Packing Measures of Homogeneous Cantor Sets

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1 Λ45ffΛ4 μ ρ Ω χ Vol.45, No fl ADVANCES IN MATHEMATICS CHINA) July, 206 doi: 0.845/sxjz b Packing Measures of Homogeneous Cant Sets QU Chengqin,, ZHU Zhiwei 2 3,, ZHOU Zuoling. School of Mathematics, South China University of Technology, Guangzhou, Guangdong, 50640, P. R. China; 2. School of Mathematics and Statistics, Zhaoqing University, Zhaoqing, Guangdong, 52606, P. R. China; 3. Lingnan College, Zhongshan University, Guangzhou, Guangdong, 50275, P. R. China) Abstract: This paper gives a fmula to calculate the packing measures of the homogeneous Cant sets. Keywds: packing measures; homogeneous Cant sets MR200) Subject Classification: 28A78; 28A80 / CLC number: O74.2 Document code: A Article ID: ) Introduction It is well known that to determine the exact Hausdff measure and packing measure of a fractal set is a difficult and imptant problem in fractal geometry. In fact, up to now, only a few results about self-similar sets are known middle-third Cant set and its variation, see [, 4, 7]). The homogeneous Cant set is a class of imptant non-self-similar sets whose dimensions and measures are usually me difficult to calculate f lacking of similarity structures. In [5], the Hausdff dimension and packing dimension were entirely determined f the first time. In [9], a fmula was given that calculates the Hausdff measures of a class of homogeneous Cant sets, and in [2], Baek obtained a lower bound of the packing measure. This paper gives a fmula to calculate the packing measures of the homogeneous Cant sets. Let I =[0, ], {n k } k be a sequence of positive integers, and {c k } k be a sequence of real numbers satisfying n k 2, 0 <n k c k < k ). F any k, let D k = {i,i 2,,i k ): i j n j, j k}, D= k 0 D k, where D 0 =. Fσ =σ,σ 2,,σ k ), τ =τ,τ 2,,τ m ), denote σ τ =σ,σ 2,,σ k,τ,τ 2,,τ m ). Let F = {I σ : σ D} be the collection of the closed sub-intervals of I satisfying i) I = I; ii) f any k andσ D k, I σ i i n k ) are sub-intervals of I σ ;meover, I σ,i σ 2,,I σ nk are equally spaced from left to right, and the left endpoints of I σ and I σ coincide, the right endpoints of I σ nk and I σ coincide; Received date: Revised date: Foundation item: Suppted by NSFC No ). chengqinqu@tom.com; lnszzl@mail.sysu.edu.cn

2 538 ß Ψ ffi 45ff iii) f any k andσ D k, j n k,wehave I σ j = c k, I σ where A denotes the diameter of A. Let E k = σ D k I σ, E = k 0 E k. We call E the homogeneous Cant set [5] determined by {n k } k, {c k } k and call F k = {I σ : σ D k } the k-th der basic intervals of E. Our main result is Theem Let E be the homogeneous Cant set determined by {n k } k, {c k } k,and sup n k <. Then ) s k P s E) =2 s ck lim sup n k n j c s j k n k, 0.) j= where P s E) isthes-dimensional packing measure of the set E. Remark There are a lot of results concerning homogeneous Cant sets. In [5], the Hausdff dimension and packing dimension were determined by the technique of net measure, that is, log n n 2 n k dim H E) = lim inf, k log c c 2 c k log n n 2 n k dim p E) = lim sup. k log c c 2 c k +logn k+ In [9], we further gave the exact Hausdff measures f a class of homogeneous Cant sets whose gaps in different levels decrease, that is, H s E) = lim inf k k n j c s j. Remark 2 In [4], Feng investigated the packing measure of the attract K of a linear iterated function system {S j x = ρ j x + t j } m j= on the real line satisfying the open set condition. It is shown that P s K) =d min, where { min{2 s R 0, 2 s R } f m =2, d min = min{2 s R 0, 2 s R,R 2 } f m>2. Here s is the self-similar dimension of K, and R 0 = min 2 j m j k= ρs k S j 0) s, m k=j+ ρs k S j ) s, R = min j m j2 k=j R 2 = min + ρs j j <j 2<m Sj2+0) S j ) 2dist Sj 2 +0)+Sj ) 2,K) ) s. If ρ j c, j =, 2,,m, and the gaps are equi-distributed, the above conclusion is a special case of Theem f c j c, j =, 2,. j=

