λm,n (m,n) (possibly different e e e { } ∈I e from the enumeration Λλ = λm,n (m,n) ) that satisfies λ0,0 = 0. To this end, let (m0,n0) be { } ∈I e ∈ I the index such that λm0,n0 = 0, and define and λm,n (m,n) as follows: e e I { } ∈I e – If (0, 0) / , set = (m0,n0) (0, 0) , and let ∈ I e I I \ { } ∪ { } e e 0, if (m,n) = (0, 0), λm,n = . λ , if (m,n) (0, 0) m,n ∈I\{ } – If (0, 0) , set = , and let e ∈ I I I e e 0, if (m,n) = (0, 0),
λm,n = λ , if (m,n) = (m ,n ), . 0,0 0 0 λ , if (m,n) (0, 0), (m ,n ) em,n ∈ ∈I\{ 0 0 } The new enumeration Λλ = λm,ne (m,n) satisfies λ0,0 ω0,0 = 0 and λm0,n0 ωm0,n0 6 { } ∈I | − | | − | λ + λ ω 6 2R , and thus we have λ ω 6 2R , for all (m,n) . The | 0,0| | m0 ,n0 − m0,n0 | ′ | m,n − m,n| ′ ∈ I e s set Λλ therefore satisfies the assumptions of Lemma 12 with ρ := ms(Λ1, Λ2) 2 , s, θ, γ, and e e ∧ R := R0 (2R′), and so the function g λ := ge ( λ), where ge is defined according to (16), e ∨ − Λλ ·− Λλ satisfies 1, if eλ′ = λ e g λ(λ′)= , for all λ′ Λ, (22) − 0, if λ = λ ∈ ′ 6 and π 2 2 1 γ− z λ +c z λ log z λ g λ(z) 6 C(ρ 1)− e 2 | − | | − | | − |, for all z C, (23) | − | ∧ ∈ where c> 0 and C > 0 depend on s, θ, γ, and R. Moreover, as λ was arbitrary, (22) and (23) hold 17 for all λ Λ. Next, following [20, p. 112], we consider the interpolation function ∈ πλz π λ 2 F (z)= βλ e − 2 | | g λ(z). (24) − λ Λ X∈ To see that F is an element of q(C), observe that F π z 2 π z λ 2 F (z) e− 2 | | 6 βλ e− 2 | − | g λ(z) | | | | | − | λ Λ ∈ X π 2 2 1 (1 γ− ) z λ +c z λ log z λ 6 C(ρ 1)− β e− 2 − | − | | − | | − | ∧ | λ| λ Λ X∈ 1 = C(ρ 1)− β f(z λ), for all z C, ∧ | λ| − ∈ j 1,2 λ Λj ∈{X } X∈ where f(z) = exp π (1 γ 2) z 2 + c z log z . Now, as γ 2 < 1, we have that f decays − 2 − − | | | | | | − exponentially, and so f W (L ,L1). Lemma 15 thus yields ∈ ∞
1 q βλ f( λ) . p,s f W (L ,L ) βλ λ Λj ℓ , for j 1, 2 , | | ·− Lq (C) k k ∞ k{ } ∈ k ∈ { } λ Λj X∈ and so
π 2 1 1 2 q F q (C) = F ( )e− |·| q C 6 C(ρ 1)− βλ f( λ) . p,s,γ C(ρ 1)− β ℓ . k kF · L ( ) ∧ | | ·− Lq (C) ∧ k k j 1,2 λ Λj ∈{X } X∈ (25)
Now, recall that ρ := ms(Λ , Λ ) s , and so ms(Λ , Λ ) 1 . ρ 1. This together with (25) 1 2 ∧ 2 1 2 ∧ s ∧ establishes (21) with some C1 > 0 depending on s, θ, γ, R0, and R′. As these quantities ultimately depend only on s and H , so does C1. Finally, (24) and the basis interpolation property (22) together yield F (λ)= eπ λ 2/2β , for all λ Λ. We have thus established (P2), thereby concluding the proof | | λ ∈ of the theorem.
V. PROOFOF THEOREM 2 In the proof of Theorem 2 we will make use of the following results from [22], as well as a combinatorial lemma about squares in the plane, whose proof can be found in the appendix.
Theorem 16 (Non-uniform Balian-Low Theorem, [22, Cor. 1.2]). Let Λ be a relatively separated R2 1 R subset of and x M ( ). If π(λ)x λ Λ is a Riesz sequence, i.e. λ Λ cλπ(λ)x L2(R) Λ,x ∈ { } ∈ k k ≍ 2 + ∈ c 2 , for all c ℓ (Λ), then D (Λ) < 1. k kℓ (Λ) ∈ P Theorem 17 ([22, Thm. 3.2]). Let Λ be a relatively separated subset of R2 and x M 1(R). Then ∈ p λ Λ cλπ(λ)x M p(R) Λ,x c ℓp(Λ), c ℓ (Λ), holds for some p [1, ] if and only if it holds k ∈ k ≍ k k ∈ ∈ ∞ for all p [1, ]. P ∈ ∞ 2 Lemma 18 ([22, Lem. 4.5]). Let Λn n N be a sequence of relatively separated subsets of R . If { } ∈ supn N rel(Λn) < , then there exists a subsequence Λnk k N that converges weakly to a relatively ∈ ∞ { } ∈ separated set.
Lemma 19. Let Y R2, n N, and suppose that K is a square in the plane of side length ⊂ ∈ n n 2n √2(2 + 1) such that #(Kn Y ) > 2 + 1. Then there exist squares K0, K1,...,Kn 1 so that, for ∩ − every j 0, 1,...,n 1 , ∈ { − } 18
(i) K K , K has sides of length √2(2j + 1) parallel to the sides of K , and K and K j ⊂ j+1 j j+1 j j+1 share a corner, (ii) #(K Y ) > 22j + 1. j ∩ We call a sequence (K0, K1,...,Kn) satisfying (i) and (ii) a sequence of nested squares.
Proof of Theorem 2. We argue by contradiction, so suppose that H ( )p is identifiable, but +( ) > L D L 1 . Define γ = √2(1 + 2 n) and R = 2(2n + 1), for n N. It then follows by Lemma 10 that, 2 n − n ∈ for every n N, there exists a µ H ( )p such that supp(µ ) is not R -uniformly close to Ω . ∈ n ∈ L n n γn Indeed, if this were not the case for some n N, we would have +( ) 6 γ 2 < 1 , contradicting ∈ D L n− 2 our assumption that +( ) > 1e. Fix such a µ for each n. e D L 2 n Now, for a fixed n N, define the sets ∈ e S = √2(2n + 1)k, √2(2n + 1)(k + 1) √2(2n + 1)ℓ, √2(2n + 1)(ℓ + 1) R2, k,ℓ × ⊂ h h for (k,ℓ) Z2, forming a partition of the plane into squares of side length √2(2n + 1). As every ∈ 2n Sk,ℓ consists of exactly 2 fundamental cells of the lattice Ωγn and the diagonal of Sk,ℓ has length R , there must exist a pair (k ,ℓ ) Z2 such that #(S supp(µ )) > 22n + 1, for otherwise n n n ∈ kn,ℓn ∩ n supp(µn) would be Rn-uniformly close to Ωγn , contradicting our choice of µn. n n e We set Kn = Skn,ℓn and apply Lemma 19 with Kn and Y = supp(µn) to obtain a sequence (Kn, Ke n,..., Kn) of nested squares. Next, let λ be the center of Kn and notee that, as is shift- 0 1 n n 0 L e e p e invariant by assumption, there exists a measure µn H ( ) with support supp(µn) λn. Therefore, e e e ∈ L e − setting Kn = Kn λ , we have that (Kn, Kn,...,Kn) is a sequence of nested squares, #(Kn j j − n 0 1 n j ∩ 2j n N e supp(µn)) > 2 + 1, and K0 = √2, √2 √2, √2 , for all n and j 0, 1,...n . e − × − ∈ ∈ { } We next need to verify the following auxiliary claim. r r nj nk Claim: Let N and suppose that r N is a subsequence of N such that , r µnk k µn n Kj = Kj ∈ { } ∈ { } ∈ for all j 1, 2, . . . , r and all k > j. Then there exists a further subsequence µ r+1 N such that nk k r+1 ∈ { r+1 } { } ∈ nr+1 nk Kr+1 = Kr+1 , for all k > r + 1. r nr r+1 Proof of Claim: Let K be the set of squares K′ Kr of side length √2(2 + 1) such that r ⊂ r r r r nr nk nk nk nk K′ and Kr have parallel sides and share a corner. As (K0 , K1 ,...,Kr , Kr+1) is a sequence of nr k K K nested squares, for all k > r + 1, we have that Kr+1 , for all k > r + 1. But # = 4, and ∈ nr K k therefore at least one element of appears infinitely often in the sequence Kr+1 k>r+1. We can r+1 { r+1 } nr+1 nk therefore extract a subsequence r+1 N of r N such that , for all , µn k µnk k Kr+1 = Kr+1 k > r +1 { k } ∈ { } ∈ establishing the claim. Now, as Kn = K0, for all n N, we can apply a diagonalization argument together with the Claim 0 0 ∈ nj n N N k N to construct a subsequence µnk k of µn n such that Kj = Kj for all j and all k > j. { } ∈ { } ∈ ∈ H p Next, as ( ) is s-regular, we have infΛ sep(Λ) > s > 0, and so supk N rel(supp(µnk )) . L ∈L ∈ 1 s− < . Therefore, by passing to a further subsequence of µnk k N if necessary, Lemma 18 implies ∞ w { } ∈ the existence of a set Λ R2 such that supp(µ ) Λ as k . Then, as #(Knj supp(µ )) > ∗ ⊂ nk −→ ∗ →∞ j ∩ nj 22j + 1, we have
+ j 2 nj 2j n Λ∗, [0, √2(2 + 1) + 2] > # Λ∗ (K + B (0)) > 2 + 1, ∩ j 1 19 for all sufficiently large j N, and so ∈ 2j + 2 + 1 1 D (Λ∗) > lim sup 2 = . (26) j √ j 2 →∞ 2(2 + 1) + 2 Now, as is CSI, we have Λ and Λ + ( s , 0) . Moreover, as sep(supp(µ )) > s, for all L ∗ ∈L ∗ 2 ∈L nk k N, we have sep(Λ ) > s, and so ms(Λ , Λ + ( s , 0)) > s . Therefore, as we assumed H ( )p to ∈ ∗ ∗ ∗ 2 2 L be identifiable, there exist C ,C > 0 depending on p and and a probing signal x M 1(R) such 1 2 L ∈ that s C1 1 ν p 6 νλπ(λ)x 6 C2 ν p , 2 ∧ k k k k λ Λ M p(R) X∈ for all ν M p supported on Λ := Λ Λ + ( s , 0) , and so by Theorems 17 and 16 we must have ∈ ∗ ∪ ∗ 2 + D (Λ) < 1. On the other hand, (26) implies
+ + D (Λ) = 2D (Λ∗) > 1, which stands in contradiction to D+(Λ) < 1. Our inital assumption must hence be false, concluding the proof of the theorem.
