61
Continuous Nowhere Differentiable Functions By: Katie G. Spurrier Submitted in Partial Fulfillment Of the requirements for Graduation with Honors from the South Carolina Honors College April 2004 Approved: Dr. Anton R. Schep Director of Thesis Dr. Ralph E. Howard Second Reader Peter C. Sederberg, Leslie S. Jones or Douglas F. Williams For South Carolina Honors College

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Page 1: Continuous Nowhere Differentiable Functionseduc.jmu.edu/~querteks/seniorthesis.pdfContinuous nowhere differentiable functions are functions that are continuous at ... It is well

Continuous Nowhere Differentiable Functions

By:

Katie G. Spurrier

Submitted in Partial Fulfillment

Of the requirements for

Graduation with Honors from the

South Carolina Honors College

April 2004

Approved:

Dr. Anton R. SchepDirector of Thesis

Dr. Ralph E. HowardSecond Reader

Peter C. Sederberg, Leslie S. Jones or Douglas F. WilliamsFor South Carolina Honors College

Page 2: Continuous Nowhere Differentiable Functionseduc.jmu.edu/~querteks/seniorthesis.pdfContinuous nowhere differentiable functions are functions that are continuous at ... It is well

Contents

Summary iv

Abstract vii

Introduction 1

Chapter 1 Nowhere Differentiable Functions 3

1.1. The Generalized van der Waerden–Takagi Function 3

1.2. Kiesswetter’s Function 11

Chapter 2 Holder Continuity 20

2.1. The Generalized van der Waerden–Takagi Function 20

2.2. Kiesswetter’s Function 25

Chapter 3 Fractional Derivatives 28

Chapter 4 Holder Continuity and Fractional Derivatives 32

4.1. The Generalized van der Waerden–Takagi Function 39

4.2. Kiesswetter’s Function 39

Chapter 5 Dimensions of Graphs of Nowhere Differentiable Functions 40

5.1. Definitions and Facts 41

5.2. The Generalized van der Waerden–Takagi Function 44

5.3. Kiesswetter’s Function 49

Conclusion 52

ii

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References 54

iii

Page 4: Continuous Nowhere Differentiable Functionseduc.jmu.edu/~querteks/seniorthesis.pdfContinuous nowhere differentiable functions are functions that are continuous at ... It is well

Summary

Continuous nowhere differentiable functions are functions that are continuous at

every point in their domains but do not have a derivative at any point in their domains.

Although continuous functions are usually presented in a way that leads students to

assume that differentiability is the norm and that nowhere differentiable functions

are the exception, one can make the case that most continuous functions are nowhere

differentiable. Two examples of continuous nowhere differentiable functions on [0, 1]

are the Generalized van der Waerden–Takagi function and Kiesswetter’s function. The

Generalized van der Waerden–Takagi function is constructed by taking the infinite

sum over n = 0, 1, 2, . . . of specially constructed functions fb,n on [0, 1]. The number

b is a fixed integer greater than or equal to 2, and the functions fb,n have bn peaks.

Since the sum of the functions converges and the peaks do not cancel each other

out in the summation, the resulting function has an “infinite number of peaks”and

is thus nowhere differentiable. Similarly, Kiesswetter’s function is constructed as a

convergent sequence of functions gn on [0, 1] for n = 0, 1, 2, . . . . For all n greater

than 0, gn has more than 4n−1 peaks. Since Kiesswetter’s function is the limit of the

sequence of gn’s as g → ∞, Kiesswetter’s function also has an “infinite number of

peaks”and is nowhere differentiable on [0, 1]. We begin by defining these functions

in detail and showing that they are in fact continuous and nowhere differentiable on

[0, 1].

Although we cannot fully visualize their graphs, we can determine certain prop-

erties of continuous nowhere differentiable functions. One quantity that we want to

consider is Holder continuity, which relates the difference in the values of a function

iv

Page 5: Continuous Nowhere Differentiable Functionseduc.jmu.edu/~querteks/seniorthesis.pdfContinuous nowhere differentiable functions are functions that are continuous at ... It is well

at two points to the distance between the two points. If there exists a positive con-

stant, M , such that |f(x) − f(y)| ≤ M |x − y|α for all x and y in the domain of

the function, then the function f is said to be Holder continuous of exponent α. We

determine the exponents for which the Generalized van der Waerden–Takagi function

and Kiesswetter’s function are Holder continuous.

Since they are nowhere differentiable, we cannot use the existence of integer–

order derivatives to determine the smoothness of continuous nowhere differentiable

functions. However, these functions do have some level of smoothness. To be able to

measure this smoothness and compare it to the smoothness of other functions, we use

fractional derivatives. The existence of fractional derivatives of a particular order can

then be used as our measure of smoothness for a nowhere differentiable function. Since

fractional derivatives can be very difficult to calculate directly, we use the connections

between fractional derivatives and Holder continuity to be able to determine the

existence of fractional derivatives of certain orders without direct calculation. Using

these connections, we are able to determine the existence of fractional derivatives for

the Generalized van der Waerden–Takagi function and Kiesswetter’s function.

We also want to be able to measure and compare the sizes of the graphs of contin-

uous nowhere differentiable functions. However, the graphs of these functions all have

infinite length. Thus, we must refine our notion of size. We use the ideas of dimension

from fractal geometry as a different way to measure the size of the graphs as subsets

of the plane. The two ideas of dimension that we use are the Hausdorff dimension

and the box–counting dimension. We apply both of these ideas to the graphs of the

Generalized van der Waerden–Takagi function and Kiesswetter’s function. Using the

connections between Holder continuity and dimension, we are able to obtain bounds,

and in some cases exact values, for the Hausdorff and box–counting dimensions of the

graphs of the two functions.

v

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Continuous nowhere differentiable functions are a fascinating topic because they

question preconceived notions about continuous functions. Although these functions

are very difficult to visualize, it is possible to learn about their behavior. The prop-

erties of Holder continuity, fractional derivatives, and dimension allow us to obtain a

better understanding of continuous nowhere differentiable functions.

vi

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Abstract

In this thesis we investigate two classes of continuous nowhere differentiable func-

tions on [0, 1] : the Generalized van der Waerden–Takagi function, f(x) =∑∞

n=0a0(bnx)

cn

where a0(x) = dist(x, Z), c > 1, b ε N, and b ≥ c, and Kiesswetter’s function. We

investigate the Holder continuity and fractional differentiability of the functions in

these classes. We also bound, and in some cases fully determine, the Hausdorff and

box–counting dimensions of the graphs of these functions.

vii

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Introduction

In this thesis, we consider continuous functions on [0, 1] that are nowhere differ-

entiable on [0, 1]. Two classic examples of continuous nowhere differentiable functions

are the van der Waerden–Takagi function and Kiesswetter’s function. The van der

Waerden–Takagi function was introduced in 1903 by Teiji Takagi [10] and reintro-

duced with different parameter values in 1930 by B. L. van der Waerden [11]. Konrad

Knopp proposed the general case of the function, but he only proved the nowhere

differentiability of the function for restricted values of the parameter [6]. We relax

the restrictions on the parameters and show that the resulting function is nowhere

differentiable. Karl Kiesswetter introduced his example of a continuous nowhere dif-

ferentiable function in 1966 [5].

Continuous nowhere differentiable functions are not Lipschitz, i.e. there does not

exist a constant M > 0 so that |f(x)−f(y)| ≤ M |x−y| for all x and y in the domain

of the function. However, there do exist relations between the absolute difference in

the values of the function at two points, |f(x)−f(y)|, and the distance between these

two points, |x − y|, raised to a power less than 1. These relations can be classified

through the concept of Holder continuity. We will study the Holder continuity of

both the Generalized van der Waerden–Takagi function and Kiesswetter’s function.

Although continuous nowhere differentiable functions do not have first-order deriva-

tives, they do have some level of smoothness. The smoothness of a continuous nowhere

differentiable function can be measured by the existence of fractional derivatives.

Since we are only interested in the existence of fractional derivatives of given ex-

ponents and not of the value of the fractional derivative, we will extend a result of

1

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Hardy and Littlewood [3] to relate Holder continuity and fractional differentiability.

We can then apply our results about Holder continuity to determine for which expo-

nents there exist fractional derivatives for the Generalized van der Waerden–Takagi

function and Kiesswetter’s function.

It can be shown that the graphs of all continuous nowhere differentiable functions

have infinite length. Thus to compare the sizes of graphs of continuous nowhere

differentiable functions with the sizes of graphs of other continuous function, we will

use the ideas of dimension from fractal geometry. We will consider the Hausdorff and

box–counting dimensions and will bound, and in some cases fully determine, these

quantities for the graphs of the Generalized van der Waerden–Takagi function and

Kiesswetter’s function.

2

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Chapter 1

Nowhere Differentiable Functions

1.1. The Generalized van der Waerden–Takagi Function

An example of a continuous nowhere differentiable function is the following function

defined on [0, 1]. Let a0(x) = dist(x, Z) and define f : [0, 1] → R by

f(x) =∞∑

n=0

a0(bnx)

cn.

where c > 1 and b ε N such that b ≥ c. In consideration of the historical context of

this function, we note that if b = c = 2, then f is the function introduced by Takagi

[10]. If b = c = 10, then f is the example given by van der Waerden [11]. Knopp

considered the generalized form of the the function but only proved that the function

was nowhere differentiable for b ≥ 4c [6]. We relax the restrictions on the values

of the parameters to those given above. We require that b ≥ c in order to obtain

a nowhere differentiable function on [0, 1]. The following fact and proposition show

that if b < c, then f is differentiable at points in [0, 1].

Fact 1.1.1. Weierstrass M-test. Let (Mn) be a sequence of positive real numbers

such that |fn(x)| ≤ Mn for all x ε D and n ε N∪{0}. If the series∑

Mn is convergent,

then∑

fn is uniformly convergent on D.