3 4 ', %!ο, #&ffi: Packing Measures of Homogeneous Cant Sets 539 Remark 3 In [6], Garcia and Zuberman determined recently the exact packing measure and Hausdff centered measure of E when n k =2fallk. Proof of Theem F any σ =σ,σ 2,,σ m ) D m,when0<k m, wedenoteσ k =σ,σ 2,,σ k ). Let x k be the length of k-th der basic interval, y k the length of the k-th der gap between two consecutive sub-intervals I σ i and I σ i+) where σ D k ). Then Set x k = c c 2 c k, y k = n kc k n k c c 2 c k. d k =n n 2 n k )x k + y k ) s, d = lim sup d k..) k Then, f any ε>0, there exists a positive integer k 0, such that d k <d+ ε f all k k 0. It is obvious that d = lim sup k n k c k n k )s k j= n jc s j. In der to prove Theem, it is sufficient to prove that P s E) =2 s d. The proof of Theem relies on the computation of the lower density of a natural measure. Let 0 <s< and ν be a measure on R. Thelowerdensityofν at x R is defined by Θ s ν[x r, x + r]) ν, x) = lim inf r 0 2r) s. There is one natural measure suppted on E denoted by μ, which is the only probability measure satisfying μi σ )=n n 2 n k ).2) f σ D k. If σ D k, τ D k+l l>0) and τ = σ p +) p 2 +) p l +) p l,denote Iσ, τ)=i σ I σ 2 I σ p I σ σ2,) I σ σ2,2) I σ σ2,p2) I σ σl,) I σ σl,2) I σ σl,pl ), where σk, j) =p +,p 2 +,,p k +,j), 0 p i n i, j p k, σ σl, p l )=τ. Lemma. Let σ D k, τ D k+l l>0), τ k = σ. Then μiσ, τ)) Iσ, τ) s min{d k+,d k+2,,d k+l }. Proof F any σ D k, τ D k+l,letaσ) be the left endpoint of I σ,andbτ) theright endpoint of I τ see Figure ). aσ) bτ) Figure The left endpoint aσ) ofi σ and the right endpoint bτ) ofi τ

4 540 ß Ψ ffi 45ff By the definition of Iσ, τ), we have μiσ, τ)) = p n n 2 n k+ + p 2 n n 2 n k p l n n 2 n k+l, and Iσ, τ) s p x k+ + y k+ )+p 2 x k+2 + y k+2 )+ + p l x k+l + y k+l )) s p x k+ + y k+ ) s + p 2 x k+2 + y k+2 ) s + + p l x k+l + y k+l ) s. Therefe, μiσ, τ)) Iσ, τ) s p p 2 p l n n 2 n k+l n n 2 n k+ + n n 2 n k p x k+ + y k+ ) s + p 2 x k+2 + y k+2 ) s + + p l x k+l + y k+l ) s min{d k+,d k+2,,d k+l }. This completes the proof of Lemma.. By Lemma., we immediately have Lemma.2 F any ε>0, there is a k 0 such that if σ D k, τ D k+l k>k 0 )and τ k = σ, then μ[aσ),bτ)]) bτ) aσ)) s d + ε). Lemma.3 Let μ be the probability measure defined by.2), and E be a homogeneous Cant set such that 0 P s E). Then P s E) = Θ s μ,x) particular Θ s μ, x) is almost everywhere constant. f μ almost all x E, in Proof Define ν =P s E)) P s E,whereP s E A) =P s E A) f any A R. Thenν and μ coincide on each I σ, and by regularity, these measures are identical. It is known see [8, Theem 6.0]) that Θ s P s E,x)=fP s and almost all x E. ThenP s E) = Θ s μ,x) f μ and almost all x E. Remark 3 By the result of Lemma.3 and the definition of d in.), if we can prove that Θ s μ, x) =2 s d f μ and almost all x E, then Theem is proved. Lemma.4 Let E be the homogeneous Cant set determined by {n k } k and {c k } k, and μ is defined as in.2). Then ) If 0 <d<, thenθ s μ, x) 2 s d f all x E. 2) If d =0,thenΘ s μ, x) = f all x E. Proof Let x E, 0<r<. Set J =[x r, x + r]. Then there exists a positive integer k, such that J contains at least a k + )-th der basic interval, but it does not contain any k-th der basic interval. Therefe, J intersects with at most two k-th der basic intervals, and r can be chosen to be sufficiently small such that k>k 0. Case J intersects two k-th der basic intervals. Let I σ),i σ2) σ),σ2) D k and bσ)) <aσ2))) be such two basic intervals, and set J = J [bσ)),aσ2))] J 2,where J = [x r, bσ))], J 2 = [aσ2)),x + r]. Without loss of generality, let x J. Then

