Báo cáo "LOCAL DIMENSION OF FRACTAL MEASURE ASSOCIATED WITH THE (0, 1, a) - PROBLEM: THE CASE a = 6 " pptx

VNU. JOURNAL OF SCIENCE, Mathematics - Physics. T.XXI, N 0 1 - 2005 LOCAL DIMENSION OF FRACTAL MEASURE ASSOCIATED WITH THE (0, 1,a) -PROBLEM:THECASEa =6 Le Xuan Son, Pham Quang Trinh Vinh Univ ersity, Nghe An Vu Hong Thanh Pedagogical College of Nghe An Abstract. Let X 1 ,X 2 , be a sequence of independent, identically distributed(i.i.d) random variables each taking values 0, 1,a with equal probabilit y 1/3.Letµ be the probability measure induced by S =  ∞ n=1 3 −n X n .Letα(s) (resp.α(s), α(s))denote the local dimension (resp. lower, upper local dimension) of s ∈ supp µ,andlet α = sup{α(s):s ∈ supp µ}; α = inf{α(s):s ∈ supp µ} E = {α : α(s)=α for some s ∈ supp µ}. In the case a =3, E =[2/3, 1], see [6]. It was hoped that this result holds true with a =3k, for any k ∈ N. We prove that it is not the case. In fact, our result shows that for k =2(a =6), α =1, α =1− log(1+ √ 5)−log 2 2log3 ≈ 0.78099 and E =[1− log(1+ √ 5)−log 2 2log3 , 1]. 1. Introduction Let X 1 ,X 2 , be a sequence of i.i.d random variables each taking values a 1 ,a 2 , ,a m with probability p 1 ,p 2 , ,p m respectively. Then the sum S = ∞  n=1 ρ n X n is well defined for 0 < ρ < 1. Let µ be the probability measure induced by S, i.e., µ(A)=Prob{ω : S(ω) ∈ A}. It is known that the measure µ is either purely singular or absolutely continuous. In 1996, Lagarias and Wang[8] showed that if m is a prime number, p 1 = p 2 = = p m =1/m and a 1 , ,a m are int egers then µ is absolutely if and only if {a 1 ,a 2 , ,a m } forms a complete system(modm), i.e., a 1 ≡ 0(modm),a 2 ≡ 1(modm), ,a m ≡ m −1(modm). An intriguing case when m =3,p 1 = p 2 = p 3 = 1 3 and a 1 =0,a 2 =1,a 3 =3, known as the ”(0, 1, 3) −P roblem”, is of great interest and has been investigated since the last decade. Typeset by A M S-T E X 31 32 Le Xuan Son, Pham Quang Trinh, Vu Hong Thanh Let us recall that for s ∈ supp µ the local dimention α(s)ofµ at s is defined by α(s)= lim h→0 + log µ(B h (s)) log h , (1) provided that the limit exists, where B h (s) denotes the ball centered at s with radius h.If the limit (1) does not exist, we define the upper and lower local dimension, denoted α(s) and α (s), by taking the upper and lower limits respectively. Observe that the local dimension is a function defined in the supp µ.Denote α =sup{α(s):s ∈ supp µ} ; α =inf{α(s):s ∈ supp µ}; and E = {α : α(s)=α for some s ∈ supp µ} be the attainable values of α(s), i.e., the range o f α. In [6], T. Hu, N. Nguyen and T. Wang have investigated the ”(0, 1, 3)- Problem” andshowedthatE =[2/3, 1]. In this note we consider the following general problem. Problem. Describle the local dimension for the (0, 1,a)- problem, where a ∈ N is a natural number. Note that the