Local time of self-affine sets of Brownian motion type and the jigsaw puzzle problem

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1 Local time of self-affine sets of Brownian motion type and the jigsaw puzzle problem (Journal of Mathematical Analysis and Applications 49 (04), pp.79-93) Yu-Mei XUE and Teturo KAMAE Abstract Let Ω [0, ] [0, ] be the solution of the set equation: Ω = k i=(ϕ Ii, ϕ Ji,τ i )(Ω), where for an interval I = [a, b] [0, ] and τ {, }, ϕ I,τ : [0, ] I is the linear map such that ϕ I, (0) = a, ϕ I, () = b, ϕ I, (0) = b, ϕ I, () = a, and {I i ; i =,, k} is a partition of [0, ] with J i = I i /. Thus, Ω is a graph of a Borel function f Ω almost surely and it is called a self-affine set of Brownian motion type. Let λ be the Lebesgue measure on [0, ] and let µ Ω = λ f Ω. The density ρ Ω = dµ Ω dλ, if it exists, is called the local time of Ω and it has been studied. It is known that dim H Ω = 3/ if ρ Ω exists. In the present study, ρ Ω is obtained by solving the so-called jigsaw puzzle on {J i, τ i ; i =,, k}, i.e., the problem of decomposing ρ Ω into a sum of its self-similar images with the support J i and the orientation τ i for i =,, k. Introduction Assume that k is an integer where k, (#) 0 = s 0 < s < < s k =, and In addition, assume further I i = [s i, s i ] (i =,, k) are intervals. J i (i =,, k) are closed intervals in [0, ] (#) such that J i = I i / (i =,, k), and τ i {, } (i =,, k), School of Mathematics and System Sciences LMIB, BeiHang University, Beijing 009, PR China (yxue@buaa.edu.cn) Advanced Mathematical Institute, Osaka City University, Japan (kamae@apost.plala.or.jp)

2 where I denotes the length b a of the interval I = [a, b]. For an interval I = [a, b] where 0 a < b and τ {, }, we define a linear map ϕ I,τ : [0, ] I by { a( x) + bx (τ = ) ϕ I,τ (x) = b( x) + ax (τ = ). We may denote ϕ I, simply by ϕ I. Therefore, ϕ I ϕ J,τ with intervals I = [a, b], J = [c, d] in [0, ] is the linear bijection [0, ] [0, ] I J such that ϕ I ϕ J, maintains the orientation, whereas ϕ I ϕ J, reverses the vertical orientation. The former map is called the upward map and the latter map is called the downward map, which are denoted by upward and downward arrows, respectively. Definition. Let Ω = Ω(I,, I k ; J,, J k ; τ,, τ k ) be the compact set in [0, ] [0, ] that satisfies the following set equation []: Ω = k (ϕ Ii ϕ Ji,τ i )(Ω). (.) We refer to Ω as a self-affine set of Brownian motion type. i= J i = I i /. τ i =..... τ = I i τ k = Figure : Set equation for Ω, where and correspond to the upward or downward linear maps, respectively, from the unit square to I i J i (i =,,, k). Self-affine sets of Brownian motion type were studied in [4], [5], and [7] from the perspective of stochastic or functional analysis. McMullen s notion of general Sierpiński carpets [3] is similar to our notion, but there are some differences, i.e., our notion has more freedom to choose the position and the

