Nonspectrality of Certain Self-Affine Measures on ℝ³

Abstract

We will determine the nonspectrality of self-affine measure corresponding to , , and in the space is supported on , where , and are the standard basis of unit column vectors in , and there exist at most mutually orthogonal exponential functions in , where the number is the best. This generalizes the known results on the spectrality of self-affine measures.

1. Introduction

Let be an expanding integer matrix; that is, all the eigenvalues of the integer matrix have modulus greater than 1. Associated with a finite subset , there exists a unique nonempty compact set such that . More precisely, is an attractor (or invariant set) of the affine iterated function system (IFS) . Denote by the cardinality of . Relating to the IFS , there also exists a unique probability measure satisfying

For a given pair , the spectrality or nonspectrality of is directly connected with the Fourier transform . From (1), we get where

The self-affine measure has received much attention in recent years. The previous research on such measure and its Fourier transform revealed some surprising connections with a number of areas in mathematics, such as harmonic analysis, number theory, dynamical systems, and others; see [1, 2] and references cited therein.

The and are all determined by the pair . So, for , in the way of examples, there are Cantor set and Cantor measure on the line. And for there is a rich variety of geometries, see Li [3–6], of which the best known example is the Sierpinski gasket. But for , it is more complex.

The problem considered below started with a discovery in an earlier paper of Jorgensen and Pedersen [7] where it was proved that certain IFS fractals have Fourier bases, and furthermore that the question of counting orthogonal Fourier frequencies (or orthogonal exponentials in ) for a fixed fractal involves an intrinsic arithmetic of the finite set of functions making up the IFS under consideration. For example, if is the middle-third Cantor example on the line, there cannot be more than two orthogonal Fourier frequencies [7, Theorem 6.1], while a similar Cantor example, using instead a subdivision scale 4 (i.e., ), turns out to have an ONB in consisting of Fourier frequencies [7, Theorem 3.4].

With the effort of Jorgensen and Pedersen [7, Example 7.1], Strichartz [8], Li [3], and Yuan [9], the related conclusions discussed that the diagonal elements of are all even or odd, If one of the diagonal elements is even, what about the result? The general case on the spectrality or nonspectrality of the self-affine measure is not known.

The present paper is motivated by these earlier results; we will determine the nonspectrality of self-affine measure ; the main result of the present paper is the following.

Theorem 1. Let correspond to and ; then the self-affine measure is a nonspectral measure, and there exist at most 4 mutually orthogonal exponential functions in , where the number 4 is the best.

By the proof of Theorem 1, we get a more general case.

Theorem 2. Let correspond to then the self-affine measure is a nonspectral measure, and there exist at most 4 mutually orthogonal exponential functions in , where the number 4 is the best.

2. Proof of Theorem 1

Firstly, we know from (2) that the zero set of the Fourier transform is For the given digit set in (4), we have where and , , and are given by

So

Since , we can verify directly that the following two lemmas hold.

Lemma 3. The sets () given by (9) satisfy the following properties:(a) ();(b);(c)if , then ;(d)if , then ;(e)if and , then and .
By proving (e), the others can be checked directly.
In (e), if , then where and , . From (10) one has (since , and ); then . In fact, if and , then is a contradiction. Since and , then .

Lemma 4. Let . Then the following statements hold:(i)if , then , where ;(ii)if , then and ;(iii)if , then and .

Suppose that are such that the five exponential functions are mutually orthogonal in , so the differences are in the zero set . Then we get

Denote by We will apply the above two lemmas to get the contradiction below.

The following ten differences belong to . Then

Claim 1. The set or or cannot contain any three differences of the form: , , , where .

Claim 1 can be checked easily. For example, if , , , so, by applying Lemma 3(c) and (12), we get and by Lemma 4(ii), which shows that at least one of the two sets and contains two differences. If contains two differences, say and , then This shows (by Lemma 3(d)) that , a contradiction with Lemma 3(b). If contains two differences, say and , then by Lemma 3(e), , a contradiction with (16).

