Bounds of Hausdorff measure of the Sierpinski gasket

Transcription

1 J Math Anal Appl wwwelseviercom/locate/jmaa Bounds of Hausdorff measure of the Sierpinski gasket Baoguo Jia School of Mathematics and Scientific Computer, Zhongshan University, Guangzhou 5075, China Received 4 January 005 Available online 4 September 006 Submitted by M Laczkovich Abstract By a new method, we obtain the lower and upper bounds of the Hausdorff measure of the Sierpinski gasket, which can approach the Hausdorff measure of the Sierpinski gasket infinitely 006 Elsevier Inc All rights reserved Keywords: Hausdorff measure; Self-similar set; Sierpinski gasket 0 Introduction The computation and estimation of the Hausdorff dimension and measure of the fractal sets are important problems in fractal geometry Generally, the computation of the Hausdorff dimension, especially the Hausdorff measure, is very difficult As a referee has previously pointed out, Hausdorff measure is an important notion in the study of fractals However there are few concrete results about computation of Hausdorff measure even for some simple fractals Part of reason is the difficulty of the problem For a self-similar set satisfying the open set condition, we know that its Hausdorff dimension equals its self-similar dimension [] However, there are not many results on the computation and estimation of Hausdorff measures for such fractal sets except for a few fractal sets on a line, like the Cantor set [] For the famous classical self-similar set, the Sierpinski gasket, its Hausdorff measure remains unknown Nevertheless, efforts have been made in order to estimate the lower and upper bounds of its Hausdorff measure [ 8] This work was supported in part by the Foundations of the National Natural Science Committee 077, Guangdong Province Natural Science Committee, China address: mcsjbg@mailsysueducn 00-47X/$ see front matter 006 Elsevier Inc All rights reserved doi:006/jjmaa

2 B Jia / J Math Anal Appl In this paper, we develop a new method of estimating the upper bounds and lower bounds of the Hausdorff measure of the Sierpinski gasket We show that the Hausdorff measure of the Sierpinski gasket can be squeezed out by sequences of lower bounds and upper bounds Precisely speaking, we show that: Theorem The Hausdorff measure of the Sierpinski gasket satisfies the estimation a n e 6 s n H s S a n, for n, where a n is defined in Proposition The above theorem provides us a way, at least in theory, to estimate the Hausdorff measure of the Sierpinski gasket as close as we want In the end of the paper, we give two conjectures about a n,forn and H s S The Hausdorff measures of the self-similar sets Let D R n be a nonempty set E R n is a self-similar set defined by m similar contracting maps S i : D D, with contracting ratios, 0 <c i < i =,,,m and satisfying open set condition, that is, there exists a nonempty open set U for which we have S i [U] S j [U]= for i j and U S i [U] for all i Then dim H E = s, 0 <H s E < + Where s satisfies m c s i =, dim H E and H s E denote the Hausdorff dimension and measure of E, respectively If S i [E] S j [E]=, 0 <i<j m, we say that E satisfies strong separate condition SSC Let J n ={i i i n : i,i,,i n m, n } and E i i i n = S i S i S in E be self-similar E It is easy to know that E = J n E i i i n Proposition [7] Suppose that E is a self-similar set satisfying the open set condition, then for any measurable set U, we have H s E U U s, where s = dim H E Proposition Suppose that E is a self-similar set satisfying the open set condition For n, k m n,letδ,δ,,δ k {E i i i n : i,i,,i n m} and μ be the common selfsimilar probability measure on the E, μe i i i n = ci s ci s ci s n Let { k Δ i s } b k = Δ i {E i in } μ, k Δ i,,,k where the imum is taken for all possible union of k elements of {E i i i n } and a n = k m n{b k } If there exists a constant A>0 such that a n An =,,, then H s E A Proof By [, p ], we can get the same values for Hausdorff measure and dimension if in the definition of Hδ s E we use δ-cover of just open set So in the mass distribution principle of [], we can replace any sets by any open sets For any open set V,letF n = E E i i i i in V i n It is obvious that F n F n+, + n= F n = E V By the property of measure μ and the definitions of a n, b k, we get