Proposition 1.1.2. If b ε N and c > 1 such that b < c, then f(x) =∑∞

n=0a0(bnx)

cn

is Lipschitz on [0, 1] and is differentiable at all points x ε [0, 1) such that x 6= k2bm for

all k, m ε N ∪ {0}.

3

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Proof. Let b ε N and c > 1 such that b < c. We let fn(x) = a0(bnx)cn on [0, 1].

Then f(x) =∑∞

n=0 fn(x). As we will show in Proposition 2.1.1, |fn(x)− fn(y)| ≤(bc

)n |x− y| for all x, y ε [0, 1] and n ε N ∪ {0}. Thus, for all x, y ε [0, 1],

|f(x)− f(y)| ≤∞∑

n=0

|fn(x)− fn(y)| ≤∞∑

n=0

(b

c

)n

|x− y| = c

b− c|x− y|

since∑∞

n=0

(bc

)nis a geometric series with b < c. Therefore, f is Lipschitz on [0, 1].

It is well known that Lipschitz functions are differentiable almost everywhere

[8, pages 108–112], but in this case, we can better determine specific points in [0, 1]

at which the function has derivatives. For all n ε N ∪ {0}, we denote the right-hand

derivative of fn by D+fn(x) = limt→x+fn(t)−fn(x)

t−x. Let g(x) =

∑∞n=0 D+fn(x). For all

n ε N ∪ {0} and x ε [0, 1), D+fn(x) is either(

bc

)nor −

(bc

)n. Thus

∑∞n=0 D+fn(x) =∑∞

n=0±(

bc

)nis uniformly convergent on [0, 1) by the Weierstrass M-Test since b < c.

Since D+fn is continuous at all points in [0, 1) except on the set {{ k2bn} : k ε Z ∩

[0, 2bn − 1]}, D+fn is Riemann integrable on [0, 1). Then

∫ x

0

g(t)dt =

∫ x

0

∞∑n=0

D+fn(t)dt =∞∑

n=0

∫ x

0

D+fn(t)dt =∞∑

n=0

fn(x) = f(x)

since∑∞

n=0 D+fn(x) is uniformly convergent on [0, 1). Then, by the Fundamental

Theorem of Calculus, f ′(x) =∑∞

n=0 D+fn(x) for all x ε [0, 1) such that∑∞

n=0 D+fn(x)

is continuous at x. Therefore, f(x) is differentiable at all points x ε [0, 1) such that

x 6= k2bm for all k,m ε N ∪ {0}. �

The preceding proof does not tell us whether or not the function is differentiable

at points in [0, 1] of the form x = k2bm for some k,m ε N ∪ {0}. The differentiability

of the function at these points is dependent on the specific values of b and c. In

fact, for certain values of b and c such that b < c, the function f(x) =∑∞

n=0a0(bnx)

cn

is everywhere differentiable on its domain. To see this, we consider the following

proposition suggested by [12, pages 33–38].

4

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Proposition 1.1.3. Let f : [0, 1] → R be defined by f(x) =∑∞

n=0a0(2nx)

4n and

g : [0, 1] → R be defined by g(x) = 2x− 2x2. Then f = g.

Proof. For all i, k ε N∪{0} such that i ≤ 2k − 1, the function a0(2nx)4n is linear on[

i2k , i+1

2k

]for all n ε N ∪ {0} such that n ≤ k − 1. Then since 2i+1

2k+1 is the midpoint of[i

2k , i+12k

],

a0

(2n(

2i+12k+1

))4n

=1

2

(a0

(2n(

i2k

))4n

+a0

(2n(

i+12k

))4n

)

for all n ε N ∪ {0} such that n ≤ k − 1. Using these facts, we observe that f and g

satisfy the same difference equation for i, k ε N ∪ {0} such that i ≤ 2k − 1

f

(2i + 1

2k+1

)− 1

2

(f

(i

2k

)+ f

(i + 1

2k

))=

k∑n=0

a0

(2i+1

2k+1−n

)4n

−1

2

(k−1∑n=0

a0

(i

2k−n

)+ a0

(i+12k−n

)4n

)

=a0

(2i+1

2

)4k

=1

2

1

4k=

2

4k+1

and

g

(2i + 1

2k+1

)− 1

2

(g

(i

2k

)+ g

(i + 1

2k

))=

2i + 1

2k− 4i2 + 4i + 1

22k+1

−(

2i + 1

2k− 2i2 + 2i + 1

22k

)=

4i2 + 4i + 2

22k+1− 4i2 + 4i + 1

22k+1

=1

22k+1=

2

4k+1.

Thus, f and g agree on the set{

i2k : i, k ε N ∪ {0}, i ≤ 2k − 1

}. Since f and g are

continuous on [0, 1],{

i2k : i, k ε N ∪ {0}, i ≤ 2k − 1

}is a dense subset of [0, 1], and

f(0) = f(1) = 0 = g(0) = g(1), then f(x) = g(x) for all x ε [0, 1]. �

5

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Now that we have established why we require that b ≥ c, we will show that the

Generalized van der Waerden–Takagi function is continuous on [0, 1].

Proposition 1.1.4. The series∑∞

n=0a0(bnx)

cn , where c > 1 and b ε N such that

b ≥ c, is uniformly convergent on [0, 1].

Proof. Since a0(x) ≤ 12

for all x ε R, then a0(bnx) ≤ 1

2for all x ε [0, 1] , n ε N∪{0}.

Let Mn = 12cn for all n ε N ∪ {0}. Then

∣∣∣a0(bnx)cn

∣∣∣ ≤ Mn for all n ε N ∪ {0}. The series∑∞n=0 Mn = 1

2

∑∞n=0

(1c

)nis a geometric series and is convergent since c > 1. Thus,

by the Weierstrass M-test,∑∞

n=0a0(bnx)

cn is uniformly convergent on [0, 1] . �

Proposition 1.1.5. The function a0(x) is continuous on R.

Proof. Since a0(x) is linear on{(

z, z + 12

)∪(z + 1

2, z + 1

): z ε Z

}, then a0(x)

is continuous on R \{Z ∪

{z2

: z ε Z}}

. It is obvious that a0(x) is continuous at t for

all t ε{Z ∪

{z2

: z ε Z}}

. Therefore, a0(x) is continuous on R. �

Proposition 1.1.6. For all n ε N∪{0}, the function fn(x) = a0(bnx)cn , where c > 1

and b ε N such that b ≥ c, is continuous on [0, 1].

Proof. Since a continuous function multiplied by a constant is a continuous

function, bnx is continuous on [0, 1] for all n ε N ∪ {0}. Since the composition of

continuous functions is a continuous function and since a0(x) is continuous on R by

Proposition 1.1.5, then a0 (bnx) is a continuous function on [0, 1] for all n ε N ∪ {0}.

Thus, for all n ε N ∪ {0}, a0(bnx)cn is continuous on [0, 1] since cn is a constant with

respect to x. �

Fact 1.1.7. Let {fn} be a sequence of continuous functions on a set A ⊆ R and

suppose that {fn} converges uniformly on A to a function f : A → R. Then f is

continuous on A.

Proposition 1.1.8. f(x) =∑∞

n=0a0(bnx)

cn is continuous on [0, 1].

6

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Proof. By Proposition 1.1.6, we know that a0(bnx)cn is continuous on [0, 1] for all

n ε N∪{0}. By Proposition 1.1.4,∑∞

n=0a0(bnx)

cn is uniformly convergent on [0, 1]. Thus,

the sequence(∑m

n=0a0(bnx)

cn

)m ε N

is uniformly convergent on [0, 1], and∑∞

n=0a0(bnx)

cn is

continuous on [0, 1] by Fact 1.1.7. �

To show that the Generalized van der Waerden–Takagi function is nowhere dif-

ferentiable on [0, 1], we need the following lemma.

Lemma 1.1.9. If f is differentiable at x = c, then if (un)n ε N is an increasing

sequence and (vn)n ε N is a decreasing sequence such that un 6= vn and un ≤ c ≤ vn for

all n ε N and such that limn→∞ vn − un = 0, then limn→∞f(vn)−f(un)

vn−un= f ′(c).

Proof. Fix ε > 0. Since f is differentiable at x = c, there exists δ > 0 such that

if 0 < |x− c| < δ, then∣∣∣f(x)−f(c)

x−c− f ′(c)

∣∣∣ < ε2.

Case 1. There exists M ε N such that un = c < vn for all n > M .

Let (un)n ε N be an increasing sequence and (vn)n ε N be a decreasing sequence such

that there exists M ε N such that un = c < vn for all n ε N such that n > M . Then

since limn→∞ vn − un = 0, there exists N ε N such that if n ε N and n > N, then

|vn − un| < δ. Let n ε N such that n > max{N, M}. Since un = c, then |vn − c| < δ.

Thus ∣∣∣∣f(vn)− f(un)

vn − un

− f ′(c)

∣∣∣∣ =

∣∣∣∣f(vn)− f(c)

vn − c− f ′(c)

∣∣∣∣ < ε

2.

Therefore, limn→∞f(vn)−f(un)

vn−un= f ′(c).

Case 2. There exists M ε N such that un < c = vn for all n > M .

By an argument similar to the one used in Case 1, it can be shown that the value

of the limit limn→∞f(vn)−f(un)

vn−unis f ′(c).

Case 3. un < c < vn for all n ε N.

