Monatsh Math (2015) 176:293–322 DOI 10.1007/s00605-014-0725-0

Spectral self-affine measures on the spatial Sierpinski gasket Jian-Lin Li

Received: 3 May 2013 / Accepted: 10 December 2014 / Published online: 30 December 2014 © Springer-Verlag Wien 2014

Abstract The self-affine measure μ M,D corresponding to a diagonal matrix M with entries p1 , p2 , p3 ∈ Z\{0, ±1} and D = {0, e1 , e2 , e3 } in the space R3 is supported on the three-dimensional Sierpinski gasket, where e1 , e2 , e3 are the standard basis of unit column vectors in R3 . In this paper we determine the spectrality of μ M,D for certain p1 , p2 and p3 . The results here generalize the corresponding results on the spectrality of self-affine measures. Keywords Iterated function system · Self-affine measure · Orthogonal exponentials · Spectrality Mathematics Subject Classification

28A80 · 42C05 · 46C05

1 Introduction Motivated by the well-known L 2 Fourier representation of periodic functions on the real line, or Rn , in recent years, a number of authors have developed the Fourier representations for general classes of self-similar measures μ; for example when selfsimilarity is defined from a fixed (finite) set of affine mappings in Rn . One asks for the existence of L 2 -Fourier series for functions in L 2 (μ) where μ is such a given selfsimilar measure, say of a particular fractal dimension. Specifically, one asks when L 2 (μ) has an orthogonal basis of complex exponentials, a Fourier basis. If so, such

Communicated by P. Friz. J.-L. Li (B) College of Mathematics and Information Science, Shaanxi Normal University, Xi’an 710062, People’s Republic of China e-mail: [email protected]

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measures μ are called spectral measures. Indeed, it is known that there is a natural family of measures μ admitting L 2 Fourier series, but there are relatively few classes of examples when the spectral property is known. This paper addresses the problem for a class of self-affine measures in the space R3 . Let M ∈ Mn (Z) be an expanding integer matrix and D ⊂ Zn be a finite digit set of the cardinality |D|. The self-affine measure μ := μ M,D considered here is the unique probability measure satisfying μ=

1  μ ◦ φd−1 , |D|

(1.1)

d∈D

and is supported on the compact set T ⊂ Rn , where T := T (M, D) is the attractor (or invariant set) of the affine iterated function system (IFS) {φd (x) = M −1 (x + d)}d∈D (the unique compact set satisfying T = ∪d∈D φd (T )). A more explicit expression of T is given by the following radix expansion

T (M, D) =

⎧ ∞ ⎨ ⎩

M− j d j : d j ∈ D

j=1

⎫ ⎬ ⎭

=

∞ 

M − j D.

(1.2)

j=1

Let S ⊂ Zn be a finite subset of the cardinality |S| = |D|. Corresponding to the dual IFS {ψs (x) = M ∗ x + s}s∈S , we use (M, S) to denote the expansive orbit of 0 under {ψs (x)}s∈S , that is (M, S) :=

⎧ k−1 ⎨ ⎩

j=0

⎫ ⎬

M ∗ j s j : k ≥ 1 and s j ∈ S , ⎭

(1.3)

where M ∗ denotes the transposed conjugate matrix of M. We call μ M,D a spectral measure if there exists a discrete set  ⊂ Rn such that E() := {eλ (x) = e2πi λ,x : λ ∈ } forms an orthogonal basis (Fourier basis) for the Hilbert space L 2 (μ M,D ). The set  is also called a spectrum for μ M,D . The question we are concerned is whether or not the measure μ M,D is a spectral measure. This question has its origin in analysis and geometry, and is started with the work of Jorgensen and Pedersen [3] who showed that for certain M, D and S, (M, S) may be a spectrum for μ M,D . Subsequently, the research on this question have received much attention. The present paper will follow the paper [8] to further studying the spectrality of self-affine measure μ M,D on the spatial Sierpinski gasket T (M, D), where ⎡

⎤ p1 0 0 M = ⎣ 0 p2 0 ⎦ ( p1 , p2 , p3 ∈ Z\{0, ±1}) and 0 0 p3 ⎧⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎫ 0 ⎬ 1 0 ⎨ 0 D = ⎝0⎠,⎝0⎠,⎝1⎠,⎝0⎠ . ⎩ ⎭ 0 0 0 1

123

(1.4)

Spectral self-affine measures

295

For the pair (M, D) in (1.4), it has been an interesting topic to examine the spectrality of self-affine measure μ M,D . With the effort of Jorgensen and Pedersen [2], [3, Example 7.1], Strichartz [10], [11, Example 2.9(e)], Dutkay and Jorgensen [1, Theorem 5.1(iii)], and the author [5, Theorem 1], [8], the spectrality or the non-spectrality of μ M,D can be summarized as the following Theorem A. Therorem A For the self-affine measure μ M,D corresponding to (1.4), the following spectrality and non-spectrality hold: (i) If p1 = p2 = p3 = p and p ∈ 2Z\{0}, then μ M,D is a spectral measure; (ii) If p j ∈ 2Z\{0, 2} for j = 1, 2, 3, then μ M,D is a spectral measure; (iii) If p j ∈ (2Z+1)\{±1} for j = 1, 2, 3, then μ M,D is a non-spectral measure, and there exist at most 4 mutually orthogonal exponential functions in L 2 (μ M,D ), where the number 4 is the best upper bound. However, if p j ∈ 2Z\{0} and p j = 2 for one or two j  s in {1, 2, 3}, then the above Theorem A(ii) cannot give us a conclusion. Also, the spectrality or the nonspectrality of μ M,D is not known if p j ( j ∈ {1, 2, 3}) have different parity. This leaves the following two open problems. Question 1. How about the spectrality of μ M,D if p j ∈ 2Z\{0}( j = 1, 2, 3) and one or two of the three numbers p1 , p2 , p3 can take the value 2? Question 2. How about the spectrality of μ M,D if p j ( j = 1, 2, 3) have different parity? In the following two sections, we discuss the above two questions respectively. The results obtained in each section answer the above two questions mostly, although we cannot answer the two questions completely. Moreover, the results here extend Theorem A in a reasonable manner. 2 On the Question 1 The expansibility of M requires | p j | > 1 for all j = 1, 2, 3. Among the three even integers “ p1 , p2 , p3 ”, the Question 1 can be considered in the following three cases: Case 1: p1 = 2 and p j ∈ 2Z\{0, ±2} for j = 2, 3; Case 2: p1 = 2, p2 = −2 and p3 ∈ 2Z\{0, ±2}; Case 3: p1 = p2 = 2 and p3 ∈ 2Z\{0, 2}. The reason is as follows. If one or two of the three numbers p1 , p2 , p3 can take the value 2, then the Question 1 can be divided into the following two cases: (a) p1 = 2 and p j ∈ 2Z\{0, 2} for j = 2, 3; (b) p1 = p2 = 2 and p3 ∈ 2Z\{0, 2}. The case (a) denotes that one of them is 2, the other two integers are in 2Z\{0, 2}; the case (b) denotes that two of them are 2, the other one integer is in 2Z\{0, 2}. Furthermore, the case (a) can be divided into the following three subcases according to the fact that p2 and p3 can take the value “ − 2”: a(i) p1 = 2 and p j ∈ 2Z\{0, ±2} for j = 2, 3;

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a(ii) p1 = 2, p2 = −2 and p3 ∈ 2Z\{0, ±2}; a(iii) p1 = 2 and p2 = p3 = −2. In the case (a), the subcase a(i) denotes that none of p2 and p3 can take the value “−2”; the subcase a(ii) denotes that one of p2 and p3 can take the value “ − 2”; the subcase a(iii) denotes that both p2 and p3 can take the value “ − 2”. Since the spectrality or the non-spectrality of μ M,D is the same as μ−M,D (see [7, Remark 3.3(ii)]), the subcase a(iii) is contained in the case (b). So, in order to answer the Question 1, one needs to deal with the above Case 1, Case 2 and Case 3 respectively. In this section, we first answer the Question 1 in the Case 1 and Case 2. We then give a remark on the Case 3. The discussion in this part leads to a solution of the Case 3 if p3 = −2. Theorem 2.1 For the self-affine measure μ M,D corresponding to (1.4), if p1 = 2 and p j ∈ 2Z\{0, ±2} for j = 2, 3, then μ M,D is a spectral measure. Theorem 2.2 For the self-affine measure μ M,D corresponding to (1.4), if p1 = 2, p2 = −2 and p3 ∈ 2Z\{0, ±2}, then μ M,D is a spectral measure. For the pair (M, D) given by (1.4) with even numbers p1 , p2 and p3 , we can choose the set ⎧⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎫ ⎞ ⎛ p1 /2 p1 /2 ⎬ 0 ⎨ 0 S = ⎝ 0 ⎠ , ⎝ p2 /2 ⎠ , ⎝ p2 /2 ⎠ , ⎝ 0 ⎠ ⊂ Z3 (2.1) ⎩ ⎭ 0 0 p3 /2 p3 /2 such that (M −1 D, S) is a compatible pair. Then, the invariant set T (M ∗ , S) is given by ⎧ ⎫ ⎛ ⎞ ⎛ ⎞ j ⎪ ⎪ s1, j / p1 ∞ s1, j ⎨ ⎬ ⎜ j⎟ ⎝ ∗ ⎠ (2.2) T (M , S) = ⎝ s2, j / p2 ⎠ : s2, j ∈ S . ⎪ ⎪ ⎩ j=1 ⎭ j s3, j s3, j / p 3

Under the condition of compatible pair, there are some results that can be applied to determine the spcetrality of self-affine measures. The first result is due to Strichartz [10] who obtained the following result which we list as Lemma 2.1 (see also [9]). Lemma 2.1 Let M ∈ Mn (Z) be expanding, D and S be finite subsets of Zn such that (M −1 D, S) is a compatible pair and 0 ∈ D ∩ S. Suppose that the zero set Z (m M −1 D (x)) of the function m M −1 D (x) is disjoint from the set T (M ∗ , S). Then (M, S) is a spectrum for μ M,D . In this Lemma 2.1, the condition that Z (m M −1 D (x)) ∩ T (M ∗ , S) = ∅ has several equivalent forms, and is known to be very difficult to check. However, if this condition holds, one can obtain an explicit spectrum (M, S). We shall manage to check this condition. For the given digit set D in (1.4), we have Z (m D (x)) := {x ∈ R3 : m D (x) = 0} = A1 ∪ A2 ∪ A3 , which yields the zero set Z (m M −1 D (x)) of the function m M −1 D (x) = m D (M ∗−1 x) by Z (m M −1 D (x)) = M ∗ Z (m D (x)) = M ∗ A1 ∪ M ∗ A2 ∪ M ∗ A3 ,

123

(2.3)

Spectral self-affine measures

297

where m D (x) =

 1 1 + e2πi x1 + e2πi x2 + e2πi x3 , x = (x1 , x2 , x3 )T ∈ R3 , 4

and A1 , A2 , A3 are given by ⎧⎛ ⎨ A1 = ⎝ ⎩

⎫ ⎞ 1/2 + k1 ⎬ ⎠ : a ∈ R, k1 , k2 , k3 ∈ Z ⊂ R3 ; a + k2 ⎭ 1/2 + a + k3

(2.4)

⎧⎛ ⎫ ⎞ ⎨ 1/2 + a + k1 ⎬ A2 = ⎝ 1/2 + k2 ⎠ : a ∈ R, k1 , k2 , k3 ∈ Z ⊂ R3 ; ⎩ ⎭ a + k3 ⎧⎛ ⎫ ⎞ a + k1 ⎨ ⎬ A3 = ⎝ 1/2 + a + k2 ⎠ : a ∈ R, k1 , k2 , k3 ∈ Z ⊂ R3 . ⎩ ⎭ 1/2 + k3

(2.5)

(2.6)

In each of the above Case 1 and Case 2, we show that T (M ∗ , S) ∩ Z (m M −1 D (x)) =

3  

 T (M ∗ , S) ∩ M ∗ A j = ∅,

(2.7)

j=1

and the desired conclusion follows from Lemma 2.1.