local dimension is an important characteristic of singular measures. For a =3k +2 the measure µ is absolutely continuous, therefore we only need to consider the case a =3k or a =3k +1,k ∈ N.Fora =3k it is conjectured that the local dimension is still the same as a =3,itmeansthatE =[2/3, 1]. Our aim in this note is to disprove this conjecture. In fact, our result is the following: Main Theorem. For a =6wehave α =1, α =1− log(1+ √ 5)−log 2 2log3 and E =[1− log(1+ √ 5)−log 2 2log3 , 1]. The proof of the Main Theorem will be given in Section 3. The next section we establish some auxiliary results used in the proof of the Main T heorem. 2. Auxiliary Results Let X 1 ,X 2 , be a sequence o f i.i.d random variables e ach taking values 0, 1, 6 with equal probability 1/3. Let S =  ∞ n=1 3 −n X n , S n =  n i=1 3 −i X i be the n-partial sum of S, and let µ, µ n be the probability measures induced by S, S n respectively. For any s =  ∞ n=1 3 −n x n ∈ supp µ, x n ∈ D: = {0, 1, 6},lets n =  n i=1 3 −i x i be it’s n-partial sum. It is easy to see that for any s n ,s  n ∈ supp µ n , |s n − s  n | = k3 −n for some k ∈ N,andfor any interval between two consecutive points in supp µ n there exists at least one point in supp µ n+1 .Let s n  = {(x 1 ,x 2 , ,x n ) ∈ D n : n  i=1 3 −i x i = s n }. Then we have µ n (s n )=#s n 3 −n for every n, (2) where # A denotes the cardinality of set A. Local dimension of fractal measure associated with 33 Two sequences (x 1 ,x 2 , ,x n )and(x  1 ,x  2 , ,x  n )inD n are said to be equivalent, denoted by (x 1 ,x 2 , ,x n ) ≈ (x  1 ,x  2 , ,x  n )if  n i=1 3 −i x i =  n i=1 3 −i x  i .Thenwehave 2.1.Claim. Assume that (x 1 ,x 2 , ,x n )and(x  1 ,x  2 , ,x  n )inD n .If(x 1 ,x 2 , ,x n ) ≈ (x  1 ,x  2 , ,x  n )andx n >x  n then x n =6,x  n =0. Proof.Since(x 1 ,x 2 , ,x n ) ≈ (x  1 ,x  2 , ,x  n ), we have 3 n−1 (x 1 −x  1 )+3 n−2 (x 2 − x  2 )+ +3(x n−1 − x  n−1 )+x n − x  n =0, which implies x n −x  n ≡ 0 (mod 3), and by virtue of x n >x  n we have x n −x  n = 6. Hence x n =6,x  n = 0. The claim is proved. Consequece 1. a) Let s n+1 ∈ supp µ n+1 and s n+1 = s n + 1 3 n+1 ,s n ∈ supp µ n .Wehave #s n+1  =#s n  for evrery n. b) For any s n ,s  n ∈ supp µ n such that s n − s  n = 1 3 n ,wehave #s n  a #s  n . Proof. Observe that a) is a directive consequence of Claim 2.1. b) It is easy to see that if s n − s  n = 1 3 n ,thens n = s n−1 + 1 3 n and s  n = s n−1 + 0 3 n , where s n−1 ∈ supp µ n−1 . Therefore from a) it follows that #s n  =#s n−1  a #s  n . Remark 1. Observe that from |s n −s  n | = k3 −n , it follows that if s n+1 ∈ supp µ n+1 and s n+1 = s n + 1 3 n+1 then s n+1 can not be represented in the forms s n+1 = s  n + 0 3 n+1 , or s n+1 = s  n + 6 3 n+1 , where s n ,s  n ,s  n ∈ supp µ n . Thus, for any s n+1 ∈ supp µ n+1 has a t most two representa- tions throught poin ts in supp µ n . 