An isodiametric problem of fractal dimension - Springer Link

Geom Dedicata (2015) 175:79–91 DOI 10.1007/s10711-014-0030-z ORIGINAL PAPER

An isodiametric problem of fractal dimension Xin-Rong Dai · Wei-Hong He · Jun Luo · Bo Tan

Received: 1 January 2013 / Accepted: 28 October 2014 / Published online: 18 November 2014 © Springer Science+Business Media Dordrecht 2014

Abstract For 0 < λ < 21 we consider the product Fλ = E λ × R, where E λ is the attractor of ln 2 the IFS {x → λx, x → 1−λ + λx} on R. The Huasdorff dimension of Fλ is s = 1 − ln λ . We s H (X ∩Fλ ) : |X | > 0 = 1 and that there is a convex compact set A (= A(λ)) show that sup |X |s with

Hs (A∩Fλ ) |A|s

= 1. Such a convex compact set A is called an “extremal set” of Fλ with

respect to s-dimensional Hausdorff measure Hs . When λ is small, say λ ≤ 15 , we further show that there exists an extremal set A with |A| ≥ √2 such that Hs (A ∩ Fλ ) = Hs (D|A| ∩ Fλ ) 3 2 for D|A| = (x, y) : x − 21 + y 2 ≤ 41 |A|2 . As an application, we can estimate the value of Hs (E λ ×[0, 1]) to any pre-set error . Keywords

Isodiametric inequality · Isodiametric problem · Hausdorff measure

Mathematics Subject Classification (2000)

28A80 · 28A75

This research is partially supported by China’s NSFC (Nos. 11371383, 11371379, 10971233 and 11171123) and by the Fundamental Research Funds for the Central Universities. X.-R. Dai · W.-H. He · J. Luo (B) School of Mathematics and Computational Science, Sun Yat-Sen university, Guangzhou 510275, China e-mail: [email protected] X.-R. Dai e-mail: [email protected] W.-H. He e-mail: [email protected] B. Tan School of Mathematics and Statistics, Huazhong University of Science and Technology, Wuhan 430074, China e-mail: [email protected]

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Geom Dedicata (2015) 175:79–91

1 Introduction Given a closed set F ⊂ Rd of Hausdorff dimension s ∈ (0, d) such that Hs (X ∩ F) ≤ |X |s for any compact set X ⊂ Rd , we introduce the (generalized) isodiametric problem on F, which asks for convex compact sets A ⊂ Rd with Hs (A ∩ F) = |A|s > 0. Such a set A is called an extremal set of F with respect to Hs . A fundamental goal in studying isodiametric problem on F is to determine the structure of an extremal set A by determining its shape, its location, and its diameter. Here, we note that the above isodiametric problem on a fractal set is different from the “isoperimetric problem” on self-similar sets like Sierpinski gasket discussed by Strichartz [8]. See Equations (3.1) and (3.2) of [8]. s d Recall that the s-dimensional Hausdorff measure of F ⊂ R is given by H (F) = s : F ⊂ lim Hδs (F), where Hδs (F) = inf |U | U | ≤ δ for any δ > 0. More, |U i i i i i δ→0

over, the Hausdorff dimension of F is dim H F := inf{t : Ht (F) = 0}, which is equal to sup{t : Ht (F) = ∞} [4, p. 29]. Let F ⊂ Rn be a self-similar set with open set condition whose Hausdorff dimension is s. It is well known that Hs (F ∩ X) ≤ |X |s for any X⊂ Rn . See, for instance, Proposition s (X ∩F) 2 of [10]. Moreover, we have sup H |X : |X | > 0 = 1 [9, Theorem 3.1]. When s ≤ 1, |s there are many examples of F on which the isodiametric problem has been solved. Example 1 For 0 < λ < 21 let E λ denote the attractor of the IFS {x → λx, x → 1 − λ + λx} on R. Then [0, 1] is an extremal set. See [1] for more examples of linear Cantor sets E, such that Hs (E) can be exactly calculated. Here , it is easy to verify that if [0, x] with λ < x ≤ 1 is an extremal set then x = 1. From this one may infer that if [0, x] with 0 < x ≤ 1 is an extremal set then x = λk for some k ≥ 0. Moreover, if [a, b] is an extremal set for 0 ≤ a ≤ λ and 1 − λ ≤ b ≤ 1 then (b − a)s = Hs (E ∩ [a, b]) = Hs (E ∩ [a, λ]) + Hs (E ∩ [1 − λ, b]) = Hs (E ∩ [0, λ − a]) + Hs (E ∩ [0, b − 1 + λ]) (by symmetry of E) ≤ (λ − a)s + (b − 1 + λ)s . Since the partial derivatives of f (x, y) = (x + y + 1 − 2λ)s − x s − y s are continuous and negative over (0, λ)2 , we have f (x, y) > f (λ, λ) = 0 for any choice of (x, y)T in [0, λ]2 \{(λ, λ)}. Setting x = λ−a and y = b−1+λ, we have (b−a)s ≥ (λ−a)s +(b−1+λ)s , where equality holds if and only if a = 0 and b = 1. Therefore, [a, b] with 0 ≤ a ≤ λ and 1 − λ ≤ b ≤ 1 is an extremal set if and only if a = 0 and b = 1. Combining this with self-similarity of E, we can see that A ∩ E λ is a similar copy of [0, 1] ∩ E λ for any other extremal set A.

