Ruelle operator for infinite conformal iterated function systems

Ruelle operator for infinite conformal iterated function systems

Chaos, Solitons & Fractals 45 (2012) 1521–1530 Contents lists available at SciVerse ScienceDirect Chaos, Solitons & Fractals Nonlinear Science, and ...

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Chaos, Solitons & Fractals 45 (2012) 1521–1530

Contents lists available at SciVerse ScienceDirect

Chaos, Solitons & Fractals Nonlinear Science, and Nonequilibrium and Complex Phenomena journal homepage: www.elsevier.com/locate/chaos

Ruelle operator for infinite conformal iterated function systems Xiao-Peng Chen a,b, Li-Yan Wu c,d, Yuan-Ling Ye c,⇑,1 a

Beijing International Center for Mathematical Research, Peking University, Beijing 100871, PR China School of Mathematical Sciences, University of Adelaide, South Australia 5005, Australia School of Mathematical Sciences, South China Normal University, Guangzhou 510631, PR China d Computer Engineering Department, Guangdong College of Industry and Commerce, Guangzhou 510510, PR China b c

a r t i c l e

i n f o

Article history: Received 18 September 2011 Accepted 6 September 2012 Available online 16 October 2012

a b s t r a c t   m Let X; fwj gm ð2 6 m < 1Þ be a contractive iterated function system (IFS), where j¼1 ; fpj gj¼1 X is a compact subset of Rd . It is well known that there exists a unique nonempty compact S set K such that K ¼ m j¼1 wj ðKÞ. Moreover, the Ruelle operator on C(K) determined by the IFS   m m X; fwj gj¼1 ; fpj gj¼1 ð2 6 m < 1Þ has been extensively studied. In the present paper, the Ruelle operators determined by the infinite conformal IFSs are discussed. Some separation properties for the infinite conformal IFSs are investigated by using the Ruelle operator.  2012 Elsevier Ltd. All rights reserved.

1. Introduction The (finite) conformal iterated function systems (IFSs) have been studied in various ways. Fan and Lau [4] considered the eigenmeasure by using the Ruelle operators. They extended Schief’s idea [19] to prove the equivalence of the open set condition (OSC) and the strong open set condition (SOSC). Peres et al. [17] and Ye [22] showed that for a conformal IFS satisfying the Hölder condition with the invariant set K, both OSC and SOSC are equivalent to 0 < Hs(K) < 1, where Hs(K) is Hausdorff measure of K and s is the zero point of the pressure function P(). Deng and Ngai [3], and then Lau et al. [8] extended the open set condition to some weak separation condition and discussed the Hausdorff dimension of the invariant set corresponding to the conformal IFS. The infinite IFSs of similarity maps were studied by Moran [13]. Moran [13] extended the classical results for finite systems of similitudes satisfying the open set condition to the infinite case. Mauldin and Urban´ski [14] estimated the dimension and Hausdorff measure of limit sets of infinite conformal IFSs. They found a



⇑ Corresponding author. E-mail addresses: [email protected] (X.-P. [email protected] (L.-Y. Wu), [email protected] (Y.-L. Ye). 1 The research was supported by the NSF of China (11171121).

way to determine the Hausdorff dimension of limit sets under some conditions such as the basic open set was connected and the system satisfying the open set condition (OSC). One of the applications of infinite conformal IFSs is the continued fractions [5]. However, it is a notoriously difficult task to get some estimates for the Hausdorf dimension of the limit set of an infinite iterated function system with overlaps [11]. Solomyak [20] studied several examples of iterated function systems with overlaps. Mihailescu and Urban´ski [12] interpreted the iterated function systems with overlaps as skew-products with overlaps in fibers. In the present paper we deal with the Ruelle operators under some weak separation conditions. Moreover, we obtain a bounded distortion property under the assumption that the basic open sets are locally connected, which is a weaker assumption than the connectedness of basic open sets  by Mauldin and  Urban´ski [14]. m Let X; fwj gm be an IFS, where wj’s are conj¼1 ; fpj gj¼1 tractive maps on a compact subset X  Rd and pj’s are positive Dini functions on X [4]. Then there exists a unique compact set K, which is invariant under the IFS, i.e. m [ wj ðKÞ: j¼1

Chen),

0960-0779/$ - see front matter  2012 Elsevier Ltd. All rights reserved. http://dx.doi.org/10.1016/j.chaos.2012.09.001

The Ruelle operator on C(K) is defined by

1522

Tðf ÞðxÞ ¼

X.-P. Chen et al. / Chaos, Solitons & Fractals 45 (2012) 1521–1530 m X

pj ðwj ðxÞÞf ðwj ðxÞÞ;

f 2 CðKÞ;

ð1:1Þ

j¼1

where C(K) is the space of all continuous functions on X. Some results about the Ruelle operator can be found in the literatures [2,6,7,9,10] and the references therein. In particular, Ruelle [18] applied the Ruelle operators to study expanding systems. Mauldin and Urban´ski [15,16] considered the Ruelle operators for the symbolic space over an infinite alphabet with Hölder continuous potentials. Recently the Ruelle operators defined by nonexpansive IFSs were considered (see e.g. [7] and [10]). We generalize the Ruelle operators to the infinite conformal IFSs with the Dini continuous potential functions. Then we obtain the PF-property (Theorem 1.1) by using the Ruelle operators defined newly. One of the difficulty is the limit sets of the infinite conformal IFSs may fail compact. The open set condition (OSC) and the strong open set condition (SOSC) are useful separation properties for studying the (finite) conformal IFSs. We study the separation properties for the infinite conformal IFSs. Moran [13] showed that self-similar sets generated by a countable system of similitudes may not be s-sets even if the open set condition (OSC) is satisfied. Szarek and Wedrychowicz [21] showed that for infinite IFSs, the OSC does not imply the SOSC. It is necessary to consider a weak separation property for the infinite conformal IFSs. We define a weak separation property for the infinite conformal IFSs. Let fsj g1 j¼1 be an infinite conformal IFS on an open set U0. We say that the infinite conformal IFS fsj g1 j¼1 satisfies the finite open set condition if for any integer n, there is a nonempty open set Un  U0 such that si(Un)  Un for any i 6 n and si(Un) \ sj(Un) = ; for any i,j 6 n with i – j. Such a weak separation property has been used to study the Hausdorff dimension of invariant set [13]. We give a sufficient condition for the infinite conformal IFS fsj g1 j¼1 to satisfy the finite (strong) open set condition (see Definition 2.1). Our basic results are:   1 Theorem 1.1. Let X; fsj g1 j¼1 ; fpj gj¼1 be an infinite conformal uniformly Dini IFS, and let J be the limit set of the IFS fsj g1 j¼1 and . = .(T) be the spectral radius of T. Then there exists a unique 0 < h 2 CðJÞ and a unique probability measure l 2 MðJÞ such that