2.2. Claim. Assume that s n ,s  n ∈ supp µ n ,n≥ 3. Then we have a) If s n − s  n = 1 3 n , then there are three following cases for the representation of s n ,s  n : 1. s n = s n−1 + 1 3 n ; s  n = s n−1 + 0 3 n , 2. s n = s n−2 + 6 3 n−1 + 1 3 n ; s  n = s  n−2 + 1 3 n−1 + 6 3 n , or 3. s n = s n−2 + 0 3 n−1 + 1 3 n ; s  n = s  n−2 + 1 3 n−1 + 6 3 n , where s n−1 ∈ supp µ n−1 and s n−2 ,s  n−2 ∈ supp µ n−2 . b) If s n −s  n = 2 3 n then there are four following cases for the representation of s n ,s  n : 1. s n = s n−2 + 0 3 n−1 + 6 3 n ; s  n = s  n−2 + 1 3 n−1 + 1 3 n , 2. s n = s n−2 + 1 3 n−1 + 0 3 n ; s  n = s  n−2 + 0 3 n−1 + 1 3 n , 3. s n = s n−2 + 6 3 n−1 + 6 3 n ; s  n = s  n−2 + 1 3 n−1 + 1 3 n , or 4. s n = s n−2 + 1 3 n−1 + 0 3 n ; s  n = s  n−2 + 6 3 n−1 + 1 3 n , 34 Le Xuan Son, Pham Quang Trinh, Vu Hong Thanh where s n−1 ∈ supp µ n−1 and s n−2 ,s  n−2 ∈ supp µ n−2 . Proof.Lets n =  n i=1 3 −i x i and s  n =  n i=1 3 −i x  i , x i ,x  i ∈ D. a) If s n −s  n = 1 3 n then 3 n−1 (x 1 −x  1 )+3 n−2 (x 2 −x  2 )+ +3(x n−1 −x  n−1 )+x n −x  n = 1, which implies s n − s  n ≡ 1(mod3),hencex n − x  n =1orx n −x  n = −5. For x n − x  n =1wehavex n =1,x  n =0.Thisisthecase1.a. For x n − x  n = −5wehavex n =1,x  n =6and 3 n−2 (x 1 − x  1 )+ +3(x n−2 − x  n−2 )+x n−1 −x  n−1 =2, which implies s n−1 − s  n−1 ≡ 2 (mod 3), hence x n−1 − x  n−1 =5(x n−1 =6,x  n−1 =1)or x n−1 − x  n−1 = −1(x n−1 =0,x  n−1 = 1) are the cases 2.a, 3.a respectively. b) The proof is similar to a). Consequence 2. Let s n <s  n <s  n be three arbitrary consecutive points in suppµ n . Then either s  n − s n or s  n − s  n is not 1 3 n . The following fact provides a useful form ula for calculating the local dimention. 2.3. Proposition. For s ∈ supp µ,wehave α(s)= lim n→∞ | log µ n (s n )| n log 3 , provided that the limit exists. Otherwise, by taking the upper and lower limits respectively we get the formulas for α(s) and α(s). We first prove: 2.4. Lemma. For any two consecutive points s n and s  n in supp µ n we have µ n (s n ) µ n (s  n ) a n. Proof. By (2) it is sufficient to show that #s n  #s  n  a n. Wewillprovetheinequalityby induction. Clearly the inequality holds for n = 1. Suppose that it is true for all n a k. Let s k+1 >s  k+1 be two arbitrary consecutive points in supp µ n+1 .Write s k+1 = s k + x k+1 3 k+1 ,s k ∈ supp µ k ,x k+1 ∈ D. We consider the following cases for x k+1 : Case 1. x k+1 =6. s k+1 = s k + 6 3 k+1 = s k + 2 3 k .Lets  k ∈ supp µ k be the smallest value larger than s k . 