On the Boundary of Self-Affine Sets

Abstract

This paper is devoted to studying the boundary behavior of self-affine sets. We prove that the boundary of an integral self-affine set has Lebesgue measure zero. In addition, we consider the variety of the boundary of a self-affine set when some other contractive maps are added. We show that the complexity of the boundary of the new self-affine set may be the same, more complex, or simpler; any one of the three cases is possible.

1. Introduction

Let be a complete matric space. Recall that a map is contractive if there exists a constant such that . We call a finite set of contractive maps an iterated function system (IFS). It is well known [1] that there exists a unique nonempty compact subset such that . We call the invariant set or attractor of the IFS. Moreover, if we associate the IFS with a set of probability weights , then there exists a unique probability measure supported on satisfying the equation We call the invariant measure.

Let be a expanding real matrix; that is, all its eigenvalues have modules larger than one. Let be the smallest absolute value of ’s eigenvalues, choose , and define for each as where is the Euclidian norm in . Then is a norm in . Let be the induced metric. It is easy to check that the map with is contractive under the metric .

Let be a expanding real matrix and . We call the family of maps on a self-affine IFS. The corresponding invariant set and invariant measure are called a self-affine set and a self-affine measure of the IFS, respectively. Furthermore, if the matrix in (3) is an orthonormal matrix multiple a constant, then such IFS is called self-similar, and the invariant set and invariant measure are called self-similar set and self-similar measure of the IFS, respectively.

Our main interests in this note are the structures and properties of the boundary of a self-affine set . For self-similar IFS, Lau and Xu [2] showed that provided that the self-similar IFS satisfies the open set condition (OSC). He et al. [3] studied the calculation of for integral self-similar IFS. Furthermore, the overlapping cases were considered by Lau and Ngai in [4]. For self-affine sets, however, less is known about and (see [5–7]). There is no method to compute the Hausdorff dimension and the Lebesgue measure of for overlapping self-affine set.

Motivated by these results, we consider the Lebesgue measures of the boundaries of integral self-affine sets. We prove that they have Lebesgue measure zero.

Theorem 1. Let be a self-affine IFS defined on . Assume that and are all integral. Let be the self-affine set of the IFS; then .

Consider two IFSs and , (they may not be self-affine). Let and be the invariant sets, respectively; then , so . We think about the natural question: what is the relationship between and ?

We prove that any one case of , , and may occur.

2. Proofs of Results

For an IFS on , we use the following notations throughout the paper. Let (or if there is no confusion), and . For any and , let and Also, we use , , and to denote the Lebesgue measure, the interior, and the boundary of a subset , respectively.

Theorem 2. Let and be two contractive IFSs on under some norm with the invariant sets and , respectively. If the invariant set contains interior points, then there exist and such that the IFSs and generate the same attractor , where and , .

Proof. Observe that This means that is the invariant set of for any and . Hence it is also the invariant set of the IFS . Now we need only to prove that is the invariant set of for some and .
Note that contains interior points; we can find a constant and a point with rational entries such that . Hence for all positive real number . Since are contractive in the norm , we can choose integers large enough such that and for all with , where is the diameter of the set under the norm . Also, we can assume that . Noting , and observing we have Therefore We see that is the invariant set of . This completes the proof.

In Theorem 2, IFS is a subset of IFS and they have the same invariant set . So do the same boundary of the invariant set. On the other hand, the invariant set of is . Obviously, either or may occur.

In the following, we consider the Lebesgue measure of for the self-affine IFS (3). We will prove Theorem 1; that is, if and are all integral. For this, we first prove some lemmas.

Lemma 3. Let the IFS in (3) be integral; that is, all entries of and are integers. Assume that the self-affine set has positive Lebesgue measure; then .

Proof. Note that the fact that and are all integral implies that the IFS is uniformly discrete, and the assertion follows from [7, Theorem 3.1].

Lemma 4. Let the IFS in (3) be integral. Suppose that contains a complete set of residues (). Then the self-affine measure in (1) is absolutely continuous with respect to the Lebesgue measure provided that

Proof. Without loss of generality, assume that is a complete set of residues () with . Then is a complete set of residues ().
For each , let , and if ; then we have Hence such probability weights satisfying (9) always exist.
To prove the absolute continuity of , by making use of [8, Theorem 3.5], we need only to show that We will prove this by induction on . By (9), the inequality (11) holds for . Assume that (11) holds for . Let with and . If , , and , then , so , and let with ; we have . Therefore Hence (11) is also true for . This completes the proof.

Remark. Lemma 4 gives a sufficient condition for the existence of -solutions of integral refinement equations: provided that contains a complete set of residues (). Condition (9) ensures that the refinement equation has a unique (up to a scalar multiple) bounded -solution with compact support if ’s satisfy (9). Condition (9) is an extension of the “sum role.”

Lemma 5. Let the IFS in (3) be integral. Suppose contains a complete set of residues (); is the corresponding self-affine set. Then .

Proof. Lemma 4 implies that there exist probability weights such that the corresponding self-affine measure is absolutely continuous with respect to the Lebesgue measure and so .
Lemma 3 implies that , so is a nonempty invariant open set (i.e., ) and . Then [8, Theorem 4.13] implies that . On the other hand, [8, Theorem 3.12] implies that the Lebesgue measure restricted on is also absolutely continuous with respect to . Hence .

Now we can prove the main theorem of the paper.

Proof of Theorem 1. If , then and Lemma 3 implies that .
Now we consider the case . Let , , , and , , where is a complete set of residues (). Making use of Theorem 2 and the notations there, there exist and such that the IFSs and have the same attractor . Let . Then . Note that contains a complete set of residues (); Lemma 5 implies that . We complete the proof.

Conflict of Interests

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

Acknowledgment

The research is supported by NSFC (nos. 11371382 and 11471075).