Th ¼ .h;

T



l ¼ .l; < l; h >¼ 1: n

Moreover, for every f 2 CðJÞ; . T f converges to h in the supremum norm, and for every n 2 MðJÞ; .n T n n converges weakly to l. n

fsj g1 j¼1

be an infinite conformal IFS, and let Theorem 1.2. Let J be the limit set of the IFS. If the Hausdorff measure Ha ðJÞ ¼ Ha ðJÞ > 0, where a is the Hausdorff dimension of J, then the system fsj g1 j¼1 satisfies the finite strong open set condition. One method for the study of Ruelle operators is to establish the semiconjugacy of an IFS to a symbolic system [4]. We do not consider the ‘‘semiconjugacy’’ of the infinite IFS and symbolic system over an infinite alphabet in the present paper. Observe that when the system is infinite,

to prove Theorem 1.1 we need to assume that the potential functions are uniformly Dini continuous and some basic properties of the limit sets. Under the assumption that both J and J are a-sets, we can prove Theorem 1.2 by making use of Schief’s idea [19]. This paper is organized as follows. In Section 2, we present some concepts for infinite conformal IFSs and some elementary facts about the Ruelle operators determined by infinite conformal IFSs. In Section 3, we study the Ruelle operators defined by the infinite conformal IFSs. In Section 4, by a technical result, we investigate some separation properties for the infinite conformal IFSs and give some simple examples. 2. Preliminaries In the paper, we always assume that X is a locally connected compact subset of Rd . Let w be a self-map on X. We call w a conformal map if there exists an open set U0 such that X  U0, w is continuously differentiable on U0 and for each x 2 U0, w0 (x) is a self-similar matrix, i.e. a scale multiple of an orthogonal matrix. We call the funtions fsj g1 j¼1 an infinite conformal iterated function system (IFS), if each sj(1 6 j < 1) is a conformal map. The family of functions fpj g1 j¼1 is said to be uniformly Dini continuous on U0 proR1 vided that 0 aðtÞ dt < þ1, where t

aðtÞ :¼ sup sup j log pj ðxÞ  log pj ðyÞj jP1

and

x;y2U 0 jxyj6t

P1

j¼1 pj ðxÞ

< 1 on U0. Under this assumption, we call   1 an infinite conformal unithe triple X; fsi g1 i¼1 ; fpj gj¼1 formly Dini IFS. Throughout the paper, we always assume that the infinite conformal IFS fsj g1 j¼1 satisfies the following conditions: (i) Each sj: X ? X is one-to-one and the system fsj g1 j¼1 is uniformly contractive, i.e., there exists 0 < s < 1 such that for any 1 6 j < 1, jsj(x)  sj(y)j 6 sjx  yj, "x,y 2 X. (ii) Each sj is conformal on U0(X) with 0 < inf js0j ðxÞj 6 sup js0j ðxÞj < 1. x2U 0 x2U 0 n o1 0 is uniformly Dini continuous on U0. (iii) jsj ðÞj j¼1

fsj g1 j¼1

be an infinite conformal IFS. Denote Let I = {1, 2, 3,    },

In ¼ fu ¼ u1 u2    un : uj 2 I; 81 6 j 6 ng; I1 ¼ fv ¼ v 1 v 2    : v j 2 I; 81 6 j < 1g; and let I ¼

S

In . For an arbitrary w = w1w2    wk 2 Ik, we

nP1

let jwj = k and sw ¼ sw1  sw2     swk . S For any w 2 I⁄ I1 and positive integer n, if n 6 jwj, we denote by wjn the word w1w2    wn. By the contractiveness of the system fsj g1 j¼1 , we have

lim diamfswjn ðXÞg ! 0;

n!1

8w 2 I 1 :

T This implies the set 1 n¼1 swjn ðXÞ is a singleton. Hence we define a continuous map p:I1 ? X by

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fpðwÞg :¼

1 \

swjn ðXÞ:

n¼1

We endow I1 with the metric d(w,u) = en(w,u), where n(w,u) is the largest integer n such that wjn = ujn. Define 1 [\

J ¼ pðI1 Þ ¼

swjn ðXÞ:

ð2:1Þ

w2I1 n¼1

We check that the set J satisfies the following equation.



[ sj ðJÞ:

ð2:2Þ

j2I

The set J is said to be the limit set of the system [14]. For an   1 infinite conformal uniformly Dini IFS X; fsi g1 i¼1 ; fpj gj¼1 , we define an operator T : CðJÞ ! CðJÞ by

Tf ðxÞ ¼

1 X pj ðxÞf ðsj ðxÞÞ:

ð2:3Þ

j¼1

T is called the Ruelle operator of the system. The dual operator T⁄ on the measure space MðJÞ is given by

T  lðEÞ ¼

1 Z X s1 ðEÞ j

j¼1

pj ðxÞdlðxÞ for any Borel set E # J ð2:4Þ

psj ðxÞ ¼ pj1 ðsj2  sj3      sjn ðxÞÞ    pjn1 ðsjn ðxÞÞpjn ðxÞ: By induction we have for any positive integer n,

X psj ðxÞf ðsj ðxÞÞ: jjj¼n

Let . = .(T) be the spectral radius of T. Since T is a positive operator, we obtain kTn1k = kTnk and 1

1

. ¼ n!1 lim kT n kn ¼ lim kT n 1kn : n!1 For the sake of convenience, we assume that

x2J

Bðx; jÞ  U 0 :

metric on the space X is given by absolute difference. Then the space X is compact on the real line. The point set {0} is a closed subset of the space X. Now we construct an infinite conformal IFS on the space  i 1 X. Denote the disjoint sets N n ¼ nþ1 ; 21n ; n ¼ 0; 1; 2; . . .. Let 2