1.a) If s  k = s k + 1 3 k then s  k+1 = s  k + 1 3 k+1 , hence by Consequence 1.a, we have #s  k+1  =#s  k .Notethatifs k+1 has a other representation, s  k+1 = s  k + 0 3 k+1 ,s  k ∈ supp µ k ,thens k and s  k are two consecutive poin ts in supp µ k and s k <s  k <s  k ,a contradiction. It follows that #s k+1  =#s k .Therefore #s k+1  #s  k+1  = #s k  #s  k  a k<k+1. Local dimension of fractal measure associated with 35 1.b) If s  k ≥ s k + 2 3 k = s k+1 . So s k+1 has at most two representations through s k and s  k ( s k+1 = s k + 6 3 k+1 and s k+1 = s  k + 0 3 k+1 ). It follows that #s k+1  a #s k  +#s  k . Since s k <s k + 1 3 k+1 <s k+1 a s  k , s  k+1 ∈ (s k ,s k+1 ). On the other hand s k ,s  k are t wo consecutive points in supp µ k ,sos  k+1 /∈ supp µ k . It follows that If s  k + 6 3 k+1 <s k + 1 3 k+1 for s  k ∈ supp µ k with s  k <s k then s  k+1 = s k + 1 3 k+1 . Therefore #s k+1  #s  k+1  a #s k  +#s  k  #s k  a k +1. If there exists s  k ∈ supp µ k suc h that s k + 1 3 k+1 <s  k + 6 3 k+1 <s k+1 (s  k <s k )then s  k+1 = s  k + 6 3 k+1 and 0 <s k − s  k < 5 3 k+1 < 2 3 k ,sos k = s  k + 1 3 k . By Consequece1.b), #s  k+1  =#s  k ≥#s k . Therefore #s k+1  #s  k+1  a #s k  +#s  k  #s k  a k +1. Case 2. x k+1 =1. s k+1 = s k + 1 3 k+1 .Thens  k+1 = s k + 0 3 k+1 .Ifthereexistss  k ∈ supp µ k such that s  k+1 = s  k + 6 3 k+1 then s  k ,s k are two consecutive points in supp µ k (because s k − s  k =2/3 k ). Therefore #s  k+1  #s k+1  = #s  k+1  #s k  a #s k  +#s  k  #s k  a k +1. Case 3. x k+1 =0. s k+1 = s k + 0 3 k+1 .Notethatifs k+1 has other represen tation, s k+1 = s  k + 6 3 k+1 then it was considered in the Case 1. So we may suppose t hat s k+1 = s k + 6 3 k+1 for all s k ∈ supp µ k . (3) Then we have #s k+1  =#s k .Write s  k+1 = s  k + x  k+1 3 k+1 ,x  k+1 ∈ D. Since s k+1 = s k + 0 3 k+1 ∈ supp µ k and x  k+1 =0,x  k+1 =1orx  k+1 = 6. Which implies #s  k+1  =#s  k . We claim that s k and s  k are two consecutive points in supp µ k . In fact, if there exists s  k ∈ supp µ k suc h that s  k <s  k <s k = s k+1 ,then s  k+1 = s  k + 6 3 k+1 (4) (If it is not the case, s  k+1 = s  k + 1 3 k+1 <s  k <s k = s k+1 ,thens  k+1 and s k+1 are not consecutive). 36 Le Xuan Son, Pham Quang Trinh, Vu Hong Thanh Since s  k+1 and s k+1 are two consecutive points, s  k <s  k+1 = s  k + 6 3 k+1 = s  k + 2 3 k , hence s  k = s  k + 1 3 k . (5) From Consequence 2 and (3), s  k + 6 3 k+1 = s  k + 2 3 k <s k = s k+1 . (6) From (4), (5) and (6) we get s  k+1 = s  k + 6 3 k+1 = s  k + 2 3 k = s  k + 1 3 k <s  k + 6 3 k+1 < s k = s k+1 , a contradiction to s  k+1 and s k+1 are two consecutive points. Therefore #s k+1  #s  k+1  = #s k  #s  k  a k<k+1. ProofofProposition2.3.We first show that for rgiven ≥ 1andforanys ∈ supp µ if there exists lim n→∞ log µ(B r3 −n (s)) log(r3 −n ) , then α(s)= lim n→∞ log µ(B r3 −n (s)) log(r3 −n ) =lim n→∞ log µ(B r3 −n (s)) log 3 −n . (7) Indeed, for 0 <ha 1taken such that 3 −n−1 < h r a 3 −n .Then log µ(B r3 −n (s)) log(r3 −n−1 ) a log µ(B h (s)) log h a log µ(B r3 −n−1 (s)) log(r3 −n ) . Since lim n→∞ log(r3 −n−1 ) log(r3 −n ) =1,wehave lim n→∞ log µ(B r3 −n (s)) log(r3 −n−1 ) =lim n→∞ log µ(B r3 −n−1 (s)) log(r3 −n ) =lim n→∞ log µ(B r3 −n (s)) log(r3 −n ) . Therefore, (7) follows. Since |S −S n | a 6 ∞  i=1 3 −n−i =3.3 −n , we have µ(B 3 −n (s)) = Prob(|S −s| a 3 −n ) a Prob(|S n − s| a 3 −n +3.3 −n =4.3 −n ) = µ n (B r3 −n (s)), (8) where r =4. Similarly, we obtain µ n (B r3 −n (s)) a µ(B (r+3)3 −n (s)). Local dimension of fractal measure associated with 37 From the latter and (8) we get log µ(B (r+3)3 −n (s)) log 3 −n a log µ n (B r3 −n (s)) log 3 −n a log µ(B 3 −n (s)) log 3 −n . Letting n →∞, by (7) we obtain α(s)= lim n→∞ log µ n (B r3 −n (s)) log 3 −n . (9) Observe that B r3 −n (s)containss n and at most six consecutive points in supp µ n (because 2r = 8 and by Consequence 2). By Lemma 2.4, log µ n (s n ) log 3 −n ≥ log µ n (B r3 −n (s)) log 3 −n ≥ log[6n 5 µ n (s n )] log 3 −n . From the latter and (9) we get α(s)= lim n→∞ log µ n (s n ) log 3 −n =lim n→∞ | log µ n (s n )| n log 3 . The proposition is proved. For each infinite sequence x =(x 1 ,x 2 , ) ∈ D ∞ defines a point s ∈ supp µ by s = S(x):= ∞  n=1 3 −n x n . Let x =(x 1 ,x 2 , )=(0, 6, 0, 6, ), i.e., x 2k−1 =0,x 2k =6,k =1, 2, (10) Then we have 2.5. Claim. For x =(x 1 ,x 2 , ) ∈ D ∞ is defined by (10), we have a) #s 2n  =#s 2n−1 ; b) #s 2(n+1)  =#s 2n  +#s 2(n−1) , (11) for every n ≥ 2, where s n denotes n- partial sum of s = S(x). Proof. a) Observe that #s 2n ≥#s 2n−1 . On the other hand, let (x  1 ,x  2 , , x  2n ) ∈ s 2n .Ifx  2n = 6, then by Claim 2.1, x  2n = 0. It follows that s  2n−1 −s 2n−1 = 2 3 2n−1 ,where s  2n−1 =  2n−1 i=1 3 −i x  i . From Claim 2.2.b), it follows that x 2n−1 = 1, a contradiction to x 2n−1 =0. Thusx  2n = 6, which implies (x  1 ,x  2 , ,x  2n−1 ) ∈s 2n−1 . That means #s 2n−1 ≥#s 2n . Therefore #s 2n  =#s 2n−1 . 38 Le Xuan Son, Pham Quang Trinh, Vu Hong Thanh b) For any element (x  1 ,x  2 , ,x  2n ,x  2n+1 ,x  2n+2 ) ∈s 2n+2 , from the proof of a) we have (x  1 ,x  2 , ,x  2n+1 ) ∈s 2n+1 .So,byClaim2.1,x  2n+1 =0orx  2n+1 =6(because x 2n+1 =0). If x  2n+1 =0then(x  1 ,x  2 , ,x  2n ) ∈s 2n . If x  2n+1 =6,since(x  1 ,x  2 , ,x  2n , 6, 6) ≈ (x 1 ,x 2 , ,x 2n−1 ,x 2n , 0, 6), s 2n −s  2n = 2 3 2n ,wheres  2n =  2n i=1 3 −i x  i . By Claim 2.2.b) and x 2n =6wehavex  2n−1 = x  2n =1, which implies (x  1 ,x  2 , ,x  2n−2 ) ∈s 2n−2  (because (0, 6, 0) ≈ (1, 1, 6)). Let A = {(x  1 ,x  2 , ,x  2n−2 ,x  2n−1 ,x  2n , 0, 6) : (x  1 ,x  2 , ,x  2n ) ∈s 2n }, B = {(y  1 ,y  2 , ,y  2n−2 , 1, 1, 6, 6) : (y  1 ,y  2 , ,y  2n−2 ) ∈s 2n−2 }. From the above arguments we have A ∪B = s 2n+2  and A ∩B = ∅. Therefore #s 2(n+1)  =#A +#B =#s 2n  +#s 2(n−1) . The lemma is proved. Consequence 3. For s ∈ supp µ is defined as in Claim 2.5 we have #s 2n  =#s 2n−1  = √ 5 5 [( 1+ √ 5 2 ) n+1 − ( 1 − √ 5 2 ) n+1 ], (12) for every n ≥ 1. Proof. It is easy to see that (12) satisfies (11). 