2.1 Proof of Theorem 2.1 In the Case 1, we consider the following compatible pair (M −1 D, S): ⎧⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎫ ⎤ 1 0 0 ⎬ 2 0 0 ⎨ 0 and M = ⎣ 0 p2 0 ⎦ , D = ⎝ 0 ⎠ , ⎝ 0 ⎠ , ⎝ 1 ⎠ , ⎝ 0 ⎠ ⎩ ⎭ 0 0 0 1 0 0 p3 ⎧⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎫ ⎞ ⎛ 1 1 0 ⎨ 0 ⎬ S = ⎝ 0 ⎠ , ⎝ p2 /2 ⎠ , ⎝ p2 /2 ⎠ , ⎝ 0 ⎠ , ⎩ ⎭ 0 p3 /2 0 p3 /2 ⎡

(2.8)

where p2 , p3 ∈ 2Z\{0, ±2}. Then, the invariant set T (M ∗ , S) is given by ⎧ ⎛ ⎞ ⎛ ⎞ ⎪ s1, j /2 j ∞ s1, j ⎨ ⎜ j⎟ T (M ∗ , S) = ⎝ s2, j / p2 ⎠ : ⎝ s2, j ⎠ ∈ ⎪ j ⎩ j=1 s / p s3, j 3, j 3

⎫ ⎪ ⎬ S

⎪ ⎭

,

(2.9)

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which shows that for any x = (x1 , x2 , x3 )T ∈ T (M ∗ , S), we have ⎧ ⎨

0 −2/3 ≤ −| p2 |/(2(| p2 | − 1)) ⎩ −2/3 ≤ −| p3 |/(2(| p3 | − 1))

≤ x1 ≤ 1, ≤ x2 ≤ | p2 |/(2(| p2 | − 1)) ≤ 2/3, ≤ x3 ≤ | p3 |/(2(| p3 | − 1)) ≤ 2/3.

(2.10)

From (2.5), the point x = (x1 , x2 , x3 )T ∈ M ∗ A2 locates on the plane x2 = ( 21 + k2 ) p2 for some k2 ∈ Z. Since |( 21 +k2 ) p2 | ≥ 1 for any k2 ∈ Z, we obtain, from (2.10), that T (M ∗ , S) ∩ M ∗ A2 = ∅. Similarly, from (2.6), the point x = (x1 , x2 , x3 )T ∈ M ∗ A3 locates on the plane x3 = ( 21 + k3 ) p3 for some k3 ∈ Z. Since |( 21 + k3 ) p3 | ≥ 1 for any k3 ∈ Z, we obtain, from (2.10), that T (M ∗ , S) ∩ M ∗ A3 = ∅. Now, from (2.4), the point x = (x1 , x2 , x3 )T ∈ M ∗ A1 locates on the plane x1 = 1 + 2k1 for some k1 ∈ Z. From (2.10), we only need to consider the subset of M ∗ A1 where k1 = 0, that is, the subset given by

A1,1

⎧⎛ ⎨ := ⎝ ⎩

⎫ ⎞ 1 ⎬ ⎠ : a ∈ R, k2 , k3 ∈ Z ⊆ M ∗ A1 ⊂ R3 . (a + k2 ) p2 ⎭ (1/2 + a + k3 ) p3

(2.11)

This reduces the the set T (M ∗ , S) ∩ M ∗ A1 to the set T (M ∗ , S) ∩ A1,1 . From (2.10) and (2.11), a necessary condition for the points of A1,1 in T (M ∗ , S) is 

|(a + k2 ) p2 | |(1/2 + a + k3 ) p3 |

≤ | p2 |/(2(| p2 | − 1)), ≤ | p3 |/(2(| p3 | − 1)).

(2.12)

It follows from (2.12) that 

|a + k2 | |1/2 + a + k3 |

≤ 1/6, ≤ 1/6,

(2.13)

which yields −5/6 ≤ k3 − k2 ≤ −1/6, a contradiction. Hence T (M ∗ , S) ∩ M ∗ A1 = T (M ∗ , S) ∩ A1,1 = ∅, and (2.7) holds. 2.2 Proof of Theorem 2.2 In the Case 2, we consider the following compatible pair (M −1 D, S): ⎧⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎫ ⎤ 2 0 0 1 0 0 ⎬ ⎨ 0 M = ⎣ 0 −2 0 ⎦ , D = ⎝ 0 ⎠ , ⎝ 0 ⎠ , ⎝ 1 ⎠ , ⎝ 0 ⎠ and ⎩ ⎭ 0 0 p3 0 0 0 1 ⎧⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎫ 1 0 1 ⎨ 0 ⎬ S = ⎝ 0 ⎠ , ⎝ −1 ⎠ , ⎝ −1 ⎠ , ⎝ 0 ⎠ , (2.14) ⎩ ⎭ 0 0 p3 /2 p3 /2 ⎡

123

Spectral self-affine measures

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where p3 ∈ 2Z\{0, ±2}. Then, the invariant set T (M ∗ , S) is given by ⎧ ⎛ ⎞ ⎛ ⎞ s1, j /2 j ∞ s1, j ⎨ ⎝ s2, j /(−2) j ⎠ : ⎝ s2, j ⎠ ∈ T (M ∗ , S) = ⎩ j s3, j j=1 s3, j / p3

⎫ ⎬ S

⎭

,

(2.15)

which shows that for any x = (x1 , x2 , x3 )T ∈ T (M ∗ , S), we have ⎧ ⎨

0 ≤ x1 ≤ 1, 2/3, −1/3 ≤ x2 ≤ ⎩ −2/3 ≤ −| p3 |/(2(| p3 | − 1)) ≤ x3 ≤ | p3 |/(2(| p3 | − 1)) ≤ 2/3.

(2.16)

With the same method as above, we know, from (2.16), that T (M ∗ , S)∩ M ∗ A2 = ∅ and T (M ∗ , S)∩ M ∗ A3 = ∅. So, we mainly deal with the set T (M ∗ , S)∩ M ∗ A1 below. Now, the point x = (x1 , x2 , x3 )T ∈ M ∗ A1 locates on the plane x1 = 1 + 2k1 for some k1 ∈ Z. From (2.16), we only need to consider the subset of M ∗ A1 where k1 = 0, that is, the subset given by

A1,2

⎧⎛ ⎨ := ⎝ ⎩

⎫ ⎞ 1 ⎬ (−2)(a + k2 ) ⎠ : a ∈ R, k2 , k3 ∈ Z ⊆ M ∗ A1 ⊂ R3 . ⎭ (1/2 + a + k3 ) p3

(2.17)

From (2.16) and (2.17), a necessary condition for the points of A1,2 in T (M ∗ , S) is 

−1/3 ≤ −2(a + k2 ) ≤ 2/3, |(1/2 + a + k3 ) p3 | ≤ | p3 |/(2(| p3 | − 1)).

(2.18)

It follows from (2.18) that  −

1 2 + 3 2(| p3 | − 1)

 ≤ k3 − k2 ≤

1 1 − . 2(| p3 | − 1) 6

(2.19)

We divide the discussion into the following two cases. Case | p3 | = 4: In the case when | p3 | = 4, there are no integer numbers k2 and k3 such that (2.19) holds. This shows that T (M ∗ , S) ∩ M ∗ A1 = T (M ∗ , S) ∩ A1,2 = ∅. Case | p3 | = 4: In the case when | p3 | = 4, the condition (2.19) gives k3 −k2 = 0. In this case, let k3 = k2 = k, the condition (2.18) shows that a+k2 = a+k3 = a+k = −1/3. So, we are limited to the point ⎞ 1 := ⎝ 2/3 ⎠ ∈ A1,2 ⊂ M ∗ A1 ⊂ R3 , (1/6) p3 ⎛

A˜ 1,2

(2.20)

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In the following, we prove that A˜ 1,2 ∈ T (M ∗ , S). This is implied in the conclusion: ⎛

⎞ 1 ⎝ 2/3 ⎠ ∈ T (M ∗ , S) ⇐⇒ x3 = 1 p3 . 30 x3

(2.21)

In fact, let T1 := {(x1 , x2 , x3 )T ∈ T (M ∗ , S) : x1 = 1}, from (2.15), we have ⎧ ⎛ ⎞ ⎛ ⎞ ⎧⎛ ⎞ ⎛ ⎞⎫⎫ s1, j /2 j ∞ s1, j 1 ⎨ 1 ⎨ ⎬⎬ ⎝ s2, j /(−2) j ⎠ : ⎝ s2, j ⎠ ∈ ⎝ −1 ⎠ , ⎝ 0 ⎠ T1 = . ⎩ ⎩ ⎭⎭ j 0 p3 /2 s3, j j=1 s3, j / p3 This is due to the fact that 0 ≤

∞ 

s1, j /2 j ≤ 1 for all s1, j ∈ {0, 1}, and

j=1

and

∞  j=1

s2, j (−2) j

=

s1, j /2 j = 1

j=1

only if all s1, j = 1. For all s2, j ∈ {0, −1}, we have −

∞ 

∞

∞

∞

j=1

j=1

j=0

 s2,2 j 1  s2,2 j+1 2 1  s2, j ≤ = − ≤ , j j 3 (−2) (4) 2 (4) j 3

2 3

(2.22)

(2.23)

only if all s2,2 j = 0 ( j = 1, 2, . . .) and s2,2 j+1 = −1 ( j =

0, 1, 2, . . .). So, if (1, 2/3, x3 )t ∈ T1 , then x3 =

∞  s3, j j j=1 p3

=

∞  s3,2 j 2j j=1 p3

+

∞  s3,2 j+1 2 j+1 j=0 p3

=

∞  p3 /2 2j j=1 p3

+

∞ 

0

2 j+1 j=0 p3

=

1 p3 . (2.24) 30

This proves (2.21), and thus A˜ 1,2 ∈ T (M ∗ , S). The above two cases give the fact that T (M ∗ , S) ∩ M ∗ A1 = ∅. Hence (2.7) holds.

2.3 Remarks on the Case 3 In the Case 3, we consider the following compatible pair (M −1 D, S): ⎧⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎫ ⎤ 1 0 0 ⎬ 2 0 0 ⎨ 0 and M = ⎣0 2 0 ⎦, D = ⎝0⎠,⎝0⎠,⎝1⎠,⎝0⎠ ⎩ ⎭ 0 0 0 1 0 0 p3 ⎧⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎫ 1 0 1 ⎨ 0 ⎬ S = ⎝0⎠,⎝1⎠,⎝ 1 ⎠,⎝ 0 ⎠ , ⎩ ⎭ 0 0 p3 /2 p3 /2 ⎡

123

(2.25)

Spectral self-affine measures

301

where p3 ∈ 2Z\{0, 2}. Then, the invariant set T (M ∗ , S) is given by ⎧ ⎛ ⎞ ⎛ ⎞ s1, j /2 j ∞ s1, j ⎨ ⎝ s2, j /2 j ⎠ : ⎝ s2, j ⎠ ∈ T (M ∗ , S) = ⎩ j s3, j j=1 s3, j / p 3

⎫ ⎬ S

⎭

.

(2.26)

It is different from the above Case 1 and Case 2 that in (2.25), we always have ⎛ ⎞ 1 ⎝ 1 ⎠ ∈ T (M ∗ , S) ∩ Z (m M −1 D (x)). 0

(2.27)

So, we cannot desire that T (M ∗ , S) ∩ Z (m M −1 D (x)) is an empty set. Also, from (M ∗ − I )−1 (1, 1, 0)T ∈ Z3 , we know that (M, S) is not a spectrum for μ M,D (see [4, Theorem 3.3]), that is, the infinite orthogonal system E((M, S)) is not complete in the Hilbert space L 2 (μ M,D ). However, for the compatible pair (2.25), we find that the set T (M ∗ , S) ∩ Z (m D (x)) is an infinite set and is independent of p3 . These facts are useful to the further research of Case 3. Theorem 2.3 Let p3 ∈ 2Z\{0, ±2}. For the compatible pair (M −1 D, S) given by (2.25), the set T (M ∗ , S) ∩ Z (m D (x)) is given by ⎧⎛ ⎧⎛ ⎫ ⎫ ⎧⎛ ⎞ ⎞⎫ ⎞ ⎬  ⎨ 1/2 + a  ⎬  ⎨ 1/2 ⎬  a ⎨ ⎝ a ⎠ : a ∈ 0, 1 ⎝ 1/2 ⎠ . ⎝ a + 1/2 ⎠ : a ∈ 0, 1 ⎩ ⎩ ⎭ ⎭ ⎭ ⎩ 2 2 1/2 0 1/2 (2.28) Proof We first have T (M ∗ , S) ∩ Z (m D (x)) =

3  

 T (M ∗ , S) ∩ A j ,

j=1

where M, D, S are given by (2.25), and A j ( j = 1, 2, 3) are given by (2.4)–(2.6) above. The proof of Theorem 2.3 can be divided into two cases: p3 > 0 and p3 < 0. The latter case is similar to the former case, so we only consider the case p3 > 0. Then, from (2.26), the point x = (x1 , x2 , x3 )T ∈ T (M ∗ , S) satisfies ⎧ ⎨ 0 ≤ x1 ≤ 1, 0 ≤ x2 ≤ 1, ⎩ 0 ≤ x3 ≤ p3 /(2( p3 − 1)) ≤ 2/3.