3 orientation of J i, whereas log J i / log I i is not necessarily / in McMullen s notion and there can be more than one rectangle on the same vertical line. In addition, continuous self-affine functions with I i = /k (i =,, k) were studied previously [6], where it was proved that functions with different coprime k and k that coincide are linear functions. The self-affine sets were studied in a general framework by Feng and Hu [9]. They considered the iterated function system {S,, S k }, where S i : R d R d (i =,, k) are contracting injections, and they considered the attractor K with the probability measure P π, where P is a shift-invariant measure on {, k} and π : {,, k} K such that π(σ σ ) = x with {x} = n=s σ S σn K. In this context, our case is P = (s s 0,, s k s k ), and S i = ϕ Ii ϕ Ji,τ i (i =,, k). We provide actual examples to demonstrate their general theory, where we obtain the local times by solving jigsaw puzzle problems (Section 3). For the solution ρ of the jigsaw puzzle {J i, τ i ; i =,, k}, we refer to a piecewise continuous function ρ : [0, ] [0, ) with ρdx = such that the closure of {(x, y); 0 y ρ(x)} is decomposed into self-similar images with the support J i and the orientation τ i modulo of the vertical isomorphism (Definition 4). This jigsaw puzzle is completely combinatorial and its solution is sensitive to the positions and the orientations of J i and τ i. Basic notions and preliminary results Let Ω = Ω(I,, I k ; J,, J k ; τ,, τ k ) be as stated in Definition with (#) and (#). For l =,, and i i l {,,, k} l, let ψ i i l = (ϕ Ii ϕ Ji,τ i ) (ϕ Iil ϕ Jil,τ il ) = (ϕ Ii ϕ Iil ) (ϕ Ji,τ i ϕ Jil,τ il ) be the linear map from [0, ] [0, ] to a small rectangle, such as I i i l J i i l, which is upward or downward depending on whether τ i τ il is or. It is known [] that Ω = l= i i l {,,,k} l ψ i i l (([0, ] [0, ]). We refer to ψ i i l as a fundamental map of level l. Moreover, it is easy to see that I i i l = J i i l. We refer to the rectangle I i i l J i i l as a 3

4 fundamental rectangle of level l. It is clear that there exist k l + points 0 = s l 0 < s l < < s l k l = such that I i i l = [s l j, sl j ], where j is such that i i l is the j-th element from below in the lexicographical order of {,,, k} l. The interval [s l j, sl j ] for j =,,, kl is referred to as the j-th fundamental interval of level l. Let C 0 := min s i s i, C := max s i s i. (.) i=,,k i=,,k Then, it is clear that 0 < C 0 C < and C l 0 sl j sl j Cl. The following facts are well known, but we give proofs to ensure that our study is self-contained. Fact. The 3/-dimensional Hausdorff measure of Ω satisfies H 3/ (Ω), hence dim H Ω 3/. Proof Take a sufficiently small δ > 0. Take a positive integer l such that C l < δ. Take the j-th fundamental rectangle from the left, such as I i i l J i i l, of level l and let ε j := I i i l. It is clear that ε j C l < δ. Since J i i l = ε / j, I i i l J i i l is covered by ε / j squares, such as U j,, U j where the diameters U j ε / j h are equal to ε j. Hence, k l j= ε / j h= j= U j h 3/ = k l j= ε / j h= ( ε j ) 3/ kl 3/4 (ε / j + )ε 3/ j ε j =. k l j= Since k l j= ε / j h= U j h Ω, this implies that H 3/ (Ω), and hence dim H Ω 3/. Fact. There exists a Borel function f Ω : [0, ] [0, ] such that Ω x = {f Ω (x)} holds for all but countably many x [0, ], where Ω x := {y; (x, y) Ω}. Proof Take any x [0, ] such that x / {s l 0, s l,, s l k }. l l= 4