√ T T , Example 2 Let F be the attractor of x → 13 x, x → 13 x + 23 , 0 , x → 13 x + 13 , 33 an IFS on R2 . As the projection onto horizontal axis sends F onto [0, 1], one can verify that √ T the equilateral triangle spanned by the points (0, 0)T , (1, 0)T and 21 , 23 is an extremal set. Here it is unclear whether A ∩ F is a similarity copy of ∩ F for any other extremal set A. Example 3 For 0 < λ ≤ 41 , let Cλ be the attractor of an IFS on R2 consisting of four contractions: x → λx, x → λx + (1 − λ, 0)T , x → λx + (0, 1 − λ)T and x → λx +

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(1 − λ, 1 − λ)T . Then the unit square [0, 1]2 is an extremal set. See [12] for further details. Here it is unclear whether A ∩ Cλ is a similarity copy of [0, 1]2 ∩ Cλ for any other extremal set A. However, when s > 1, except for cases that Hs | E is Lebesgue measure restricted to E, no extremal sets are known for the isodiametric problem on a self-similar set E satisfying open set condition and having Hausdorff dimension s. Even for a simple fractal such as Sierpinski’s triangle or Sierpinski’s carpet, little is known about the shape and location of its extremal sets. This paper considers the isodiametric problem on Fλ = E λ × R for 0 < λ < 21 , where E λ denotes the attractor of the iterated function system { f 1 (x) = λx, f 2 (x) = 1−λ + λx} on R. We also call it the “middle (1 − 2λ) Cantor set” in the sequel. Clearly, Fλ is a closed set. Firstly, we obtain some properties of Fλ that are useful in studying the isodiametric problem on Fλ . s H (X ∩Fλ ) ln 2 Theorem 1.1 Given λ ∈ (0, 21 ) and s = 1 − ln : |X | > 0 = 1 and λ , then sup |X |s there exists an extremal set A that intersects the two lines x = λ and x = 1 − λ at the same time; moreover, we have Hs (E λ1×[0,1]) Hs | Fλ = Hs−1 E × H1 . λ

Recall that the Steiner symmetrization of a convex compact set X ⊂ R2 about the horizontal axis, denoted as S (X ), consists of all the intervals t × [−at , at ] with at = 21 H1 (X ∩ L t ), for t × 0 running through the projection of X onto horizontal axis and the lines L t transversal to horizontal axis at t × 0. Clearly, |S (A)| ≤ |A| and Hs (A ∩ Fλ ) = Hs (S (A) ∩ Fλ ) for any convex compact set A ⊂ R2 . So, S (A) is an extremal set whenever A is. Therefore, we may just consider the extremal sets that are symmetric about the horizontal axis. Secondly, we continue to discuss thestructure of the specific extremal set A by comparing 1 2 1 2 2 A ∩ Fλ with D|A| ∩ Fλ , where D|A| = (x, y) : x − 2 + y ≤ 4 |A| is a disk centered at ( 21 , 0). Theorem 1.2 If A ⊂ R2 is an extremal set whose projection onto the horizontal axis is symmetric about the line x = 21 and if M ◦ S (A) denotes the Minkowski symmetrization Fig. 1 A rough depiction for F1/5 and D|A| for an extremal set A with |A| > √2 3

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: p1 , p2 ∈ S (A) of S (A), then D|A| is an extremal set with 21 , 0 + M ◦ S (A) ∩ Fλ = D|A| ∩ Fλ .