VI. PROOFSOF THEOREMS 3 AND 4 We start with the following proposition that quantifies the behavior of Riesz sums of time-frequency shifts of the probing signal x under perturbation of the individual time-frequency shifts. We do so under a mild condition on the time-frequency spread of the probing signal. Concretely, x will be assumed to be an element of the weighted modulation space M 1 (R)= f : f L1 (R) . m { ∈ S′ Vϕ ∈ m } Proposition 3. Let p [1, ), Λ C a separated set, and let α ℓp(Λ). Suppose that x M 1 (R) ∈ ∞ ⊂ ∈ ∈ m with the weight function m(z)=1+ z . Then | | (i) there exists a Φ W (L ,L1) depending only on x such that ∈ ∞ (x π(ǫ)x) (u, v) 6 ǫ Φ(u, v), |Vϕ − | | | for all (u, v) R2 and ǫ C with ǫ 6 1. ∈ ∈ | | (ii)
2πiRe(λ)Im(ελ) αλπ(λ)x αλe π(λ + ελ)x . p,s,x ε ℓ (Λ) α ℓp(Λ), − M p(R) k k ∞ k k λ Λ λ Λ X∈ X∈ ε ε for all = ελ λ Λ ℓ∞(Λ) such that ℓ (Λ) 6 1. { } ∈ ∈ k k ∞ Proof. (i) Fix an ǫ C with ǫ 6 1. We split ∈ | | (x π(ǫ)x) 6 (x x) + (x x) (27) |Vϕ − | |Vϕ − MIm(ǫ) | |VϕMIm(ǫ) −TRe(ǫ) | and bound each term on the right-hand side separately, beginning with the second term. To this end, we first define the auxiliary quantity F (u, v) = ( x)(u, v)e2πiuv . Then, for all (u, v) R2 and Vϕ ∈ τ R, we have ∈ 2πiτv (x x) (u, v)= ( x)(u, v) e− ( x)(u τ, v) = F (u, v) F (u τ, v) |Vϕ −Tτ | | Vϕ − Vϕ − | | − − | 0 = (∂ F )(u + r, v)dr 6 τ sup (∂ F )(u + r, v) . u | | | u | Z τ r [ τ,0] − ∈ −
20
Therefore, for all (u, v) R2, ∈ (x x) (u, v)= (x x) (u, v Im(ǫ)) |VϕMIm(ǫ) −TRe(ǫ) | |Vϕ −TRe(ǫ) | − 6 Re(ǫ) sup (∂uF )(u + r, v Im(ǫ)) | | r [ Re(ǫ),0] | − | ∈ − 6 ǫ sup (∂uF )(u + r1, v + r2) , (28) | |( ·r1,r2) 61 | | k k∞ where we used the assumption ǫ 6 1. Next, recalling the definition of F , we have | | (∂ F )(u, v) = (∂ ( x)(u, v)) e2πiuv + ( x)(u, v) ∂ e2πiuv | u | u Vϕ Vϕ · u = ∂ x, ϕ + 2πiv x, ϕ | uh MvTu i h Mv Tu i| 6 ( x)(u, v) + 2π v ( x)(u, v) , (29) | Vϕ′ | | || Vϕ | where interchanging ∂ with the inner product in the last step is justified due to x L1 (R2) u Vϕ′ ∈ m (which follows from [7, Prop. 12.1.2]). Therefore, (28) and (29) together yield
ϕ Im(ǫ)(x Re(ǫ)x) (u, v) 6 ǫ sup Ψ(u + r1, v + r2), (30) |V M −T | | | ·(r1,r2) 61 k k∞ for all (u, v) R2, where we set ∈ Ψ(u, v) := ( x)(u, v) + 2π ( u + v ) ( x)(u, v) . | Vϕ′ | | | | | | Vϕ | We bound the first term in (27) in a similar manner, this time using another auxiliary quantity, namely G(u, v) = ( x)(u, v)e2πiuv . Then, using the fundamental identity of time-frequency analysis [7, eq. Vϕ (3.10)] b 2πiuv 2 ( g)(u, v)= e− ( bg)(v, u), f , g ′, (u, v) R , Vf Vf − ∈ S ∈ S ∈ and the fact that ϕ = ϕ, we obtain b
(x x) (u, v)= b(x x) (v, u) |Vϕb − MIm(ǫ) | |Vϕ −TIm(ǫ) | − 6 ǫ sup (∂uG)(v + r1, (u + r2)) , (31) | |( ·rb1,r2) 61 |b − | k k∞ and
(∂ G)(v, u) 6 ( x)(v, u) + 2π u ( x)(v, u) (32) | u − | | Vϕ′ − | | || Vϕ − | 6 ( c x)(v, u) + 2π u ( ϕb x)(v, u) (33) | Viϕ′b − | | || V b − | = ( ϕ x)(u, v) + 2π u ( ϕx)(u, v) (34) | V ′ b | | || V b | for all (u, v) R2, where (32) is obtained analogously to (29), in (33) we used ϕ = iϕ , and in (34) ∈ ′ ′ we again used the fundamental identity of time-frequency analysis. b Combining (31) and (34) thus yields
ϕ(x Im(ǫ)x) (u, v) 6 ǫ sup Ψ(u + r1, v + r2), (35) |V − M | | | ·(r1 ,r2) 61 k k∞ 21 and so (30) and (35) together give
ϕ(x π(ǫ)x) (u, v) 6 2 ǫ sup Ψ(u + r1, v + r2). |V − | | | ·(r1,r2) 61 k k∞ Therefore, in order to complete the proof of item (i), it suffices to take
Φ(u, v)= sup 2Ψ(u + r1, v + r2), (r1,r2) 61 k k∞ and show that Φ W (L ,L1). In fact, as ∈ ∞
1 Φ W (L∞,L ) . sup Φ(u + y1, v + y2)dudv 6 2 sup Ψ(u + r1, v + r2)dudv 2 2 k k R (y1,y2) 61 R (r1,r2) 62 Z k k∞ Z k k∞
. Ψ 1 , k kW (L∞,L ) 1 it suffices to establish that Ψ W (L ,L ). To this end, note that x 1 R2 x 1 R (see [7, ∈ ∞ kVϕ′ kL ( ) ≍ k kM ( ) Prop. 11.4.2]), and so by [7, Prop. 12.1.11] we have x W (L ,L1). Next, as ϕ M 1 (R) and Vϕ′ ∈ ∞ ∈ m x M 1 (R), we have by [7, Prop. 12.1.11] that x W (L ,L1 ). Therefore, using u + v . ∈ m Vϕ ∈ ∞ m | | | | 1+ u + iv = m(u + iv), we have | |
1 1 1 Ψ W (L ,L ) . ϕ′ x W (L ,L ) + ϕx W (L ,L ) < , k k ∞ kV k ∞ kV k ∞ m ∞ as desired.
