Sign retrieval in shift-invariant spaces with totally positive generator

We show that a real-valued function $f$ in the shift-invariant space generated by a totally positive function of Gaussian type is uniquely determined, up to a sign, by its absolute values $\{|f(\lambda)|: \lambda \in \Lambda \}$ on any set $\Lambda \subseteq \mathbb{R}$ with lower Beurling density $D^{-}(\Lambda)>2$.

We consider a totally positive function of Gaussian type, i.e., a function g ∈ L 2 (R) whose Fourier transform factors as with δ 1 , . . . , δ m ∈ R, C 0 , γ > 0, m ∈ N ∪ {0}, and the shift-invariant space generated by its integer shifts within L ∞ (R). As a consequence of (1), each f ∈ V ∞ (g) is continuous, the defining series converges unconditionally in the weak * topology of L ∞ , and the coefficients c k are unique.
The shift-invariant space V ∞ (g) enjoys the following sampling property [4]: every separated set Λ ⊆ R with lower Beurling density strictly larger than 1 provides the norm equivalence (A similar property holds for all L p norms, 1 ≤ p ≤ ∞.) In this article, we show the following uniqueness property for the absolute values of realvalued functions in V ∞ (g). Theorem 1 (Sign retrieval). Let g be a totally positive function of Gaussian type, as in (1), and Λ ⊆ R with lower Beurling density Assume that f 1 , f 2 ∈ V ∞ (g) are real-valued and |f 1 | ≡ |f 2 | on Λ. Then either f 1 ≡ f 2 or See [2] for motivation for sign retrieval in shift-invariant spaces. When g is a Gaussian function -corresponding to m = 0 in (1) -Theorem 1 was recently obtained in [3,Theorem 1]. Here, that result is extended to all totally positive functions of Gaussian type.
The intuition behind Theorem 1 is as follows. Suppose that |f 1 | ≡ |f 2 | on Λ, and split Λ into Then f 1 − f 2 vanishes on Λ 1 while f 1 + f 2 vanishes on Λ 2 . Under (3), one may expect one of two subsets Λ j to have lower Beurling density larger than 1. The sampling inequalities (2) would then imply that either f 1 − f 2 or f 1 + f 2 are identically zero. This argument breaks down, however, because Beurling's lower density is not subadditive. For example, Λ 1 := Λ ∩ (−∞, 0] and Λ 2 := Λ ∩ (0, ∞) have always zero lower Beurling density. The proof of the sign retrieval theorem for Gaussian generators in [3] resorts instead to a special property of the Gaussian function, namely that V ∞ (g)·V ∞ (g) is contained in a dilation of V ∞ (g) by a factor of 2, whereg is another Gaussian function. Thus, in the Gaussian case, the sampling theorem can be applied after rescaling to the set Λ to conclude that (f 1 − f 2 ) · (f 1 + f 2 ) ≡ 0, and, by analyticity, that either f 1 ≡ f 2 or f 1 ≡ −f 2 . A similar argument applies to Paley-Wiener spaces [1, Theorem 2.5]. We are unaware of an analogous dilation property for general totally positive generators.
To prove Theorem 1 for all totally positive generators of Gaussian type we take a different route. We define the upper average circular density of a set Λ ⊆ R as with the convention that D + circ (Λ) = ∞, if Λ is uncountably infinite. The density is named circular because, in (5) The upper average circular density can be alternatively described as follows: for any lattice αZ, α > 0, From here, it follows easily that D + circ dominates Beurling's lower density: see Lemma 1 below. We call D + circ an upper density because, due to the sublinearity of lim sup, for any two sets Λ 1 , Λ 2 ⊆ R.
Below we prove the following uniqueness result formulated in terms of the upper average circular density of the zero set {f = 0} of a function f (counted without multiplicities). The uniqueness theorem allows one to carry out the natural proof the sign retrieval theorem.
As the proof shows, Theorem 1 remains valid if (3) is relaxed to D + circ (Λ) > 2. Sign retrieval also holds under the same density condition for the shift-invariant spaces V p (g) defined with respect to L p norms, 1 ≤ p ≤ ∞, as these are contained in V ∞ (g).
