Arbitrarily sparse spectra for self-affine spectral measures
aa r X i v : . [ m a t h . F A ] J un ARBITRARILY SPARSE SPECTRA FOR SELF-AFFINE SPECTRALMEASURES
LIXIANG AN AND CHUN-KIT LAI
Abstract.
Given an expansive matrix R ∈ M d ( Z ) and a finite set of digit B taken from Z d /R ( Z d ).It was shown previously that if we can find an L such that ( R, B, L ) forms a Hadamard triple, thenthe associated fractal self-affine measure generated by (
R, B ) admits an exponential orthonormalbasis of certain frequency set Λ, and hence it is termed as a spectral measure. In this paper, we showthat if
B < | det( R ) | , not only it is spectral, we can also construct arbitrarily sparse spectrum Λin the sense that its Beurling dimension is zero. In memory of Dr. Tian-You Hu Introduction
Definitions and main results.
Let R ∈ M d ( Z ) be an expansive matrix (i.e. all of itseigenvalues have modulus strictly greater than 1). Let B, L ⊂ Z d be finite sets of integer vectorswith q := L = B . We say that the system ( R, B, L ) forms a
Hadamard triple if the matrix H = 1 √ q h e − πi h R − b,l i i b ∈ B,l ∈ L is unitary, i.e., H ∗ H = I .Given an expansive matrix R ∈ M d ( Z ) and given B ⊂ Z d . By the result of Hutchinson [12], wecan define the affine iterated function system (IFS) { τ b ( x ) = R − ( x + b ) , x ∈ R d , b ∈ B } which hasa unique compact attractor T ( R, B ), called self-affine set , satisfying T ( R, B ) = [ b ∈ B τ b ( T ( R, B )) . The self-affine measure (with equal weights) is the unique probability measure µ = µ ( R, B ) satis-fying µ ( E ) = 1 q X b ∈ B µ ( τ − b ( E ))for all Borel subsets E of R d . This measure is supported on the attractor T ( R, B ). Mathematics Subject Classification.
Key words and phrases.
Beurling dimension, spectral measure, self-affine measures.The research of Lixiang An is supported by NSFC grant 11601175. T.-Y Hu was a dedicated mathematician. He was also a great mentor and a dear friend of the authors of thispaper. Academically, he introduced us into the field of spectral measures through his paper [11]. It was sad to hearthat he passed away due to Covid-19.
In a previous work, Dutkay, Hausserman and the second-named author proved the followingtheorem ([8], see [17] for the proof on R ). Theorem 1.1.
Suppose that ( R, B, L ) forms a Hadamard triple on R d . Then the self-affine measure µ ( R, B ) admits an exponential orthonormal basis E (Λ) = { e πiλ · x : λ ∈ Λ } for some countable set Λ ⊂ R d . We say that a Borel probability measure µ on R d is called a spectral measure if we can find acountable set Λ ⊂ R d such that the set of exponential functions E (Λ) := { e πiλ · x : λ ∈ Λ } forms anorthonormal basis for L ( µ ). If such Λ exists, then Λ is called a spectrum for µ . The above theoremsaid that µ ( R, B ) is a spectral measure if (
R, B, L ) forms a Hadamard triple with some L ⊂ Z d .In this paper, We study in more detail about the sparseness of the spectrum as measured byBeurling dimension. Definition 1.2.
Denote Q dh ( x ) = x + [ − h, h ] d be the cube centered at x on R d .(1) Let Λ be a discrete subset of R d . For r >
0, the upper Beurling density corresponding to r (or r -Beurling density ) is defined by D + r (Λ) = lim sup h →∞ sup x ∈ R d ∩ Q dh ( x )) h r . (2) The upper Beurling dimension (or simply the Beurling dimension ) of Λ is defined bydim + (Λ) := sup { r > D + r (Λ) > } = inf { r > D + r (Λ) < ∞} . Our main result is the following.
Theorem 1.3. If ( R, B, L ) forms a Hadamard triple on R d with B < | det R | , then the spectralmeasure µ ( R, B ) admits a spectrum Λ with Beurling dimension zero. Historical overview.
Historically, spectral measure was first studied by Fuglede who intro-duced the notion of spectral sets and explored its relationship with translational tile [9]. The firstsingularly continuous spectral measure was found by Jorgensen and Pedersen [15]. They showedthat the standard middle-fourth Cantor measures µ is spectral, while the middle-third Cantormeasure µ is not spectral. µ is generated by R = 4 and B = { , } . ( R, B, L ) forms a Hadamardtriple with L = { , } . Using L , they found that one of the spectra of µ is given byΛ = n − X j =0 j ǫ j : ǫ j ∈ { , } , n ≥ . This is, however, not the only spectrum for µ . We can also see that (4 , { , } , L n ) with L n = { , n } also form Hadamard triples. Indeed, one can also show that 5 n Λ all are spectra of µ . A directcalculation shows that all these spectra have Beurling dimension ln 2 / ln 4.In an attempt to capture the right density condition for the spectra of µ , Dutkay, Han, Sunand Weber [5] proposed the notion of Beurling dimension, and they brought this notion from thestudy of Gabor pseudo-frame in [2]. In [5], the authors showed that all spectra of µ must have RBITRARILY SPARSE SPECTRA FOR SELF-AFFINE SPECTRAL MEASURES 3
Beurling dimension at most ln 2 / ln 4 which is the Hausdorff dimension of the attractor. Under atechnical condition on the spectrum Λ, a spectrum of µ must have a Beurling dimension ln 2 / ln 4.It used to be a conjecture that the technical condition can be removed. However, Dai, He andthe second-named author disproved the conjecture by exhibiting a spectrum of Beurling dimensionzero for µ [3]. The existence of sparse spectra with Beurling dimension zero is also true for otherone-dimensional self-similar measures whose digit sets are consecutive { , , ..., q − } . Our mainTheorem 1.3 now further generalizes the behavior of arbitrarily sparseness of spectra in Beurlingdimension to all singular self-affine measures generated by Hadamard triples.The arbitarily sparseness behavior was in stark contrast with the classical cases. The classicalresult of Landau [19] showed that if Λ is a spectrum for L (Ω) and Ω ⊂ R d (or more generally E (Λ) is a Fourier frame for L (Ω)), then the d -Beurling density of Λ must be at least the Lebesguemeasure of Ω and thus its Beurling dimension must be d . Therefore arbitrarily sparse spectrumdoes not exist for Lebesgue measure. For a simple proof of Landau’s theorem, one can consult [21,Chapter 5]. Landau’s theorem is now fundamental in modern sampling theory (see e.g. [1, 21] fordetails).The arbitrarily sparseness makes us ask naturally if there is any lower bound for the Beurlingdimension of the spectra of a spectral measure. The following proposition provides a simple butuseful answer. Proposition 1.4.
