An example concerning Fourier analytic criteria for translational tiling
aa r X i v : . [ m a t h . C A ] S e p AN EXAMPLE CONCERNING FOURIER ANALYTIC CRITERIAFOR TRANSLATIONAL TILING
NIR LEV
Abstract.
It is well-known that the functions f ∈ L ( R d ) whose translates alonga lattice Λ form a tiling, can be completely characterized in terms of the zero set oftheir Fourier transform. We construct an example of a discrete set Λ ⊂ R (a smallperturbation of the integers) for which no characterization of this kind is possible:there are two functions f, g ∈ L ( R ) whose Fourier transforms have the same set ofzeros, but such that f + Λ is a tiling while g + Λ is not. Introduction f be a function in L ( R ) and let Λ ⊂ R be a discrete set. We say that f tiles R at level w with the translation set Λ, or that f + Λ is a tiling of R at level w (where w is a constant), if X λ ∈ Λ f ( x − λ ) = w a.e. (1.1)and the series in (1.1) converges absolutely a.e.In the same way one can define tiling of R d by translates of a function f ∈ L ( R d ).For example, if f = Ω is the indicator function of a set Ω, and f + Λ is a tiling atlevel 1, then this means that the translated copies Ω + λ , λ ∈ Λ, fill the whole spacewithout overlaps up to measure zero. To the contrary, for tiling by a general real orcomplex-valued function f , the translated copies may have overlapping supports.Tilings by translates of a function have been studied by several authors, see, inparticular, [LM91], [KL96], [Kol04], [KL16], [Liu18], [KL20].1.2. It is well-known that in the study of translational tilings, the set Z ( b f ) := { t : b f ( t ) = 0 } (1.2)of the zeros of the Fourier transform b f ( t ) = Z f ( x ) exp( − πitx ) dx (1.3)plays an important role. For example, let Λ be a lattice in R d , then f + Λ is a tilingif and only if the set Z ( b f ) contains Λ ∗ \ { } , where Λ ∗ is the dual lattice. This meansthat the functions f that tile by a lattice Λ can be completely characterized in terms ofthe zero set Z ( b f ). (One can show that the tiling level is given by w = b f (0) det(Λ) − .) Date : September 22, 2020.2010
Mathematics Subject Classification.
Key words and phrases.
Tiling, translates, Fourier transform, distributions, spectral synthesis.Research supported by ISF Grant No. 227/17 and ERC Starting Grant No. 713927.
The necessity of the condition for tiling in the last example can be generalized asfollows. For a discrete set Λ ⊂ R we consider the measure δ Λ := X λ ∈ Λ δ λ . (1.4)We will assume that Λ has bounded density , which means thatsup x ∈ R ∩ [ x, x + 1)) < ∞ . (1.5)In particular (1.5) implies that the measure δ Λ is a temperate distribution on R , so ithas a well-defined Fourier transform b δ Λ in the distributional sense. Theorem 1.1 ([KL16]) . Let f ∈ L ( R ) , and Λ ⊂ R be a discrete set of bounded density.If f + Λ is a tiling at some level w , then supp( b δ Λ ) \ { } ⊂ Z ( b f ) . (1.6)A similar result is true also in R d . In the earlier works [KL96], [Kol00a], [Kol00b] thisresult was proved under various extra assumptions.If Λ is a lattice, then b δ Λ = det(Λ) − · δ Λ ∗ by the Poisson summation formula. Thisimplies that supp( b δ Λ ) = Λ ∗ . Hence in this case the condition (1.6) is not only necessary,but also sufficient, for f + Λ to be a tiling at some level w .However for a general discrete set Λ of bounded density, the sufficiency of the condition(1.6) for tiling has remained an open problem. In this paper, we settle this problem inthe negative. Our main result is the following: Theorem 1.2.
There is a discrete set Λ ⊂ R of bounded density (a small perturbationof the integers) with the following property: given any real scalar w there are two real-valued functions f, g ∈ L ( R ) whose Fourier transforms have the same set of zeros, butsuch that f + Λ is a tiling at level w while g + Λ is not a tiling at any level. Moreover, we will show that if the given scalar w is positive , then the functions f, g can be chosen positive as well.It follows that the necessary condition (1.6) is generally not sufficient for tiling: Corollary 1.3.
There exist a set Λ ⊂ R of bounded density and a positive function f ∈ L ( R ) , such that (1.6) is satisfied however f + Λ is not a tiling at any level. But even stronger, Theorem 1.2 shows that also no other condition can be given interms of the Fourier zero set Z ( b f ) that would characterize the functions f ∈ L ( R )such that f + Λ is a tiling, even under the extra assumption that f is positive.Our approach is based on the relation of the problem to Malliavin’s non-spectralsynthesis example [Mal59c]. The proof involves the implicit function method due toKargaev [Kar82], who proved the existence of a set Ω ⊂ R of finite measure such thatthe Fourier transform of its indicator function vanishes on some interval.2. Preliminaries. Notation.
In this section we recall some preliminary background and fix notation that will beused later on. For further details we refer the reader to [Kah70].