2.6. Claim. For s ∈ supp µ is defined as in Claim 2.5 we have α(s)=1− log(1 + √ 5) −log 2 2log3 . Proof.Forn ≥ 2takek ∈ N such that 2k a n<2(k +1). By (12), √ 5 5 (a k+1 1 − a k+1 2 ) a #s n  a √ 5 5 (a k+2 1 − a k+2 2 ), where a 1 = 1+ √ 5 2 ,a 2 = 1− √ 5 2 . It follows that | log √ 5 5 (a k+2 1 − a k+2 2 )3 −2k | 2(k +1)log3 a | log µ n (s n )| n log 3 a | log √ 5 5 (a k+1 1 −a k+1 2 )3 −2k−2 | 2k log 3 . Since lim k→∞ | log √ 5 5 (a k+2 1 −a k+2 2 )3 −2k | 2(k +1)log3 =lim k→∞ | log √ 5 5 (a k+1 1 −a k+1 2 )3 −2k−2 | 2k log 3 =1− log a 1 2log3 , Local dimension of fractal measure associated with 39 by Proposition 2.3 we get α(s)=1− log(1 + √ 5) −log 2 2log3 . The claim is proved. 2.7. Claim. Let x =(x 1 ,x 2 , )beasequencedefined by (10). Then we have 3#s 2n−1  < 2#s 2n+1  for every n, where s = S(x)ands n denotes n-partial sum of s. Proof. Observe that the assertion holds for n =1, 2. For n ≥ 3, by Claim 2.5 we have 2#s 2n+1  =2#s 2n−1  +2#s 2n−3  =3#s 2n−1 −#s 2n−1  +2#s 2n−3  =3#s 2n−1 −#s 2n−3 −#s 2n−5  +2#s 2n−3  =3#s 2n−1  +#s 2n−3 −#s 2n−5  > 3#s 2n−1 . The claim is proved. 2.8. Claim. Assume that s n+1 ∈ supp µ n+1 has two representations through poin ts in supp µ n (n>3). Then, either #s n+1  =#s n−1  +#s n−3  for some s n−1 ∈ supp µ n−1 and some s n−3 ∈ supp µ n−3 , or #s n+1  a 2# s n−2  for some s n−2 ∈ supp µ n−2 . Proof. Let s n+1 = s n + 0 3 n+1 = s  n + 6 3 n+1 , which implies s n −s  n = 2 3 n , so by Claim 2.2.b), x  n =1,x n =0orx n = 6. We consider the case x n =0. Thecasex n = 6 is similar. We have s n+1 = s n−1 + 0 3 n + 0 3 n+1 = s  n−1 + 1 3 n + 6 3 n+1 . (13) We claim that s n has only one representation through point s n−1 ∈ supp µ n−1 .Infact,if it is not the case, s n = s n−1 + 0 3 n = s  n−1 + 6 3 n ,then s n+1 = s n−1 + 0 3 n + 0 3 n+1 = s  n−1 + 6 3 n + 0 3 n+1 = s  n−1 + 1 3 n + 6 3 n+1 , which implies s n−1 − s  n−1 = s  n−1 − s  n−1 = 1 3 n−1 . a contradiction to Consequence 2. Hence, #s n+1  =#s n−1  +#s  n−1 . From (13) yield s n−1 −s  n−1 = 1 3 n−1 , by Claim 2.2.a), x n−1 = 1. So that , by Consequence 1.a), #s n−1  =#s n−2 .Therefore #s n+1  =#s n−2  +#s  n−1 . 40 Le Xuan Son, Pham Quang Trinh, Vu Hong Thanh Consider the following cases. 1. If s  n−1 has o nly one representation through some point s  n−2 ∈ supp µ n−2 then #s  n−1  =#s  n−2 . Without loss of generality we may assume that #s n−2 ≥#s  n−2 . Then #s n+1  =#s n−2  +#s  n−2  a 2# s n−2 . 