Hadamard triples generate self-affine spectral measures

# Hadamard triples generate self-affine spectral measures

Dorin Ervin Dutkay [Dorin Ervin Dutkay] University of Central Florida
Department of Mathematics
4000 Central Florida Blvd.
P.O. Box 161364
Orlando, FL 32816-1364
U.S.A.
John Haussermann [John Haussermann] University of Central Florida
Department of Mathematics
4000 Central Florida Blvd.
P.O. Box 161364
Orlando, FL 32816-1364
U.S.A.
and  Chun-Kit Lai [Chun-Kit Lai] Department of Mathematics, San Francisco State University, 1600 Holloway Avenue, San Francisco, CA 94132.
###### Abstract.

Let be an expanding matrix with integer entries and let be finite integer digit sets so that form a Hadamard triple on in the sense that the matrix

 1√|detR|[e2πi⟨R−1b,ℓ⟩]ℓ∈L,b∈B

is unitary. We prove that the associated fractal self-affine measure obtained by an infinite convolution of atomic measures

 μ(R,B)=δR−1B∗δR−2B∗δR−3B∗...

is a spectral measure, i.e., it admits an orthonormal basis of exponential functions in . This settles a long-standing conjecture proposed by Jorgensen and Pedersen and studied by many other authors. Moreover, we also show that if we relax the Hadamard triple condition to an almost-Parseval-frame condition, then we obtain a sufficient condition for a self-affine measure to admit Fourier frames.

###### Key words and phrases:
Hadamard triples, quasi-product form, self-affine sets, spectral measure
###### 2010 Mathematics Subject Classification:
Primary 42B05, 42A85, 28A25.

## 1. Introduction

### 1.1. Fuglede’s Problems

As it is well known, Fourier discovered that the exponential functions form an orthonormal basis for and his discovery is now one of the fundamental pillars in modern mathematics. It is natural to ask what other measures have this property, that there is a family of exponential functions which form an orthonormal basis for their -space?

Let be a Borel probability measure on and let denote the standard inner product on . We say that is a spectral measure if there exists a countable set called the spectrum of such that is an orthonormal basis for . Suppose that the Fourier transform of is defined to be

 ˆμ(ξ)=∫e−2πi⟨ξ,x⟩dμ(x).

It is straightforward to verify that a measure is a spectral measure with spectrum if and only if the following two conditions are satisfied:

1. (Orthogonality) for all distinct and

2. (Completeness) If for , for all , then .

Furthermore, we say that a Lebesgue measurable subset of is a spectral set if the corresponding Lebesgue measure supported on , , is a spectral measure. In this paper, we are interested in the following question

(Q1) When is a Borel probability measure spectral?

The question was first studied by Fuglede [14] in 1974 while he was working on a problem by Segal on the existence of commuting extensions of the partial differential operators on domains of . Fuglede proved that the domains under some regularity condition for which such extensions exist are exactly those with the property that there exists an orthogonal exponential basis for , with Lebesgue measure. The regularity condition was removed later by Pedersen [47]. In the same paper, Fuglede proposed his famous conjecture:

Fuglede’s Conjecture:  A measurable set is a spectral set in if and only if tiles by translation.

Fuglede’s Conjecture has been studied by many authors, e.g., Jorgensen, Pedersen, Lagarias, Łaba, Kolountzakis, Matolcsi, Iosevich, Tao, Wang and others ([23, 24, 20, 21, 27, 29, 31, 33, 35, 37, 54]), but it had baffled experts for 30 years until Terence Tao [54] constructed the first counterexample, a spectral set which is not a tile in , . The example and technique were refined later to disprove the conjecture in both directions on for [44, 29, 28]. The conjecture is still open in dimensions and .