V1 = N0 [ N2 [ . . . [ N2n . . . , V2 = N1 [ N5 [ . . . [ N4n+1 . . . , V3 = N3 [ N11 [ . . . [ N8n+3 . . . 2 V1  V2, in the same way we select Vi such that Vi \ Vj = ; for any i – j and Vi includes infinite points of the space X, i,j P 1. Let ¼ fs : X ! X : j P 1g be an infinite conformal IFS, fsj g1 j j¼1

(

1 y ¼ 1; . . . ; 2j1 ; 0;

0;

2 V j \ X; y ¼ 21j ; . . . ;

1 j P 1, such that sj(y) < y for any y ¼ 2i1 for all i P 1. Extend the infinite conformal IFS fsj g1 ¼ fsj : X !   j¼1 X : j P 1g to the open set U 0 :¼ 1; 32 by a similar   construction on Nj1 and let the map sj : 1; 32 [ ð1; 0Þ # f0g; j ¼ 1; 2;   . Then for any positive integer n, there   exists an open set U n :¼ 0; 21n on the space U0 such that

sj(Un)  Un and si(Un) \ sj(Un) = ; for each 1 6 i,j 6 n and i – j. Thus the system satisfies the finite open set condition. 1 , then If the open sets in U0 include the point j1 2     1 1 ¼ si j1 ¼ 0 for any j < i. So from the construction sj j1 2

Let J be the limit set of the system fsj g1 j¼1 . For any r > 0, S denote BðJ; rÞ ¼ Bðx; rÞ. Choose j > 0 such that

[

n o Example 2.2. Let the set X :¼ 1; 12 ; 14 ; . . . ; 21n ; . . . ; 0 . The

2

we can not find an open set U  U0 such that si(U)  U and si(U) \ sj(U) = ; for each 1 6 i,j < 1 and i – j.

diamðU 0 Þ ¼ supfjx  yj : x; y 2 U 0 g 6 1:

X 0 :¼

Definition 2.1. We say an infinite conformal IFS fsj g1 j¼1 satisfies the finite open set condition if for any integer n, there is a nonempty open set Un  U0 such that sj(Un)  Un for each 1 6 j 6 n and si(Un) \ sj(Un) = ; for each 1 6 i,j 6 n and i – j. The finite strong open set condition holds if T furthermore Un J – ; for all n P 1. Obviously the open set condition implies the finite open set condition. We present a simple example to show that the converse is not true.

sj ðyÞ ¼

(For the finite case see e.g. [1]). For j = j1j2    jn, we let

T n f ðxÞ ¼

To study the separation properties of the infinite conformal IFS, we introduce the finite open set condition as follows.

3. Ruelle operator

ð2:5Þ

x2X

From the contractiveness of sj’s, we conclude that sj(X0)  X0 for any j 2 I. Hence we assume that

BðJ; jÞ  X: We say the infinite conformal IFS fsj g1 j¼1 satisfies the open set condition (OSC) if there exists a nonempty bounded open set U  U0 such that

sj ðUÞ  U and si ðUÞ \ sj ðUÞ ¼ ; for any i; j 2 I and i – j: Such an open set U is called a basic open set for the system T fsj g1 J – ;, then the system fsj g1 j¼1 . If moreover U j¼1 is said to satisfy the strong open set condition (SOSC).

Proposition 3.1. Let fsj g1 j¼1 be an infinite conformal IFS, and let J be the limit set of the system fsj g1 j¼1 . Then

fsu ðxÞ : u 2 I g ¼ J;

8x 2 J:

Proof. For any x 2 J and u 2 I⁄,

su ðxÞ 2 su ðJÞ  su ðJÞ  J: This implies that fsu ðxÞ : u 2 I g  J. For any y 2 J, by (2.1) and the contractiveness of the 1 system fsj g1 such that j¼1 , there exists w 2 I

lim swjn ðxÞ ¼ y;

n!1

8x 2 J:

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X.-P. Chen et al. / Chaos, Solitons & Fractals 45 (2012) 1521–1530

This implies that J  fsu ðxÞ : u 2 I g. Hence

T n0 f ðxÞ ¼

X

psu ðxÞf ðsu ðxÞÞ P psu ðxÞf ðsu0 ðxÞÞ > 0: 0

juj¼n0



J  fsu ðxÞ : u 2 I g: The assertion follows.

h

Proposition 3.2. Let the operator T be defined as in (2.3). Then (i) 1 6 maxx2J .n T n 1ðxÞ; 8n > 0; (ii) if there exist k > 0 and 0 < h 2 CðJÞ such that T h = kh, then k = . and there exist A,B > 0 such that

A 6 .n T n 1ðxÞ 6 B;

8x 2 J;

8n > 0:

This implies that T is irreducible. For the dimension of the eigensubspace, we suppose that there exist two independent strictly positive .-eigenfunctions h1 ; h2 2 CðJÞ. Without loss of generality we assume that 0 < h1 6 h2 and h1(x0) = h2(x0) for some x0 2 J. Then h = h2  h1(P0) is a .-eigenfunction of T and h(x0) = 0. It follows that Tnh(x0) = .nh(x0) = 0, which contradicts the irreducibility of T. Hence the dimension of the .-eigensubspace is at most 1. The strict positivity of h follows directly from the irreducibility of T. h

Proof. (i) Suppose it is not true, then there exists a k such that kTk1k < .k. Hence 1

1

1

. ¼ ð.ðT k ÞÞk 6 kT k kk ¼ kT k 1kk < .; which it is a contradiction. (ii) Let a1 ¼ minx2J hðxÞ; a2 ¼ maxx2J hðxÞ. Then for all x 2 J and all n > 0

a1 hðxÞ kn n 0< 6 ¼ T hðxÞ 6 kn T n 1ðxÞ: a2 a2 a2

  1 be an infinite conProposition 3.4. Let X; fsi g1 i¼1 ; fpj gj¼1 formal uniformly Dini IFS and . be the spectral radius of T. Then (i) there exists 0 < h 2 CðJÞ such that Th = .h; (ii) there exist A,B > 0 such that

A 6 .n T n 1ðxÞ 6 B;

8x 2 J; 8n > 0:

ð3:1Þ Proof.