(ii) Recalling the definition of p R , we have k · kM ( )
2πiRe(λ)Im(ελ) αλπ(λ)x αλe π(λ + ελ)x − M p(R) λ Λ λ Λ X∈ X∈ 2πiRe(λ)Im(ελ) = αλ ϕ(π(λ)x) αλe ϕ(π(λ + ελ)x) V − V Lp(R2) λ Λ λ Λ X∈ X∈ = αλ ϕ(π(λ)(x π(ελ)x)) , V − Lp(R2) λ Λ X∈ 2πiRe(λ)Im(ελ) where we used the commutation relation π(λ + ελ) = e− π(λ)π(ελ). Now, by item (i) we have
ϕ(π(λ)(x π(ε )x)) = ϕ(x π(ε )x) ( λ) 6 ε Φ( λ) |V − λ | |V − λ | ·− k kℓ∞(Λ) ·− pointwise, for all λ Λ, and so, by Lemma 15, we find that ∈
αλ ϕ(π(λ)(x π(ελ)x)) 6 αλ ε ℓ (Λ)Φ( λ) . p,s,x ε ℓ (Λ) α ℓp(Λ). V − Lp(R2) | |k k ∞ ·− Lp(R2) k k ∞ k k λ Λ λ Λ X∈ X∈ This establishes (ii) and completes the proof.
p We next show that weak-* convergence of measures µ Ms implies weak-* convergence of the n ∈ measurements x M p(R). Hµn ∈ Proposition 4. Let p (1, ), x M 1 (R) with the weight function m(z)=1+ z , and let ∈ ∞ ∈ m | | p p q µn n N Ms be a sequence converging to some µ Ms in the weak-* topology W (C0,L ). { } ∈ ⊂ ∈ Then x x in the weak-* topology of M p(R). Hµn →Hµ Proof. As p (1, ), M p(R) is reflexive and so its weak and weak-* topologies coincide [7, ∈ ∞ 22
Thm. 11.3.6]. Due to the dual pairing (7), this topology is generated by the linear functionals q R ,y M p(R) M q (R), for y M ( ), and so we have to show that h · i × ∈ q p q p q R lim µn x,y M (R) M (R) = µx,y M (R) M (R), for all y M ( ). (36) n × × →∞hH i hH i ∈ q Now, for y M (R), set fy(λ) := y,π(λ)x M q (R) M p(R), for λ = τ + iν. If we show that fy C0, ∈ h i × ∈ we will then have
p q p q p q µn x,y M (R) M (R) = µn( λ ) π(λ)x,y M (R) M (R) = µn,fy W ( ,L ) W (C ,L ), hH i × { } h i × h i M × 0 λ supp(µn) ∈ X (37)
w∗ since the dual pairing is continuous in its first argument, and so, as µ µ by assumption, (37) n −−→ will imply (36). Therefore, in order to complete the proof it suffices to show that f C , for all y ∈ 0 y M q(R). ∈ To this end, fix an arbitrary y M q(R) and note that then y W (L ,Lq) by [7, Thm. 12.2.1]. ∈ Vx ∈ ∞ On the other hand, H¨older’s inequality yields
fy(λ) = y,π(λ)x M q (R) M p(R) 6 ( ϕy)(u, v) ( ϕπ(λ)x)(u, v) dudv | | h i × R2 | V | | V | Z
6 ( ϕy)(u, v) ( ϕx)(u τ, v ν) dudv R2 | V | | V − − | Z = ( y x( ) )(τ,ν). (38) |Vϕ |∗|Vϕ − · | Next, as x M 1(R), we have x W (L ,L1) by [7, Thm. 12.2.1], which, together with y ∈ Vϕ ∈ ∞ Vϕ ∈ Lq(R2) and (38), implies by [7, Thm. 11.1.5] that f W (L ,Lq). This, in particular, shows that y ∈ ∞ f (z) 0 as z . Consider now arbitrary λ C and ǫ C with ǫ 6 1. Then, by applying y → | |→∞ ∈ ∈ | | item (i) of Proposition 3 with x replaced by π(λ)x, we find a Φ W (L ,L1) W (L ,Lp) λ ∈ ∞ ⊂ ∞ depending only on x and λ such that
(π(λ)x π(ǫ)π(λ)x) 6 ǫ Φ |Vϕ − | | | · λ pointwise. We thus have
2πiRe(ǫ)Im(λ) π(λ)x π(λ + ǫ)x p R = π(λ)x e π(ǫ)π(λ)x p R k − kM ( ) k − kM ( ) 2πiRe(ǫ)Im(λ) = π(λ)x p R 1 e + π(λ)x π(ǫ)π(λ)x p R k kM ( ) − k − kM ( ) 2πiRe(ǫ)Im(λ) 6 π(λ)x p R 1 e + ǫ Φ q R2 , k kM ( ) − | | · k λkL ( )
and so limǫ 0 π(λ + ǫ)x π(λ)x M p(R). Therefore, by the continuity of the dual pairing in the → k → k second argument, we get
fy(λ + ǫ)= y,π(λ + ǫ)x M q (R) M p(R) y,π(λ)x M q (R) M p(R) = fy(λ) as ǫ 0, h i × → h i × → and so, as λ was arbitrary, we deduce that f is continuous. We have hence established that f C , y y ∈ 0 completing the proof.
We are now ready to prove Theorems 3 and 4. In addition to Propositions 3 and 4, we will need the Banach-Alaoglu theorem as well as the inequality (14). 23
H p Proof of Theorem 3. (i) Let µ, µ ( ) be such that µe x = µx, write µ = λ Λ αλδλ, ∈ L H H ∈ µ = e e αeδe, and for R> 0 define λ Λ λ λ P ∈ e s P δ = min λ λ : λ Λ B (0), λ Λ Λ 1, and e R | − | ∈ ∩ R ∈ \ ∧ 2 ∧ Λ = λ nΛ Λ : λ > R + δ , d(λ, Λ) 6 δ o. R { ∈ \ e | | R e e R}
Informally, δR is thee distancee e betweeneΛ and Λ restrictede to the disk BR(0) (not counting the points in Λ Λ), and ΛR is the part of the support of µ which is at least R + δR away from the origin and ∩ e everywhere within δR of Λ. Fix an R> 0, and, for λ ΛR, write λ(λ) for the point of Λ such that e e e ∈ λ λ(λ) 6 δR. Note that this point is unique, as δR 6 s/2. Next, define the measures | − | e e e e e e 1 2πi Im(λ(λ) λ)Re(λ) e e e e µR = α λ e − δλ(λ), and e e λ ΛR X∈ 2 e e e µR = α λ δ λ e e e λ Λ ΛR ∈X\ 1 e and note that µ is the measure obtained by “shifting the support” of the restricted measure µ1e R ΛR 2 onto Λ, and µR is the remaining part of µ. Next, we havee ms(supp(µ µ1 ), supp(µ2 )) > δ . Now, as supp(µ µ1 ) supp(µ)e and − R R R − R ⊂ supp(µ2 ) esupp(µ), we have that µ µe1 and µ2 are elements of H ( )p, and so the identifiability R ⊂ − R R L e e 1 2 e condition (10) can be applied to the measures µ µR and µR, yielding e e e e− 1 2 C1(δR 1) µ µ µ p 6 µx µe1 x µe2 x M p(R) ∧ k − R − Rk kH −H R −He R k e = ex e1 x e2 x p R µ µR µR M ( ) e e kH −H −H k e e e 2πi Im(λ(λ) λ)Re(λ) = α e π(λ)x α e e − π(λ(λ))x λ λ M p(R) e e − e e λXΛR λXΛR ∈ e ∈ e . δ α e p , p,s,x R λ λ ΛR ℓ · k{ } ∈ k where in the last step we used item (ii) of Proposition 3 with ε = λ(λ) λ e e , noting that λ ΛR { − } ∈ λ(λ) λ 6 δR 6 1, for all λ ΛR. Therefore, by dividing both sides by δR, we obtain | − | ∈ e e 1 2 µ µ µ . H p α e p . e e e eR R p p, ( ) ,x λ λ ΛR ℓ k − − k L k{ } ∈ k 1 2 Now, as R> 0 was arbitrary and α e p 0 as R , we deduce that µ µ µ 0 e λ λe ΛR ℓ R R p k{ } ∈ k → →∞ k − − k → as R . Moreover, as (µ µ)1 6 µ µ1 µ2 , we obtain (µ µ)1 0 → ∞ k − BR(0)kp k − R − Rkp k − BR(0)kp → as R , which implies µ = µ and hence completes the proof of (i). e e →∞ (ii) The “if” direction follows immediatelye by Propositione 4.e To show the “only if” direction,e suppose that x x in thee weak-* topology of M p(R). It then suffices to establish that every Hµn → Hµ subsequence of µn n N has a further subsequence that converges to µ in the weak-* topology { } ∈ M p of s . To this end, fix an arbitrary subsequence µnk k N of µn n N and let Λ = supp(µ) and { } ∈ { } ∈ 2 Λk = supp(µnk ). We then have lim supn rel(Λk) . s− < , and so by [22, Lem. 4.5], Λk k N ∞ { } ∈ has a subsequence Λkℓ ℓ N that converges weakly to a relatively separated set Λ. Note that, as { } ∈ sep(Λk) > s, for all k N, we also have sep(Λ) > s. Now, using (14) and the identifiability ∈ e e 24 condition (10), we have
sup µ p . sup µ . H p sup x p R < , nkℓ W ( ,L ) p,s nkℓ p p, ( ) ,x µnk M ( ) k N k k M k N k k L k N kH ℓ k ∞ ∈ ∈ ∈ and therefore, by the Banach-Alaoglu theorem [31, Thm. 3.15], µnk ℓ N has a subsequence, which ℓ ∈ { } w∗ we will w.l.o.g. also denote by µ N to lighten notation, such that µ µ, for some µ nkℓ ℓ nkℓ { } ∈ w −−→ ∈ W ( ,Lp). Note that then supp(µ) Λ as Λ Λ, and so µ H ( )p. It remains to show that M ⊂ kℓ −→ ∈ L µ = µ. To this end, note that by Proposition 4 we have x ex in the weak-*e topologye of Hµn → Hµ p e e M (R), and so by uniqueness ofe weak-* limits we deduce thate ex = x. From this it follows Hµ Hµ bye item (i) of the theorem that µ = µ, which finishes the proof.