Although Fuglede’s Conjecture in its original form has been disproved, there is a clear connection between spectral sets and tilings, but the precise correspondence is still a mystery. Furthermore, spectral sets are a particular case of a broader class of problems concerning the existence and construction of families of complex exponential functions that form either Riesz bases or, more generally, Fourier frames [26, 30, 45]. Also, it is known that Fuglede’s conjecture is true under some additional assumptions and in some other groups [22], and it is related to the construction of Gabor and wavelet bases [42, 55]. We will refer to the problems concerning spectral measures and their relation to translational tilings as the Fuglede problem.

### 1.2. Fractal Spectral measures and Main Results

Another major advance in the study of the Fuglede problem was the discovery that fractal singular measures can also be spectral. This opened up a new possibility of applying the well-developed Fourier analysis techniques to certain classes of fractals.

In 1998, Jorgensen and Pedersen [23] constructed the first example of a singular, non-atomic spectral measure. The measure is the Hausdorff measure supported on a Cantor set, where the scaling factor is 4 and the digits are 0 and 2; we call them the one-fourth Cantor measure/set. The spectrum for this measure is the set

 Λ:={n∑k=04klk:lk∈{0,1},n∈N}.

They also proved that the usual Middle Third Cantor measure is non-spectral. The Fourier series on the one-fourth Cantor measure were studied by Strichartz who proved in [53] that they have much better convergence properties than their classical counterparts: the Fourier series associated to continuous functions converge uniformly and the Fourier series of -functions converge in the -norm.

Following this discovery, many other examples of singular measures have been constructed, and the spectral property of various classes of fractal measures have been analyzed, see, e.g., [23, 32, 51, 52, 10, 40, 56, 41, 18, 4, 1, 15] and the references therein. To the best of our knowledge, all these constructions have been based on the central idea of Hadamard matrices and Hadamard triples:

###### Definition 1.1.

Let be an expansive matrix (i.e., all eigenvalues have modulus strictly greater than 1) with integer entries. Let be finite sets of integer vectors with ( denotes the cardinality). We say that the system forms a Hadamard triple (or forms a compatible pair, as it is called in [32] ) if the matrix

 (1.1) H=1√N[e2πi⟨R−1b,ℓ⟩]ℓ∈L,b∈B

is unitary, i.e., .

The system forms a Hadamard triple if and only if the Dirac measure is a spectral measure on with spectrum . Infinite convolutions of rescaled discrete measures produce self-affine measures, which we define below.

###### Definition 1.2.

For a given expansive integer matrix and a finite set of integer vectors with , we define the affine iterated function system (IFS)

 τb(x)=R−1(x+b),(x∈Rd,b∈B).

The self-affine measure (with equal weights) is the unique probability measure satisfying

 (1.2) μ(E)=∑b∈B1Nμ(τ−1b(E)), for all Borel % subsets E of Rd.

This measure is supported on the attractor which is the unique compact set that satisfies

 T(R,B)=⋃b∈Bτb(T(R,B)).

The set is also called the self-affine set associated with the IFS. It can also be described as

 T(R,B)={∞∑k=1R−kbk:bk∈B}.

One can refer to [19] and [13] for a detailed exposition of the theory of iterated function systems.

For a given integral expanding matrix and a simple digit set for . We let

 (1.3) Bn:=B+RB+R2B+...+Rn−1B={n−1∑j=0Rjbj:bj∈B}.
 (1.4) LTn:=L+RTL+(RT)2L+...+(RT)n−1L={n−1∑j=0(RT)jℓj:ℓj∈L}.

Another important description of the self-affine measure is as the infinite convolution of discrete measures

 (1.5) μ(R,B)= δR−1B∗δR−2B∗δR−3B∗... = μn∗μ>n,

where

 μn=δR−1B∗δR−2B∗...∗δR−nB=δR−n(Bn)

and by self-similarity. For a finite set in ,

 δA:=1#A∑a∈Aδa,

where is the Dirac measure at .

Suppose that is a Hadamard triple. Then are Hadamard triples for all . Hence, each factor is a spectral measure. Moreover, because and have integer entries, we can see that all are spectral. Hence, it is natural to conjecture that the weak limit of is spectral:

Conjecture:   Suppose that forms a Hadamard triple. Then the self-affine measure is a spectral measure.

