Key words and phrases. Hausdorff measure, self-similar set, Sierpinski triangle The research of Móra was supported by OTKA Foundation #TS49835

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1 Key words and phrases. Hausdorff measure, self-similar set, Sierpinski triangle The research of Móra was supported by OTKA Foundation #TS49835 Department of Stochastics, Institute of Mathematics, Budapest University of Technology and Economics, Budapest H-1521 B.O.box 91, Hungary, ESTIMATE OF THE HAUSDORFF MEASURE OF THE SIERPINSKI TRIANGLE Abstract. It is well-known that the Hausdorff dimension of the Sierpinski triangle Λ is s = log 3/ log 2. However, it is a long standing open problem to compute the s-dimensional Hausdorff measure of Λ denoted by H s (Λ). In the literature the best existing estimate is H s (Λ) In this paper we improve significantly the lower bound. We also give an upper bound which is weaker than the one above but everybody can check it easily. Namely, we prove that 0.77 H s (Λ) holds. 1

2 2 1. Introduction In this paper we consider the Sierpinski triangle or gasket Λ. This is constructed as follows: take an equilateral triangle of side length equal to one, remove the inverted equilateral triangle of half length having the same center, then repeat this process for the remaining triangles infinitely many times as showed on Figs. 1, 2. E 3 E 33 E 31 E 32 E 1 E 2 E 13 E 11 E 12 E 23 E 21 E 22 Figure 1. The triangles at the 1st, 2nd and the 3rd level Figure 2. The Sierpinski triangle In this paper we assumed that the diameter of the Sierpinski triangle is equal to 1. If the Sierpinski triangle is rescaled in such a way that its diameter is equal to t then the lower and upper bounds should be multiplied by t log 3/ log 2.

3 ESTIMATE OF THE HAUSDORFF MEASURE OF THE SIERPINSKI TRIANGLE3 The Sierpinski triangle is one of the most famous fractals, and the Hausdorff dimension and measure are the most important characteristics of a fractal sets. The Sierpinski triangle is defined by an iterated function system, which satisfies a technical condition called the open set condition (OSC). Thus it follows from Hutchinson s Theorem [1] that the Hausdorff dimension is equal to s = log 3/ log 2, and the s- dimensional Hausdorff measure H s (Λ) of Λ is positive and finite. Since the Sierpinski triangle has an important role in many applications, it would be desirable to get a better understanding of its size. Therefore in the last two decades there have been a considerable attention paid to the computation of the s-dimensional Hausdorff measure of the Sierpinski triangle: In 1987 Marion [2] showed that is an upper bound. In 1997 this was improved to 0.915, and later to 0.89 by Z. Zhou [3], [4]. In 2000 Z. Zhou and Li Feng proved that H s (Λ) in [5]. The best upper bound is , which was given by Wang Heyu and Wang Xinghua [6] in 1999 (in Chinese) with a computer algorithm. In 2002 B. Jia, Z. Zhou and Z. Zhu [7] showed that 0.5 is a lower bound on the s-dimensional Hausdorff measure of the Sierpinski triangle. In 2004 R. Houjun and W. Weiyi [8] improved it to Finally, in 2006 B. Jia, Z. Zhou and Z. Zhu [9] proved that is a lower bound. The main result of this paper is that H s (Λ) The difficulty comes from geometry. The s-dimensional Hausdorff measure of Λ is defined by

4 4 (1.1) H s (Λ) = lim inf { A k s, where A k < δ and δ 0 k A k is a countable cover of Λ }, where A k denotes the diameter of the set A k. When we estimate the Hausdorff measure we need to understand what is the most economical (in the sense of (1.1)) system of covers. Our most natural guess for this system is the covers by the level n triangles (the equilateral triangles on Fig. 1). However, this system of covers would result that the s- dimensional Hausdorff measure of Λ was equal to 1. On the other hand it is known that H s (Λ) < Therefore the best system of covers cannot possibly be the trivial one and this makes the problem difficult. To improve the existing best estimate on H s (Λ) we use a Theorem of B. Jia. [10]. To state this Theorem we need to introduce some definitions. It is well known (see [11]) that 3 Λ = S i (Λ), i=1 where ( 1 S 1 (x, y) = 2 x, 1 ) 2 y, ( 1 S 2 (x, y) = x, ) 2 y, ( 1 S 3 (x, y) = ) 3 2 x, y.