p1 − p2 2

Theorem 1.3 If λ ≤ 15 and if an extremal set A intersects both the two lines x = λ and x = 1 − λ then |A| ≥ √2 and 21 , 0 + M ◦ S (A) ∩ Fλ = D|A| ∩ Fλ . (See Fig. 1 for a 3 rough depiction of D|A| , when λ = 1/5). 2 ln 2 Then, for every fixed λ ∈ (0, 15 ], Dt = (x, y) : x − 21 + y 2 ≤ 41 t 2 and s = 1 − ln λ, we consider the maximum of Hs−1 E × H1 (Dt ∩ Fλ ) 2 λ (1.1) ϕ(λ, t) := , t≥ √ . ts 3 As limt→∞ ϕ(λ, t) = 0, the result of Theorem 1.3 indicates the existence of some tλ > √2 3 such that ϕ(λ, tλ ) = max ϕ(λ, t) : t ≥ √2 . The number tλ is even unique, by the following 3 theorem. Theorem 1.4 For every λ ∈ (0, 21 ) there is a unique tλ > 1 with ϕ(λ, tλ ) = max{ϕ(λ, t) : t ≥ 1}. Therefore, for 0 < λ ≤ 15 , the diameter of an extremal set A is equal to tλ λk for some integer k ≥ 0. The result of Theorem 1.4 says that the diameter of an extremal set is determined by λ. Combining this with Theorem 1.3, the structure of extremal sets for isodiametric problem on Fλ is essentially unique in the sense that the relative location, the shape (after a Steiner symmetrization and a Minkowski symmetrization), and the diameter are all determined by λ. Combining Theorems 1.1 and 1.4, we can see that Hs (E λ × [0, 1]) is the reciprocal of ϕ(λ, tλ ); so we can estimate Hs (E λ×[0, 1]) by calculating lower and upper bounds of ϕ(λ, tλ ), based on good approximations of the function t → Hs (Dt ∩ Fλ ) for λ ≤ 15 . We present our paper in the following order. In Sect. 2 we prove Theorem 1.1. In Sect. 3 we prove Theorems 1.2–1.4. As an application, in Sect. 4 we provide a concrete algorithm to estimate Hs (E λ ×[0, 1]) to any pre-set error. 2 Isodiametric problem on Fλ We will prove Theorem 1.1 by showing a series of claims, under the assumptions of Theorem 1.1. Claim 1. Hs (Fλ ) > Hs (E λ ×[0, 1]) > 0.

Proof For A, B ⊂ R with Ht (A) and Ht (B) positive, Marstrand shows that Ht+t (A× B) ≥

cHt (A)Ht (B) for some positive constant c independent of A, B. See [7, Corollary 1] or [4, p.92, Proposition 7.1]. Since Hs−1 (E λ ) = H1 ([0, 1]) = 1 > 0, we have Hs (E λ × [0, 1]) > 0. s s Claim 2. H (E λ ×[0, 1]) ≤ |U j | for any countable cover {U j : j} of E λ ×[0, 1]. In √ js s particular, H (E λ ×[0, 1]) ≤ 2 . Proof We may assume that j |U j |s < ∞ and that |U j | ≤ δ for some constant δ > 0. For each integer k ≥ 1 and each word α = i 1 i 2 · · · i k ∈ {1, 2}k , let f α = f i1 ◦ f i2 ◦ · · · ◦ f ik . Then

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1 k k k f α (E λ ) × iλ , (i + 1)λ : α ∈ {1, 2} , 0 ≤ i ≤ k λ

is a cover of E λ ×[0, 1]. As gα,i defined by gα,i f α (x) x −−−→ y λk (y + i) is a similarity contraction on R2 with ratio λk , gα,i (U j ) : j is a cover of f α (E λ )×[iλk , (i + 1)λk ] for 0 ≤ i ≤ λ1k and hence

1 λk

Uk :=

gα,i (U j ) : j

i=0 α∈{1,2}k

is a cover of E λ ×[0, 1], whose elements are of diameter no greater than λk δ. Therefore, we have s 1 s k k s + 1 λ H (E λ ×[0, 1]) ≤ lim inf 2 × |U | = |U j |s . j k→∞ λk j

j

Claim 3. Hs (X ∩ Fλ ) ≤ |X ∩ Fλ |s (≤ |X |s ) for any compact set X . Proof For any compact set X ⊂ R2 with diameter |X | > 0, we can fix an integer k ≥ 1 such that λk |X ∩ Fλ | < 1. Then we may choose a similarity contraction g : R2 → R2 with contraction ratio λk such that g(X ∩ Fλ ) is a subset of E λ × [0, 1]. As Hs (g(X ∩ Fλ )) = λks Hs (X ∩ Fλ ), Hs (X ∩ Fλ ) ≤ |X ∩ Fλ |s if and only if Hs (g(X ∩ Fλ )) ≤ |g(X ∩ Fλ )|s . So, we may assume that X is a subset of E λ×[0, 1]. Then we can find a cover {U j } of E λ×[0, 1]\X for an arbitrary ε > 0 such that j |U j |s is smaller than Hs (E λ ×[0, 1] \ X ) + ε, which is just Hs (E λ ×[0, 1]) − Hs (X ) + ε. Since {X } ∪ {U j } is a cover of E λ ×[0, 1], we can infer from Claim 2 that Hs (E λ ×[0, 1]) ≤ |X |s + |U j |s < |X |s + Hs (E λ ×[0, 1]) − Hs (X ) + ε. j

