A class of Sierpinski carpets with overlaps

J. Math. Anal. Appl. 340 (2008) 1422–1432 www.elsevier.com/locate/jmaa

A class of Sierpinski carpets with overlaps Yuru Zou, Yuanyuan Yao, Wenxia Li ∗,1 Department of Mathematics, East China Normal University, Shanghai 200062, PR China Received 25 April 2007 Available online 3 October 2007 Submitted by B. Bongiorno

Abstract Based on a scheme by Lau and Ngai, a sufficient and necessary condition for an IFS of contractive similitudes to satisfy the open set condition is given. A class of Sierpinski carpets with parameters (λ1 , λ2 ) is then investigated. © 2007 Elsevier Inc. All rights reserved. Keywords: Hausdorff dimension; Generalized finite type; (No) complete overlap

1. Introduction d Let {Sj }N j =1 be an iterated function system (IFS) of contractive similitudes on R defined by

Sj (x) = ρj Rj x + bj

for j = 1, . . . , N,

(1)

Rd

where ρj ∈ (0, 1), bj ∈ and Rj is a d × d orthogonal matrix for each 1  j  N . Then there exists a unique nonempty compact set F ⊂ Rd invariant under Sj s, i.e., F=

N 

Sj (F ).

j =1

F is also called the self-similar set corresponding to (S1 , . . . , SN ). It is well known that if Sj s satisfy the open set condition (OSC) [2,3], i.e., there exists a nonempty bounded open set O such that N 

Sj (O) ⊆ O,

with Si (O) ∩ Sj (O) = ∅ for i = j,

j =1

then we have dimH F = dimB F = ξ

and 0 < Hξ (F ) < +∞,

* Corresponding author.

E-mail address: [email protected] (W. Li). 1 Supported by the National Natural Science Foundation of China (#10571058) and Shanghai Priority Academic Discipline.

0022-247X/$ – see front matter © 2007 Elsevier Inc. All rights reserved. doi:10.1016/j.jmaa.2007.09.057

Y. Zou et al. / J. Math. Anal. Appl. 340 (2008) 1422–1432

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where ξ is called the similarity dimension of F and satisfies N 

ξ

ρi = 1.

(2)

i=1

However, in the absence of the OSC the images of F under the Sj s have overlaps and the above dimension formula fails in general. In this case it is much harder to compute dimH F and it has been the subject of several studies. The Hausdorff dimension for some self-similar sets with overlap structure has been computed by some authors [1,4,7, 9,10]. Ngai and Wang [6] introduced the notion of finite type and described a scheme for the computation of the Hausdorff dimension when the finite type occurs. However, the finite type condition forces the ratios {ρi }N i=1 to be exponentially commensurable. Recently, Lau and Ngai [5] proposed the concept of the generalized finite type which is an extension of the finite type. It allows us to determine the Hausdorff dimension of F when the generalized finite type condition is satisfied. The Lau–Ngai’s  scheme can be outlined as follows. One can refer to [5] for details. Let ΣN = {1, 2, . . . , N} and ΣN∗ = n0 ΣNn be the set of all finite words, where ΣNn is the set of all words of length n, with ΣN0 containing only the empty word ∅. For j ∈ ΣNn let |j| = n denote the length of j. For i ∈ ΣNm and j ∈ ΣNn , let ij ∈ ΣNm+n be the concatenation of i and j, and call i an initial segment of ij. For two indices i, j ∈ ΣN∗ , we write i  j if i is an initial segment of j (including i = j). For j = (j1 , . . . , jm ) ∈ ΣNm , denote Sj := Sj1 ◦ · · · ◦ Sjm ,

ρj := ρj1 · · · ρjm

and Rj := Rj1 · · · Rjm .

{Mk }∞ k=0

of nested index sets which generalize the notion of “level of In [5], the authors introduced a sequence iteration.” Therefore, with ξ as defined by (2), we have  ξ  ξ ρj = 1 and ρj = 1 for any fixed i ∈ Mk . (3) j∈Mk

Let V :=



∗ , ij∈M j∈ΣN k+1

k0 Vk

where     V0 := (I, 0) and Vk := (Si , k): i ∈ Mk

for all k  1.