Let (un)n ε N and (vn)n ε N be sequences such that un < c < vn for all n ε N. Then

since limn→∞ vn − un = 0, there exists N ε N such that if n ε N and n > N, then

|vn − un| < δ. Let n ε N such that n > N . Since |vn − un| < δ and un < c < vn,

7

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|vn − c| < δ and |un − c| < δ. Then∣∣∣∣f(vn)− f(un)

vn − un

− f ′(c)

∣∣∣∣ =

∣∣∣∣f(vn)− f(un)− (vn − un)f ′(c)

vn − un

∣∣∣∣=

∣∣∣∣f(vn)− f(un)− f(c) + f(c)− (vn + c− c− un)f ′(c)

vn − un

∣∣∣∣≤

∣∣∣∣f(vn)− f(c)− (vn − c)f ′(c)

vn − un

∣∣∣∣+

∣∣∣∣f(un)− f(c)− (un − c)f ′(c)

un − vn

∣∣∣∣≤

∣∣∣∣f(vn)− f(c)− (vn − c)f ′(c)

vn − c

∣∣∣∣+

∣∣∣∣f(un)− f(c)− (un − c)f ′(c)

un − c

∣∣∣∣=

∣∣∣∣f(vn)− f(c)

vn − c− f ′(c)

∣∣∣∣+ ∣∣∣∣f(un)− f(c)

un − c− f ′(c)

∣∣∣∣<

ε

2+

ε

2

= ε.

Therefore, limn→∞f(vn)−f(un)

vn−un= f ′(c). �

We now return to the Generalized van der Waerden–Takagi function to show that

the function is nowhere differentiable on [0, 1].

Proposition 1.1.10. The function f(x) =∑∞

n=0a0(bnx)

cn , where c > 1 and b ε N

such that b ≥ c, is nowhere differentiable on [0, 1].

Proof. Let t ε [0, 1]. We want to show that f(x) is not differentiable at x = t.

Case 1. t = 1.

Let (um)m ε N and (vm)m ε N be sequences such that um = 1− 1bm for all m ε N and

vm = 1 for all m ε N. Clearly, limm→∞ vm − um = 0. Then

f(vm)− f(um)

vm − um

=

∑∞n=0

a0(bn)cn −

∑∞n=0

a0((1− 1bm )bn)

cn

1bm

8

Page 16: Continuous Nowhere Differentiable Functionseduc.jmu.edu/~querteks/seniorthesis.pdfContinuous nowhere differentiable functions are functions that are continuous at ... It is well

= bm

(−

m−1∑n=0

a0

(bn − bn

bm

)cn

)

= −m−1∑n=0

bm+n

bmcn

= −m−1∑n=0

(b

c

)n

.

Since b ≥ c > 0, limn→∞(

bc

)n 6= 0. Thus the sequence(

f(vm)−f(um)vm−um

)m ε N

, which is

equivalent to the sequence(∑m−1

n=0 −(

bc

)n)m ε N , does not converge as m → ∞. By

Lemma 1.1.9, f(x) is not differentiable at x = 1.

Case 2. t ε [0, 1).

Since t ε [0, 1), then for all m ε N there exists km ε Z ∩ [0, 2bm − 1] such that km

2bm ≤

t < km+12bm . Let um = km

2bm for all m ε N and let vm = km+12bm for all m ε N. Then (vm)m ε N is

a decreasing sequence and (um)m ε N is an increasing sequence such that limm→∞ vm−

um = 0.

Let m ε N. Since um = km

2bm and vm = km+12bm , then for all n such that 0 ≤ n ≤ m− 1

there exists z ε Z such that bnvm and bnum are either both contained in[z, z + 1

2

]or

are both contained in[z + 1

2, z + 1

]. Thus for all n such that 0 ≤ n ≤ m− 1, a0(bnx)

cn

is linear on [um, vm] and has a slope of either(

bc

)nor −

(bc

)n. Then for all n such that

0 ≤ n ≤ m− 1,a0(bnvm)

cn − a0(bnum)cn

vm − um

= ±(

b

c

)n

.

For n ≥ m, we must consider 2 subcases.

Subcase 1. b is even.

Let n = m. If km is even, then a0

(bnkm

2bm

)= 0 and a0

(bn(km+1)

2bm

)= 1

2. Thus,

a0(bnvm)cn − a0(bnum)

cn

vm − um

=1

2cm − 01

2bm

=

(b

c

)m

.

By a similar argument, if km is odd, then

a0(bnvm)cn − a0(bnum)

cn

vm − um

= −(

b

c

)m

.

9

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Now let n > m. Then bn−mkm and bn−m(km + 1) are both even since b is even.

Thus, a0

(bnkm

2bm

)= 0 and a0

(bn(km+1)

2bm

)= 0. Then, for all n ε N such that n > m,

a0(bnvm)cn − a0(bnum)

cn

vm − um

= 0.

Therefore if b is even, then

f(vm)− f(um)

vm − um

=m∑

n=0

±(

b

c

)n

. (1)

Since b ≥ c > 0, limn→∞±(

bc

)n 6= 0. Thus limm→∞∑m

n=0±(

bc

)ndoes not con-

verge. Then limm→∞f(vm)−f(um)

vm−umdoes not exist. By Lemma 1.1.9, f(x) is not differ-

entiable at x = t.

Subcase 2. b is odd.

Let n ≥ m. If km is even, then a0

(bnkm

2bm

)= 0 and a0

(bn(km+1)

2bm

)= 1

2. Thus if km is

even, thena0(bnvm)

cn − a0(bnum)cn

vm − um

=1

2cn − 01

2bm

=bm

cn.

for all n ε N such that n ≥ m. If km is odd, then by a similar argument

a0(bnvm)cn − a0(bnum)

cn

vm − um

= −bm

cn

for all n ε N such that n ≥ m. Thus,

f(vm)− f(um)

vm − um

=m−1∑n=0

±(

b

c

)n

± bm

∞∑n=m

(1

c

)n

.

Since c > 1,∑∞

n=m

(1c

)n= c

c−1

(1c

)m. Therefore if b is odd, then

f(vm)− f(um)

vm − um

=m−1∑n=0

±(

b

c

)n

± c

c− 1

(b

c

)m

. (2)

Since b ≥ c > 1, then limn→∞±(

bc

)n 6= 0 and limm→∞± cc−1

(bc

)m 6= 0. Thus

limm→∞∑m−1

n=0 ±(

bc

)n ± cc−1

(bc

)mdoes not converge. Then limm→∞

f(vm)−f(um)vm−um

does

not exist. By Lemma 1.1.9, f(x) is not differentiable at x = t. �

10

Page 18: Continuous Nowhere Differentiable Functionseduc.jmu.edu/~querteks/seniorthesis.pdfContinuous nowhere differentiable functions are functions that are continuous at ... It is well

1.2. Kiesswetter’s Function

Another example of a continuous nowhere differentiable function is Kiesswetter’s

function. The original construction of the function proposed by Kiesswetter defined

the function using base 4 expansion [5]. The following alternate construction from

[1, pages 201–202] uses Kiesswetter’s curve to define the function. Kiesswetter’s curve

is constructed using the functions

f1

x

y

=

14

0

0 −12

x

y

f2

x

y

=

14

0

0 12

x

y

+

14

−12

f3

x

y

=

14

0

0 12

x

y

+

12

0

f4

x

y

=

14

0

0 12

x

y

+

34

12

.

To construct Kiesswetter’s curve, let L0 be the line segment from (0, 0) to (1, 1)

in the Cartesian plane. At each stage of the construction, replace Ln with Ln+1 =

f1[Ln] ∪ f2[Ln] ∪ f3[Ln] ∪ f4[Ln]. The sequence (Ln)nεN is a convergent sequence of

graphs, and the limit is the graph of Kiesswetter’s function.

Let gn : [0, 1] → R be the function whose graph is Ln. Then for all t ε [0, 1],

the point (t, gn(t)) ε Ln. We note that, from the definition of Kiesswetter’s curve, the

function g0 : [0, 1] → R is defined by g0(t) = t. To define gn(t) for n ε N, we develop

the following recurrence relation.

11

Page 19: Continuous Nowhere Differentiable Functionseduc.jmu.edu/~querteks/seniorthesis.pdfContinuous nowhere differentiable functions are functions that are continuous at ... It is well

Proposition 1.2.1. For all t ε [0, 1] and all n ε N,

gn(t) =

−12

gn−1(4t), 0 ≤ t ≤ 14

12gn−1(4t− 1)− 1

2, 1

4< t ≤ 1

2

12gn−1(4t− 2), 1

2< t ≤ 3

4

12gn−1(4t− 3) + 1

2, 3

4< t ≤ 1.

(3)

Proof. Let t ε [0, 1] and n ε N. Then (t, gn−1(t)) ε Ln−1.

By the definition of the Kiesswetter’s curve,

f1

t

gn−1(t)

=

14

0

0 −12

t

gn−1(t)

=

14t

−12

gn−1(t)

ε Ln.

Thus if 0 ≤ t ≤ 14, then

(t, −1

2gn−1(4t)

)ε Ln.

Similarly, since

f2

t

gn−1(t)

=

14

0

0 12

t

gn−1(t)

+

14

−12

=

14t + 1

4

12gn−1(t)− 1

2

ε Ln,

then, for 14

< t ≤ 12,(t, 1

2gn−1(4t− 1)− 1

2

)ε Ln.

Applying f3, we show that

f3

t

gn−1(t)

=

14

0

0 12

t

gn−1(t)

+

12

0

=

14t + 1

2

12gn−1(t)

ε Ln.

Thus if 12

< t ≤ 34, then

(t, 1

2gn−1(4t− 2)

)ε Ln.

Similarly, since the definition of Kiesswetter’s curve states that

f4

t

gn−1(t)

=

14

0

0 12

t

gn−1(t)

+

34

12

=

14t + 3

4

12gn−1(t) + 1

2

ε Ln,

then, for 34

< t ≤ 1,(t, 1

2gn−1(4t− 3) + 1

2

)ε Ln.

