# On Characterizations of Fourier Frames and Tilings

Authors: Dao-Xin Ding; Hai-Xiong Li
Journal: Journal of Applied Mathematics
 (2013)

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## Abstract

We give some characterizations of Fourier frames and tilings and obtain a more general form of characterizations of spectra and tilings.


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## Body

## 1. Introduction

A countable family of elements{fn}n∈ℕ in a separable Hilbert space H is called a frame if there are positive constants A,B such that
(1)A∥f∥2≤∑n∈ℕ‍|〈f,fn〉|2≤B∥f∥2
for all f∈H. A and B are called frame bounds. The sequence is called a tight frame if A=B. The sequence is called Bessel if the second inequality above holds. In this case, B is called the Bessel bound. Frames were first introduced by Duffin and Schaeffer [1] in the context of nonharmonic Fourier series, and today they have applications in a wide range of areas. A frame can be considered as a generalized basis in the sense that every element in H can be written as a linear combination of the frame elements.In this paper, we consider Fourier frames for a special separable Hilbert space. LetΩ⊂ℝd have positive Lebesgue measure m(Ω)>0 and let Λ be a discrete subset of ℝd. The inner product and the norm on L2(Ω) are
(2)〈f(x),g(x)〉Ω=1m(Ω)∫Ω‍f(x)g(x)¯dx,∥f∥Ω2=1m(Ω)∫Ω‍|f(x)|2dx.
We write
(3)eλ(x):=e2πi〈λ,x〉forx∈ℝd,ℰ(Λ):={eλ(x):λ∈Λ}.
If ℰ(Λ) is a frame or an orthonormal basis for L2(Ω), then ℰ(Λ) and (Ω,Λ) are called a Fourier frame and a spectral pair, respectively. In the case of the spectral pair, the Λ is then called a spectrum for Ω and Ω is called a spectral set. We follow the terminology of [2] and consider the packing and tiling in ℝd by compact set Ω of the following kind.A compact setΩ in ℝd is a regular region if it has positive Lebesgue measure, is the closure of its interior Ω∘, and has a boundary ∂Ω=Ω∖Ω∘ of measure zero. If Ω is a regular region, then a discrete set Λ is a packing set for Ω if the sets {Ω+λ:λ∈Λ} have disjoint interiors or the intersections (Ω+λ)∩(Ω+μ) for λ≠μ in Λ have measure zero. It is a tiling set if, further, the translates {Ω+λ:λ∈Λ} cover ℝd up to measure zero. In these cases, we say that Ω+Λ is a packing or tiling of ℝd, respectively. Equivalently, we call (Ω,Λ) a packing pair or a tiling pair, respectively.It is well known that spectral sets and tilings are connected by the following conjecture of Fuglede [3].Spectral Set ConjectureA setΩ in ℝd is a spectral set if and only if it tiles ℝd by translations.Many people attempt to prove the spectral set conjecture for some special sets, although the conjecture is false in many cases (see [4–7]). For example, Jorgensen and Pedersen [8] conjectured that ([0,1]n,Λ) is a spectral pair if and only if ([0,1]n,Λ) is a tiling pair. They established the conjecture for dimension n≤3 and for all n when Λ is a discrete periodic set. Iosevich and Pedersen [9] simultaneously and independently established the above-mentioned conjecture by a different approach based on a geometric argument. Kolountzakis [10] gave an alternative proof of this fact, which is based on a characterization of translational tiling by a Fourier analytic criterion. Lagarias et al. [2] related the spectra of sets Ω to tiling in the Fourier space and obtained the following characterization of spectra and tilings.Theorem 1.
LetΩ be a regular region in ℝd and let Λ be such that the set of exponentials ℰ(Λ) is orthogonal for L2(Ω). Suppose that D is a regular region with m(Ω)m(D)=1 such that D+Λ is a packing of ℝd. Then, Λ is a spectrum for Ω if and only if D+Λ is a tiling of ℝd.Li [11] presented an elementary approach to obtain a more general form of Theorem 1. Enlightened by the ideas from [11, 12], we give some characterizations of Fourier frames and tilings and extend several results in [2] and [11].