A directed graph G is then constructed with V as the set of vertices and j ∈ ΣN∗ as the directed edge connecting from (Si , k) to (Sij , k + 1) (this implies that i ∈ Mk , (Si , k) ∈ Vk and ij ∈ Mk+1 , (Sij , k + 1) ∈ Vk+1 ). The vertex (Sij , k + 1) is called an offspring of the vertex (Si , k), and conversely, the vertex (Si , k) is called the parent of the vertex (Sij , k +1). Thus each vertex in Vk has at least one offspring in Vk+1 , and each vertex in Vk+1 has at least one parent in Vk . The reduced graph GR is obtained from G by first removing all but the smallest (in the lexicographical order) directed edge going to a vertex, then removing all vertices that do not have offsprings, together with all the vertices and edges  leading only to them. The set of the kth order vertices of the reduced graph GR is denoted by VR,k and VR := k0 VR,k . N Fix any nonempty bounded open set Ω ⊂ Rd which is invariant under {Sj }N j =1 Sj (Ω) ⊂ Ω. Two vertices j =1 , i.e., v, v ∈ Vk (allowing v = v ) are neighbors (with respect to Ω) if Sv (Ω) ∩ Sv (Ω) = ∅ (where for v = (Si , k) ∈ Vk we use the convenient notation Sv := Si ). The set of vertices Ω(v) := {v : v is a neighbor of v} is called the neighborhood of v (with respect to Ω). Two vertices v ∈ Vk and v ∈ Vk  are equivalent, denoted by v ∼Ω v (or simply v ∼ v ), if, for τ := Sv ◦ Sv−1 : Rd → Rd , the following conditions are satisfied: (i) {Su : u ∈ Ω(v )} = {τ ◦ Su : u ∈ Ω(v)}; (ii) for u ∈ Ω(v) and u ∈ Ω(v ) such that Su = τ ◦ Su , and for any positive integer   1, an index i ∈ ΣN∗ satisfies (Su ◦ Si , k + ) ∈ Vk+ if and only if it satisfies (Su ◦ Si , k  + ) ∈ Vk  + . One can check that ∼ is an equivalent relation on V. The equivalence class containing the vertex v ∈ V is denoted by [v] and is called the neighborhood type of v. The IFS {Sj }N j =1 is said to be of generalized finite type if V/ ∼= {[v]: v ∈ V} is a finite set (cf. Lau and Ngai [5, Definition 2.2]). Now suppose that the IFS {Sj }N j =1 is of generalized finite type with T1 , . . . , Tq as its neighborhood types. For each α  0, a weighted incidence matrix

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Y. Zou et al. / J. Math. Anal. Appl. 340 (2008) 1422–1432 q

Aα = (Aα (i, j ))i,j =1 is defined as follows. Fix i (1  i  q) and a vertex v ∈ VR such that [v] = Ti . Let u1 , . . . , um be the offsprings of v in GR and let ik , 1  k  m, be the unique edge in GR connecting v to uk . Then   ik  ρiαk  v − (4) → uk , [uk ] = Tj . Aα (i, j ) = d Lau–Ngai’s Theorem. (See [5, Theorem 1.2].) Assume that an IFS {Sj }N j =1 of contractive similitudes on R with attractor F is of generalized finite type, and let λα be the spectral radius of the associated weighted incidence matrix Aα . Then

dimH F = dimB F = α, where α is the unique number such that λα = 1. Moreover, 0 < Hα (F ) < +∞. As we can see above, for general case it is hard to classify the neighborhood types and determine the associated weighted incidence matrix. The difficulty mainly comes from the fact that #VR,k < #Mk often happens where, throughout this paper, #A denotes the cardinality of a finite set A. This is equivalent to that there exist distinct i, j ∈ Mk such that Si ≡ Sj . Thus, if it is the case, then dimH F < ξ and overlaps between Si (F ), i = 1, . . . , N , occur. We say N ∗ that the IFS {Si }N i=1 is of complete overlap if there exist distinct i, j ∈ ΣN such that Si ≡ Sj . Otherwise, the IFS {Si }i=1 is said to be of no complete overlap. In this paper, we make use of Lau–Ngai’s scheme to analyze a class of Sierpinski carpets with parameters. Let λ1 , λ2 ∈ [0, 1) and take φi : R2 → R2 as φi (x, y) =

1 (x, y) + (ai , bi ) , 4

i = 1, . . . , 5,

(5)