Since Tn is a positive operator, we conclude that for all n>0

a1 6 kn kT n k: a2

UðtÞ ¼

a2 k T 1ðxÞ 6 : a1 n n

ð3:2Þ

This implies that for all n > 0

a2 : a1

aðsj tÞ; for all t 2 ½0; 1;

ð3:3Þ

where a(t) = supjP1maxjxyj6tj logpj(x)  log pj(y)j. The uniformly Dini continuity of fpj g1 j¼1 shows that U() is well-defined and U(t) < + 1, for all t 2 [0,1]. Let C þ ðJÞ :¼ ff 2 CðJÞ : f > 0g, and set

F :¼ ff 2 C þ ðJÞ : f ðxÞ 6 f ðyÞeUðjxyjÞ g; 1 n

Hence . ¼ lim kT n k ¼ k. This, together with (3.1) and n!1 (3.2), implies that

A :¼

1 X j¼0

Similarly for all x 2 J and all n > 0

kn kT n k 6

(i) Define

a1 a2 6 .n T n 1ðxÞ 6 B :¼ ; a2 a1

For any function f 2 F and any x; y 2 J, we have

Tf ðxÞ ¼

8x 2 J;

8n > 0:

  1 Proposition 3.3. Let X; fsi g1 i¼1 ; fpj gj¼1 be an infinite conformal uniformly Dini IFS. Then the Ruelle operator T is irreducible and

dimfh 2 CðJÞ : Th ¼ .h; h P 0g 6 1; if h P 0 is a .-eigenfunction of T, then h > 0. Proof. For any given f 2 CðJÞ with f P 0 and f X 0, let V ¼ fx 2 J : f ðxÞ > 0g. For any x 2 J, by Proposition 3.1, there exists a u0 2 I⁄ such that su0 ðxÞ 2 V. Let n0 = ju0j. Then

1 X pj ðxÞf ðsj ðxÞÞ

ð3:4Þ

j¼1



We call the operator T : CðJÞ ! CðJÞ irreducible if for any non-trivial, non-negative f 2 CðJÞ and for any x 2 J, there exists n > 0 such that Tnf(x) > 0.

8x; y 2 J:

6

1 X pj ðyÞf ðsj ðyÞÞeaðjxyjÞþUðsjxyjÞ j¼1

¼ Tf ðyÞeaðjxyjÞþUðsjxyjÞ 6 Tf ðyÞeUðjxyjÞ : So T(F)  F. We define L : F ! CðJÞ by

Lf ðxÞ :¼

Tf ðxÞ : kTf k

By (3.4) we deduce

Lf ðxÞ 6 Lf ðyÞeUðjxyjÞ : So L(F)  F. Let

F0 ¼ F

\ feUð1Þ 6 f 6 1g:

It is easy to show that F0 is a convex compact subset of CðJÞ, and L(F0)  F0. The Schauder fixed point theorem yields an

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X.-P. Chen et al. / Chaos, Solitons & Fractals 45 (2012) 1521–1530

h 2 F0 such that Lh = h. This implies that Th = kThkh. We conclude kThk = . by Proposition 3.2. Hence Th = .h. (ii) The proof comes from Proposition 3.2 immediately. h

We claim that ~f is a constant function and limn!1 kPn f  ~f k ¼ 0. For this we let sðgÞ ¼ minx2J gðxÞ. P1 Since j¼1 qj ðxÞ ¼ 1 for all x 2 J. It is easy to see that sð~f Þ 6 sðP~f Þ and

From the above basic properties, we are now ready to prove our first result by modified Lau and Ye [10]. We include the details here for the sake of completeness.

sðf Þ 6 sðPf Þ 6    6 sð~f Þ:

Proof of Theorem 1.1. By Proposition 3.4, there exists 0 < h 2 CðJÞ and constants B P A > 0 such that Th = .h and n n

A 6 . T 1ðxÞ 6 B;

8x 2 J; 8n > 0:

ð3:5Þ

þ

Let C ðJÞ :¼ ff 2 CðJÞ : f > 0g, and let

D ¼ ff 2 C þ ðJÞ : there exists c > 0 such that f ðxÞ

~f ðs ðy ÞÞ ¼ gð~f Þ :¼ max~f ðxÞ; j n

8j 2 I ; jjj ¼ n:

x2J

n!1

where U is given in (3.3). It is easy to show that e :¼ S1 Dk , hence D e is dense in C þ ðJÞ. D#D k¼1 For any f 2 Dk and x; y 2 J, we have 1 X pj ðxÞf ðsj ðxÞÞ j¼1 1 X pj ðyÞf ðsj ðyÞÞekaðjxyjÞþkUðsjxyjÞ

n!1

sð~f Þ ¼ n!1 lim ~f ðsjn ðxn ÞÞ ¼ ~f ðzÞ ¼ lim ~f ðsjn ðyn ÞÞ ¼ gð~f Þ: n!1 We deduce ~f ðxÞ sð~f Þ is a constant function. By (3.6) and the dual version for gð~f Þ, we have lim kPn f  ~f k ¼ 0. n!1 In particular, by taking f = h1, we see that Pn(h1) converges uniformly, then .nTn1 converges uniformly. To prove that

lim k.n T n 1  hk ¼ 0;

j¼1

¼ Tf ðyÞe

n!1

¼ Tf ðyÞekUðjxyjÞ :

we let

kaðjxyjÞþkUðsjxyjÞ

It follows that TDk  Dk. For any g 2 C þ ðJÞ; f 2 Dk and n > 0, we have n

j.n T n gðxÞ  .n T n gðyÞj 6 k.n T n f kj1  TT n ff ðyÞ j þ 2k.n T n kkf  gk ðxÞ 6 Bkf kðekUðjxyjÞ  1Þ þ 2kf  gk;

for any x; y 2 J. By the assumptions on D and U, we deduce that for any g 2 C þ ðJÞ; f.n T n gg1 n¼1 is a bounded equicontinuous sequence. For any f 2 CðJÞ, we choose a > 0 such that f + a > 0. Then from the above process, it follows that the sequences 1 n n f.n T n ðf þ aÞg1 n¼1 and f. T agn¼1 are bounded equicontinuous, hence the sequence f.n T n f g1 n¼1 is also bounded equicontinuous. We let

pj ðxÞhðsj ðxÞÞ .hðxÞ

and define an operator P : CðJÞ ! CðJÞ by

Pf ðxÞ ¼

Similarly there exists yn 2 J such that

Hence

Dk :¼ ff 2 C þ ðJÞ : f ðxÞ 6 f ðyÞekUðjxyjÞ g;

qj ðxÞ ¼

8j 2 I ; jjj ¼ n:

z :¼ lim sjn ðxn Þ ¼ lim sjn ðyn Þ 2 J:

Then D is dense in C þ ðJÞ. Let

6

~f ðs ðx ÞÞ ¼ sð~f Þ; j n

We assume jn = 11    1, jjnj = n. Then

6 f ðyÞecjxyj g:

Tf ðxÞ ¼

ð3:6Þ

By taking the limit, we have sðP~f Þ 6 sð~f Þ, hence sðP~f Þ ¼ sð~f Þ. For any P n > 0, we select xn 2 J such that Pn ~f ðxn Þ ¼ sðP n ~f Þ. Then jjj¼n qsj ðxn Þ ¼ 1 implies that

1 X qj ðxÞf ðsj ðxÞÞ:

fn ðxÞ ¼

n1 1X .i T i 1ðxÞ: n i¼0

From (3.5), it follows that ffn g1 n¼1 is bounded by A and B and is an equicontinuous subset of CðJÞ. By Arzelà–Ascoli theo~ 2 CðJÞ such that rem, we assume that there exists a h ~ lim kfn  hk ¼ 0. Then n!1

~ ¼ lim kTfn  .fn k 6 lim ~  .hk kT h 6 lim

.

n!1 n

.

n!1 n

n!1

k1  .n T n 1k

ð1 þ BÞ ¼ 0;

~ and h ~ P A > 0. Proposition 3.3 implies that ~ ¼ .h i.e. T h ~ ¼ ch for some c > 0. Without loss of generality, we ash ~ This implies that lim k.n T n 1  hk ¼ 0. sume that h ¼ h. n!1 1 So lim kPn ðh Þ  1k ¼ 0. n!1 Now we define a function t : CðJÞ ! R; ht; f i ¼ sð~f Þð¼ ~f ðxÞ; x 2 JÞ. Then t is a bounded linear functional on CðJÞ, and ht,1i = 1, h t,h1i = 1. From

ht; Pf i ¼ sðP~f Þ ¼ sð~f Þ;

j¼1

For any f 2 CðJÞ, we have Tnf = .nh Pn (f  h1). This is a bounded equicontinuous implies that fPn f g1 n¼1 sequence in CðJÞ for every f 2 CðJÞ. We know from the Arzelà–Ascoli theorem that there exists ~f 2 CðJÞ and a subni ~ sequence fPni f g1 i¼1 such that lim kP f  f k ¼ 0. i!1

we obtain P⁄t = t. Let l : CðJÞ ! R be defined by hl,fi = h t,f h1i. Then hl,1i = ht,h1i = 1 and l is a probability measure. So T⁄l = .l and hl,hi = h t,1i = 1. Hence for any f 2 CðJÞ; .n T n f converges to hl,fih in the supremum norm. Also it follows that for every n 2 MðJÞ; .n T n n converges weakly to hn,h il.

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X.-P. Chen et al. / Chaos, Solitons & Fractals 45 (2012) 1521–1530

By Proposition 3.3, we conclude that the eigen function h is unique. For the uniqueness of the eigen measure, we note that if r 2 MðJÞ satisfies T⁄r = . r and hr,hi = 1, then for every f 2 CðJÞ,

Since Bðy0 ; dy0 Þ conclude

hr; f i ¼ lim h.n T n r; f i ¼ lim hr; .n T n f i ¼ hr; hl; f ihi

Consequently,

n!1

rj jx  yj 6 jsj ðxÞ  sj ðyÞj 6 c1 r j jx  yj:

¼ hl; f i:

(iii) By the contractiveness of the infinite conformal IFS, we select an integer k0 such that

h

jsj ðxÞ  sj ðyÞj 6 d;

4. Separation properties

ri ¼ inf js0i ðxÞj;

jsj ðxÞ  sj ðyÞj 6 Rj jx  yj 6 c3 r j jx  yj:

Ri ¼ supjs0i ðxÞj:

x2U 0

x2U 0





1 Lemma 4.1. Let X; fsi g1 i¼1 ; fpj gj¼1 be an infinite conformal

uniformly Dini IFS. Then

Let j be as given by (2.5). We 0 < e < c1 2  minfj; dg. From the assumption on X,

take

BðJ; c2 eÞ  X:

ð4:5Þ

For j 2 I , let Gj = sj (B(J,e)). By (4.3) and (4.5), we conclude that for any x 2 J,

8j 2 I  ;

c1 1 r i r j 6 r ij 6 c 1 r i r j ;



8i; j 2 I ;

ð4:1Þ

Bðsj ðxÞ; c1 2 er j Þ  sj ðBðx; eÞÞ  Bðsj ðxÞ; c 2 er j Þ:

ð4:2Þ

It follows that

(ii) there exist c2 P c1 and d > 0 such that for any x,y 2 X with jx  yj 6 d,

c1 2 rj 6





(i) there exists c1 > 1 such that

Rj 6 c1 r j ;

8jjj > k0 :

The choice of the d (the Lebesgue number) and the selfsimilar property of sj implies that

For any i 2 I⁄, let

J i ¼ si ðJÞ;

sj ðBðx0 ; dx0 ÞÞ is convex and connected, we

juj ðsj ðxÞÞ  uj ðsj ðyÞÞj 6 r 1 j jsj ðxÞ  sj ðyÞj:

n!1

This implies that r = l.