N EXAMPLE CONCERNING FOURIER ANALYTIC CRITERIA FOR TILING 3
The closed support of a Schwartz distribution S , or a function φ , on the real line R or on the circle T = R / Z , is denoted by supp( S ) or supp( φ ) respectively.If S is a Schwartz distribution on T , its Fourier coefficients b S ( n ) are defined by b S ( n ) = h S, e − πint i , n ∈ Z . The action of S on a function φ ∈ C ∞ ( T ) is denoted by h S, φ i . We have h S, φ i = X n ∈ Z b S ( n ) b φ ( − n ) . (2.1)Let A ( T ) be the Wiener space of continuous functions φ on T whose Fourier seriesconverges absolutely. It is a Banach space endowed with the norm k φ k A ( T ) = X n ∈ Z | b φ ( n ) | . A distribution S on T is called a pseudomeasure if S can be extended to a continuouslinear functional on A ( T ). This is the case if and only if the Fourier coefficients b S ( n )are bounded. The space P M ( T ) of all pseudomeasures is a Banach space with the norm k S k P M ( T ) = sup n ∈ Z | b S ( n ) | . The duality between the spaces A ( T ) and P M ( T ) is given by h S, φ i = X n ∈ Z b S ( n ) b φ ( − n ) , S ∈ P M ( T ) , φ ∈ A ( T ) , (2.2)which is consistent with (2.1).In a similar way, we will denote by A ( R ) the space of Fourier transforms of functionsin L ( R ), that is, φ ∈ A ( R ) if and only if φ ( t ) = Z R b φ ( x ) e πitx dx, b φ ∈ L ( R ) , k φ k A ( R ) = k b φ k L ( R ) . The Banach space dual to A ( R ) is then the space P M ( R ) of temperate distributions S on R whose Fourier transform b S is in L ∞ ( R ). The space P M ( R ) is normed as k S k P M ( R ) = k b S k L ∞ ( R ) , and the duality between the spaces A ( R ) and P M ( R ) is given by h S, φ i = Z R b S ( x ) b φ ( − x ) dx, S ∈ P M ( R ) , φ ∈ A ( R ) . (2.3)The elements of the space P M ( R ) are called pseudomeasures on R .The product φψ of two functions φ, ψ ∈ A (on either T or R ) is also in A , and k φψ k A k φ k A k ψ k A . If S ∈ P M and φ ∈ A , then the product Sφ is a pseudomeasure defined by h Sφ, ψ i = h S, φψ i , ψ ∈ A, and we have k Sφ k P M k S k P M k φ k A . NIR LEV If S ∈ P M , φ ∈ A and if φ vanishes in a neighborhood of supp( S ), then Sφ = 0. Thisis obvious from the definition of supp( S ) if φ is a smooth function of compact support,while for a general φ ∈ A this follows by approximation.If S is a Schwartz distribution on R supported on a compact interval I = [ a, b ], thenits Fourier transform b S is an infinitely smooth function on R given by b S ( x ) = h S, e − πixt i , x ∈ R . (In fact, b S is the restriction to R of an entire function of exponential type).If S is a distribution on R supported on an interval I of length | I | <
1, then S maybe considered also as a distribution on T , and in this case we have S ∈ P M ( T ) if andonly if S ∈ P M ( R ). If, in addition, φ is a function on R such that supp( φ ) ⊂ I , then φ ∈ A ( T ) if and only if φ ∈ A ( R ), and the action h S, φ i then has the same value withrespect to either definition (2.2) or (2.3).3. Malliavin’s non-spectral synthesis phenomenon spectral synthesis problem , posed by Beurling, asks the following: Let V bea closed, linear subspace of the space ℓ ∞ ( Z ) endowed with the weak* topology (asthe dual of ℓ ). We say that V is translation-invariant if whenever a sequence { c ( n ) } belongs to V , then so do all of the translates of { c ( n ) } . Define the spectrum σ ( V ) ofa translation-invariant subspace V to be the (closed) set of points t ∈ T such that thesequence e t := { exp(2 πint ) } is in V . Is it true that V is generated by the exponentials e t , t ∈ σ ( V ), i.e. is V the weak* closure of the linear span of these exponentials?There are also other, equivalent formulations of the spectral synthesis problem, see[KS94, Chapter IX]. One of them is the following: Let S ∈ P M ( T ), φ ∈ A ( T ), andassume that φ vanishes on supp( S ). Does it follows that h S, φ i = 0?The answer to the last question is affirmative if φ is smooth, or, more generally, if φ ∈ A ( T ) ∩ Lip( ). This result is due to Beurling and Pollard, see e.g. [Kah70, ChapterV, Section 5]. However, it was proved by Malliavin that in the general case, the questionadmits a negative answer: Theorem 3.1 (Malliavin [Mal59a], [Mal59b]) . There exist a pseudomeasure S ∈ P M ( T ) and a function φ ∈ A ( T ) such that φ vanishes on supp( S ) , but h S, φ i 6 = 0 . The spectral synthesis problem can be posed more generally in any locally compactabelian group G . For compact groups the problem admits a positive answer; whileMalliavin showed [Mal59c] that the answer is negative for all non-compact groups G .For more details on the subject we refer the reader to [KS94, Chapter IX], [Kah70,Chapter V], [Rud62, Chapter 7], [GM79, Chapter 3].3.2. Let S ∈ P M ( T ) and φ ∈ A ( T ) be given by Malliavin’s theorem (Theorem 3.1),that is, φ vanishes on supp( S ) while h S, φ i 6 = 0. Since φ does not vanish everywhere onthe circle T , there is an open interval I of length | I | < S ) ⊂ I . Hencewe may regard S also as a distribution on R , and we have S ∈ P M ( R ). By multiplying φ on a smooth function supported on I and which is equal to 1 in a neighborhood ofsupp( S ), we may assume that supp( φ ) ⊂ I as well, and consequently φ ∈ A ( R ). N EXAMPLE CONCERNING FOURIER ANALYTIC CRITERIA FOR TILING 5
Furthermore, by applying a linear change of variable to S and φ , we may actuallysuppose that I is an arbitrary open interval on R . We shall take I = ( a, b ) where a, b are any two numbers satisfying 0 < a < b < .For each r > T r ∈ P M ( R ) by T r := δ + r ( S + e S ) , (3.1)where e S ( t ) := S ( − t ). We will prove the following result:
Theorem 3.2.