, b3 ) = (3, 0), (a4 , b4 ) = (3, 3) and (a5 , b5 ) = (0, 3). The unique nonwhere (a1 , b1 ) = (0, 0), (a2 , b2 ) = (λ1 , λ2 ), (a3 empty compact set Fλ1 ,λ2 satisfying Fλ1 ,λ2 = 5i=1 φi (Fλ1 ,λ2 ) is called as a (λ1 , λ2 )-Sierpinski carpet. In this case, the sequence {Mk }∞ k=0 of nested index sets is given by letting       Mk = Σ5k and so Vk := (φj , k)  j ∈ Σ5k is identical with φj (0), k  j ∈ Σ5k , (6) and except in the case that (λ1 , λ2 ) = (0, 0), each vertex v ∈ Vk has just 5 offsprings in Vk+1 which are connected from v by the directed edges k ∈ Σ51 in the graph G, respectively. However, for a vertex v ∈ Vk+1 there may be more than one parents. This happens if and only if #Vk+1 < #Σ5k+1 (= 5k+1 ). When the IFS {φi }5i=1 is of no complete overlap, each vertex v ∈ Vk just has 5 offsprings in Vk+1 and each vertex v ∈ Vk+1 has just one parent in Vk in the graph G. Therefore, the reduced graph GR is identical with the graph G. In this paper, we prove that dimH Fλ1 ,λ2 = log 5 5 log 4 if the IFS {φi }i=1 is of generalized finite type and no complete overlap. This actually gives a sufficient and necessary condition for the IFS {φi }5i=1 to satisfy the open set condition. The proof is given in Section 2 for the 5 general IFS {Si }N i=1 (see Theorem 2.1). Note that the IFS {φi }i=1 is of finite type (and so generalized finite type) when (λ1 , λ2 ) ∈ Q × Q by Theorem 2.4 and the following remark in [6]. Thus, we then give a characterization of the rational pairs (λ1 , λ2 ) for which the corresponding IFS {φi }5i=1 are of complete overlap (see Theorem 3.1). In addition, we also give an explicit criterion for a class of rational pairs (λ1 , λ2 ) for which the corresponding IFS {φi }5i=1 are of no complete overlap (see Theorem 3.4). 2. Hausdorff dimension and the open set condition It is clear that the IFS {Sj }N j =1 is of no complete overlap if it satisfies the open set condition. In fact, when the IFS {Sj }N is of complete overlap, there exist distinct i, j ∈ ΣN∗ such that Si = Sj . Take an index set M ⊂ ΣN∗ such that j =1

ξ i, j ∈ M and k∈M ρk = 1 (such index set always exists). So   Sk (F ) = Sk (F ), F= k∈M

k∈M\{i}

Y. Zou et al. / J. Math. Anal. Appl. 340 (2008) 1422–1432

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giving dimH F < ξ , i.e., the IFS {Sj }N j =1 fails to satisfy the open set condition. Moreover, Lau and Ngai [5, ExamN ple 2.5] proved that the IFS {Sj }j =1 is of generalized finite type if it satisfies the open set condition. In the following theorem, we prove that the converse is also true. Theorem 2.1. Let Sj (x), j = 1, . . . , N , be given by (1). Then the IFS {Sj }N j =1 satisfies the open set condition if and only if it is of generalized finite type and no complete overlap. Proof. We only need to prove the sufficiency. Suppose that the IFS {Sj }N j =1 is of generalized finite type with respect ∞ to the sequence {Mk }k=0 of nested index sets. Let T1 , . . . , Tq be the neighborhood types. Since the IFS {Sj }N j =1 is of no complete overlap, we have that #VR,k = #Mk and so the reduced graph GR is identical with the graph G. Thus, for a fixed i (1  i  q) and a vertex v ∈ VR such that v = (Si , ) for some i ∈ M and [v] = Ti , its offsprings, say u1 , . . . , um , are just uk = (Siik ,  + 1)

with iik ∈ M+1 , k = 1, . . . , m.

q

Let Aα = (Aα (i, j ))i,j =1 be the weighted incidence matrix where Aα (i, j ) =

  ik  ρiαk  v − → uk , [uk ] = Tj ,

by (4). Therefore, for each 1  i  q, q 

Aξ (i, j ) =

j =1

m 



ξ

ρ ik =

k=1

ξ

∗ , ij∈M j∈ΣN +1

ρj = 1,

by (3), where ξ , defined as (2), is the similarity dimension of F . Therefore, dimH F = dimB F = ξ with 0 < Hξ (F ) < +∞ by Lau–Ngai’s Theorem [5, Theorem 1.2]. The desired result is then obtained by the main theorem in [8]. 2 The above theorem motivates one to investigate the no complete overlap condition for a given IFS. By Q we denote the set of rational numbers throughout this paper. For the IFS {φi }5i=1 defined as in (5), the following is direct. Proposition 2.2. Let {φi }5i=1 be as defined in (5). If (λ1 , λ2 ) ∈ / Q × Q, then the IFS {φi }5i=1 is of no complete overlap. Proof. Suppose there exists a complete overlap. Then there are two different sequences of indices (i1 , . . . , ik ) and (j1 , . . . , jk ) such that φ(i1 ,...,ik ) (0) = φ(j1 ,...,jk ) (0), i.e., k  ai =1



