On a class of fractal functions with graph Hausdorff dimension 2

Chaos, Solitons and Fractals 32 (2007) 1625–1630 www.elsevier.com/locate/chaos

On a class of fractal functions with graph Hausdorff dimension 2 q T.F. Xie a, S.P. Zhou b

b,*

a China Institute of Metrology, Hangzhou, Zhejiang 310034, China Institute of Mathematics, Zhejiang Sci-Tech University, Xiasha Economic Development Area, Hangzhou, Zhejiang 310018, China

Accepted 28 December 2005

Abstract We prove that the graph of the continuous function 1  b  X a y ¼ UðxÞ ¼ kk / kk x ; 0 6 x 6 1 k¼1

has Hausdorff dimension 2, where k > 1, b > a > 1, /(x) = 2x, 0 6 x 6 1/2, /(x) = /(x) and /(x + 1) = /(x). Ó 2006 Elsevier Ltd. All rights reserved.

1. Introduction Fractal curves is an interesting topic in the study of fractal geometry. Researchers are very interested in investigating dimensions of continuous curves y = f(x) (or graphs of the continuous functions), a 6 x 6 b. It seems that it is motivated by some problems recently raised in quantum physics by El Naschie [9]. Fractal dimensions is the most important attribute of fractals [2–5,9,12,13,15], and the Hausdorff dimension as well as the box counting dimension is widely used [16]. For s 2 (1, 2), we have some standard techniques to construct continuous curves y = f(x) with a given box dimension or a Hausdorff dimension s (see, [1,3,6–8,11]). However, these construction techniques are not usually applicable for the case s = 2. We know that the well-known Peano curve fills a square in the plane, and its graph obviously has box dimension or has Hausdorff dimension equal 2, but it cannot be formulated by an explicit expression y = f(x). Hence, it is an interesting subject to find out how to construct curves of the form y = f(x) which has box dimension or Hausdorff dimension 2. The history of the subject can be traced back to 1993, when Sun and Wen [10] obtained a continuous function, whose graph possesses box dimension 2. Using a simple constructive approach, our recent work [14] demonstrated a class of continuous functions with graph box dimension 2. One result shows that, when for an even number k P 2, if P 1 n¼1 an < 1; an > 0, log an = o(n), n ! 1, then the graph of the function q *

Supported in part by NSF of China (10471130). Corresponding author. E-mail address: [email protected] (S.P. Zhou).

0960-0779/$ - see front matter Ó 2006 Elsevier Ltd. All rights reserved. doi:10.1016/j.chaos.2005.12.038

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y ¼ f ðxÞ ¼

1 X

an cos kn x;

06x6p

n¼1

has box dimension 2. It is also clear that, there is an essential difference between box dimension and Hausdorff dimension. It is naturally to ask that if we are able to, by an explicit and simple way, construct a class of continuous functions with exact graph Hausdorff dimension 2. This paper will give an affirmative answer. Let /(x) = 2x, 0 6 x 6 1/2, and extend it to the interval [1/2, 1/2] as an even function (by defining /(x) = /(x)) and to the whole real line as a periodic function of period 1 (by defining /(x + 1) = /(x)). In other words, /(x) is a so-called ‘‘zigzag’’ function. We will prove that, for k > 1 and 1 < a < b, the Hausdorff dimension of the graph C(U) = {(x, y) : y = U(x), 0 6 x 6 1} of the function 1  b  X a UðxÞ ¼ kk / kk x ; 0 6 x 6 1 k¼1

is dimHC(U) = 2. In fact, with notation, namely, Hausdorff measure of C(U), ( ) 1 X Hs ðCðUÞÞ ¼ lim inf juk js : fuk g1  CðUÞ; ju j 6 d ; k k¼1 d!0

k¼1

where jukj indicates the diameter of uk, we have inffs : Hs ðCðUÞÞ ¼ 0g ¼ supfs : Hs ðCðUÞÞ ¼ 1g ¼ 2. It is easy to see that U(x) is a continuous but nowhere differentiable function. Since C(U) is given via y = U(x), 0 6 x 6 1, it cannot be dense in any region of the plane. Consequently, it is not difficult to construct a continuous surface with Hausdorff dimension 3, thus a tool for potential applications in physics is provided [9]. We will establish a probability measure in Section 2, which serves as a basic tool to prove dimHC(U) = 2. Notations and lemmas will be given in Section 3, and finally the main theorem will be proved in Section 4.