5 ESTIMATE OF THE HAUSDORFF MEASURE OF THE SIERPINSKI TRIANGLE5 Let E be the equilateral triangle of side length one with vertices: (0, 0), ) (1, 0),. Now we define the level n triangles ( 1 2, 3 2 E i1...i n := S i1...i n (E) = S i1 S in (E) for all (i 1... i n ) {1, 2, 3} n. Let µ be the uniform distribution measure on the Sierpinski triangle that is for all n and for all i 1... i n µ(e i1...i n ) = 1 3 n. After B. Jia we introduce the sequence (1.2) a n = min k n j=1 (n) j s k n /3 n = min kn j=1 (n) j s µ( k n j=1 (n) j ), where the minimum is taken for all non-empty sets of distinct level n { triangles (n) 1,..., (n) k n }. It is easy to see that a n is non-increasing (see [10]). Further B. Jia showed ([10]) that a n is an upper bound on the Hausdorff measure of the Sierpinski triangle, and he also gave a lower bound using a n : Theorem 1 (B. Jia). The Hausdorff measure of the Sierpinski triangle satisfies: (1.3) a n e s ( 1 2) n H s (Λ) a n This Theorem implies that a n tends to H s (Λ). Unfortunately there seems to be no way to compute a n for n 6. B. Jia [10] calculated a 1 and a 2. We can calculate a 3, a 4, a 5, but by (1.3) it results only that H s (Λ) > 0.54, which is not an improvement on the

6 6 already existing lower bound. So instead of this direct approach we give a lower bound on a n for every n. By using (1.3) this lower bound is also a lower bound on H s (Λ). Using some complicated algorithm ribed in Sec. 4, 5 we point out that a n 0.77 for all n N. With Theorem 1 this implies that H s (Λ) We remind the reader that the best existing lower bond in the literature [9] was given in 2006: H s (Λ) Using the second inequality of Theorem 1, in Sec. 2 an upper bound is given on H s (Λ) as follows: we provide a carefully selected collection of level 30 triangles (30) 1,..., (30) k 30 { }. This collection results an upper bound on a 30 which in return gives the upper bound H s (Λ) In 1999 two Chinese mathematicians [6] published an upper bound which is better than this but their paper was published in Chinese giving in this way limited opportunity to check if their algorithm was correct. Remark 1. I want to thank my supervisor, Károly Simon for his support writing this article. 2. Upper bound In the definition of a n (1.2) the minimum is taken for all non-empty sets of distinct level n triangles. We provide a collection of level n

7 ESTIMATE OF THE HAUSDORFF MEASURE OF THE SIERPINSKI TRIANGLE7 triangles for all n, which gives an upper bound on a n by definition, and an upper bound on H s (Λ) by Theorem 1. Take the following 6 points: {(1/4, 0), (3/4, 0), (1/8, 3/8), (3/8, 3 3/8), (5/8, 3 3/8), (7/8, 3/8)}. Let D 1, D 2,..., D 6 be the closed discs centered at these six points with radius We write D := D 1 D 2 D 6. Take all those level n triangles, which are contained in D (see Fig. 3 for an example). It is easy to see that the maximum distance between the chosen triangles will be exactly Let us denote c n = 0.75log 3/ log 2 k n /3 n, where k n is the number of the chosen level n triangles, which are in the region of intersection of the six discs. The values for the c n for small n are given by the following table:

8 8 Figure 3. The black triangles are the chosen 174 level 0.75log 3/ log 2 5 triangles, so c 5 =. Six arrows show the six 174/3 5 given points. Number of chosen n 0.75log 3/ log 2 n triangles (k n ) k n /3 k n/3 n Therefore using Theorem 1 we obtain that