12

Page 20: Continuous Nowhere Differentiable Functionseduc.jmu.edu/~querteks/seniorthesis.pdfContinuous nowhere differentiable functions are functions that are continuous at ... It is well

Thus,

gn(t) =

−12

gn−1(4t), 0 ≤ t ≤ 14

12gn−1(4t− 1)− 1

2, 1

4< t ≤ 1

2

12gn−1(4t− 2), 1

2< t ≤ 3

4

12gn−1(4t− 3) + 1

2, 3

4< t ≤ 1.

We now prove some properties of the functions, gn for n ε N ∪ {0} that will be

used to show that the function whose graph is Kiesswetter’s curve is continuous and

nowhere differentiable.

Proposition 1.2.2. Let g0(t) = t on [0, 1] and let gn(t) be defined on [0, 1] by (3)

for all n ε N. Then, for all n ε N ∪ {0}, gn(0) = 0.

Proof. Since g0(t) = t for all t ε [0, 1], g0(0) = 0. Let k ε N ∪ {0} and assume

gk(0) = 0. Then

gk+1(0) =−1

2gk(4(0)) =

(−1

2

)(0) = 0.

Therefore, gn(0) = 0 for all n ε N ∪ {0}. �

Proposition 1.2.3. Let g0(t) = t on [0, 1] and let gn(t) be defined on [0, 1] by (3)

for all n ε N. Then, for all n ε N ∪ {0}, gn(1) = 1.

Proof. Since g0(t) = t for all t ε [0, 1], g0(1) = 1. Let k ε N ∪ {0} and assume

gk(1) = 1. Then

gk+1(1) =1

2gk(4(1)− 3) +

1

2=

1

2gk(1) +

1

2=

1

2+

1

2= 1.

Thus, gn(1) = 1 for all n ε N ∪ {0}. �

Proposition 1.2.4. Let g0(t) = t on [0, 1] and let gn(t) be defined on [0, 1] by

(3) for all n ε N. If j, k ε N ∪ {0} such that j ≥ k, then gj

(m4k

)= gk

(m4k

)for all

m ε Z ∩ [0, 4k].

13

Page 21: Continuous Nowhere Differentiable Functionseduc.jmu.edu/~querteks/seniorthesis.pdfContinuous nowhere differentiable functions are functions that are continuous at ... It is well

Proof. Let k = 0. Then m = 0 or m = 1. By Proposition 1.2.2, gj

(040

)=

g0

(040

)= 0 for all j ε N∪{0} such that j ≥ 0. Similarly by Proposition 1.2.3, gj

(140

)=

g0

(140

)= 1 for all j εN∪{0} such that j ≥ 0. Thus, for all m ε Z such that 0 ≤ m ≤ 40,

gj

(m40

)= g0

(m40

)for all j ε N ∪ {0} such that j ≥ 0.

Now let k ε N such that k ≥ 0 and assume that for all m ε Z ∩ [0, 4k], gj

(m4k

)=

gk

(m4k

)for all j ε N such that j ≥ k.

We now consider k + 1. Fix m ε Z such that 0 ≤ m ≤ 4k+1 and let j ε N such that

j ≥ k + 1.

Case 1. 0 ≤ m4k+1 ≤ 1

4.

Using (3), we show that

gj

( m

4k+1

)=−1

2gj−1

(4m

4k+1

)=−1

2gj−1

(m

4k

).

Since 0 ≤ m4k+1 ≤ 1

4, then 0 ≤ m ≤ 4k. Also since j ≥ k + 1, then j − 1 ≥ k. Thus, by

the inductive hypothesis, gj−1

(m4k

)= gk

(m4k

). So

gj

( m

4k+1

)=−1

2gk

(m

4k

)= gk+1

( m

4k+1

).

Case 2. 14

< m4k+1 ≤ 1

2.

Since 14

< m4k+1 ≤ 1

2, then 0 < m−4k ≤ 4k. Using (3) and the inductive hypothesis

as in case 1, we show that

gj

( m

4k+1

)=

1

2gj−1

(m− 4k

4k

)− 1

2=

1

2gk

(m− 4k

4k

)− 1

2= gk+1

( m

4k+1

).

Case 3. 12

< m4k+1 ≤ 3

4.

Since 12

< m4k+1 ≤ 3

4, then 0 < m − 2(4k) ≤ 4k. Then by (3) and the inductive

hypothesis,

gj

( m

4k+1

)=

1

2gj−1

(m− 2(4k)

4k

)=

1

2gk

(m− 2(4k)

4k

)= gk+1

( m

4k+1

).

Case 4. 34

< m4k+1 ≤ 1.

14

Page 22: Continuous Nowhere Differentiable Functionseduc.jmu.edu/~querteks/seniorthesis.pdfContinuous nowhere differentiable functions are functions that are continuous at ... It is well

Since 34

< m4k+1 ≤ 1, then 0 < m − 3(4k) ≤ 4k. Thus using (3) and the inductive

hypothesis, we show that

gj

( m

4k+1

)=

1

2gj−1

(m− 3(4k)

4k

)+

1

2=

1

2gk

(m− 3(4k)

4k

)+

1

2= gk+1

( m

4k+1

).

Thus, for all m ε Z ∩ [0, 4k+1], gj

(m

4k+1

)= gk+1

(m

4k+1

)for all j ε N such that j ≥

k + 1. Therefore for all j, k ε N ∪ {0} such that j ≥ k, gj

(m4k

)= gk

(m4k

)for all

m ε Z ∩ [0, 4k]. �

Proposition 1.2.5. Let g0(t) = t on [0, 1] and let gn(t) be defined on [0, 1] by (3)

for all n ε N. If k ε N∪{0}, then, for all n, m ε Z such that n ≥ k and 0 ≤ m ≤ 4k− 1,∣∣gn

(m4k

)− gn

(m+14k

)∣∣ = 12k

Proof. By Proposition 1.2.4, if k, n ε N ∪ {0} such that n ≥ k, then gn

(m4k

)=

gk

(m4k

)for all m ε Z ∩ [0, 4k]. Thus, we need only show that if k ε N ∪ {0}, then∣∣gk

(m4k

)− gk

(m+14k

)∣∣ = 12k for all m ε Z such that 0 ≤ m ≤ 4k − 1.

First let k = 0. Let m ε [0, 40 − 1]. Thus, m = 0. Then∣∣∣∣g0

(0

40

)− g0

(1

40

)∣∣∣∣ = |0− 1| = 1

20.

Now let k = j ε N and assume∣∣gj

(m4j

)− gj

(m+14j

)∣∣ = 12j for all m ε Z∩[0, 4j−1]. Let

s ε Z such that 0 ≤ s ≤ 4j+1 − 1. We want to show that∣∣gj+1

(s

4j+1

)− gj+1

(s+14j+1

)∣∣ =

12j+1 .

Case 1. 0 ≤ s4j+1 ≤ 1

4− 1

4j+1 .

Since 0 ≤ s4j+1 ≤ 1

4− 1

4j+1 , then 0 ≤ s ≤ 4j−1. Thus, by the inductive hypothesis,∣∣gj

(s4j

)− gj

(s+14j

)∣∣ = 12j . Then using (3), we show that∣∣∣∣gj+1

( s

4j+1

)− gj+1

(s + 1

4j+1

)∣∣∣∣ =

∣∣∣∣−1

2gj

( s

4j

)− −1

2gj

(s + 1

4j

)∣∣∣∣=

1

2

∣∣∣∣gj

( s

4j

)− gj

(s + 1

4j

)∣∣∣∣=

1

2

(1

2j

)=

1

2j+1.

15

Page 23: Continuous Nowhere Differentiable Functionseduc.jmu.edu/~querteks/seniorthesis.pdfContinuous nowhere differentiable functions are functions that are continuous at ... It is well

Case 2. 14− 1

4j+1 < s4j+1 ≤ 1

2− 1

4j+1 .

Since 14− 1

4j+1 < s4j+1 ≤ 1

2− 1

4j+1 and s ε Z, then 14≤ s

4j+1 ≤ 12− 1

4j+1 . Thus

0 ≤ s− 4j ≤ 4j − 1. Then by the inductive hypothesis and (3),∣∣∣∣gj+1

( s

4j+1

)− gj+1

(s + 1

4j+1

)∣∣∣∣ =

∣∣∣∣12gj

( s

4j− 1)− 1

2− 1

2gj

(s + 1

4j+ 1

)+

1

2

∣∣∣∣=

1

2

∣∣∣∣gj

(s− 4j

4j

)− gj

(s− 4j + 1

4j

)∣∣∣∣=

1

2

(1

2j

)=

1

2j+1.

The other two cases are similar and are left to the reader. �

Now that we have established some of the properties of the functions whose limit

is Kiesswetter’s function, we will now show that Kiesswetter’s function is continuous

on [0, 1]. We first show that the sequence of functions (gn)n ε N∪{0} is a sequence of

continuous functions on [0, 1] and that the sequence converges uniformly on [0, 1].

Proposition 1.2.6. Let g0(t) = t on [0, 1] and let gn(t) be defined on [0, 1] by (3)

for all n ε N. Then the function gn(t) is continuous on [0, 1] for all n ε N ∪ {0}.

Proof. By the definition of Kiesswetter’s curve, g0(t) is continuous on [0, 1]. Let

k ε N ∪ {0} and assume that gk(t) is continuous on [0, 1]. Then, by (3), gk+1(t) is

continuous on[0, 1

4

)∪(

14, 1

2

)∪(

12, 3

4

)∪(

34, 1]. Thus to show that gk+1(t) is continuous

on [0, 1], we need only show that gk+1(t) is continuous at t = 14, t = 1

2, and t = 3

4. We

will show that gk+1(t) is continuous at t = 14. The other proofs are similar and are

left to the reader.