2. If s  n−1 has two representations through points in supp µ n−2 , s  n−1 = s n−2 + 0 3 n−1 = s  n−2 + 6 3 n−1 ,then s n+1 = s n−2 + 1 3 n−1 + 0 3 n + 0 3 n+1 = s n−2 + 0 3 n−1 + 1 3 n + 6 3 n+1 = s  n−2 + 6 3 n−1 + 1 3 n + 6 3 n+1 . Since (1, 0, 0) ≈ (0, 1, 6) ,s n−2 = s  n−2 , and so s n−2 −s  n−2 = 2 3 n−2 . Hence, by Claim 2.2.b), x  n−2 =1. Thus,s  n−2 = s  n−3 + 1 3 n−2 . We check that s n−2 has only one representation through some point s n−3 ∈ supp µ n−3 . If it is not the cases s n−2 = s n−3 + 0 3 n−2 = s  n−3 + 6 3 n−2 ,then s n+1 = s n−3 + 0 3 n−2 + 1 3 n−1 + 0 3 n + 0 3 n+1 = s  n−3 + 6 3 n−2 + 1 3 n−1 + 0 3 n + 0 3 n+1 = s  n−3 + 1 3 n−2 + 6 3 n−1 + 1 3 n + 6 3 n+1 , which implies s n−3 −s  n−3 = s  n−3 −s  n−3 = 1 3 n−3 . Which is a contradiction to Consequence 2. So, #s n−2  =#s n−3 . Therefore #s n+1  =#s n−3  +#s  n−1 . The claim is proved. 2.9. Claim. Let k ≥ 3 be a natural number such that #t 2n+1  a #s 2n+1  for all n a k and for every t 2n+1 ∈ supp µ 2n+1 . Then 2#t 2n  a #s 2n+1  +#s 2n−1  for all n a k and for every t 2n ∈ supp µ 2n , where s is definedasinClaim2.5ands n denotes n-partial sum of s. Proof. Observe that, if t 2n has only one representation through point t 2n−1 ∈ supp µ 2n−1 then the claim is true. Suppose that t 2n has two representations through points in supp µ 2n−1 , by Claim 2.8, either #t 2n  =#t 2n−2  +#t 2n−4  or #t 2n  a 2# t 2n−3 . 1. Let #t 2n  =#t 2n−2  +#t 2n−4 . Putting t 2n+1 = t 2n−2 + 0 3 2n−1 + 6 3 2n + 0 3 2n+1 ,t 2n−1 = t 2n−4 + 0 3 2n−3 + 6 3 2n−2 + 0 3 2n−1 [...]... Ngai, Iterated function systems with overlaps, Asian J Math 4(2000), 527 - 552 4 T Hu, The local dimensions of the Bernoulli convolution associated with the golden number, Trans Amer Math Soc 349(1997), 2917 - 2940 5 T Hu and N Nguyen, Local dimensions of fractal probability measures associated with equal probability weight, Preprint 6 T Hu, N Nguyen and T Wang, Local dimensions of the probability measure. .. probability measure associated with the (0, 1, 3) - problem, Preprint 7 T Hu, Some open questions related to Probability, Fractal, Wavelets, East - West J of Math Vol 2, No 1(2000), 5 5-7 1 8 J C Lagarias and Y Wang, Tiling the line with translates of one tile, Inventions Math 124(19 96) , 341 - 365 9 S M Ngai and Y Wang, Hausdorff dimention of the self - similar sets with overlaps,J London Math Soc (to appear) ... is odd; ki = 2i(1−r) [ r ] if i is even, where [x] denotes the largest integer Let nj = j ki and let i=1 Ej = {i : i x j and i is even} ; Oj = {i : i j and i is odd}, Local dimension of fractal measure associated with ej = ki ; oj = i∈Ej 43 ki i∈Oj Then nj = oj + ej 3.2 Claim With the above notation we have j nj−1 oj = 0 ; lim = 1 and lim = r j→∞ nj j→∞ nj j→∞ nj lim Proof The proof of the first limit... α(s) = lim The Main Theorem is proved Acknowledgements The authors are grateful to Professor To Nhu Nguyen for his helpful suggestions and valuable discussions during the preparation of this paper References 1 K J Falconer, Techniques in Fractal Geometry, John Wiley & Sons, 1997 2 K J Falconer, Fractal