Given any ε > there exists a real sequence Λ = { λ n } , n ∈ Z , satisfying | λ n − n | ε for all n , such that b δ Λ = T r in the interval ( − b, b ) for some r > . The proof of this theorem will be given in the next section. Our goal in the presentsection is to complete the proof of Theorem 1.2 based on this result. We will show thatΛ has the property from the statement of the theorem: given any real scalar w thereare two real-valued functions f, g ∈ L ( R ) whose Fourier transforms have the same setof zeros, but such that f + Λ is a tiling at level w while g + Λ is not a tiling at anylevel. Moreover, if the given scalar w is positive , then the functions f, g can be chosenpositive as well.3.3. Since the set Λ has bounded density, for any h ∈ L ( R ) the convolution h ∗ δ Λ isa locally integrable function satisfyingsup x ∈ R Z x +1 x | ( h ∗ δ Λ )( y ) | dy < + ∞ , (3.2)see [KL96, Lemma 2.2]. This implies that h ∗ δ Λ is a temperate distribution on R . Lemma 3.3.
Let h be a function in L ( R ) such that supp( b h ) ⊂ ( − b, b ) . Then theFourier transform of h ∗ δ Λ is the pseudomeasure T r · b h .Proof. The assertion means that for any Schwartz function β we have Z R ( h ∗ δ Λ )( x ) b β ( x ) dx = h T r , b h · β i . (3.3)Let χ be a Schwartz function whose Fourier transform b χ is nonnegative, has compactsupport, R b χ ( t ) dt = 1, and for each ε > χ ε ( x ) := χ ( εx ). Let q ε := ( b h · β ) ∗ b χ ε , then q ε is an infinitely smooth function with compact support. As ε →
0, the function q ε remains supported on a certain closed interval J contained in ( − b, b ), and q ε convergesto b h · β in the space A ( R ). The assumption that b δ Λ = T r in ( − b, b ) thus implies thatlim ε → h b δ Λ , q ε i = lim ε → h T r , q ε i = h T r , b h · β i . (3.4)The function β is the Fourier transform of some function α in the Schwartz class. Let p ε := ( h ∗ α ) · χ ε , then p ε is a smooth function in L ( R ) and we have b p ε = q ε . Since q ε belongs to the Schwartz space, the same is true for p ε , and it follows that The distribution e S can be more formally defined by h e S, ψ i := h S, e ψ i where e ψ ( t ) := ψ ( − t ). NIR LEV h b δ Λ , q ε i = h δ Λ , b q ε i = X λ ∈ Λ p ε ( − λ ) = X λ ∈ Λ ( h ∗ α )( − λ ) χ ε ( − λ )= X λ ∈ Λ χ ε ( − λ ) Z R α ( − x ) h ( x − λ ) dx. (3.5)Now we need the following: Claim.