Let µ be a finite Borel and singluar measure on R d such that its Fourier transformsatisfies that for | ξ | large enough, (1.1) | b µ ( ξ ) | ≤ C | ξ | − γ and let E (Λ) be a set of exponentials such that there exists A > , (1.2) A k f k ≤ X λ ∈ Λ (cid:12)(cid:12)(cid:12)(cid:12)Z f ( x ) e − πiλ · x dµ ( x ) (cid:12)(cid:12)(cid:12)(cid:12) , ∀ f ∈ L ( µ ) . Then γ ≤ dim + (Λ) . It is known that supremum of γ such that (1.1) holds is called the Fourier dimension of µ .This proposition implies that the Beurling dimension is at least the Fourier dimension of µ . Forself-affine measure of our consideration, it is easy to prove that all such measures have Fourierdimension zero.Indeed, a stronger result was proved in [14] in which if (1.1) holds for a spectral measure, thenits spectrum must satisfy(1.3) X λ ∈ Λ \{ } | λ | − γ = ∞ . One can show that (1.3) implies that γ ≤ dim + (Λ). We here will provide another independentproof of Proposition 1.4. In private communication with Y. Wang and B. Strichartz, they also have noticed such arbitrarily sparse behaviorof spectra of µ around 2000 LIXIANG AN AND CHUN-KIT LAI
Using (1.3), one can show that the surface measure of any convex body with everywhere positiveGaussian curvature does not admit any Fourier frame, and is therefore not spectral. In view of thisresult, an interesting problem that arises but not yet appeared to have a simple answer is that:
Question:
Does there exist a singular spectral measure whose Fourier dimension is positive?For some more results about Beurling dimension, Fourier decay and spectral measures, one canalso refer to [13, 20, 22]. In [22], the relationship of different dimensions were studied and the abovequestion was also mentioned.It is also worth mentioning that a widely open problem is to determine if µ admits a Fourierframe or Riesz basis. Beurling dimension has been an indicator to see if such frame is possible toexist [6]. It was recently found that it is possible to construct an exponential Riesz sequence (notethat a complete Riesz sequence will be a Riesz basis) with maximal Beurling dimension log Sketch of the proof.
We now sketch the proof of Theorem 1.3. First, it is known thatin the Hadamard triple, B must be a distinct respresentative in the group Z d /R ( Z d ). Therefore, B ≤ | det( R ) | . When B = | det R | , then B is a distinct respresentative in the group Z d /R ( Z d ). µ is just the Lebesgue measure supported on the fundamental domain T ( R, B ). Hence the spectraof spectral measure µ has Beurling dimension d . Therefore, B < | det R | is necessary in theassumption. In this case, µ is singular to the Lebesgue measure.Throughout the paper, we will assume, without loss of generality, 0 ∈ B ∩ L . Otherwise, we do atranslation of the measure. Similar the strategy of the proof in [8], for any singular to the Lebesguemeasure µ on R d , its periodic zero set is defined as follows: Z ( µ ) = { ξ ∈ R d : b µ ( ξ + k ) = 0 , ∀ k ∈ Z d } . Our proof of Theorem 1.3 is divided into two cases Z ( µ ) = ∅ or Z ( µ ) = ∅ . Definition 1.5.
We say that a countable set Λ = { λ n } ∞ n =0 ⊂ R d is called b -lacunary , if λ = 0, | λ | ≥ b and for all n ≥ | λ n +1 | ≥ b | λ n | . As we will see, lacunary sequences must have Beurling dimension (See Proposition 2.2). Ourtheorem in this case is as follows:
Theorem 1.6. If ( R, B, L ) forms a Hadamard triple with B < | det R | and Z ( µ ( R, B )) = ∅ , thenfor all b > , the spectral measure µ ( R, B ) admits a b-lacunary spectrum Λ . The case Z ( µ ( R, B )) = ∅ is more complicated. Our strategy is to reduce the self-affine pair( R, B ) to a pair ( e R, e B ) which has quasi product-form structure. The self-affine measure µ ( e R, e B )projects on R r is a self-affine measure µ ( f R , f B ) satisfies Z ( µ ( f R , f B )) = ∅ . Then we can constructa spectrum of µ ( R, B ) has zero Beurling dimension.We organize our paper as follows: In section 2, we present some preliminaries. We will reviewthe property of Beurling dimension of a discrete set and the condition Z ( µ ) = ∅ . In section 3 , wewill prove Theorem 1.6. In section 4, we will conjugate with some matrix so that ( R, B ) are of thequasi-product form and complete the proof of Theorem 1.3. We will finally prove Proposition 1.4in the last section.
RBITRARILY SPARSE SPECTRA FOR SELF-AFFINE SPECTRAL MEASURES 5 Preliminaries
In this section, we will set up some basic propositions for the rest of our paper. These resultsthat serve as the basis for our proofs.2.1.
Beurling dimension.
We will establish some basic properties of Beurling dimension in thissubsection.
Proposition 2.1.
Let R be an invertible matrix in M d ( R ) and Λ ⊂ R d is a discrete set. Then dim + (Λ) = dim + ( R Λ) . Proof.
For the invertible matrix R , there exist constants c > c > Q dc (0) ⊂ R − Q d (0) ⊂ Q dc (0) . As hQ d (0) = Q dh (0), it implies that for all h > Q dc h (0) ⊂ R − Q dh (0) ⊂ Q dc h (0) . Note that ∩ Q dc h ( y )) = − y ) ∩ Q dc h (0)) . ∩ Q dc h ( R − x )) = − R − x ) ∩ Q dc h (0)) ≤ − R − x ) ∩ R − Q dh (0)) ≤ ∩ Q dc h ( R − x )) . As R Λ ∩ Q ) = ∩ R − Q ), we have − R − x ) ∩ R − Q dh (0)) = R Λ ∩ Q dh ( x )) . We have thus obtained ∩ Q dc h ( R − x )) ≤ R Λ ∩ Q dh ( x )) ≤ ∩ Q dc h ( R − x )) . Dividing by h r and taking supremum and limsup, we have c r D + r (Λ) ≤ D + r ( R Λ) ≤ c r D + r (Λ) . Therefore, dim + (Λ) = dim + ( R Λ) follows. (cid:3)
As one can imagine that a b-lacunary set must be very sparse in R d , the following propositiongives us the affirmative answer. We will use this frequently in the rest of the paper. Proposition 2.2.