5 4 ', %!ο, #&ffi: Packing Measures of Homogeneous Cant Sets 54 aσ2)) bσ)) J < J + J 2. Therefe, μj) J s = μj )+μj 2 ) J + J 2 + aσ2)) bσ))) s μj )+μj 2 ) 2 s J s + J 2 s ) 2 s min { μj ) J s, μj 2) J 2 s Let u = x + r, i.e., J 2 =[aσ2)),u]. If u = bσ2)), by Lemma.2, we have μj 2 ) J 2 s d + ε..4) If u E = l σ D l I σ and u bσ2)), then there exists τ D, such that {u} = l I τ l. Therefe, J 2 =[aσ2)),u]= l [aσ2)),bτ l)], and [aσ2)),u] [aσ2)),bτ l + ))] [aσ2)),bτ l)]. Therefe, μj 2 ) = lim l μ[aσ2)),bτ l)]). On the other hand, we can choose l sufficiently large, such that I τ l I σ2),thatis,τ k = σ2). Then by Lemma.2, we have μj 2 ) J 2 s = lim μ[aσ2)),bτ l)]) μ[aσ2)),bτ l)]) l J 2 s lim l bτ l) aσ2))) s d + ε, }. 0 <d<, μj 2 ) =, d =0. J 2 s If u E, i.e., u I l σ D l I σ = l I σ D l I σ ), then there exists a positive integer l>0, such that u I σ D k+l I σ, i.e., u belongs to the gap between two consecutive sub-intervals I σ and I σ, whereσ,σ D k+l. Let u be the right endpoint of I σ, u 2 be the left endpoint of I σ, and denote J =[aσ2)),u ], J =[aσ2),u 2 )]. Then J J 2 J and μj ) = μj 2 ) = μj ). Following the notations of Lemma., it is easy to see that J = Iσ2),σ ), and J 2 J = p x k+ + y k+ )+p 2 x k+2 + y k+2 )+ + p l x k+l + y k+l ). Similar to the proof of Lemma., we have μj 2 ) J 2 s μj ) J s, 0 <d<,.5) d + ε μj 2 ) =, d =0..6) J 2 s F the interval J, similar to the above argument, we have μj ) J s, 0 <d<,.7) d + ε

6 542 ß Ψ ffi 45ff μj ) =, J s d =0..8) Therefe, μj) J s 2 s, d + ε 0 <d<,.9) μj) =, J s d =0..0) Case 2 J intersects only one k-th der basic interval. Let I σ σ D k ) be such a basic interval. If the left endpoint of J lies in the left of aσ), set J = J I σ. Since x I σ,then aσ) x r) < J, and thus μj) J s μj ) 2 s J s. Similar to the proof of Case, we have.9) and.0). If the right endpoint of J lies in the right of bσ), J I σ, we also have.9) and.0). Since ε is arbitrary, we complete the proof of Lemma.4. Lemma.5 Let E be the homogeneous Cant set determined by {n k } k>0 and {c k } k>0, sup n k <, and μ be defined as in.2). Then ) If 0 <d<, thenθ s μ, x) 2 s d μ-a.e. on E. 2) If d =, thenθ s μ, x) = 0 f all x E. Proof F any σ D k,letτ D k and I τ be the first k-th der basic interval to the left of I σ. Since 0 <n k c k <, we have aσ) bτ) > 0. Hence, there exists l = lk) >ksuch that r l = x l + y l <aσ) bτ), and μ[aσ) r l,aσ)+r l ]) = n n 2 n l. It follows that Θ s μ[aσ) r l,aσ)+r l ]) μ, aσ)) lim inf k 2r l ) s = lim inf k 2 s =2 s d, d k 0 <d<,.) Θ s μ, aσ)) = 0, d =..2) Now, f fixed k>2, put σ) = D k,and A p = I σ σ), A = A p. l=p σ D l p=0 Case p =0. SetD 0 = { }, D k = {σ D k : σ σ)}, and D 2k = {σ D 2k : σ k D k,σ D k σ)}, D 3k = {σ D 3k : σ 2k D 2k, σ D 2k σ)}, D lk = {σ D lk : σ l )k D l )k,σ D l )k σ)}, where D lk σ) = {τ σ) : τ D lk }. By the definition of A 0 we have μa 0 ) μ I σ σ) ), l=0 σ D lk