We have X λ ∈ Λ Z R | α ( − x ) | · | h ( x − λ ) | dx < + ∞ . (3.6)We observe that | χ ε ( − λ ) | χ ε ( − λ ) → ε → λ . Hence the claimallows us to apply the dominated convergence theorem to the sum (3.5), which yieldslim ε → h b δ Λ , q ε i = X λ ∈ Λ Z R α ( − x ) h ( x − λ ) dx. (3.7)The claim also allows us to exchange the sum and integral in (3.7), and it follows thatlim ε → h b δ Λ , q ε i = Z R α ( − x ) X λ ∈ Λ h ( x − λ ) dx = Z R ( h ∗ δ Λ )( x ) b β ( x ) dx. (3.8)Comparing (3.4) and (3.8), we see that (3.3) holds.It remains to prove the claim. Indeed, we have X λ ∈ Λ Z R | α ( − x ) | · | h ( x − λ ) | dx = Z R | h ( − x ) | X λ ∈ Λ | α ( x − λ ) | dx. (3.9)The inner sum on the right hand side of (3.9) is a bounded function of x , since α is aSchwartz function and Λ has bounded density, while h is a function in L ( R ). Hencethe integral in (3.9) converges, and this completes the proof of the lemma. (cid:3) S ∈ P M ( R ), supp( S ) ⊂ ( a, b ) where 0 < a < b < , φ ∈ A ( R ) isa function with supp( φ ) ⊂ ( a, b ), φ vanishes on supp( S ), and h S, φ i 6 = 0. Let ψ bea smooth function whose zero set Z ( ψ ) is the same as Z ( φ ). In particular, we havesupp( ψ ) ⊂ ( a, b ) and ψ vanishes on supp( S ) as well. Let also τ be a smooth functionsatisfying τ ( − t ) = τ ( t ), supp( τ ) ⊂ ( − a, a ), and τ (0) = 1.Given a real scalar w we define two functions f, g ∈ L ( R ) by the conditions b f ( t ) = w · τ ( t ) + ψ ( t ) + ψ ( − t ) , (3.10) b g ( t ) = w · τ ( t ) + φ ( t ) + φ ( − t ) , (3.11)then f, g are real-valued and their Fourier transforms have the same set of zeros.By Lemma 3.3 the Fourier transform of f ∗ δ Λ is the pseudomeasure b f · T r = wδ + r ( Sψ + ( g Sψ )) = wδ , where the first equality is due to (3.1) and (3.10), while the second equality is true since ψ is smooth and vanishes on supp( S ), hence Sψ = 0. We conclude that f ∗ δ Λ = w a.e.,which means that f + Λ is a tiling at level w . N EXAMPLE CONCERNING FOURIER ANALYTIC CRITERIA FOR TILING 7
In the same way, Lemma 3.3 implies that the Fourier transform of g ∗ δ Λ is b g · T r = wδ + r ( Sφ + ( g Sφ )) . However in this case, Sφ is not the zero distribution, since h Sφ, i = h S, φ i 6 = 0. Thisshows that the Fourier transform of g ∗ δ Λ is not a scalar multiple of δ , and it followsthat g + Λ is not a tiling at any level.3.5. The above construction yields real-valued functions f and g , but these two func-tions need not be positive. We will now show that if the given scalar w is positive, thenthe construction can be modified so as to yield everywhere positive functions f, g .In what follows, φ and ψ continue to denote the same two functions as above. Step 1 : We show that there is a nonnegative sequence { c ( k ) } ∈ ℓ ( Z ), such that | b φ ( x ) | c ( k ) , k ∈ Z , | x − k | . (3.12)Indeed, we have φ ∈ A ( R ) and supp( φ ) ⊂ ( a, b ). Considered as a function in A ( T ), φ may be expressed on ( a, b ) as the sum of an absolutely convergent Fourier series. Hencethere is a finite (complex) measure µ supported on Z such that φ ( t ) = b µ ( − t ), t ∈ ( a, b ).Let Φ be an infinitely smooth function such that Φ( t ) = 1 if t ∈ supp( φ ), while Φ( t ) = 0for t ∈ R \ ( a, b ). Then φ ( t ) = b µ ( − t )Φ( t ) for every t ∈ R , which implies that b φ ( x ) = ( µ ∗ b Φ)( x ) = X n ∈ Z µ ( n ) b Φ( x − n ) , x ∈ R . Since the Fourier transform b Φ has fast decay, there is a sequence { γ ( k ) } ∈ ℓ ( Z ) suchthat | b Φ( x ) | | γ ( k ) | whenever | x − k | . It follows that | b φ ( x ) | c ( k ) := X n ∈ Z | µ ( n ) | · | γ ( k − n ) | , | x − k | , which establishes (3.12). Step 2 : We may assume that the same sequence { c ( k ) } also satisfies | b ψ ( x ) | c ( k ) , k ∈ Z , | x − k | . (3.13)Indeed, we may apply the same procedure from Step 1 also to the function ψ , andthen define { c ( k ) } to be the maximum of the two sequences obtained from both steps. Step 3 : Let { d ( k ) } be any positive sequence in ℓ ( Z ) such that d ( k ) > c ( k ) , k ∈ Z . (3.14)We show that there is τ ∈ A ( R ) such that supp( τ ) ⊂ ( − a, a ) and b τ ( x ) > d ( k ) , k ∈ Z , | x − k | . (3.15)Let χ be an infinitely smooth function, supp( χ ) ⊂ ( − a, a ), whose Fourier transform b χ is nonnegative and satisfies b χ ( x ) > − , ]. Let τ ( t ) := b ν ( − t ) χ ( t ),where ν is a positive, finite measure supported on Z defined by ν := P n d ( n ) δ n . Thenthe function b τ = ν ∗ b χ is in L ( R ), so that we have τ ∈ A ( R ), and b τ ( x ) = X n ∈ Z d ( n ) b χ ( x − n ) > d ( k ) b χ ( x − k ) > d ( k ) , | x − k | , NIR LEV which gives (3.15).