By arbitrariness of ε, this completes the proof for Claim 3. s ∩Fλ ) : |X | > 0 = 1. Claim 4. sup H (X |X |s

s ∩Fλ ) : |X | > 0 ≤ 1; so we just need to show that Proof By Claim 3, we have sup H (X |X |s

s

H (X ∩ Fλ ) sup : |X | > 0 ≥ 1. |X |s √ 2 Recall that 0 < Hs (E λ ×[0, 1]) ≤ 2 by Claims 1 and 2; in other words, E λ ×[0, 1] is an s-set. Thus the upper convex density of E λ ×[0, 1],

s

H (U ∩ (E λ ×[0, 1])) s D c (E λ ×[0, 1], p) := lim sup : p ∈ U, 0 < |U | ≤ r , r →0 |U |s is equal to one at Hs -almost all p ∈ E λ ×[0, 1]. See [3, p.24, Theorem 2.3]. This ends the proof.

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ln 2 Claim 5. For every 0 < λ < 21 and s = 1 − ln λ , let kλ be the smallest integer with √ s s kλ ≥ 2(kλ + 1) 2 . Then a convex compact subset X of [0, 1] × R with |X | ≥ kλ is not an extremal set of Fλ . Proof Let a = min y| (x, y)T ∈ X for some x ∈ [0, 1] . We may cover X by [|X |] + 1 sets of the form p + (0, a + i)T p ∈ E λ × [0, 1] , for integers i with 0 ≤ i ≤ [|X |]. Here √ s [|X |] is the integral part of |X |. Therefore, we have Hs (X ∩ Fλ ) ≤ ([|X |] + 1) 2 by Claim s x 2. As the function f (x) = 2(x+1) is increasing on (0, ∞), we have

√ s kλs [|X |]s > ≥ 2. 2([|X |] + 1) 2(kλ + 1) √ s Thus we further have |X |s ≥ [|X |]s > 2([|X |] + 1) 2 ≥ 2Hs (X ∩ Fλ ). This ends our proof. Claim 6. There is an extremal set A intersecting the two lines x = λ and x = 1 − λ at the same time. Proof As the upper convex density at a point (x, y) does not depend on the second coordinate s y, we may fix a point p = (x0 , 0) in E λ × [0, 1] with x0 ≤ λ such that D c (E λ ×[0, 1], p) = 1. Then we may choose for any k ≥ 1 a convex compact set Ck containing x such that Hs (Ck ∩ (E λ ×[0, 1])) > 1 − 2−k |Ck |s . Let Ck0 = Ck ∩ (E λ ×[0, 1]) and carry out the following procedure: • if the convex hull of Cki−1 , i.e. the smallest convex compact set containing Cki−1 , intersects the two lines x = λ and x = 1 − λ at the same time, then let Bk = Cki−1 and stop the procedure; • if the convex hull of Cki−1 lies to the left of the line x = λ, let Cki be the image of Cki−1 under the mapping x → λ−1 x on R2 ; • if the convex hull of Cki−1 lies to the right of the line x = 1 − λ, let Cki be the image of Cki−1 under the mapping x → λ−1 x + (1 − λ1 , 0)T . After finitely many steps, the above procedure will stop and provide a compact set Bk , a similarity copy of Ck ∩ (E λ × [0, 1]). Clearly, Bk hence its convex hull Ak intersects the two lines x = λ and x = 1 − λ at the same time. Moreover, Bk is a subset of Fλ intersecting the horizontal axis such that for all k ≥ 1 we have Hs (Bk )

|Bk |s

=

Hs (Ck ∩ Fλ )

|Ck |s

> 1 − 2−k .

Using the proof of Claim 5, we can infer that the diameter of its convex hull Ak is smaller than kλ in Claim 5. Therefore, {Ak } is a sequence of nonempty compact subsets of [0, 1] × [−kλ , kλ ], and we may choose a subsequence of {Ak : k} which is convergent under Hausdorff distance between compact sets. Clearly, the limit set A∞ intersects both the two lines x = λ and x = 1 − λ, and satisfies the following inequality Hs (A∞ ∩ Fλ ) ≥ lim sup Hs (Bk ∩ Fλ ) ≥ lim sup 1 − 2−k |Bk |s = |A∞ |s ≥ 1 − 2λ. k→∞

k→∞

That is to say, the convex set A∞ is an extremal set we need. Claim 7. Hs (E λ1×[0,1]) Hs | Fλ = Hs−1 E × H1 . λ