Since gk(t) is continuous on [0, 1] and gk+1(t) = 12gk(4t) for 0 ≤ t ≤ 1

4, thus

limt→ 1

4

− gk+1(t) = gk+1(14). So we only need to show that the limit lim

t→ 14

− gk+1(t) =

limt→ 1

4

+ gk+1(t). Since

limt→ 1

4

−gk+1(t) = lim

t→ 14

−1

2gk(4t) = lim

t→1−

−1

2gk(t) = −1

2

16

Page 24: Continuous Nowhere Differentiable Functionseduc.jmu.edu/~querteks/seniorthesis.pdfContinuous nowhere differentiable functions are functions that are continuous at ... It is well

and

limt→ 1

4

+gk+1(t) = lim

t→ 14

+

1

2gk(4t− 1)− 1

2= lim

t→0+

1

2gk(t)−

1

2= −1

2,

then limt→ 1

4

− gk+1(t) = limt→ 1

4

+ gk+1(t). Therefore, gk+1(t) is continuous at t = 14. �

Proposition 1.2.7. Let g0(t) = t on [0, 1] and let gn(t) be defined on [0, 1] by (3)

for all n ε N. Then the sequence {gn}n ε N∪{0} converges uniformly on [0, 1].

Proof. Fix ε > 0. Pick N ε N such that 2N+1 > 1ε. Let j, n ε N such that j > n ≥

N. Then if x ε [0, 1], there exists m ε Z ∩ [0, 4n − 1] such that m4n ≤ x ≤ m+1

4n . Since

j > n, then gj

(m4n

)= gn

(m4n

)by Proposition 1.2.4. Then

|gj(x)− gn(x)| =∣∣∣gj(x)− gj

(m

4n

)+ gn

(m

4n

)− gn(x)

∣∣∣≤

∣∣∣gj(x)− gj

(m

4n

)∣∣∣+ ∣∣∣gn(x)− gn

(m

4n

)∣∣∣=

∣∣∣∣gj(x)− gj

(4j−nm

4j

)∣∣∣∣+ ∣∣∣gn(x)− gn

(m

4n

)∣∣∣ .Since m

4n ≤ x ≤ m+14n and gn is linear on [ m

4n , m+14n ], then, by Proposition 1.2.5,∣∣∣gn(x)− gn

(m

4n

)∣∣∣ ≤ ∣∣∣∣gn

(m

4n

)− gn

(m + 1

4n

)∣∣∣∣ ≤ 1

2n.

Since 4j−nm4j ≤ x ≤ 4j−nm+4j−n

4j , then there exists s ε Z ∩ [0, 4j−n − 1] so that

4j−nm+s4j ≤ x ≤ 4j−nm+s+1

4j . Since gj(x) is linear on the interval [4j−nm+s

4j , 4j−nm+s+14j ]

and x ε [4j−nm+s

4j , 4j−nm+s4j ], then, by Proposition 1.2.5,∣∣∣∣gj(x)− gj

(4j−nm

4j

)∣∣∣∣ ≤ ∣∣∣∣gj

(4j−nm + s

4j

)− gj

(4j−nm + s + 1

4j

)∣∣∣∣ ≤ 1

2j.

Thus,

|gj(x)− gn(x)| ≤∣∣∣∣gj(x)− gj

(4j−nm

4j

)∣∣∣∣+ ∣∣∣gn(x)− gn

(m

4n

)∣∣∣≤ 1

2j+

1

2n<

1

2n−1< ε.

17

Page 25: Continuous Nowhere Differentiable Functionseduc.jmu.edu/~querteks/seniorthesis.pdfContinuous nowhere differentiable functions are functions that are continuous at ... It is well

We recall that for all n ε N ∪ {0} the function gn(t) is the function whose graph

is Ln. Since the graph of Kiesswetter’s curve is the limit of the sequence (Ln)n ε N∪{0},

we can define Kiesswetter’s function, g : [0, 1] → R, by g(t) = limn→∞ gn(t) since

the sequence {gn}n ε N∪{0} converges. Now we can show that Kiesswetter’s function is

continuous and nowhere differentiable on [0, 1].

Proposition 1.2.8. The function g(t) = limn→∞ gn(t), where g0(t) = t and gn(t)

is defined by (3) for all n ε N, is continuous on [0, 1].

Proof. By Proposition 1.2.6, gn(x) is continuous on [0, 1] for all n ε N ∩ {0}.

Then by Proposition 1.2.7, the sequence {gn}n ε N∪{0} converges uniformly on [0, 1] to

g. Thus, g(x) is continuous on [0, 1] by Fact 1.1.7. �

Proposition 1.2.9. The function g(t) = limn→∞ gn(t), where g0(t) = t and gn(t)

is defined by (3) for all n ε N, is nowhere differentiable on [0, 1].

Proof. Let c ε [0, 1]. We want to show that g(x) is not differentiable at c.

Case 1. c = 1.

Let (um)m ε N and (vm)m ε N be sequences such that um = 4m−14m for all m ε N and

vm = 4m

4m = 1 for all m ε N. Clearly, limm→∞ vm− um = 0. Then by Propositions 1.2.4

and 1.2.5, ∣∣∣∣g(vm)− g(um)

vm − um

∣∣∣∣ =

∣∣∣∣∣ limn→∞ gn

(4m

4m

)− limn→∞ gn

(4m−14m

)1

4m

∣∣∣∣∣= 4m

∣∣∣∣gm

(4m

4m

)− gm

(4m − 1

4m

)∣∣∣∣= 4m 1

2m= 2m.

Then∣∣∣g(vm)−g(um)

vm−um

∣∣∣ → ∞ as m → ∞ so the sequence(

g(vm)−g(um)vm−um

)m ε N

does not

converge. Thus, by Lemma 1.1.9, g(x) is not differentiable at x = c.

Case 2. c ε [0, 1).

18

Page 26: Continuous Nowhere Differentiable Functionseduc.jmu.edu/~querteks/seniorthesis.pdfContinuous nowhere differentiable functions are functions that are continuous at ... It is well

Since c ε [0, 1), then for all m ε N there exists km ε Z∩ [0, 4m−1] such that km

4m ≤ c <

km+14m . Let um = km

4m for all m ε N and let vm = km+14m for all m ε N. Then (um)m ε N is an

increasing sequence and (vm)m ε N is a decreasing sequence. Clearly, limm→∞ vm−um =

limm→∞1

4m = 0.

Then by Propositions 1.2.4 and 1.2.5,∣∣∣∣g(vm)− g(um)

vm − um

∣∣∣∣ =

∣∣∣∣∣ limn→∞ gn

(km+14m

)− limn→∞ gn

(km

4m

)1

4m

∣∣∣∣∣= 4m

∣∣∣∣gm

(km + 1

4m

)− gm

(km

4m

)∣∣∣∣= 4m 1

2m= 2m.

Since limm→∞

∣∣∣g(vm)−g(um)vm−um

∣∣∣ = limm→∞ 2m = ∞, the sequence(

g(vm)−g(um)vm−um

)m ε N

does not converge. Therefore, g(x) is not differentiable at x = c by Lemma 1.1.9. �

19

Page 27: Continuous Nowhere Differentiable Functionseduc.jmu.edu/~querteks/seniorthesis.pdfContinuous nowhere differentiable functions are functions that are continuous at ... It is well

Chapter 2

Holder Continuity

A function f : [0, 1] → R is said to be a Holder continuous function of exponent

α if, for some constant M ,

|f(x)− f(y)| ≤ M |x− y|α

for all x, y ε [0, 1]. We note that if f is Holder continuous of exponent 1, then f is

Lipschitz. The maximal value of α such that f is a Holder continuous function

of exponent α is related to the existence of the derivative of f on [0, 1]. We will

investigate the Holder continuity of the examples given in Chapter 1. Since the

Generalized van der Waerden-Takagi function and Kiesswetter’s function are nowhere

differentiable on [0, 1], we know that the two functions are not Holder continuous

of exponent α for α ≥ 1. Lipschitz functions are differentiable almost everywhere

[8, pages 108–112], and functions that are Holder continuous of exponent α > 1

have zero as a derivative everywhere. Thus to fully determine the exponents, α, for

which the Generalized van der Waerden–Takagi function and Kiesswetter’s function

are Holder continuous, we need only consider α < 1.

2.1. The Generalized van der Waerden–Takagi Function

We first consider the Generalized van der Waerden–Takagi function. In order to

be able to consider∣∣∣∑∞

n=0a0(bnx)

cn −∑∞

n=0a0(bny)

cn

∣∣∣ for x, y ε [0, 1], we need to obtain a

bound for∣∣∣a0(bnx)

cn − a0(bny)cn

∣∣∣ for x, y ε [0, 1] and n ε N ∪ {0}.

20

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Proposition 2.1.1. For all x, y ε [0, 1] and n ε N ∪ {0},∣∣∣∣a0 (bnx)

cn− a0 (bny)

cn

∣∣∣∣ ≤ (b

c

)n

|x− y|.

Proof. Let n ε N∪{0} and x, y ε [0, 1]. Let k ε Z such that k is the closest integer

to bny. Then

a0 (bnx) ≤ |bnx− k| ≤ |bnx− bny|+ |bny − k| = |bnx− bny|+ a0 (bny) .

Thus,

a0 (bnx)− a0 (bny) ≤ |bnx− bny| .

By a similar argument, it can be shown that

a0 (bny)− a0 (bnx) ≤ |bny − bnx| .

Thus,

|a0 (bnx)− a0 (bny)| ≤ bn |x− y| .

Therefore, ∣∣∣∣a0 (bnx)

cn− a0 (bny)

cn

∣∣∣∣ ≤ (b

c

)n

|x− y| .

We first consider the Holder continuity of the Generalized van der Waerden–Takagi

function with b = c. The following proof is along the lines of the proof given by Shidfar

and Sabetfakhri in [9].

Theorem 2.1.2. If c ε N such that c ≥ 2, then f(x) =∑∞

n=0a0(cnx)

cn is Holder

continuous of exponent α on [0, 1] if and only if α < 1.

