A Simple Crystalline Measure
aa r X i v : . [ m a t h . C A ] J un A Simple Crystalline Measure
Alexander Olevskii and Alexander Ulanovskii
Abstract
We prove that every pair of exponential polynomials with imaginaryfrequencies generates a Poisson-type formula.
Following Yves Meyer [M16], we say that µ is a crystalline measure if µ is a purely atomic measure supported by a locally finite set and itsdistributional Fourier transform is also a purely atomic measure supportedby a locally finite set. This concept is intimately connected with analysisof adequate mathematical models for the quasicrystal phenomenon.The classical example of a crystalline measure is given by the Poissonmeasure µ = X k ∈ Z δ k , which assigns to each integer a unit mass. This measure, often called theDirac comb, corresponds to crystals.The existence of crystalline measures whose support does not containan infinite arithmetic progression is a non-trivial fact. At present severalexamples of such measures are known based on different constructions.The reader can find information on the history of problem and relatedresults updated to 2019 in: Yves Meyer [M94], Jeffry Lagarias [L00], NirLev and Alexander Olevskii [LO15], [LO16] and [LO17], Yves Meyer [M16]and [M17]. See also the references in these papers.Recently a new progress was obtained by Pavel Kurasov and Peter Sar-nak [KS20], who constructed positive crystalline measures with uniformlydiscrete support.Even more recently, Yves Meyer has given a geometric approach toconstruction of crystalline measures based on the concept of curve modelset and used inner holomorphic functions, see [M20]. The example from[KS20] is obtained in this work.Inspired by [KS20] and [M20], we present an approach to the prob-lem which gives simple examples of crystalline measures satisfying theproperties mentioned above. Poisson-type Formulas
Recall that a set Λ of real or complex numbers is called uniformly discrete(u.d.) if inf λ,λ ′ ∈ Λ ,λ = λ ′ | λ − λ ′ | > . (1)The classical Poisson summation formula states X k ∈ Z f ( k ) = X k ∈ Z ˆ f ( k ) , f ∈ S ( R ) , where S ( R ) is the Schwartz space and ˆ f is the Fourier transform of f .We present an approach which yields a number of Poisson-type formu-las X λ ∈ Λ c λ f ( λ ) = X s ∈ S a s ˆ f ( s ) , (2)where Λ is a u.d. set in C or R , S ⊂ R is a locally finite set satisfying S ∩ ( − r, r ) ≤ Cr m , r > , (3)with some C and m , and the sequence of coefficients c λ is bounded. IfΛ ⊂ R then the sequence a s is bounded, too. The formula holds for certainclasses of functions f specified below.We now introduce three classes of exponential polynomials. Let Exp denote the set of all exponential polynomials ψ ( z ) = l X j =1 b j e πiα j z , where l ∈ N , α j ∈ R , b j ∈ C . We denote byΛ ψ := { z ∈ C : ψ ( z ) = 0 } the zero set of ψ. One may easily check that for every ϕ ∈ Exp there is apositive r such that Λ ϕ lies in the strip | Im z | < r .Let Exp s denote the subset of Exp of exponential polynomials ϕ suchthat the zeros of ϕ are simple and the zero set Λ ϕ is u.d.Finally, let Exp r denote the subset of Exp s of exponential polynomials ϕ such that Λ ϕ ⊂ R .It turns out that every pair of functions ψ ∈ Exp, ϕ ∈ Exp s satisfying Λ ϕ Λ ψ . (4)gives rise to a Poisson-type formula: Theorem 1
Given any pair ψ ∈ Exp, ϕ ∈ Exp s satisfying (4).(i) Formula (2) holds with Λ = Λ ϕ , some set S ⊂ R satisfying (3) anda bounded sequence c λ defined by c λ = ψ ( λ ) ϕ ′ ( λ ) , λ ∈ Λ . (5) Observe that by a theorem of Ritt [R29], condition Λ ϕ ⊂ Λ ψ holds if and only if thereexists ϕ ∈ Exp such that ψ ( z ) = ϕ ( z ) ϕ ( z ) . ormula (2) holds for every entire function f satisfying f ( x + iy ) ∈ S ( R ) , for every fixed y .(ii) If ϕ ∈ Exp r , then the sequence a s is bounded and (2) holds forevery f ∈ S ( R ) . The set S and coefficients a s in (2) will also be explicitly defined. Wealso remark that the classes of functions f in Theorem 1 (i) and (ii) canbe substantially enlarged, see the construction below.We will prove Theorem 1 in sec. 4.Observe that Theorem 1 admits an extension: In fact every pair ofexponential polynomials ψ, ϕ ∈ Exp satisfying (4) generates a Poisson-type formula. If the roots of ϕ are not simple and the set Λ ϕ is not u.d.,then the left hand-side of this formula may contain derivatives of f andthe coefficients may no longer be bounded. For simplicity of presentation,we will not consider this case. The following is an immediate corollary of Theorem 1:
Corollary 1
Given any pair ϕ ∈ Exp r , ψ ∈ Exp satisfying (4). Let c λ be defined in (5). Then µ := X λ ∈ Λ ϕ c λ δ λ is a crystalline measure whose Fourier transform is given by ˆ µ = X s ∈ S a s δ s . Choose ψ = ϕ ′ . Then in (5) we have c λ = 1 , λ ∈ Λ. This gives
Corollary 2
Assume ϕ ∈ Exp r . Then µ := X λ ∈ Λ ϕ δ λ (6) is a crystalline measure. We see that every exponential polynomial ϕ ∈ Exp r gives rise tocrystalline measures supported by Λ ϕ . A ‘non-trivial’ crystalline measureis a measure which is not a linear combination of Poisson measures. Hence,one wishes to find ϕ ∈ Exp r such that Λ ϕ does not contain any arithmeticprogression. A simple examples is given by ϕ ( z ) := sin πz + δ sin z. (7) Example 1
Let < δ ≤ / , and ϕ be defined in (7). Then(i) Λ ϕ = { k + δ k : k ∈ Z } , where δ k ∈ [ − / , / satisfy | sin πδ k | ≤ δ, k ∈ Z ; (8) (ii) Λ ϕ does not contain any arithmetic progression. e will prove this in sec. 5.The proof below shows that statements (i) and (ii) in the example holdfor every function ϕ ( z ) = sin πz + n X j =1 d j sin α j z, n ∈ N , d j ∈ R , < α j < π, provided at least one ratio π/α j is irrational and δ := P nj =1 | d j | ≤ / . Let ϕ be given in (7). By (8), we see that the numbers δ k in Example 1satisfy δ k → δ →
0. Therefore, the set Λ ϕ ‘approaches’ the set ofintegers Z . By Corollary 2, this gives Corollary 3
For every ε > there is a set Λ = { k + δ k : k ∈ Z } , ≤ | δ k | < ε, k ∈ Z , (9) which does not contain any arithmetic progression, and the correspondingmeasure in (6) is crystalline. On the other hand, it follows from [F20], that if Λ satisfies (9) where δ k → , | k | → ∞ , and the measure µ in (6) is crystalline, then Λ = Z andso µ is the Dirac comb. Denote by C different positive constants.Fix two functions ψ ( z ) = l X j =1 b j e πiβ j z ∈ Exp, ϕ ( z ) = m X j =1 d j e πiα j z ∈ Exp s , where l ∈ N , m ≥ β < ... < β l , α < ... < α m . We assume that thecoefficients b j , d j are different from zero. We also assume that condition(4) is true.The proof consists of several steps.1. Denote by γ the contour which consists of two parallel lines γ := R − iR − and γ := R + iR + , where the latter line is oriented from ∞ to −∞ and the numbers R ± > E the set of all entire functions f of finite exponential type satisfying for every r > z ∈ C , | Im z |
We have inf λ ∈ Λ ϕ | ϕ ′ ( λ ) | > . Indeed, set ϕ λ ( z ) := ϕ ( z + λ ) , λ ∈ Λ ϕ . Then ϕ λ (0) = 0.Recall that the zeros of ϕ are simple and the zero set Λ ϕ is u.d. Hence,there exists ǫ > ϕ λ has exactly one root in thecircle | z | ≤ ǫ. By the Argument Principle, I | z | = ǫ ϕ ′ λ ( z ) ϕ λ ( z ) dz = 2 πi, λ ∈ Λ ϕ . Assume the claim is not true: there is a sequence λ j ∈ Λ ϕ such that ϕ ′ ( λ j ) → , j → ∞ . Clearly, there is a subsequence λ j ( k ) such that ϕ λ j ( k ) converge uniformly on compacts to some exponential polyniomial ˜ ϕ , sothat I | z | = ǫ ˜ ϕ ′ ( z )˜ ϕ ( z ) dz = 2 πi. On the other hand, clearly ˜ ϕ has a double zero at the origin, and so theabove formula contradicts the Argument Principle.3. Let us calculate I j , j = 1 , . Write1 ϕ ( z ) = 1 d e πiα z + ... + d m e πiα m z = (1 /d ) e − πiα z ... + ( d m /d ) e πi ( α m − α ) z . Since α j − α > , j > , there exists R + > (cid:12)(cid:12)(cid:12) ( d /d ) e πi ( α − α ) z + ... + ( d m /d ) e πi ( α m − α ) z (cid:12)(cid:12)(cid:12) < , Im z ≥ R + . Hence, for Im z ≥ R + we have1 ϕ ( z ) = e − πiα z d ∞ X k =0 ( − k (cid:18) d d e πi ( α − α ) z + ... + d m d e πi ( α m − α ) z (cid:19) k . It easily follows that there is a discrete set S + and coefficients p s , s ∈ S + , such that ψ ( z ) ϕ ( z ) = X s ∈ S + p s e πisz , Re z ≥ R + . (11) ne may check that S + ⊂ { β − α , ..., β l − α } + ∞ [ k =0 [ n j ≥ , P n j = k { n ( α − α )+ ... + n m ( α m − α ) } . Claim 2 S + ⊂ [ β − α , ∞ ) and satisfies (3). We omit the simple proof.Similarly to above, there exists R − > S − ⊂ [ α m − β l , ∞ ) and coefficients q s such that ψ ( z ) ϕ ( z ) = X s ∈ S − q s e − πisz , Re z ≤ − R − . (12)Again, one may check that S − satisfies (3).Let us now calculate the integrals I j . Using (11) and (12) we get I = X s ∈ S − q s ˆ f ( s ) , I = − X s ∈ S + p s ˆ f ( − s ) , f ∈ E, where each series converges absolutely, since it contains only a finite num-ber of elements.3. Comparing the above calculations of integral I , we see that formula (2)is valid for every f ∈ E , where Λ = Λ ϕ , the sequence c λ is defined in (5)and bounded, the set S := S − ∪ ( − S + ) satisfies (3), a s = q s / πi, s ∈ S − ,and a s = − p s / πi, s ∈ − S + .Let us prove that if Λ ϕ ⊂ Z then the sequence of coefficients a s isbounded. Lemma 1
Assume formula (2) holds for all functions f ∈ E , where Λ ⊂ R is a u.d. set, S ⊂ R is a locally finite set and the sequence c λ is bounded.Then the sequence a s is bounded, too. Proof. For every 0 < β < s ∈ S set ψ β,s ( z ) := e πisz β (cid:18) sin( πβz ) πβz (cid:19) . It is clear that ψ β,s ∈ E and that the L ( R )-norm L := k ψ β,s k is finite and does not depend on β and s . The Fourier transform ˆ ψ β,s isequal to L at the point s and vanishes outside the interval ( s − β, s + β ).Since S is locally finite, we may choose β so small that S ∩ ( s − β, s + β ) = { s } . Apply (2) with f = ψ β,s . By Bessel’s inequality (see i.e.Proposition 2.7 in [OU16]), the left hand-side of (2) admits an estimate X λ ∈ Λ | c λ || ψ β,w ( λ ) | ≤ C, where C does not depend on β . The right hand-side of (2) contains onlyone term a s ˆ ψ β,s ( s ) = La s . Hence, | a s | < C, s ∈ S , which proves thelemma. f the set Λ ϕ does not lie on R , a similar argument shows that thecoefficients a s have at most exponential growth.4. To prove Theorem 1 (i), it remains to check that (2) holds for all entirefunctions f satisfying f ( x + iy ) ∈ S ( R ) for every fixed y ∈ R . Indeed, theFourier transform ˆ f of such a function satisfies | ˆ f ( t ) | ≤ C r e − r | t | , for every r > , where C r depends on r . Hence, both sides of (2) absolutely converge andso the proof follows by a usual approximation argument. In fact, onemay check that (2) remains true for all functions f analytic in the strip | Im z | ≤ r and satisfying (10), where r = max { R − , R + } .5. Finally, let ϕ ∈ Exp r . Then Λ = Λ ϕ ⊂ R . Theorem 1 (ii) can bededuced from Theorem 1 (i) by a usual approximation argument, takingin mind that both sides of (2) converge absolutely for every f ∈ S ( R ).Again, one may check that (2) remains true for all f satisfyingsup x ∈ R (1 + | x | ) | f ( x ) | < ∞ , sup x ∈ R | x | m +2 | ˆ f ( x ) | < ∞ . Let ϕ be given in (7).(i) It is clear that ϕ changes the sign on each interval ( k − γ, k + γ ) , k ∈ Z , k = 0 , where γ = γ ( δ ) ∈ (0 , /
6] is defined by | sin πγ | = δ. It remains to check that it has exactly one zero on every such interval,and has no other zeros.Denote by Γ n ⊂ C the squareΓ n := { z = x + iy : max( | x | , | y | ) = n + 1 / } , n ∈ N . Clearly, sin πz has exactly 2 n + 1 zeros inside Γ n . By above, ϕ ( z ) =sin πz + δ sin z has at least 2 n + 1 zeros inside Γ n . It is easy to see that | sin πz | > δ | sin z | on Γ n , provided n is large enough. So, the result followsfrom Rouch´e’s theorem.(ii) Consider the ‘projection’ Λ of Λ onto [ − / , /
2) defined asΛ := [ − / , / ∩ [ k ∈ Z (Λ + k ) ! . Then Λ ⊂ ( − / , / α + β Z ⊂ Λ for some α and β >
0. If β isirrational, the projection of α + β Z onto [ − / , /
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