9 ESTIMATE OF THE HAUSDORFF MEASURE OF THE SIERPINSKI TRIANGLE9 holds. This implies: c 30 a 30 H s (Λ) Theorem 2. The Hausdorff measure of the Sierpinski triangle is less than One can show we cannot get a better upper bound on the s-dimension Hausdorff measure of the Sierpinski triangle than Lower bound, basic idea For the convenience of the reader after giving the necessary definitions we are going to present a strongly simplified rough version of the idea of the algorithm. In Sec. 5. we will present the algorithm itself. Definition 1. Let g > h be positive integers, and let (g) 1, (g) 2,... (g) k be a set of distinct level g triangles, (h) 1, (h) 2,... (h) l be a set of distinct level h triangles. We say that the set { (g) i } k i=1 is a endant of the set { (h) j } l j=i and we write { (h) the following conditions hold: j } l j=i { (g) } k i=1, if both of i For all i {1, 2,... k} there is a j, such that (g) i (h) j. For all j {1, 2,... l} there is at least one i, such that (g) i (h) j. See Fig. 4 for an example. This relation naturally defines a tree T for which the equilateral triangle E is the root. The set of level n nodes is equal to the set of all (non-empty) union of level n triangles. A level

10 10 Figure 4. The set in the middle is a endant of the left one, but the set on the right is NOT a endant of the left one. n node { (n) j { (n+1) i the tree. } l j=1 is connected to a level (n + 1) node { (n+1) } k i=1 if } k i=1 is a endant of { (n) j } l j=1. Figure 5 shows the top of i Figure 5. The top of the tree T. Let v = { (n) i } k i=1 be a level n node. Then we write v 0 = n and we denote T v the sub tree of T having v as root. (T v consists of v and all those nodes w, which are endant of v.) Let E v := k i=1 (n) i.

11 ESTIMATE OF THE HAUSDORFF MEASURE OF THE SIERPINSKI TRIANGLE11 We define a v := E v s k/3 n = E v µ(e v ). Our purpose is to give a lower bound on a n (defined in (1.2)) for sufficiently large n, so we obtain a lower bound on its limit, H s (Λ). It comes directly from the definitions that a n = min v =n a v. By using a n H s (Λ) and taking infimum on both sides on n we obtain (3.1) inf v T a v = lim n a n = H s (Λ). Let v = { (n) i } k i=1. We write b v := max 1 i,j k x (n) i Observe that for these x, y we have min,y (n) j x y s k/3 n. x y = max 1 i,j k dist( i, j ). Lemma 1. The value b v is a lower bound for a w whenever v w holds. Namely, Proof. For v = { (n) i b v inf w T v a w. } k i=1 let w = { (g) t } l t=1 T v be arbitrary. To give a lower bound on a w first we give a lower bound on the diameter of E w, then we give an upper bound on µ(e w ). We consider (n) i some 1 i, j k. w is a endant of v, so E w (n) i and (n) j for and E w (n) j are non-empty (see Fig. 6 for example). Thus the diameter of E w is at

12 12 All the endants will have a part in these places, so the diameters will be at least this long. Diameter of a endant. Figure 6. The set on the right is a endant of the left one. least E w > min x (n) i,y (n) j x y = dist( (n) i, (n) j ). This inequality holds for all 1 i, j k, so we can take the maximum over these pairs: E w > max 1 i,j k x (n) i min,y (n) j x y. Because v w, we have E w E v. This yields l/3 g = µ(e w ) µ(e v ) = k/3 n, therefore b v = max 1 i,j k x (n) i min,y (n) j x y s k/3 n E w s l/3 g = a w.

13 ESTIMATE OF THE HAUSDORFF MEASURE OF THE SIERPINSKI TRIANGLE13 First of all we give a lower bound only on a subtree defined by a finite set of nodes A. We define the set T A as follows: v T A if v A, or v is a endant of a node w, which is in A. Namely, We write T A = w A T w. B A = min v A b v. We can apply the previous Lemma for every node in the set A, thus we have (3.2) B A inf v T A a v. We are going to apply this inequality for the so called cross-sections. These are some subsets C T of the nodes such that the lower bound on a v, v T C is a lower on the Hausdorff measure of the Sierpinski triangle. To make this definition precise first we define the set of the parents of C called P C as P C = {v v w, w C}, namely P C is the set of nodes, which have a endant in C. Definition 2. We call a finite set C T a cross-section, if there exists a function ϕ, ϕ : T \ (T C P C ) T C P C such that for every node v T \ (T C P C ) we have a ϕ(v) a v,

14 14 and µ(e ϕ(v) ) 3µ(E v ). Let v be a level n node. For a k > n we write Γ k (v) for that level k endant of v which has maximal µ measure. That is Γ k (v) = {E i1,...i k E i1,...i k E v }. See Fig. 7. We remark that (3.3) a v = a Γk (v). Namely, E v = E Γk (v) and µ(e v ) = µ(e Γk (v)) hold. Fact 1. Let H be an arbitrary subset of T. Then for every k 0 we have (3.4) inf a v = T inf a v. v T H v T H {w w k} Proof. It is enough to verify that inf a v T inf a v T H v T H {w w v k} holds. To do so, let u T H \ {w w k} be arbitrary. By using Γ k (u) T H {w w k} and (3.3) we have a u = a Γk (u) T inf a v, v T H {w w k} which completes the proof.