3 08 B Jia / J Math Anal Appl μv = μe V= μ = lim μ n + + n= E i i in V E i i i n F n = lim n + μf n a n E i i in V E i i i n By the mass distribution principle of [], we have H s E A Proposition For n,thea n decreases and lim n + a n = H s E Proof Suppose that a n = k m n Δ i {E i in },,,k { k Δ i s μ k Δ i } = U k 0 μu k0, s V s a n A V s where the U k0 is the union of some k 0 elements of {E i i i n } By Proposition, we get H s E U k0 U k0 s So μu k0 H s E U k0 s Therefore H s E U k 0 s μu k0 = a n Next we prove that a n decreases Because U k0 is the union of some k 0 elements of {E i i i n }, U k0 is the union of some k 0 m elements of {E i i i n i n+ } By the definition of a n, we obtain a n+ U k 0 s μu k0 = a n Let L = lim n + a n,soa n L By Proposition, H s E L In, let n + We get H s E LSoH s E = lim n + a n = L Corollary If c = c = =c m = c, then b k = Δi {E i in },,,,k { k Δ i s kc ns }, where the imum is taken for all possible union of k elements of {E i i i n } a n = {b k } k m n Then for n, a n decreases and lim n a n = H s E The Hausdorff measure of the Sierpinski gasket Take an equilateral triangle including its inside with side length in the Euclidean plane R Call it S 0 and delete all but the three corner equilateral triangles including their boundary with side length to obtain S see Fig Continue in this way, replacing at the each equilateral triangle of S n by the three corner equilateral triangles with side length n to get S n We obtain S 0 S S n The nonempty set S = 0 S n is called the Sierpinski gasket For each n 0, S n consists of n equilateral triangles with side length n Any one of such equilateral triangles is called a n -basic equilateral triangle The Hausdorff dimension of S is s = dim H S = log Theorem The Hausdorff measure of the Sierpinski gasket satisfied the estimation a n e 6 s n H s S a n, for n

4 B Jia / J Math Anal Appl Fig Construction of the Sierpinski gasket Fig The first two steps of the Sierpinski gasket construction By a simple calculation, we can get that a =, a = s see Fig 5 By the inequality, e x x, forx 0, we have an error bound of our estimation, error a n e 6 s n e 6 s n, for n Proof Let n, k n, Δ,Δ,,Δ k S n Case If Δ i Δ 0 j j =,,, Δ 0 j for j =,,, is defined by Fig, let k Δ i =d, by [5, Proposition ] see Fig, we have k Δ i Δ i Δ EFD 4 For each Δ i, there exists Δ n j S n such that Δ i Δ n j j =,,,k n and Δ n,δ n,,δ n k n are different from each other It is easy to know that k k n and k n Δ n j j= n d + = d + n

5 00 B Jia / J Math Anal Appl So 4 k Δ i k n j= Δn j d d + n 4 + = n + 4 n Therefore k Δ i s s k n j= k Δn j s n + 4 n k n s k n j= Δn j s + 4 n k n n It is similar to the a n of Corollary Let { k a n Δ i s k } = k n k : for each l =,,, Δ i Δ 0 n l Δ i S n,,,k It is obvious that a n a n From, we have a n +4 s a n n Therefore, for any l, a l+n s + 4 a l+n l+n s s s + 4 l+n + 4 l+n + 4 a n n Take logarithm on two sides and use inequality, ln + x < x, x>0, we get ln a l+n l+n n ] l+n [ln s ln ln ln a n [ l+n s l+n + +4 n ] + ln a n It is similar to the proof of Proposition It is easy to get that {a n } decreases Suppose lim n a n = α and let l + Wehave ln α s 4 n + ln a n = ln a n e 8s n So α a n e 8s n a n e 8s n Case If k Δ i only intersects two basic equilateral triangles of S, with no loss generality, suppose k Δ i intersects Δ 0, Δ0 see Fig a If Δ i Δ Δ and Δ i Δ 0 or