Proof of Theorem 4. Fix ǫ> 0 and an arbitrary finite subset Λ Λ such that µ µ <ǫ, where e ⊂ k − kp 1e e e p we set µ = µ Λ, and write µ = λ Λ αλδλ. Then µx µx M (R) (n) 2πi Re(λ)Im(ελ)e νn = αλ e− δλ. λ Λ˜ X∈ e ε e (n) e Then, as Λ is finite, we have limn n ℓ (Λ) = 0 and supn N α ℓp(Λ) < , which together →∞ k k ∞ ∈ k k ∞ with item (ii) of Proposition 3 yields e (n) 2πi Re(λ)Im(ελ) (n) p ε νen x νn x M (R) = α e− π(λ)x α π(λ + n(λ))x λ λ p R kH −H k e − e M ( ) λ Λ λ Λ X∈ X∈ ε (n) . p,s,x n e α p e 0 as n . (39) k kℓ∞(Λ)k kℓ (Λ) → →∞ To bound ρ , we employ the identifiability condition (10) with the measures ρ and µ k nkp n − ν . Concretely, we note that supp(ρ ) supp(µ ) and supp(µ ν ) supp(µ) supp(µ), n n ⊂ n − n ⊂ ⊂ and so ρ and µ ν are indeed elements of H ( )p. Now, ms(supp(µ ν ), supp(ρ )) > es n − n L − n n − eε e e e n(λ) e > s/2, for sufficiently large n, and so, by combining (39), µex µx M p(R) VII. PROOFSOF PROPOSITION 2 AND COROLLARIES 5, 6, AND 7 sep fin Ray Proof of Proposition 2. Let be any of the classes s , , and . It then follows directly L L Ls,N Ls,θ,R from the definitions of these sets that Λ+ z := λ + z : λ Λ , for all Λ and z R2, so { ∈ }∈L ∈L ∈ it remains to verify closure under weak convergence. To this end, let Λn n N be a sequence in { } ∈ L w 2 such that Λn Λ as n , for some Λ R . In all three cases we have that infn N sep(Λn) > s −→ →∞ ⊂ sep ∈ implies sep(Λ) > s, which establishes that s is CSI. L Suppose now that = fin . It then suffices to show that #Λ 6 N. To this end, let Λ be an L Ls,N arbitrary finite subset of Λ, and consider a point λ Λ. Then, for all sufficiently large n, there exists ∈ e a λn Λn such that λ λn < (sep(Λ) s)/2. We deduce that, for all sufficiently large n, there ∈ | − | ∧ e exists a Λn Λn such that #Λn = #Λ. But #Λn 6 N, for all n N, by definition of the class ⊂ e ∈ fin , so we must have #Λ 6 N. Now, as Λ was arbitrary, we must have #Λ 6 N, and so Λ fin . Ls,N ∈Ls,N e fin e e This establishes that s,N is CSI. L e e For = Ray , fix an arbitrary translate K = (x,x + R) (y,y + R) of (0,R)2, and consider L Ls,θ,R x,y◦ × a point λ Λ K . Then, as K is open, we have that for all sufficiently large n there exists ∈ ∩ x,y◦ x,y◦ a λ Λ such that λ λ < s/2 and λ K . Therefore, as λ was arbitrary, and Λ K n ∈ n | − n| n ∈ x,y◦ ∩ x,y◦ is finite, we have #(Λ K ) 6 #(Λ K ) 6 θR2, for sufficiently large n. Thus, as K was ∩ x,y◦ n ∩ x,y◦ x,y◦ arbitrary, we obtain n+ Λ, (0,R)2 6 θR2, and so Λ Ray . This establishes that Ray is CSI ∈ Ls,γ,R Ls,γ,R and thereby completes the proof. Proof of Corollary 5. The proof is effected by verifying the conditions of Theorem 1. As s := fin + fin 1 infΛ fin sep(Λ) > 0, by definition of s,N , it suffices to show that ( s,N ) < 2 . To this end, ∈Ls,N L D L fix a Λ = λ , . . . , λ fin , where n 6 N. Set R = 2√2 N+1 , and for k 1,...,n let { 1 n}∈Ls,N ⌈ 2 ⌉ ∈ { } S = ξ Ω : ξ λ 6 R , where we recall that Ω = 2(m + in) : m,n Z . Now, for every k { ∈ 2 | − k| } 2 { ∈ } k 1,...,n , choose a ξk Sk ξ1,...,ξk 1 . Note that this is possible as #Sk > N, for all ∈ { } ∈ \ { − } k. We thus have λ ξ 6 R for all k, and so, as Λ was arbitrary, we have that Λ is R-uniformly | k − k| close to Ω , for all Λ fin . Lemma 10 therefore implies +( fin ) 6 2 2 < 1 , completing the 2 ∈ Ls,N D Ls,N − 2 proof. 1 Proof of Corollary 6. First assume that s> 2 3− 4 . In view of Theorem 1, it again suffices to verify · + sep 1 that ( s ) < . To do this, we will need a special case of the plane packing inequality by D L 2 Folkman and Graham [32]: Let K be a compact convex set. Then any subset of K whose any two points are at least 1 apart (in the Euclidean metric) has cardinality at most 2 Area(K)+ 1 Per(K) + 1. √3 2 For our purposes K will be a square of side length R, to be specified later. By scaling, we see that then any subset of [0,R]2 whose any two points are at least s apart has cardinality at most 2 R2 2R 2 2 1 1/2 1 2 2 + + 1. Let θ s , , γ (√2,θ ), and R > 0 such that 2(sR) + R 6 √3 s s ∈ √3 − 2 ∈ − − − 2 2 sep θ s . Then, for all Λ s , we have − √3 − ∈L 2 + 2 2 R 2R 2 2 1 2 2 2 n (Λ, (0,R) ) 6 + + 1 6 s− + 2(sR)− + R− R 6 θR , √ s2 s √ 3 3 26 and so, by Lemma 9, there exists an R′ = R′(θ, γ, R) > 0 such that Λ is R′-uniformly close to Ωγ. + sep 2 1 sep p Therefore, as Λ was arbitrary, it follows by Lemma 10 that ( s ) 6 γ < , and so H ( s ) D L − 2 L is identifiable by the probing signal ϕ. 1 1 √3 4 4 1 Now consider the case s 6 2 3− and let r = 2 3− , Λ′ = rm(1, 0) + rn 2 , 2 : m,n · 1 · ∈ sep + √3 2 − 1 + sep + 1 Z s . Then D (Λ )= r = , and so by Lemma 8 we have ( s ) > D (Λ )= . ∈L ′ 2 2 D L ′ 2 sep p It thus follows by Theorem 2 that H ( s ) is not identifiable. L Proof of Corollary 7. Consider first θ < 1 . Fix γ (√2,θ 1/2), and let Λ Ray . Then, by 2 ∈ − ∈ Ls,θ,R Lemma 9, there exists an R′ = R′(θ, γ, R) > 0 such that Λ is R′-uniformly close to Ωγ. Therefore, as Λ was arbitrary, it follows by Lemma 10 that +( Ray ) 6 γ 2 < 1 , and so Theorem 1 implies D Ls,θ,R − 2 that H ( Ray )p is identifiable by the probing signal ϕ. Ls,θ,R 1 For θ > , suppose by way of contradiction that there exists a sequence Rn n N of positive 2 { } ∈ Ray p numbers such that limn Rn = and H ( ) is identifiable, for all n N. Let γn n N be →∞ ∞ Ls,θ,Rn ∈ { } ∈ 1/2 a sequence of positive numbers such that γn >θ− >s and 2 1 1 2 γn− + 4 Rn− γn− + Rn− 6 θ, 1/2 for all n N, and limn γn = θ− . We then have ∈ →∞ + 2 1 2 1 n (Ωγn , (0,Rn) ) 6 Rnγn− + 4 Rnγn− + 1 6 θRn, 2 where the second term takes into account the points of Ωγn along the four sides of the square (0,Rn) . 