This conjecture has been proposed since Jorgensen and Pedersen’s first discovery of spectral singular measures. It was first proved on by Laba and Wang [32] and later refined in [10]. The situation becomes more complicated when . Dutkay and Jorgensen showed that the conjecture is true if satisfies a technical condition called the reducibility condition [11]. The conjecture is true under some additional assumptions, introduced by Strichartz [51, 52]. Some low-dimensional special cases were also considered by Li [40, 41]. In this paper, one of our main objectives is to prove that this conjecture is true, Hadamard triples always generate self-affine spectral measures.

###### Theorem 1.3.

Let be a Hadamard triple. Then the self-affine measure is spectral.

### 1.3. Outline of the Proof

Throughout the paper, we will assume, without loss of generality, that . The proof of Theorem 1.3 involves three main steps and each individual step is of independent interest.

Step 1: The No-Overlap condition

###### Definition 1.4.

We say that the self-affine measure in Definition 1.2 satisfies the no-overlap condition or measure disjoint condition if

 μ(τb(T(R,B))∩τb′(T(R,B)))=0, for all b≠b′∈B.

We say that is a simple digit set for if distinct elements of are not congruent .

It is easy to verify that must be a simple digit set for if is a Hadamard triple. We will prove that if the digit set is simple, then the no-overlap condition is satisfied. The no-overlap condition is related to the open set condition (OSC) and strong open set condition (SOSC).

###### Definition 1.5.

We say that the iterated function system satisfies the open set condition (OSC) if there exists a non-empty open set such that

 τb(U)∩τb′(U)=∅, and ⋃b∈Bτb(U)⊂U.

The iterated function system satisfies the strong open set condition (SOSC) if we can furthermore choose the open set such that .

These conditions have been well-studied in the case of self-similar measures for which for some and orthogonal matrix (see e.g. [48, 39]), but we did not find any such results in the literature for the case self-affine measures. We will first prove the no-overlap condition for the self-affine measure in our interest.

###### Theorem 1.6.

Let be a expansive integer matrix and let be a simple digit set for . Then the affine iterated functions system associated to and satisfies the OSC, SOSC and the no-overlap condition.

In the study of spectral measures, the no overlap condition for a self-affine measures is particularly important since it guarantees that and its -th level iterates will have measure . With the help of this theorem, we can also compute for the set of step functions on the self-affine set .

After establishing the no-overlap condition, we can start the proof of Theorem 1.3. The mutual orthogonality of the exponential functions is not difficult to show. The main challenge is to establish the completeness of the set of exponential functions. We consider the following periodic zero set of the Fourier transform:

 (1.6) Z:={ξ∈Rd:ˆμ(ξ+k)=0, for all%  k∈Zd}

We will divide our proof into two cases: (i) and (ii) .

Step 2:

This case is easier to handle. For a Hadamard triple and a sequence of positive integers , we let . The self-affine measure can be rewritten as

 μ(R,B)=δR−m1Bn1∗δR−m2Bn2∗...∗δR−mkBnk∗...

Then we note that if we have another set of integer vectors, with (mod ), then still form Hadamard triples. Using this, we can produce many mutually orthogonal sets of exponential functions with frequencies given by:

 (1.7) Λk=Jn1+(RT)m1Jn2+(RT)m2Jn3+...+(RT)mk−1Jnk,
 (1.8) Λ=∞⋃k=1Λk.

We will show that under the assumption , we can pick such a set that is indeed also complete, so it is a spectrum. In fact, we have

###### Theorem 1.7.

Suppose that forms a Hadamard triple and is the associated self-affine measure. Then the following are equivalent

1. ,

2. has a spectrum in .

In particular, if , then is a spectral measure.