Step 4 : Now suppose that we are given a positive scalar w . We then define the twofunctions f, g by the conditions b f ( t ) = w · τ (0) − · h τ ( t ) + ψ ( t ) + ψ ( − t )2 i , (3.16) b g ( t ) = w · τ (0) − · h τ ( t ) + φ ( t ) + φ ( − t )2 i . (3.17)We observe that by the definition of the function τ we have τ (0) = b ν (0) χ (0) = ( R dν )( R b χ ( x ) dx ) > , and in particular τ (0) is nonzero. Then f, g are in L ( R ), their Fourier transforms havethe same set of zeros, and by the same argument as before one can verify that f + Λ isa tiling at level w , while g + Λ is not a tiling at any level.Finally we check that f and g are everywhere positive functions. Indeed, we have f ( x ) = w · τ (0) − · hb τ ( − x ) + Re( b ψ ( − x )) i ,g ( x ) = w · τ (0) − · hb τ ( − x ) + Re( b φ ( − x )) i , and by (3.12), (3.13), (3.14) and (3.15) it follows that f ( x ) , g ( x ) > x ∈ R .This completes the proof of Theorem 1.2 based on Theorem 3.2. (cid:3) It remains to prove Theorem 3.2. This will be done in the next section.4.
Kargaev’s implicit function method ⊂ R of positive and finite measure, such that the Fourier transform of its indicator function Ω vanishes on some open interval ( a, b )?The question was answered in the affirmative by Kargaev [Kar82]. The solution wasbased on an innovative application of the infinite-dimensional implicit function theorem,which established the existence of a set of the form Ω = S n ∈ Z [ n + α n , n + β n ], where { α n } , { β n } are two real sequences in ℓ , that has the above mentioned property.The approach was later used in [KL16] in order to prove the existence of non-periodictilings of R by translates of a function f . In that paper, a self-contained presentation ofthe method was given in a simplified form, that does not invoke the infinite-dimensionalimplicit function theorem.In this section, we use an adapted version of Kargaev’s method in order to proveTheorem 3.2 (and more, in fact). The presentation below generally follows the lines of[KL16, Sections 2, 3], but the proof also requires some additional arguments.4.2. Let { α n } , n ∈ Z , be a bounded sequence of real numbers. To such a sequence weassociate a function F on the real line, defined by F ( x ) = X n ∈ Z F n ( x ) , x ∈ R , (4.1)where F n is the function [ n,n + α n ] if α n >
0, or − [ n + α n ,n ] if α n < N EXAMPLE CONCERNING FOURIER ANALYTIC CRITERIA FOR TILING 9
Since the sequence { α n } is bounded, the series (4.1) is easily seen to converge in thespace of temperate distributions to a bounded function F on R . In particular, F is atemperate distribution. Theorem 4.1.
Given two numbers b ∈ (0 , ) and ε > , there is δ > with the followingproperty: Let S be a Schwartz distribution on R satisfying S ( − t ) = S ( t ) , supp( S ) ⊂ ( − b, b ) , sup k ∈ Z | b S ( k ) | δ. (4.2) Then there is a bounded, real sequence α = { α n } , n ∈ Z , such that k α k ∞ ε and b F = S in ( − b, b ) , (4.3) where F is the function defined by (4.1) . The proof of Theorem 4.1 is given below. It is divided into a series of lemmas.4.3. Given a number b ∈ (0 , ) we choose l = l ( b ) such that b < l < . We also choosean infinitely smooth function Φ satisfying the conditions Φ( − t ) = Φ( t ) for all t ∈ R ,Φ( t ) = 1 for t ∈ [ − b, b ], and Φ( t ) = 0 for t ∈ R \ ( − l, l ).Let I := [ − , ]. Denote by b ψ ( k ) the k ’th Fourier coefficient of a function ψ on I : b ψ ( k ) = Z I ψ ( t ) e − πikt dt, k ∈ Z . (4.4)The following lemma is inspired by [KV92, Lemma 2.2]. Lemma 4.2.
Let ϕ ∈ C ∞ ( R ) , ϕ (0) = 0 , and denote ψ s ( t ) := ϕ ( st )Φ( t ) . Then | b ψ s ( k ) | C | s | | k | m (4.5) for every s ∈ [ − , and k ∈ Z , where C = C (Φ , ϕ, m ) > is a constant which dependsneither on s nor on k .Proof. First suppose that k = 0. We have | ϕ ( t ) | C | t | for t ∈ I , hence | b ψ s (0) | C | s | Z I | t Φ( t ) | dt = C | s | . (4.6)Next we assume that k = 0. We integrate by parts m times and use the fact that thefunction ψ s vanishes in a neighborhood of the points ± . This yields b ψ s ( k ) = 1(2 πik ) m Z I ψ ( m ) s ( t ) e − πikt dt. (4.7)By the product rule for the m ’th derivative we have ψ ( m ) s ( t ) = m X j =0 (cid:18) mj (cid:19) s j ϕ ( j ) ( st )Φ ( m − j ) ( t ) . (4.8)Combining (4.7) and (4.8) yields the estimate | b ψ s ( k ) | π | k | ) m m X j =0 (cid:18) mj (cid:19) | s | j Z I | ϕ ( j ) ( st )Φ ( m − j ) ( t ) | dt. Since the derivatives ϕ ′ , ϕ ′′ , . . . , ϕ ( m ) are bounded on I , each one of the terms in thesum corresponding to j = 1 , , . . . , m is bounded by C | s | , while the term correspondingto j = 0 can be estimated using | ϕ ( t ) | C | t | , t ∈ I , which again yields C | s | . (cid:3) Lemma 4.3.