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Proof Given a compact X ⊂ R2 , let X n denote the n1 -neighborhood of X for n ≥ 1, i.e. the union of the open balls B(x, n1 ) with x ∈ X . Then {X n } is a sequence of open sets decreasing to X , and hence limn→∞ μ(X n ) = μ(X ) for μ = Hs | Fλ and μ = Hs−1 E × H1 . λ Recall that [0, 1] \ E λ is the disjoint union of countably many open intervals {Un }, such that U1 = (λ, 1− λ) is of length 1 − 2λ, U2 and U3 are of length λ(1 − 2λ); and in general, for 2k ≤ n ≤ 2k+1 − 1 , Un is of length λk (1 − 2λ). For k ≥ 1, let Ik be the difference 2k −1 between [0, 1] and n=1 Un . Then Ik is the union of 2k many disjoint closed intervals each of length λk and Fλ ⊂ Ik × R for any k ≥ 1. 1 Choose an increasing sequence of integers kn such that λkn < 2n , and fix a tiling Tn of k k n n Ikn × R by tiles of the form J × pλ , ( p + 1)λ , where J is one of the 2kn closed intervals in Ikn . Let Yn be the union of all the tiles in Tn that intersect X . Since s−1 1 s kn kn ) = λkn H (J × pλ , ( p + 1)λ λkn Fλ Hs (E λ ×[0, 1]) = Hs−1 (J ) H1 pλkn , ( p + 1)λkn , Eλ

we can infer that 1 Hs F (J × pλkn , ( p+1)λkn ) = Hs−1 E × H1 J × pλkn , ( p+1)λkn λ λ λ ×[0, 1]) Hs | F (Yn ) = Hs−1 × for any of interval J in Ik . So, for each Yn , we have s 1 Hs (E

H (E λ×[0,1])

n

λ

Eλ

H1 (Yn ). Combining this with the containments (X ∩ Fλ ) ⊂ (Yn ∩ Fλ ) ⊂ (X n ∩ Fλ ), we

have

1 Hs (E λ ×[0, 1])

Hs F (X ) = λ

1

lim Hs−1 E × H1 (Yn ) = Hs−1 E × H1 (X ).

Hs (E λ ×[0, 1]) n→∞

λ

λ

3 Shapes and locations of extremal sets for small λ In this section we prove Theorems 1.2–1.4. 3.1 Proof for Theorem 1.2 Proof By the definition of Steiner symmetrization Hs (S (X ) ∩ Fλ ) = Hs (X ∩ Fλ ) holds for any convex compact set X ⊂ R2 and its Steiner symmetrization S (X ) about horizontal axis. For each x ∈ E λ with {(x, y) : y ∈ R} ∩ S (X ) = ∅, let h(x) be the largest number such that {(x, y) : 0 ≤ y ≤ h(x)} ⊂ S (X ). If the projection of an extremal set A onto the horizontal axis is symmetric about the line x = 21 then the common part of S (A) and [0, 1] × {0} is a segment centered at the point 1 2 , 0 . Moreover, for each x ∈ E λ with {(x, y) : y ∈ R} ∩ S (A) = ∅, we may choose two points p1 =(x1 , ±h(x1 )) and p 2 = (1 − x1 , ∓h(1 − x1 )) in S (A) and verify that the common part of (x − 21 , y) : y ∈ R and

p1 − 21 , 0 p2 − 21 , 0 p1 − p 2 M ◦ S (A) = : p1 , p2 ∈ S (A) = − : p1 , p2 ∈ S (A) 2 2 2

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contains the vertical segment

1 h(x) + h(1 − x) h(x) + h(1 − x) x− × − , . 2 2 2 Since Hs (E λ1×[0,1]) Hs | Fλ = Hs−1 E × H1 by Theorem 1.1, we can infer that λ 1 Hs F , 0 + M ◦ S (A) ≥ Hs F (S (A)). λ λ 2 As M ◦ S (A) ⊂ p ∈ R2 : | p| ≤ 21 |A| , we can see that 21 , 0 + M ◦ S (A) is a subset of D|A| . Therefore, s 1 Hs F (D|A| ) ≥ Hs F , 0 + M ◦ S (A) ≥ Hs (A ∩ Fλ ) = |A|s = D|A| λ λ 2

hence D|A| is an extremal set that we need. Here, we want to note that if A is an extremal set with |A| ≥ S (A) about the horizontal axis implies that |y| ≤