Let b > and Λ is a b − lacunary set. Then dim + (Λ) = 0 .Proof. Denote B ( x, r ) be the open ball centered at x ∈ R d with radius r >
0. Let A = B (0 , b ) and A n = B (0 , b n ) \ B (0 , b n − ) be the annuli regions centered at the origin for any n ≥
2. We claim that(Λ \ { } ) ∩ A n has at most one element. In fact, if λ, λ ′ ∈ (Λ \ { } ) ∩ A n , then b n − ≤ | λ | , | λ ′ | < b n .It follows that | λ ′ || λ | < b, | λ || λ ′ | < b. This contradicts to the definition of b -lacunary set. Claim:
If Λ is b -lacunary and given any cubes Q dh ( x ), ∩ Q dh ( x )) ≤ log b (4 √ dh ) + 2 . LIXIANG AN AND CHUN-KIT LAI
Proof.
For any h > x ∈ R d , we can find integers n x,h ≥ k x,h ≥ Q dh ( x ) ⊂ k x,h [ i =1 A n x,h + i and Q dh ( x ) ∩ A n x,h + i = ∅ for all i = 1 , ..., k x,h . Moreover, diam( Q dh ( x )) ≥ dist( A n x,h +1 , A n x,h + k x,h ).When k x,h >
2, that is 2 √ dh ≥ b n x,h + k x,h − − b n x,h +1 ≥ b k x,h − − . So(2.1) k x,h ≤ log b (2 √ dh + 1) + 2 ≤ log b (4 √ dh ) + 2 . It is clear that when k x,h ≤
2, the above inequality (2.1) also holds. Combining with the claim inthe first paragraph, we have that ∩ Q dh ( x )) ≤ Λ ∩ k x,h [ i =1 A n x,h + i ≤ k x,h ≤ log b (4 √ dh ) + 2 . This justifies the claim.Hence for any r >
0, we havelim sup h →∞ sup x ∈ R d (cid:0) Λ ∩ Q dh ( x ) (cid:1) h r ≤ lim h →∞ log b (4 √ dh ) + 2 h r = 0 . (2.2)That is to say D + r (Λ) = 0. From the definition of Beurling dimension, we have dim + (Λ) = 0. (cid:3) Periodic Zero set Z ( µ ) = ∅ . Throughout the paper, the Fourier transform of a Borel prob-ability measure µ on R d is defined to be b µ ( ξ ) = Z e − πiξ · x dµ ( x ) . In this subsection, we will be devoted to understanding the condition Z ( µ ) = { ξ ∈ R d : b µ ( ξ + k ) = 0 , ∀ k ∈ Z d } = ∅ when µ is a singular measure. The following is an important observation and we will strength theconclusion further when µ is a self-affine measures in the next section. Proposition 2.3.
Suppose the periodic zero set Z ( µ ) is empty and µ is singular. Then for all ξ ∈ R d , the set K ξ = { k ∈ Z d : b µ ( ξ + k ) = 0 } is an infinite set. RBITRARILY SPARSE SPECTRA FOR SELF-AFFINE SPECTRAL MEASURES 7
Proof.
Note that K ξ + k = K ξ + k for any ξ ∈ R d and k ∈ Z d . So we just need to consider the set K ξ for ξ ∈ [0 , d . Let µ ξ be the complex measure e − πiξ · x dµ ( x ). Consider the complex measure on T d , which we identify as [0 , d , ν T d ,ξ ( E ) = X n ∈ Z d µ ξ ( E + n )for all Borel set E ∈ T d . Then ν T d ,ξ is a measure on T d and its Fourier coefficients equal [ ν T d ,ξ ( n ) = c µ ξ ( n ) = b µ ( ξ + n )(For details, see [16]). Since Z ( µ ) is empty, K ξ is not an empty set and therefore, ν T d ,ξ is not azero measure on T d . We establish the following claim: Claim: ν T d ,ξ is singular to the Lebesgue measure on T d Proof of claim:
We first note that ν T d ,ξ is absolutely continuous with respect to ν T d , . Indeed, if ν T d , ( E ) = 0, then µ ( E + n ) = 0 for all n ∈ Z d . Hence, µ ξ ( E + n ) = 0 and ν T d ,ξ ( E ) = 0 follows.Therefore, we just need to show that ν T d , is singular with respect to the Lebesgue measure.Let m R d and m T d be the Lebesgue measure on R d and T d respectively. Since µ is singular to m R d , we can find a set A such that m R d ( A ) = 0 and µ is supported on A . For each n ∈ Z d , let A n = (cid:16) A ∩ ([0 , d + n ) (cid:17) − n, so that A = [ n ∈ Z d ( A n + n ) . Let E = S n ∈ Z d A n and let T d \ E be its complement. Then ν T d , ( T d \ E ) = X n ∈ Z d µ \ n ∈ Z d (cid:16) T d \ A n (cid:17) + n ≤ X n ∈ Z d µ (cid:16)(cid:16) T d \ A n (cid:17) + n (cid:17) = 0since µ is supported on A n + n on [0 , d + n . Hence, we know ν T d , is supported on E . But weknow that m T d ( E ) ≤ X n ∈ Z d m T d ( A n ) = X n ∈ Z d m ( A n + n ) = 0 . This shows that ν T d , is a singular measure with respect to m T d . This justifies the claim.We finally argue by contradiction. Suppose that K ξ < ∞ . Then we can find some N suchthat the measure [ ν T d ,ξ ( n ) = 0 for all | n | > N . Then we know that the Fourier coefficients { [ ν T d ,ξ ( n ) : n ∈ Z d } is square-summable. By the unitary isomorphism of L ( T d ) and ℓ ( Z d ), we canfind f ∈ L ( T d ), in fact a trigonometric polynomial, such that [ ν T d ,ξ ( n ) = b f ( n ) . This means that ν T d ,ξ = f ( x ) dx , which is a contradiction since ν T d ,ξ is singular to the Lebesguemeasure. (cid:3) LIXIANG AN AND CHUN-KIT LAI
We remark that the above proposition is clearly false if µ is not singular. For example, if µ is theLebesgue measure on [0 , K = { } only since the Fourier transform on the characteristicfunction of the unit interval is equal to zero on all non-zero integers. Proposition 2.4.