From (15) and Claim 1, we only need to deal with the following four typical cases.

Case 1. , , , and .

Case 2. , , , and .

Case 3. , and .

Case 4. , and , .

Note that Case 1 denotes the or distribution in (15), Case 2 denotes the distribution in (15), and Case 3 denotes the distribution in (15), while Case 4 denotes the (or ) distribution in (15). If the four cases can be proved, then the other cases can be proved similarly.

2.1. Case 1

In this case, we have Applying Lemma 3(c), (12), and Lemma 4(ii), we get From , Case 1 can be divided into two subcases.

Case 1.1. , , , , and .

Case 1.2. , , , , and .

Step 1. We will prove Case 1.1; since , Case 1.1 can be divided into three cases.

Case 1.1.1. , , , , , and .

Case 1.1.2. , , , , , and .

Case 1.1.3. , , , , , and .

We will give a method to deal with each case by considering the remainder differences in (14), and each case is concluded with a contradiction.

Case 1.1.1. In this case, we have by applying Lemma 3(c), (12), and Lemma 4(ii), we get

By Lemma 3(a) and Claim 1 we know that . So or , so Case 1.1.1 can be divided into two subcases.

Case 1.1.1.1. , , , , , , and .

Case 1.1.1.2. , , , , , , and .

By considering the remainder differences in (14), we apply Lemmas 3 and 4 to deal with each case.(I)In Case 1.1.1.1, we have by applying Lemma 3(d), (12) and Lemma 4(iii), we have (i)If , so ; by applying Lemma 3(c), (12), and Lemma 4(ii), we get It follows from (23), (26), and that which shows a contradiction with Lemma 3(b).(ii)If , so ; by applying Lemma 3(e) and (12), we get Now, consider the remainder difference in (14): by Lemma 3(a) and Claim 1, we have , so . If , then by applying Lemma 3(c), (12), and Lemma 4(ii) we get It follows from (21), (30), and that , a contradiction with Lemma 3(b). If , then by applying Lemma 3(d), (12), and Lemma 4(iii), we get It follows from (25), (32), and that and , a contradiction with Lemma 3(b).

From parts (i), (ii), and (25), Case 1.1.1.1 is proved.(II)In Case 1.1.1.2, we have by applying Lemma 3(e) and (12), we get

We consider the remainder three differences , , and in (14). By Claim 1, we have , so .(i)If , we have by applying Lemma 3(c), (12), and Lemma 4(ii), we get It follows from (21), (36), and that a contradiction with Lemma 3(b).(ii)If , we have by applying Lemma 3(d), (12), and Lemma 4(iii) we have We consider the remainder difference in (14). By Claim 1, we have , so . If , we have by applying Lemma 3(c), (12), and Lemma 4(ii), we get , a contradiction with (39). If , we have by applying Lemma 3(d), (12), and Lemma 4(iii), we get It follows from (39), (42), and that , a contradiction with (34).

The above (i) and (ii) in (II) illustrate that Case 1.1.1.2 is proved. Hence Case 1.1.1 is proved.

Case 1.1.2. In this case, we have by applying Lemma 3(d), (12), and Lemma 4(iii), we get By applying Lemma 3(a) and Claim 1, we get

If , then ; by Lemma 3(e), and (12), we get , , a contradiction with (44); hence We consider ; if , we get ; by Lemma 3(c), (12), and Lemma 4(ii), we have which, combined with (21) and , shows that , , a contradiction with Lemma 3(b); hence

Since , by Lemma 3(e) and (12), we get

Combined with (21), (44), and , we get which, combined with (50), (51), and , shows that , , a contradiction of (45). This proves Case 1.1.2.

Case 1.1.3. From , this Case can be divided into the three cases.

Case 1.1.3.1. , , , , , , and .

Case 1.1.3.2. , , , , , , and .

Case 1.1.3.3. , , , , , , and .

The above three cases denote the or or distribution. In each case, we have (21). The discussion of the first two cases is similar. In the following, we use the above method to deal with Cases 1.1.3.1 and 1.1.3.3.