Geometry, Mathematical Foundations and Applications,John Wiley & Sons, 1993 3 A Fan, K S Lau and S... where sn = n 3−i xi i=1 The following fact given an estimation for the greatest lower bound of local dimension 2.10 Proposition For every n, (x1 , x2 , , x2n+1 ) = (0, 6, 0, 6, , 0, 6, 0) is a maximal sequence Proof We will prove the proposition by induction By Claim 2.8, it is straightforward to check that the assertion holds for n = 1, 2, 3 (# s3 = 2, # s5 = 3, # s7 = 5) Suppose that it is... µ2k−1 Then, by Claim 2.5 we get # t2k+3 # s2k+1 + # s2k−1 = # s2k+3 Case 2 # t2k+3 2# t2k for some t2k ∈ supp µ2k Then, by Claims 2.5 and 2.9 we have # t2k+3 2# t2k # s2k+1 + # s2k−1 = # s2k+3 Therefore, (x1 , x2 , , x2n+1 ) = (0, 6, 0, 6, , 0, 6, 0) is a maximal sequence The proposition is proved Le Xuan Son, Pham Quang Trinh, Vu Hong Thanh 42 3 Proof of The Main Theorem We call an infinite... 3) + 2(1−r) k(k r + 1) − k oj = r j→∞ nj lim The claim is proved We define s ∈ supp µ by s = S(x), where x = (0, 6, 0, 1, 1, , 1, 0, 6, 0, 6, 0, 1, 1, , 1, ) k1 =3 k2 k3 =5 ( 16) k3 Note that, for i ∈ Oj , from (12), # ski ⎧ √ ⎨> +3 ki +3 5 ki2 (a1 = − a2 2 ) = ⎩ 5 < For s ∈ supp µ is defined ( 16) and for nj−1 # ski √ ki +3 5 a1 2 5 √ ki +3 +1 5 a1 2 5 n < nj we have # sn i∈Oj−1 # ski i∈Oj Hence,... = (x1 , x2 , ) ∈ D∞ a prime sequence if # sn = 1 for every n, where sn = n 3−i xi i=1 √ 5)−log log a 3.1 Claim α = 1, α = 1 − log(1+ log 3 2 = 1 − 2 log 1 2 3 Proof For any prime sequence x = (x1 , x2 , ) we have # sn = 1 for every n, where sn = n 3−i xi Therefore, by Proposition 2.3 we get i=1 α = α(s) = lim n→∞ where s = S(x) From Claim 2 .6 we have 1− α | log µn (sn )| = 1, n log 3 log a1 ... Combinating (14) and (15) we get α≥1− Therefore log a1 2 log 3 √ log(1 + 5) − log 2 log a1 =1 − =1 − 2 log 3 2 log 3 The claim is proved To complete the proof of our Main Theorem it remains to show that E = [1 − √ log(1+ 5)−log 2 log a , 1], i.e., for any β ∈ (1− 2 log 1 , 1) there exists s ∈ supp µ for which α(s) = β 2 log 3 3 log 3 Let r = 2(1 − β) log a1 It is easy to see that 0 < r < 1 For i = 1,. ..Local dimension of fractal measure associated with 41 we have # t2n+1 ≥ 2# t2n−2 , # t2n−1 ≥ 2# t2n−4 It follows that 2# t2n = 2# t2n−2 + 2# t2n−4 # t2n+1 + # t2n−1 2 # t2n # s2n+1 + # s2n−1 2# t2n−3 By Claims 2.5 and 2.7 we get 2# t2n 4# t2n−3 4# s2n−3 = # s2n−3 + 3# s2n−3 # s2n−3 + 2# s2n−1 = # s2n+1 + # s2n−1 The claim is proved We say that x = (x1 , x2 , , xn ) ∈ Dn is a maximal sequence . JOURNAL OF SCIENCE, Mathematics - Physics. T.XXI, N 0 1 - 2005 LOCAL DIMENSION OF FRACTAL MEASURE ASSOCIATED WITH THE (0, 1 ,a) -PROBLEM:THECASEa =6 Le Xuan. local dimension for the (0, 1 ,a )- problem, where a ∈ N is a natural number. Note that the local dimension is an important characteristic of singular measures. For