Abstract
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 \(g \in L^2({\mathbb {R}})\) whose Fourier transform factors as
with \(\delta _1,\ldots ,\delta _m\in {\mathbb {R}}, C_0, \gamma >0, m \in {\mathbb {N}} \cup \{0\}\), and the shift-invariant space
generated by its integer shifts within \(L^\infty ({\mathbb {R}})\). As a consequence of (1), each \(f \in V^\infty (g)\) is continuous, the defining series converges unconditionally in the weak\(^*\) topology of \(L^\infty \), and the coefficients \(c_k\) are unique [6, Theorem 3.5].
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The shift-invariant space \(V^\infty (g)\) enjoys the following sampling property [4]: every separated set \(\Lambda \subseteq {\mathbb {R}}\) with lower Beurling density
strictly larger than 1 provides the norm equivalence
(A similar property holds for all \(L^p\) norms, \(1 \le p \le \infty \); see [4, Theorems 5.2 and 3.1].)
In this article, we show the following uniqueness property for the absolute values of real-valued functions in \(V^\infty (g)\).
(Sign retrieval) Let g be a totally positive function of Gaussian type, as in (1), and \(\Lambda \subseteq {\mathbb {R}}\) with lower Beurling density
Assume that \(f_1, f_2 \in V^\infty (g)\) are real-valued and \(\left| f_1 \right| \equiv \left| f_2 \right| \) on \(\Lambda \). Then either \(f_1 \equiv f_2\) or \(f_1 \equiv - f_2\).
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.Footnote 1
The intuition behind Theorem 1 is as follows. Suppose that \(\left| f_1 \right| \equiv \left| f_2 \right| \) on \(\Lambda \), and split \(\Lambda \) into
Then \(f_1 - f_2\) vanishes on \(\Lambda _1\) while \(f_1 + f_2\) vanishes on \(\Lambda _2\). Under (3), one may expect one of the two subsets \(\Lambda _j\) 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, \(\Lambda _1 := \Lambda \cap (-\infty ,0]\) and \(\Lambda _2 := \Lambda \cap (0,\infty )\) 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 \(V^\infty (g) \cdot V^\infty (g)\) is contained in a dilation of \(V^\infty (\tilde{g})\) by a factor of 2, where \(\tilde{g}\) is another Gaussian function. Thus, in the Gaussian case, the sampling theorem can be applied after rescaling to the set \(\Lambda \) to conclude that \((f_1-f_2) \cdot (f_1+f_2) \equiv 0\), and, by analyticity, that either \(f_1 \equiv f_2\) or \(f_1 \equiv - f_2\). 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 \(\Lambda \subseteq {\mathbb {R}}\) as
with the convention that \(D^+_{\mathrm {circ}}(\Lambda )=\infty \), if \(\Lambda \) is uncountably infinite. The density is named circular because, in (5), each point \(\lambda \in [-t,t]\) is weighted with the measure of the largest vertical segment \(\{\lambda \} \times (-a,a)\) contained in the two-dimensional open disk \(B_t(0) \subseteq {\mathbb {R}}^2\).
The upper average circular density can be alternatively described as follows: as shown in Lemma 1 below, for any lattice \(\alpha {\mathbb {Z}}\), \(\alpha >0\),
From here, it follows easily that \(D^+_{\mathrm {circ}}\) dominates Beurling’s lower density:
see Lemma 1 below. We call \(D^+_{\mathrm {circ}}\) an upper density because, due to the sublinearity of \(\limsup \),
for any two sets \(\Lambda _1, \Lambda _2 \subseteq {\mathbb {R}}\).
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).
(Uniqueness theorem) Let g be a totally positive function of Gaussian type, as in (1). Let \(f \in V^\infty (g)\) be non-zero. Then \(D^+_{\mathrm {circ}}(\{f=0\}) \le 1\).
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 of Theorem 1, assuming Theorem 2) Assume that \(D^{-}(\Lambda )>2\) and write \(\Lambda = \Lambda _1 \cup \Lambda _2\) as in (4). Then, by (7) and (8), either \(D^+_{\mathrm {circ}}(\Lambda _1) >1\) or \(D^+_{\mathrm {circ}}(\Lambda _2) >1\) (possibly both). In the first case, Theorem 2 shows that \(f_1 \equiv f_2\), while in the second, \(f_1 \equiv -f_2\).\(\square \)
As the proof shows, Theorem 1 remains valid if (3) is relaxed to \(D^+_{\mathrm {circ}}(\Lambda )>2\). Sign retrieval also holds under the same density condition for the shift-invariant spaces \(V^p(g)\) defined with respect to \(L^p\) norms, \(1 \le p \le \infty \), as these are contained in \(V^\infty (g)\).
Let \(\Lambda \subseteq {\mathbb {R}}\) and \(\alpha >0\). Then (6) and (7) hold true.
FormalPara ProofWe assume that \(\Lambda \) has no accumulation points, since, otherwise, both sides of (6) are infinite. Denote provisionally the right hand side of (6) by \({\tilde{D}}^+_{\mathrm {circ}}(\Lambda )\), and set \(\Lambda ' := \Lambda \setminus \{0\}\).
Step 1. Let \(\varepsilon \in (0,1)\), and \(C=C_{\alpha ,\varepsilon }>0\) a constant to be specified. We claim:
To prove the first claim, we first note that \(D^+_{\mathrm {circ}}(\Lambda )=D^+_{\mathrm {circ}}(\Lambda ')\), since, if \(0 \in \Lambda \), the contribution of the point \(\lambda =0\) is
and does not affect the limit on r. Similarly, since \(\sqrt{t^2 - \lambda ^2} \le t\),
The last quantity is finite because \(\Lambda \) 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 \(\Lambda \) has no accumulation points, there exists \(\eta >0\) such that \(\Lambda '\cap (-\eta ,\eta )=\emptyset \). Therefore, we can estimate the part of the integral excluded in (10) as
which is finite because \(\eta >0\). This proves (10).