4

=

k  aj =1



4

and

k  bi =1



4

=

k  bj =1



4

(7)

,

where ai , aj ∈ {0, λ1 , 3}, bi , bj ∈ {0, λ2 , 3}. Without loss of generality, we assume λ1 ∈ / Q. Let I = {1    k: ai = λ1 }, J = {1    k: aj = λ1 }. Rewriting (7), we have     a j ai  − − = λ1 4 − 4 − (8)  4 4 ∈I

and λ2

 ∈I

∈J

4− −

 ∈J

∈{1,...,k}\J

4− =

 ∈{1,...,k}\J

∈{1,...,k}\I

b j − 4

 ∈{1,...,k}\I

bi  . 4

(9)

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Y. Zou et al. / J. Math. Anal. Appl. 340 (2008) 1422–1432

Fig. 1. The second iteration with (λ1 , λ2 ) = (1/3, 1/3) for the left, and (λ1 , λ2 ) = (2/5, 2/5) for the right.





From (8) it follows that ∈I 4− = ∈J 4− and ∈{1,...,k}\J aj 4− = ∈{1,...,k}\I ai 4− , yielding I = J and ai = aj ,  = 1, 2, . . . , k. Thus, the equality (9) leads to bi = bj ,  = 1, 2, . . . , k. Therefore, we have (ai , bi ) = (aj , bj ),  = 1, 2, . . . , k, yielding (i1 , . . . , ik ) = (j1 , . . . , jk ). 2 By Theorem 2.1 and Proposition 2.2 we have Corollary 2.3. Suppose that (λ1 , λ2 ) ∈ / Q × Q. Then dimH Fλ1 ,λ2 = dimB Fλ1 ,λ2 =

log 5 log 4

log 5

and 0 < H log 4 (Fλ1 ,λ2 ) < ∞

if the IFS {φi }5i=1 is of generalized finite type. Since the Sierpinski carpets Fλ1 ,λ2 considered here are relatively “thin,” the following example shows that the IFS {φi }5i=1 indeed satisfy the open set condition for many choices of the pairs (λ1 , λ2 ). Example 2.4. Let {φj }5j =1 be defined as in (5). The IFS {φi }5i=1 satisfies the open set condition when (λ1 , λ2 ) ∈ ([1/3, 2/5] × [0, 1)) ∪ ([0, 1) × [1/3, 2/5]). Proof. Let the sequence {Mk }∞ sets be taken as in (6). Let Ω = (0, 1) × (0, 1). Then Ω is k=0 of nested index  an open set invariant under the IFS {φj }5j =1 , i.e., 5i=1 φi (Ω) ⊆ Ω. Let T1 denote the neighborhood type of the root vertex (I, 0). It is easy to see that vertices (φ(3) , 1), (φ(4) , 1), (φ(5) , 1) ∈ V1 := {(φi , 1), i ∈ Σ51 } have the same neighborhood type as (I, 0). While vertices (φ(1) , 1), (φ(2) , 1) have new neighborhood types, denoted by T2 and T3 , respectively, when (λ1 , λ2 ) = (0, 0). From Fig. 1 one can see that the vertices in V2 := {(φi , 2), i ∈ Σ52 } have no new neighborhood types other than Tk , k = 1, 2, 3, if    Projx φ(2,1) (Ω) ∪ φ(2,2) (Ω) ⊆ (0, 1) \ Projx φ(1,1) (Ω) ∪ φ(1,2) (Ω) ∪ Projx φ(1,3) (Ω) and Projx φ(2,3) (Ω) ∩ Projx φ(1,3) (Ω) = ∅, where Projx denotes the projection on x-axis. This is equivalent to ⎧λ λ1 1 1 ⎪  + , ⎨ 4 16 16 ⎪ ⎩ λ1 + λ1 + 1  3 , 4 16 16 16 which leads to 13  λ1  25 . Therefore, the IFS {φi }5i=1 is of generalized finite type when (λ1 , λ2 ) ∈ [1/3, 2/5] × [0, 1). The same argument by considering the projection on y-axis shows that the IFS {φi }5i=1 is of generalized finite type

Y. Zou et al. / J. Math. Anal. Appl. 340 (2008) 1422–1432

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Fig. 2. The second iteration with (λ1 , λ2 ) = (1/3, 1/5) for the left, (λ1 , λ2 ) = (1/2, 1/5) for the central, and (λ1 , λ2 ) = (1/2, 2/5) for the right.