Suppose that the periodic zero set Z ( µ ) is empty. Then there exists ǫ > and δ > such that for all ξ ∈ [0 , d , there exists k ξ ∈ Z d such that | b µ ( ξ + y + k ξ ) | ≥ ǫ whenever | y | < δ . Specially, for ξ = 0 , we can take k = 0 .Proof. As Z ( µ ) is empty, for any ξ ∈ [0 , d , we can find k ξ ∈ Z d and ǫ ξ > | b µ ( ξ + k ξ ) | ≥ ǫ ξ > . By the continuity of b µ . we can find δ ξ > | y | ≤ δ ξ , we have | b µ ( ξ + y + k ξ ) | ≥ ǫ ξ . As [0 , d ⊂ S ξ ∈ [0 , d B ( ξ, δ ξ / , d , we can find ξ , ..., ξ N ∈ [0 , d suchthat [0 , d ⊂ B ( ξ , δ ξ / ∪ ... ∪ B ( ξ N , δ ξ N / δ = min (cid:26) δ ξ j j = 1 , ..., N (cid:27) , ǫ = min n ǫ ξ j j = 1 , ..., N o . Now, δ and ǫ are positive and independent of ξ ∈ [0 , d . We claim that the stated property holds.Indeed, for any ξ ∈ [0 , d , ξ ∈ B ( ξ j , δ ξ j /
2) for some j = 1 , ..., N . Hence, | b µ ( ξ + k ξ j ) | = | b µ ( ξ j + ( ξ − ξ j ) + k ξ j ) | ≥ ǫ ξ j ≥ ǫ. Therefore, we just redefine k ξ = k ξ j to obtain our desired conclusion. (cid:3) Proof of Theorem 1.6
In this section, we first outline how one can construct a Fourier basis for the self-affine measure µ ( R, B ) and then prove Theorem 1.6 that we can find b -lacunary spectra if the periodic zero set isempty.Recall that for the self-affine measure µ = µ ( R, B ), by iterating the invariance identity, its Fouriertransform can be expressed as an infinite product b µ ( ξ ) = ∞ Y n =1 c δ B (( R t ) − n ξ ) = ∞ Y n =1 \ δ R − n B ( ξ ) . Here δ A denotes the equal-weighted Dirac mass supported on the finite set A . From this infiniteproduct, we obtain another expression of the self-affine measure through an infinite convolution ofatomic measures µ ( R, B ) = δ R − B ∗ δ R − B ∗ .... = w − lim n →∞ ( δ R − B ∗ δ R − B ∗ ... ∗ δ R − n B )where w-lim is the weak limit of the probability measures. RBITRARILY SPARSE SPECTRA FOR SELF-AFFINE SPECTRAL MEASURES 9
Given a subsequence of positive integers { n k } , we define B n k = B + RB + · · · + R n k − B . Letting m k = n + · · · + n k . Then the self-affine measure µ = µ ( R, B ) can be factorized along thissubsequence as(3.1) µ = δ R − m B n ∗ δ R − m B n ∗ · · · . Define also µ >k = δ R − mk +1 B nk +1 ∗ δ R − mk +2 B nk +2 ∗ · · · . The following lemma is known, whose proof can be found in ([18], Proposition 3.1).
Lemma 3.1.
Suppose ( R, B, L ) forms a Hadamard triple, for any n ≥ , (i) then ( R n , B n , L n ) is also a Hadamard triple; (ii) if f L n ≡ L n (mod ( R t ) n ) , then ( R n , B n , f L n ) is also a Hadamard triple. We note that (
R, B, L ) forms a Hadamard triple if and only if { e πiℓ · x : ℓ ∈ L } will form anorthonormal basis for L ( δ R − B ). Hence, since we know { ( R n k , B n k , L n k ) } form Hadamard triples,we define(3.2) Λ k = L n + ( R t ) m L n + · · · + ( R t ) m k − L n k , and Λ = ∞ [ k =1 Λ k , and Λ forms a mutually orthogonal set for L ( µ ( R, B )). The following is the main theorem givinga sufficient condition for an orthgonal set to be complete [8, 18].
Theorem 3.2 ([8, 18]) . Let ( R, B, L ) be a Hadamard triple. Let Λ k and Λ be defined as in (3.2).Suppose that (3.3) δ (Λ) := inf k ≥ inf λ k ∈ Λ k | b µ >k ( λ k ) | > . Then the self-affine measure µ is a spectral measure with a spectrum Λ in Z d . Condition (3.3) is a sufficient condition guaranteeing the mutually orthogonal sets to be complete.This condition was first proposed by Strichartz [23, 24]. In general, it cannot be removed. On theother hand, this condition is also not necessary [3]. The following theorem provides a strengthenedresult of Proposition 2.3 in the case of self-affine measures.
Theorem 3.3.
Suppose a self-affine measure µ := µ ( R, B ) satisfies Z ( µ ) = ∅ . Then there is a ǫ > such that for any ξ ∈ [0 , d , K ξ,ǫ = { k ∈ Z d : | b µ ( ξ + k ) | ≥ ǫ } is an infinite set.Proof. Take ǫ > δ > ξ = 0, fromProposition 2.3, K is an infinite set, we can choose a t ∈ K \ { } . Then there exists ǫ ′ > δ ′ > | b µ ( y + t ) | ≥ ǫ ′ , ∀ | y | < δ ′ . Let ǫ = ǫ · ǫ ′ and δ = min { δ, δ ′ } which are positive constants independent of ξ ∈ [0 , d . From Proposition 2.4, for any ξ ∈ [0 , d , there is a k ξ ∈ Z d such that(3.5) | b µ ( ξ + k ξ ) | ≥ ǫ. Fix ξ ∈ [0 , d and k ξ , we can find an integer n ≥ | ( R t ) − n ( ξ + k ξ ) | < δ . When n ≥ n ,since | ( R t ) − n ( ξ + k ξ ) | < δ , the inequality (3.4) implies that | b µ (( R t ) − n ( ξ + k ξ ) + t ) | ≥ ǫ ′ . It together with inequality (3.5), we have | b µ ( ξ + k ξ + ( R t ) n t ) | = ∞ Y k =1 | M B (( R t ) − k ( ξ + k ξ + ( R t ) n t )) | = n Y k =1 | M B (( R t ) − k ( ξ + k ξ )) | · ∞ Y k =1 | M B (( R t ) − k (( R t ) − n ( ξ + k ξ ) + t )) |≥ | b µ ( ξ + k ξ ) | · | b µ (( R t ) − n ( ξ + k ξ ) + t ) |≥ ǫ · ǫ ′ = ǫ . The above inequality implies that { k ξ + ( R t ) n t } ∞ n = n ⊂ K ξ,ǫ . As ( R t ) k is an expanding matrix, 1 is not an eigenvalue of ( R t ) k . So ( R t ) n t = ( R t ) m t when n = m . Hence K ξ,ǫ is an infinite set. (cid:3) Next we prove Theorem 1.6.