T

jsj ðxÞ  sj ðyÞj 6 c2 rj ; jx  yj

8j 2 I  ;

ð4:3Þ

[   [   B J j ; c1 B sj ðxÞ; c1 Bðx; eÞÞ 2 er j ¼ 2 er j  Gj ¼ sj ð x2J

8j 2 I ; jjj > k0 :

ð4:4Þ

ð4:7Þ

x2J

[  Bðsj ðxÞ; c2 er j Þ ¼ BðJ j ; c2 erj Þ: x2J

For any two compact subsets E,F of Rd , we define

(iii) there exist c3 P c2 and k0 such that for any x,y 2 X,

jsj ðxÞ  sj ðyÞj 6 c3 rj jx  yj;

ð4:6Þ

jEj ¼ supfjx  yj : x; y 2 Eg; DðE; FÞ ¼ inffjx  yj : x 2 E; y 2 Fg; dðE; FÞ ¼ inffe : E  BðF; eÞ; F  BðE; eÞg:

Proof. (i) For each j 2 I, let pj ðÞ ¼ js0j ðÞj and still denote (3.3) by U(t). For any x,y 2 U0, j 2 I⁄, jjj = n, we have

j logjs0j ðyÞj  logjs0j ðxÞk 6

n   X   logjs0ji ðyiþ1 Þj  logjs0ji ðxiþ1 Þj i¼1

6

We say u,v 2 I⁄ are comparable if there exists w 2 I⁄ such that u = vw or v = uw. S Denote Fn = {1,2,    ,n}. Let F n ¼ kP1 F kn , where

F kn ¼ fu ¼ u1 u2    uk : ui 2 F n ; 1 6 i 6 kg: For any n 2 I and 0 < r 6 1, we let

n X

Q n ðrÞ ¼ fv ¼ v 1 v 2    v m 2 F n : rv < r 6 rv 1 v 2 v m1 g:

i¼1

Let k0 be as given by Lemma 4.1(iii). For any w 2 F n with jwj = k0 + 1, we define

aðsni jy  xjÞ 6 Uð1Þ < 1:

where yi ¼ sji      sjn ðyÞ, yn+1 = y, y 2 U0. Consequently, we deduce (4.1). And the chain rule yields (4.2). (ii) For any x 2 X, there exists dx > 0 such that B(x,dx)  U0. Since X is locally connected, we assume that T B(x,dx) X is connected. Let d be the Lebesgue number of {B(x,dx)}x2X. Then for any x,y 2 X, if jx  yj 6 d, there exists x0 2 X such that x; y 2 Bðx0 ; dx0 Þ. For such x and y, we know that sj ðxÞ; sj ðyÞ 2 Bðy0 ; dy0 Þ for some y0 2 X. The self-similar property of sj implies that

jsj ðxÞ  sj ðyÞj 6 Rj jx  yj 6 c1 r j jx  yj: On the other hand, T sj ðBðx0 ; dx0 ÞÞ. Then

R1 6 ju0j ðxÞj 6 r 1 j ; j

let

0 0 uj ðxÞ :¼ s1 j ðxÞ; 8x 2 Bðy ; dy Þ

8x 2 Bðy0 ; dy0 Þ

\

In ðwÞ ¼ fv 2 Q n ðjGw jÞ : J v

Gw – ;g:

n

Suppose I (w) is defined, and then for any 1 6 j 6 n, we define

In ðjwÞ ¼ M

[

N;

where M = {jv:v 2 In(w)} and

N ¼ fv 2 Q n ðjGjw jÞ : v 1 – j; J v

\

Gjw – ;g:

Lemma 4.2. For any fixed n, there exist constants kn and ðnÞ c4 > 0 such that

1 sj ðBðx0 ; dx0 ÞÞ:

\

ðnÞ c4

6

ru ðnÞ 6 c4 ; rv

8u 2 In ðv Þ; jv j P kn :

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X.-P. Chen et al. / Chaos, Solitons & Fractals 45 (2012) 1521–1530

Proof. Let k0 be as given in Lemma 4.1(iii). For any positive integer n, we select an integer kn > k0 such that

minfjuj : u 2 In ðv Þ and jv j P kn g > k0 :

(ii) Let k denote the Lebesgue measure on Rd . For any positive integer n, by definition of the finite open set condition, there exists a nonempty bounded open set Un such that si(Un)  Un," i 6 n, and

(1) If v1 – u1, by the construction of N, we have u 2 Qn(jGvj). Then

It follows that

kðU n Þ P

n n Z X X kðsi ðU n ÞÞ ¼

ðnÞ 1

ru < jGv j 6 ru1 u2 um1 6 c1 ðr Þ r u ;

i¼1

where r(n) = min16j6n{rj}. It follows from (4.3) that

P cd 1

Hence

ru < jGv j 6 c2 er v þ jJ v j 6 c3 ð2e þ jJjÞr v : From the arguments above, we conclude that there exists an > 0 such that

1 ru 6 6 an : an r v

r u0 6 an : rv0

h

P For any t P 0, let wðtÞ ¼ i2I Rti , and let h = inf{t:w(t) < 1}. For any s > h, we define the Ruelle operator T s : CðJÞ ! CðJÞ by 1 X js0j ðxÞjs  f ðsj ðxÞÞ:

Let .(Ts) denote the spectral radius of Ts. Proposition 4.4. Assume wðhÞ ¼ limþ wðtÞ > 1 and t!h wð1Þ ¼ lim wðtÞ < 1, then there exists a unique a > h such t!1 that .(Ta) = 1.

