At which points exactly has Lebesgue’s singular function the derivative zero ?

# At which points exactly has Lebesgue’s singular function the derivative zero ?

Kiko Kawamura
University of North Texas 111Supported in part by Japanese GCOE Program G08: “Fostering Top Leaders in Mathematics - Broadening the Core and Exploring New Ground”.  222Address: Department of Mathematics, University of North Texas, 1155 Union Circle #311430, Denton, TX 76203-5017, USA; E-mail: kiko@unt.edu
July 27, 2019
###### Abstract

Let be Lebesgue’s singular function with a real parameter (). As is well known, is strictly increasing and has a derivative equal to zero almost everywhere. However, what sets of actually have or ? We give a partial characterization of these sets in terms of the binary expansion of . As an application, we consider the differentiability of the composition of Takagi’s nowhere differentiable function and the inverse of Lebesgue’s singular function.

AMS 2000 subject classification: 26A27 (primary); 26A15, 26A30, 60G50 (secondary)

Key words and phrases: Takagi’s function, Lebesgue’s singular function, Nowhere-differentiable function, Dini derivatives.

## 1 Introduction

Imagine flipping an unfair coin with probability of heads and probability of tails. Note that . Let the binary expansion of : be determined by flipping the coin infinitely many times. More precisely, if the -th toss is heads and if it is tails. We define Lebesgue’s singular function as the distribution function of :

 La(x):=prob{t≤x},0≤x≤1.

It is well-known that is strictly increasing, but the derivative is almost everywhere. This distribution function was also defined in different ways and studied by a number of authors: Cesaro (1906), Faber (1910), Lomnicki and Ulam(1934), Salem (1943), De Rham (1957) and others. For instance, De Rham [3] studied as a unique continuous solution of the functional equation

 La(x)={aLa(2x),0≤x≤12,(1−a)La(2x−1)+a,12≤x≤1, (1)

where , and .

From (1), it is clear that the graph of is self-affine. Because of its connection with fractals, several applications have been found in recent years: for instance, in physics [12, 13], real analysis [5, 6], digital sum problems [7, 9] and complex dynamical systems [10]. There is even a connection with the Collatz conjecture [2].

Reconsider the differentiability of . It is known that for any , is either zero, or , or it does not exist. Then, it is natural to ask at which points exactly we have or .

In fact, De Rham [3] gave the following partial answer to this question. Let the binary expansion of be , where . For those having two binary expansions, we choose the expansion which is eventually all zeros. As an exception, fix for every if .

Define

 In:=n∑k=1εk. (2)

Note that is the number of ’s occurring in the first binary digits of .

Suppose that tends to a limit as , and let

 l0:=log2aloga−log(1−a). (3)

Then the derivative exists and is zero, when . An English translation of De Rham’s original paper is included in Edgar’s book [4].

Unfortunately, De Rham’s paper did not contain a proof. The main purpose of this note is to give a proof of De Rham’s statement and extend his result. The paper is organized as follows. Section 2 states and proves the main results. The key to the proof is to use Lomnicki and Ulam’s expression from 1934 [8]. De Rham might have had a different proof in mind, as he did not mention Lomnicki and Ulam’s paper. In Section 3, as an application, we consider a question about the differentiability of the composition of Takagi’s nowhere differentiale function and the inverse of Lebesgue’s singular function.

## 2 The main result

For convenience, define the right-hand and left-hand derivatives of as follows.

 L′a+(x): =limh→0+La(x+h)−La(x)h, L′a−(x): =limh→0−La(x+h)−La(x)h,

provided the limits exist.

From the self-affinity of the graph, we have

###### Lemma 2.1.

For any for which exists,

 L′a+(x)=L′(1−a)−(1−x).

Define

 D1(x):=limn→∞Inn=limn→∞1nn∑k=1εk, (4)

provided the limit exists, and put . In other words, is the density of the digit in the binary expansion of , for .

###### Theorem 2.2.
1. If is dyadic, then .

2. If is not dyadic and , then

###### Remark 2.3.

De Rham’s statement is equivalent to the following. For a value of for which exists, when .

###### Remark 2.4.

If is a binary normal, that is, , then Theorem 2.2 gives , since .

Proof of Theorem 2.2. First, suppose is a dyadic point, say . Let where . Since is increasing, this implies that

 La(x+2−(k+1))−La(x)2−k≤La(x+h)−La(x)h≤La(x+2−k)−La(x)2−(k+1). (5)

The key to the proof is to use the following expression for , given by Lomnicki and Ulam [8]:

 La(x)=a1−a∞∑n=1εnan−In(1−a)In, (6)

where is defined by (2). By (6), we have

 La(x+2−k)−La(x)=ak−IN(1−a)IN.

Since is a positive constant,

 limk→∞La(x+2−k)−La(x)2−k =limk→∞(2a)k(1−aa)IN ={0,if 0

By (5), it follows that

 L′a+(x)={0,if 0

Since is also a dyadic, the left-hand derivative follows from Lemma 2.1:

 L′a−(x)=L′(1−a)+(1−x)={+∞,if 0

Therefore, does not exist if is dyadic.

Next, suppose is not dyadic and . Let be the address of the -th in the binary expansion of , and . Since is increasing, this implies that

 La(x+2−pk+1)−La(x)2−pk≤La(x+h)−La(x)h≤La(x+2−pk)−La(x)2−pk+1. (7)

Using (6), we have

 La(x+2−pk)−La(x)=ak(1−a)pk−k+(1−a1−a)∞∑n=pk+1εnan−In(1−a)In. (8)

Let be the address of the -th appearing after position in the binary expansion of . Then we have

 ∞∑n=pk+1εnan−In(1−a)In=ak(1−a)pk−k∞∑l=1an(l)−pk−l(1−a)l.