In Case 1.1.3.1, since by applying Lemma 3(c), (12), and Lemma 4(ii), we have

We consider the remainder three differences , , and . From (21) and (53) we have

If , by Lemma 3, so ; we get , a contradiction with (53). If , so by Lemma 3(e); we get , a contradiction with (21). Hence (54) and (55) hold.

According to (54), we deal with the following two cases.(i)If , then, from , by applying Lemma 3(c), (12), and Lemma 4(ii), we get

From (21), (56), and , we get a contradiction with Lemma 3(b).(ii)If , since and , by applying Lemma 3(e) and (12), we get

We consider (55); if , by Lemma 3 and , we get , a contradiction with (58). If , since , by applying Lemma 3(d), (12), and Lemma 4(iii), we get From (58), (59) and , we have , a contradiction with Lemma 4(i) (for gives ).

The above (i), (ii), and (54) indicae that Case 1.1.3.1 is proved.

In Case 1.1.3.3, we get by applying Lemma 3(e) and (12), we have By Lemma 3(a) and , , we know from Claim 1 that . So or .

From (61), we consider the following two cases.(i)If , we consider , if . From , by applying Lemma 3(c), (12), and Lemma 4(ii), we get From (21), (63), and , we get , a contradiction with Lemma 3(b). If , since , by applying Lemma 3(c), (12), and Lemma 4(ii), we get From (21), (64), and , we get that , combined with (61) and , yields , a contradiction with Lemma 3(b).(ii)If , so ; by applying Lemma 3(d), (12), and Lemma 4(iii), we get If , by Lemma 3(c) and , we get , a contradiction with (65). If , so , by applying Lemma 3(d), (12), and Lemma 4(iii), we get

From (65), (66), and , we get , a contradiction with (62).

The above (i), (ii), and (61) illustrate that Case 1.1.3.3 is proved and Case 1.1 is proved.

Step 2. In Case 1.2, from , this case can be divided into three cases.

Case 1.2.1. , , , , , and .

Case 1.2.2. , , , , , and .

Case 1.2.3. , , , , , and .

Case 1.2.2 is similar to Case 1.1.3; we will deal with the other two cases by considering the remainder differences in (14), and each case is concluded with a contradiction.

Case 1.2.1. In this case, we get by applying Lemma 3(c), (12), and Lemma 4(ii), we get

By Lemma 3(a) and , we know from Claim 1 that . So or , so Case 1.2.1 can be divided into two cases.

Case 1.2.1.1. , , , , , , .

Case 1.2.1.2. , , , , , , .

The above two cases denote the distribution and distribution. Case 1.2.1.1 is similar to Case 1.1.3.1. So we only need to deal with Case 1.2.1.2. By considering the remainder differences in (14), we apply Lemmas 3 and 4 to deal with the case.

In Case 1.2.1.2, we get by applying Lemma 3(e) and (12), we get

By Claim 1, ; then or .

From (70), we consider the following two cases.(i)If , since , by applying Lemma 3(c), (12), and Lemma 4(ii) we get

It follows from (68), (71), and that which shows a contradiction with Lemma 3(b).(ii)If , we consider , if . Since , by applying Lemma 3(c), (12), and Lemma 4(ii), we get

From (21), (73), and , we get , , a contradiction with Lemma 3(b). If , then ; by applying Lemma 3(d), (12), and Lemma 4(iii), we get

It follows from (68), (74), and that we get

From (70), (75), and , we get , , a contradiction with Lemma 3(b).

Parts (i), (ii), and (70) indicate that Case 1.2.1.2 is proved and Case 1.2.1 is proved.

Case 1.2.3. In this case, since by applying Lemma 3(e) and (12), we get

From (77) Case 1.2.3 can be divided into two cases.

Case 1.2.3.1. , , , , and .

Case 1.2.3.2. , , , , and .

The above two cases denote the distribution and distribution. Case 1.2.3.1 is similar to Case 1.2.1.2. Case 1.2.3.2 is similar to Case 1.1.3.3. So Case 1.2.3 is proved, and so Case 1.2 is proved.