Our method for the proof of the completeness of the set of exponential functions differs from all the other existing proofs in literature, see, e.g., [32, 51, 52, 5, 4], where the completeness is established by checking the Jorgensen-Pedersen criterion (i.e. ). Our proof of this theorem with relies on an approach from matrix analysis which exploits the isometry property of Hadamard matrices, i.e. . This allows us to show that the frame inequalities are satisfied for the set of all step functions, and then, by a density argument, the collection of the exponential functions has to be complete. This argument also gives us sufficient conditions to consider another famous question as to whether Fourier frames can exist for non-spectral self-affine measures, such as the Middle Third Cantor measure (See Subsection 1.4).

Step 3:

To complete the proof of Theorem 1.3 we have to consider the case . When , there is an exponential function that is orthogonal to every exponential function with integer frequencies. This implies that none of the subsets of integers can be complete and hence none of the sets in (1.8) can be complete.

It is possible that . The simplest example is to consider the interval which is generated by the IFS and . In this case, . However, this is rather trivial since the greatest common divisor (gcd) of is not 1. In fact, by some conjugation, we can assume the smallest -invariant lattice containing all sets , denoted by is . On , it is equivalent to gcd and we can settle this case using the result for . However, this simple situation ceases to exist when and we can find spectral self-affine measures with and .

To settle this case, our strategy is to identify as an invariant set of some dynamical system, and use the techniques in [2]. By doing so, we are able to show that in the case when the digit set will be reduced to a quasi product-form. Our methods are also similar to the ones used in [36]. However, as is not a complete set of representatives (mod ) (as it was in [36]), several additional adjustments will be needed. From the quasi-product form structure obtained, we construct the spectrum directly by induction on the dimension with the help of the Jorgensen-Pedersen criterion.

### 1.4. Fourier frames

We say that the self-affine measure admits a Fourier frame if there exists such that

 A∥f∥2≤∑λ∈Λ|∫f(x)e−2πi⟨λ,x⟩dμ(x)|2≤B∥f∥2, ∀ f∈L2(μ).

It is clear that the concept of Fourier frames is a natural generalization of exponential orthonormal bases. Whenever Fourier frames exist, is called a frame spectral measure and is called a frame spectrum. Frames on a general Hilbert space were introduced by Duffin and Schaeffer [7] and are now a fundamental research area in applied harmonic analysis, which is developing rapidly both in theory and in applications. In theory, Fourier frames are related to de Brange’s theory in complex analysis [46, 50]. In application, people regard frames as “overcomplete bases” and because of their redundancy, the reconstruction process is more robust to errors in data and it is now widely used in signal transmission and reconstruction. Reader may refer to [3] for the background of the general frame theory.

For the measures which are non-spectral, it is natural to ask the following question.

(Q2) Can a non-spectral fractal measure still admit some Fourier frames?

Some of the fundamental properties of Fourier frames were investigated in [17, 8, 9, 12]. This question was first proposed by Strichartz [52, p.212]. In particular, there have been discussions whether, specifically, the one-third Cantor measure is frame spectral. We introduce the following condition as a natural generalization of Hadamard triples.

###### Definition 1.8.

We say that the pair satisfies the almost-Parseval-frame condition if for any , there exists and a subset such that

 (1.9) (1−ϵ)∑b∈Bn|wb|2≤∑λ∈Jn∣∣ ∣∣1√Nn∑b∈Bnwbe−2πi⟨R−nb,λ⟩∣∣ ∣∣2≤(1+ϵ)∑b∈Bn|wb|2

for all . Equivalently,

 (1−ϵ)∥w∥2≤∥Fnw∥2≤(1+ϵ)∥w∥2

where and denotes the Euclidean norm.

Hadamard triples do satisfy this condition (even with ) and we prove:

###### Theorem 1.9.

Suppose that is simple digit set for and that satisfies the almost-Parseval-frame condition. Suppose that . Then the self-affine measure admits a Fourier frame with .