15 ESTIMATE OF THE HAUSDORFF MEASURE OF THE SIERPINSKI TRIANGLE15 Figure 7. The node v on the left is a level 2 node, the node on the right is the node Γ 4 (v). For a cross-section C we define B C = min v C b v. Lemma 2. For every cross-section C we have (3.5) inf v T a v = inf v T C a v, and (3.6) B C inf n a n = H s (Λ). Proof. It is easy to see that (3.6) is an immediately consequence of (3.5). Namely, using (3.1) and (3.2) we have B C inf a v = inf a v = inf a n = H s (Λ) v T C v T n

16 16 Let M C be the maximum level of the nodes which are contained in the set C. We define K MC = { } v v 0 M C. Using Fact 1 we have inf a v = v T inf a v, and inf a v = v T K MC v T C Thus to prove (3.5) it is enough to verify that inf a v. v T C K MC (3.7) inf a v = inf a v v T K MC v T C K MC holds. We fix a v K MC \ T C. To verify (3.7) we will show that there exists a node t K MC T C such that (3.8) a v a t holds. Since K MC P C =, thus v T \(T C P C ). C is a cross-section, by definition there exists ϕ such that ϕ(v) T C P C. If ϕ(v) K MC, then ϕ(v) T C as well, so (3.8) follows from choosing t = ϕ(v) and by using a v a ϕ(v). If ϕ(v) P C, then let us consider Γ MC (ϕ(v)). If Γ MC (ϕ(v)) T C then t := Γ MC (ϕ(v)) yields (3.8). If Γ MC (ϕ(v)) / T C then by (3.3) and by

17 ESTIMATE OF THE HAUSDORFF MEASURE OF THE SIERPINSKI TRIANGLE17 the definition of ϕ and Γ MC we have a v a ϕ(v) = a ΓMC (ϕ(v)) (3.9) µ(e ΓMC (ϕ(v))) 3µ(E v ) Γ MC (ϕ(v)) K MC \ T C So, we can repeat the same for the node w 1 := Γ MC (ϕ(v)) instead of v. If Γ MC (ϕ(w 1 )) T C then we are ready as we saw above. If not then (3.9) holds for w 1 instead of v. Note that this follows that 0 < 9µ(E v ) µ(e ΓMC (ϕ(w 1 ))) 1. This shows that there must exists a finite N such that Γ MC (ϕ(w N )) T C, where w k+1 := Γ MC (ϕ(w k )). This completes the proof of (3.8) Take the following set: (3.10) C 0 = {v v 0 = 2, v T {E1,E 2 } T {E1,E 3 } T {E2,E 3 }, v {E 1,2, E 2,1 }, v {E 1,3, E 3,1 }, v {E 2,3, E 3,2 }} {{E 1, E 2, E 3 }}. See Fig. 1 for labelling. There are 7 7 = 49 endants of the node {E 1, E 2 } at level 2. Counting the same for {E 1, E 3 } and {E 2, E 3 } we have 3 49 = 147 nodes. Let us remove the nodes {E 1,2, E 2,1 }, {E 1,3, E 3,1 }, {E 2,3, E 2,3 }, and take the node {E 1, E 2, E 3 }, so we get the set C 0. Thus C 0 consists of = 145 nodes. Proposition 1. The set C 0 is a cross-section.