Since and , the series in the right hand side above converges, say to .

For convenience, define

 C1(x,k):=1+(1−a1−a)C(x,k).

Then we can write (8) as

 La(x+2−pk)−La(x)=ak(1−a)pk−kC1(x,k). (9)

Since , it follows that

 min{1,1−aa}≤C1(x,k)≤max{1,1−aa}. (10)

By (9), we have

 La(x+2−pk)−La(x)2−pk+1 ={2pk+1pkakpk(1−a)1−kpk}pkC1(x,k), La(x+2−pk+1)−La(x)2−pk ={2pkpk+1ak+1pk+1(1−a)1−k+1pk+1}pk+1C1(x,k).

Since tends to a nonzero limit as , we have as . Therefore, it follows from (7) and (10) that

Finally, for the left-hand derivative, it follows from Lemma 2.1 that

since when . This concludes the proof.

###### Remark 2.5.

A careful study of the above proof shows that the existence of the full limit is not necesary. The following generalization is straightforward:

1. Suppose . Then

 L′a(x)={0,if limn→∞supIn/nl0.
2. Suppose . Then

 L′a(x)={0,if limn→∞infIn/n>l0,+∞,if limn→∞supIn/n

where is defined by (3).

Note that Theorem 2.2 left out the boundary case; that is, those numbers for which ; in other words, numbers which have the following densities:

 D1(x)=log2aloga−log(1−a),D0(x)=log2(1−a)log(1−a)−loga.

Let us define some additional notation. As a complement of , define to be the number of ’s occurring in the first binary digits of :

 On:=n∑k=1(1−εk).

Let be the address of the -th in the binary expansion of as a complement of . Observe that

 qk≤n if and only if In≥k, pk≤n if and only if On≥k.

Then, it is easy to prove the following lemma by contradiction.

###### Lemma 2.6.

Let and . If as , then .

###### Theorem 2.7.

Suppose satisfies . Let and suppose .

1. If as , then

 L′a(x)= {+∞,if 0
2. If as , then

 L′a(x)= {0,if 0

Proof of Theorem 2.7.

We follow the same argument for non-dyadic points as in the proof of Theorem 2.2. Let . Since tends to a nonzero limit as , is of smaller order than . Then, it follows from (9) that

 La(x+2−pk)−La(x)2−pk =[{2(1−a)}1D0(x)a(1−a)−1]k{2(1−a)}f(k)C1(x,k) ={2(1−a)}f(k)C1(x,k),

because

Thus,

 La(x+2−pk)−La(x)2−pk+1 ={2f(k+1)f(k)(1−a)}f(k)⋅21D0(x)C1(x,k), La(x+2−pk+1)−La(x)2−pk ={2f(k)f(k+1)(1−a)}f(k+1)⋅21D0(x)C1(x,k+1).

Since as , it follows from (7) and (10) that if ,

 L′a+(x) ={+∞,if 0

Similary, if as , then

 L′a+(x) ={0,if 0

Next, consider the left-hand derivative. From Lemma 2.1, we have . It is clear that also satisfies , since for . Let . Since , we have if . It follows from Lemma 2.6 that if or , then . This concludes the proof.

## 3 Application

We apply the main result to the following simple question. In classical calculus, the chain rule is used to compute the derivative of the composition of two differentiable functions. However, what can we say, for example, about the differentiability of the composition of a nowhere differentiable function and a singular function? For instance, let be Takagi’s nowhere differentiable function, which is defined by

 T(x)=∞∑k=012k|2kx−⌊2kx+12⌋|,0≤x≤1.

Is nowhere differentiable? See Figure . If , the figure of the graph looks somewhat like Takagi’s function; on the other hand, if , the shape of the graph is more like Lebesgue’s singular function. Thus, we can guess that might not be nowhere differentiable if is close to .

Although does not have a finite derivative anywhere, it is known to have an improper infinite derivative at many points. In fact, Allaart and Kawamura [1] proved that the set of points where or has Hausdorff dimension one. Note that the inverse of Lebesgue’s singular function is also singular. Hence, if we try to (naively) use the chain rule to compute the derivative of , we may run into one of the indeterminate products or .

The following theorem gives an answer to this concrete question: has a finite but vanishing derivative at uncountably many points.

###### Theorem 3.1.

Let , and put . If and , then

 (T∘L−1a)′(x)=0. (11)

Proof.

Define . Then we can write

 T(L−1a(x+h))−T(L−1a(x))h=T(y+~h)−T(y)~hlog2(1/|~h|)⋅~hlog2(1/|~h|)h. (12)

Allaart and Kawamura [1] proved that if exists and , then

 limh→0T(x+h)−T(x)hlog2(1/|h|)=D0(x)−D1(x).

Therefore, we have

 −1≤limh→0T(y+~h)−T(y)~hlog2(1/|~h|)≤1.

A slight modification of the proof of Theorem 2.2 yields

Substituting these results into (12) gives (11).

## Acknowledgment

This research was done mainly during my visit to RIMS, Kyoto university. I am grateful to Prof. H. Okamoto for his support and warm-hearted hospitality. Also, I would like to thank Prof. P. Allaart for his helpful comments and suggestions in preparing this paper.

Lastly, I wish to dedicate this paper to the memory of Prof. H. Shinya, who taught me a deeper understanding of calculus.

## References

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