Proof. Let α < 1. Let x and y ε [0, 1] such that x 6= y. Thus 0 < |x − y| ≤ 1.

Then there exists k ε Z such that k ≥ 0 and 1ck+1 < |x− y| ≤ 1

ck .

Then by Proposition 2.1.1,

k−1∑n=0

∣∣∣∣a0(cnx)

cn− a0(c

ny)

cn

∣∣∣∣ ≤ k−1∑n=0

(c

c

)n

|x− y| = k|x− y| ≤ k1

ck(1−α)|x− y|α. (4)

21

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Since 0 ≤ a0(cnt)cn ≤ 1

cn for all n ε N ∪ {0} and t ε [0, 1], then∣∣∣∣a0(cnx)

cn− a0(c

ny)

cn

∣∣∣∣ ≤ 1

cn

for all n ε N ∪ {0}. Thus

∞∑n=k

∣∣∣∣a0(cnx)

cn− a0(c

ny)

cn

∣∣∣∣ ≤ ∞∑n=k

1

cn≤ 1

ck−1≤ c2|x− y| ≤ c2|x− y|α (5)

since 1ck+1 < |x− y| ≤ 1.

Then combining (4) and (5), we obtain

∞∑n=0

∣∣∣∣a0(cnx)

cn− a0(c

ny)

cn

∣∣∣∣ ≤ k1

ck(1−α)|x− y|α + c2|x− y|α =

(c2 +

k

ck(1−α)

)|x− y|α.

Since kck(1−α) =

(k(

11−α)ck

)(1−α)

for all k ε N ∪ {0} and c > 1, limk→∞k

ck(1−α) = 0.

Thus, limk→∞ c2 + kck(1−α) = c2. Then the sequence

(c2 + k

ck(1−α)

)k ε N∪{0} is bounded,

and there exists Mα > 0 such that∣∣c2 + k

ck(1−α)

∣∣ ≤ Mα for all k ε N ∪ {0}.

Then∞∑

n=0

∣∣∣∣a0(cnx)

cn− a0(c

ny)

cn

∣∣∣∣ ≤ Mα|x− y|α.

Thus, f is Holder continuous of exponent α for all α < 1. To prove the other direction

of the theorem, we recall that f is nowhere differentiable on [0, 1] by Proposition

1.1.10. Therefore, f is not Holder continuous of order α for all α ≥ 1. �

We now consider the Holder continuity of the Generalized van der Waerden-Takagi

function with b > c.

Proposition 2.1.3. If c > 1 and b ε N such that b > c, then f(x) =∑∞

n=0a0(bnx)

cn

is Holder continuous of exponent α on [0, 1] for all α ≤ log clog b

.

Proof. Let α ≤ log clog b

< 1. Let x and y ε [0, 1] such that x 6= y. Thus 0 < |x−y| ≤ 1.

Then there exists k ε Z such that k ≥ 0 and 1bk+1 < |x− y| ≤ 1

bk . Thus by Proposition

22

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2.1.1,

k−1∑n=0

∣∣∣∣a0(bnx)

cn− a0(b

ny)

cn

∣∣∣∣ ≤k−1∑n=0

(b

c

)n

|x− y|

[1

bk(1−α)

k−1∑n=0

(b

c

)n]|x− y|α

=

[((bc

)k − 1(bc

)− 1

)1

bk(1−α)

]|x− y|α

=

[c

b− c

((bα

c

)k

− 1

bk(1−α)

)]|x− y|α.

Since 0 ≤ a0(bnt)cn ≤ 1

cn for all n ε N ∪ {0} and t ε [0, 1], then∣∣∣∣a0(bnx)

cn− a0(b

ny)

cn

∣∣∣∣ ≤ 1

cn.

for all n ε N ∪ {0}. Thus, by the same argument as in (5) in Proposition 2.1.2,

∞∑n=k

∣∣∣∣a0(bnx)

cn− a0(b

ny)

cn

∣∣∣∣ ≤ c2|x− y|α.

Then

∞∑n=0

∣∣∣∣a0(bnx)

cn− a0(b

ny)

cn

∣∣∣∣ ≤[

c

b− c

((bα

c

)k

− 1

bk(1−α)

)+ c2

]|x− y|α .

Since b > 1, limk→∞1

bk(1−α) = 0. Thus, the limit limk→∞c

b−c

((bα

c

)k − 1bk(1−α)

)+ c2

= limk→∞c

b−c

(bα

c

)k+c2 exists if and only if limk→∞

(bα

c

)kexists. Since α ≤ log c

log b, then

c≤ 1. Thus limk→∞

(bα

c

)kis finite. The sequence

(c

b−c

((bα

c

)k − 1bk(1−α)

)+ c2

)k ε N∪{0}

is then bounded, and there exists Mα > 0 such that∣∣∣ cb−c

((bα

c

)k − 1bk(1−α)

)+ c2

∣∣∣ ≤ Mα

for all k ε N ∪ {0}.

Then∞∑

n=0

∣∣∣∣a0(bnx)

cn− a0(b

ny)

cn

∣∣∣∣ ≤ Mα|x− y|α.

Proposition 2.1.4. If c > 1 and b ε N such that b > c, then f(x) =∑∞

n=0a0(bnx)

cn

is not Holder continuous of exponent α on [0, 1] for α > log clog b

.

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Proof. Since b is an integer, c > 1, and b > c, f is a continuous nowhere differ-

entiable function by Propositions 1.1.8 and 1.1.10. Thus, f is not Holder continuous

of exponent greater than or equal to 1.

To determine the Holder continuity of f of exponent α for α < 1, let log clog b

< α < 1.

Let k ε N. Since b ≥ 2, 1bk−n ≤ 1

2for all n ε N ∪ {0} such that n < k. Thus,∣∣∣∣f ( 1

bk

)− f (0)

∣∣∣∣ = f

(1

bk

)=

∞∑n=0

a0

(bn 1

bk

)cn

=k−1∑n=0

(bc

)nbk

=

∑k−1n=0

(bc

)nbk(1−α)

∣∣∣∣ 1

bk− 0

∣∣∣∣α

=

((bc

)k − 1)

c

bk(1−α)(b− c)

∣∣∣∣ 1

bk− 0

∣∣∣∣α=

c

b− c

((bα

c

)k

− 1

bk(1−α)

)∣∣∣∣ 1

bk− 0

∣∣∣∣α .

Thus, limk→∞c

b−c

((bα

c

)k − 1bk(1−α)

)must exist in order for f to be Holder continu-

ous of exponent α. However, since α > log clog b

, bα > c and(

c

)k →∞ as k →∞. Thus,

cb−c

((bα

c

)k − 1bk(1−α)

)is unbounded as k → ∞ since limk→∞

1bk(1−α) = 0. Therefore, f

is not Holder continuous of exponent α on [0, 1]. �

Theorem 2.1.5. Let c > 1 and b ε N such that b > c. Then f(x) =∑∞

n=0a0(bnx)

cn

is Holder continuous of exponent α on [0, 1] if and only if α ≤ log clog b

.

Proof. This theorem is the immediate result of Propositions 2.1.3 and 2.1.4. �

Corollary 2.1.6. For all m ε N such that m ≥ 2, the function,

f(x) =∞∑

n=0

a0 ((cm)n x)

cn,

with c > 1 such that cm ε N, is Holder continuous on [0, 1] if and only if α ≤ 1m

.

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Proof. Fix m ε N \ {1}. Let c > 1 such that cm ε N. Since m ≥ 2, cm > c. Thus,

f satisfies the hypotheses for Theorem 2.1.5. Therefore, f is Holder continuous of

exponent α on [0, 1] for α ≤ log clog cm = 1

m, and f is not Holder continuous of exponent

α on [0, 1] for α > log clog cm = 1

m. �

2.2. Kiesswetter’s Function

We now consider Kiesswetter’s function.

Proposition 2.2.1. The function g(t) = limn→∞ gn(t), where g0(t) = t and gn(t)

is defined by (3) for all n ε N, is Holder continuous of exponent α on [0, 1] for all

α ≤ 12.

Proof. Let α ≤ 12. Let x, y ε [0, 1] so that x 6= y. Then there exists k ε N ∪ {0}

such that 14k+1 < |x− y| ≤ 1

4k .

Then by Proposition 1.2.7,

|gj(x)− gk(x)| ≤ 1

2k+

1

2j

for all j ε N such that j > k. Thus,

|g(x)− gk(x)| = limj→∞

|gj(x)− gk(x)| ≤ limj→∞

1

2k+

1

2j=

1

2k.

Similarly

|g(y)− gk(y)| ≤ 1

2k.

Then

|g(x)− g(y)| = |g(x)− gk(x) + gk(x)− gk(y) + gk(y)− g(y)|

≤ |g(x)− gk(x)|+ |gk(x)− gk(y)|+ |gk(y)− g(y)|

≤ 1

2k+ |gk(x)− gk(y)|+ 1

2k.

25

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In order to bound |gk(x) − gk(y)|, we assume, without loss of generality, that

y > x. Then there exists m ε Z ∩[0, 4k − 1

]such that m

4k ≤ x < m+14k .

Case 1. x < y ≤ m+14k .

Then by Proposition 1.2.5,

|gk(x)− gk(y)| ≤∣∣∣∣gk

(m

4k

)− gk

(m + 1

4k

)∣∣∣∣ =1

2k.

Case 2. m+14k < y < m+2

4k .

Then by Proposition 1.2.5,

|gk(x)− gk(y)| ≤∣∣∣∣gk(x)− gk

(m + 1

4k

)∣∣∣∣+ ∣∣∣∣gk

(m + 1

4k

)− gk(y)

∣∣∣∣≤

∣∣∣∣gk

(m

4k

)− gk

(m + 1

4k

)∣∣∣∣+ ∣∣∣∣gk

(m + 1

4k

)− gk

(m + 2

4k

)∣∣∣∣= 2

(1

2k

).