2. A probability measure We adopt repeated subdivision method (cf. [4]) to define a probability measure M (by making a mass distribution). For any given number k > 1 and b > 1, since b

lim

kðnþ1Þ b

kn

n!1

¼ 1.

Choose a natural number n0 such that, for all n P n0, b

b

kðnþ1Þ P 4kn .

ð1Þ

First make partition 0 ¼ t0 < t1 < t2 <    < tk0 6 1 < tk0 þ1 ; h bi where tk ¼ knb . Hence k 0 ¼ 2kn0 . Let 2k

0

En0 ;k ¼ ½tk1 ; tk ;

k ¼ 1; 2; . . . ; k 0 ;

and define MðEn0 ;k Þ ¼

1 ; E# n0

ð2Þ

k0 # where E# n0 indicates the number of elements of En0 ¼ fE n0 ;k gk¼1 . Obviously, E n0 ¼ k 0 . Next, define En,k and MðEn;k Þ by induction for n > n0. Suppose that En1,k and MðEn1;k Þ are well defined, put En1 = {En1,k}, and let E# n1 be the number of elements of En1. We make a partition of the interval [0, 1] as follows:

0 ¼ t00 < t01 <    < t0k0 6 1 < t0k 0 þ1 ; where t0k ¼ knb , k = 0, 1, . . . , k 0 + 1. Let I n;k ¼ ½t0k1 ; t0k , denote En,n1,j = {In,k : In,k  En1,j}, j ¼ 1; 2; . . . ; E# n1 , and rear2k range the elements of [jEn,n1,j from left to right to be written as En,k. In case En,k 2 En,n1,j, define

T.F. Xie, S.P. Zhou / Chaos, Solitons and Fractals 32 (2007) 1625–1630

MðEn;k Þ ¼

1627

MðEn1;j Þ . E# n;n1;j

We see that En,k is a closed interval, and the interiors of any two En,k and En;k0 are not intersect for k 5 k 0 . We also note that E# n;n1;j , the number of the elements of En,n1,j, satisfies (cf. (1)) b

0<

b

kn

kn

 2 6 E# n;n1;j 6

b ; kðn1Þ kðn1Þ thus, from the definition of MðEn;k Þ we know that, for En,k 2 En,n1,j, !1 b b b kðn1Þ 2kðn1Þ kðn1Þ MðE Þ 6 MðE Þ 6 MðE Þ 1  . n1;j n;k n1;j b b b kn kn kn b

Therefore, for En;k  En0 ;j , b

b

kn0

MðEn0 ;j Þ 6 MðEn;k Þ 6

b

kn

kn0

b

kn

n Y

MðEn0 ;j Þ

1

2kðk1Þ kk

k¼n0 þ1

b

!1

b

.

With (2), it immediately yields that b

kn0

b

kn0

1

1

h b i 6 MðEn;k Þ 6 b h b i kn 2kn0 2kn0

b

kn

n Y

1

k¼n0 þ1

2kðk1Þ kk

b

!1

b

or 1 b

2kn

6 MðEn;k Þ 6

1 kn

n Y

b

1

k¼n0 þ1

2kðk1Þ kk

b

b

!1 .

ð3Þ

Denote En = [kEn,k, then En is a closed set, and En+1  En from the construction. Therefore, E = \nEn is closed too. For any set A in the real line, define ( ) X MðV j Þ : A \ E  [j V j ; V j 2 E ; MðAÞ ¼ inf j

where E is the class of all En,k for n = n0, n0 + 1, . . ., k ¼ 1; 2; . . . ; E# n . Hence, M is a measure with support in E (cf. [3]).