## 2. Main Results and Their Proofs

Throughout this section, letΩ and D be two regular regions in ℝd. By the definition of frames, we may get that the following lemma.Lemma 2.
LetΔ⊂ℝd be a discrete set. If ℰ(Δ) is a frame for L2(Ω) with frame bounds A,B, then
(4)A(m(Ω))2≤∑δ∈Δ‍|χΩ^(t-δ)|2≤B(m(Ω))2,∀t∈ℝd,
where
(5)χΩ^(u)=∫ℝd‍χΩ(x)e-2πi〈u,x〉dx,u∈ℝd
is the Fourier transform of the characteristic function χΩ(x).Proof.
By the frame inequality (1), for any t∈ℝd, we have
(6)∑δ∈Δ‍|χΩ^(t-δ)|2=∑δ∈Δ‍|m(Ω)〈et(x),eδ(x)〉Ω|2≤B(m(Ω))2∥et(x)∥Ω2=B(m(Ω))2.
Similarly, we get A(m(Ω))2≤∑δ∈Δ‍|χΩ^(t-δ)|2.Remark 3.
In the case forA=B, if ℰ(Δ) is a tight frame for L2(Ω) with the frame bound B, then
(7)∑δ∈Δ‍|χΩ^(t-δ)|2=B(m(Ω))2,∀t∈ℝd.
If ℰ(Δ) is a Bessel sequence for L2(Ω) with the Bessel bound B, then
(8)∑δ∈Δ‍|χΩ^(t-δ)|2≤B(m(Ω))2,∀t∈ℝd.
Moreover, ℰ(Δ) is an orthonormal basis for L2(Ω) if and only if
(9)∑δ∈Δ‍|χΩ^(t-δ)|2=(m(Ω))2,∀t∈ℝd.
Since |χΩ^(u)|=|χΩ^(-u)|,  for all u∈ℝd, if we substitute “−” for “+” in (4), (7), (8), and (9), all the above results also hold.In the remainder of this paper, we assume thatΘ⊂ℝd is a discrete subset and Λ and Γ are two finite subsets of ℝd such that Θ+Λ and Θ+Γ are two direct sums.Theorem 4.
Ifℰ(Θ+Λ) is a frame for L2(Ω) with frame bounds A,B, and (D,Θ+Γ) is a tiling pair, then
(10)#ΛB#Γ≤m(Ω)m(D)≤#ΛA#Γ,
where # denotes the cardinality of some set.Proof.
LetΛ:={λ1,λ2,…,λp} and Γ:={γ1,γ2,…,γq} with #Λ=p and #Γ=q. Since ℰ(Θ+Λ) is a frame for L2(Ω) with frame bounds A,B, it follows from Lemma 2 that
(11)A(m(Ω))2≤∑i=1p‍∑θ∈Θ‍|χΩ^(t+θ+λi)|2≤B(m(Ω))2,≤∑i=1p‍∑θ∈Θ‍|χΩ^(t+θ+λi)|2≤B(m(Ω))265∀t∈ℝd.
Note that (D,Θ+Γ) is a tiling pair, from the Plancherel’s formula on L2(ℝd), we have the following:
(12)m(Ω)=∥χΩ∥ℝd2=∥χΩ^∥ℝd2=∫ℝd‍|χΩ^(t)|2dt=1p∫ℝd‍∑i=1p‍|χΩ^(t+λi)|2dt=1p∫∪θ∈Θ,1≤j≤q(D+θ+γj)‍∑i=1p‍|χΩ^(t+λi)|2dt=1p∑j=1q‍∑θ∈Θ‍∫D‍∑i=1p‍|χΩ^(t+θ+λi+γj)|2dt=1p∑j=1q‍∫D‍∑θ∈Θ‍∑i=1p‍|χΩ^(t+θ+λi+γj)|2dt=1p∑j=1q‍∫D‍∑i=1p‍∑θ∈Θ‍|χΩ^(t+θ+λi+γj)|2dt≤qpB(m(Ω))2m(D).
The bottom third equality holds by Lebesgue dominated convergence theorem and the last inequality follows from (11). Thus, p/qB≤m(Ω)m(D). Similarly, we get m(Ω)m(D)≤p/qA. Hence, the proof is completed.Since an orthonormal basis is also a tight frame with frame boundsA=B=1, we get the following corollary.Corollary 5.
Ifℰ(Θ+Λ) is an orthonormal basis for L2(Ω), and (D,Θ+Γ) is a tiling pair, then m(Ω)m(D)=#Λ/#Γ.Lemma 6.