18 18 Proof. Note that T \ (T C0 P C0 ) = T E1 T E2 T E3 T {E1,2,E 2,1 } T {E1,3,E 3,1 } T {E2,3,E 2,3 }. To define the function ϕ in the Definition 2, first we define an auxiliary function ψ : T \ (T C0 P C0 ) T. (See Figs. 8 and 9.) For v T {Ej }, where j = 1, 2, 3 let ψ(v) := {E i2,i 3,...,i n E j,i2,i 3,...,i n v}, for v T {E1,2,E 2,1 } let ψ(v) := {E 1,i1,i 2,...,i n E 1,2,i1,i 2,...,i n v} {E 2,i1,i 2,...,i n E 2,1,i1,i 2,...,i n v}, for v T {E1,3,E 3,1 } let ψ(v) := {E 1,i1,i 2,...,i n E 1,3,i1,i 2,...,i n v} {E 3,i1,i 2,...,i n E 3,1,i1,i 2,...,i n v}, for v T {E2,3,E 3,2 } let ψ(v) := {E 2,i1,i 2,...,i n E 2,3,i1,i 2,...,i n v} {E 3,i1,i 2,...,i n E 3,1,i1,i 2,...,i n v}. Clearly, E ψ(v) = 2 E v and µ(e ψ(v) ) = 3µ(E v ). Thus we have a ψ(v) = a v.

19 ESTIMATE OF THE HAUSDORFF MEASURE OF THE SIERPINSKI TRIANGLE19 Figure 8. The node v on the left is a endant of the node {E 1 }, the node on the right is ψ(v). The arrow shows the fix point of rescaling. Figure 9. The node v on the left is a endant of the node {E 2,3, E 3,2 }, the node on the right is ψ(v). The arrow shows the fix point of rescaling. This follows that for every v T \ (T C0 P C0 ) there exists an N such that C 0 is a cross-section with the function (See Fig. 10.) ϕ(v) := ψ N (v) = ψ ψ(v) T }{{} C0 P C0. N

20 20 Figure 10. The node v on the left is a endant of the node {E 1,3, E 3,1 }. The node in the middle is Ψ(v). The node on the right is ϕ(v) = Ψ(Ψ(v)). For the convenience of the reader we present a simplified algorithm for choosing cross-sections C n in the next Sec. Finally, in Sec. 5 we improve this algorithm significantly by using symmetries and a convexity argument. 4. Algorithm Our purpose is to choose cross-sections C n in such a way that B Cn gets as large as possible, but a computer can check it in acceptable length of time. It is a natural idea to choose a starting cross-section, and modify it in hope to get a better lower bound. For n = 0 take the set C 0 defined in (3.10). For every n in the n-th step find a node v C n where b v = min w C n b w. To obtain C n+1 from C n we throw away v from C n and we add to C n all the next level endants of v. It follows from the definition of b v that B Cn+1 B Cn.

21 ESTIMATE OF THE HAUSDORFF MEASURE OF THE SIERPINSKI TRIANGLE21 The following algorithm consists of three steps. It gives a lower bound on the s-dimensional Hausdorff measure of the Sierpinski triangle every time it reaches Step 2. It will run forever, but during its running it will give better and better lower bounds. Algorithm 1. Step 1. Start with the set C 0 from the previous Section. Let n := 0. Step 2. Find min v Cn b v. Below we prove that C n is a cross-section. So, it follows from Lemma 2 that we have min b v H s (Λ). v C n Step 3. Find a node v C n for which b v = min w Cn b w (if such a v is not unique, then choose any of them). Let us suppose v is level m node. We define S n as the set of all of those level m + 1 nodes, which are endants of the node v. That is S n := {w w 0 = m + 1, v w}, Let C n+1 := S n C n \ {v}. Increase n by 1. Go to Step 2. Above we used the fact that C n is a cross-section for every n. This is so because we have already seen that C 0 is a cross-section and T Cn+1 P Cn+1 = T Cn P Cn holds for all n.

22 22 5. Lower bound, making the algorithm faster Our aim here is to improve the algorithm presented in the previous Section. To do so, for every n we define a cross section Q n. Namely, let Q 0 := C 0. Assume that Q n is already defined. To define Q n+1 first we define a certain set of nodes D n Q n as it is detailed later in the Section. It is important that the set D n is much smaller than Q n. We choose a v D n for which (5.1) b v = min w D n b w. Then the special choice of D n will guarantee that (5.2) b v H s (Λ). To get Q n+1 we replace v (defined in (5.1)) with its next level endants. To define D n we need to introduce the notion of the convexity of a node. We remark that D n will consist only of convex nodes. Definition 3. Let v be a level n node. We write conv(v) = {E i1,i 2,...,i n E i1,i 2,...,i n is contained in the convex hull of E v }. See Fig. 11. for an example. We call a node v convex, if v = conv(v), otherwise we call it non-convex. Lemma 3. Let v be a non-convex level n node. If v is a level m endant of the node v, and Θ conv(v) \ v is a level n triangle, then the closed convex hull of E v intersects Θ.