Proof. We define the function

Together with (4.2), this implies that

6

6 cd1 .

j¼1

where vl+1 – ul+1. From the construction of M, it follows that u0 2 In(v0 ). Similarly to the case of (1), we deduce that

Let

d i¼1 Ri

T s f ðxÞ ¼

v ¼ v 1 v 2    v l v lþ1    v n :¼ v 1 v 2    v l v 0 ; u ¼ v 1 v 2    v l ulþ1    un :¼ v 1 v 2    v l u0 :

ðnÞ c4

P1

v1 = u1, we write

an c1

n X Rdi kðU n Þ:

i¼1

Note that

 2 1

js0i ðxÞjd dkðxÞ

n X Rdi 6 cd1 :

ðnÞ 1 c1 2 er v 6 jGv j 6 c 1 ðr Þ r u :



Un

Since k(Un) > 0, we have

Hence

a1 n 6

i¼1

i¼1

c1 2 er v 6 jGv j:

(2) If

8 i; j 6 n; i – j:

si ðU n Þ \ sj ðU n Þ ¼ ;;

We consider the following two cases:

1

.ðT s Þ ¼ n!1 lim kT ns 1kn :

ru 6 an c21 : rv

¼ an c21 . Then the result follows.

h

Then it is continuous and strictly decreasing on (h, +1). Since

T ns 1ðxÞ ¼

X js0j ðxÞj;

8x 2 J:

jjj¼n

Proposition 4.3. Let fsj g1 j¼1 be an infinite conformal IFS. (i) The system fsj g1 j¼1 satisfies the finite strong open set condition if it satisfies the finite open set condition; P1 d 1 d (ii) i¼1 Ri 6 c 1 if the system fsj gj¼1 satisfies the finite open set condition.

This implies that

kT ns 1k ¼ sup

x2J jjj¼n

From Lemma 4.1,

X X s X X r sj 6 js0j ðxÞjs 6 Rj 6 c1 r sj : jjj¼n

Proof.

X js0j ðxÞjs :

jjj¼n

jjj¼n

jjj¼n

It follows that

(i) For any positive integer n, from the assumption that the system fsj gnj¼1 satisfies the open set condition, the result of Schief [19] implies that the system fsj gnj¼1 satisfies the finite strong open set condition. Hence there exists a nonempty bounded open set Un such that si(Un)  Un," i 6 n, and

si ðU n Þ \ sj ðU n Þ ¼ ;; T

8 i; j 6 n; i – j:

Moreover U n J F n – ;, where J F n is the invariant set of the T system fsi gni¼1 . It is easy to see that J F n  J. Hence Un J – ;.

1 lim kT ns 1kn n!1

¼ lim

n!1

X

!1n rsj

:

jjj¼n

Let

fn ðsÞ :¼

X 1 log r sj : n jjj¼n

For any 0 6 k 6 1, s1,s2 2 inequality,

(h, + 1), by the Hölder’s

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X.-P. Chen et al. / Chaos, Solitons & Fractals 45 (2012) 1521–1530

fn ðks1 þ ð1  kÞs2 Þ ¼ 1n log

X ks þð1kÞs 2 rj 1

P a~ 0 ~ :¼ hj . Then 0 < h ~ 2 CðJÞ and hðxÞ ~ Let h ¼ 1 J j¼1 jsj ðxÞj hðsj ðxÞÞ for all x 2 J. Hence

jjj¼n

X ks ð1kÞs 2 ¼ log rj 1 rj

1 Z X

1 n

jjj¼n

X s 6 log rj 1

!k

1 n

jjj¼n

X s rj 2

!ð1kÞ

a ~ hðxÞdH ðxÞ P

Jj

j¼1

jjj¼n

(h, +1) is continuous, we confirm that .(Ts) is continuous. If s 2 (h, +1), then there exists i0 2 I such that rsi0 P rsj for every j 2 I. We have

X X X r sþt 6 r sj r tj 6 r sj c1 r nt i0 ; j jjj¼n

jjj¼n

Z X 1 ~ ðxÞÞdHa ðxÞ js0j ðxÞja hðs j J

¼ kfn ðs1 Þ þ ð1  kÞfn ðs2 Þ:

n!1

ð4:8Þ

jjj¼n

for t > 0. We check that .(Ts) is strictly decreasing by using (4.8). Combining the assumption, we see that there exists a unique a > h such that .(Ta) = 1. h

a ~ hðxÞdH ðxÞ

J

¼

We conclude that fn() is convex. This implies that P lim 1n log jjj¼n r sj is convex. Since the convex function on

Z

¼

j¼1

1 Z X

a ~ hðxÞdH ðxÞ:

Jj

j¼1

(The last equality follows from Lemma 4.6.) This implies T that Ha(Ji Jj) = 0 for any i – j. It follows immediately that T a H (Ju Jv) = 0 for any incomparable u,v 2 I⁄. h Proof of Theorem 1.2. Let c1, c2, c3 and d be as given in Lemma 4.1. And let i > 0 satisfy the condition: 21 ca 1 > i. There exists a covering V1,    ,Vn of J such that

V :¼

n [ V i  J;

d0 :¼ DðJ; V c Þ < d;

ð4:9Þ

i¼1

In the following we always let a be the constant such that .(Ta) = 1. We know that a is the Hausdorff dimension of J [14]. Proposition 4.5. Ha ðJÞ 6 Ha ðJÞ < 1. Proof. For any fixed z 2 J, by (4.4), there exist k0 and jjj > k0 such that jJ j j 6 cjs0j ðzÞj, so we have

X a X jJ j j 6 ca js0j ðzÞja : jjj¼n

From Theorem 1.1, we know that there exists a unique h 2 CðJÞ such that

lim

X js0j ðzÞja ¼ lim T na 1ðzÞ ¼ hðzÞ:

This implies Ha ðJÞ 6 Ha ðJÞ < 1.

h

Lemma 4.6. [4] Let a > 0 be a constant. Let w be conformal and invertible and D be a Borel subset in the domain of w such that 0 < Ha(D) < 1. Then we have the following change of variable formula:

Z

a

jw0 ðxÞja dH ðxÞ:

D

Lemma 4.7. Suppose 0 < H (J) < 1. Then

Ha ðJ u

J v Þ ¼ 0 for any incomparable u; v 2 I :

n o1 Proof. Note that js0j ðÞj is uniformly Dini continuous. j¼1

Since Ta has spectral radius 1, by Theorem 1.1, there exists 0 < h 2 CðJÞ such that

hðxÞ ¼

1 X js0j ðxÞja hðsj ðxÞÞ; j¼1

0 1 0 that dðJ u ; J v Þ P c1 2 d r u . Otherwise dðJ u ; J v Þ < c2 d r u , and by (4.3) and (4.9),

This implies that J v  J v  su ðVÞ. By Lemma 4.7,

ð1 þ eÞHa ðJ u Þ < Ha ðJ u Þ þ Ha ðJ v Þ ¼ Ha ðJ u

J v Þ 6 Ha ðsu ðVÞÞ:

(The first inequality follows from Lemma 4.6.) This implies that e < ca1 i, which contradicts the choice of e. Hence, there exists d0 > 0 such that for any w 2 I⁄,

8 u; v 2 In ðwÞ; u – v :

dðJ u ; J v Þ P d0 rw ; Let d1 ¼ such that

[



ðnÞ 3c3 c4

1

ð4:10Þ

d0 . Then there is a finite set Z  J

Bðx; d1 Þ:

x2Z

For any w 2 I⁄ with jwj > kn, by (4.10) there exists x 2 J such that

jsu ðxÞ  sv ðxÞj P d0 r w ;

8 u; v 2 In ðwÞ; u – v :

For such x there exists a z such that jx  zj < d1. By (4.4) and the choice of kn in Lemma 4.2,

jsu ðxÞ  su ðzÞj 6

8x 2 J:

[

erau Ha ðJÞ 6 eHa ðJu Þ 6 ðc1 ru Þa Ha ðV n JÞ < ðc1 ru Þa iHa ðJÞ:

J a

\

For any given u,v 2 In(w), we let J u ¼ su ðJÞ and J v ¼ sv ðJÞ. We assume that Ha(Ju) 6 Ha(Jv). Then for any given e > 0 satisfying ca1 i < e < 1, we have eHa(Ju) 6 Ha(Jv). We claim

Thus

n!1

Ha ðwðDÞÞ ¼

Ha ðVÞ < ð1 þ iÞHa ðJÞ 6 ð1 þ iÞHa ðJÞ:

0 DðJ u ; su ðVÞc Þ P c1 2 d ru :

jjj¼n

n!1 jjj¼n

and

1 d0 rw ; 3

jsv ðxÞ  sv ðzÞj 6

1 d0 r w : 3

This implies that

jsu ðzÞ  sv ðzÞj P d0 r w =3:

ð4:11Þ

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X.-P. Chen et al. / Chaos, Solitons & Fractals 45 (2012) 1521–1530

For each z 2 Z, let

T nz

n

n

¼ fu 2 I ðwÞ : 9v 2 I ðwÞ such that ð4:11Þ holdsg:

Together with (4.11), this implies that

In ðwÞ ¼

[

T nz :

jy  y1 j < c2 

1 c1 e e  riuw0 6 2 riuw0 ; 2 2 2c2

jy  y2 j < c2 

1 c1 e e  rjuw0 6 2 rjuw0 : 2 2 2c2

Hence

z2Z

DðJ iuw0 ; J j Þ < c1 2 er iuw0 ;

For each z, the sets

  Bðsu ðzÞ; d0 rw =6Þ : u 2 T nz are disjoint and contained in BðGw ; jJ u j þ d0 r w =6Þ. By (4.4) and Lemma 4.2, there exist x 2 J and c > 0 such that

which contradicts (4.13). And it is known that Un This completes the proof. h

T

J – ;.

Example 4.8. Let X = [0,1] and let fsj g1 j¼1 ¼ fsj : X ! X : j P 1g be an infinite conformal IFS consisting of similarities

BðGw ; jJ u j þ d0 rw =6Þ  Bðx; crw Þ: We deduce that there exists k such that max ]T nz 6 k, then

If w0 2 I⁄ with jw0j > kn and

sj(x) = 22jx + 2j  22j. Thus js0j ðÞj ¼ 22j and the system satisfies the OSC. It is easy to see that J is compact and a ¼ 12. So from Theorem 1.1 we know that there exist h 2 C(J) and l 2 M(J) such that

]In ðw0 Þ ¼ sup ]In ðwÞ;

Th ¼ h;

z2Z

n

]I ðwÞ 6 ]Z 

max]T nz z2Z

6 ]Z  k:

jwjPkn

T  l ¼ l;

< l; h >¼ 1:

ð4:12Þ

S 5 Example 4.9. Let X ¼ ½1; 2 and let fsj g1 j¼1 ¼ fsj : X 2 ! X : j P 1g be an infinite conformal IFS consisting of sim-

By the maximality of w0, we only need prove the following.

ilarities sj(x) = 22jx + 1. Thus js0j ðÞj ¼ 22j and the system

we prove that n

n

I ðiw0 Þ ¼ fij : j 2 I ðw0 Þg;

8i2

F n :

satisfies the OSC. It is easy to see that J is compact and

fil : l 2 In ðwÞg  In ðiw0 Þ:

a ¼ 12. So from Theorem 1.1 we know that there exist

The definition of In(w0) implies that

In ðjw0 Þ  fjv : v 2 In ðw0 Þg;

h 2 C(J) and l 2 M(J) such that

j ¼ 1; 2;    ; n: Th ¼ h;

We conclude that

fil : l 2 In ðw0 Þg  In ðiw0 Þ:

n

Jl ¼ fJ l : l 2 Q ðjvw0 jÞ; l1 ¼ lg: Then Jl is a cover of Jl. Since v1 – l, then l R In(vw0). By the T construction of N, we have J l Gvw0 ¼ ;. Hence by (4.7)

DðJ l ; J vw0 Þ P c1 2 er vw0 ; which implies

l – v 1:

ð4:13Þ S

Now we let Un ¼ where Gu ¼    u2F n . We claim that the set Un satisfies the su B J; 21 c2 2 e

Guw0 ,

condition of finite SOSC. Indeed, Un is an open set and for each i 2 Fn,

si ðU n Þ ¼

[    [  si Guw0 ¼ Giuw0  U n :

u2F n

< l; h >¼ 1:

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For any fixed 1 6 l 6 n, v = v1    vn 2 I⁄, v1 – l, we consider the family

DðJ l ; J vw0 Þ P c1 2 er vw0 ;

T  l ¼ l;

u2F n

T We claim that for each i,j 2 Fn, i – j, si(Un) sj(Un) = ;.  Otherwise, there exist u; v 2 F n such that T Giuw0 Gjvwbf 0 – ;. Assume that

r iuw0 P r jvw0 : Let y be in the intersection. Then there exist y1 2 J iuw0 and y2 2 J jvw0 such that

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