Proof. ( Proof of Theorem 1.6 ). Let ǫ > b µ implies that there exists a δ > ξ ∈ [0 , d and | y | < δ ,we have(3.6) | b µ ( ξ + y + k ) | ≥ ǫ / k ∈ K ξ,ǫ .We now construct inductively Λ k as in (3.2) so that Λ k is b − lacunary and δ (Λ) ≥ ǫ / >
0. FromLemma 3.1 (ii), without loss of generality, we assume L = { l = 0 , l , · · · , l q − } ⊂ R t [0 , d ∩ Z d .Then ( R t ) − n L n ⊂ [0 , d . Denote n = 1. Let λ = l + ( R t ) k ( R t ) − l k ( R t ) − l ∈ K ( R t ) − l ,ǫ . For 2 ≤ i ≤ q −
1, since K ( R t ) − l i ,ǫ is infinite, we can take a k ( R t ) − l i ∈ K ( R t ) − l i ,ǫ such that λ i = l i + R t k ( R t ) − l i and | λ i | ≥ b | λ i − | . Then Λ = { λ = 0 , λ , λ , · · · , λ q − } is b -lacunary and from (3.6) | b µ > ( λ i ) | = | b µ (( R t ) − l i + k ( R t ) − l i ) | ≥ ǫ / , ≤ i ≤ q − . Suppose that Λ k − has been constructed which is a b − lacunary set andinf λ k − ∈ Λ k − | b µ > ( k − ( λ k − ) | ≥ ǫ / . RBITRARILY SPARSE SPECTRA FOR SELF-AFFINE SPECTRAL MEASURES 11
We can take a large enough n k in the subsequence with the following happen:sup λ k − ∈ Λ k − k ( R t ) − m k λ k − k < δ . (Recall that m k = n + ... + n k ) We now defineΛ k = Λ k − + { ( R t ) m k − l k + ( R t ) m k k x lk : l k ∈ L n k } where x l k = ( A t ) − n k l k ∈ [0 , d and k = 0. Then Λ k − ⊂ Λ k . As K x l ,ǫ is an infinite set, bychoosing k x l ∈ K x l ,ǫ as large as we wanted, we can ensure that Λ k is b -lacunary. Now writing λ k = λ k − + ( R t ) m k − l k + ( R t ) m k k x lk , for some λ k − ∈ Λ k − , we have | d µ >k ( λ k ) | = | b µ (( R t ) − m k λ k ) | = | b µ (( R t ) − m k λ k − + x l k + k x lk ) | (using (3.6)) ≥ ǫ / > . Hence, δ (Λ) > b -lacunary set Λ = S ∞ k =1 Λ k is a spectrum of µ according toTheorem 3.2. (cid:3) As a corollary, we settle the case for the self-similar measure on R . Corollary 3.4.
Let
R > be an integer and B ⊂ Z be a digit set with B < R and gcd ( B ) = 1 .Suppose that ( R, B, L ) forms a Hadamard triple. Then Z ( µ ( R, B )) = ∅ and µ ( R, B ) admits aspectrum Λ with dim + (Λ) = 0 .Proof. It has been proved in [8, Section 5] that if gcd( B )=1, then the periodic zero set of µ ( R, B )is empty and therefore it has a b − lacunary spectrum in Z by Theorem 1.6, which has Beurlingdimension zero by Proposition 2.2. (cid:3) Proof of Theorem 1.3.
We first discuss some preliminary reduction that we can perform in order to prove our maintheorem.
Definition 4.1.
Let R , R be d × d integer matrices, and the finite sets B , B , L , L be in Z d .We say that two triples ( R , B , L ) and ( R , B , L ) are conjugate (through the matrix M ) if thereexists an integer unimodular matrix M such that R = M R M − , B = M B and L = ( M t ) − L .(Here, unimodular matrix means its determinant is 1).The following proposition is obtained from some simple computations. Proposition 4.2.
Suppose that ( R , B , L ) and ( R , B , L ) are two conjugate triples, through thematrix M . Then (i) If ( R , B , L ) is a Hadamard triple then so is ( R , B , L ) . (ii) The measure µ ( R , B ) is spectral with spectrum Λ if and only if µ ( R , B ) is spectral withspectrum ( M t ) − Λ . (iii) Spectral measures µ ( R , B ) and µ ( R , B ) have spectrum Λ with dim + (Λ) = 0 simultane-ously.Proof. The proof of (i), (ii) can be found in e.g. ([7], Proposition 3.4). The (iii) follows from thefact that dim + (Λ) = dim + (( M t ) − Λ) which is proved in Proposition 2.1 (cid:3)
We use E × F to denote the Cartesian product of E and F so that E × F = { ( e, f ) t : e ∈ E, f ∈ F } .We first introduce the following notations. Definition 4.3.
