_{1}

^{*}

The main purpose of this note is to estimate the size of the set
T_{μλ} of points, at which the Cantor function is not differentiable and we find that the Hausdorff dimension of
T_{μλ} is [log2/log3]
^{2}. Moreover, the Packing dimension of
T_{μλ} is log2/log3. The log2 = log
_{e}2 is that if
a
^{x} =
N (
a >0, and
a≠1), then the number
x is called the logarithm of
N with a base, recorded as
x = log
_{a}
N, read as the logarithm of
N with a base, where a is called logarithm Base number,
N is called true number.

Let T be the Cantor middle thirds set (Start with the closed interval [ 0,1 ] . Remove the open interval ( 1 3 , 2 3 ) to obtain [ 0, 1 3 ] ∪ [ 2 3 ,1 ] , i.e. two disjoint

closed segments. Remove the middle thirds of those two segments, and you end up with four disjoint segments. After infinitely many steps, the result is called the Cantor set.) and θ be the Cantor ternary function (The function can be defined in the following way, given x ∈ [ 0,1 ] consider its ternary expansion a i , i.e. a choice of coefficients a i ∈ { 0,1,2 } such that

x = ∑ i = 1 ∞ a i 3 i . (1.1)

Define n ( x ) to be ∞ if none of the coefficients a i takes the value 1 and the smallest integer n such that a n = 1 otherwise.

Then

f ( x ) = ∑ i = 1 n ( x ) − 1 a i 2 i + 1 + 1 2 n ( x ) . (1.2)

Let N * be the set of non-differentiable points of Cantor function θ .

Denote

θ − = lim inf r → 0 − θ ( t + r ) − θ ( t ) r . (1.3)

and

θ + = lim inf r → 0 + θ ( t + r ) − θ ( t ) r . (1.4)

See [

lim inf r → 0 θ ( t + r ) − θ ( t ) r = ∞ . (1.5)

So,

N * = { t ∈ T : θ − ( t ) < ∞ or θ + ( t ) < ∞ } .

Let t = . t 1 t 2 ⋯ be the ternary expansion of a point in T, which is not an endpoint of a complementary interval. Let z ( n ) denote the position of the nth zero in this ternary expansion, and t ( n ) denote the position of the nth two in the expansion of t.

For each μ ≥ 0 and λ ≥ 0 , define

T μ λ = { t ∈ T : θ − ( t ) = μ and θ + ( t ) = λ } .

Firstly, several authors [

Let the Cantor T in R be defined by T = ∪ j = 1 r h j ( T ) with a disjoint union, where

the h j are similitude mappings with ratios 0 < a j < 1 . Let μ be the self-similar Borel probability measure on T corresponding to the probability vector ( p 0 , p 1 , ⋯ , p r ) . Let S be the set of points at which the probability distribution function F ( x ) of μ has no derivative, finite or infinite. On the one hand, for the case where p j > a j , Wen xia [

Since about the dimension of the set T μ λ of points, Eidswick didn’t study it. So, our main job in this paper is to give its dimension, that is, the following theorem.

Theorem 1.1. dim H T μ λ = ( log 2 / log 3 ) 2 , and dim P T μ λ = log 2 / log 3 .

First, we give a proposition, which simplifies the calculation of θ + and θ − .

Proposition 2.1. Let t = . t 1 t 2 ⋯ t n be ternary expansion of t, and define

e n = . t 1 t 2 ⋯ t z ( n ) − 1 2 , (2.1)

d n = . t 1 t 2 ⋯ t z ( n ) − 1 1. (2.2)

Let

Q n : = θ ( e n ) − θ ( t ) e n − t and P n : = θ ( d n ) − θ ( t ) d n − t ,

we have

θ + ( t ) = lim inf n → ∞ Q n , (2.3)

and

θ − ( t ) = lim inf n → ∞ P n . (2.4)

Proof. We just need to prove (2.3), since (2.4) is proved in a similar way.

If

lim inf δ → 0 θ ( t + δ ) − θ ( t ) δ = + ∞ ,

then, the proposition is apparently true.