Let ϕ ∈ C ∞ ( R ) , ϕ ′ (0) = 0 , and denote ψ u,v ( t ) := ϕ ( vt ) − ϕ ( ut ) t · Φ( t ) . (4.9) Then | b ψ u,v ( k ) | max {| u | , | v |} · C | v − u | | k | m (4.10) for every u, v ∈ [ − , and k ∈ Z , where C = C (Φ , ϕ, m ) > is a constant which doesnot depend on u , v or k .Proof. We may suppose that u < v . We observe that ψ u,v ( t ) = Z vu ϕ ′ ( st )Φ( t ) ds = Z vu ψ s ( t ) ds, (4.11)where we define ψ s ( t ) := ϕ ′ ( st )Φ( t ). Hence b ψ u,v ( k ) = Z vu b ψ s ( k ) ds, k ∈ Z . (4.12)By Lemma 4.2 the estimate (4.5) is valid for every s ∈ [ u, v ], where C = C (Φ , ϕ, m ) > s or k . Hence | b ψ u,v ( k ) | C | k | m Z vu | s | ds (4.13)from which (4.10) follows. (cid:3) T is a Schwartz distribution supported on [ − l, l ], then b T ( k ) denotes the k ’thFourier coefficients of T : b T ( k ) = h T, e − πikt i , k ∈ Z . Lemma 4.4.
Let T be a distribution supported on [ − l, l ] . Then the series X n ∈ Z b T ( n ) e πint (4.14) converges unconditionally in the distributional sense to T in the open interval ( − , ) . This follows from the unconditional convergence of the series (4.14) to T consideredas a distribution on the circle T .4.5. Let X be the space of all bounded sequences of real numbers α = { α n } , n ∈ Z ,endowed with the norm k α k X := sup n ∈ Z | α n | that makes X into a real Banach space.Let Y be the space of distributions T supported on [ − l, l ] whose Fourier coefficients b T ( k ), k ∈ Z , are real and bounded. If we endow Y with the norm k T k Y := sup k ∈ Z | b T ( k ) | (4.15) N EXAMPLE CONCERNING FOURIER ANALYTIC CRITERIA FOR TILING 11 then also Y is a real Banach space, which can be viewed as a closed subspace of P M ( T ).We observe that a distribution T supported on [ − l, l ] has real Fourier coefficients(that is, b T ( k ) ∈ R for every k ∈ Z ) if and only if T ( − t ) = T ( t ). Lemma 4.5.
Let { T n } , n ∈ Z , be a sequence of elements of Y . Assume that there is asequence γ ∈ X such that | b T n ( k ) | | γ n | | k − n | (4.16) for every n and k in Z . Then the series T = X n ∈ Z T n (4.17) converges unconditionally in the distributional sense to an element T ∈ Y satisfying k T k Y K k γ k X (4.18) where K is an absolute constant.Proof. Indeed, the condition (4.16) implies that for any φ ∈ A ( T ) we have |h T n , φ i| = (cid:12)(cid:12)(cid:12) X k ∈ Z b T n ( k ) b φ ( − k ) (cid:12)(cid:12)(cid:12) | γ n | X k ∈ Z | b φ ( − k ) | | k − n | , and hence X n ∈ Z |h T n , φ i| K k γ k X k φ k A ( T ) , K := X n ∈ Z
11 + | n | . This shows that the series (4.17) converges unconditionally in the weak* topology ofthe space
P M ( T ) (the dual of A ( T )) to an element T ∈ Y satisfying (4.18). (cid:3) α = { α n } , n ∈ Z , be a sequence in X such that k α k X
1. Define( Rα )( t ) := X n ∈ Z e πint · e πiα n t − − πiα n t πit · Φ( t ) . (4.19)Let T n be the n ’th term of the series (4.19). We observe that T n ∈ Y . If we applyLemma 4.2 to the function ϕ ( t ) := ( e πit − − πit ) / (2 πit ) with s = α n and m = 2,then it follows from the lemma that condition (4.16) is satisfied with γ n := Cα n , where C > α , k or n . Hence by Lemma 4.5 the series (4.19) convergesin the distributional sense to an element of the space Y , and we have k Rα k Y C k α k X , α ∈ X, k α k X , (4.20)where the constant C does not depend on α .We note that the mapping R defined by (4.19) is nonlinear . r > U r denote the closed ball of radius r around the origin in X : U r := { α ∈ X : k α k X r } . (4.21) Lemma 4.6.