3. Notations and a lemma Let UðxÞ ¼ Uðx; k; a; bÞ ¼

1 X

 b  a kk / kk x ;

k¼n0

where n0 is defined as in Section 2, and n  b  X a kk / kk x ; n P n0 . S n ðU; xÞ ¼ k¼n0

For d > 0, denote Sd = Sd(u, v) to be a square parallel to the axes in the plane with center (u, v) and side-length d. For a set A in the plane, define P x ðAÞ ¼ fx : ðx; yÞ 2 Ag. Since a > 0, it is not difficult to deduce that a

jUðxÞ  S n ðU; xÞj 6 C k;a;n kðnþ1Þ ; where the positive constant Ck,a,n is independent of x, and Ck,a,n ! 1 as n ! 1. By setting n o a MðEn;k Þ ¼ ðx; yÞ : x 2 En;k ; jy  S n ðU; xÞj < C k;a;n kðnþ1Þ ;

ð4Þ

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T.F. Xie, S.P. Zhou / Chaos, Solitons and Fractals 32 (2007) 1625–1630

we have CðU; En;k Þ  MðEn;k Þ; where C(f, A) = {(x, y) : x 2 A, y = f(x)}. Lemma. Let k > 0, b > a > 1, then for any given Sd = Sd(u, v) and all n P n0, Px(Sd \ M(En,k)) is contained in an interval of length not exceeding   a a b 1 kn n d þ 2C k;a;n kðnþ1Þ ; 0 2C k;a;b;n provided C 0k;a;b;n ! 1, Ck,a,n ! 1 as n ! 1. Proof. By adopting the method used in Section 2, we take k 0 ; k 1 ; . . . ; k nn0 such that En0 ;k 0  En0 þ1;k1      En1;knn0  En;k . It is clear that Sn(U, x) is a continuous and differentiable function in En,k, and jS 0n ðU; xÞj P 2C 0k;a;b;n kn

b na

ð5Þ C 0k;a;b;n

holds for x 2 En,k. Furthermore, inequality (4), it is easy to verify that

is a positive constant independent of x tending to 1 as n ! 1. Applying

P x ðS d ðu; vÞ \ MðEn;k ÞÞ  P x ðS dþ2Ck;a;n kðnþ1Þa ðu; vÞ \ CðS n ðUÞ; En;k ÞÞ. In view of (5), C(Sn(U), En,k) is a monotone curve in the interval En,k, consequently, P x ðS dþ2Ck;a;n kðnþ1Þa ðu; vÞ \ CðS n ðUÞ; En;k ÞÞ is either an empty set, or a set containing only a single point, or an interval [x1, x2]. The first two cases are trivial, and for the last, we calculate that a

jS n ðU; x1 Þ  S n ðU; x2 Þj 6 d þ 2C k;a;n kðnþ1Þ . By applying the mean value theorem to (5) we have b na

jS n ðU; x1 Þ  S n ðU; x2 Þj P 2C 0k;a;b;n kn

jx2  x1 j.

Altogether, we obtain that jx2  x1 j 6

1 2C 0k;a;b;n

kn

a nb



 a d þ 2C k;a;n kðnþ1Þ ;

and this completes the proof. h

4. Theorem and proof Now we can establish our main result1: Theorem. Let k > 0, b > a > 1, and UðxÞ ¼ Uðx; k; a; bÞ ¼

1 X

 b  a kk / kk x ;

k¼n0

where n0 is defined as in Section 2, then dimHC(U) = 2. Proof. Choose a natural number N P n0 such that for all n P N, 3 C k;a;n < ; 2 1

1 C 0k;a;b;n > . 2

For convenience, we set the summation starting from n0 without loss of generality.

ð6Þ

T.F. Xie, S.P. Zhou / Chaos, Solitons and Fractals 32 (2007) 1625–1630

1629

b

Also, for d 2 ð0; ð2kN Þ1 Þ, take natural numbers n and q such that    b 1 b 1 a a 6 d < 2kn ; 3kðnþqÞ 6 d < 3kðnþq1Þ . 2kðnþ1Þ

ð7Þ

Obviously n P N holds in this case. We already pointed out in Section 2 that, En+1,k is a closed interval of length b ð2kðnþ1Þ Þ1 . Moreover, the interiors of any two of these intervals are not intersect. Therefore, for any given square Sd(u, v), the number of M(En+1,k) intersecting Sd is at most b

M 6 2dkðnþ1Þ þ 2.