So, we suppose that

lim inf δ → 0 θ ( t + δ ) − θ ( t ) δ = c < + ∞ ,

then, we can find { δ n } n = 1 + ∞ such that

lim n → ∞ θ ( t + δ n ) − θ ( t ) δ n = lim inf δ → 0 θ ( t + δ ) − θ ( t ) δ . (2.5)

Note that e n + 1 + 1 3 Z ( n + 1 ) = d n and lim n → + ∞ d n = t , we have

( t , e 1 + 1 3 z ( 1 ) ) = ∪ n = 1 + ∞ [ d n , e n + 1 3 z ( n ) ) .

Therefore, for any k ≥ 1 , there is an integer n k , such that t + δ k ∈ [ d n k , e n k + 1 3 z ( n k ) ) .

By (2.5), for any ε > 0 , there is an integer N > 0 such that: when k > N ,

θ ( t + δ k ) − θ ( t ) δ k < c + ε .

Denote η k = t + δ k − e n k , then 0 < η k < 3 − z ( n k ) , we have

θ ( t + δ k ) = θ ( e n k + η k ) = θ ( e n k ) + θ ( η k ) .

So,

θ ( e n k ) − θ ( t ) e n k − t = θ ( t + δ k ) − θ ( η k ) − θ ( t ) δ k − η k ,

and since lim k → + ∞ θ ( η k ) η k = + ∞ , it follows that

lim inf k → + ∞ θ ( e n k ) − θ ( t ) e n k − t ≤ lim k → + ∞ θ ( t + δ k ) − θ ( t ) δ k .

On the other hand, it is obvious that:

lim inf k → + ∞ θ ( e n k ) − θ ( t ) e n k − t ≥ lim inf δ → 0 θ ( t + δ ) − θ ( t ) δ ,

so we complete the proof of (2.3).

☐

Besides, we need the following Lemma due to Billingsley:

Lemma 2.2. Let b ≥ 2 be an integer and for x ∈ [ 0,1 ] , let I n ( x ) be the nth generation half-open b-adic interval of the form [ j − 1 b n , j b n ] cintaining x, let

A ⊂ [ 0,1 ] be Borel and let μ be a finite Borel measure on [0, 1], suppose μ ( A ) > 0 . If

α ≤ lim inf n → ∞ log μ ( I n ( x ) ) log | I n ( x ) | ≤ β ,

then

α ≤ dim H ( A ) ≤ β .

For the proof of this lemma, we refer readers to Section 1.4 of [

Corollary 2.3. For S ⊂ N , define

A S = { x : ∑ k ∈ S x k 3 k : x k ∈ { 0,2 } } . (2.6)

We have

dim H ( A S ) = lim _ N → ∞ # ( S ∩ { 1, ⋯ , N } ) N ⋅ log 2 log 3 . (2.7)

Proof. To see this, let μ be the probability measure on A S that give equal measure to nth generation covering intervals. This measure makes the digits { x k } k ∈ S in (2.6) independent identically distributed uniform random bits. For any x ∈ A S ,

log μ ( I n ( x ) ) log | I n ( x ) | = log 2 − # ( S ∩ { 1, ⋯ , N } ) log 3 − n = # ( S ∩ { 1, ⋯ , N } ) N ⋅ log 2 log 3 .

Thus the liminf of the left-hand side is the liminf of the right-hand side. By Lemma 2.2, this proves the corollary.

Last, we need to introduce a theorem and a formula:

Theorem 2.4. Let t be a point of T which is not an endpoint of a comple-mentary interval, let z ( n ) denote the position of the nth zero in its ternary

expansion, and let λ = lim inf n → ∞ 3 z ( n ) 2 ( n + 1 ) . Then λ ≤ θ + ( t ) ≤ 2 λ . Furthermore, if z ( n + 1 ) − z ( n ) → ∞ , then, θ + ( t ) = λ .

Let, Q n = θ ( e n ) − θ ( t ) e n − t . Then,

Q n = 3 z ( n ) 2 z ( n + 1 ) ∑ 1 ∞ 2 − ( z ( n + k ) − z ( n + 1 ) ) 1 + 2 ∑ 1 ∞ 3 − ( z ( n + k ) z ( n ) ) (2.8)

The proof of theorem 2.4 and formula (2.8), we see [

Proof of Theorem. First, let’s construct a subset of T μ λ .

