# On the Boundary of Self-Affine Sets

Authors: Qi-Rong Deng; Xiang-Yang Wang
Journal: Abstract and Applied Analysis
 (2015)

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## 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.


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## Body

## 1. Introduction

Let(X,ρ) be a complete matric space. Recall that a map S:X→X is contractive if there exists a constant 0<r<1 such that ρ(S(x),S(y))≤rρ(x,y). We call a finite set of contractive maps {Sj}j=1m aniterated function system (IFS). It is well known [1] that there exists a unique nonempty compact subset K⊂X such that K=⋃j=1m‍Sj(K). We call K theinvariant set orattractor of the IFS. Moreover, if we associate the IFS with a set of probability weights {pi>0:i=1,…,m}, then there exists a unique probability measure μ supported on K satisfying the equation
(1)μ(·)=∑j=1m‍pjμ(Sj-1(·)).
We call μ theinvariant measure.LetA be a d×d expanding real matrix; that is, all its eigenvalues have modules larger than one. Let λ be the smallest absolute value of A’s eigenvalues, choose c∈(1,λ), and define x for each x∈Rd as
(2)x=∑n=1∞‍cnA-nx,
where · is the Euclidian norm in Rd. Then · is a norm in Rd. Let ρ(x,y)=x-y be the induced metric. It is easy to check that the map S(x)=A-1(x+c) with x,c∈Rd is contractive under the metric ρ.LetA be a d×d expanding real matrix and D={d1,d2,…,dm}⊂Rd. We call the family of maps on Rd(3)Six=A-1x+di,i=1,2,…,m
aself-affine IFS. The corresponding invariant set K and invariant measure μ are called aself-affine set and aself-affine measure of the IFS, respectively. Furthermore, if the matrix A in (3) is an orthonormal matrix multiple a constant, then such IFS is calledself-similar, and the invariant set and invariant measure are calledself-similar set andself-similar measure of the IFS, respectively.Our main interests in this note are the structures and properties of the boundary∂K of a self-affine set K. For self-similar IFS, Lau and Xu [2] showed that dim⁡H(∂K)<d provided that the self-similar IFS satisfies theopen set condition (OSC). He et al. [3] studied the calculation of dim⁡H(∂K) 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 K and ∂K (see [5–7]). There is no method to compute the Hausdorff dimension and the Lebesgue measure L(∂K) of ∂K 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{A-1(x+dj)}j=1m be a self-affine IFS defined on Rd. Assume that A and dj are all integral. Let K be the self-affine set of the IFS; then L(∂K)=0.Consider two IFSs{Sj}j=1m and {Sj}j=1n, m<n (they may not be self-affine). Let K1 and K2 be the invariant sets, respectively; then K1⊆K2, so dim⁡(K1)≤dim⁡(K2). We think about the natural question: what is the relationship between ∂K1 and ∂K2?We prove that any one case ofdim⁡H(∂K2)=dim⁡H(∂K1), dim⁡H(∂K2)<dim⁡H(∂K1), and dim⁡H(∂K2)>dim⁡H(∂K1) may occur.