ð8Þ

At the same time, for any M(En+1,k) intersecting Sda , by the lemma and (6) we know that, the set Px(M(En+1,k)) is cona b tained in an interval of length ðd þ 3kðnþ2Þ Þkðnþ1Þ ðnþ1Þ , and the number of Px(M(En+2,k)) intersecting this interval does not exceed   a a b b d þ 3kðnþ2Þ kðnþ1Þ ðnþ1Þ 2kðnþ2Þ þ 2 b

since the length of En+2,k is ð2kðnþ2Þ Þ1 . In this way, by (8), the number of M(En+2,k) intersecting Sd does not exceed          a a b b b a a b b M 2 d þ 3kðnþ2Þ kðnþ1Þ ðnþ1Þ kðnþ2Þ þ 2 6 2dkðnþ1Þ þ 2 2 d þ 3kðnþ2Þ kðnþ1Þ ðnþ1Þ kðnþ2Þ þ 2 . Write the number of M(En+j,k) intersecting Sd as Dj, j = 2, 3, . . . , then ( a b a d P 3kðnþ2Þ ; 24d2 kðnþ1Þ kðnþ2Þ ; D2 6 a a b b a 24Mkðnþ1Þ ðnþ2Þ ðnþ1Þ þðnþ2Þ ; d < 3kðnþ2Þ .

ð9Þ

a

In case d < 3kðnþ2Þ , or q P 3, by this kind of technique we apply the lemma repeatedly to achieve that a

Dq 6 24q1 Mkðnþ1Þ

ðnþ1Þb ðnþqÞa ðnþqÞb

k

k

;

then it follows that a

Dq 6 24q d2 kðnþ1Þ kðnþqÞ

b

ð90 Þ

by (8) and (7). To prove dimHC(U) = 2, we define a measure l by the following technique. For a set A in the plane, let lðAÞ ¼ MðP x ðA \ CðUÞÞ;

ð10Þ

where M is the probability measure established in Section 2. It is easy to verify that l is a probability measure with b support contained in C(U). Obviously, for d 2 ð0; ð2kN Þ1 Þ, after selecting n and q that satisfy (7), by (10) and (9), 0 (9 ), we get 8 2 ðnþ1Þa ðnþ2Þb > k max MðEnþ2;k Þ; q 6 2;   < 24d k k ð11Þ lðS d Þ ¼ M P x ðS d \ ð[k MðEnþq;k ÞÞÞ 6 a b q 2 ðnþ1Þ ðnþqÞ > : 24 d k k max MðEnþq;k Þ; q P 3. k

However, inequality (3) yields that  m b k km . MðEm;k Þ 6 k1 With (11), ( lðS d Þ 6

 a k nþ2 ðnþ1Þ k ; k1   q 2 k nþq ðnþ1Þa k ; 24 d k1 24d2



q 6 2; q P 3;

that is to say, there is a constant C > 0 depending upon k only such that a

lðS d Þ 6 C nþq d2 kðnþ1Þ . On the other hand, (7) implies that b

a

kðnþ1Þ 6 6kðnþq1Þ ; so that n + q 6 12(n + 1)b/a. It is easy to see that

ð12Þ

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T.F. Xie, S.P. Zhou / Chaos, Solitons and Fractals 32 (2007) 1625–1630 a

b=a

b

lim kn C 12ðnþ1Þ kðnþ1Þ ¼ 0

n!1

for any  2 (0, 2), thus there is a d0 > 0 depending only upon  such for all 0 < d < d0 that b

b=a

a

kn C 12ðnþ1Þ kðnþ1Þ 6 1. b

Together with (12) (noting that d 6 kn by (7)) we get lðS d Þ 6 d2 ;

0 < d 6 d0 .

Therefore, applying mass distribution principle and considering arbitrariness of Sd, we have 2   6 dimH CðUÞ 6 2. Finally, because of arbitrariness of , the result dimH CðUÞ ¼ 2 follows directly. This completes the proof.

h

5. Conclusion Let k > 1, b > a > 1, define /(x) = 2x, 0 6 x 6 1/2, and /(x) = /(x) and /(x + 1) = /(x). Then the graph of the continuous function 1 X a b kk /ðkk xÞ; 0 6 x 6 1 y ¼ UðxÞ ¼ k¼1

has Hausdorff dimension equal 2.

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