Fig. 3. The second iteration with (λ1 , λ2 ) = (1/4, 1/3) for the left, (λ1 , λ2 ) = (2/5, 1/3) for the central, and (λ1 , λ2 ) = (2/5, 2/3) for the right.

Fig. 4. The second iteration with (λ1 , λ2 ) = (4/5, 1/4) for the left, and (λ1 , λ2 ) = (4/5, 1/2) for the right.

when (λ1 , λ2 ) ∈ [0, 1) × [1/3, 2/5]. Now, by (4) the weighted incidence matrix Aα = (Aα (i, j ))3i,j =1 is given by ⎛ 3 1 1 ⎞ 4α

⎜ Aα = ⎝ 43α 3 4α

4α 1 4α 1 4α

4α 1 4α 1 4α

⎟ ⎠. log 5

Thus for (λ1 , λ2 ) ∈ ([1/3, 2/5] × [0, 1)) ∪ ([0, 1) × [1/3, 2/5]), we have 0 < H log 4 (Fλ1 ,λ2 ) < +∞ by Lau–Ngai’s Theorem [5, Theorem 1.2]. The conclusion follows by an application of Schief’s theorem [8]. 2 Remark. The same argument as above shows that the IFS {φi }5i=1 satisfies the open set condition when (λ1 , λ2 ) ∈ ([1/3, 1/2] × [1/5, 2/5]) ∪ ([1/5, 2/5] × [1/3, 1/2]) (see Fig. 2), (λ1 , λ2 ) ∈ ([1/4, 2/5] × [1/3, 1)) ∪ ([1/3, 1) × [1/4, 2/5]) (see Fig. 3) and (λ1 , λ2 ) ∈ ([4/5, 1) × [1/4, 1/2]) ∪ ([1/4, 1/2] × [4/5, 1)) (see Fig. 4). In all above cases, no newer neighborhood types occur in the second iteration. The corresponding weighted incidence matrices Aα are the same as that in the above example. 3. Classification of the rational pairs (λ1 , λ2 ) Note that from Theorem 2.4 and the following remark in [6] we obtain that the IFS {φi }5i=1 is of finite type (so generalized finite type by [5, Example 2.6]) for any (λ1 , λ2 ) ∈ (Q × Q) ∩ ([0, 1) × [0, 1)). In this section, we focus on the classification of the rational pairs (λ1 , λ2 ). For the IFS {φi }5i=1 defined as in (5), it is easy to see that it is of complete overlap if and only if there exist (i1 , . . . , ik ), (j1 , . . . , jk ) ∈ Σ5k for some k  1 such that φ(1,i1 ,...,ik ) (0) = φ(2,j1 ,...,jk ) (0), i.e.,

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Y. Zou et al. / J. Math. Anal. Appl. 340 (2008) 1422–1432 k k  ai  λ 1  a j + = 4 4+1 4+1 =1

and

=1

k k  bi  λ 2  b j + = . 4 4+1 4+1 =1

=1

The following theorem gives a description of (λ1 , λ2 ) ∈ Q × Q for which the corresponding IFS {φi }5i=1 are of complete overlap. Theorem 3.1. Let P be the set of (λ1 , λ2 ) ∈ (Q × Q) ∩ ([0, 1) × [0, 1)) such that the corresponding IFS {φi }5i=1 be of complete overlap. Let Ak = {((α )k=1 , (β )k=1 , (β∗ )k=1 ) ∈ {0, −1, 1}k × {0, −1, 1}k × {0, −1, 1}k : α β = −1 and α β∗ = −1 for 1    k}. Then



 ∞     ∗ k 3 k=1 β∗ 4k− 3 k=1 β 4k−  k k (α ∈ [0, 1) × [0, 1) P= ∈ A , ) , (β ) , β 

k

k  =1  =1 k .  =1 k+ k− 4k + k− 4 α 4 α 4   =1 =1 k=1

(We remark that the denominator 4k + k=1 α 4k− > 0 for any (α )k=1 ∈ {0, −1, 1}k .) Proof. Suppose (λ1 , λ2 ) ∈ P. Then there exist (i1 , . . . , ik ), (j1 , . . . , jk ) ∈ Σ5k for some k  1 such that φ(1,i1 ,...,ik ) (0) = φ(2,j1 ,...,jk ) (0), i.e., k k  ai  λ 1  a j + = 4 4+1 4+1 =1

and

=1

k k  bi  λ 2  b j + = , 4 4+1 4+1 =1

(10)