For a vector x ∈ R d , we write it as x = ( x , x ) t with x ∈ R r and x ∈ R d − r . Wedenote by π ( x ) = x , π ( x ) = x . For a subset E of R d , and x ∈ R r , x ∈ R d − r , we denote by E ( x ) := { y ∈ R d − r : ( x , y ) t ∈ E } , E ( x ) := { x ∈ R r : ( x, x ) t ∈ E } . We define Z [ R, B ] to be the smallest R − invariant lattice containing all P n − j =0 R j B . To provetheorem 1.3, there is no loss of generality to assume that Z [ R, B ] = Z d since we can always conjugatethe Hadamard triple to produce a Hadamard triple with Z [ R, B ] = Z d . If ( R, B ) and ( e R, e B ) areconjugate through an integer unimodular matrix M , then Z [ e R, e B ] = M Z [ R, B ] = M Z d = Z d . By studying the dynamical system underlying the self-affine system, the following decompositionwas proved in [8, Section 6 and 7]. If (
R, B, L ) is a Hadamard triple such that Z [ R, B ] = Z d and Z ( µ ) = ∅ , we can always conjugate with some integer unimodular matrix so that ( R, B ) are of thefollowing quasi-product form :(4.1) R = (cid:18) R C G (cid:19) (4.2) B = { ( u i , d j ( u i )) t : 1 ≤ i ≤ N , ≤ j ≤ N := | det G |} , and { d j ( u i ) : 1 ≤ j ≤ N } is a complete set of representatives (mod G Z d − r ). Note that R − k canbe written in the following form R − k = (cid:18) R − k C k G − k (cid:19) for some C k ∈ M d − r,r ( R ). For the self-affine set T ( R, B ), we can express it as a set of infinite sums T ( R, B ) = ( ∞ X k =1 R − k b k : b k ∈ B ) . RBITRARILY SPARSE SPECTRA FOR SELF-AFFINE SPECTRAL MEASURES 13
Therefor any element ( x , x ) t ∈ T ( R, B ) can be written as x = ∞ X k =1 R − k u i k , x = ∞ X k =1 C k u i k + ∞ X k =1 G − k d j k ( u i k ) . Hence π ( T ( R, B )) = T ( R , π ( B ))is a self-affine set where π ( B ) = { u i : 1 ≤ i ≤ N } . For each x = ∞ X k =1 R − k u i k ∈ T ( R , π ( B )) , we have ( T ( R, B )) ( x ) = ∞ X k =1 C k u i k + ( ∞ X k =1 G − k d j k ( u i k ) : 1 ≤ j k ≤ N ) almost surely tiles R d − r . Moreover, we can find a L ′ ≡ L (mod R t ) such that ( R , π ( B ) , L ′ ( l )) isa Hadamard triple on R r for all l ∈ π ( L ′ ). Let µ ( R , π ( B )) be the self-affine measure supportedon T ( R , π ( B )) and µ ( x )2 be the infinite convolution product δ G − B ( i ) ∗ δ G − B ( i ) ∗ . . . , where B ( i k ) := { d j ( u i k ) : 1 ≤ j ≤ N } We call µ ( x )2 the Cantor-Moran measure supported on( T ( R, B )) ( x ) − P ∞ k =1 C k u i k . the measures µ ( R , π ( B )) and µ ( x )2 are called a quasi-product formdecomposition of µ ( R, B ). Theorem 4.4.
Let ( R, B, L ) be a Hadamard triple on R d such that Z [ R, B ] = Z d and Z ( µ ) = ∅ .Then we can conjugate with an integer unimodular matrix M so that µ ( M RM − , M B ) has a quasi-product form µ and µ ( x )2 on R r × R d − r with Z ( µ ) = ∅ .Proof. We will prove the result by using induction on dimension d . When d = 1, the assumption Z [ R, B ] = Z forces gcd( B ) = 1. In this case, it has been proved in [8, Section 5] that Z ( µ ) = ∅ .Hence, if Z ( µ ) = ∅ , then d ≥ d = 2, as we have discussed, we can conjugate with an integer unimodular matrix M ∈ M ( Z )so that ( M RM − , M B ) are of the quasi-product form on R × R as following: M RM − = (cid:18) R C G (cid:19) M B = { ( u, d j ( u )) t : u ∈ B , ≤ j ≤ | G |} ⊂ Z , and { d j ( u ) : 1 ≤ j ≤ | G |} ⊂ Z is a complete set of representatives (mod | G | ) (Here | G | ≥ B ) = 1 since Z [ M RM − , M B ] = Z . Then µ ( M RM − , M B )has a quasi-product form µ and µ ( x )2 on R × R where µ = µ ( R , B ) is the self-similar measuresupported on T ( R , B ) ⊂ R . Since gcd( B ) = 1 and ( R , π ( B ) , L ′ ( l )) forms a Hadamard tripleon R for some L ′ , l ∈ π ( L ′ ), we have Z ( µ ) = ∅ . This justifies the result on dimension two. Assume our statement is true for any dimensions less than d . On dimension d , we can conjugatewith an integer unimodular matrix M ∈ M d ( Z ) so that ( M RM − , M B ) has quasi-product form M RM − = (cid:18) R C G (cid:19) M B = { ( u, d j ( u )) t : u ∈ B , ≤ j ≤ | det G |} ⊂ Z d , where Z [ R , B ] = Z r and { d j ( u ) : 1 ≤ j ≤ | det G |} ⊂ Z d − r is a complete set of representatives(mod G Z d − r ). If Z ( µ ( R , B )) = ∅ , then µ = µ ( R , B ) and µ ( x )2 is the desired quasi-productform on R r × R d − r .If Z ( µ ( R , B )) = ∅ , by the assumption on r < d , we can conjugate with an unimodular matrix M ∈ M r ( Z ) such that M R M − = (cid:18) R C G (cid:19) M B = { M u : u ∈ B } = { ( v, w j ( v )) t : v ∈ B , ≤ j ≤ | det G |} ⊂ Z r , where { w j ( v ) : 1 ≤ j ≤ | G |} ⊂ Z r − r is a complete set of representatives (mod G Z r − r ) and Z ( µ ( R , B )) = ∅ . Denote M = (cid:16) M
00 I d − r (cid:17) M where I d − r is the identity matrix and rewrite d i ( u ) as d i ( v, w j ( v )) if M u = ( v, w j ( v )) t . Then( M RM − , M B ) has quasi-product form on R r × R d − r as following M RM − = (cid:18) R C ′ G ′ (cid:19) where G ′ = (cid:18) G C ′′ G (cid:19) for some matrix C ′′ ∈ M d − r,r − r ( Z ) and M B = n ( v, w j ( v ) , d j ( v, w j ( v )) t : v ∈ B , ≤ j ≤ | det G | , ≤ j ≤ | det G | o , where { ( w j ( v ) , d j ( v, w j ( v )) t : 1 ≤ j ≤ | det G | , ≤ j ≤ | det G |} ⊂ Z d − r is a complete set of representatives (mod G ′ Z d − r ). As Z ( µ ( R , B )) = ∅ , µ = µ ( R , B ) and µ ( x )2 is the desired quasi-product form decomposition of the self-affine measure µ ( R, B ) on R r × R d − r . (cid:3) The spectrality for µ and µ ( x )2 in the quasi-product-form decomposition was proved in [8] Theorem 4.5. [8, Proposition 8.4]
Suppose ( R, B, L ) is a Hadamard triple and Z ( µ ) = ∅ . Let µ , µ ( x )2 be a quasi-product form of µ on R r × R d − r , then µ is spectral and for µ − almost every x ∈ T ( R , B ) , µ ( x )2 admits a spectrum Γ which is a full-rank lattice in R d − r . RBITRARILY SPARSE SPECTRA FOR SELF-AFFINE SPECTRAL MEASURES 15
The following lemma is proved in [7].