In fact, we will see that a natural geometric condition will guarantee that . This condition is satisfied for the one-third Cantor measure. The construction of Fourier frames now turns into a problem of matrix analysis, which is to construct finite sets so that the almost-Parseval-frame condition holds. At this time, we were unable to give a full solution. However, the recent solution of the Kadison-Singer conjecture [43] enabled Nitzan, Olevskii, Unlanovskii [45] to construct Fourier frames on unbounded sets of finite measures. One of their lemmas gives us a weak solution:

###### Proposition 1.10.

Suppose that is simple digit set for . There exist universal constants such that for all , there exists such that

 c0∑b∈Bn|wb|2≤∑λ∈Jn∣∣ ∣∣1√Nn∑b∈Bnwbe−2πi⟨R−nb,λ⟩∣∣ ∣∣2≤C0∑b∈Bn|wb|2

for all .

In the proof in Theorem 1.9, the idea is to concatenate the sets in order to obtain the frame spectrum , but for this we need in Definition 1.8 to be arbitrarily small . We cannot use the same method using just Proposition 1.10. It would be nice if we could construct an increasing sequence of sets , because then the frame spectrum can be obtained as the union of the sets . In any case, this proposition sheds some light on the plausibility of the almost-Parseval-frame condition. In fact, if we consider fractal measures that are not self-affine, and we allow some flexibility in the choice if the contraction ratios at different levels, then it can be proved that non-spectral fractal measure with Fourier frames do exist via the almost-Parseval-frame condition [38].

We organize our paper as follows: In Section 2, we prove the no-overlap condition for self-affine measures. In Section 3, we study the almost-Parseval-frame condition and concatenation of Hadamard triples. In Section 4, we will prove Theorem 1.3 under the assumption . In Section 5, we further reduce our problem to and we prove Theorem 1.3 on . In Section 6, we introduce the techniques from [2]. In Section 7, we use these techniques to show that must be of quasi-product form if . In Section 8, we prove Theorem 1.3 in full generality. In Section 9, We study Fourier frames on self-affine measures using the almost-Parseval-frame condition. We end the paper with some open problems in Section 10. Finally, an appendix is given to sharpen the frame bound in Section 9.

## 2. The no-overlap condition

This section is devoted to study the no-overlap condition from fractal geometry point of view and note that no Hadamard triple assumption is imposed. Throughout the section, we will fix the affine IFS given by an expansive matrix with integer entries and a simple digit set for and . The map is defined by

 τb(x)=R−1(x+b),(b∈B,x∈Rd).

and is its attractor.

We introduce some multi-index notation to describe our IFS. Let ( copies) and . For each ,

 τb(x)=τb1∘...∘τbn(x).

Also for any set , we define . Given a set of probabilities , , (), the associated self-affine measure is the unique Borel probability measure supported on satisfying the invariance identity

 (2.1) μ=∑b∈Bpbμb,

where we define , for all Borel sets , see [19]. By iterating this identity, we have

 μ=∑b∈Bnpbμb,

where and for all Borel sets and . With the definition of OSC and SOSC in Definition 1.5, the following theorem was proved by He and Lau [16, Theorem 4.4], see also [48] for self-similar IFSs.

###### Theorem 2.1.

[16] For a self-affine IFS, the OSC and SOSC are equivalent.

For any set , we denote by , , the closure, interior and its boundary respectively. The following theorem shows that the strong open set condition implies the no-overlap condition. Its proof is motivated by [6, Lemma 2.2].

###### Theorem 2.2.

Suppose that the IFS satisfies the strong open set condition with the open set . Then for the self-affine measure in (2.1), one has and . Moreover, satisfies the no-overlap condition.

###### Proof.

As , we can find and such that . In particular, there exists , for some such that . Let and let

 Ek=⋃b∈Bnk∖Ckτb(T(R,B)).

For any , there exists at least one such that . Then

 τb(T(R,B))⊂τb1...bs−1(τb0(T(R,B)))⊂τb1...bs−1(U).

