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(λ)|:λ∈Λ}\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\{|f(\lambda )|: \lambda \in \Lambda \}$$\end{document} on any set Λ⊆R\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\Lambda \subseteq {\mathbb {R}}$$\end{document} with lower Beurling density D-(Λ)>2\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$D^{-}(\Lambda )>2$$\end{document}. We consider a totally positive function of Gaussian type, i.e., a function g∈L2(R)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$g \in L^2({\mathbb {R}})$$\end{document} whose Fourier transform factors as 1g^(ξ)=∫Rg(x)e-2πixξdx=C0e-γξ2∏ν=1m(1+2πiδνξ)-1,ξ∈R,\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \hat{g}(\xi )= \int _{{\mathbb {R}}} g(x) e^{-2\pi i x \xi } dx = C_0 e^{- \gamma \xi ^2}\prod _{\nu =1}^m (1+2\pi i\delta _\nu \xi )^{-1}, \quad \xi \in {\mathbb {R}}, \end{aligned}$$\end{document}with δ1,…,δm∈R,C0,γ>0,m∈N∪{0}\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\delta _1,\ldots ,\delta _m\in {\mathbb {R}}, C_0, \gamma >0, m \in {\mathbb {N}} \cup \{0\}$$\end{document}, and the shift-invariant spaceV∞(g)={f=∑k∈Zckg(·-k):c∈ℓ∞(Z)},\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} V^\infty (g) = \Big \{ f=\sum _{k \in {\mathbb {Z}}} c_k\, g(\cdot -k): c \in \ell ^\infty ({\mathbb {Z}}) \Big \}, \end{aligned}$$\end{document}generated by its integer shifts within L∞(R)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$L^\infty ({\mathbb {R}})$$\end{document}. As a consequence of (1), each f∈V∞(g)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$f \in V^\infty (g)$$\end{document} is continuous, the defining series converges unconditionally in the weak∗\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$^*$$\end{document} topology of L∞\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$L^\infty $$\end{document}, and the coefficients ck\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$c_k$$\end{document} are unique [6, Theorem 3.5].

The shift-invariant space V ∞ (g) enjoys the following sampling property [4]: every separated set ⊆ R with lower Beurling density 2r strictly larger than 1 provides the norm equivalence In this article, we show the following uniqueness property for the absolute values of real-valued functions in V ∞ (g). Theorem 1 (Sign retrieval) Let g be a totally positive function of Gaussian type, as in (1), and ⊆ R with lower Beurling density D − ( ) > 2. (3) Assume that f 1 , f 2 ∈ V ∞ (g) are real-valued and | f 1 | ≡ | f 2 | on . Then either 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. 1 The intuition behind Theorem 1 is as follows. Suppose that | f 1 | ≡ | f 2 | on , and split into Then f 1 − f 2 vanishes on 1 while f 1 + f 2 vanishes on 2 . Under (3), one may expect one of the two subsets 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, 1 := ∩ (−∞, 0] and 2 := ∩ (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 to conclude that ( 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 ⊆ R as with the convention that D + circ ( ) = ∞, if is uncountably infinite. The density is named circular because, in (5) The upper average circular density can be alternatively described as follows: as shown in Lemma 1 below, 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 1 , 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).

Theorem 2 (Uniqueness theorem) Let g be a totally positive function of Gaussian type, as in
The uniqueness theorem (which applies also to non-real-valued functions) allows one to carry out the following natural proof of the sign retrieval theorem.
Proof (Proof of Theorem 1, assuming Theorem 2) Assume that D − ( ) > 2 and write = 1 ∪ 2 as in (4). Then, by (7) and (8), either D + circ ( 1 ) > 1 or D + circ ( 2 ) > 1 (possibly both). In the first case, Theorem 2 shows that f 1 ≡ f 2 , while in the second, As the proof shows, Theorem 1 remains valid if (3) is relaxed to D + circ ( ) > 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 ( ) = D + circ ( ), since, if 0 ∈ , the contribution of the point λ = 0 is and does not affect the limit on r . Similarly, since The last quantity is finite because 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 has no accumulation points, there exists η > 0 such that ∩ (−η, η) = ∅. Therefore, we can estimate the part of the integral excluded in (10) as 2α 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 × Z is D − ( ).

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 ⊆ 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 \ {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 ( ) ≤ 1, as claimed. Note, in particular, that has no accumulation points. Inductive step. Let g be as in (1) and f = k c k g(· − k) ∈ V ∞ (g) non-zero. Without loss of generality we assume that, in (1), δ 1 , . . . , δ m are non-zero. Since g is realvalued, by replacing c k with (c k ) or (c k ) if necessary, we may assume that f is real-valued. Let g 1 be 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 Suppose that 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 ∪ {0, ∞}, 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 ∩ [0, ∞ The set + 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 longer than the corresponding segment through λ k : k>0,λ k ∈(0,t] see Fig. 1.
We now add the term corresponding to λ 0 , and bound t 2 − λ 2 0 ≤ t to obtain λ∈ ∩[0,t] 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 = { f = 0} has an accumulation point, Rolle's theorem applied as before shows that so does = { f 1 = 0}, and, therefore, D + circ ({ f 1 = 0}) = ∞ contradicting the inductive hypothesis.