Thus, the proof of Case 1 is completed.

2.2. Case 2

In this case, since by applying Lemma 3(e) and (12), we get From , the discussion here can be divided into two cases: and . That is, we have the following two subcases.

Case 2.1. , , , , and .

Case 2.2. , , , , and .

The discussion of Case 2.1 is analogous to Case 2.2; it denotes the or distribution. So we only need to deal with Case 2.1.

From , Case 2.1 can be divided into three cases.

Case 2.1.1. , , , , , and .

Case 2.1.2. , , , , , and .

Case 2.1.3. , , , , , and .

The above three cases denote the distribution, distribution, and distribution. Case 2.1.1 is similar to Case 1.2.1, Case 2.1.2 is similar to Case 1.1.3, and Case 2.1.3 is similar to Case 1.2.3.

So the proof of Case 2 is completed.

2.3. Case 3

In this case, since , and , by Lemmas 3 and 4, we get

From , Case 3 can be divided into the following two cases.

Case 3.1. and .

Case 3.2. and .

Case 3.2 is similar to Case 1.1, so we only need to deal with Case 3.1.

From , the Case 3.1 can be divided into the following two cases.

Case 3.1.1. , , and , , and .

Case 3.1.2. , , , , , and .

Case 3.1.2 is similar to Case 1.1.2, so we only need to deal with Case 3.1.1.

Consider the remainder difference in (14). From , Case 3.1.1 can be divided into three cases.

Case 3.1.1.1. , , , , , , and .

Case 3.1.1.2. , , , , , , and .

Case 3.1.1.3. , , , , , , and .

Case 3.1.1.2 is similar to Case 3.1.1.1, and Case 3.1.1.3 is similar to Case 1.1.1.1. So we only need to deal with Case 3.1.1.1. In this case, since , by applying Lemma 3(c), (12), and Lemma 4(ii), we get Combined with (80), (82) yields a contradiction with Lemma 3(b). Case 3.1.1.1 is proved.

So the proof of Case 3 is completed.

2.4. Case 4

In this case, since , and , by Lemmas 3 and 4, we get

From , Case 4 can be divided into the following two cases.

Case 4.1. , and , , .

Case 4.2. , , , and , .

Case 4.2 is similar to Case 1.2, so we only need to deal with Case 4.1.

From , Case 4.1 can be divided into the following two cases.

Case 4.1.1. , , and , , and .

Case 4.1.2. , , , , , and .

Case 4.1.2 is similar to Case 1.2.3, so we only need to deal with Case 4.1.1.

Consider the remainder difference in (14). According to , Case 4.1.1 can be divided into three cases.

Case 4.1.1.1. , , , , , , and .

Case 4.1.1.2. , , , , , , and .

Case 4.1.1.3. , , , , , , and .

Case 4.1.1.2 is similar to Case 1.2.1.2. We will deal with the other two cases by considering the remainder differences in (14), and each case is concluded with a contradiction.

Case 4.1.1.1. In this case, since , by Lemma 3(c), (12), and Lemma 4(ii), we get combined with (84), (86), and yields , , a contradiction with Lemma 3(b).

Case 4.1.1.3. In this case, since , by Lemma 3(e) and (12), we get combined with (85), (87), and , yields , , a contradiction with Lemma 3(b). So Case 4.1.1.3 is proved, and Case 4.1.1 is proved.

So the proof of Case 4 is completed.

Summing up the above discussion, we know that there exist at most mutually orthogonal exponential functions in . We can find many such orthogonal systems which contain elements, for example, the exponential function system with given by This shows that the number is the best. The proof of Theorem 1 is complete.

The proof of Theorem 2 is similar, it is so omitted.

Corollary 5. For the self-affine measure corresponding to if , then is a nonspectral measure, and there exist at most 4 mutually orthogonal exponential functions in , where the number 4 is the best.

The corollary improved Yuan [9, Theorem 1].

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of the paper.

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).