(2.29)

In the following, we give an explicit formula for the sets T (M ∗ , S) ∩ A j ( j = 1, 2, 3).   Claim 1 T (M ∗ , S) ∩ A1 = {(1/2, 1/2, 0)T , (1/2, 1, 1/2)T , (1/2, 0, 1/2)T }.

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Proof From (2.29), we only need to consider the subset of A1 where k1 = 0, that is, the subset given by ⎧⎛ ⎫ ⎞ 1/2 ⎨ ⎬ ⎠ : a ∈ R, k2 , k3 ∈ Z ⊆ A1 ⊂ R3 . (2.30) A1,3 := ⎝ a + k2 ⎩ ⎭ 1/2 + a + k3 A necessary condition for the points of A1,3 in T (M ∗ , S) is 

0 ≤ a + k2 ≤ 1, 0 ≤ 1/2 + a + k3 ≤ p3 /(2( p3 − 1)) ≤ 2/3.

(2.31)

From (2.31), we have k2 − k3 = 0 or k2 − k3 = 1. So, the discussion is limited to the following two sets: ⎧⎛ ⎫ ⎞ 1/2 ⎨ ⎬ 1 ⎠ : a ∈ R, k ∈ Z and 0 ≤ a + k ≤ B3P := ⎝ a + k ⊆ A1,3 , (2.32) ⎩ 6⎭ 1/2 + a + k and B3Q

⎫ ⎧⎛ ⎞ 1/2 ⎬ ⎨ −1 ≤ a + k ≤ 0 ⊆ A1,3 . (2.33) := ⎝ a + k + 1 ⎠ : a ∈ R, k ∈ Z and ⎭ ⎩ 2 1/2 + a + k

Moreover, we have T (M ∗ , S)∩ A1 = T (M ∗ , S)∩ A1,3 = (T (M ∗ , S)∩ B3P )∪(T (M ∗ , S)∩ B3Q ). (2.34) On the other hand, to look for points (x1 , x2 , x3 )T ∈ T (M ∗ , S) with x1 = 1/2, we obtain, from (2.26), that T2 := {(x1 , x2 , x3 )T ∈ T (M ∗ , S) : x1 = 1/2} = T2a ∪ T2b ∪ T2c ∪ T2d ,

(2.35)

where

T2a

T2b

T2c

⎧⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎧⎛ ⎞ ⎛ ⎞ ⎫⎫ s1, j /2 j ∞ s1, j 1 ⎨ 0 ⎨ 1 ⎬⎬  ⎝ s2, j /2 j ⎠ : ⎝ s2, j ⎠ ∈ ⎝ 1 ⎠ , ⎝ 0 ⎠ := ⎝ 0 ⎠ + ; ⎩ ⎩ ⎭⎭ j 0 0 p s /2 j=2 s3, j / p 3, j 3 3

(2.36)

⎧⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎧⎛ ⎞ ⎛ ⎞⎫⎫ s1, j /2 j ∞ s1, j 0 ⎨ 1/2 ⎨ 0 ⎬⎬  ⎝ s2, j /2 j ⎠ : ⎝ s2, j ⎠ ∈ ⎝ 0 ⎠ , ⎝ 1 ⎠ := ⎝ 1/2 ⎠ + ; (2.37) ⎩ ⎩ ⎭⎭ j 0 0 p s /2 j=2 s3, j / p 3, j 3 3 ⎧⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎧⎛ ⎞ ⎛ ⎞⎫⎫ j s1, j /2 ∞ s1, j 1 ⎨ 0 ⎨ 1 ⎬⎬  ⎝ s2, j /2 j ⎠ : ⎝ s2, j ⎠ ∈ ⎝ 1 ⎠ , ⎝ 0 ⎠ := ⎝ 1/2 ⎠ + ; (2.38) ⎩ ⎩ ⎭⎭ j 1/2 0 p s /2 j=2 s3, j / p 3, j 3 3

123

Spectral self-affine measures

T2d

303

⎧⎛ ⎛ ⎞ ⎛ ⎞ ⎧⎛ ⎞ ⎛ ⎞ ⎞⎫⎫ s1, j /2 j ∞ s1, j 0 ⎨ 0 ⎨ 1/2 ⎬⎬  ⎝ s2, j /2 j ⎠ : ⎝ s2, j ⎠ ∈ ⎝ 0 ⎠ , ⎝ 1 ⎠ := ⎝ 0 ⎠ + . (2.39) ⎩ ⎩ ⎭⎭ j 0 p 1/2 s /2 j=2 s3, j / p 3, j 3 3

(I) For any (x1 , x2 , x3 )T ∈ T2a , it follows from (2.36) that 0 ≤ x2 ≤ 1/2 and 0 ≤ x3 ≤ 1/(2( p3 − 1)) ≤ 1/6. If x2 − x3 = −1/2, then x3 = x2 + 1/2 ≥ 1/2, a contradiction of 0 ≤ x3 ≤ 1/6. This shows that T2a ∩B3P = ∅; If x2 −x3 = 1/2, then x2 = 1/2 and x3 = 0, which shows that T2a ∩ B3Q = {(1/2, 1/2, 0)T }. Hence (T2a ∩ B3P ) ∪ (T2a ∩ B3Q ) = {(1/2, 1/2, 0)T }. (II) For any (x1 , x2 , x3 )T ∈ T2b , it follows from (2.37) that 1/2 ≤ x2 ≤ 1, which shows that T2b ∩ B3P = ∅. The condition x2 − x3 = 1/2 holds only if s2, j = s3, j = 0 for all j = 2, 3, . . .. This shows that T2b ∩ B3Q = {(1/2, 1/2, 0)T }. Hence (T2b ∩ B3P ) ∪ (T2b ∩ B3Q ) = {(1/2, 1/2, 0)T }. (III) For any (x1 , x2 , x3 )T ∈ T2c , we write x2 = 1/2 + α2 and x3 = 1/2 + α3 , it follows from (2.38) that 0 ≤ α2 ≤ 1/2 and 0 ≤ α3 ≤ 1/(2( p3 − 1)) ≤ 1/6. The condition x2 − x3 = −1/2 gives α3 = 1/2 + α2 ≥ 1/2, a contradiction, hence T2c ∩ B3P = ∅. The condition x2 − x3 = 1/2 gives α2 − α3 = 1/2, which leads to α2 = 1/2 and α3 = 0. This shows that T2c ∩ B3Q = {(1/2, 1, 1/2)T }. Hence (T2c ∩ B3P ) ∪ (T2c ∩ B3Q ) = {(1/2, 1, 1/2)T }. (IV) For any (x1 , x2 , x3 )T ∈ T2d , we write x2 = 0 + β2 and x3 = 1/2 + β3 , it follows from (2.39) that 0 ≤ β2 ≤ 1/2 and 0 ≤ β3 ≤ 1/(2( p3 − 1)) ≤ 1/6. The condition x2 − x3 = 1/2 gives β2 = 1 + β3 ≥ 1, a contradiction, hence T2d ∩ B3Q = ∅. The condition x2 − x3 = −1/2 gives β2 − β3 = 0, which leads to β2 = β3 = 0. This shows that T2d ∩ B3P = {(1/2, 0, 1/2)T }. Hence (T2d ∩ B3P ) ∪ (T2d ∩ B3Q ) = {(1/2, 0, 1/2)T }. It follows from (2.34), (2.35) and the above four cases (I)(II)(III)(IV) that the set T (M ∗ , S)∩ A1 = {(1/2, 1/2, 0)T , (1/2, 1, 1/2)T , (1/2, 0, 1/2)T } is finite. This completes the proof of Claim 1.   Claim 2 T (M ∗ , S) ∩ A2 = {(1/2, 1/2, 0)T , (0, 1/2, 1/2)T , (1, 1/2, 1/2)T }. Proof From (2.29), we only need to consider the subset of A2 where k2 = 0, that is, the subset given by

A1,4

⎧⎛ ⎫ ⎞ ⎨ 1/2 + a + k1 ⎬ ⎠ : a ∈ R, k1 , k3 ∈ Z ⊆ A2 ⊂ R3 . 1/2 := ⎝ ⎩ ⎭ a + k3

(2.40)

A necessary condition for the points of A1,4 in T (M ∗ , S) is 

0 ≤ 1//2 + a + k1 ≤ 1, 0 ≤ a + k3 ≤ p3 /(2( p3 − 1)) ≤ 2/3.

(2.41)

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From (2.41), we have k1 − k3 = 0 or k1 − k3 = −1. So, the discussion is limited to the following two sets:

B4P

⎧⎛ ⎫ ⎞ ⎨ 1/2 + a + k ⎬ ⎠ : a ∈ R, k ∈ Z and 0 ≤ a + k ≤ 1 ⊆ A1,4 , (2.42) 1/2 := ⎝ ⎩ 2⎭ a+k

and ⎧⎛ ⎫ ⎞ ⎨ 1/2 + a + k ⎬ ⎠ : a ∈ R, k ∈ Z and −1 ≤ a + k ≤ −1 ⊆ A1,4 . 1/2 B4Q := ⎝ ⎩ 2 3 ⎭ a+k+1 (2.43) Moreover, we have T (M ∗ , S)∩ A2 = T (M ∗ , S)∩ A1,4 = (T (M ∗ , S)∩B4P )∪(T (M ∗ , S)∩B4Q ). (2.44) On the other hand, to look for points (x1 , x2 , x3 )T ∈ T (M ∗ , S) with x2 = 1/2, we obtain, from (2.26), that T3 := {(x1 , x2 , x3 )T ∈ T (M ∗ , S) : x2 = 1/2} = T3a ∪ T3b ∪ T3c ∪ T3d ,

(2.45)

where

T3a

T3b

T3c

T3d

⎧⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎧⎛ ⎞ ⎛ ⎞⎫⎫ s1, j /2 j ∞ s1, j 0 ⎨ 0 ⎨ 1 ⎬⎬  ⎝ s2, j /2 j ⎠ : ⎝ s2, j ⎠ ∈ ⎝ 1 ⎠ , ⎝ 1 ⎠ := ⎝ 0 ⎠ + ; ⎩ ⎩ ⎭⎭ j 0 0 p s /2 j=2 s3, j / p 3, j 3 3

(2.46)

⎧⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎧⎛ ⎞ ⎛ ⎞⎫⎫ s1, j /2 j ∞ s1, j 1 ⎨ 1/2 ⎨ 0 ⎬⎬  ⎝ s2, j /2 j ⎠ : ⎝ s2, j ⎠ ∈ ⎝ 0 ⎠ , ⎝ 0 ⎠ := ⎝ 1/2 ⎠ + ; (2.47) ⎩ ⎩ ⎭⎭ j 0 0 p s /2 j=2 s3, j / p 3, j 3 3

⎧⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎧⎛ ⎞ ⎛ ⎞⎫⎫ s1, j /2 j ∞ s1, j 1 ⎨ 0 ⎨ 0 ⎬⎬  ⎝ s2, j /2 j ⎠ : ⎝ s2, j ⎠ ∈ ⎝ 0 ⎠ , ⎝ 0 ⎠ := ⎝ 1/2 ⎠ + ; (2.48) ⎩ ⎩ ⎭⎭ j 1/2 0 p s /2 j=2 s3, j / p 3, j 3 3

⎧⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎧⎛ ⎞ ⎛ ⎞⎫⎫ s1, j /2 j ∞ s1, j 0 ⎨ 1/2 ⎨ 1 ⎬⎬  ⎝ s2, j /2 j ⎠ : ⎝ s2, j ⎠ ∈ ⎝ 1 ⎠ , ⎝ 1 ⎠ := ⎝ 0 ⎠ + . (2.49) ⎩ ⎩ ⎭⎭ j 1/2 0 p s /2 j=2 s3, j / p 3, j 3 3

(I) For any (x1 , x2 , x3 )T ∈ T3a , it follows from (2.46) that 0 ≤ x1 ≤ 1/2 and 0 ≤ x3 ≤ 1/(2( p3 − 1)) ≤ 1/6. If x1 − x3 = −1/2, then x3 = x1 + 1/2 ≥ 1/2, a contradiction of 0 ≤ x3 ≤ 1/6. This shows that T3a ∩B3Q = ∅; If x1 −x3 = 1/2, then x1 = 1/2 and x3 = 0, which shows that T3a ∩ B3P = {(1/2, 1/2, 0)T }. Hence (T3a ∩ B3P ) ∪ (T3a ∩ B3Q ) = {(1/2, 1/2, 0)T }.