5 Then, for any l =,,, Ω x is contained in a unique interval J i i l for some i i l {,,, k} l. Since these closed intervals are decreasing in l and J i i l C l/, it holds that #Ω x =. For a general x [0, ], we define f Ω (x) = lim max J i i l. l i i l ; x I i i l Then, f Ω is a Borel function such that Ω x = {f Ω (x)} holds for all but countably many x [0, ]. Definition. Let λ be the Lebesgue measure on [0, ]. We define a probability Borel measure λ Ω on [0, ] [0, ] and a probability Borel measure µ Ω on [0, ], so for any Borel sets U [0, ] [0, ] and V [0, ], λ Ω (U) = λ{x [0, ]; (x, f Ω (x)) U} and µ Ω (V ) = λ(f Ω V ). Then by Fact, λ Ω (Ω) = holds. Moreover, the marginal distribution of λ Ω is λ to the first coordinate and µ Ω to the second coordinate. Definition 3. If µ Ω is absolutely continuous with respect to λ, the density dµ Ω dλ is denoted by ρ Ω, which is called the local time of Ω. Let Ω = Ω(I,, I k ; J,, J k ; τ,, τ k ). prove the following results. In the present study, we Lemma. The measure µ Ω is the unique probability Borel measure µ that satisfies µ = J i µ ϕ. (.) i= In particular, µ Ω depends only on the totality of (J, τ ),, (J k, τ k ) such that J + + J k =, and it does not depend on their order or I,, I k. Lemma. Ω has the local time ρ if and only if a Borel function ρ : [0, ] [0, ) satisfies the following functional equation: ρdλ =, and (.3) ρ(x) = i= J i x Ji ρ(ϕ x) for almost all x [0, ]. (.4) Theorem. ([9]) For any Ω, one of the following three cases holds: () µ Ω is supported by a one-point set, () µ Ω is a continuous and singular measure, (3) Ω has a local time. 5

6 Theorem. If Ω has a bounded local time, then dim H Ω = 3/. Moreover, 0 < H 3/ (Ω) < holds and λ Ω is absolutely continuous with respect to H 3/. Remark. It was proved in [9] that dim H λ Ω = 3/ if dim H µ Ω =, where dim H µ was defined as inf{dim H S; µ(s) = } for a probability measure µ. However, it is not generally true that dim H λ Ω = dim H Ω (Example 3). 3 Jigsaw puzzle Definition 4. For a set S in [0, ] [0, ] and x [0, ], denote S x = {y [0, ]; (x, y) S}. We say that S is vertically convex if S is a regular set (in the sense that the closure of the interior of S is S) such that S x is either the empty set, one-point set, or a closed interval for any x [0, ]. Let U and V be vertically convex sets in [0, ] [0, ]. They are said to be vertically isomorphic if U x = V x holds for almost all x [0, ] (w.r.t. the Lebesgue measure). The vertically isomorphic class that contains U is denoted by [U]. In the following figures of the solutions of the jigsaw puzzle, we use elements in [U], which we denote by [U]. Definition 5. Let ρ : [0, ] [0, ) be a piecewise continuous function in the sense that it is either left continuous or right continuous at any point in [0, ], where limits from both sides exist at any point in (0, ), and it is continuous except for finitely many points. For a piecewise continuous function ρ : [0, ] [0, ) with ρdλ =, denote Γ ρ = the closure of {(x, y); 0 y ρ(x)}. Then, Γ ρ is vertically convex and ρ(x) = (Γ ρ ) x holds for almost all x [0, ]. Let J = [a, b] be an interval in [0, ]. Denote J Γ ρ = {(ϕ J (x), J y); (x, y) Γ ρ }, J Γ ρ = {(ϕ J, (x), J y); (x, y) Γ ρ } (see Figure ). Let {J i ; i =,, k} be a family of intervals in [0, ] such that k i= J i = and τ τ k {, } k. In the above, ρ is called the solution of the jigsaw puzzle for {J,, J k ; τ,, τ k } if the vertically convex sets Λ i [J τ i Γ ρ ] (i =,, k) exist such that: () Γ ρ = k i= Λ i, and () the interiors of Λ i and Λ j are disjoint for any i, j with i j. In this case, (Γ ρ ) x = k i= Λ x holds for almost all x [0, ]. Theorem 3. Let ρ : [0, ] [0, ) be a piecewise continuous function with ρdλ =. For any Ω = Ω(I,, I k ; J,, J k ; τ,, τ k ), ρ is the local time of Ω if and only if ρ is the solution of the jigsaw puzzle for {J,, J k ; τ,, τ k }. 6