So in both cases |gk(x)− gk(y)| ≤ 2(

12k

). Thus

|g(x)− g(y)| ≤ 4

(1

2k

)= 8

(1

2k+1

)< 8 |x− y|

12 ≤ 8|x− y|α.

Proposition 2.2.2. The function g(t) = limn→∞ gn(t), where g0(t) = t and gn(t)

is defined by (3) for all n ε N, is not Holder continuous of exponent α on [0, 1] for all

α > 12.

Proof. Let α > 12. Let xk = m

4k and yk = m+14k where k ε N ∪ {0} and m ε Z ∩

[0, 4k − 1]. Then xk, yk ε [0, 1] and |xk − yk| = 14k . By Propositions 1.2.4 and 1.2.5,

|g(xk)− g(yk)| = |gk(xk)− gk(yk)| =1

2k= 4k(α− 1

2)|xk − yk|α.

However 4k(α− 12) is not bounded, so there does not exist a constant M such that

|g(xk)− g(yk)| ≤ M |xk − yk|α for all k ε N ∪ {0}.

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Theorem 2.2.3. The function g(t) = limn→∞ gn(t), where g0(t) = t and gn(t) is

defined by (3) for all n ε N, is a Holder continuous function of exponent α on [0, 1] if

and only if α ≤ 12.

Proof. This theorem is an immediate consequence of Propositions 2.2.1 and

2.2.2. �

27

Page 35: Continuous Nowhere Differentiable Functionseduc.jmu.edu/~querteks/seniorthesis.pdfContinuous nowhere differentiable functions are functions that are continuous at ... It is well

Chapter 3

Fractional Derivatives

Although continuous nowhere differentiable functions do not have first-order deriva-

tives at any point in their domains, they do have some level of smoothness. We need

a way to be able to measure this smoothness and to compare the smoothness of dif-

ferent examples of continuous nowhere differentiable functions. Fractional derivatives

will provide us with such a measure of smoothness for these functions. We begin by

defining fractional integrals and fractional derivatives.

If we denote the n-fold integral of a function f as D−nf , then

D−1f(t) =

∫ t

0

f(ξ)dξ. (6)

Proposition 3.0.1. For all n ε N,

D−nf(t) =1

(n− 1)!

∫ t

0

(t− ξ)n−1f(ξ)dξ. (7)

Proof. Let n = 1. Then by (6),

D−1f(t) =

∫ t

0

f(ξ)dξ =

∫ t

0

(t− ξ)1−1f(ξ)dξ.

Thus, (7) holds for n = 1. Now let n = k ε N and assume (7) holds. Then let n = k+1

and consider

D−(k+1)f(t) =

∫ t

0

D−kf(x)dx

=

∫ t

0

1

(k − 1)!

∫ x

0

(x− ξ)k−1f(ξ)dξdx

=1

(k − 1)!

∫ t

0

f(ξ)

∫ t

ξ

(x− ξ)k−1dxdξ

28

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=1

(k − 1)!

∫ t

0

f(ξ)

[(x− ξ)k

k

]x=t

x=ξ

=1

k!

∫ t

0

f(ξ)(t− ξ)kdξ.

So (7) holds for n = k + 1. Thus, by induction, (7) holds for all n ε N. �

The definition of the n-fold integral can then be generalized to define fractional

integrals of order υ > 0 by

D−υf(t) =1

Γ(υ)

∫ t

0

(t− ξ)υ−1f(ξ)dξ,

where Γ(υ) is the Gamma function. The fractional derivative of f(t) of order µ > 0,

for t such that it exists, can be defined as

Dµf(t) = Dm[D−(m−µ)f(t)

]where m ε N such that m ≥ dµe, where dxe is the smallest integer greater than or equal

to x. To show that the value of Dµf(t) does not depend on the choice of m, provided

m ≥ dµe, we need the following facts from [4, pages 220–221] and [7, page 16].

Fact 3.0.2. Let a(t) and b(t) be defined and have continuous derivatives for t1 <

t < t2. Let f(x, t) be continuous and have a continuous derivative ∂f∂t

in a domain of

the x− t plane which includes {(x, t) : t ε [t1, t2], x ε [a(t), b(t)]}. Then for t1 < t < t2,

d

dt

∫ b(t)

a(t)

f(x, t)dx = f(b(t), t)b′(t)− f(a(t), t)a′(t) +

∫ b(t)

a(t)

[∂

∂tf(x, t)

]dx.

Fact 3.0.3. Let Γ(x) be the Gamma function. Then Γ(x + 1) = xΓ(x) for all

x ε R such that x is not a negative integer.

Proposition 3.0.4. Let µ > 0. Then for all m ε N such that m ≥ dµe,

Dm[D−(m−µ)f(t)

]= Ddµe [D−(dµe−µ)f(t)

]. (8)

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Proof. Let m = dµe. Clearly, (8) holds for m = dµe. Now let k ε N such that

k ≥ dµe. Let m = k and assume (8) holds. Then let m = k + 1 and consider

Dk+1[D−(k+1−µ)f(t)

].

By the inductive hypothesis and Facts 3.0.2 and 3.0.3,

Dk+1[D−(k+1−µ)f(t)

]= Dk+1

[1

Γ(k + 1− µ)

∫ t

0

(t− ξ)k−µf(ξ)dξ

]= Dk

[1

Γ(k + 1− µ)

d

dt

∫ t

0

(t− ξ)k−µf(ξ)dξ

]= Dk

[1

Γ(k + 1− µ)

∫ t

0

(k − µ)(t− ξ)k−µ−1f(ξ)dξ

]= Dk

[1

Γ(k − µ)

∫ t

0

(t− ξ)k−µ−1f(ξ)dξ

]= Dk

[D−(k−µ)f(t)

]= Ddµe [D−(dµe−µ)f(t)

]since k ≥ dµe. So (8) holds for m = k + 1. Thus, (8) holds for all m ε N such that

m ≥ dµe. �

To illustrate that fractional derivatives are indeed different from classical deriva-

tives, we now calculate the fractional derivative of order 12

of f(t) = c, where c is a

constant.

Since d12e = 1, we write the fractional derivative as

D12 c = D1

[D−12 c]

and show that

D−12 c =

1

Γ(12)

∫ t

0

(t− ξ)−12 cdξ =

2c√

t√π

.

Then the fractional derivative of f of order 12

for t 6= 0 is

D12 c = D1

(2c√π

√t

)=

c√πt

, (9)

30

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which is not equal to zero provided c 6= 0. This indicates the difference between

fractional derivatives and classical derivatives since all integer order derivatives of the

constant function are 0.

31

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Chapter 4

Holder Continuity and Fractional Derivatives

Although fractional derivatives are very useful in determining the smoothness of

functions that are not first–order differentiable, they can be very difficult to calculate

directly. However, since we are only using fractional derivatives as a measure of

smoothness, we are only concerned with the existence of fractional derivatives of a

specific order and not of the value of the quantity. Thus, we only need a way to

determine if a fractional derivative of a particular order exists for a function. To do

this, we use the connections between Holder continuity and fractional derivatives. We

want to show that a function, f ∗(x), that is Holder continuous of exponent k ≤ 1 on

[0, 1] has fractional derivatives of order β, where 0 < β < k, at all points in [0, 1].

However, this fact is not true without a normalization of the function. The constant

function, g(t) = c, is Holder continuous of exponent k for all k ≥ 0, but, as we showed

in (9), g does not have a fractional derivative of order 12

at t = 0 unless c = 0. Thus,

we must normalize f ∗ so that f ∗(0) = 0. In order to obtain the uniform convergence

necessary to complete the proof, we must also extend the function f ∗ on [0, 1] to the

function f on (−∞, 1], where f(x) = f ∗(x) on (0, 1] and f(x) = f ∗(0) on (−∞, 0].

Proposition 4.0.1. Let α ≥ 0. If f ∗ is a Holder function for the exponent α on

[0, 1], then the extension of f ∗ to (−∞, 1] defined by

f(x) =

f ∗(x), 0 < x ≤ 1

f ∗(0), x ≤ 0

is Holder continuous for α on (−∞, 1].

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Proof. Since f ∗ is Holder continuous for α, there exists M > 0 such that for all

x, y ε [0, 1]

|f ∗(x)− f ∗(y)| ≤ M |x− y|α.

Let x, y ε (−∞, 1].

Case 1. x, y ε (−∞, 0].

Then |f(x)− f(y)| = |f ∗(0)− f ∗(0)| = 0 ≤ M |x− y|α.

Case 2. x, y ε (0, 1].

Then |f(x)− f(y)| = |f ∗(x)− f ∗(y)| ≤ M |x− y|α.

Case 3. x ε (0, 1], y ε (−∞, 0].

Then |f(x)− f(y)| = |f ∗(x)− f ∗(0)| ≤ M |x− 0|α ≤ M |x− y|α.

Case 4. x ε (−∞, 0], y ε (0, 1].

If we substitute y = x and x = y in Case 3, we find that |f(x)−f(y)| ≤ M |x−y|α.

Thus, |f(x)− f(y)| ≤ M |x− y|α for all x, y ε (−∞, 1]. �

Using Proposition 4.0.1 and the ideas of Hardy and Littlewood [3], we are now able

to establish the connection between Holder continuity and fractional derivatives that

will allow us to consider the smoothness of the Generalized van der Waerden–Takagi

function and Kiesswetter’s function without directly calculating fractional derivatives.

Proposition 4.0.2. Let 0 < β < k ≤ 1. Let f ∗ be a Holder continuous function

of exponent k on [0, 1] such that f ∗(0) = 0. If f : (−∞, 1] → R is the extension of f ∗

defined by

f(x) =

f ∗(x), 0 < x ≤ 1

f ∗(0) = 0, x ≤ 0,

then Dβf(x) exists for all x ε [0, 1].