=1

where (a1 , b1 ) = (0, 0), (a2 , b2 ) = (λ1 , λ2 ), (a3 , b3 ) = (3, 0), (a4 , b4 ) = (3, 3) and (a5 , b5 ) = (0, 3). Let I = {1  p  k: ip = 2} and J = {1  q  k: jq = 2}. From (10) it follows that ⎧     k k−q k−p ⎪ ⎪ 4 λ1 = + 4 − 4 4k−p aip − 4k−q ajq , ⎪ ⎪ ⎨ q∈J p∈I p∈{1,2,...,k}\I q∈{1,2,...,k}\J  (11)     ⎪ ⎪ k k−q k−p k−p k−q ⎪ λ2 = 4 − 4 4 bi p − 4 b jq . ⎪ ⎩ 4 + q∈J

p∈I

p∈{1,2,...,k}\I

q∈{1,2,...,k}\J

Set S1 = {1, 2, . . . , k} \ (S2 ∪ S3 ), S2 = I \ J and S3 = J \ I. Then 

4k−q −

q∈J



4k−p =

p∈I

k 

α 4k− ,

=1

where (α )k=1 ∈ {0, −1, 1}k with α = 0 for  ∈ S1 , α = −1 for  ∈ S2 and α = 1 for  ∈ S3 . Note that   {1, 2, . . . , k} \ I = S3 ∪ S1 \ (I ∩ J ) and {1, 2, . . . , k} \ J = S2 ∪ S1 \ (I ∩ J ) . Thus we have  p∈{1,2,...,k}\I

=3



∈S3

and

 

∈S3

4k−q ajq

q∈{1,2,...,k}\J

ai k−  −aj k− 4 4 + + 3 3 ∈S2



4k−p bip −

p∈{1,2,...,k}\I

=3



4k−p aip −

q∈{1,2,...,k}\J

∈S1 \(I ∩J )

ai − aj k− 4 3



4k−q bjq

bi k−  −bj k− 4 4 + + 3 3 ∈S2



 ∈S1 \(I ∩J )

bi − bj k− . 4 3

Y. Zou et al. / J. Math. Anal. Appl. 340 (2008) 1422–1432

Now take ⎧ ai bi ⎪ β = 3 , β∗ = 3 ⎪ ⎪ ⎪ ⎪ −aj −bj ⎨ β = 3  , β∗ = 3  ⎪ ai −aj bi −bj ⎪ ⎪ β =  3  , β∗ =  3  ⎪ ⎪ ⎩ β = β∗ = 0

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for  ∈ S3 , for  ∈ S2 , for  ∈ S1 \ (I ∩ J ), for  ∈ I ∩ J .

Then, it is easy to verify that β , β∗ ∈ {0, 1} for  ∈ S3 , β , β∗ ∈ {0, −1} for  ∈ S2 and β , β∗ ∈ {0, −1, 1} for  ∈ S1 \ (I ∩ J ). So ((α )k=1 , (β )k=1 , (β∗ )k=1 ) ∈ Ak . Finally, by (11) we have ⎧

3 k=1 β 4k− ⎪ ⎪ ⎪ ⎪ ⎨ λ1 = 4k + k α 4k− , =1 

k ⎪ 3 =1 β∗ 4k− ⎪ ⎪ ⎪ . ⎩ λ2 = k k 4 + =1 α 4k− Conversely, suppose that for some ((α )k=1 , (β )k=1 , (β∗ )k=1 ) ∈ Ak we have

 k 3 k=1 β∗ 4k− 3 =1 β 4k− , ∈ [0, 1) × [0, 1).



4k + k=1 α 4k− 4k + k=1 α 4k−

(12)

Take (λ1 , λ2 ) as the pair of real numbers in (12). Denote S2 = {1    k | α = −1}, Now take (a) (b) (c) (d) (e)

S3 = {1    k | α = 1} and S1 = {1, 2, . . . , k} \ (S2 ∪ S3 ).

(i1 , . . . , ik ), (j1 , . . . , jk ) ∈ Σ5k

where

i = 2 if  ∈ S2 ; j = 2 if  ∈ S3 ; i ∈ {1, 3, 4, 5} satisfying (ai , bi ) = (3β , 3β∗ ) if  ∈ S3 ; j ∈ {1, 3, 4, 5} satisfying (aj , bj ) = (−3β , −3β∗ ) if  ∈ S2 ; i , j ∈ {1, 3, 4, 5} satisfying (ai , bi ) − (aj , bj ) = (3β , 3β∗ ) if  ∈ S1 .