Lemma 4.6 ([7], Lemma 4.4) . If Λ is a spectrum for the measure µ , then for all x ∈ R d − r X λ ∈ Λ | b µ ( x + λ , x ) | = Z T ( A ,π ( B )) | b µ ( s )2 ( x ) | dµ ( s ) . We recall also the Jorgensen-Pedersen Lemma for checking when a countable set is a spectrumfor a measure.
Lemma 4.7. [15]
Let µ be a compactly supported probability measure on R d . Then a countable set Λ is a spectrum for L ( µ ) if and only if X λ ∈ Λ | b µ ( ξ + λ ) | ≡ , ξ ∈ R d . We now state a general class of spectrum for the quasi product-form.
Theorem 4.8.
Suppose Γ is a spectrum of µ ( x )2 for µ − almost every x and { Λ γ } γ ∈ Γ is a class ofspectra of µ . Then S γ ∈ Γ (Λ γ × { γ } ) is a spectrum of µ .Proof. Since Λ γ is a spectrum of µ , by Lemma 4.6, we have X γ ∈ Γ X λ ∈ Λ γ | b µ ( x + λ, x + γ ) | = Z T ( A ,π ( B )) X γ ∈ Γ | b µ ( s )2 ( x + γ ) | dµ ( s ) = Z T ( A ,π ( B )) dµ ( s ) = 1 . This means that S γ ∈ Γ (Λ γ × { γ } ) is a spectrum of µ by Lemma 4.7. (cid:3) In [8], only the case that all Λ γ are the same was considered. However, to construct zero Beurlingdimension spectra, Λ × Γ is not enough since { } × Γ is contained in the spectrum and this willcontribute to the Beurling dimension d − r since Γ is lattice on R d − r . To overcome this problem,we will need different sparse spectra for each γ . Now we can prove our main result. Proof. ( Proof of Theorem 1.3 ) If Z ( µ ) = ∅ , then the result follows from Theorem 1.6. Supposenow that Z ( µ ) = ∅ . Since conjugation maintains the Beurling dimension of Λ, without loss ofgenerality, we assume µ has quasi-product form µ = µ ( R n , B n ) , µ ( x )2 on R r × R d − r with Z ( µ ) = ∅ from Theorem 4.4. Moreover, ( R n , B n , L n ) forms a Hadamard triple. From Theorem 1.6, for any b >
1, we can construct a b − lacunary spectrum of µ .Let the full-rank lattice Γ = A Z d − r is a spectrum of µ ( x )2 for µ − almost every x ∈ T ( R , B ).We choose a sequence of vectors { m n } ∞ n =0 ⊂ Z d − r such that m = 0 and | m n | = 2 n − . Let N ,r = 0and N n,r = { m ∈ Z d − r : 2 n − ≤ | m | < n } , n ≥ . For any n ≥
0, let 0 ∈ Λ m n be a spectrum of µ satisfies Λ m n is 8 N n,r +1 − lacunary . By a reversetriangle inequality, it implies that(4.3) min {| λ − λ ′ | : λ = λ ′ ∈ Λ m n } ≥ N n,r +1 . We now enumerate the set { m ∈ Z d − r : 2 n − ≤ | m | < n } = { m n, , ..., m n,N n,r } . We then take spectrum on each of integers m n,j as follows:Λ m n,j = Λ m n + j · N n,r e , j = 1 , ..., N n,r , where e is the unit vector in the x -direction. For any m ∈ Z d − r \ { } , there is an unique integer n ≥ n ≤ | m | < n +1 . Therefore, by Theorem 4.8,Λ = [ m ∈ Z d − r (Λ m × { Am } ) =: (cid:18) I r A (cid:19) Λ ′ is a spectrum of µ where I r is the r × r identity matrix andΛ ′ = [ m ∈ Z d − r (Λ m × { m } ) . By Proposition 2.1, dim + (Λ) = dim + (Λ ′ ). Next we will prove that dim + (Λ ′ ) = 0. Claim 1:
For any n ≥ ≤ j , j ≤ N n,r ,Λ n,m j ∩ Λ n,m j = ∅ . Proof.
Suppose there is an positive integer n and 1 ≤ j , j ≤ N n,r such thatΛ n,m j ∩ Λ n,m j = ∅ , and choose an element x in the intersection. Then it has two expressions x = λ + j · N n,r e = λ + j · N n,r e where λ , λ are distinct elements in Λ m n as j = j . It implies that λ − λ = ( j − j )2 N n,r e . The right hand of the equation implies(4.4) | λ − λ | < N n,r · N n,r ≤ N n,r · N n,r = 4 N n,r . But from (4.3), | λ − λ | ≥ N n,r +1 , which is a contradiction. Hence, the claim is true. Claim 2: δ n := inf | λ − λ | : λ = λ ∈ N n,r [ j =1 Λ n,j = 2 N n,r . Proof.