Step 2. For \(t>0\), we note that
Next, we choose the constant \(C_{\alpha , \varepsilon }\) so that the following estimates hold for \(t \ge C_{\alpha , \varepsilon }\):
where the last estimate requires choosing \(C_{\alpha , \varepsilon }\) large. Similarly,
where the last estimate requires choosing \(C_{\alpha , \varepsilon }\) large. Combining the last two estimates with (9) and (10), and letting \(\varepsilon \longrightarrow 0\), we deduce (6).
Step 3. To prove (7), let \(r>s>0\) and use (6) with \(\alpha =1\) to estimate
Letting \(s \longrightarrow \infty \), we see that
where we used that the two-dimensional lower Beurling density of \(\Lambda ' \times {\mathbb {Z}}\) is \(D^{-}(\Lambda )\).
\(\square \)
We can now prove the uniqueness Theorem. The proof builds on [4, Lemma 5.1] and [5, Theorems 4.3 and 4.9].
We proceed by induction on m in (1).
In the case \(m=0\), the function g is a Gaussian \(g(x) = C'_0 e^{-a x^2}\), with \(a=\frac{\pi ^2}{\gamma }>0\). Let \(f=\sum _k c_k g(\cdot -k)\in V^\infty (g)\) with c bounded be non-zero, and denote by \(\Lambda \subseteq {\mathbb {R}}\) its zero set.
As shown in [5, Lemma 4.1], f possesses an extension to an entire function on \(\mathbb {C}\), satisfying the growth estimate \(|f(x+iy)| \lesssim e^{a y^2}\) for \(x,y \in {\mathbb {R}}\), where the implied constant depends on f. Let \(n \ge 0\) be the order of f at \(z=0\), and consider the analytic function
where \(C_1 \in \mathbb {C}\) 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 \(i\tfrac{\pi }{a}{\mathbb {Z}}\) [5, Lemma 4.2]. Therefore, the set of complex zeros of F contains \(\big (\Lambda \setminus \{0\}\big ) + i\tfrac{\pi }{a}{\mathbb {Z}}\):
By (6) with \(\alpha =\tfrac{\pi }{a}\), the zero-counting function of F,
satisfies
On the other hand, by Jensen’s formula combined with (11), for all \(r>0\),
By combining (12) and (13) we conclude that \(D^+_{\mathrm {circ}}(\Lambda ) \le 1\), as claimed. Note, in particular, that \(\Lambda \) has no accumulation points.
Inductive step. Let g be as in (1) and \(f=\sum _k c_k g(\cdot -k) \in V^\infty (g)\) non-zero. Without loss of generality we assume that, in (1), \(\delta _1, \ldots , \delta _m\) are non-zero. Since g is real-valued, by replacing \(c_k\) with \(\mathfrak {R}(c_k)\) or \(\mathfrak {I}(c_k)\) 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 \(V^\infty (g_1)\) 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 \(D^+_{\mathrm {circ}}(\{f_1=0\}) \le 1\).
Let \(\Lambda =\{f=0\}\) and \(\Gamma =\{f_1=0\}\). Suppose that \(\Lambda \) 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 \(N \in {\mathbb {N}}\cup \{0,\infty \}\), and we do not count multiplicities. Hence, \(\lambda _k >0\), if \(k>0\).
By Rolle’s Theorem applied to the differential operator \(I + \delta _m \partial _x\), there is a sequence of zeros of \(f_1\) that interlaces \(\Lambda \cap [0,\infty )\). (Indeed, we note that \(\frac{d}{dx} \big [ e^{x/\delta _m} f(x)\big ] = \frac{1}{\delta _m} e^{x/\delta _m} f_1(x)\) and apply the standard version of Rolle’s theorem; see [4, Lemma 5.1] or [5, Lemma 4.8] for details.) We parameterize that sequence as \(\Gamma ^+=\big \{\gamma _k : k=1,\ldots ,N\}\), with
The set \(\Gamma ^+\) is empty if \(N=0\). For \(t>0\), this ordering associates with each \(\lambda _k \in (0,t]\), \(k \not =0\), a distinct zero of \(f_1\), \(\gamma _k \in (0,t]\), such that the vertical segment through \(\gamma _k\) contained in the disk \(B_t(0)\) is longer than the corresponding segment through \(\lambda _k\):
see Fig. 1.
We now add the term corresponding to \(\lambda _0\), and bound \(\sqrt{t^2 - \lambda _0^2} \le t\) 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 \(D^+_{\mathrm {circ}}(\{f=0\}) \le D^+_{\mathrm {circ}}(\{f_1=0\})\). The latter quantity is bounded by 1 by inductive hypothesis. Finally, if \(\Lambda =\{f=0\}\) has an accumulation point, Rolle’s theorem applied as before shows that so does \(\Gamma =\{f_1=0\}\), and, therefore, \(D^+_{\mathrm {circ}}(\{f_1=0\}) = \infty \) contradicting the inductive hypothesis. \(\square \)
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Acknowledgements
J. L. R. gratefully acknowledges support from the Austrian Science Fund (FWF): Y 1199 and P 29462.
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Romero, J.L. Sign Retrieval in Shift-Invariant Spaces with Totally Positive Generator. J Fourier Anal Appl 27, 27 (2021). https://doi.org/10.1007/s00041-020-09804-z
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DOI: https://doi.org/10.1007/s00041-020-09804-z