|A| ≥

1 1 + |A|2 ≥ 4

|A| 2

√2 3

then the symmetry of

for each point (x, y) ∈ S (A). As

1 + y2 ≥

(x − x )2 + y 2

for every (x, y)T ∈ S (A) and every point (x , 0)T with x ∈ [0, 1], we can see that S (A) ∪ {(x, 0) : 0 ≤ x ≤ 1} is a set of diameter smaller than or equal to |A|. Therefore, the convex hull A of S (A) ∪ {(x, 0) : 0 ≤ x ≤ 1} is an extremal set with |A | = |A|. As S (A) is extremal, the set S (A) ∪ {(x, 0) : 0 ≤ x ≤ 1} and hence the interval [0, 1] × {0} is contained in S (A). This provides useful hints to construct the proof for Theorem 1.3 as follows. Proof Since the Steiner symmetrization does not increase the diameter of a set, we only need to show that the diameter of S (A) is greater than √2 . As S (A) is symmetric about 3 the horizontal axis, the interior of S (A) is bounded by two curves y = ± f (x) for some continuous map f : [0, 1] → [0, ∞). (If (x, 0) ∈ / S (A)we just put f (x) = 0). 2 Let h = max { f (x)+ f (x +1−λ) : 0 ≤ x ≤ λ}, D = (x, y) : x − 21 + y 2 ≤ 41 + h 2 , 2 and ψ(x) = 41 + h 2 − x − 21 . Then Hs−1 E × H1 (A) ≤ [ f (x) + f (x + 1 − λ)] d Hs−1 E ≤ h λ

and H

s−1

× H (D) = 4 1

Eλ

λ

[0,λ]

[0,λ]

ψ(x) d H

s−1

Eλ

=2

[0,λ]

(ψ(x) + ψ(λ − x)) d Hs−1 E λ .

As ψ(x) is concave on [0, λ], the above integral is greater than ψ (λ) d Hs−1 E ≥ ψ(0) + ψ(λ) = h + h 2 + λ − λ2 . 4 [0,λ]

λ

Given two points p1 := (x, f (x))T and p2 := (x + 1 − λ, − f (x + 1 − λ))T in S (A) with 0 ≤ x ≤ λ, we have (1 − λ)2 + ( f (x) + f (x + 1 − λ))2 = | p1 − p2 |2 ≤ |S (A)|2 ≤ |A|2 .

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AN ISODIAMETRIC PROBLEM OF FRACTAL DIMENSION 1.2 2.115

1.0

2.110

0.8 0.6

2.105

0.4 0.05

0.10

0.15

0.199

0.20

0.200

0.201

0.202

0.203

0.204

0.205

Fig. 2 The graphs of f 1 (λ) (thick) and f 2 (λ) (light)

So we have h 2 ≤ |A|2 − (1 − λ)2 and Hs−1 E × H1 (D) ≥ h + h 1 + λ

Then we may end our proof by claiming that |A| > |D|s |A|s

⎛ 3 − 3λ > f 1 (λ) We can infer that g(λ) = f 1 (λ) − f 2 (λ) > 0 and hence that Consequently, we have extremal.

Hs (A∩Fλ ) |A|s

|D|s |A|s ,

(3.1)

for λ ≤ 15 .

≤ 1, which contradicts to the fact that A is

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3.2 Proof for theorem 1.4. f (λ, t) = Hs−1 × Proof Writing Fλ (x) = Hs−1 (E λ ∩ [0, x]), we can represent 2 λ t2 1 − − x d Fλ (x) = H1 (Dt ∩ Fλ ) as a Stieltjes integral f (λ, t) = 4 4 2 0 λ 2 t 2 − (1 − 2x)2 d Fλ (x). Recall that 0

* ) λ td Fλ (x) f t (λ, t) s f (λ, t) s f (λ, t) −s ϕt (λ, t) = − s+1 = t − 2 ts t t 0 t 2 − (1 − 2x)2

is continuously differentiable with respect to t on (1, ∞). If we put λ λ 1 H1 (λ, t) = d Fλ (x), H2 (λ, t) = 1−(1−2x)2 /t 2 d Fλ (x), 1 − (1 − 2x)2 /t 2 0 0 H1 (λ,t) H1 (λ,t) then ϕt (λ, t) > 0 if and only if s < H . Clearly, limt→∞ H = 1 for all λ < 21 , so 2 (λ,t) 2 (λ,t) we can infer that ϕt (λ, t) < 0 for large enough t. On the other hand, λ2 λ 1 1 d Fλ (x) + d Fλ (x) H1 (λ, t) = 2 2 2 1 − (1 − 2x) /t 1 − (1 − 2x)2 /t 2 λ−λ 0 1 1 √ , and + is larger than 41 √ 2 2 2 2 2 1−(1−2λ ) /t