As satisfies the open set condition for the IFS , we have Hence, . Now,

 1≥μ(U)≥μ(Ek)= ∑b∈Bnkpbμ(τ−1b(Ek)) ≥ ∑b∈Bnk∖Ckpbμ(τ−1b(Ek)) ≥ ∑b∈Bnk∖Ckpbμ(τ−1b(τb(T(R,B))) = ∑b∈Bnk∖Ckpb = ∑b∈Bnkpb−∑b∈Ckpb = 1−(∑b∈Cpb)k=1−(1−pb0)k.

As , tends to 0 as tends to infinity. This shows that . As (because ), we must have and .

For the no-overlap condition, we note that . Then . Hence,

 τb(T(R,B))∩τb′(T(R,B))⊆¯¯¯¯¯¯Ub∩¯¯¯¯¯¯¯Ub′=(Ub∩∂(Ub))∪(Ub′∩∂(Ub′)).

But is an open set satisfying the OSC, so . The no overlap condition will follow if we can show that for all .

Suppose on the contrary that , we apply (2.1)) and obtain

 0<μ(∂Ub)=∑b′∈Bpbμ(τ−1b′(∂Ub))

This implies that for some , . But and is supported essentially on , so we have

 μ((∂U+Rb−b′)∩U)>0.

As is open, . This implies that and this contradicts to the open set condition for (by a translation we can always assume ). Hence, and this completes the proof. ∎

###### Proposition 2.3.

Let be a integer expansive matrix and be a simple digit set for . Suppose that is a complete representative class (). Then the open set condition for the IFS is satisfied with open set .

###### Proof.

The statement that the open set condition is satisfied for the IFS with open set is probably known, but we present the proof for completeness. Let and note that . By taking the interior, we have . Also is non-empty, by [34]. To see that , we take Lebesgue measure on the invariance identity and obtain

 Leb(T)=Leb⎛⎝⋃b∈¯¯¯¯Bτb(T)⎞⎠≤∑b∈¯¯¯¯BLeb(τb(T))=#¯¯¯¯B|det(R)|Leb(T)=Leb(T).

Here Leb denotes the Lebesgue measure of and because is a set of complete representatives (mod ). is non-zero, by [34]. Hence,

 Leb⎛⎝⋃b∈¯¯¯¯Bτb(T)⎞⎠=∑b∈¯¯¯¯BLeb(τb(T))

and Leb()=0. This implies that since is an open set. ∎

Theorem 1.6 follows readily from the results above.

###### Proof of Theorem 1.6.

Proposition 2.3 shows that the OSC is satisfied, Theorem 2.1 shows that the SOSC is satisfied and then Theorem 2.2 shows that the no-overlap condition holds. ∎

In the end of this section, we mention that unequally-weighted self affine measures do not admit any Fourier frames, using one of our previous results.

###### Theorem 2.4.

Let be an expansive matrix with integer entries and let be a simple digit set for . Suppose that defined in (2.1) admits a Fourier frames. Then all are equal.

###### Proof.

Note that all these measures satisfies the no-overlap condition by Theorem 1.6. By [12, Theorem 1.5], all are equal. ∎

Because of the previous theorem, for the remainder of the paper, we will assume that the self-affine measures have equal weights .

## 3. The almost-Parseval-frame conditions and Hadamard triples.

In this section, we study the almost-Parseval-frame condition in Definition 1.8. First of all, we note that there is no loss of generality to assume , because we can replace by , and (1.9) is satisfied with replaced by .

###### Proposition 3.1.

Suppose that the pair satisfies the almost Parseval frame condition and is the set satisfying (1.9), with . We have the following:

(i) The elements in have distinct residues modulo .

(ii) Let (), then also satisfies (1.9).

###### Proof.

(i) Suppose on the contrary that we can find such that and are in the same equivalence class modulo . Let , for all , and plug it in (1.9). From the upper bound, we have

 2Nn+∑λ∈Jn∖{λ′,λ′′}∣∣ ∣∣1√Nn∑b∈Bnwbe−2πi⟨R−nb,λ⟩∣∣ ∣∣2≤(1+ϵ)Nn.