23 ESTIMATE OF THE HAUSDORFF MEASURE OF THE SIERPINSKI TRIANGLE23 Figure 11. The node v consists of the black triangles. The convex hull of E v is showed with dashed lines. The black and the gray triangles together form the node conv(v). Proof. We assume that v = { i } k i=1. By definition of convexity we have { k } k Θ α i x i x i i, α i 0, α i = 1. i=1 i=1 For i = 1, 2,..., k let t i i be arbitrary points. To verify the assertion of the Lemma it is enough to show that { k } k (5.3) Θ α i t i α i 0, α i = 1 i=1 i=1 holds for every choice of t 1,..., t k. We prove it by contradiction. Let us suppose there exist t 1,..., t k such that (5.3) does not hold. Then there exists a line e, such that e separates Θ and the convex hull of t 1,..., t k. Let a be one of the normal unit vectors of e. Put r := z a, where z e arbitrary, and dot means the scalar product. Let us define (5.4) q := max (x y) a. x,y Θ

24 24 Without loss of generality we may assume that and {( k ) max α i t i a α i 0, i=1 min x a > r x Θ } k α i = 1 < r hold, otherwise take a instead of a. The last inequality and (5.4) implies that i=1 max x Θ x a > q + r, let us denote x 0 where the maximum is attained. Since Θ conv(v)\v, thus for i = 1, 2,..., k there exist u i i, and β 0 such that x 0 = k β i u i i=1 holds. Using the fact all level n triangles are translations of each others and using (5.4), for i = 1, 2,..., k we have (u i t i ) a q. Observe that ( k ) ( k ) q + r < x 0 a = β i t i a + β i (u i t i ) a < r + q, i=1 i=1 which is a contradiction, and completes the proof.

25 ESTIMATE OF THE HAUSDORFF MEASURE OF THE SIERPINSKI TRIANGLE25 The next Lemma shows that for any endant w of a non-convex node v the value of a w can be at most slightly bigger than some of the same level endants of conv(v). We will need this to verify (5.2). Lemma 4. Let v be a non-convex level n node and let m > n be arbitrary. If v conv(v) v, v = m, then there exists a node w, w = m, w, such that (5.5) E w E v m and hold. E v E w Proof. We write v = { i } k i=1, and conv(v) \ v = {Θ j } l j=1. Let us define the polygon H as the closed convex hull of E v. We proved in Lemma 3 that Θ j intersects H for 1 j l. If the polygon H intersects a triangle Θ j, then for all j there exists at least one level m triangle Θ j Θ j, such that Θ j intersects H as well. We write t j for a point where H intersects the triangle Θ j. Let w = v {Θ j} l j=1. Let q 0, w 0 E w be some points where the maximum E w = max q w, q,w E w is attained. If q 0, w 0 E v then we have E w = q 0 w 0 E v, thus the inequality (5.5) holds. If one of them is not in E v, let say q 0,

26 26 then there exists a j such that q 0 Θ j and w 0 E v. Using triangle inequality we have E w = q 0 w 0 q 0 t j + t j w 0 Θ j + E v = 1 2 m + E v because q 0, t j Θ j. If both q 0 and w 0 are not in E v, then using triangle inequality twice we have E w 2 2 m + E v. The following Lemma helps us to reduce the number of cases to be checked in an analogous way to the previous Lemma. Lemma 5. Let v = { i } k i=1 be a level n node, = E i 1,...,i n be a level n triangle such that v. Further, let x be one of the vertices of the triangle. We write D(x, r) for the closed disc centered at x with radius r. If E v D(x, max 1 i,j k dist( i, j )) holds then for all level m endant v of the node v there exists a level m triangle, such that E v E v m Proof. Let be that level m triangle, which has x as one of its vertices, and. As we saw in the proof of Lemma 1, max 1 i,j k dist( i, j )) is a lower bound on E v. Furthermore, E v = max q w. q,w E v Let q 0, w 0 be those points where this maximum is attained. Either q 0, w 0 E v, or one of the points, let us say q 0, and w 0 E v.