H2 (λ, t) = 0

λ2

1−(1−2λ) /t

1 − (1 − 2x)2 /t 2 d Fλ (x) +

λ

λ−λ2

1 − (1 − 2x)2 /t 2 d Fλ (x)

is smaller than 41 1 − (1 − 2λ2 )2 /t 2 + 1 − (1 − 2λ)2 /t 2 . Since t > 1 and λ < 21 , we have H1 (λ, t) 1 2 1 lim ≥√ >√ . = √ t1 H2 (λ, t) 3 4λ − 4λ2 4λ2 − 4λ4 1 − (1 − 2λ)2 /t 2 1 − (1 − 2λ2 )2 /t 2 Summing up the above arguments, we can infer that ϕt (λ, t) > 0 for t slightly larger than H1 (λ,t) 1 and that the equation s = H has at least one solution in (1, ∞). Since H1 (λ, t) and 2 (λ,t)

H1 (λ,t) H2 (λ, t) are respectively decreasing and increasing with respect to t, the ratio H is 2 (λ,t) decreasing with respect to t on [1, ∞). This indicates that, for any λ ∈ (0, 21 ), ϕt (λ, t) = 0 has a unique solution tλ in (1, ∞), such that

(1) ϕt (λ, t) > 0 on (1, tλ ) and hence ϕ(λ, t) strictly increases on [1, tλ ]; (2) ϕt (λ, t) < 0 on (tλ , ∞) and hence ϕ(λ, t) strictly decreases on [tλ , ∞). This means that tλ is the unique point at which ϕ takes its maximum over [1, ∞), which ends the proof for Theorem 1.4.

4 An application ln 2 We give an application of our results, in estimating the s (= 1 − ln λ ) dimensional Hausdorff s s−1 1 measure H (E λ×[0, 1]) of E λ×[0, 1]. Actually, H × H (Dt ∩ Fλ ) = Hs (E λ1×[0,1]) Hs (Dt ∩

123


89

Fλ ) holds for any t ≥ 1 and any λ ∈ (0, 21 ), so we may combine the results of Theorems 1.3 and 1.4 to infer that Hs (E λ ×[0, 1]) is equal to the reciprocal of ϕ(λ, tλ ) = max{ϕ(λ, t) : t ≥ 1}, for each λ ∈ (0, 15 ]. We will describe a concrete algorithm to calculate lower and upper bounds of the function f (λ, t), the difference between which can be as small as we wish. This helps us to efficiently estimate Hs (E λ ×[0, 1]). Let g1,λ (x) = λx, g2,λ (x) = λx+1−λ. Then E λ is the attractor of the IFS {g1,λ , g2,λ }, and H− ln 2/ ln λ (E λ ) = 1 for λ ∈ (0, 21 ) [3, Theorem 1.15, p.16]. Let Fλ (x) = Hs−1 (E λ ∩ [0, x]) for 0 ≤ x ≤ 1. Then we can rewrite f (λ, t) = Hs−1 × H1 (Dt ∩ Fλ ) as the following Stieltjes integral 2 λ 1 t2 (4.1) f (λ, t) = 4 − − x d Fλ (x). 4 2 0 Let g,λ = g1 ,λ ◦· · ·◦ gn ,λ for = 1 2 · · · n ∈ {0, 1}n . Then {g,λ ([0, 1]) : = 1 2 · · · n ∈ {0, 1}n } is a collection of 2n closed intervals of length λn . Let b,λ denote the na,λ andp−1 left and end points of g,λ ([0, 1]) respectively. Then a,λ = λ (1 − λ) and p p=1 b,λ = a,λ + λn for any = 1 2 · · · n ∈ {0, 1}n . 2 2 When t ≥ 1 is fixed, t4 − 21 − x is an increasing function of x on [0, 21 ], thus on [0, λ]. So, the following simple functions ⎧ ⎨ S (t, λ, n) = t 2 − 1 − a 2 x ∈ a,λ , b,λ , = 02 · · · n ∈ {0, 1}n L ,λ 4 2 ⎩ S (t, λ, n) = t 2 − 1 − a − λn 2 x ∈ a , b , = 0 · · · ∈ {0, 1}n U

4

2

,λ

,λ

,λ

2

n

(4.2)

λ

SL (t, λ, n)d Fλ and an upper bound fU (t, λ, n) = give a lower bound f L (t, λ, n) = 4 0 λ SU (t, λ, n)d Fλ of f (λ, t) for each n ≥ 2. As Fλ b,λ − Fλ a,λ = 2−n for any 4 0