123

Spectral self-affine measures

305

(II) For any (x1 , x2 , x3 )T ∈ T3b , it follows from (2.47) that 1/2 ≤ x1 ≤ 1, which shows that T3b ∩ B4Q = ∅. The condition x1 − x3 = 1/2 holds only if s2, j = s3, j = 0 for all j = 2, 3, · · · . This shows that T3b ∩ B4P = {(1/2, 1/2, 0)T }. Hence (T3b ∩ B4P ) ∪ (T3b ∩ B4Q ) = {(1/2, 1/2, 0)T }. (III) For any (x1 , x2 , x3 )T ∈ T3c , we write x1 = 0 + γ1 and x3 = 1/2 + γ3 , it follows from (2.48) that 0 ≤ γ1 ≤ 1/2 and 0 ≤ γ3 ≤ 1/(2( p3 − 1)) ≤ 1/6. The condition x1 − x3 = 1/2 gives γ1 = 1 + γ3 ≥ 1, a contradiction, hence T3c ∩ B4P = ∅. The condition x2 − x3 = −1/2 gives γ1 − γ3 = 0, which leads to γ1 = γ3 = 0. This shows that T3c ∩ B4Q = {(0, 1/2, 1/2)T }. Hence (T3c ∩ B4P ) ∪ (T3c ∩ B4Q ) = {(0, 1/2, 1/2)T }. (IV) For any (x1 , x2 , x3 )T ∈ T3d , we write x1 = 1/2 + γ1 and x3 = 1/2 + γ3 , it follows from (2.49) that 0 ≤ γ1 ≤ 1/2 and 0 ≤ γ3 ≤ 1/(2( p3 − 1)) ≤ 1/6. The condition x1 − x3 = −1/2 gives γ3 = 1/2 + γ1 ≥ 1/2, a contradiction, hence T3d ∩ B4Q = ∅. The condition x1 − x3 = 1/2 gives γ1 − γ3 = 1/2, which leads to γ1 = 1/2 and γ3 = 0. This shows that T3d ∩ B4P = {(1, 1/2, 1/2)T }. Hence (T3d ∩ B4P ) ∪ (T3d ∩ B4Q ) = {(1, 1/2, 1/2)T }. It follows from (2.44), (2.45) and the above four cases (I)(II)(III)(IV) that the set T (M ∗ , S)∩ A2 = {(1/2, 1/2, 0)T , (0, 1/2, 1/2)T , (1, 1/2, 1/2)T } is finite. This completes the proof of Claim 2.   Claim 3 The set T (M ∗ , S) ∩ A3 is infinite. Proof From (2.29), we only need to consider the subset of A3 where k3 = 0, that is, the subset given by

A1,5

⎧⎛ ⎫ ⎞ a + k1 ⎨ ⎬ := ⎝ 1/2 + a + k2 ⎠ : a ∈ R, k1 , k2 ∈ Z ⊆ A3 ⊂ R3 . ⎩ ⎭ 1/2

(2.50)

A necessary condition for the points of A1,5 in T (M ∗ , S) is 

0 ≤ a + k1 ≤ 1, 0 ≤ 1/2 + a + k2 ≤ 1.

(2.51)

From (2.51), we have k1 − k2 = 0 or k1 − k2 = 1. So, the discussion is limited to the following two sets:

B5P

⎧⎛ ⎫ ⎞ a+k ⎨ 1⎬ := ⎝ 1/2 + a + k ⎠ : a ∈ R, k ∈ Z and 0 ≤ a + k ≤ ⊆ A1,5 , (2.52) ⎩ 2⎭ 1/2

and

B5Q

⎧⎛ ⎫ ⎞ ⎨ a+k+1 ⎬ −1 := ⎝ 1/2 + a + k ⎠ : a ∈ R, k ∈ Z and ≤ a + k ≤ 0 ⊆ A1,5 . (2.53) ⎩ ⎭ 2 1/2

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Moreover, we have T (M ∗ , S)∩ A3 = T (M ∗ , S)∩ A1,5 = (T (M ∗ , S)∩ B5P )∪(T (M ∗ , S)∩ B5Q ). (2.54) On the other hand, to look for points (x1 , x2 , x3 )T ∈ T (M ∗ , S) with x3 = 1/2, we divide the set T (M ∗ , S) into the following four sets: T (M ∗ , S) = T˜4a ∪ T˜4b ∪ T˜4c ∪ T˜4d , ⎫ ⎧⎛ ⎞ ⎛ ⎞ ⎛ ⎞ s1, j /2 j ∞ s1, j ⎬ ⎨ 0  ⎝ s2, j /2 j ⎠ : ⎝ s2, j ⎠ ∈ S ; T˜4a := ⎝ 0 ⎠ + ⎭ ⎩ j 0 s3, j j=2 s3, j / p 3 ⎧⎛ ⎫ ⎞ ⎛ ⎞ ⎛ ⎞ s1, j /2 j ∞ s1, j ⎨ 1/2 ⎬  ⎝ s2, j /2 j ⎠ : ⎝ s2, j ⎠ ∈ S ; T˜4b := ⎝ 1/2 ⎠ + ⎩ ⎭ j 0 s3, j j=2 s3, j / p 3 ⎧⎛ ⎫ ⎞ ⎛ ⎞ ⎛ ⎞ s1, j /2 j ∞ s1, j ⎨ 0 ⎬  ⎝ s2, j /2 j ⎠ : ⎝ s2, j ⎠ ∈ S ; T˜4c := ⎝ 1/2 ⎠ + ⎩ ⎭ j 1/2 s3, j j=2 s3, j / p 3 ⎧⎛ ⎫ ⎞ ⎛ ⎞ ⎛ ⎞ s1, j /2 j ∞ s1, j ⎨ 1/2 ⎬  ⎝ s2, j /2 j ⎠ : ⎝ s2, j ⎠ ∈ S . T˜4d := ⎝ 0 ⎠ + ⎩ ⎭ j 1/2 s3, j j=2 s3, j / p 3

where

(2.55)

(2.56)

(2.57)

(2.58)

(2.59)

It can be proved from (2.56) and (2.57) that if (x1 , x2 , x3 )T ∈ T˜4a ∪ T˜4b , then 0 ≤ x3 ≤ 1/(2( p3 − 1)) ≤ 1/6. So, from (2.55), we have T4 := {(x1 , x2 , x3 )T ∈ T (M ∗ , S) : x3 = 1/2} = T4c ∪ T4d ,

(2.60)

where

T4c

⎧⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎧⎛ ⎞ ⎛ ⎞⎫⎫ s1, j /2 j ∞ s1, j 1 ⎬⎬ ⎨ 0 ⎨ 0  ⎝ s2, j /2 j ⎠ : ⎝ s2, j ⎠ ∈ ⎝ 0 ⎠ , ⎝ 1 ⎠ : = ⎝ 1/2 ⎠ + ⎩ ⎩ ⎭⎭ j 1/2 0 0 s3, j j=2 s3, j / p 3 ⎧⎛ ⎫ ⎞ ⎛ ⎞ a ⎨ 0 ⎬ = ⎝ 1/2 ⎠ + ⎝ a ⎠ : 0 ≤ a ≤ 1/2 , (2.61) ⎩ ⎭ 1/2 0

and

T4d

123

⎧⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎧⎛ ⎞ ⎛ ⎞⎫⎫ s1, j /2 j ∞ s1, j 1 ⎬⎬ ⎨ 1/2 ⎨ 0  ⎝ s2, j /2 j ⎠ : ⎝ s2, j ⎠ ∈ ⎝ 0 ⎠ , ⎝ 1 ⎠ : = ⎝ 0 ⎠+ ⎩ ⎩ ⎭⎭ j 1/2 0 0 s3, j j=2 s3, j / p 3

Spectral self-affine measures

307

⎧⎛ ⎫ ⎞ ⎛ ⎞ a ⎨ 1/2 ⎬ = ⎝ 0 ⎠ + ⎝ a ⎠ : 0 ≤ a ≤ 1/2 . ⎩ ⎭ 1/2 0

(2.62)

For any (x1 , x2 , x3 )T ∈ T4c , we write x1 = 0 + η1 and x2 = 1/2 + η2 , it follows from (2.61) that 0 ≤ η1 ≤ 1/2 and 0 ≤ η2 ≤ 1/2. The condition x1 − x2 = 1/2 gives η1 = 1+η2 ≥ 1, a contradiction, hence T4c ∩ B5Q = ∅. The condition x1 −x2 = −1/2 gives η1 − η2 = 0. However, from (2.61), we have ⎞ ⎞ ⎛ 0 1/2 j ⎝ 1/2 ⎠ + ⎝ 1/2 j ⎠ ∈ T4c ∩ B5P , 1/2 0 ⎛

j = 1, 2, 3, . . . .

(2.63)

In fact, by taking a+k ∈

⎧ ∞ ⎨ ⎩

j=2

⎫ ⎬

s1, j /2 j : all s1, j ∈ {0, 1} , ⎭

(2.64)

we know, from (2.52) and (2.61), that T4c ⊆ B5P , hence T4c ∩ B5P = T4c is an infinite set. It follows from (2.54) and (2.60) that the set T (M ∗ , S) ∩ A3 is infinite. For any (x1 , x2 , x3 )T ∈ T4d , we write x1 = 1/2 + η1 and x2 = 0 + η2 , it follows from (2.62) that 0 ≤ η1 ≤ 1/2 and 0 ≤ η2 ≤ 1/2. The condition x1 −x2 = −1/2 gives η2 = 1 + η1 ≥ 1, a contradiction, hence T4d ∩ B5P = ∅. The condition x1 − x2 = 1/2 gives η1 − η2 = 0. However, from (2.62), we have ⎞ ⎞ ⎛ 1/2 1/2 j ⎝ 0 ⎠ + ⎝ 1/2 j ⎠ ∈ T4d ∩ B5Q , 1/2 0 ⎛

j = 1, 2, 3, . . . .

(2.65)

In fact, T4d ⊆ B5Q , hence T4d ∩ B5Q = T4d is an infinite set. It follows from (2.54) and (2.60) that the set T (M ∗ , S) ∩ A3 is infinite. Finally, it follows from (2.54), (2.55) and (2.60) that T (M ∗ , S) ∩ A3 = T4c ∪ T4d , where T4c and T4d are given by (2.61) and (2.62) respectively. From Claims 1, 2 and 3, we get Theorem 2.3.   Note that, with the same method as above, one can show that for the compatible pair (M −1 D, S) given by (2.25), if p3 ∈ 2Z\{0, ±2} and p3 > 0, then T (M ∗ , S) ∩ Z (m M −1 D (x)) =

3  

 T (M ∗ , S) ∩ M ∗ A j = {(1, 1, 0)T }.

(2.66)

j=1

Under the condition of compatible pair (M −1 D, S), it follows from Theorem 2.3 and (2.66) that we cannot expect the condition that T (M ∗ , S) ∩ Z (m D (x)) is a finite set and the condition that T (M ∗ , S) ∩ Z (m M −1 D (x)) is an empty set. The two conditions

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play an important role in the study of spectrality of self-affine measure μ M,D . The previous known methods depend upon these conditions heavily. Even so, for the Case 3, we can prove the following. Theorem 2.4 For the self-affine measure μ M,D corresponding to (1.4), if p1 = p2 = 2 and p3 ∈ 2Z\{0, ±2}, then μ M,D is a spectral measure. Proof Consider the following compatible pair (M1−1 D, S1 ): ⎧⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎫ ⎤ −2 0 0 1 0 0 ⎬ ⎨ 0 M1 = ⎣ 0 −2 0 ⎦ , D = ⎝ 0 ⎠ , ⎝ 0 ⎠ , ⎝ 1 ⎠ , ⎝ 0 ⎠ and ⎩ ⎭ 0 0 − p3 0 0 0 1 ⎧⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎫ −1 0 −1 ⎨ 0 ⎬ S1 = ⎝ 0 ⎠ , ⎝ −1 ⎠ , ⎝ −1 ⎠ , ⎝ 0 ⎠ . (2.67) ⎩ ⎭ 0 0 − p3 /2 − p3 /2 ⎡

Then, the invariant set T (M1∗ , S1 ) is given by ⎧ ⎫ ⎛ ⎞ ⎛ ⎞ ∞ s1, j /(−2) j s1, j ⎨ ⎬ ⎝ s2, j /(−2) j ⎠ : ⎝ s2, j ⎠ ∈ S1 , T (M1∗ , S1 ) = ⎩ ⎭ s3, j j=1 s3, j /(− p3 ) j

(2.68)

which shows that for any x = (x1 , x2 , x3 )T ∈ T (M1∗ , S1 ), we have ⎧ ⎨

−1/3 ≤ x1 ≤ 2/3, 2/3, −1/3 ≤ x2 ≤ ⎩ −2/3 ≤ −| p3 |/(2(| p3 | − 1)) ≤ x3 ≤ | p3 |/(2(| p3 | − 1)) ≤ 2/3.