7 J Γ ρ /3 J /3 J Γ ρ /3 J /3 Figure : J Γ ρ and J Γ ρ for J = [/3, /3] and ρ with ρ(x) = 6x (0 x /3), = 3 3x (/3 < x ) Example. If J = [0, /3], J = [/3, /3], J 3 = [/3, ], then the local time ρ = ρ Ω exists and it satisfies { 4x 0 x / ρ(x) = 4 4x / < x for any τ, τ, τ 3 {, } since it is symmetric and the solution of the jigsaw puzzle is as shown in Figure 3. In a special case where Ω = Ω([0, 4/9], [4/9, 5/9], [5/9, ]; [0, /3], [/3, /3], [/3, ];,, ), Ω = {(t, f Ω (t)); t [0, ]} holds with a continuous function f Ω, which is called a deterministic Brownian path and this was studied in [7] (Figure 4). Example. If J = J = [0, /], J 3 = J 4 = [/, ], and τ, τ, τ 3, τ 4 {, } are arbitrary, then the local time ρ = ρ Ω exists and it satisfies ρ(x) = for almost all x [0, ] since it is the solution to the jigsaw puzzle for {J, J, J 3, J 4,,,, }, as shown in Figure 5. Example 3. If J = [0, /], J = J 3 = J 4 = [/, ] and τ = τ = τ 3 = τ 4 =, then µ Ω is a singular measure such that µ Ω ([m/ n, (m + )/ n )) = n i=0 θ(m i ) for any n =,, and m = 0,,, n, where m = n i=0 m i i with m i {0, } and θ(0) = /4, θ() = 3/4. It was proved in [3] that dim H Ω = log ( + 3), while it was proved in [9] that dim H λ Ω = (3/8) log 3. 7

8 is vertically J3 Γ ρ isomorphic to [J3 Γ ρ] x x 4/3 [J Γ ρ] [J 3 Γ ρ] [J Γ ρ] /3 Figure 3: The solution ρ of the jigsaw puzzle for {J, J, J 3 ;,, } with J = [0, /3], J = [/3, /3], J 3 = [/3, ] Figure 4: Ω([0, 4/9], [4/9, 5/9], [5/9, ]; [0, /3], [/3, /3], [/3, ];,, ) 8

9 [J Γ ρ] [J 4 Γ ρ] / [J Γ ρ] [J 3 Γ ρ] / Figure 5: Solution of the jigsaw puzzle in Example. [J4 Γ ρ] [J 3 Γ ρ] [J Γ ρ] [J Γ ρ] / Figure 6: Solution of the jigsaw puzzle for {J, J, J 3, J 4 ;,,, } with J = [0, /], J = J 3 = J 4 = [/, ]. Example 4. If J = J = J 3 = J 4 = [/, ] and τ = τ = τ 3 = τ 4 =, then µ Ω is supported by the one-point set {}, since Ω = [0, ] {}. Example 5. Let J = [0, /], J = J 3 = J 4 = [/, ], as in Example 3, but let τ = τ =, τ 3 =, τ 4 =. Then, µ Ω is an absolutely continuous measure since ρ(x) = x (x [0, ]) is the solution of the jigsaw puzzle shown in Figure 6. In this case, f Ω is a continuous function, which is related to the Rudin-Shapiro sequence [4]. 9