= 02 · · · · n ∈ {0, 1}n , one may further check that ⎧ 2 n ⎪ 4 t2 1 p−1 (1 − λ) ⎪ f (t, λ, n) = ⎪ p=2 p λ 2n 4 − 2 − ⎨ L p ∈{0,1} 2 ⎪ ⎪ 4 t2 ⎪ f (t, λ, n) = − 1 − np=2 p λ p−1 (1 − λ) − λn n ⎩ U 2

For t ≥ than

√2 , 3

1

p ∈{0,1}

4

(4.3)

2

0 < λ < 21 , and n ≥ 2, we can infer that |SU (t, λ, n) − SL (t, λ, n)| is smaller

n 2 : = 0 · · · ∈ {0, 1}n−1 − a − λ ,λ 2 n √ 2 2 < 3λn . (4.4) 2 13 − 41 √ So 0 < fU (t, λ, n)− f L (t, λ, n) < 2 3λn for each n ≥ 2. Let An := {a0,λ : ∈ {0, 1}n−1 } for any n ≥ 2. Then A2 = {0, λ(1−λ)}, and An+1 = {λx : x ∈ An }∪{λ(x +1−λ) : x ∈ An } for any n ≥ 2. Therefore, ⎧ 1 2 t2 ⎪ ⎪ f L (t, λ, n) = 24n ⎨ 4 − 2 −x x∈An (4.5) 1 ⎪ 4 t2 n 2 ⎪ − − x − λ ⎩ fU (t, λ, n) = 2n 4 2 max

− a,λ

2

−

1

x∈An

123

90


Considering the case n = 4, we have 2 2 t2 t2 t2 1 1 1 2 3 4 2 f L (t, λ, 4) = − + − −λ +λ − − λ2 + λ3 + 4 4 4 2 4 2 2 2 1 1 t2 t2 − − λ2 + λ4 + − − λ + λ2 + 4 2 4 2 2 2 1 1 t2 t2 2 3 4 3 − −λ+λ −λ +λ − −λ+λ + + 4 2 4 2 2 1 t2 4 + − −λ+λ 4 2 2 2 2 2 2 1 1 1 t t t2 22 fU (t, λ, 4) = − − λ4 + − −λ3 + − − λ2 + λ3 − λ4 4 2 4 2 4 2 2 2 1 1 t2 t2 + − − λ2 + − − λ + λ2 − λ4 4 2 4 2 2 2 1 1 t2 t2 + − − λ + λ2 − λ3 + − − λ + λ3 − λ4 4 2 4 2 2 1 t2 + − −λ 4 2 We use NMaximize of Mathematica 8.0 to “maximize” two functions f L (t,λ,4) and fU (t,λ,4) , ts ts f L (t,λ,4) fU (t,λ,4) 2 1 such that < ϕ(λ, t) < . Recall that tλ > √ for λ ≤ 5 . We only maximize s ts 3 t 2 ϕ(λ, t) over √ , ∞ , whence the difference between f L (t,λ,4) and fU (t,λ,4) is smaller than ts ts 3

3λ4 . Consequently, for t ≥

obtain a pair of lower and upper bounds for ϕ(λ, tλ )

with an error smaller than

the following table for the estimates of ϕ(λ, tλ ) and

√2 , we 3 3λ4 . See

H1−ln 2/ ln λ (E λ ×[0, 1]), for specific choices of λ ∈ (0, 15 ]. Value of λ

Upper bound of ϕ(λ, tλ )

Lower bound of ϕ(λ, tλ )

Error estimates

Interval containing Hs (E λ ×[0, 1])

1 5 1 6 1 7 1 8 1 9 1 10

0.702626

0.701483

0.001143

(1.423232, 1.425551)

0.706784

0.706297

0.000487

(1.414859, 1.415835)

0.711554

0.711314

0.000240

(1.405375, 1.405849)

0.716226

0.716096

0.000130

(1.396207, 1.396461)

0.720599

0.720522

0.000077

(1.387734, 1.387825)

0.724629

0.724581

0.000048

(1.3800165, 1.38010795)

As ϕ(λ, tλ ) > ϕ λ, √2 > 3

√1 3

√ 1+ln 2/ ln 5 3 2

ing Hs (E λ ×[0, 1]) has a length smaller than

123

≈ 0.469965 for λ ≤ 15 , the interval contain-

3λn 2×0.469965

≈ 3.19172λn and decreases to zero


91

very quickly. This is really an efficient algorithm to estimate the value of Hs (E λ ×[0, 1]), which is independent of the choice of λ ≤ 15 . Acknowledgments The authors are grateful to the referee for many critics which improve clarity and readability of this paper. They also owe their thanks to Dr. Li Feng at Albany State University (Georgia) for discussions on related problems and for suggestions on English writing when they are making revisions of the paper.

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