## 2. Proofs of Results

For an IFS{Sj}j=1n on Rd, we use the following notations throughout the paper. Let Σm={1,…,m} (or Σ if there is no confusion), and Σ*=⋃n≥1‍Σn. For any I=i1i2⋯in∈Σn and J=j1j2⋯jk∈Σk, let IJ=i1i2⋯inj1j2⋯jk and
(4)pI=pi1pi2⋯pin,SI=Si1∘Si2∘⋯∘Sin,dI=din+Adin-1+⋯+An-1di1,Dn=D+AD+⋯+An-1D.
Also, we use L(E), Eo, and ∂E to denote the Lebesgue measure, the interior, and the boundary of a subset E⊂Rd, respectively.Theorem 2.
Let{ϕj}j=1m and {ψi}i=1k be two contractive IFSs on Rd under some norm · with the invariant sets K1 and K2, respectively. If the invariant set K1 contains interior points, then there exist a,n∈N and α∈Zd such that the IFSs F={φi1i2⋯in:1≤ij≤m} and F∪G generate the same attractor aK1+α, where G={ψj1j2⋯jn:1≤ji≤k} and φj(x)=aϕj(a-1(x-α))+α, j=1,…,m.Proof.
Observe that(5)⋃j=1m‍φjaK1+α=⋃j=1m‍aϕjK1+α=a(⋃j=1m‍ϕj(K1))+α=aK1+α.
This means that aK1+α is the invariant set of {φj}j=1m for any a>0 and α∈Rd. Hence it is also the invariant set of the IFS F. Now we need only to prove that aK1+α is the invariant set of F∪G for some a,n∈N and α∈Zd.
Note thatK1 contains interior points; we can find a constant r>0 and a point x0∈K1 with rational entries such that B2r(x0)⊂K1. Hence B2ar(0)⊂aK1-ax0 for all positive real number a>0. Since {ψi}i=1k are contractive in the norm ·, we can choose integers a,n∈N large enough such that K2⊂Bar(0) and |ψJ(aK1+α)|<ar for all J∈Σk* with |J|≥n, where |E| is the diameter of the set E⊂Rd under the norm ·. Also, we can assume that α=-ax0∈Zd. Noting K2⊆Bar(0)⊆B2ar(0)⊆aK1+α, |ψj1j2⋯jn(aK1+α)|<ar and observing
(6)ψj1j2⋯jn(aK1+α)∩K2⊇ψj1j2⋯jn(K2)≠∅,
we have
(7)ψj1j2⋯jn(aK1+α)⊆aK1+α.
Therefore
(8)aK1+α=⋃f∈F‍f(aK1+α)⊆⋃f∈F∪G‍f(aK1+α)⊆aK1+α.
We see that aK1+α is the invariant set of F∪G. This completes the proof.In Theorem2, IFS F is a subset of IFS F∪G and they have the same invariant set aK1+α. So do the same boundary of the invariant set. On the other hand, the invariant set of G is K2. Obviously, either dim⁡H(∂(aK1+α))<dim⁡H(∂K2) or dim⁡H(∂(aK1+α))>dim⁡H(∂K2) may occur.In the following, we consider the Lebesgue measure of∂K for the self-affine IFS (3). We will prove Theorem 1; that is, L(∂K)=0 if A and dj are all integral. For this, we first prove some lemmas.Lemma 3.
Let the IFS in (3) be integral; that is, all entries of A and dj are integers. Assume that the self-affine set K has positive Lebesgue measure; then Ko≠∅.Proof.
Note that the fact thatA and dj 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 {dj}j=1m contains a complete set of residues (mod⁡AZd). Then the self-affine measure μ in (1) is absolutely continuous with respect to the Lebesgue measure provided that
(9)∑j:(di-dj)∈AZd‍pj=1|det⁡(A)|,i=1,…,m.Proof.
Without loss of generality, assume thatD~={d1,…,dl} is a complete set of residues (mod⁡AZd) with det⁡(A)=l. Then D~n:=D~+AD~+⋯+An-1D~ is a complete set of residues (mod⁡AnZd).
For eachi∈{1,…,l}, let Ii={j:1≤j≤m, (dj-di)∈AZd} and pj=1/l#Ii if j∈Ii; then we have
(10)∑j∈Ii‍pj=1|det⁡(A)|,i=1,…,l.
Hence such probability weights {pi}i=1m satisfying (9) always exist.
To prove the absolute continuity ofμ, by making use of [8, Theorem 3.5], we need only to show that
(11)∑J∈Σn,dJ=z‍pJ≤det⁡A-n,∀n>0,z∈Zd.
We will prove this by induction on n. By (9), the inequality (11) holds for n=1. Assume that (11) holds for n=k. Let z=di+Az1 with di∈D~ and z1∈Zd. If J∈Σk, j∈Σ, and dJj=z, then dj+AdJ=di+Az1, so (dj-di)∈AZd, and let dj=di+Aej with ej∈Zd; we have ej+dJ=z1. Therefore
(12)∑Jj∈Σk+1,dJj=z‍pJj≤∑j∈Σ,(dj-di)∈AZd‍pj∑J∈Σk,dJ=z1-ej‍pJ≤det⁡A-k∑j∈Σ,(dj-di)∈AZd‍pj≤det⁡A-(k+1).
Hence (11) is also true for n=k+1. This completes the proof.Remark. Lemma 4 gives a sufficient condition for the existence of L1-solutions of integral refinement equations:
(13)f(x)=det⁡A∑j=1m‍pjf(Ax-dj)
provided that {d1,…,dm}⊂Zd contains a complete set of residues (mod⁡AZd). Condition (9) ensures that the refinement equation has a unique (up to a scalar multiple) bounded L1-solution with compact support if pj’s satisfy (9). Condition (9) is an extension of the “sum role.”Lemma 5.
Let the IFS in (3) be integral. Suppose {dj}j=1m contains a complete set of residues (mod⁡AZd); K is the corresponding self-affine set. Then L(∂K)=0.Proof.
Lemma4 implies that there exist probability weights {pj}j=1m such that the corresponding self-affine measure μ is absolutely continuous with respect to the Lebesgue measure and so L(K)>0.
Lemma3 implies that Ko≠∅, so Ko is a nonempty invariant open set (i.e., ⋃j=1mSj(Ko)⊆Ko) and μ(Ko)>0. Then [8, Theorem 4.13] implies that μ(∂K)=0. On the other hand, [8, Theorem 3.12] implies that the Lebesgue measure restricted on K is also absolutely continuous with respect to μ. Hence L(∂K)=0.Now we can prove the main theorem of the paper.Proof of Theorem1.
IfKo=∅, then ∂K=K and Lemma 3 implies that L(∂K)=0.
Now we consider the caseKo≠∅. Let ϕj(x)=A-1(x+dj), φj(x)=aϕj(a-1(x-α))+α=A-1(x-α+adj+Aα), j=1,…,m, and ψi(x)=A-1(x+zi), i=1,…,k, where Z={z1=0,…,zk} is a complete set of residues (mod⁡AZd). Making use of Theorem 2 and the notations there, there exist a,n∈N and α∈Zd such that the IFSs F and F∪G have the same attractor aK+α. Let D~=aD-α+Aα. Then F∪G={A-n(x+d):d∈D~n∪Zn}. Note that D~n∪Zn contains a complete set Zn of residues (mod⁡AZd); Lemma 5 implies that L(∂K)=a-dL(∂(aK+α))=0. We complete the proof.



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*Source: 573604-2015-02-24.xml*