Given any ρ > there is < r < such that k Rβ − Rα k Y ρ k β − α k X , α, β ∈ U r . (4.22) In particular, if r is small enough then R is a contractive (nonlinear) mapping on U r .Proof. Let α, β ∈ U r (0 < r < Rβ − Rα )( t ) = X n ∈ Z e πint · ( e πiβ n t − πiβ n t ) − ( e πiα n t − πiα n t )2 πit · Φ( t ) . (4.23)Let T n be the n ’th element of the series (4.23). We apply Lemma 4.3 to the function ϕ ( t ) := ( e πit − πit ) / (2 πi ) with u = α n , v = β n and m = 2. The lemma implies thatthe condition (4.16) is satisfied with γ n := Cr · ( β n − α n ), where the constant C doesnot depend on r , α , β , k or n . It therefore follows from Lemma 4.5 that we have theestimate k Rβ − Rα k Y Cr k β − α k X where C is a constant not depending on r , α or β . Hence it suffices to choose r small enough so that Cr ρ . (cid:3) T ∈ Y we denote by F ( T ) the sequence of Fourier coefficientsof T , namely, the sequence { b T ( k ) } , k ∈ Z . This defines a linear mapping F : Y → X satisfying kF ( T ) k X = k T k Y . Lemma 4.7.
Given any ε > there is δ > with the following property: Let S ∈ Y , k S k Y δ . Then one can find an element T ∈ Y , k T − S k Y ε k S k Y , which solves theequation T + R ( F ( T )) = S .Proof. Fix S ∈ Y such that k S k Y δ , and let B = B ( S, ε ) := { T ∈ Y : k T − S k Y ε k S k Y } . We observe that if T ∈ B then k T k Y (1 + ε ) k S k Y . Define a map H : B → Y by H ( T ) := S − R ( F ( T )) , T ∈ B, and notice that an element T ∈ B is a solution to the equation T + R ( F ( T )) = S ifand only if T is a fixed point of the map H .Let us show that if δ is small enough then H ( B ) ⊂ B . Indeed, if T ∈ B then using(4.20) we have k H ( T ) − S k Y = k R ( F ( T )) k Y C kF ( T ) k X = C k T k Y C (1 + ε ) k S k Y . Hence if we choose δ such that C (1 + ε ) δ ε then we obtain k H ( T ) − S k Y ε k S k Y , and it follows that H ( B ) ⊂ B .It also follows from Lemma 4.6 that if δ is small enough, then H is a contractivemapping from the closed set B into itself. Indeed, let T , T ∈ B , then we have k H ( T ) − H ( T ) k Y = k R ( F ( T )) − R ( F ( T )) k Y ρ kF ( T ) − F ( T ) k X = ρ k T − T k Y , where 0 < ρ <
1. Then the Banach fixed point theorem implies that H has a (unique)fixed point T ∈ B , which yields the desired solution. (cid:3) N EXAMPLE CONCERNING FOURIER ANALYTIC CRITERIA FOR TILING 13
Proof of Theorem 4.1 . Let S be a Schwartz distribution satisfying (4.2). Then S ∈ Y and k S k Y δ . Define S ( t ) := S ( − t ), then also S is a distribution in Y andwe have k S k Y = k S k Y . By Lemma 4.7, if δ is small enough then there is an element T ∈ Y , k T − S k Y ε k S k Y , which solves the equation T + R ( F ( T )) = S .Let α ∈ X be the sequence defined by α n = b T ( n ), n ∈ Z , then k α k X ε providedthat δ is small enough. Let F be the function given by (4.1) that is associated to thissequence α = { α n } . We have b F ( − t ) = lim N →∞ X | n | N b F n ( − t )in the sense of distributions, and b F n ( − t ) = e πint · e πiα n t − πit . Hence b F ( − t ) = lim N →∞ h X | n | N α n e πint + X | n | N e πint · e πiα n t − − πiα n t πit i . The first sum converges to T in ( − b, b ) according to Lemma 4.4; while the second sumconverges to Rα in ( − b, b ), which is due to (4.19) and the fact that Φ( t ) = 1 on ( − b, b ).We conclude that b F ( − t ) = ( T + Rα )( t ) = S ( t ) in ( − b, b ).This means that b F = S in ( − b, b ) and thus Theorem 4.1 is proved. (cid:3) Theorem 4.8.
Given two numbers a, b such that < a < b < , and given ε > , thereis δ > with the following property: Let S be a distribution on R satisfying S ( − t ) = S ( t ) , supp( S ) ⊂ ( − b, − a ) ∪ ( a, b ) , sup k ∈ Z | b S ( k ) | δ. (4.24) Then there is a real sequence
Λ = { λ n } , n ∈ Z , such that | λ n − n | ε for all n , and b δ Λ = δ + S in ( − b, b ) . (4.25) Proof.