This implies that

 ∑λ∈Jn∖{λ′,λ′′}∣∣ ∣∣1√Nn∑b∈Bnwbe−2πi⟨R−nb,λ⟩∣∣ ∣∣2≤(ϵ−1)Nn<0

which is a contradiction. (ii) follows immediately from for all , and . ∎

Assuming that the almost-Parseval-frame condition is satisfied, we consider sequences such that and let and be the associated quantities satisfying

 (1−ϵk)∑b∈Bnk|wb|2≤∑λ∈Jnk∣∣ ∣∣∑b∈Bnk1√Nnkwbe−2πi⟨R−nkb,λ⟩∣∣ ∣∣2≤(1+ϵk)∑d∈Dnk|wd|2.

Letting , we consider the and defined in (1.7) and (1.8), i.e.,

 Λk=Jn1+(RT)m1Jn2+(RT)m2Jn3+...+(RT)mk−1Jnk, Λ=∞⋃k=1Λk.

Note that the digit sets satisfy

 Bmk+1=Rnk+1Bmk+Bnk+1,Bm1=Bn1.
###### Proposition 3.2.

With the notations above, we have

 ck∥w∥2≤∑λ∈Λk∣∣ ∣∣1√Nmk∑b∈Bmkwbe−2πi⟨R−mkb,λ⟩∣∣ ∣∣2≤Ck∥w∥2

where and .

###### Proof.

We prove this by induction on . The inequality for is the almost-Parseval-frame condition with and . Assuming the inequality is proved for , we now establish it for . We consider the upper bound inequality. If and , we can write uniquely and where , , and . For any vectors , we have

 ∑λ∈Λk+1∣∣ ∣∣1√Nmk+1∑b∈Bmk+1wbe−2πi⟨R−mk+1b,λ⟩∣∣ ∣∣2 = ∑λ1∈Λk∑λ2∈Jnk+1∣∣ ∣∣1√Nmk+1∑b2∈Bnk+1∑b1∈Bmkwb1b2e−2πi⟨R−mk+1(Rnk+1b1+b2),λ1+(RT)mkλ2⟩∣∣ ∣∣2 = ∑λ1∈Λk∑λ2∈Jnk+1∣∣ ∣∣1√Nnk+1∑b2∈Bnk+1e−2πi⟨R−nk+1b2,λ2⟩∑b1∈Bmk1√Nmkwb1b2e−2πi⟨R−mkb1+R−mk+1b2,λ1⟩∣∣ ∣∣2 ≤ (1+ϵk+1)∑λ1∈Λk∑b2∈Bnk+1∣∣ ∣∣1√Nmk∑b1∈Bmkwb1b2e−2πi⟨R−mkb1+R−mk+1b2,λ1⟩∣∣ ∣∣2 = ≤ (1+ϵk+1)Ck∑b2∈Bnk+1∑b1∈Bmk|wb1b2|2=Ck+1∥w∥2.

The proof for the lower bound is similar. ∎

Now, we turn to study Hadamard triples as defined in (1.1) in the introduction. We first remark that the elements of must be in distinct residue classed modulo , because must have mutually orthogonal rows. This implies that

 (3.1) ∑ℓ∈Le2πi⟨R−1(b−b′),ℓ⟩=0, % if b≠b′.

If for some , the sum above is equal to . Similarly, the elements must be in distinct residue class modulo . As is a unitary matrix, it is clear that we have for all . i.e.

 ∑ℓ∈L∣∣ ∣∣∑b∈Bwb1√Ne−2πi⟨R−1b,ℓ⟩∣∣ ∣∣2=∑b∈B|wb|2.

From this, we will conclude in Corollary 3.3 that satisfies the almost-Parseval-frame condition (with !). We also need to consider towers of Hadamard triples. Using the definition of in (1.3) and