It is easy to see that the (i1 , . . . , ik ), (j1 , . . . , jk ) ∈ Σ5k are well defined according to (a)–(e). Its verification is left for the readers. With these choices for (i1 , . . . , ik ), (j1 , . . . , jk ) and (λ1 , λ2 ) taken as in (12), one can check that φ(1,i1 ,...,ik ) (0) = φ(2,j1 ,...,jk ) (0), as desired. 2 For k ∈ N, denote

 k    ∗ k 3 k=1 β∗ 4k− 3 =1 β 4k−  k k Pk = (α ∈ A , ) , (β ) , β ∈ [0, 1) × [0, 1) 



 =1  =1 k ,  =1 4k + k=1 α 4k− 4k + k=1 α 4k− k+1 ∗ k+1 where Ak is defined as in Theorem 3.1. Now, for ((α )k+1 =1 , (β )=1 , (β )=1 ) ∈ Ak+1 , we consider the pair of numbers



 ∗ k+1− k+1− 3 k+1 3 k+1 =1 β 4 =1 β 4 . ,

k+1 k+1− 4k+1 + k+1− 4k+1 + k+1 =1 α 4 =1 α 4

Note that

k+1

k+1− =1 β 4

k+1− 4k+1 + k+1 =1 α 4

3

and

k+1

∗ k+1− =1 β 4

k+1− 4k+1 + k+1 =1 α 4

3

=

=

4(3

k

=1 β 4

4(4k +

k

k

k− ) + 3β k+1

=1 α 4

k− ) + α k+1

∗ k− ) + 3β ∗ =1 β 4 k+1 .

4(4k + k=1 α 4k− ) + αk+1

4(3

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Y. Zou et al. / J. Math. Anal. Appl. 340 (2008) 1422–1432

The following statements can be verified easily: (I) If 0 < 3

k

=1 β 4

< 4k +

k−

k

=1 α 4

k− ,

then

k+1

k+1− =1 β 4

k+1− 4k+1 + k+1 =1 α 4

3

∈ (0, 1),

for all

αk+1 βk+1 = −1. (II) If 3 k=1 β 4k− = 0, equivalently, β1 = · · · = βk = 0, then

k+1

k+1− =1 β 4

k+1 4k+1 + =1 α 4k+1−

3

=

4(4k +

k

3βk+1

=1 α 4

k− ) + α k+1

∈ (0, 1),

for α

k+1 βk+1 = −1 with βk+1 = −1, i.e., (αk+1 , βk+1 ) ∈ {(0, 0), (0, 1), (−1, 0), (1, 0), (1, 1)}. (III) If 3 k=1 β 4k− < 0, then

k+1

k+1− =1 β 4

k+1− 4k+1 + k+1 =1 α 4

3

=

4(3

k

4(4k +

=1 β 4

k

k− ) + 3β k+1

=1 α 4

k− ) + α k+1

< 0,

for all αk+1 βk+1 = −1.



(IV) Finally, we point out that 3 k=1 β 4k− < 4k + k=1 α 4k− always holds for any (α )k=1 , (β )k=1 ∈ {0, −1, 1}k with α β = −1, 1    k. In fact, we have 3

k 

β 4k− −

=1

k 

α 4k− = 2

=1

k 

β 4k− +

=1

k k   (β − α )4k−  3 4k− < 4k . =1

=1

The same situation in (I)–(IV) occur when the (β )k=1 is replaced by (β∗ )k=1 . Therefore, we can obtain the following corollary which gives an alternative description for the set P. Corollary 3.2. P =

∞

k=1 Pk





Pk+1 = n

n

( m1 , m2 )∈Pk

where Pk , k ∈ N, is increasing and  4n1 + 3β(n1 ) 4n2 + 3β ∗ (n2 )   ∗ ∗ ,  α, β(n1 ), β (n2 ) ∈ A1 with β(0), β (0) = −1 , 4m + α 4m + α

where A1 is defined as in Theorem 3.1. In the above formula, each ( nm1 , nm2 ) ∈ Pk is taken as its original form,



i.e., n1 and n2 are, respectively, of form 3 k=1 β 4k− and 3 k=1 β∗ 4k− , m of form 4k + k=1 α 4k− (so we 0 0 n 4n and 4m+1 ,m and 4m , etc.). distinguish, e.g., 4m Remark. Note that (β )k=1 is independent of (β∗ )k=1 . Thus if for (α )k=1 ∈ {0, −1, 1}k , we denote  P k,(α )k