27 ESTIMATE OF THE HAUSDORFF MEASURE OF THE SIERPINSKI TRIANGLE27 By using = 1/2 m and triangle inequality both cases implies the statement. The following Theorem will show (with D = D n ) how the sequence of sets {D n } n=0 mentioned in the introduction of this Section gives us a lower bound on the Hausdorff measure H s (Λ). Then after this theorem we will construct {D n } n=0. Theorem 3. Let Q T be a cross-section. We choose an arbitrary D Q which satisfies the following assumption: For all v Q \ D there exists a node w T Q P Q, such that w = v, E v E w, for v v there exists a w w with v = w =: m such that E v E w and E w E v m. Then B D = min t D b t H s (Λ). Proof. Let us denote the finitely many elements of Q \ D by: Q \ D = {v 1, v 2,..., v k }. Let ε > 0 be arbitrary. Choose an M > max v Q v which also satisfies ( ) k s < 1 + ε, δ2 M

28 28 where δ := inf v τ C0 E v. We remind the reader that C 0 was defined in (3.10). It is easy to see that δ > 0. Recall that K M = {v v 0 M}. Fix an arbitrary v 0 K M T Q. To prove the assertion of the Theorem, it is enough to show that (5.6) a v 0 B D 1 + ε. Namely, H s (Λ) = inf v T a v = inf v T Q a v = inf a v v K M T Q B D 1 + ε, here we used first (3.1) then (3.5) and at the third equality we used Fact 1. Now we define by mathematical induction a finite (at least one and at most k elements) sequence of nodes v 0, v 1,..., v l, where v l T D and v 0, v 1,..., v l 1 T Q \ T D. Namely, assume that we have already defined v n for an n 0. If v n T D, then let v l = v n be the last element of the sequence. Otherwise v n T Q \ T D, so there exists a node v in {v 1, v 2,..., v k } such that v in v n. By the assumptions of the Theorem there exist nodes w in w in and w i n, such that w i n, v n = w i n, E v n E w in and E w in E v n m. Now we

29 ESTIMATE OF THE HAUSDORFF MEASURE OF THE SIERPINSKI TRIANGLE29 let v n+1 := w i n. From this it immediately follows that E v n E v n+1 E v n E v n M δ δ M = So, using that µ ( E v n ) µ ( E v n+1 ) we obtain that δ2 M. (5.7) a v n = E v n s µ(e v n ) E v µ(e v n+1 )(1 + 2 ) = a v n+1 δ2 s (1 + 2 ). M δ2 s M n+1 s Note that for n = 0, 1, 2,..., l 1 we have E vin E win, v in v n, w in v n+1 and E v 1 E v 2 E v l. This yields that v i0, v i1,..., v il 1 are all different. This follows that l k holds and v l T D. By applying (5.7) l times we get / ( a v 0 a v l ) / l s ( B δ2 M D ) k s B D δ2 M 1 + ε, which gives (5.6) and completes the proof. In the following we present the Algorithm. We remark that the starting set can be reduced by using symmetry. We will consider it at the end of this Section. Algorithm 2. Step 1. Let Q 0 := C 0 (which was defined in (3.10)). Step 2. Let D 0 = {v v C 0, v is convex}. Let n := 0. Step 3. Find min v Dn b v. Below we prove that (5.8) min v D n b v H s (Λ)

30 30 holds. Step 4. Find a node v D n for which b v = min w Dn b w (if such a v is not unique, then choose any of them). Let U n be the set of non-convex endants of v in one generation. That is U n := {w w = v + 1, v w, w is non-convex }. V n := {w w = v + 1, v w, a level v + 1 triangle / w, such that the conditions of Lemma 5 holds by replacing n with w and v with w in Lemma 5.} Moreover, we define W n := {w w = v + 1, v w} \ (U n V n ). Note that the set U n V n W n contains all of those nodes which are endants of the node v in one generation. Let D n+1 := W n (D n \ {v}). Increase n by 1. Go to Step 3. The only thing remained to be done is to verify (5.8). To do so, we will use Theorem 3. Let us fix n, and consider the set Q n = D n (C 0 \ D 0 ) n 1 k=0 U k V k.