10 4 Proof of the results Proof of Lemma By (.) and Fact, it holds that ϕ Ji,τ i (f Ω (x)) = f Ω (ϕ Ii (x)) for any i =,,, k and for all but countably many x [0, ]. Moreover, since Ii λ = I i λ ϕ I i for any i =,,, k, we have λ = k i= I i λ ϕ I i. Hence, we have µ Ω = λ f Ω = I i λ ϕ I i = i= I i λ f Ω i= ϕ = i= f Ω J i µ Ω ϕ Therefore, µ Ω satisfies (.). If a probability Borel measure µ satisfies (.), then by applying (.) l times, we have µ = i,,i l = J i i l µ ϕ J i i l,τ i τ il, (4.) Let g : [0, ] [0, ] be any continuous function such that g(x) g(y) C x y for any x, y [0, ] with a constant C. Then, since µ ϕ J i i l,τ i τ il is supported by the interval J i i l with a length of at most C l/ gdµ ϕ J i i l,τ i τ il for any probability Borel measures µ and ν. satisfy (.), and hence (4.), then gdµ gdν i,,i l = CC l/ J i i l i,,i l = gdν ϕ J i i l,τ i τ il CC l/ gdµ ϕ J i i l,τ i τ il J i i l = CC l/., we have Therefore, if both µ and ν gdν ϕ J i i l,τ i τ il Thus, letting l, we have gdµ = gdν. Since this holds for any Hölder continuous function g, we have µ = ν. Thus, the probability Borel measure µ that satisfies (.) is unique. 0

11 Proof of Lemma If Ω has the local time ρ, then by (.), we have ρ = dµ Ω dλ = = i= i= J i dµ Ω ϕ dλ J i Ji dµ Ω ϕ dλ ϕ = i= J i Ji ρ ϕ. Conversely, if ρ satisfies (.3) and (.4), then the probability Borel measure µ on [0, ] with dµ(x) = ρ(x)dλ(x) satisfies (.). Thus, ρ is the local time of Ω. Assume that Ω has a bounded local time ρ Ω such that ρ Ω (x) C for any x [0, ]. Lemma 3. Let I be a fundamental interval and J be any closed interval. Then, it holds that λ Ω (I J) C I / J. Proof Let I J be the fundamental rectangle that corresponds to I. To estimate λ Ω (I J) from the above, we may assume that J J since λ Ω (I ([0, ] \ J )) = 0. Let ϕ I ϕ J,τ be the fundamental mapping where τ = or correspond to whether J is upward or downward. Since (ϕ I ϕ J,τ ) maps I J to a rectangle [0, ] [a, b] with b a = J I / and we have λ Ω (I J) = I λ Ω ((ϕ I ϕ J,τ ) (I J)) = I λ Ω ([0, ] [a, b]), λ Ω (I J) = I µ Ω ([a, b]) I C J I / = C I / J. Lemma 4. For any rectangle U, which is a product of the intervals on the x-axis and the y-axis, it holds that λ Ω (U) C / 0 C U 3/, where U is the diameter of U. Proof We may take U = [a, b] [c, d] U such that d c b a U. Take at most two fundamental intervals I i (i =, ) (one of them can be the empty set) such that [a, b] is contained in I I and I i C0 (b a) (i =, ). Since U (I [c, d]) (I [c, d]), by Lemma 3, we have λ Ω (U) λ Ω (I [c, d]) + λ Ω (I [c, d]) C (C 0 (b a))/ (b a) C / 0 C U 3/.

12 Proof of Theorem For any δ > 0 and rectangles U i R (i =,, ), which are products of the intervals on the x-axis and y-axis such that U i < δ (i =,, ) and i= U i Ω, we have i= U i 3/ (/)C / 0 C i= λ Ω (U i ) (/)C / 0 C λ Ω(Ω) = (/)C / 0 C by Lemma 4. Hence, H 3/ (Ω) (/)C / 0 C. Since H3/ (Ω) was proved in Fact, we have 0 < H 3/ (Ω) < and dim H Ω = 3/. By Lemma 4, Λ Ω is absolutely continuous with respect to H 3/. Assume that a function ρ : [0, ] [0, ) is the solution of the jigsaw puzzle for {J,, J k ; τ,, τ}. Thus, the vertically convex sets Λ i (i =,,, k) exist such that for each i =,,, k, Λ i is vertically isomorphic to J τ i i Γ ρ and ()() in Definition 5 hold. Then, we have (Γ ρ ) x = (Λ i ) x = i= i= (J τ i i Γ ρ) x for almost all x [0, ]. Therefore, we have ρ(x) = i= J i x Ji ρ(ϕ x) for almost all x [0, ]. Hence, by Lemma, ρ is the local time of Ω. Λ Λ Λ Λ Γ 3 Λ 3. Figure 7: Piling up Γ 3 above Λ Λ Conversely, if ρ is the local time of Ω, then we have ρ(x) = i= J i x Ji ρ(ϕ x)