6 B Jia / J Math Anal Appl Δ i Δ Δ and Δ i Δ 0, by the symmetry, we can suppose that Δ i Δ Δ and Δ i Δ 0 At that time, k Δ i 4 Let a n = k n Δ i S n,,,k = 8 { k Δ i k n : It is similar to a n of case } k Δ i satisfies Like case, it is easy to prove that {a n } decreases Suppose that lim n a n = α,wehave a n n s a n = + 8 n s a n So α a b If n e 6 s n a n e 6 Δ i Δ Δ, s n 4 by the similarity, there exist a positive integer t and Δ,Δ S t see Figs, such that k t Δ = Δ0, t Δ = Δ0, Δ i Δ Δ and Δ i Δ k Δ, Δ i Δ or Δ i Δ k Δ, Δ i Δ Because k Δ i s = k t Δ i s, k k n k t Δ i satisfies case b Note that Δ Δ contains n t n elements of S n It is similar to a n of case a Let a n = k n Δ i S n,,,k { k Δ i s k } k : Δ i Δ Δ n

7 0 B Jia / J Math Anal Appl Fig Enlargement of Δ, Δ n e 6 The {a n } decreases and suppose that lim n a n = α,wehaveα a s n a n e 6 s n Case If k Δ i only intersects one basic equilateral triangle of S, by the similarity, there exist a positive integer t 0 and i 0,i0,,i0 t 0 {,, } such that k Δ i E i 0 i i0 0 = f t i 0 f 0 i 0 f i 0 t0 S and k Δ i intersects at least two of E i 0 i 0 i0 t 0,E i 0 i0 i0 t 0,E i 0 i0 i0 t 0 Therefore f i 0 t 0 and f it 0 0 Let f i 0 k Δ i s μ k Δ i = f i 0 a 4 n = k n f k i 0 Δ i intersects at least two of f S, f S, f S Note that f i 0 t 0 f i 0 μ k f i 0 t 0 f k i 0 Δ i s f i 0 f Δ i 0 i f k i 0 Δ i satisfies case or case It is similar to a n of case Δ i S n,,,k { k Δ i s k k : Δ i Δ 0 n i 0,i } 0 {,, } The {a n 4 } decreases and suppose that lim n a n 4 = α 4,wehaveα 4 a n 4 {e 6 e 8s n } a n e 6 s n From cases, we obtain H s S = lim a n = lim { a } n n n,a n,a n,a4 n = {α,α,α,α 4 } { a n e 8s n,a n e 6 s n,a n e 6 s n,a n e 6 s } n = an e 6 So, a n e 6 s n H s S a n,forn Two conjectures about Sierpinski gasket s n, s n For the Sierpinski gasket, by a simple calculation and the definition of a n in Corollary, it is easy to get a = seefig4,a = s see Fig 5 When n is bigger than, a good and efficient algorithm of a n still needs to be found We can only make the following conjectures

8 B Jia / J Math Anal Appl Fig 4 Fig 5 Fig 6 Fig 7 Conjecture For the Sierpinski gasket, a = 7s see Fig 6, a = s see Fig 7, a 5 = 5s , a 6 = 49s , a 7 = 97s , a 8 = 9s Conjecture The Hausdorff measure of the Sierpinski gasket satisfies a 8 e 6 s 8 H s S a 8 Acknowledgment The author is grateful to the referee for reading the manuscript carefully and his/her valuable suggestions for the improvement of the paper References [] KJ Falconer, Fractal Geometry Mathematical Foundations and Applications, Wiley, New York, 990 [] K Falconer, The Geometry of Fractal Sets, Cambridge Univ Press, Cambridge, UK, 985

9 04 B Jia / J Math Anal Appl [] ZL Zhou, Feng Li, A new estimate of the Hausdorff measure of the Sierpinski gasket, Nonlinearity [4] ZL Zhou, The Hausdorff measure of Sierpinski gasket, Sci China Ser A ; Sci China Ser A [5] BG Jia, ZL Zhou, ZW Zhu, A lower estimate for the Hausdorff measure of the Sierpinski gasket, Nonlinearity [6] ZL Zhou, The Hausdorff measure of the Koch curve and Sierpinski, Progr Natur Sci [7] ZL Zhou, The Hausdorff measure of the self-similar set the Koch curve, Sci China Ser A [8] Yuan Huojun, Nanjin University dissertation, 00