Abstract and Applied Analysis

Volume 2014 (2014), Article ID 294182, 9 pages

http://dx.doi.org/10.1155/2014/294182

## Nonspectrality of Certain Self-Affine Measures on

^{1}College of Mathematics and Information Science, Shaanxi Normal University, Xi’an 710062, China^{2}Applied Mathematics Department, Yuncheng University, Yuncheng Shanxi 044000, China

Received 11 September 2013; Accepted 30 November 2013; Published 8 January 2014

Academic Editor: Natig M. Atakishiyev

Copyright © 2014 Gui-Bao Gao. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### 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*

*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*

*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*

*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*

*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*

*Conflict of Interests*

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

*Acknowledgments*

*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).*

*References*

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