31 ESTIMATE OF THE HAUSDORFF MEASURE OF THE SIERPINSKI TRIANGLE31 In the following we will check the assumptions of Theorem 3 by replacing Q with Q n and D with D n. It is easy to see that Q n is a cross-section, because T Qn P Qn = T C0 P C0. For v Q n \ D n there exists an i = 0, 1, 2,..., n 1 such that an v U i V i, or v C 0 \ D 0. If v U i or v C 0 \ D 0, then v is non-convex. Let w = conv(v). We have v = w, and E v E w. Let v v be arbitrary. By using Lemma 4 for v and m = v, there exist w, w = v = m, such that E v E w and E w E v m. If v V i, then the conditions of Lemma 5 holds for v, n = v and for a level n triangle / v. Let w = v { }. We have v = w, and E v E w. Let v v be arbitrary. By using Lemma 5 there exists a level m := v triangle, such that E v E v m. By choosing w = v { }, we obtain that Using Theorem 3 we get E v E w and E w E v m. B Dn H s (Λ)

32 32 which completes the proof of (5.8). By symmetry we can assume that for every level 4 endants v of the node {E 1, E 2, E 3 } we have #(v (T {E11 } T {E12 } T {E13 })) #(v (T {E21 } T {E22 } T {E23 })) #(v (T {E31 } T {E32 } T {E33 })). To reduce the usage of the computer memory we modify the Algorithm 2. First we fix a constant Z. We store only those nodes, which are necessary to prove that a fixed constant Z is a lower bound on the Hausdorff measure of the Sierpinski triangle. Let D n = {v v D n, b v Z}. During the modified Algorithm we store D n instead of D n. We use D n to find a node v D n such that b v = min w Dn b w. If D n is the empty set, then otherwise we have Z < min v D n b v H s (Λ), min v D n b v = min v D n b v. If this modified Algorithm reaches a state where D n =, then by using inequality (5.8) we have Z < H s (Λ).

33 ESTIMATE OF THE HAUSDORFF MEASURE OF THE SIERPINSKI TRIANGLE33 6. Running results I wrote the program in C++ language. For Z = 0.73 the program runs for half an hour, for Z = 0.77 it runs for a 4 days. The best result, what I managed to reach, is The program is available as an electric supplement at my homepage: morap/sierpinski.zip References [1] Hutchinson, J. E., Fractals and Self-similarity, Indiana Univ. Math. J., 30(1981), [2] Marion J., Mesures de Hausdorff d ensembles fractals, Ann. Sci. Math. Quebec, 11 (1987), [3] Zuoling Zhou, The Hausdorff measures of the Koch curve and Sierpiński gasket, Prog. Nat. Sci., 7 (1997), [4] Zuoling Zhou, Hausdorff measures of Sierpiński gasket, Sci. China, A 40 (1997), [5] Zuoling Zhou, Li Feng, A new estimate of the Hausdorff measure of the Sierpiński gasket, Nonlinearity, 13 (2000), [6] Wang Heyu, Wang Xinghua, Computer search for the upper estimation of Hausdorff measure of classical fractal sets II Analysis of the coding and lattice tracing techniques for Sierpinski gasket as a typical example, Chinese J. Num. Math. and Appl., 21(1999): 4, pp [7] Baoguo Jia, Zuoling Zhou, Zhiwei Zhu, A lower bound for the Hausdorff measure of the Sierpinski gasket, Nonlinearity, 15 (2002), [8] R. Houjun, W. Weiyi, An Approximation Method to Estimate the Hausdorff Measure of the Sierpinski Gasket, Analysis in Theory and Applications, vol 20 (2004), no 2,

34 34 [9] Baoguo Jia, Zuoling Zhou, Zhiwei Zhu, B., A New Lower Bound of the Hausdorff Measure of the Sierpinski Gasket, Analysis In Theory And Applications, 22(2006),no. 1, [10] Baoguo Jia, Bounds of Hausdorff measure of the Sierpinski gasket, Journal of Mathematical Analysis and Applications, 330 (2007), no. 2, [11] K. Falconer, Fractal geometry - Mathematical Foundations and Applications, 2. ed. (Wiley, 2003).