(2.69)

With the same method as the above Sects. 2.1 or 2.2, we know, from (2.69), that T (M1∗ , S1 ) ∩ M1∗ A1 = ∅, T (M1∗ , S1 ) ∩ M1∗ A2 = ∅ and T (M1∗ , S1 ) ∩ M1∗ A3 = ∅. So, by Lemma 2.1, (M1 , S1 ) is a spectrum for μ M1 ,D , equivalently, 

|μˆ M1 ,D (ξ + λ)|2 = 1, ∀ ξ ∈ Rn .

(2.70)

λ∈(M1 ,S1 )

Since |μˆ M1 ,D (ξ )| = |μˆ M,D (ξ )| for all ξ ∈ Rn (see [7, Remark 3.3(ii)]), we know, from (2.70), that (M1 , S1 ) is also a spectrum for μ M,D . This completes the proof of Theorem 2.4.   In the end of this section, it should be pointed out that the above process of proving Theorem 2.4 also provides a method of looking for spectrum. For example, consider the M and D given by (1.4) with p1 = p2 = p3 = p ∈ 2Z\{0}, we have the following conclusions: (i) if p < 0, then (M, S) is a spectrum for μ M,D , where ⎧⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎫ ⎞ ⎛ p/2 0 p/2 ⎬ ⎨ 0 S = ⎝ 0 ⎠ , ⎝ p/2 ⎠ , ⎝ p/2 ⎠ , ⎝ 0 ⎠ ; ⎩ ⎭ 0 p/2 p/2 0

123

Spectral self-affine measures

309

(ii) if p > 0, then (M1 , S1 ) is a spectrum for μ M,D , where M1 = −M and ⎧⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎫ − p/2 0 − p/2 ⎬ ⎨ 0 S1 = ⎝ 0 ⎠ , ⎝ − p/2 ⎠ , ⎝ − p/2 ⎠ , ⎝ 0 ⎠ . ⎩ ⎭ 0 0 − p/2 − p/2 This gives a proof of Theorem A(i), which simplifies the corresponding result of [1, Theorem 5.1(iii)]. 3 On the Question 2 For the Question 2, we can modify the techniques of [8] to obtain the following interesting result. Theorem 3.1 If M and D are given by (1.4) with p1 ∈ 2Z and p2 = p3 ∈ 2Z + 1, then μ M,D is a non-spectral measure, and there exist at most 4 mutually orthogonal exponential functions in L 2 (μ M,D ), where the number 4 is the best upper bound. Proof Firstly, we know, from [8], that the zero set Z (μˆ M,D (ξ )) of the Fourier transform μˆ M,D (ξ ) is ∞  M ∗ j Z (m D (ξ )) := B1 ∪ B2 ∪ B3 , (3.1) Z (μˆ M,D (ξ )) = j=1

where ⎧⎛ ∞ ⎪ ⎨  ⎜ B1 = ⎝ ⎪ j=1 ⎩

⎫ ⎞ j ⎪ (1/2 + k1 ) p1 ⎬ ⎟ j 3 (a + k2 ) p2 ⎠ : a ∈ R, k1 , k2 , k3 ∈ Z ⊂ R , ⎪ ⎭ j (1/2 + a + k3 ) p2

(3.2)

⎫ ⎧⎛ ⎞ j ⎪ ⎪ ⎬ ⎨ (1/2 + a + k1 ) p1 ⎜ ⎟ j ⊂ R3 , : a ∈ R, k , k , k ∈ Z B2 = ⎝ (1/2 + k2 ) p2 ⎠ 1 2 3 ⎪ ⎪ ⎭ j j=1 ⎩ (a + k3 ) p2 ⎫ ⎧⎛ ⎞ j ⎪ (a + k1 ) p1 ∞ ⎪ ⎬ ⎨  ⎜ j⎟ 3 B3 = ⎝ (1/2 + a + k2 ) p2 ⎠ : a ∈ R, k1 , k2 , k3 ∈ Z ⊂ R . ⎪ ⎪ ⎭ ⎩ j j=1 (1/2 + k3 ) p2 ∞ 

(3.3)

(3.4)

Since p1 ∈ 2Z\{0} and p2 ∈ (2Z + 1)\{±1}, we can verify that the following two Lemmas hold.   Lemma 3.1 The sets B j ( j = 1, 2, 3) given by (3.2), (3.3) and (3.4) satisfy the following properties: (a) ξ ∈ B j ⇐⇒  −ξ ∈ B j ( j = 1, 2, 3);  (b) Z (μˆ M,D (ξ )) Z3 = Z (μˆ M,D (ξ )) ((1/2, 1/2, 1/2)T + Z3 ) = ∅; (c) If ξ = (ξ1 , ξ2 , ξ3 )T ∈ B1 , then ξ1 ∈ Z, ξ2 − ξ3 ∈ 21 + Z and ξ3 − ξ2 ∈

1 2

+ Z;

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(d) If ξ = (ξ1 , ξ2 , ξ3 )T ∈ (B1 ± B1 ), then ξ1 ∈ Z and ξ3 − ξ2 ∈ Z; (e) If ξ = (ξ1 , ξ2 , ξ3 )T ∈ B2 ± B2 , then ξ2 ∈ Z; (f) If ξ = (ξ1 , ξ2 , ξ3 )T ∈ B3 ± B3 , then ξ3 ∈ Z. Lemma 3.2 Let ξ = (ξ1 , ξ2 , ξ3 )T ∈ Z (μˆ M,D (ξ )) = B1 ∪ B2 ∪ B3 . Then the following statements hold: (i) If ξ = (ξ1 , ξ2 , ξ3 )T ∈ (B1 ± B1 ), then ξ ∈ B2 ∪ B3 ; (ii) If ξ ∈ B j , then ξ j ∈ 21 + Z, where j = 2, 3; (iii) If ξ1 ∈ Z, then ξ ∈ B1 ∪ B2 ∪ B3 ; If ξ1 , ξ2 ∈ Z, then ξ ∈ B1 ∪ B3 ; If ξ1 , ξ3 ∈ Z, then ξ ∈ B1 ∪ B2 ; (iv) If ξ2 ∈ Z, then ξ ∈ B1 ∪ B3 and ξ3 ∈ Z; (v) If ξ3 ∈ Z, then ξ ∈ B1 ∪ B2 and ξ2 ∈ Z. Secondly, if λ j ( j = 1, 2, 3, 4, 5) ∈ R3 are such that the five functions e2πi λ1 ,x , e2πi λ2 ,x , e2πi λ3 ,x , e2πi λ4 ,x , e2πi λ5 ,x

are mutually orthogonal in L 2 (μ M,D ), then λ j − λk ∈ Z (μˆ M,D (ξ )) = B1 ∪ B2 ∪ B3 (1 ≤ j = k ≤ 5).

(3.5)

We write the difference λ j − λk = (x j,k , y j,k , z j,k )T ∈ R3 for 1 ≤ j = k ≤ 5, and apply the above two Lemmas to deduce a contradiction below. Now, the following ten differences: λ2 − λ1 , λ3 − λ1 , λ4 − λ1 , λ5 − λ1 λ3 − λ2 , λ4 − λ2 , λ5 − λ2 λ4 − λ3 , λ5 − λ3 λ5 − λ4

(3.6)

belong to the union of the three sets B1 , B2 and B3 . In particular, we have λ2 − λ1 , λ3 − λ1 , λ4 − λ1 , λ5 − λ1 ∈ B1 ∪ B2 ∪ B3 .

(3.7)

Claim 1: Each set B1 (or B2 or B3 ) can not contain any three differences of the form λ j1 − λ j , λ j2 − λ j , λ j3 − λ j , where j1 = j, j2 = j, j3 = j, 1 ≤ j ≤ 5 and j1 , j2 , j3 ∈ {1, 2, 3, 4, 5} are three different numbers. Proof of Claim 1. (i) If λ j1 − λ j , λ j2 − λ j , λ j3 − λ j ∈ B1 , then we have λ j1 − λ j3 = (λ j1 − λ j ) − (λ j3 − λ j ) ∈ (B1 − B1 ) ∩ Z (μˆ M,D (ξ )), λ j2 − λ j3 = (λ j2 − λ j ) − (λ j3 − λ j ) ∈ (B1 − B1 ) ∩ Z (μˆ M,D (ξ )), λ j1 − λ j2 = (λ j1 − λ j ) − (λ j2 − λ j ) ∈ (B1 − B1 ) ∩ Z (μˆ M,D (ξ )),

123

Spectral self-affine measures

311

x j1 , j3 , x j2 , j3 , x j1 , j2 , z j1 , j3 − y j1 , j3 , z j2 , j3 − y j2 , j3 , z j1 , j2 − y j1 , j2 ∈ Z, (3.8) and by Lemmas 3.2(i), λ j1 − λ j3 , λ j2 − λ j3 , λ j1 − λ j2 ∈ B2 ∪ B3 , which shows that at least one of the two sets B2 and B3 , say B3 , contains two differences, say λ j2 − λ j3 and λ j1 − λ j2 . Then λ j1 − λ j3 = (λ j1 − λ j2 ) + (λ j2 − λ j3 ) ∈ B3 + B3 . This shows (by Lemma 3.1(f)) z j1 , j3 ∈ Z, combined with (3.8), we get a contradiction of Lemma 3.1(b). (ii) If λ j1 − λ j , λ j2 − λ j , λ j3 − λ j ∈ B2 , then y j1 , j3 , y j2 , j3 , y j1 , j2 ∈ Z and λ j1 − λ j3 , λ j2 − λ j3 , λ j1 − λ j2 ∈ (B2 − B2 ) ∩ Z (μˆ M,D (ξ )) ⊆ B1 ∪ B3 , which shows that at least one of the two sets B1 and B3 contains two differences: if B1 contains two differences, say λ j2 − λ j3 and λ j1 − λ j2 , then λ j1 − λ j3 = (λ j1 − λ j2 ) + (λ j2 − λ j3 ) ∈ B1 + B1 , and by Lemma 3.1(d) λ j1 − λ j3 ∈ Z3 , a contradiction of Lemma 3.1(b); if B3 contains two differences, say λ j2 − λ j3 and λ j1 − λ j2 , then λ j1 − λ j3 = (λ j1 − λ j2 ) + (λ j2 − λ j3 ) ∈ B3 + B3 , and by Lemma 3.1(f) z j1 , j3 ∈ Z, a contradiction of Lemma 3.2(iv); (iii) If λ j1 − λ j , λ j2 − λ j , λ j3 − λ j ∈ B3 , then z j1 , j3 , z j2 , j3 , z j1 , j2 ∈ Z and λ j1 − λ j3 , λ j2 − λ j3 , λ j1 − λ j2 ∈ (B3 − B3 ) ∩ Z (μˆ M,D (ξ )) ⊆ B1 ∪ B2 , which shows that at least one of the two sets B1 and B2 contains two differences: if B1 contains two differences, say λ j2 − λ j3 and λ j1 − λ j2 , then λ j1 − λ j3 = (λ j1 − λ j2 ) + (λ j2 − λ j3 ) ∈ B1 + B1 , and by Lemma 3.1(d) λ j1 − λ j3 ∈ Z3 , a contradiction of Lemma 3.1(b); if B2 contains two differences, say λ j2 − λ j3 and λ j1 − λ j2 , then λ j1 − λ j3 = (λ j1 − λ j2 ) + (λ j2 − λ j3 ) ∈ B2 + B2 , and by Lemma 3.1(e) y j1 , j3 ∈ Z, a contradiction of Lemma 3.2(v). This completes the proof of Claim 1.  

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The set B1 has a little more properties than the sets B2 and B3 . But the sets B2 and B3 has the similar property. Also, the following symmetrical property (ξ1 , ξ2 , ξ3 )T ∈ B2 ⇐⇒ (ξ1 , ξ3 , ξ2 )T ∈ B3

(3.9)

holds. From (3.7) and the above Claim 1, we need to consider the three cases: the set B1 contains no difference in (3.7); the set B1 contains one difference in (3.7); the set B1 contains two differences in (3.7). That is, we only need to deal with the following four typical cases: Case 1: λ2 − λ1 , λ3 − λ1 ∈ B2 and λ4 − λ1 , λ5 − λ1 ∈ B3 . Case 2: λ2 − λ1 ∈ B1 and λ3 − λ1 , λ4 − λ1 ∈ B2 , λ5 − λ1 ∈ B3 ; Case 3: λ2 − λ1 , λ3 − λ1 ∈ B1 and λ4 − λ1 ∈ B2 , λ5 − λ1 ∈ B3 . Case 4: λ2 − λ1 , λ3 − λ1 ∈ B1 and λ4 − λ1 , λ5 − λ1 ∈ B2 . 3.1 Proof of Case 1 In Case 1, we have λ3 − λ2 = (λ3 − λ1 ) − (λ2 − λ1 ) ∈ B2 − B2 ; λ5 − λ4 = (λ5 − λ1 ) − (λ4 − λ1 ) ∈ B3 − B3 , which yield (by applying Lemma 3.1(e)(f) and Lemma 3.2(iv)(v)) λ3 − λ2 = (x3,2 , y3,2 , z 3,2 )T ∈ B1 ∪ B3 and y3,2 ∈ Z, z 3,2 ∈ Z,

(3.10)

λ5 − λ4 = (x5,4 , y5,4 , z 5,4 )T ∈ B1 ∪ B2 and y5,4 ∈ Z, z 5,4 ∈ Z.