We choose an infinitely smooth function Ψ such that Ψ( − t ) = Ψ( t ) for all t ∈ R ,Ψ( t ) = − / (2 πit ) in ( − b, − a ) ∪ ( a, b ), and Ψ( t ) = 0 for t ∈ R \ ( − l, l ).Let S be a distribution satisfying (4.24), then S ∈ Y and k S k Y δ . Define a newdistribution S := S · Ψ, then also S ∈ Y . We have k S k Y M k S k Y , M := k Ψ k A ( T ) . By Theorem 4.1, if δ is small enough then there is a sequence α ∈ X , k α k X ε , suchthat the function F defined by (4.1) satisfies b F = S in ( − b, b ). It follows that thedistributional derivative F ′ of the function F satisfies c F ′ ( t ) = 2 πit b F ( t ) = 2 πitS ( t ) = 2 πit Ψ( t ) S ( t ) = − S ( t ) in ( − b, b ) , (4.26)which is true since 2 πit Ψ( t ) = − S ). Let Λ = { λ n } , n ∈ Z , be defined by λ n := n + α n . Then we have | λ n − n | ε for all n . It follows from the definition (4.1) of F that F ′ = X n ∈ Z ( δ n − δ λ n ) = δ Z − δ Λ , that is, δ Λ = δ Z − F ′ . By Poisson’s summation formula b δ Z = δ Z , hence we have b δ Λ = δ Z − c F ′ = δ + S in ( − b, b )due to (4.26). The proof of Theorem 4.8 is thus concluded. (cid:3) S is apseudomeasure on R such that supp( S ) ⊂ ( a, b ), then the distribution r ( S + e S ) satisfiesthe conditions (4.24) if r > { λ n } with the properties as in the statement of Theorem 3.2.5. Addendum: A problem of Kolountzakis
The following question was posed to us by Kolountzakis: Does there exist a realsequence Λ = { λ n } , n ∈ Z , satisfying A λ n +1 − λ n B, n ∈ Z , (5.1)where A, B > f + Λ is a tiling for some nonzero f ∈ L ( R ),but there is no nonnegative f with this property?We will prove the following result, which answers the above question affirmatively: Theorem 5.1.
There is a real sequence
Λ = { λ n } , n ∈ Z , satisfying (5.1) for whichthere exist tilings f + Λ with nonzero f ∈ L ( R ) , but any such a tiling is necessarily atiling at level zero. In particular Λ cannot tile with any nonnegative (nonzero) f .Proof. Let a, b be two numbers such that 0 < a < b < . Let ψ be a smooth evenfunction, ψ ( t ) > − a, a ), and ψ ( t ) = 0 outside ( − a, a ). By Theorem 4.1, given any ε > α = { α n } , n ∈ Z , satisfying | α n | ε for all n , and suchthat b F ( t ) = rψ ( t ) in ( − b, b ) for some r >
0, where F is the function defined by (4.1).Let the sequence Λ = { λ n } , n ∈ Z , be defined by λ n := n + α n . Then b δ Λ = δ − πirtψ ( t ) in ( − b, b ). (5.2)This can be shown in the same way as done in the proof of Theorem 4.8 above.In particular we have b δ Λ = 0 in the open set G := ( − b, − a ) ∪ ( a, b ), so there existnonzero real-valued Schwartz functions f such that f + Λ is a tiling (at level zero). Itsuffices to choose f such that supp( b f ) is contained in G .On the other hand, suppose that there is f ∈ L ( R ) such that f + Λ is a tiling atsome nonzero level w . Then f must have nonzero integral, see [KL96, Lemma 2.3(i)].In turn this implies [KL16, Section 4] that b δ Λ = c · δ in some neighborhood ( − η, η ) ofthe origin, where c is a (nonzero) scalar. This contradicts (5.2), hence no such f exists.In particular, if f is nonnegative and f + Λ is a tiling, then the tiling level must bezero and f vanishes a.e. (cid:3) N EXAMPLE CONCERNING FOURIER ANALYTIC CRITERIA FOR TILING 15
Remark 5.2.
We note that for a set Λ ⊂ R of bounded density to admit no tilings f + Λ with a nonnegative (nonzero) f , it is not only sufficient, but also necessary, thatany tiling f + Λ be a tiling at level zero. Indeed, suppose that Λ admits a tiling at somenonzero level. Then, as in the proof of Theorem 5.1, this implies that b δ Λ = c · δ in someinterval ( − η, η ), where c is a nonzero scalar. It follows that f + Λ is a tiling whenever f is a Schwartz function with supp( b f ) ⊂ ( − η, η ). In particular, there exist tilings f + Λwith f nonnegative. 6. Remarks ⊂ R be a discrete set of bounded density. One can show, see [KL20,Theorem 2.2], that if the temperate distribution b δ Λ is a measure on R , then condition(1.6) is not only necessary, but also sufficient, for a function f ∈ L ( R ) to tile at somelevel w with the translation set Λ. In this case, the tiling level is given by w = c (Λ) b f (0),where c (Λ) is the mass that the measure b δ Λ assigns to the origin.For example, if Λ is a periodic set then b δ Λ is a (pure point) measure, and f + Λ is atiling if and only if (1.6) holds. It follows that the set Λ in Theorem 1.2 is not periodic,nor can it be represented as a finite union of periodic sets.6.2. The proof of Theorem 1.2 shows that the functions f, g can be chosen smooth,and moreover such that f has fast decay, | f ( x ) | = o ( | x | − N ) as | x | → + ∞ for every N .However the function g in the theorem cannot decay fast.6.3. Theorem 1.2 also holds in R d for any d >
1. This can be deduced from theone-dimensional result by taking cartesian products.
Acknowledgement.
The author is grateful to Mihalis Kolountzakis for motivatingdiscussions and for posing the problem discussed in Section 5, as well as the problemabout positivity of the functions in Theorem 1.2.
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