Let μ 0 = log μ log 2 , λ 0 = log λ log 2 . We can find increasing sequences of positive integers ( a n ) and ( b n ) such that

a 2 n ξ − b 2 n → λ 0 , (2.9)

and

a 2 n + 1 ξ − b 2 n + 1 → μ 0 , (2.10)

when n is large enough, where ξ = log 3 log 2 . At the same time, we may request that a n < b n , a n + 1 − b n ≥ 4 n and b n ≥ 3 n hold for all n ≥ 1 . Moreover, denote M = [ a 2 n + 1 − b 2 n 2 n ] and M * = [ a 2 n + 2 − b 2 n + 1 2 n ] , we insert these points ( b 2 n , i ) i = 1 M and ( b 2 n + 1 , j ) j = 1 M * in to ( b 2 n , a 2 n + 1 ) and ( b 2 n + 1 , a 2 n + 2 ) , respectively, such that

b 2 n , i + 1 − b 2 n , i = 2 n , (2.11)

and

b 2 n + 1 , j + 1 − b 2 n + 1 , j = 2 n , (2.12)

when n is large enough, for i = 1 , 2 , ⋯ , M and j = 1 , 2 , ⋯ , M * ; and let b 2 n = b 2 n , 0 , b 2 n + 1 , 0 = b 2 n + 1 . We will show the selection of { a n } and { b n } later.

Define t m = 0 , if m ∈ { a 2 n , b 2 n , b 2 n ,1 , ⋯ , b 2 n , M , b 2 n + 1,1 + 1, b 2 n + 1,2 + 1, ⋯ , b 2 n + 1, M * + 1 } , or a 2 n + 1 < m < b 2 n + 1 .

Define t m = 0 , if m ∈ { a 2 n + 1 , b 2 n + 1 , b 2 n + 1,1 , b 2 n + 1,2 , ⋯ , b 2 n + 1, M * , b 2 n ,1 + 1, b 2 n ,2 + 1, ⋯ , b 2 n , M + 1 } , or a 2 n < m < b 2 n .

And restrict t m ∈ { 0,2 } (That is, t m is free to take a value of 0 or 2 for all other values of m).

And we claim that t = ∑ m = 1 ∞ t m 3 m ∈ T μ λ . We’ll give a simple proof after the selection of ( a n ) n = 1 + ∞ and ( b n ) n = 1 + ∞ , let Ω be the set of all such t.

For the selection of ( a n ) n = 1 + ∞ , ( b n ) n = 1 + ∞ , first of all, take a 0 = b 0 = 1 . Assume that a 2 n and b 2 n is taken, we choose a 2 n + 1 , such that

(a) a 2 n + 1 ≥ 8 n + b 2 n ;

(b) [ ξ a 2 n + 1 ] − [ μ 0 ] > a 2 n + 1 (note that ξ > 1 );

(c) | { ξ a 2 n + 1 } − { μ 0 } | < 1 n (note that ( ξ n ) is uniformly distributed modulo 1).

where [ x ] represents the integer part that represents taking x, and { y } represents the fractional portion of the y. And take b 2 n + 1 = [ ξ a 2 n + 1 ] − [ μ 0 ] . We choose a 2 n + 2 such that

(d) a 2 n + 2 ≥ 4 ( 2 n + 1 ) + b 2 n + 1 ;

(e) [ ξ a 2 n + 2 ] − [ λ 0 ] > a 2 n + 2 ;

(f) | { ξ a 2 n + 2 } − { λ 0 } | < 1 n + 1 .

and take b 2 n + 2 = [ ξ a 2 n + 2 ] − [ λ 0 ] .

These ( a n ) n = 1 + ∞ and ( b n ) n = 1 + ∞ meet the desired conditions.

From a 2 n ξ − b 2 n → λ 0 and a 2 n + 1 ξ − b 2 n + 1 → μ 0 , we can get that

3 a 2 n 2 b 2 n → λ and 3 a 2 n + 1 2 b 2 n + 1 → μ .

To see that t = ∑ m = 1 ∞ t m 3 m ∈ T μ λ , by the Theorem 2.4 and formula (2.8), we only

need to show that b 2 n and a 2 n are adjacent positions of zeros, whose spacing is largest; and b 2 n + 1 and a 2 n + 1 are adjacent positions of twos, whose spacing is largest.