(3.11)

From (3.10) and (3.11), there are four subcases: Case 1.1: λ3 − λ2 , λ5 − λ4 ∈ B1 ; Case 1.2: λ3 − λ2 ∈ B1 and λ5 − λ4 ∈ B2 ; Case 1.3: λ3 − λ2 ∈ B3 and λ5 − λ4 ∈ B1 ; Case 1.4: λ3 − λ2 ∈ B3 and λ5 − λ4 ∈ B2 . The Case 1.3 is similar to the Case 1.2, so we omit the proof of Case 1.3. (i) In Case 1.1, we have (by Lemma 3.1(c)) x3,2 ,

y3,2 ∈ Z, z 3,2 ∈

1 + Z and x5,4 ∈ Z, 2

y5,4 ∈

1 + Z, z 5,4 ∈ Z. (3.12) 2

Note that the remainder four differences λ4 − λ2 , λ5 − λ2 and λ4 − λ3 , λ5 − λ3 in (3.6) are also in the set B1 ∪ B2 ∪ B3 . So, the Case 1.1 can be divided into three subcases: Case 1.1.1: λ4 − λ2 ∈ B1 ; Case 1.1.2: λ4 − λ2 ∈ B2 ; Case 1.1.3: λ4 − λ2 ∈ B3 .

Case 1.1.1

123

B1

B2

B3

λ3 − λ2 , λ5 − λ4 , λ4 − λ2

λ2 − λ1 , λ3 − λ1

λ4 − λ1 , λ5 − λ1

Spectral self-affine measures

313

In Case 1.1.1, we have λ4 − λ3 = (λ4 − λ2 ) − (λ3 − λ2 ) ∈ B1 − B1 ; λ5 − λ2 = (λ5 − λ4 ) + (λ4 − λ2 ) ∈ B1 + B1 , which yield (by applying (3.12) and Lemma 3.1(c)(d)) 1 + y4,2 + Z. 2 (3.13) It follows from Lemma 3.2(i) that λ4 − λ3 , λ5 − λ2 ∈ B2 ∪ B3 . Combined with (3.13) and Lemma 3.2(ii), we see that λ4 − λ3 ∈ B2 ∪ B3 yields y4,2 ∈ 21 + Z, while λ5 − λ2 ∈ B2 ∪ B3 yields y4,2 ∈ Z, a contradiction. Hence the Case 1.1.1 is proved. x4,3 ∈ Z,

y4,3 , z 4,3 ∈ y4,2 + Z and x5,2 ∈ Z,

y5,2 , z 5,2 ∈

B1

B2

B3

λ3 − λ2 , λ5 − λ4

λ2 − λ1 , λ3 − λ1 , λ4 − λ2

λ4 − λ1 , λ5 − λ1

Case 1.1.2 In Case 1.1.2, we have λ4 − λ1 = (λ4 − λ2 ) + (λ2 − λ1 ) ∈ B2 + B2 and y4,1 ∈ Z, z 4,1 ∈

1 + Z, (3.14) 2

combined with (3.12), λ5 −λ1 = (λ5 −λ4 )+(λ4 −λ1 ) and x5,1 = x4,1 +Z,

y5,1 , z 5,1 ∈

1 +Z. (3.15) 2

Now, the remainder three differences λ5 − λ2 , λ4 − λ3 , λ5 − λ3 in (3.6) are also in the set B1 ∪ B2 ∪ B3 . From Lemma 3.1(a) and Claim 1, λ5 − λ2 ∈ B2 . Also λ5 − λ2 ∈ B1 , otherwise, if λ5 − λ2 ∈ B1 , then λ5 − λ3 = (λ5 − λ2 ) − (λ3 − λ2 ) ∈ B1 − B1 ⊆ B2 ∪ B3 ⇒ y5,3 , z 5,3 ∈

1 + Z, 2

combined with (3.15), we have y3,1 , z 3,1 ∈ Z, a contradiction of λ3 − λ1 ∈ B2 . So, λ5 − λ2 ∈ B3 . In this case, we see that λ5 − λ3 ∈ B1 ∪ B2 ∪ B3 , a contradiction. In fact, by Claim 1 and Lemma 3.1(a), λ5 − λ3 ∈ B3 . If λ5 − λ3 ∈ B2 , then λ5 − λ1 = (λ5 − λ3 ) + (λ3 − λ1 ) ∈ B2 + B2 ⇒ y5,1 ∈ Z, a contradiction of (3.15). If λ5 − λ3 ∈ B1 , then λ5 − λ2 = (λ5 − λ3 ) + (λ3 − λ2 ) ∈ B1 + B1 ⊆ B2 ∪ B3 ⇒ y5,2 ∈ 21 +Z, and from (3.15), y2,1 = y5,1 − y5,2 ∈ Z, a contradiction of λ2 −λ1 ∈ B2 . Hence the Case 1.1.2 is proved.

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B2

B3

λ3 − λ2 , λ5 − λ4

λ2 − λ1 , λ3 − λ1

λ4 − λ1 , λ5 − λ1 , λ4 − λ2

Case 1.1.3 In Case 1.1.3, we have 1 + Z, z 2,1 ∈ Z, (3.16) 2

λ2 − λ1 = (λ4 − λ1 ) − (λ4 − λ2 ) ∈ B3 − B3 and y2,1 ∈ combined with (3.12), λ3 −λ1 = (λ3 −λ2 )+(λ2 −λ1 ) and x3,1 = x2,1 +Z,

y3,1 , z 3,1 ∈

1 +Z. (3.17) 2

Now, the remainder three differences λ5 − λ2 , λ4 − λ3 , λ5 − λ3 in (3.6) are also in the set B1 ∪ B2 ∪ B3 . From Lemma 3.1(a) and Claim 1, λ4 − λ3 ∈ B3 . Also λ4 − λ3 ∈ B1 , otherwise, if λ4 − λ3 ∈ B1 , then λ5 − λ3 = (λ5 − λ4 ) + (λ4 − λ3 ) ∈ B1 + B1 ⊆ B2 ∪ B3 ⇒ y5,3 , z 5,3 ∈

1 + Z, 2

combined with (3.17), we have y5,1 , z 5,1 ∈ Z, a contradiction of λ5 − λ1 ∈ B3 . So, λ4 − λ3 ∈ B2 . In this case, it follows from λ4 − λ1 = (λ4 − λ3 ) + (λ3 − λ1 ) ∈ B2 + B2 and λ4 − λ1 ∈ B3 that y4,1 ∈ Z and z 4,1 ∈ 21 + Z, which combines with (3.12) and λ5 − λ1 = (λ5 − λ4 ) + (λ4 − λ1 ), yield y5,1 , z 5,1 ∈ 21 + Z. From (3.17) and λ5 − λ3 = (λ5 − λ1 ) − (λ3 − λ1 ), we have y5,3 , z 5,3 ∈ Z, a contradiction of Lemma 3.2(iv)(v). Hence the Case 1.1.3 is proved. The proof of Case 1.1 is completed. (ii) In Case 1.2, it follows from (3.10), (3.11), Lemmas 3.1(c) and 3.2(ii) that x3,2 ,

y3,2 ∈ Z, z 3,2 ∈

1 +Z and x5,4 ∈ Q, 2

y5,4 ∈

1 +Z, z 5,4 ∈ Z. (3.18) 2

Note that the remainder four differences λ4 − λ2 , λ5 − λ2 and λ4 − λ3 , λ5 − λ3 in (3.6) are contained in the set B1 ∪ B2 ∪ B3 . We divide the Case 1.2 into three subcases: Case 1.2.1: λ4 − λ2 ∈ B1 ; Case 1.2.2: λ4 − λ2 ∈ B2 ; Case 1.2.3: λ4 − λ2 ∈ B3 . The Case 1.2.1 is similar to the Case 1.1.2, so, we only need to consider the last two cases.

B1

B2

B3

λ3 − λ2

λ2 − λ1 , λ3 − λ1 , λ5 − λ4 , λ4 − λ2

λ4 − λ1 , λ5 − λ1

Case 1.2.2 In Case 1.2.2, we have λ5 − λ2 = (λ5 − λ4 ) + (λ4 − λ2 ) ∈ B2 + B2 ⇒ y5,2 ∈ Z, λ5 − λ2 ∈ B1 ∪ B3 .

123

Spectral self-affine measures

315

If λ5 − λ2 ∈ B1 , then x5,2 ∈ Z, y5,2 ∈ Z and z 5,2 ∈ 21 + Z, which combines with (3.18) and λ5 − λ3 = (λ5 − λ2 ) − (λ3 − λ2 ), yields λ5 − λ3 ∈ Z3 , a contradiction of Lemma 3.1(b). If λ5 − λ2 ∈ B3 , then x5,2 ∈ Q, y5,2 ∈ Z and z 5,2 ∈ 21 + Z, which combines with (3.18) and λ5 − λ3 = (λ5 − λ2 ) − (λ3 − λ2 ), yields y5,2 ∈ Z and z 5,2 ∈ Z, a contradiction of Lemma 3.2(iv)(v). Hence the Case 1.2.2 is proved. B1

B2

B3

λ3 − λ2

λ2 − λ1 , λ3 − λ1 ,λ5 − λ4

λ4 − λ1 , λ5 − λ1 , λ4 − λ2

Case 1.2.3 In Case 1.2.3, we have λ2 − λ1 = (λ4 − λ1 ) − (λ4 − λ2 ) ∈ B3 − B3 and x2,1 ∈ Q, 1 y2,1 ∈ + Z, z 2,1 ∈ Z, 2 combined with (3.18) and λ3 − λ1 = (λ3 − λ2 ) + (λ2 − λ1 ), 1 x3,1 ∈ x2,1 + Z, and y3,1 , z 3,1 ∈ + Z. 2

(3.19)

By Lemma 3.1(a) and Claim 1, we know that λ4 − λ3 ∈ B3 , and hence λ4 − λ3 ∈ B1 ∪ B2 . If λ4 −λ3 ∈ B1 , then, from (3.18) and λ4 −λ2 = (λ4 −λ3 )+(λ3 −λ2 ) ∈ B1 + B1 ⊆ B2 ∪ B3 , we have x4,2 ∈ Z, y4,2 , z 4,2 ∈ 21 + Z and x4,3 ∈ Z,

y4,3 ∈

1 + Z and z 4,3 ∈ Z, 2

combined with (3.18) and λ5 − λ3 = (λ5 − λ4 ) + (λ4 − λ3 ), we have y5,3 , z 5,3 ∈ Z, a contradiction of Lemma 3.2(iv)(v). If λ4 −λ3 ∈ B2 , then, from λ4 −λ1 = (λ4 −λ3 )+(λ3 −λ1 ) ∈ B2 +B2 , we have y4,1 ∈ Z, z 4,1 ∈ 21 + Z, which combines with (3.18) and λ5 − λ1 = (λ5 − λ4 ) + (λ4 − λ1 ), yields y5,1 , z 5,1 ∈ 21 + Z. It follows from λ5 − λ3 = (λ5 − λ1 ) − (λ3 − λ1 ) and (3.19) that y5,3 , z 5,3 ∈ Z, a contradiction of Lemma 3.2(iv)(v). Hence the Case 1.2.3 is proved. The proof of Case 1.2 is completed. (iii) In Case 1.4, it follows from (3.10), (3.11), Lemmas 3.1(c) and 3.2(ii) that 1 1 x3,2 ∈ Q, y3,2 ∈ Z, z 3,2 ∈ + Z and x5,4 ∈ Q, y5,4 ∈ + Z, z 5,4 ∈ Z. (3.20) 2 2 Note that the remainder four differences λ4 − λ2 , λ5 − λ2 and λ4 − λ3 , λ5 − λ3 in (3.6) are contained in the set B1 ∪ B2 ∪ B3 . We divide the Case 1.4 into three subcases: Case 1.4.1: λ4 − λ2 ∈ B1 ; Case 1.4.2: λ4 − λ2 ∈ B2 ; Case 1.4.3: λ4 − λ2 ∈ B3 . The Case 1.4.1 is similar to the Case 1.2.3, and the Case 1.4.3 is similar to the Case 1.4.2, so, we only need to consider the Case 1.4.2.