Note that 2 N n,r e ∈ Λ n, , 2 · N n,r e ∈ Λ n, and (cid:12)(cid:12) N n,r e − · N n,r e (cid:12)(cid:12) = 2 N n,r , so δ n ≤ N n,r . On the other hand, for any distinct elements λ ∈ Λ n,j , λ ∈ Λ n,j , we can writethem as λ = λ ′ + j · N n,r e , λ = λ ′ + j · N n,r e for some λ ′ , λ ′ ∈ Λ m n . If λ ′ = λ ′ , then j = j and | λ − λ | = (cid:12)(cid:12) ( j − j ) · N n,r e (cid:12)(cid:12) ≥ N n,r . If λ ′ = λ ′ , then from (4.3), we have | λ ′ − λ ′ | ≥ N n,r +1 . Hence, with the inequality obtained in (4.4), we have the following | λ − λ | ≥ | λ ′ − λ ′ | − N n,r · | j − j | ≥ N n,r +1 − N n,r > N n,r . This justifies the claim.Now we return to the proof of theorem. For any h > x , x ) t ∈ R r × R d − r , choose n ≥ n > n be the smallest integer such that Q d − rh ( x ) ∩ Z d − r ⊂ B (0 , n ) \ B (0 , n ) ∪ { } . Then we have 2 n − − n +1 ≤ h √ r ≤ n − n . Then(4.5) n − n ≤ (4 h √ r ) . Note that Q dh (( x , x ) t ) = Q rh ( x ) × Q d − rh ( x ). This allows us to have the following: ′ ∩ Q dh (( x , x ) t )) = X m ∈ Q d − rh ( x ) ∩ Z d − r m ∩ Q rh ( x )) ≤ n X n = n +1 X n − ≤| m | < n m ∩ Q rh ( x )) + ∩ Q rh ( x )) . (4.6)We now decompose the first summand into two parts n X n = n +1 ( .... ) = X n < n ≤ n , Nn,r > h √ r ( .... ) + X n < n ≤ n , Nn,r ≤ h √ r ( .... )In the first case, 2 N n,r > h √ r = diam( Q rh ( x )) . From the Claim 1 and Claim 2, we have S n − ≤| m | < n Λ m has at most one element in Q rh ( x ). So X n < n ≤ n , Nn,r > h √ r X n − ≤| m | < n m ∩ Q rh ( x )) ≤ P n < n ≤ n , Nn,r > h √ r ≤ n − n ≤ (4 h √ r ) . (4.7)In the second case, for the integer n < n ≤ n with2 N n,r ≤ h √ r, we have N n,r ≤ log (2 h √ r ) . For 2 n − ≤ | m | < n , Λ m is a 8 N n,r +1 − lacunary set, so by the Claim in Proposition 2.2, m ∩ Q rh ( x )) ≤ log Nn,r +1 (2 h √ r ) + 2 ≤ (2 h √ r ) . For Λ , it is a 8 − lacunary set,(4.8) ∩ Q rh ( x )) ≤ log (2 h √ r ) + 2 ≤ (2 h √ r ) . We have X { n : n
0. As for any γ > h → + ∞ (log (4 h √ r )) h γ = 0 , we have D + γ (Λ ′ ) = lim sup h → + ∞ sup ( x ,x ) ∈ R d ′ ∩ Q dh (( x , x ) t )) h γ = 0 . This shows dim + (Λ ′ ) = 0. (cid:3) Fourier decay
In this section, we will prove Proposition 1.4 and some results about Fourier decay of self-affinemeasures.
Proof. ( Proof of Proposition 1.4 ) As µ is a singular measure, by [10, Proposition 2.1], we havethat the lower ( d -)Beurling density of Λ is zero. i.e. D − (Λ) = lim h →∞ inf x ∈ R d ∩ Q dh ( x )) h d = 0 . Note that this implies that for all h >
0, there exists ξ h such that(5.1) Q dh ( ξ h ) ∩ Λ = ∅ . Indeed, if (5.1) is not true, then one can find h > Q dh ( x ) ∩ Λ = ∅ for all x ∈ R d . Butthen we partition R d into disjoint union of cubes with side length h and each cube has at leastone element in Λ. This implies that D − (Λ) ≥ h − >
0, a contradiction.
RBITRARILY SPARSE SPECTRA FOR SELF-AFFINE SPECTRAL MEASURES 19
Now, using (5.1), for any k >
0, we can find ξ k such that(5.2) Q d k ( ξ k ) ∩ Λ = ∅ . Suppose that D + α (Λ) < ∞ . Then there exists a constant C ′ > ∩ Q dh ( x )) ≤ C ′ h α , ∀ x ∈ R d . We now take f = e πi h ξ k ,x i into (1.2), A ≤ X λ ∈ Λ | b µ ( ξ k − λ ) | = X λ ∈ Λ ∩ ( Q d k ( ξ k )) C | b µ ( ξ k − λ ) | (by (5.2))= ∞ X j = k X λ ∈ Q d j +1 ( ξ k ) \ Q d j ( ξ k ) | b µ ( ξ k − λ ) | ≤ C ∞ X j = k ∩ ( Q d j +1 ( ξ k ) \ Q d j ( ξ k )))2 − jγ (by (1.1)) ≤ C ′ · C ∞ X j = k (2 j +1 ) α − jγ = C ′ · C ∞ X j = k j ( α − γ ) . Suppose that α < γ . Then the right hand side above will tend to zero as k tends to infinity whichmeans that Λ cannot satisfy (1.2). Hence, α ≥ γ . In particular, this implies that γ ≤ dim + (Λ) bytaking infimum of α such that D + α (Λ) < ∞ . (cid:3) Because of Proposition 1.4 and Theorem 1.3, we can show that all self-affine spectral measureswe considered have Fourier dimension zero. However, much stronger can be proved easily as follows.
Proposition 5.1.
Let R ∈ M d ( Z ) be an expansive matrix and B ∈ Z d be a finite set. Suppose that B < | det( R ) | . Then the Fourier transform self-affine measure µ ( R, B ) does not decay to zero.Proof. We think this result is probably well-known. We just present here for completeness. Notethat if
B < | det( R ) | . µ = µ ( R, B ) must be singular. By Proposition 2.3, we can find k = 0 and k ∈ Z d such that b µ ( k ) = 0. For all integers n >
0, noting that B are all integer vectors, b µ (( R t ) n k ) = ∞ Y j =1 c δ B (( R t ) − j (( R t ) n k )) = ∞ Y j = n +1 c δ B (( R t ) n − j k ) = b µ ( k ) = 0 . This shows that all such self-affine measures do not decay. (cid:3)
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E-mail address : [email protected] [Chun-Kit Lai]Department of Mathematics, San Francisco State University, 1600 Holloway Avenue,San Francisco, CA 94132. E-mail address ::