LetΘ+Λ be such that the set of exponentials ℰ(Θ+Λ) is a Bessel sequence for L2(Ω) with the Bessel bound B. If D+Θ+Γ is a tiling of ℝd with m(Ω)m(D)≤#Λ/B#Γ, then
(13)∑λ∈Λ‍∑θ∈Θ‍|χΩ^(t+θ+λ)|2=B(m(Ω))2,∀t∈ℝd.Proof.
Keep the assumptions onΛ and Γ in the above proof. Since ℰ(Θ+Λ) is a Bessel sequence for L2(Ω) with the Bessel bound B, it follows from Remark 3 that
(14)∑i=1p‍∑θ∈Θ‍|χΩ^(t+θ+λi)|2≤B(m(Ω))2,∀t∈ℝd.
Since D+Θ+Γ is a tiling of ℝd and m(Ω)m(D)≤#Λ/B#Γ=p/qB, for any y∈ℝd, then we have
(15)m(Ω)=∫ℝd‍|χΩ^(t)|2dt=1p∫ℝd‍∑i=1p‍|χΩ^(t+λi)|2dt=1p∫∪θ∈Θ,1≤j≤q(D+y+θ+γj)‍∑i=1p‍|χΩ^(t+λi)|2dt=1p∑j=1q‍∑θ∈Θ‍∫D+y‍∑i=1p‍|χΩ^(t+θ+λi+γj)|2dt=1p∑j=1q‍∫D+y‍∑θ∈Θ‍∑i=1p‍|χΩ^(t+θ+λi+γj)|2dt=1p∑j=1q‍∫D+y‍∑i=1p‍∑θ∈Θ‍|χΩ^(t+θ+λi+γj)|2dt≤qpB(m(Ω))2m(D)≤m(Ω),
which yields
(16)∑i=1p‍∑θ∈Θ‍|χΩ^(t+θ+λi)|2=B(m(Ω))2
for almost every t in D+y. Since y is arbitrary, (16) holds for almost every t in ℝd. By the continuity of the function on the left side of (16), we see that (16) holds for every t in ℝd.Theorem 7.
Ifℰ(Θ+Λ) is orthogonal in L2(Ω) and (D,Θ+Γ) is a tiling pair with m(Ω)m(D)=#Λ/#Γ, then ℰ(Θ+Λ) is an orthonormal basis for L2(Ω).Proof.
The proof is straightforward by the above lemma.Theorem 8.
Suppose thatD+Θ+Γ is a packing of ℝd with #Λ/A#Γ≤m(Ω)m(D). If ℰ(Θ+Λ) is a frame for L2(Ω) with the frame bounds A,B, then D+Θ+Γ is a tiling of ℝd.Proof.
Sinceℰ(Θ+Λ) is a frame for L2(Ω) with the frame bounds A,B, then (11) holds. If D+Θ+Γ is a packing of ℝd, then it follows from (11) and #Λ/A#Γ≤m(Ω)m(D) that
(17)m(Ω)=∫ℝd‍|χΩ^(t)|2dt=1p∫ℝd‍∑i=1p‍|χΩ^(t+λi)|2dt≥1p∫∪θ∈Θ,1≤j≤q(D+θ+γj)‍∑i=1p‍|χΩ^(t+λi)|2dt=1p∑j=1q‍∑θ∈Θ‍∫D‍∑i=1p‍|χΩ^(t+θ+λi+γj)|2dt=1p∑j=1q‍∫D‍∑θ∈Θ‍∑i=1p‍|χΩ^(t+θ+λi+γj)|2dt=1p∑j=1q‍∫D‍∑i=1p‍∑θ∈Θ‍|χΩ^(t+θ+λi+γj)|2dt≥qpA(m(Ω))2m(D)≥m(Ω).
Thus, D+Θ+Γ is a tiling of ℝd.It is clear that the above theorem yields the following corollary.Corollary 9.
Suppose thatD+Θ+Γ is a packing of ℝd with #Λ/#Γ=m(Ω)m(D). If ℰ(Θ+Λ) is an orthonormal basis for L2(Ω), then D+Θ+Γ is a tiling of ℝd.Combining Theorem7 with Corollary 9, we obtain a more general form of the theorem in [11] and Theorem 1.Theorem 10.
Suppose thatℰ(Θ+Λ) is orthogonal in L2(Ω), and (D,Θ+Γ) is a packing pair with m(Ω)m(D)=#Λ/#Γ. Then, (Ω,Θ+Λ) is a spectral pair if and only if (D,Θ+Γ) is a tiling pair.Example 11.
Letp,q be two positive integers. Take the following:
(18)Ω=[0,1],D=[0,pq],Θ={pk:k∈ℤ},Λ={0,1,…,p-1},Γ={0,pq,2pq,…,(q-1)pq}.
We see that m(Ω)m(D)=#Λ/#Γ, (Ω,Θ+Λ) is a spectral pair and (D,Θ+Γ) is a tiling pair.



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