123

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B2

B3

λ2 − λ1 , λ3 − λ1 , λ5 − λ4 , λ4 − λ2

λ4 − λ1 , λ5 − λ1 ,λ3 − λ2

Case 1.4.2 In Case 1.4.2, we have λ4 − λ1 = (λ4 − λ2 ) + (λ2 − λ1 ) ∈ B2 + B2 ⇒ x4,1 ∈ Q, 1 y4,1 ∈ Z, z 4,1 ∈ + Z, 2 λ5 − λ2 = (λ5 − λ4 ) + (λ4 − λ2 ) ∈ B2 + B2 ⇒ y5,2 ∈ Z, λ5 − λ2 ∈ B1 ∪ B3 . If λ5 − λ2 ∈ B1 , then x5,2 , y5,2 ∈ Z and z 5,2 ∈ 21 + Z, which combines with (3.20) and λ5 − λ3 = (λ5 − λ2 ) − (λ3 − λ2 ), yields y5,3 , z 5,3 ∈ Z, a contradiction of Lemma 3.2(iv)(v). So, λ5 − λ2 ∈ B3 . In this case, λ5 − λ3 = (λ5 − λ2 ) − (λ3 − λ2 ) ∈ B3 − B3 ⇒ z 5,3 ∈ Z, λ5 − λ3 ∈ B1 ∪ B2 . In the case when λ5 −λ3 ∈ B1 , we have x5,3 ∈ Z, y5,3 ∈ 21 +Z and z 5,3 ∈ Z, which combines with (3.20) and λ4 − λ3 = (λ5 − λ3 ) − (λ5 − λ4 ), yields y4,3 , z 4,3 ∈ Z, a contradiction of Lemma 3.2(iv)(v). In the case when λ5 −λ3 ∈ B2 , we have x5,3 ∈ Q, y5,3 ∈ 21 +Z and z 5,3 ∈ Z, which combines with (3.20) and λ4 − λ3 = (λ5 − λ3 ) − (λ5 − λ4 ), yields y4,3 , z 4,3 ∈ Z, a contradiction of Lemma 3.2(iv)(v). This illustrates that the Case 1.4.2 is proved, and the Case 1.4 is proved. Thus, the proof of Case 1 is completed. 3.2 Proof of Case 2 In Case 2, we have λ4 − λ3 = (λ4 − λ1 ) − (λ3 − λ1 ) ∈ B2 − B2 and y4,3 ∈ Z, λ4 − λ3 ∈ B1 ∪ B3 . The discussion here can be divided into two cases: λ4 − λ3 ∈ B1 and λ4 − λ3 ∈ B3 . That is, we have the following two subcases: Case 2.1: λ2 − λ1 , λ4 − λ3 ∈ B1 , λ3 − λ1 , λ4 − λ1 ∈ B2 , λ5 − λ1 ∈ B3 ; Case 2.2: λ2 − λ1 ∈ B1 , λ3 − λ1 , λ4 − λ1 ∈ B2 , λ5 − λ1 , λ4 − λ3 ∈ B3 . In the following, we consider the above two cases respectively. (I) Case 2.1: In this case, we first have (by Lemma 3.1(c)) λ4 − λ3 = (x4,3 , y4,3 , z 4,3 ) ∈ B1 and x4,3 ∈ Z, y4,3 ∈ Z, z 4,3 ∈

1 + Z. (3.21) 2

From λ3 − λ2 ∈ B1 ∪ B2 ∪ B3 , the Case 2.1 can be divided into three cases:

123

Spectral self-affine measures

317

Case 2.1.1: λ2 −λ1 , λ4 −λ3 , λ3 −λ2 ∈ B1 , λ3 −λ1 , λ4 −λ1 ∈ B2 , λ5 −λ1 ∈ B3 ; Case 2.1.2: λ2 −λ1 , λ4 −λ3 ∈ B1 , λ3 −λ1 , λ4 −λ1 , λ3 −λ2 ∈ B2 , λ5 −λ1 ∈ B3 ; Case 2.1.3: λ2 −λ1 , λ4 −λ3 ∈ B1 , λ3 −λ1 , λ4 −λ1 ∈ B2 , λ5 −λ1 , λ3 −λ2 ∈ B3 . The Case 2.1.3 is similar to the Case 1.1, so we only need to deal with the first two cases. (i) Case 2.1.1: In this case, we first have (by Lemma 3.1(d) and (3.21)) λ3 −λ1 = (λ3 −λ2 )+(λ2 −λ1 ) ∈ B1 + B1 and x3,1 ∈ Z, y3,1 , z 3,1 ∈

1 +Z, (3.22) 2

1 +Z, z 3,1 ∈ Z, (3.23) 2 and λ4 − λ2 = (λ4 − λ3 ) + (λ3 − λ2 ) ∈ B1 + B1 ⊆ B2 ∪ B3 . There are two cases: Case 2.1.1.1: λ4 − λ2 ∈ B2 and Case 2.1.1.2: λ4 − λ2 ∈ B3 . λ4 −λ1 = (λ4 −λ3 )+(λ3 −λ1 ) ∈ B2 and x3,1 ∈ Z, y3,1 ∈

B1

B2

B3

λ2 − λ1 , λ4 − λ3 , λ3 − λ2

λ3 − λ1 , λ4 − λ1 , λ4 − λ2

λ5 − λ1

Case 2.1.1.1 In Case 2.1.1.1, we have x4,2 ∈ Z, y4,2 , z 4,2 ∈

1 2

+Z

λ3 − λ2 = (λ4 − λ2 ) − (λ4 − λ3 ) and x3,2 ∈ Z, y3,2 ∈

1 + Z, z 3,2 ∈ Z (3.24) 2

1 +Z. (3.25) 2 By Lemma 3.1(a) and Claim 1, λ5 − λ2 ∈ B1 and λ5 − λ2 ∈ B2 ∪ B3 . If λ5 − λ2 ∈ B3 , then λ2 −λ1 = (λ5 −λ1 )−(λ5 −λ2 ) ∈ B3 − B3 and z 2,1 ∈ Z, a contradiction of (3.25). So, λ5 − λ2 ∈ B2 and from λ5 − λ4 = (λ5 − λ2 ) − (λ4 − λ2 ) ∈ B2 − B2 ⇒ λ5 − λ4 ∈ B1 ∪ B3 . In the case when λ5 − λ4 ∈ B1 , we have x5,4 , y5,4 ∈ Z, z 5,4 ∈ 21 + Z, which combines with (3.21), yields λ5 − λ3 = (λ5 − λ4 ) + (λ4 − λ3 ) ∈ Z3 , a contradiction of Lemma 3.1(b). In the case when λ5 − λ4 ∈ B3 , we have x5,4 ∈ Q, y5,4 ∈ Z, z 5,4 ∈ 1 2 + Z, which combines with (3.21) and λ5 − λ3 = (λ5 − λ4 ) + (λ4 − λ3 ), yields y5,3 , z 5,3 ∈ Z, a contradiction of Lemma 3.2(iv)(v). λ2 −λ1 = (λ4 −λ1 )−(λ4 −λ2 ) ∈ B2 − B2 and x2,1 , y2,1 ∈ Z, z 2,1 ∈

B1

B2

B3

λ2 − λ1 , λ4 − λ3 , λ3 − λ2

λ3 − λ1 , λ4 − λ1

λ5 − λ1 , λ4 − λ2

Case 2.1.1.2 In Case 2.1.1.2, we have x4,2 ∈ Z, y4,2 , z 4,2 ∈ 21 + Z. By Lemma 3.1(a) and Claim1, λ5 − λ2 ∈ B1 and λ5 − λ2 ∈ B2 ∪ B3 . If λ5 − λ2 ∈ B3 , then λ2 − λ1 = (λ5 − λ1 ) − (λ5 − λ2 ) ∈ B3 − B3 ⇒ x2,1 ∈ Z, y2,1 ∈ 21 + Z, z 2,1 ∈ Z, and from

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λ4 −λ1 = (λ4 −λ2 )+(λ2 −λ1 ), y4,1 = y4,2 +y2,1 ∈ Z, a contradiction of λ4 −λ1 ∈ B2 . So, λ5 −λ2 ∈ B2 . By Lemma 3.1(a) and Claim 1, λ5 −λ3 ∈ B1 and λ5 −λ3 ∈ B2 ∪ B3 . If λ5 − λ3 ∈ B3 , then λ3 − λ1 = (λ5 − λ1 ) − (λ5 − λ3 ) ∈ B3 − B3 ⇒ z 3,1 ∈ Z, a contradiction of (3.22). If λ5 − λ3 ∈ B2 , then λ3 − λ2 = (λ5 − λ2 ) − (λ5 − λ3 ) ∈ B2 − B2 ⇒ x3,2 , y3,2 ∈ Z, z 3,2 ∈ 21 + Z, which combines with (3.21), yields λ4 − λ2 = (λ4 − λ3 ) + (λ3 − λ2 ) ∈ Z3 , a contradiction of Lemma 3.1(b). (ii) Case 2.1.2: In this case, we have λ2 −λ1 = (λ3 −λ1 )−(λ3 −λ2 ) ∈ B2 − B2 and x2,1 , y2,1 ∈ Z, z 2,1 ∈

1 +Z. (3.26) 2

There are three cases: Case 2.1.2.1: λ4 − λ2 ∈ B1 , Case 2.1.2.2: λ4 − λ2 ∈ B2 , and Case 2.1.2.3: λ4 − λ2 ∈ B3 . The Case 2.1.2.1 is similar to the Case 2.1.1.1. The Case 2.1.2.3 is similar to the Case 1.1.2. So, we only need to consider the Case 2.1.2.2.

B1

B2

B3

λ2 − λ1 , λ4 − λ3 , λ5 − λ2

λ3 − λ1 , λ4 − λ1 , λ3 − λ2 , λ4 − λ2

λ5 − λ1

Case 2.1.2.2 In Case 2.1.2.2, it follows from Claim 1 that λ5 − λ2 ∈ B2 . Also λ5 − λ2 ∈ B3 (otherwise, λ2 −λ1 = (λ5 −λ1 )−(λ5 −λ2 ) ∈ B3 − B3 ⇒ z 2,1 ∈ Z, a contradiction of (3.26)). So, λ5 − λ2 ∈ B1 . By Lemma 3.1(c)(d) and λ5 − λ1 = (λ5 − λ2 ) + (λ2 − λ1 ) ∈ B1 + B1 , x5,1 ∈ Z,

y5,1 , z 5,1 ∈

1 + Z and x5,2 ∈ Z, 2

y5,2 ∈

1 + Z, z 5,2 ∈ Z. (3.27) 2

It follows from Claim 1 and Lemma 3.1(a) that λ5 − λ3 , λ5 − λ4 ∈ B2 and λ5 − λ3 , λ5 −λ4 ∈ B1 ∪ B3 . Again, by Claim 1, λ5 −λ3 , λ5 −λ4 ∈ B1 and λ5 −λ3 , λ5 −λ4 ∈ B3 can not hold. So, λ5 −λ3 ∈ B1 and λ5 −λ4 ∈ B3 (or λ5 −λ3 ∈ B3 and λ5 −λ4 ∈ B1 , which can be discussed in the same way). From λ5 − λ4 = (λ5 − λ3 ) − (λ4 − λ3 ) ∈ B1 − B1 and λ5 − λ4 ∈ B3 , we have x5,4 ∈ Z,

y5,4 , z 5,4 ∈

1 + Z, 2

(3.28)

which combines with (3.27), yields λ4 − λ1 = (λ5 − λ1 ) − (λ5 − λ4 ) ∈ Z3 , a contradiction of Lemma 3.1(b). This completes the proof of Case 2.1. (II) Case 2.2: In this case, we first have (by Lemma 3.1(e) and Lemma 3.2(ii)) λ4 − λ3 = (x4,3 , y4,3 , z 4,3 ) ∈ B3 and x4,3 ∈ Q, y4,3 ∈ Z, z 4,3 ∈

1 + Z. (3.29) 2

From λ3 − λ2 ∈ B1 ∪ B2 ∪ B3 , the Case 2.2 can be divided into three cases: Case 2.2.1: λ2 −λ1 , λ3 −λ2 ∈ B1 , λ3 −λ1 , λ4 −λ1 ∈ B2 , λ5 −λ1 , λ4 −λ3 ∈ B3 ;