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Note that is, in fact, the synthesis operator treated in [22], and so items HΛ (i) and (ii) follow immediately from [22, §2.5]. We proceed to establish (iii). To this end, recall that 28 1 πt2 ϕ(t) = 2 4 e , fix an arbitrary (x,ω) Λ, and let Λ be a finite subset of Λ such that (x,ω) Λ. − ∈ ∈ Then, for a, b R and (τ,ν) Λ, we have ∈ ∈ e e ν τ eω b aϕ, x+a bϕ M q (R) M p(R) hM T M− − T T M− i × = ϕ, a ω+b τ ν x+a bϕ L2 (R) h T− M T− M− T M− i 2πi(x+a)b = e− ϕ, a ω+b τ ν b x+aϕ L2(R) h T− M T− M− − T i 2πi(x+a)b 2πiτ(ν+b) = e− e ϕ, a (ω+b)+( ν b) (x+a) τ ϕ L2(R) h T− M − − T − i 2πi(x+a)b 2πiτ(ν+b) 2πia(ω ν) = e− e e− − ϕ, ω ν (x+a) τ aϕ L2(R) h M − T − − i 2πiab 2πiτν 2πia(ω ν) 2πib(x τ) = e− e e− − − − ( ϕ)(x τ,ω ν), Vϕ − − and therefore 2πiab e e Λ (α, ω b aϕ), x+a bϕ M q(R) M p(R) H M− − T T M− × 2πiτν 2πia(ω ν) 2πib(x τ) (41) = ατ,ν e e− − − − ( ϕϕ)(x τ,ω ν). e V − − (τ,ν) Λ X∈ πǫ(a2+b2) Now, let ǫ > 0 be arbitrary. We multiply both sides of (41) by ǫ e− and integrate over (a, b) R2. Then, as the Fourier transform of √ǫϕ( √ǫ) is ϕ( /√ǫ), the integral of the right-hand ∈ · · side equals 2 2 1 2πiτν π[(ω ν) +(x τ) ]ǫ− ατ,νe e− − − ( ϕϕ)(x τ,ω ν), e V − − (τ,ν) Λ X∈ and the integral of the left-hand side satisfies πǫ(a2+b2) 2πiab e ǫe− e Λ (α, ω b aϕ), x+a bϕ M q (R) M p(R) da db R2 H M− − T T M− Z × πǫ(a2+b2) 6 ǫe e (α, ϕ), ϕ da db − Λ ω b a x+a b M q (R) M p(R) R2 H M− − T T M− × Z πǫ(a2+b2) e 1 q 6 ǫe− Λ(α, ) ω b aϕ M (R) x+a bϕ M (R ) da db R2 kH · kkM− − T k kT M− k Z 6 e (α, ) ϕ 1 R ϕ q R . kHΛ · kk kM ( )k kM ( ) We hence deduce that 2 2 1 2πiτν π[(ω ν) +(x τ) ]ǫ− α e e− − − ( ϕ)(x τ,ω ν) 6 e (α, ) ϕ 1 R ϕ q R , τ,ν Vϕ − − kHΛ · kk kM ( )k kM ( ) e (τ,νX) Λ ∈ and so, upon letting ǫ 0, we obtain → 2πixω α e ( ϕ)(0, 0) 6 e (α, ) ϕ 1 R ϕ q R . x,ω Vϕ kHΛ · kk kM ( )k kM ( ) e Next, note that we can write Λe = Λ Λ Λ, and so, by item (i) of the proposition, there exists H H −H \ a universal constant C > 0 such that e e p Λ(α, ) 6 Λ(α, ) + C αλ λ Λ Λ ℓ . kH · k kH · k k{ } ∈ \ k 2 Therefore, since ( ϕ)(0, 0) = ϕ 2 = 1, we have Vϕ k kL (R) e p 1 q αx,ω 6 Λ(α, ) + C αλ λ Λ Λ ℓ ϕ M (R) ϕ M (R). | | kH · k k{ } ∈ \ k k k k k 29 e p The term αλ λ Λ Λ ℓ can now be made arbitrarily small by choosing a sufficiently large Λ, and k{ } ∈ \ k hence we deduce that e α 6 (α, ) ϕ 1 R ϕ q R . | x,ω| kHΛ · kk kM ( )k kM ( ) As (x,ω) Λ was arbitrary, we obtain ∈ 1 q α ℓ∞ = sup αx,ω 6 Λ(α, ) ϕ M (R) ϕ M (R), k k (x,ω) Λ | | kH · kk k k k ∈ establishing (9). Proof of Lemma 8. Item (i) is a direct consequence of Definition 2 and the fact that n+(Λ, (0,R)2) 6 n+(Λ, [0,R]2). To show item (ii), let ǫ> 0 be arbitrary, and set θ = +( )+ǫ. Then, by Definition D L 2, we have n+(Λ, [0,R]2) 6 +( )+ ǫ R2 D L for every Λ and sufficiently large R, and thus ∈L + 2 + n (Λ, [0,R] ) + D (Λ) = lim sup 2 6 ( )+ ǫ. (42) R R D L →∞ + + Now, as (42) holds for every Λ , we deduce that supΛ D (Λ) 6 ( )+ ǫ, and hence, as ∈ L ∈L D L + + ǫ> 0 was arbitrary, we obtain supΛ D (Λ) 6 ( ). ∈L D L Proof of Lemma 9. We identify C with R2 for ease of exposition. For a positive integer q, define the set according to Sq = [qkR,q(k + 1)R] [qℓR,q(ℓ + 1)R] : k,ℓ Z , Sq { × ∈ } and note that is a collection of squares in R2 of side length qR tessellating the plane. A simple Sq counting argument now yields, for all K , ∈ Sq 1 2 1 #(Ω K) > (qRγ− ) 4 qRγ− + 1 , γ ∩ − where the subtracted term accounts for the qRγ 1 points adjacent to each of the edges of the square ⌈ − ⌉ K, but which might not be inside it. On the other hand, as every element of can be covered by Sq (q + 1)2 translates of (0,R)2, using n+(Λ, (0,R)2) 6 θR2 we have #(Λ K) 6 (q + 1)2 θR2 = ∩ · ((q + 1)Rθ1/2)2 for all K . Therefore, as γ 1 > θ1/2 by assumption, there exists a positive ∈ Sq − integer q′ = q′(θ, R, γ) such that 1 2 1 1/2 2 (q′Rγ− ) 4 q′Rγ− + 1 > (q′ + 1)Rθ . − We thus have #(Ω K) > #(Λ K), for all K , and can therefore enumerate Λ = γ ∩ ∩ ∈ Sq′ λm,n (m,n) so that, for every (m,n) , λm,n and ωm,n are contained in the same square { } ∈I ∈ I K . Setting R = √2q R to be the length of the diagonal of the squares in now yields m,n ∈ Sq′ ′ ′ Sq′ λ ω 6 R , for all (m,n) , as desired. | m,n − m,n| ′ ∈I 2 + 2 Proof of Lemma 10. We again identify C with R for ease of exposition. Note that we have n (Λ, [0,R′] ) 6 30 n+(Ω , [0,R + 2R]2), for all R > 0 and Λ , by the uniform closeness assumption, and so γ ′ ′ ∈L + 2 + 2 2 + n (Λ, [0,R′] ) n (Ωγ, [0,R′ + 2R] ) 2R 2 ( ) = limsup sup 2 6 lim sup 2 1+ = γ− . D L R′ Λ R′ R′ (R′ + 2R) R′ →∞ ∈L →∞ Proof of Lemma 12. Note that the terms z λ and z in the defining expressions of g and g − 0,0 Λ Λ cancel owing to λ0,0 = 0, so the interpolation property (a) follows immediately. We may hence proceed to establishing statement (b). To this end, we begin by observing that the assumptions (i),e (ii), and (iii) remain valid and the conclusion of the lemma unchanged if we replace ρ by ρ 1 and R ∧ by R 1 (3γ), so we may assume w.l.o.g. that ρ 6 1 and R > 1 (3γ). Now, set := (0, 0) ∨ ∨ ∨ I′ I \ { } and write π 2 2 γ− z π 2 2 2 γ− z σγ(z)e− | | d(z, Λ) gΛ(z)e− 2 | | = h(z), (43) d(z, Ωγ ) z where d is the Euclideane distance from a point to a subset of C and d(z, Ω ) z z (1 z/λ ) exp(z/λ ) h(z)= γ exp − m,n m,n . d(z, Λ) ωm,n − λm,n (1 z/ωm,n) exp(z/ωm,n) (m,n) s (m,n) ′ Y∈I Y∈I − π 2 2 γ− z Note that d(z, Λ)/z 6 1 owing to 0 Λ, and by Lemma 11 we have σ (z) e− 2 | | /d(z, Ω ) . | | ∈ | γ | γ γ 1, so in order to complete the proof it suffices to establish 1 c z log z h(z) . ρ− e | | | |, z C, (44) | | s,θ,γ,R ∀ ∈ for some c = c(s, θ, γ, R) > 0. This will be effected by bounding various auxiliary quantities associated with h. To this end, we begin by bounding (1 z/λm,n) exp(z/λm,n) haux(z) := − , (1 z/ωm,n) exp(z/ωm,n) (m,n) A(z) Y∈ − from above, where A(z) is given by ⊂I A(z)= (m,n) ′ : λ > 2 z 6R . ∈I | m,n| | |∨ To this end, we first establish the following basic bounds valid for all z C and (m,n) A(z): ∈ ∈ (A1) ω > 5R, ω / λ (5/6, 5/4), | m,n| | m,n| | m,n|∈ 4 1 4 1 (A2) z/ωm,n < 5 z 64R + 7 z >4R , and | | {| | } {| | } (A3) z/λ < 1/2. | m,n| The inequality in (A1) follows from ω > λ R> 6R R = 5R, and the upper and lower | m,n| | m,n|− − bounds on ω / λ are due to ω > λ R > 5 λ and λ > ω R > | m,n| | m,n| | m,n| | m,n|− 6 | m,n| | m,n| | m,n|− 4 ω . To show (A2), consider the cases z 6 4R and z > 4R separately. If z 6 4R, then 5 | m,n| | | | | | | z/ω < 4R/5R = 4/5, whereas if z > 4R, then ω > λ R> 2 z R> 7 z and so | m,n| | | | m,n| | m,n|− | |− 4 | | z/ω < 4/7. Finally, (A3) follows directly by the definition of A(z). | m,n| 31 Now, using (A2), we have 1 z/λm,n exp(z/λm,n) haux(z) 6 | − || | | | 1 z/ωm,n exp(z/ωm,n) (m,n) A(z) Y∈ | − || | 1 1 1+ e 2 1 z/λ exp(z/λ ) 6 1 + 1 2 | − m,n|| m,n | z >4R z 64R 4 4 {| | } {| | } 1 e 5 1 z/ωm,n exp(z/ωm,n) (m,n) A(z) 5 − ! (m,n) A(z) | − || | Y∈7 − Y∈ 7 ωm,n 6 z 67R ωm,n > z | | 4 | | | | 4 | | (1 z/λm,n) exp(z/λm,n) . γ,R − , z C, (45) e (1 z/ωm,n) exp(z/ωm,n) ∀ ∈ (m,n) A(z) − Y∈ where we set A(z)= (m,n) A(z) : ω > 7 z . { ∈ | m,n| 4 | |} Next, for z C and (m,n) eA(z), it follows by (A2) that 1 z/ω and 1 z/z lie in the ∈ ∈ − m,n − m,n domain C R6 of the complex logarithm, which we denote by Log. We can thus write \ 0 (1 z/λ ) exp(z/λ ) − m,n m,n e (1 z/ωm,n) exp(z/ωm,n) (m,n) A(z) − Y∈ z z z z = exp Log 1 + Log 1 e − λm,n λm,n − − ωm,n − ωm,n (m,n) A(z) Y∈ k k ∞ 1 z ∞ 1 z = exp + e − k λm,n k ωm,n (m,n) A(z) " k=2 k=2 # Y∈ X X ∞ zk 1 1 = exp k k , e k ωm,n − λm,n " (m,n) A(z) k=2 # X∈ X which together with (45) yields ∞ z k 1 1 h (z) . exp | | , z C. (46) | aux | γ,R k ωk − λk ∀ ∈ " e k=2 m,n m,n # (m,nX) A(z) X ∈ Using (A1), we further have k 1 k 1 ℓ ℓ 1 1 λm,n ωm,n ℓ=0− λm,n − − ωm,n k k = | − | k | |k | | ωm,n − λm,n ω λ | Pm,n| | m,n| k 1 k 1 6 ℓ R ω − − 6 · | m,n| ℓ=0 5 2k 4 k ωm,n | | P5 k (k+1) 6 5R 1.5 ωm,n − , (47) · | | 7 z for all z C, (m,n) A(z), and all k N. Next, defining c (z, R) := | | 5R, (A1) and (A2) ∈ ∈ ∈ aux 4 ∨ imply that A(z) (m,n) Z2 : ω > c (z, R) , for all z C, and so (46) and (47) together ⊂ { ∈ | m,n| aux } ∈ e 32 yield k ∞ 5R z k (k+1) haux(z) 6 exp | | 1.5 ωm,n − | | e k · | | " (m,n) A(z) k=2 # X∈ X k ∞ 5R z k (k+1) 6 exp | | 1.5 ω − k · | m,n| " (m,n) Z2 k=2 # X∈ X ωm,n >c (z,R) | | aux k ∞ 5R z k (k+1) = exp | | 1.5 ω − . (48) k · · | m,n| " k=2 (m,n) Z2 # X X∈ ωm,n >c (z,R) | | aux 7 z 7 z 20γ Recall that R > 3γ, and so c (z, R)= | | 5R > | | implies aux 4 ∨ 4 ∨ √2 c (z, R) γ/√2 > 0.95 c (z, R) > 1.6 z . (49) aux − aux | | Now, write Kγ for the square of side length γ centered at (0, 0) and use R > 3γ and (49) to obtain the following bound γ k+1 ωm,n + (k+1) √2 (k+1) ω − 6 | | w − dw (50) m,n ω | | m,n ! (mγ,nγ)+Kγ | | | | ωm,n >caux(z,R) ωm,n >caux(z,R) Z | |X | |X | | k+1 γ (k+1) 6 1+ w − dw 5R√2 w >c (z,R) γ | | | | Z| | aux − √2 k+1 1 k 1 2π − 6 1+ caux(z, R) γ/√2 15√2 k 1 − − k+1 1 k 6 1.04 2π (1.6 z ) − · | | k 1 k < 11 0.65 z − , (51) · | | for all z C 0 and k > 2. We now use (51) in (48) to obtain ∈ \ { } k ∞ 5R z k k 1 k haux(z) . γ,R exp | | 1.5 11 0.65 z − | | " k · · · | | # Xk=2 55 ∞ 6 R z 0.975k " | | · 2 # Xk=2 1100R z = e | |, z C. (52) ∀ ∈ We are now ready to bound h, and we do so by treating the cases z > 3R and z 6 3R separately. | | | | 33 Case z > 3R : We analyze h(z) as a product h(z)= 5 h (z), where | | k=1 k 1 1 Q d(z, Ωγ ) 1 z/λm,n h1(z) = exp z , h2(z)= − , " λm,n − ωm,n # d(z, Λ) 1 z/ωm,n (m,n) ′ s (m,n) ′ X∈I \I Y∈I − λm,n 62 z λm,n z 62R | | | | | − | 1 z/λm,n 1 z/λm,n h3(z)= − , h4(z)= − , and 1 z/ωm,n 1 z/ωm,n (m,n) ′ (m,n) ′ Y∈I − Y∈I − λm,n z >2R λm,n z >2R | − | | − | ωm,n 62R ωm,n >2R | | | | λm,n 62 z | | | | (1 z/λm,n) exp(z/λm,n) h5(z)= − , (1 z/ωm,n) exp(z/ωm,n) (m,n) ′ Y∈I − λm,n >2 z | | | | and bound the functions hj in order. Bounding h : Note that λ > 3R implies ω > 2R, and hence λ > ω R> 1 ω . 