13 for almost all x [0, ]. Since ρ is piecewise continuous, the functions J i x Ji ρ(ϕ x) of x for i =,,, k are also piecewise continuous. Moreover, we have (Γ ρ ) x = k i= (J τ i i Γ ρ) x for almost all x [0, ]. We denote Γ i = J τ i i Γ ρ. We define the vertically convex and mutually disjoint (in their interiors) subsets Λ,, Λ k of Γ ρ so Λ i is vertically isomorphic to Γ i for i =,,, k. Let Λ = Γ. We pile up the amount of Γ above Λ and define Λ as the piled part. Next, we pile up the amount of Γ 3 above Λ Λ and define Λ 3 as the piled part (Figure 7). In this manner, we can define a partition Λ,, Λ k of Γ ρ, as in Definition 5, with respect to {J,, J k ; τ,, τ}. Hence, ρ is the solution of the jigsaw puzzle for {J,, J k ; τ,, τ}. 5 Further examples Example 6. (Dai, Feng, and Wang [8]) Let J = [0, /], J = [( )/, ] and τ, τ {, } be arbitrary. Then, ρ(x) = 4+3 x 0 x < x ( x) < x is the solution of the jigsaw puzzle shown in Figure 8. Example 7. For any 0 a, let J = [0, /], J = J 3 = [a, a + (/)], J 4 = [/, ] and either τ = and τ = τ 3 = τ 4 = or τ 3 = and τ = τ = τ 4 =. Then, { x/a (0 x a) ρ(x) = ( x)/( a) (a < x ), where ρ(0) = if a = 0 and ρ() = if a = is the solution of the jigsaw puzzle shown in Figure 9. If a =, this is the same as Example 5. Example 8. If J = [0, /], J = [/4, 3/4], J 3 = J 4 = [/, ] and τ =, τ = τ 3 = τ 4 =, then the local time ρ := ρ Ω exists and it satisfies { /3 (0 x /) ρ(x) = 4/3 (/ < x ). This is the solution of the jigsaw puzzle for {J, J, J 3, J 4 ;,,, }, which is shown in Figure 0. 3

14 Example 9. If J = J = J 3 = J 4 = [/, ] and only two of τ, τ, τ 3, τ 4 are and the others are, then the local time ρ := ρ Ω exists and it satisfies { 0 (0 x /) ρ(x) = (/ < x ) This is the solution of the jigsaw puzzle for {J, J, J 3, J 4 ; τ, τ, τ 3, τ 4 } (Figure ). Example 0. Let J = [0, /3], J = [/9, 4/9], J 3 = [/3, /3], J 4 = J 5 = [4/9, 7/9], J 6 = [5/9, 8/9], J 7 = J 8 = J 9 = [/3, ], and τ =, τ =, τ 3 =, τ 4 =, τ 5 =, τ 6 =, τ 7 = τ 8 = τ 9 = (τ 4 =, τ 5 = is also possible) Then, the local time ρ := ρ Ω exists and it satisfies / (0 x /3) ρ(x) = (/3 < x /3) 3/ (/3 < x ) since it is the solution of the jigsaw puzzle for {J, J, J 3, J 4, J 5, J 6, J 7, J 8, J 9 ; τ, τ, τ 3, τ 4, τ 5, τ 6, τ 7, τ 8, τ 9 }, as it is shown in Figure. Example. If J = [0, 4/7], J = [/7, 4/7], J 3 = [/7, 5/7], J 4 = [3/7, 5/7], J 5 = [3/7, ], then the local time ρ := ρ Ω exists and it satisfies { 4x 0 x / ρ(x) = 4 4x / < x for any τ, τ, τ 3, τ 4, τ 5 {, }, since ρ is the solution of the jigsaw puzzle for {J, J, J 3.J 4.J 5 ; τ, τ, τ 3, τ 4, τ 5 }, as it is shown in Figure 3. 6 Open problems Problem : Is there a self-affine set of Brownian motion type with an unbounded local time? Problem : Is there a self-affine set of Brownian motion type with a bounded local time that is not piecewise linear? 4