Since a 2 n → b 2 n ξ + λ ξ , for any ϵ > 0 , when n is large enough, we have

| a 2 n − b 2 n ξ − λ ξ | < ϵ ,

then, when n is large enough

b 2 n − a 2 n > b 2 n − b 2 n ξ − λ 0 ξ − ε = ( 1 − 1 ξ ) b 2 n − λ 0 ξ − ε ≈ 0.36 b 2 n > 2 n .

Then, according to our above-mentioned construction of t = ∑ m = 1 ∞ t m 3 m , other

spacings of zeros are at most 2n, so b 2 n − a 2 n is the largest spacing of zeros. Similarly, we have b 2 n + 1 − a 2 n + 1 > 2 n , and b 2 n + 1 − a 2 n + 1 is the largest spacing of twos.

Denote

A n = ( { k ∈ N : b 2 n ≤ k ≤ a 2 n + 1 } ∪ { k ∈ N : b 2 n + 1 ≤ k ≤ a 2 n + 1 } ) ,

B n = ( ( ∪ i = 1 M { b 2 n , i , b 2 n , i + 1 } ) ∪ ( ∪ j = 1 M * { b 2 n + 1 , j , b 2 n + 1 , j + 1 } ) ) .

Let

B : = ∪ n = 1 + ∞ ( A n \ B n )

in fact, the set B is all positions that t m can freely select 0 or 2 in the above construction. Using Corollary 2.3, we have

dim p Ω = dim B Ω = lim sup N → ∞ # { k ∈ B , k ≤ N } N ⋅ log 2 log 3 ,

and

dim H Ω = lim inf N → ∞ # { k ∈ B , k ≤ N } N ⋅ log 2 log 3 .

And we calculate these as follows: It’s not hard to see that, when N = a 2 n + 1 , we can get the superior limit:

lim sup N → ∞ # { k ∈ B , k ≤ N } N ⋅ log 2 log 3 = lim n → ∞ a 2 n + 1 − b 2 n − 2 M a 2 n + 1 ⋅ log 2 log 3 = log 2 log 3

where lim n → ∞ b 2 n a 2 n + 1 = 0 , lim n → ∞ M a 2 n + 1 = 0 .

On the other hand, when N = b n , we can get the inferior limit:

dim H Ω = liminf N → ∞ # { k ∈ B , k ≤ N } N ⋅ log 2 log 3 = lim n → ∞ # { k ∈ B , k ≤ b n } n ⋅ log 2 log 3 ,

and by calculating

lim n → ∞ # { k ∈ B , k ≤ b 2 n + 1 } b 2 n + 1 ⋅ log 2 log 3 ≥ lim n → ∞ a 2 n + 1 − b 2 n − 2 M b 2 n + 1 ⋅ log 2 log 3 = ( log 2 / log 3 ) 2

where lim n → ∞ a 2 n + 1 b 2 n + 1 = log 2 log 3 , lim n → ∞ b 2 n b 2 n + 1 = 0 and lim n → ∞ M b 2 n + 1 = 0 ;

lim n → ∞ # { k ∈ B , k ≤ b 2 n + 2 } b 2 n + 2 ⋅ log 2 log 3 ≥ lim n → ∞ a 2 n + 2 − b 2 n + 1 − 2 M * b 2 n + 2 ⋅ log 2 log 3 = ( log 2 / log 3 ) 2

where lim n → ∞ a 2 n + 2 b 2 n + 2 = log 2 log 3 , lim n → ∞ b 2 n + 1 b 2 n + 2 = 0 , lim n → ∞ M * b 2 n + 2 = 0 .

In Section 1, we already know dim p N * = log 2 / log 3 , and dim H N * = ( log 2 / log 3 ) 2 .

Since Ω ⊂ T μ λ ⊂ N * , we have dim H Ω = dim H T μ λ = ( log 2 / log 3 ) 2 and dim P Ω = dim P T μ λ = log 2 / log 3 .

Finally, through the above proof, we solved dimension of the set T μ λ of points that is Eidswick didn’t study it. In other words, we finally get that dim H T μ λ = ( log 2 / log 3 ) 2 , and dim P T μ λ = log 2 / log 3 .

The author declares no conflicts of interest regarding the publication of this paper.

Yan, M.Q. (2020) Dimension of the Non-Differentiability Subset of the Cantor Function. Journal of Applied Mathematics and Physics, 8, 107-114. https://doi.org/10.4236/jamp.2020.81009