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Case 2.2.2: λ2 −λ1 ∈ B1 , λ3 −λ1 , λ4 −λ1 , λ3 −λ2 ∈ B2 , λ5 −λ1 , λ4 −λ3 ∈ B3 ; Case 2.2.3: λ2 −λ1 ∈ B1 , λ3 −λ1 , λ4 −λ1 ∈ B2 , λ5 −λ1 , λ4 −λ3 , λ3 −λ2 ∈ B3 . The Case 2.2.1 is similar to the Case 2.1.3 and Case 1.1, so we only need to deal with the last two cases. From λ4 − λ2 ∈ B1 ∪ B2 ∪ B3 , the Case 2.2.2 can be further divided into three cases: Case 2.2.2.1: λ4 − λ2 ∈ B1 , Case 2.2.2.1: λ4 − λ2 ∈ B2 , and Case 2.2.2.1: λ4 − λ2 ∈ B3 , which are similar to the Case 1.1.2, Case 1.2.2, and Case 1.2.3 respectively. Thus, the proof of Case 2 is completed. 3.3 Proof of Case 3 In Case 3, we have λ3 − λ2 = (λ3 − λ1 ) − (λ2 − λ1 ) ∈ B1 − B1 ⊆ B2 ∪ B3 and x3,2 ∈ Z, 1 y3,2 , z 3,2 ∈ + Z. 2 The discussion here can be divided into two cases: λ3 − λ2 ∈ B2 and λ3 − λ2 ∈ B3 . That is, we have the following two subcases: Case 3.1: λ2 − λ1 , λ3 − λ1 ∈ B1 , λ4 − λ1 , λ3 − λ2 ∈ B2 , λ5 − λ1 ∈ B3 ; Case 3.2: λ2 − λ1 , λ3 − λ1 ∈ B1 , λ4 − λ1 ∈ B2 , λ5 − λ1 , λ3 − λ2 ∈ B3 . Since the sets B2 and B3 have the similar property (see also (3.9)), the Case 3.2 can be treated as the Case 3.1. Both Cases are similar to the Case 2.1. 3.4 Proof of Case 4 In Case 4, we have λ3 − λ2 = (λ3 − λ1 ) − (λ2 − λ1 ) ∈ B1 − B1 ; λ5 − λ4 = (λ5 − λ1 ) − (λ4 − λ1 ) ∈ B2 − B2 , which yield (by applying Lemmas 3.1(c)(d)(e) and 3.2(iv)(v)) λ3 − λ2 = (x3,2 , y3,2 , z 3,2 )T ∈ B2 ∪ B3 and x3,2 ∈ Z, y3,2 , z 3,2 ∈

1 + Z, (3.30) 2

λ5 − λ4 = (x5,4 , y5,4 , z 5,4 )T ∈ B1 ∪ B3 and y5,4 ∈ Z.

(3.31)

From (3.30) and (3.31), there are four subcases: Case 4.1: λ3 − λ2 ∈ B2 and λ5 − λ4 ∈ B1 ; Case 4.2: λ3 − λ2 ∈ B2 and λ5 − λ4 ∈ B3 ; Case 4.3: λ3 − λ2 ∈ B3 and λ5 − λ4 ∈ B1 ; Case 4.4: λ3 − λ2 ∈ B3 and λ5 − λ4 ∈ B3 . The Case 4.2, Case 4.3, and Case 4.4 are similar to the Case 2.1.2, Case 2.1.1, and Case 2.1.3 respectively. So, we only need to consider the Case 4.1 below.

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In the Case 4.1, (3.31) becomes the following λ5 − λ4 = (x5,4 , y5,4 , z 5,4 )T ∈ B1 and x5,4 ,

y5,4 ∈ Z, z 5,4 ∈

1 + Z. (3.32) 2

According to λ4 − λ2 ∈ B1 ∪ B2 ∪ B3 , the Case 4.1 can be further divided into three subcases: Case 4.1.1: λ4 − λ2 ∈ B1 ; Case 4.1.2: λ4 − λ2 ∈ B2 ; Case 4.1.3: λ4 − λ2 ∈ B3 . The Case 4.1.3 is similar to the Case 2.1.1.1. In the Case 4.1.1, we have λ5 − λ2 = (λ5 − λ4 ) + (λ4 − λ2 ) ∈ B1 + B1 and λ5 − λ2 = (x5,2 , y5,2 , z 5,2 )T ∈ B2 ∪ B3 and x5,2 ∈ Z, y5,2 , z 5,2 ∈

1 + Z, (3.33) 2

which combines with (3.30) and λ5 −λ3 = (λ5 −λ2 )−(λ3 −λ2 ), yields λ5 −λ3 ∈ Z3 , a contradiction of Lemma 3.1(b). So, we need to consider the Case 4.1.2. B1

B2

λ2 − λ1 , λ3 − λ1 , λ5 − λ4

λ4 − λ1 , λ5 − λ1 , λ3 − λ2 , λ4 − λ2

B3

Case 4.1.2 In Case 4.1.2, we have λ4 − λ3 = (λ4 − λ2 ) − (λ3 − λ2 ) ∈ B2 − B2 ⇒ λ4 − λ3 ∈ B1 ∪ B3 and y4,3 ∈ Z. From (3.32), we have λ4 − λ3 ∈ B3 (otherwise, λ5 − λ3 = (λ5 − λ4 ) + (λ4 − λ3 ) ∈ Z3 ). From λ5 − λ3 = (λ5 − λ4 ) + (λ4 − λ3 ), we have y5,3 = y5,4 + y4,3 ∈ Z and λ5 − λ3 ∈ B1 ∪ B3 . From (3.32), we have λ5 − λ3 ∈ B3 (otherwise, λ4 − λ3 = (λ5 − λ3 ) − (λ5 − λ4 ) ∈ Z3 ). Hence, λ5 − λ4 = (λ5 − λ3 ) − (λ4 − λ3 ) ∈ B3 − B3 ⇒ z 5,4 ∈ Z, a contradiction of (3.32). This completes the proof of Theorem 3.1. 3.5 Remarks on the Question 2 For the Question 2, we would like to point out that if “ p1 , p2 , p3 ” in (1.4) have different parity, the determination of spectrality or non-spectrality of self-affine measure μ M,D will become more complicated to deal with. In this case, we cannot obtain a set S ⊂ Z3 so that (M −1 D, S) is a compatible pair, and there are no desired result available. However, from [6, Theorem 2], we can only show that the following conclusion holds. Proposition 3.2 For p1 , p2 , p3 ∈ Z\{0, ±1} and p4 , p5 , p6 ∈ Z, if ⎧⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎫ ⎡ ⎤ p1 p4 p6 1 0 0 ⎬ ⎨ 0 M = ⎣ 0 p2 p5 ⎦ and D = ⎝ 0 ⎠ , ⎝ 0 ⎠ , ⎝ 1 ⎠ , ⎝ 0 ⎠ ⎩ ⎭ 0 0 0 1 0 0 p3

(3.34)

satisfies the condition that any two of the three numbers p1 , p2 , p3 are in the set 2Z, then there are infinite families of orthogonal exponentials E() in L 2 (μ M,D ) with  ⊆ Z3 .

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Proof Let α1 := (1/2, 1/2, 0)T , α2 = (1/2, 0, 1/2)T and α3 = (0, 1/2, 1/2)T . Then α j ∈ Z (m D (x)) for j = 1, 2, 3. There are three cases such that two of the three numbers p1 , p2 , p3 are in the set 2Z\{0}: Case 1: p1 , p2 ∈ 2Z\{0}; Case 2: p2 , p3 ∈ 2Z\{0}; Case 3: p1 , p3 ∈ 2Z\{0}. The Case 1 can be divided into two subcases: Case 1.1: p3 ∈ 2Z\{0} and Case 1.2: p3 ∈ (2Z + 1)\{±1}. In Case 1.1, we have (M ∗ )2 α3 ∈ Z3 , which implies the conclusion. So, we only need to consider the Case 1.2, which can be further divided into two subcases: Case 1.2.1: p5 ∈ (2Z + 1) and Case 1.2.2: p5 ∈ 2Z. In Case 1.2.1, we have M ∗ α3 ∈ Z3 , which implies the conclusion. So, we only need to consider the Case 1.2.2, which can be further divided into two subcases: Case 1.2.2.1: p6 ∈ 2Z and Case 1.2.2.2: p6 ∈ (2Z + 1). In Case 1.2.2.1, we have (M ∗ )2 α1 ∈ Z3 , which implies the conclusion. So, we only need to consider the Case 1.2.2.2, which can be further divided into two subcases: Case 1.2.2.2.1: p4 ∈ (2Z + 1) and Case 1.2.2.2.2: p4 ∈ 2Z. In the final two cases, we have (M ∗ )2 α2 ∈ Z3 and M ∗ α2 ∈ Z3 respectively. This proves the Case 1. In Case 2, we have (M ∗ )2 α3 ∈ Z3 , which implies the conclusion. The Case 3 can be divided into two subcases: Case 3.1: p2 ∈ 2Z\{0} and Case 3.2: p2 ∈ (2Z + 1)\{±1}. In Case 3.1, we have (M ∗ )2 α3 ∈ Z3 , which implies the conclusion. So, we only need to consider the Case 3.2, which can be further divided into two subcases: Case 3.2.1: p4 ∈ 2Z and Case 3.2.2: p4 ∈ (2Z + 1). In Case 3.2.1, we have (M ∗ )2 α2 ∈ Z3 , which implies the conclusion. So, we only need to consider the Case 3.2.2, which can be further divided into two subcases: Case 3.2.2.1: p5 ∈ 2Z and Case 3.2.2.2: p5 ∈ (2Z + 1). In Case 3.2.2.1, we have (M ∗ )2 α1 ∈ Z3 , which implies the conclusion. So, we only need to consider the Case 3.2.2.2, which can be further divided into two subcases: Case 3.2.2.2.1: p6 ∈ (2Z + 1) and Case 3.2.2.2.2: p6 ∈ 2Z. In the final two cases, we have M ∗ α1 ∈ Z3 and (M ∗ )2 α1 ∈ Z3 respectively. This proves the Case 3. The proof of Proposition 3.2 is completed.   In the same way, one can prove that for the pair (M, D) given by (3.34), if one of the following eight conditions holds: (i) (ii) (iii) (iv) (v) (vi) (vii) (viii)

p1 , p4 , p5 ∈ 2Z and p2 , p3 , p6 p1 , p4 ∈ 2Z and p2 , p3 , p5 , p6 p1 , p5 , p6 ∈ 2Z and p2 , p3 , p4 p1 ∈ 2Z and p2 , p3 , p4 , p5 , p6 p2 ∈ 2Z and p5 ∈ 2Z + 1; p1 , p2 + p4 , p5 + p6 ∈ 2Z; p1 , p4 , p3 + p6 ∈ 2Z; p2 , p3 + p5 ∈ 2Z

∈ 2Z + 1; ∈ 2Z + 1; ∈ 2Z + 1; ∈ 2Z + 1;

then there are infinite families of orthogonal exponentials E() in L 2 (μ M,D ) with  ⊆ Z3 . Acknowledgments The author would like to thank the anonymous referees for their valuable suggestions. The present research is supported by the National Natural Science Foundation of China (No.11171201).

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References 1. Dutkay, D.E., Jorgensen, P.E.T.: Analysis of orthogonality and of orbits in affine iterated function systems. Math. Z. 256, 801–823 (2007) 2. Jorgensen, P.E.T., Pedersen, S.: Harmonic analysis of fractal measures. Constr. Approx. 12, 1–30 (1996) 3. Jorgensen, P.E.T., Pedersen, S.: Dense analytic subspaces in fractal L 2 −spaces. J. Anal. Math. 75, 185–228 (1998) 4. Li, J.-L.: Spectral self-affine measures in Rn . Proc. Edinb. Math. Soc. 50, 197–215 (2007) 5. Li, J.-L.: The cardinality of certain μ M,D −orthogonal exponentials. J. Math. Anal. Appl. 362, 514–522 (2010) 6. Li, J.-L.: On the μ M,D -orthogonal exponentials. Nonlinear Anal. 73, 940–951 (2010) 7. Li, J.-L.: Spectra of a class of self-affine measures. J. Funct. Anal. 260, 1086–1095 (2011) 8. Li, J.-L.: Spectrality of self-affine measures on the three-dimensional Sierpinski gasket. Proc. Edinb. Math. Soc. 55, 477–496 (2012) 9. Li, J.-L.: On a criterion of Strichartz for spectral pairs. Chin. J. Contemp. Math. 34, 1–12 (2013) 10. Strichartz, R.: Remarks on “Dense analytic subspaces in fractal L 2 −spaces”. J. Anal. Math. 75, 229– 231 (1998) 11. Strichartz, R.: Mock Fourier series and transforms associated with certain Cantor measures. J. Anal. Math. 81, 209–238 (2000)

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