1 | m,n| | m,n| | m,n| | m,n|− 2 | m,n| We thus have the following bound for (m,n) : ∈I′ \Is 1 1 1 1 1 1 λm,n ωm,n 6 λm,n 63R (2s− + γ− )+ λm,n >3R | − | λ − ω {| | } {| | } ω λ m,n m,n | m,n|| m,n| 1 1 1 2 6 λm,n 63R (2s− + γ− ) + 2R ωm,n − , {| | } | | and therefore, as # (m,n) : λ 6 3R 6 9θR2, { ∈I′ | m,n| } 1 1 h1(z) 6 exp z , | | | | zm,n − ωm,n " (m,n) ′ # X∈I zm,n 62 z | | | | 2 1 1 2 6 exp 9θR (2s− + γ− ) z + 2R z ω − . (53) | | | | · | m,n| " (m,n) ′ # X∈I ωm,n 62 z +R | | | | Recalling that log z > log(3R) > log(3) > 0, we can bound in a manner similar to (51) to obtain | | γ + γ 2 2 √2 2 ωm,n − 6 w − dw | | γ γ γ | | | | ! √ 6 w 62 z +R+ √ (m,n) ′ Z 2 2 X∈I | | | | ωm,n 62 z +R | | | | . log z . (54) γ,R | | Using this in (53) thus yields c z log z h (z) 6 e 1| | | |, (55) | 1 | for some c = c (s, θ, γ, R) > 0 and all z C such that z > 3R. 1 1 ∈ | | Bounding h : We write h = hnum/hden, where 2 | 2| 2 2 num 1 λm,n z den 1 ωm,n z h2 (z)= | − |, and h2 (z)= | − |. d(z, Λ) λm,n d(z, Ωγ ) ωm,n (m,n) ′ (m,n) ′ Y∈I | | Y∈I | | λm,n z 62R λm,n z 62R | − | | − | num In order to bound h2 , we observe that one of the following two circumstances occurs: – The distance from z to Λ is minimized at λ , and so d(z, Λ) = z > 3R> 1. 0,0 | | 34 – The distance from z to Λ is minimized at a point λ , where (m,n) , and so the term d(z, Λ) m,n ∈I′ cancels with one of the factors λ z in the product over (m,n) : λ z 6 2R . | m,n − | { ∈I′ | m,n − | } These facts lead to the following bound num λm,n z 1 16θR2 h2 (z) 6 | − |∨ 6 2 . (56) z λm,n z (m,n) ′ Y∈I | | − | − | λm,n z 62R | − | For hden, we similarly observe that either (Re z, Im z) [ γ , γ ]2, or d(z, Ω ) cancels with a 2 ∈ − 2 2 γ factor ω z in the product over (m,n) : λ z 6 2R . In either case the numerators | m,n − | { ∈I′ | m,n − | } of the terms remaining in the product satisfy ω z > γ , and we thus have | m,n − | 2 16θR2 den γ (γ/2) 1 γ (γ/2) 1 h2 (z) > 1 ∧ > 1 ∧ . √2 ∧ ωm,n λm,n + λm,n z + z √2 ∧ 3R + z (m,n) ′ Y∈I | − | | − | | | | | λm,n z 62R | − | (57) The inequalities (56) and (57) together yield 16θR2 z h (z) . z . e| |. (58) | 2 | θ,γ,R | | θ,R Bounding h : Recall that λ > s for all (m,n) , and there is at most one (m ,n ) 3 | m,n| 2 ∈ I′ \Is ′ ′ ∈ (0, 0) , and for this (m ,n ) we have λ > ρ, by assumption (i). We thus get Is \ { } ′ ′ | m′,n′ | 1 z/λm,n ωm,n λm,n z h3(z) = | − | 6 | | | − | | | 1 z/ωm,n λm,n λm,n z R (m,n) ′ (m,n) ′ Y∈I | − | Y∈I | | | − |− λm,n z >2R λm,n z >2R | − | | − | ωm,n 62R ωm,n 62R | | | | 2 2 16γ− R 1 2R 2R 1 4R 6 ρ− 6 ρ− (s/2) 1 2R R (s/2) 1 ωm,n 62R | Y| ∧ − ∧ 1 . s,γ,R ρ− . (59) Bounding h : We write h = hnum/hden, where 4 | 4| 4 4 num ωm,n λm,n den ωm,n λm,n h4 (z)= 1 − , and h4 (z)= 1 − . − ωm,n z − ωm,n (m,n) ′ (m,n) ′ Y∈I − Y∈I λm,n z >2R λm,n z >2R | − | | − | ωm,n >2R ωm,n >2R | | | | λm,n 62 z λm,n 62 z | | | | | | | | Now, fix a z C with z > 3R and write z = z + ω , where (k,ℓ) Z2 and (Re z , Im z ) ∈ | | ′ k,ℓ ∈ ′ ′ ∈ [ γ , γ ]2. Then, as − 2 2 2 (m,n) ′ : λ z >2R, λ 6 2 z (m,n) Z : ω z >R, ω z 6 3 z +R , { ∈I | m,n− | | m,n| | |} ⊂ { ∈ | m,n− | | m,n− | | | } 35 and Ω ω =Ω , we have the following bound γ − k,ℓ γ num ωm,n λm,n R h4 (z) 6 1+ | − | 6 1+ | | ωm,n z 2 ωm,n z (m,n) ′ (m,n) Z Y∈I | − | Y∈ | − | λm,n z >2R R< ωm,n z 63 z +R | − | | − | | | ωm,n >2R | | λm,n 62 z | | | | R = 1+ ωm,n ωk,ℓ z (m,n) Z2 ′ Y∈ | − − | R< ωm,n ωk,ℓ z′ 63 z +R | − − | | | R 6 exp , (60) ωm,n z " (m,n) Z2 ′ # X∈ | − | R< ωm,n z′ 63 z +R | − | | | where in the last inequality we used log(1 + x) 6 x for x > 0. Now, ω z > R and R > 3γ | m,n − ′| imply ω > R γ > √2γ and so ω z > ω γ > 1 ω . We therefore have | m,n| − √2 | m,n − ′| | m,n|− √2 2 | m,n| 1 2 6 ωm,n z′ γ γ ωm,n R< ωm,n z 63 z +R R < ωm,n 63 z +R+ | X− ′| | | | − | − √2 | X| | | √2 | | γ ωm,n + √2 1 6 2| | w − dw γ γ ωm,n (mγ,nγ)+Kγ | | | | R < ωm,n 63 z +R+ Z − √2 | X| | | √2 | | R . w 1 dw γ,R √ − R 2 R √2γ< ω 63 z +R+√2γ | | | | − 2 Z − | | | | . z . (61) γ,R | | As z was arbitrary, (61) holds for all z C with z > 3R. Using (61) in (60) thus yields ∈ | | num cnum z h (z) 6 e 4 | | (62) | 4 | for some cnum = cnum(γ, R) > 0 and all z C with z > 3R. 4 4 ∈ | | den The quantity h4 is bounded from below in a similar fashion: den ωm,n λm,n R h4 (z) = 1 − > 1 , | | − ωm,n 2 − ωm,n (m,n) ′ (m,n) Z Y∈I Y∈ | | λm,n z >2R 2R< ωm,n 62 z +R | − | | | | | ωm,n >2R | | λm,n 62 z | | | | 2R > exp , (63) − ωm,n " 2R< ωm,n 62 z +R # | X| | | | | where in the last inequality we used log(1 + x) > 2x for x 1 , 0 . Another integral bound yields ∈ − 2 1 . γ,R z , ωm,n | | 2R< ωm,n 62 z +R | X| | | | | which together with (63) gives den cden z h (z) > e− 4 | |, (64) | 4 | 36 for some cden = cden(γ, R) > 0 and all z C with z > 3R. Combining (62) and (64) thus yields 4 4 ∈ | | (cnum+cden) z h (z) 6 e 4 4 | |, (65) | 4 | for all z C with z > 3R. ∈ | | Bounding h : Note that λ > 2 z implies λ > 2 z > 6R, and so (m,n) : λ > 5 | m,n| | | | m,n| | | { ∈ I′ | m,n| 2 z = A(z). We thus have h = h , which satisfies (52). | |} 5 aux Bounding h: We combine (55), (58), (59), (65), and (52) to obtain 1 (1+cnum+cden+1100R) z +c z log z 1 d z log z h(z) . ρ− e 4 4 | | 1| | | | 6 ρ− e | | | |, (66) | | s,θ,γ,R for some d = d(s, θ, γ, R) > 0 and all z C with z > 3R. This completes the derivation of the ∈ | | desired upper bound on h in the case z > 3R. | | Case z 6 3R: We write h(z)= h (z)h (z), where | | 6 7 d(z, Ωγ ) 1 z/λm,n (1 z/λm,n) exp(z/λm,n) h6(z)= − , and h7(z)= − . d(z, Λ) 1 z/ωm,n (1 z/ωm,n) exp(z/ωm,n) (m,n) ′ (m,n) ′ Y∈I − Y∈I − λm,n 66R λm,n >6R | | | | Note that λ > 6R and z 6 3R together imply λ > 2 z , and so h = h . Hence it only | m,n| | | | m,n| | | 7 aux remains to bound h . To this end, write h = hnum/hden, where 6 | 6| 6 6 num z λm,n z den z ωm,n z h6 (z)= − and h6 (z)= − . d(z, Λ) λm,n d(z, Ωγ ) ωm,n (m,n) ′ (m,n) ′ Y∈I Y∈I λm,n 66R λm,n 66R | | | | Now, the term d(z, Λ) cancels with either z or one of the factors λ z, and similarly, d(z, Ω ) m,n − γ cancels with either z or one of the factors ω z. In either case the numerators of the terms remaining m,n− in the product satisfy ω z > γ . We again recall that λ > s for all (m,n) , and | m,n − | 2 | m,n| 2 ∈I′ \Is there is at most one (m ,n ) (0, 0) , and for this (m ,n ) we have λ > ρ. These ′ ′ ∈ Is \ { } ′ ′ | m′,n′ | observations together yield the following bounds: 144θR2 num 1 ( λm,n + z ) 1 1 (9R) 1 h (z) 6 3Rρ− | | | | ∧ 6 3Rρ− ∧ | 6 | (s/2) 1 (s/2) 1 (m,n) ′ Y∈I ∨ ∨ λm,n 66R | | 144θR2 den (γ/2) 1 (γ/2) 1 C h6 (z) > ∧ > ∧ , for z s.t. z 6 3R. | | ωm,n λm,n + λm,n 7R ∈ | | (m,n) ′ Y∈I | − | | | λm,n 66R | | Therefore h (z) . ρ 1, for z with z 6 3R, which together with (52) gives | 6 | s,θ,γ,R − | | 1 1100R h(z) . ρ− e z , for z C s.t. z 6 3R. (67) | | s,θ,γ,R | | ∈ | | The inequalities (66) and (67) can now be combined to yield the bound (44), concluding the proof. Proof of Lemma 13. Consider first the case when y (R) is a Schwartz function. We then have ∈ S 37 y L2(R), and thus ∈ y,π(λ)ϕ M q (R) M p(R) = ϕy, ϕπ(λ)ϕ Lq (R2) Lp(R2) h i × hV V i × = ( ϕy)(s,ξ) ( ϕπ(λ)ϕ)(s,ξ)dsdξ R2 V V ZZ = y(t)(π(λ)ϕ) (t)dt (68) R Z = ( y)(λ) Vϕ πiτν π λ 2/2 = e− e− | | (By)(λ¯), (69) where (68) follows from [7, Thm. 3.2.1], and (69) is [7, Prop. 3.4.1]. Now take an arbitrary y M q(R). As (R) is dense in M q(R) for q [1, ) (see [7, Prop. ∈ S ∈ ∞ 11.3.4]), we can take a sequence y (R) such that y y in M q(R). The calculation { n}n∞=1 ⊂ S n → above thus shows that πiτν π λ 2/2 B ¯ yn,π(λ)ϕ M q (R) M p(R) = e− e− | | ( yn)(λ), n N. (70) h i × ∀ ∈ Furthermore, as the dual pairing is continuous, we have yn,π(λ)ϕ M q (R) M p(R) y,π(λ)ϕ M q (R) M p(R) as n . h i × → h i × →∞ B B On the other hand, by the isometry property (18) we also have yn y q (C) = yn y M q (R) 0 k − kF k − k → as n . Thus, as the evaluation functional F F (λ) is continuous on q(C) (see [28, Lem. → ∞ 7→ F 2.32]), we obtain (Bf )(λ) (Bf)(λ), which together with (70) and (69) establishes the claim of n → the proposition. Proof of Lemma 14. (i) Suppose that is bounded below. Then the operator ˜ : X Im( ) given A A → A by ˜(x) = (x), for x X, is a continuous map between Banach spaces, and has a continuous A A ∈ inverse. In other words, ˜ is an isomorphism between Banach spaces. Thus ˜ : (Im( )) X A A∗ A ∗ → ∗ is also an isomorphism between Banach spaces, and so, by the inverse mapping theorem [31, Cor. 2.12], so is ( ˜ ) 1 : X (Im( )) . Consider now an arbitrary f X , and set h = ( ˜ ) 1f. A∗ − ∗ → A ∗ ∈ ∗ A∗ − As h is a continuous linear functional on Im( ) Y , it follows by the Hahn-Banach theorem [31, A ⊂ Thm. 3.6] that h can be extended to a continuous linear functional hY defined on Y . Now, since hY Im( )= h, we have | A ∗h ,x = h , x = h, x = h, ˜x = hA Y i h Y °A i h A i h A i Im( ) ∈ A = ˜∗h, x = f,x for x X, hA i h i ∈ and thus, as x X was arbitrary, we deduce that h = f. Finally, since f was arbitrary, we have ∈ A∗ Y that is surjective. A∗ (ii) Let f be an arbitrary element of X with f = 1, and let g Y be such that g = f ∗ k k ∈ ∗ A∗ and a g 6 1. Note that then g = 0, and so g/ g is a well-defined element of Y of unit norm. k k 6 k k ∗ Therefore, g 1 x > x, = g − x, ∗g > a x,f , kA k A g k k |h A i| |h i| k k 38 for all x X. Taking the supremum of the right-hand side over f X and using the fact that ∈ ∈ ∗ supf X , f =1 x,f = x yields x > a x , as desired. ∈ ∗ k k |h i| k k kA k k k Proof of Lemma 15. For m,n Z write ∈ K = s m 1 , s m + 1 s n 1 , s n + 1 , m,n √2 λ − 2 √2 λ 2 × √2 λ − 2 √2 λ 2 h h and, for λ Λ, let (m ,n ) be the (unique) element of Z 2 such that λ K . Note that, as ∈ λ λ ∈ m,n sep(Λ) = s, every Km,n contains at most one element of Λ. Next, define the functions 2 1 ˜ A(z)= aλ Km ,n (z) and f(z) = max f(z + w) . s λ λ w 6s | | λ Λ | | X∈ We then have θλ f(z λ) 6 θλ f˜ z s(mλ + inλ)/√2 6 A(w)f˜(z w) dw = (A f˜)(z), | || − | | | − C − | | ∗ λ Λ λ Λ Z X∈ X∈ for all z C. Therefore, using [7, Prop. 11.1.3], we obtain ∈ 2/p A f˜ q C 6 f˜ 1 C A q C . s− f 1 θ p , k ∗ kL ( ) k kL ( )k kL ( ) k kW (L∞,L )k kℓ (Λ) where the last equality follows by computing the norm of A explicitly. Proof of Lemma 19. Note that it suffices to prove the claim for j = n 1, as the general statement then − √ n 1 1 follows by induction. To this end, divide Kn into four disjoint squares of side length 2 2 − + 2 . By the pigeonhole principle, one of these squares must contain at least 22(n 1) + 1 points of K Y . − n ∩ Denote this square by K′, and Let Kn 1 be the square which contains K′ and satisfies property (i) − 2(n 1) in the statement of the Lemma. Then #(Kn 1 Y ) > #(K′ Y ) > 2 − + 1, and so Kn 1 − ∩ ∩ − satisfies (ii), as desired.