15 + + [J Γ ρ] [J Γ ρ] a b c d Figure 8: a =, b =, c =, d = [a, ] and J = [0, d], J = [J 3 Γ ρ ] [J Γ ρ] [J Γ ρ] [J4 Γ ρ] / Figure 9: Solution of the jigsaw puzzle for {J, J, J 3, J 4 ;,,, } with J = [0, /], J = J 3 = [a, a + (/)], J 4 = [/, ]. 5

16 4/3 [J 4 Γ ρ] /3 [J Γ ρ] [J Γ ρ] [J 3 Γ ρ] / Figure 0: Solution of the jigsaw puzzle for {J, J, J 3, J 4 ;,,, } with J = [0, /], J = [/4, 3/4], J 3 = J 4 = [/, ]. [J 3 Γ ρ] [J 4 Γ ρ] [J Γ ρ] [J Γ ρ] / 3/4 Figure : Solution of the jigsaw puzzle for {J, J, J 3, J 4 ;,,, } with J = J = J 3 = J 4 = [/, ]. 6

17 3/ [J7 Γ ρ] [J 9 Γ ρ] / [J Γ ρ] [J Γ ρ] [J 3 Γ ρ] [J 4 Γ ρ] [J 5 Γ ρ] [J 6 Γ ρ] [J8 Γ ρ] /3 Figure : Solution of the jigsaw puzzle in Example 0. [J [J Γ 4 Γ ρ] ρ] [J Γ ρ] [J 3 Γ ρ] 4/7 [J 5 Γ ρ] Figure 3: Solution of the jigsaw puzzle for {J, J, J 3, J 4, J 5 ;,,,, } with J = [0, 4/7], J = [/7, 4/7], J 3 = [/7, 5/7], J 4 = [3/7, 5/7], J 5 = [3/7, ]. 7

18 Acknowledgments: The authors thank Professor Wen Zhi-Ying (Tsinghua University) for his useful suggestions. The authors also thank Beijing University of Aeronautics and Astronautics for inviting one of the authors and providing the opportunity for collaboration. This research was supported partly by NSFC (No. 904). The quality of English of the paper is improved by the suggestions of the language editing staffs. References [] K.J. Falconer, The dimension of self-affine fractals II, Math. Proc. Camb. Phil. Soc. (99), pp [] J.E. Hutchinson, Fractal and self-similarity, Indiana Univ. Math. J. 30 (98), pp [3] C. McMullen, The Hausdorff dimension of general Sierpiński carpets. Nagoya Math. J. 96 (984), pp. -9 [4] T. Kamae and M. Keane, A class of deterministic self-affine processes, Japan J. Applied Math. 7- (990), pp [5] T. Bedford and T. Kamae, Stieltjes integration and stochastic calculus with respect to self-affine functions, Japan J. Industrial & Applied Math. 8-3 (99), pp [6] T. Kamae, A characterization of self-affine functions, Japan J. Applied Math. 3- (986), pp [7] T.K amae, Stochastic analysis based on deterministic Brownian motion, Israel J. Math. 5 (00), pp [8] X-R. Dai, D-J. Feng, and Y. Wang, Classification of refinable splines, Constr. Approx. 4 (006), pp [9] D-J. Feng and H. Hu, Dimension theory of iterated function systems, Communications on Pure and Applied Mathematics, Vol. LXII (009), pp