7 4 ', %!ο, #&ffi: Packing Measures of Homogeneous Cant Sets 543 and I σ σ) σ D lk,l=0,, 2, ) are pairwise disjoint. Therefe, where μa 0 ) l=0 σ D lk μi σ σ) ) = +n n 2 n k ) + n n 2 n k n n 2 n 2k +n n 2 n k )n k+ n k+2 n 2k ) n l )k+ n l )k+2 n lk ) + n n 2 n l+)k ) ) = + + n n 2 n k n n 2 n k n k+ n k+2 n 2k ) ) + n n 2 n k n k+ n k+2 n 2k ) + n l )k+ n l )k+2 n lk n lk+ n lk+2 n l+)k = Ωl, k), ) ) Ωl, k) = n n 2 n k n k+ n k+2 n 2k n lk+ n lk+2 n l+)k Since sup n k <, wehaveωl, k) 0asl and μa 0 )=. Case 2 p. F any τ D p,setãτ = l=p A p we have A p = Ã τ. τ D p F any τ D p, similar to the calculation of μa 0 ), we have μãτ )= n n 2 n p, ). σ D l,σ p=τ I σ σ). By the definition of so μa p ) =, and μa) =. On the other hand, f any x A there are infinitely many n such that there exists a σ D n with x aσ) <x n c n+ c n+2 c n+k < x 2 k n.taker = r n x 2 k n. Then [x r, x + r] [aσ) r n,aσ)+r n ], which implies Θ s μ, x) 2 k ) s 2 s d, μ-a.e. on E, f0<d<. Taking k we obtain Θ s μ, x) 2 s d μ-a.e. on E. If d =, from.2), Θ s μ, x) = 0 f all x E. This completes the proof of Lemma.5. ProofofTheem If d =0,thenΘ s μ, x) = f all x E by Lemma.4. So Theem follows from Theem 6. 2) in [8]. If d =, thenθ s μ, x) = 0 f all x E by Lemma.5. So Theem follows from [8, Theem 6. )]. And if 0 d,theem follows from Lemmas.3.5.

8 544 ß Ψ ffi 45ff Example Let E be the middle-third Cant set. It is well known that dim H E) = dim p E) =s = log 2 log 3,andHs E) =,wheredim H E), dim p E) denote respectively the Hausdff dimension and the packing dimension of the set E, andh s E) is the Hausdff measure of the set E. On the other hand, by Theem we obtain P s E) =4 s, which can also be found in [4]. Acknowledgements The auths express their gratitude to Profess Wen Zhiying f many good suggestions during the first auth s visit to the Mningside Center of Mathematics, and we also thank Profess Rao Hui f some advices. References [] Ayer, E. and Strichartz, R.S., Exact Hausdff measure and intervals of maximum density f Cant sets, Trans. Amer. Math. Soc., 999, 359): [2] Baek, H.K., Packing dimension and measure of homogeneous Cant sets, Bull. Aust. Math. Soc., 2006, 743): [3] Cutler, C.D., The density theem and Hausdff inequality f packing measure in general metric spaces, Illinois J. Math., 995, 394): [4] Feng, D.J., Exact packing measure of linear Cant sets, Math. Nachr., 2003, 248/249: [5] Feng, D.J., Wen, Z.Y. and Wu, J., Some dimensional results f homogeneous Man sets, Sci. China Math., 997, 405): [6] Garcia, I. and Zuberman, L., Exact packing measure of central Cant sets in the line, J. Math. Anal. Appl., 202, 3862): [7] Marion, J., Mesures de Hausdff et théie de Perron-Frobenius des matrices non-négatives, Ann. Inst. Fourier Grenoble), 985, 354): in French). [8] Mattila, P., Geometry of Sets and Measures in Euclidean Spaces: Fractals and Rectifiability, Cambridge Stud. Adv. Math., Vol. 44, Cambridge: Cambridge Univ. Press, 995. [9] Qu, C.Q., Rao, H. and Su, W.Y., Hausdff measure of homogeneous Cant set, Acta Math. Sin., Engl. Ser., 200, 7): -. [0] Tayl, S.J. and Tricot, C., Packing measure, and its evaluation f a Brownian path, Trans. Amer. Math. Soc., 985, 2882): [] Tayl, S.J. and Tricot, C., The packing measure of rectifiable subsets of the plane, Math. Proc. Cambridge Philos. Soc., 986, 992): [2] Zhou, Z.L. and Wu, M., The Hausdff measure of a Sierpinski carpet, Sci. China Math., 999, 427): Cant /,2)- =57, ;98 2, :<6 3. fflfiπ ß ßßfi, ±$, ±Ξ, 50640; 2. ψ ßfi ßffνΦßfi, ψ, ±Ξ, 52606; 3. "» ßflfflßfi, ±$, ±Ξ, 50275) 43 >OFAJPEIMKB Cant HCN@?DCGL..0* N@?D; KB Cant H