=1

 =

k

k− =1 β 4

4k + k=1 α 4k−

3

   k k ∈ [0, 1)  (β )=1 ∈ {0, −1, 1} with β α = −1, 1    k ,

then Pk =

 (α )k=1 ∈{0,−1,1}k

 P k,(α )k

=1

 ×P k,(α )k . =1

Although Theorem 3.1 shows that each (λ1 , λ2 ) ∈ (Q × Q) ∩ ([0, 1) × [0, 1)) \ P makes the corresponding IFS {φi }5i=1 be of no complete overlap, it is still necessary to find out some classes of these parameters explicitly. To do this, we first prove a proposition. For a, b ∈ N, by a, b we denote their greatest common divisor.

Y. Zou et al. / J. Math. Anal. Appl. 340 (2008) 1422–1432

1431

Proposition 3.3. Let φˆ j : R → R, j = 1, 2, 3, be given by φˆ 1 (x) =

1 φˆ 2 (x) = (x + λ1 ) 4

x , 4

1 and φˆ 3 (x) = (x + 3), 4

where λ1 ∈ (0, 1). Let λ1 = ab ∈ Q ∩ (0, 1) with a, b = 1. If a ≡ 0 (mod 4), b ≡ 0 (mod 4) and a + b ≡ 0 (mod 4), then the IFS {φˆ i }3i=1 is of no complete overlap. Proof. Suppose that the IFS {φˆ i }3i=1 is of complete overlap. Then there are two distinct (i1 , . . . , ik ), (j1 , . . . , jk ) ∈ Σ3k for some k  1 such that k  ci =1



4

=

k  cj =1



4

,

where (c1 , c2 , c3 ) = (0, λ1 , 3). Multiplying both sides of the above equality by 4k a, we have k 

4k− ci =

=1

k 

4k− cj  ,

(13)

=1

where (c1 , c2 , c3 ) = (0, b, 3a). Note that a ≡ 0 (mod 4) and a + b ≡ 0 (mod 4) imply that 3a ≡ 0 (mod 4) and 3a ≡ b (mod 4), respectively. From (13) it follows that cik ≡ cj k (mod 4), implying ik = jk . Again from (13) we also have i = j for 1    k − 1 by the same reason. This leads to a contradiction. 2 ni where mi ∈ N, ni ∈ N, mi > ni and mi , ni  = 1, i = 1, 2. If both m1 , n1 , m1 + n1 ≡ 0 Theorem 3.4. Let λi = m i (mod 4) and m2 , n2 , m2 + n2 ≡ 0 (mod 4), then the IFS {φi }5i=1 is of no complete overlap.

Proof. Suppose the IFS {φi }5i=1 is of complete overlap. Then there are distinct (i1 , . . . , ik ), (j1 , . . . , jk ) ∈ Σ5k for some k  1 such that k  ai =1



4

=

k  aj =1



4

k  bi

and

=1



4

=

k  bj =1



4

(14)

,

where (a1 , b1 ) = (0, 0), (a2 , b2 ) = (λ1 , λ2 ), (a3 , b3 ) = (3, 0), (a4 , b4 ) = (3, 3) and (a5 , b5 ) = (0, 3). Let γ , γ ∗ : Σ5 → Σ3 be such that γ (n) = n for n = 1, 2, 3,

γ (4) = 3

and γ (5) = 1,

and γ ∗ (n) = n for n = 1, 2,

γ ∗ (3) = 1

and γ ∗ (4) = γ ∗ (5) = 3.

Then (14) can be rewritten as k  aγ (i =1

4

)

=

k  aγ (j =1

4

)

and

k  bγ ∗ (i =1

4

)

=

k  bγ ∗ (j =1

4

)

,

with (γ (i1 ), . . . , γ (ik )), (γ (j1 ), . . . , γ (jk )), (γ ∗ (i1 ), . . . , γ ∗ (ik )), (γ ∗ (j1 ), . . . , γ ∗ (jk )) ∈ Σ3k . Note that i = j implies that γ (i ) = γ (j ) or γ ∗ (i ) = γ ∗ (j ). Thus, the IFS {φˆ j }3j =1 is of complete overlap. This contradicts Proposition 3.3. 2 Acknowledgments The authors would like to thank the anonymous referees for their careful reading and helpful comments that led to the improvement of the manuscript.

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