Step 1. Let ε ∈ (0, 1), and C = C α,ε > 0 a constant to be specified. We claim: To prove the first claim, we first note that D + circ (Λ) = D + circ (Λ ′ ), since, if 0 ∈ Λ, the contribution of the point λ = 0 is 4 πr 2 r 0 dt = 4 πr , and does not affect the limit on r. Similarly, since √ t 2 − λ 2 ≤ t, The last quantity is finite because Λ has no accumulation points. This proves (9), because it shows that limiting the integral to [C, r] does not affect the limit on r. Second, a change of variables shows that (10) holds with C replaced by 0. In addition, since Λ has no accumulation points, there exists η > 0 such that Λ ′ ∩ (−η, η) = ∅. Therefore, we can estimate the part of the integral excluded in (10) as which is finite because η > 0. This proves (10).
Step 2. For t > 0, we note that Next, we choose the constant C α,ε so that the following estimates hold for t ≥ C α,ε : where the last estimate requires choosing C α,ε large. Similarly, where the last estimate requires choosing C α,ε large. Combining the last two estimates with (9) and (10), and letting ε −→ 0, we deduce (6).
Step 3. To prove (7), let r > s > 0 and use (6) with α = 1 to estimate Letting s −→ ∞, we see that where we used that the two-dimensional lower Beurling density of Λ ′ × Z is D − (Λ).
Proof of Theorem 2. We proceed by induction on m in (1).
In the case m = 0, the function g is a Gaussian g(x) = C ′ 0 e −ax 2 , with a = π 2 γ > 0. Let f = k c k g(· − k) ∈ V ∞ (g) with c bounded be non-zero, and denote by Λ ⊆ R its zero set.
As shown in [5,Lemma 4.1], f possesses an extension to an entire function on C, satisfying the growth estimate |f (x + iy)| e ay 2 for x, y ∈ R, where the implied constant depends on f . Let n ≥ 0 be the order of f at z = 0, and consider the analytic function where C 1 ∈ C is chosen so that F (0) = 1. Then F satisfies for some constant C > 0. Moreover, the zero set of f is invariant under addition of i π a Z [5, Lemma 4.2]. Therefore, the set of complex zeros of F contains Λ \ {0} + i π a Z: By (6) with α = π a , the zero-counting function of F , n F (t) := #{z ∈ C : F (z) = 0, |z| ≤ t}, satisfies lim sup On the other hand, by Jensen's formula combined with (11), for all r > 0, By combining (12) and (13) we conclude that D + circ (Λ) ≤ 1, as claimed. Note, in particular, that Λ has no accumulation points.
Inductive step. Let g be as in (1) and f = k c k g(· − k) ∈ V ∞ (g) non-zero. Since g is realvalued, by replacing c k with ℜ(c k ) or ℑ(c k ) if necessary, we may assume that f is real-valued. Let g 1 is given by (1), except that the product runs only up to m − 1. Then belongs to V ∞ (g 1 ) and is real-valued and non-zero, because the coefficients c k are real and at least one of them is non-zero. We assume by inductive hypothesis that D + circ ({f 1 = 0}) ≤ 1. Let Λ = {f = 0} and Γ = {f 1 = 0}. Suppose that Λ has no accumulation points. Assume for the moment that, in addition, f has a non-negative zero, and order the set of non-negative zeros of f increasingly: where N ∈ N ∪ {∞}, and we do not count multiplicities. Hence, λ k > 0, if k > 0. By Rolle's Theorem applied to the differential operator I + δ m ∂ x , there is a sequence of zeros of f 1 that interlaces Λ ∩ [0, ∞) (see [4,Lemma 5.1] or [5,Lemma 4.8] for details). We parameterize that sequence as Γ + = γ k : k = 1, . . . , N}, with The set Γ + is empty if N = 0. For t > 0, this ordering associates with each λ k ∈ (0, t], k = 0, a distinct zero of f 1 , γ k ∈ (0, t], such that the vertical segment through γ k contained in the disk B t (0) is larger than the corresponding segment through λ k : k>0,λ k ∈(0,t] t 2 − λ 2 k ≤ k>0,γ k ∈(0,t] t 2 − γ 2 k ≤ γ∈Γ∩(0,t] t 2 − γ 2 ; see Figure 1. We now add the term corresponding to λ 0 , and bound t 2 − λ 2 0 ≤ t to obtain The previous estimate is also trivially true is f has no non-negative zeros. Arguing similarly with the non-positive zeros of f we conclude that This shows that D + circ ({f = 0}) ≤ D + circ ({f 1 = 0}). The latter quantity is bounded by 1 by inductive hypothesis. Finally, if Λ = {f = 0} has an accumulation point, Rolle's theorem applied as before shows that so does Γ = {f 1 = 0}, and, therefore, D + circ ({f 1 = 0}) = ∞ contradicting the inductive hypothesis.