Proof. For ε > 0 and x ε [0, 1], we define

f1−β,ε(x) =1

Γ(1− β)

∫ x−ε

0

f(t)(x− t)−βdt.

33

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Claim 4.0.3. For x ε [0, 1], the limit limε→0+

∫ x−ε

0f(t)(x− t)−βdt exists.

Proof. Fix λ > 0. Since f is continuous on (−∞, 1] and constant on (−∞, 0],

f is bounded. Thus, there exists R > 0 such that |f(t)| ≤ R for all t ε (−∞, 1]. Let

(εn)n εN be a sequence of positive numbers that converges to 0. Then there exists N ε N

such that if n ε N so that n > N, then εn <(

λ(1−β)2R

) 11−β

.

Let n, m ε N such that n > m > N and let x ε [0, 1]. Then since β − 1 < 0

∣∣∣∣∫ x−εm

x−εn

f(t)(x− t)−βdt

∣∣∣∣ ≤∣∣∣∣∫ x−εm

x−εn

|f(t)|(x− t)−βdt

∣∣∣∣≤

∣∣∣∣∫ x−εm

x−εn

R(x− t)−βdt

∣∣∣∣=

∣∣∣∣R(x− t)−β+1

β − 1

∣∣∣∣t=x−εm

t=x−εn

=

∣∣∣∣R(εm)−β+1

β − 1− R(εn)−β+1

β − 1

∣∣∣∣≤ R(εm)−β+1

1− β+

R(εn)−β+1

1− β

2+

λ

2= λ.

Thus,(∫ x−εn

0f(t)(x− t)−βdt

)n ε N

is a Cauchy sequence and is uniformly conver-

gent. Therefore, the uniform limit limε→0+

∫ x−ε

0f(t)(x− t)−βdt exists for x ε [0, 1] and

is denoted∫ x

0f(t)(x− t)−βdt. �

Thus, f1−β can be defined on [0, 1] as the limit of 1Γ(1−β)

∫ x−ε

0f(t)(x − t)−βdt as

ε → 0+ which can be denoted as f1−β(x) = 1Γ(1−β)

∫ x

0f(t)(x − t)−βdt. We note that,

by the definition of the fractional derivative of order β, f ′1−β = Dβf.

Then for ε > 0

Γ(1− β)f ′1−β,ε(x) = f(x− ε)ε−β − β

∫ x−ε

0

f(t)(x− t)−β−1dt

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for x ε [0, 1] by Fact 3.0.2. If x ε (0, 1], then

Γ(1−β)f ′1−β,ε(x) = β

∫ x−ε

0

(f(x)−f(t))(x−t)−β−1dt−ε−β(f(x)−f(x−ε))+f(x)x−β.

Since f is a Holder continuous function of exponent k on (−∞, 1], there exists

M > 0 so that

|f(x)− f(t)| ≤ M |x− t|k

for all x, t ε (−∞, 1].

Claim 4.0.4. As ε → 0+, ε−β(f(x)− f(x− ε)) converges uniformly to 0 on [0, 1].

Proof. Fix λ > 0. Let δ <(

λM

) 1k−β . Let x ε [0, 1]. If 0 < ε < δ, then∣∣∣∣f(x)− f(x− ε)

εβ− 0

∣∣∣∣ ≤ M |ε|k

εβ= Mεk−β < Mδk−β = λ.

Claim 4.0.5. The limit limε→0+

∫ x−ε

0(f(x)−f(t))(x− t)−β−1dt exists for x ε [0, 1],

and the convergence is uniform.

Proof. Fix λ > 0. Let (εn)n εN be a sequence of positive numbers that converges

to 0. Then there exists N ε N such that if n ε N so that n > N , then εn <(

λ(k−β)2M

) 1k−β

.

Let n, m ε N such that n > m > N . Let x ε [0, 1]. Then∣∣∣∣∫ x−εm

x−εn

(f(x)− f(t))(x− t)−β−1dt

∣∣∣∣ ≤∣∣∣∣∫ x−εm

x−εn

|f(x)− f(t)||x− t|−β−1dt

∣∣∣∣≤

∣∣∣∣∫ x−εm

x−εn

M |x− t|k−β−1dt

∣∣∣∣=

∣∣∣∣ −M

k − β(x− t)k−β

∣∣∣∣t=x−εm

t=x−εn

=M

k − β

∣∣(εm)k−β − (εn)k−β∣∣

≤ M

k − β

(∣∣(εm)k−β∣∣+ ∣∣(εn)k−β

∣∣)<

M

k − β

(λ(k − β)

2M+

λ(k − β)

2M

)= λ.

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So(∫ x−εn

0(f(x)− f(t))(x− t)−β−1dt

)n ε N

is a Cauchy sequence and is uniformly

convergent. Thus, the uniform limit limε→0+

∫ x−ε

0(f(x) − f(t))(x − t)−β−1dt exists

and is denoted∫ x

0(f(x)− f(t))(x− t)−β−1dt. �

Thus, for x ε (0, 1], we define

g(x) = β

∫ x

0

(f(x)− f(t))(x− t)−β−1dt + f(x)x−β

and note that g is the uniform limit of Γ(1− β)f ′1−β,ε on (0, 1]. Since for ε > 0,

Γ(1− β)f ′1−β,ε(0) = Γ(1− β)

(f(−ε)ε−β − β

∫ −ε

0

f(t)(−t)−β−1dt

)= Γ(1− β)

(0− β

∫ −ε

0

0dt

)= 0,

we define g(0) = 0. Since Γ(1 − β)f ′1−β,ε converges uniformly to g(x) on (0, 1] as

ε → 0+ and Γ(1−β)f ′1−β,ε(0) = g(0) for all ε > 0, then f ′1−β,ε(x) converges uniformly

to 1Γ(1−β)

g(x) on [0, 1].

Claim 4.0.6. g is continuous on [0, 1].

Proof. Since f is continuous on [0, 1], g is continuous on (0, 1] by definition. To

show that g is continuous at x = 0, we show that limx→0+ g(x) = g(0) = 0.

Fix λ > 0. Then there exists 0 < δ <(

λ(k−β)Mk

) 1k−β

. If x ε [0, 1] such that |x−0| < δ,

then

|g(x)− 0| =

∣∣∣∣β ∫ x

0

(f(x)− f(t))(x− t)−β−1dt + f(x)x−β

∣∣∣∣≤ β

∫ x

0

|f(x)− f(t)||x− t|−β−1dt + |f(x)− f(0)|x−β

≤ β

∫ x

0

M(x− t)k−β−1dt + Mxk−β

≤[−Mβ

k − β(x− t)k−β

]t=x

t=0

+ Mxk−β

=Mβ

k − βxk−β + Mxk−β

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=

(Mk

k − β

)xk−β

<

(Mk

k − β

)δk−β = λ.

Thus, g is continuous at x = 0 and is therefore continuous on [0, 1]. �

Then for x ε [0, 1],

f1−β(x)− f1−β(0) = limε→0+

(f1−β,ε(x)− f1−β,ε(0))

= limε→0+

∫ x

0

f ′1−β,ε(t)dt

=

∫ x

0

limε→0+

f ′1−β,ε(t)dt

=1

Γ(1− β)

∫ x

0

g(t)dt.

So

f1−β(x) =1

Γ(1− β)

∫ x

0

g(t)dt− f1−β(0).

Then by the Fundamental Theorem of Calculus,

f ′1−β(x) =1

Γ(1− β)g(x)

for 0 < x < 1. Thus, f ′1−β(x) = Dβf(x) exists for all x ε (0, 1). However, we want

to show the existence of the β–order fractional derivative of f on [0, 1], so we must

consider the points x = 0 and x = 1.

To show that f1−β has a right–hand derivative at x = 0, we consider the conver-

gence off1−β(x)−f1−β(0)

xas x → 0+.

Claim 4.0.7. limx→0+f1−β(x)−f1−β(0)

x= g(0)

Γ(1−β)= 0.

Proof. Fix λ > 0. Since g is continuous at x = 0, there exists δ > 0 such that

if x ε [0, 1] and |x − 0| < δ, then |g(x) − g(0)| < Γ(1 − β)λ. Let x ε [0, 1] such that

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|x− 0| < δ, then∣∣∣∣f1−β(x)− f1−β(0)

x− g(0)

Γ(1− β)

∣∣∣∣ =

∣∣∣∣ 1

xΓ(1− β)

∫ x

0

g(x)dx

∣∣∣∣=

1

xΓ(1− β)

∫ x

0

|g(x)− g(0)|dx

<1

xΓ(1− β)

∫ x

0

Γ(1− β)λdx

=

(1

xΓ(1− β)

)(Γ(1− β)λx)

= λ.

Similarly to show that f1−β has a left–hand derivative at x = 1, we consider the

convergence off1−β(1)−f1−β(x)

1−xas x → 1−.

Claim 4.0.8. limx→1−f1−β(1)−f1−β(x)

1−x= g(1)

Γ(1−β).

Proof. Fix λ > 0. Since g is continuous at x = 1, there exists δ > 0 such that

if x ε [0, 1] and |x − 1| < δ, then |g(x) − g(1)| < Γ(1 − β)λ. Let x ε [0, 1] such that

|x− 1| < δ, then∣∣∣∣f1−β(1)− f1−β(x)

1− x− g(1)

Γ(1− β)

∣∣∣∣ =1

Γ(1− β)

∣∣∣∣ 1

1− x

∫ 1

x

g(x)dx− g(1)

∣∣∣∣=

1

Γ(1− β)

∣∣∣∣∣∫ 1

xg(x)− g(1)dx

1− x+ g(1)− g(1)

∣∣∣∣∣≤ 1

(1− x)Γ(1− β)

∫ 1

x

|g(x)− g(1)|dx

<1

(1− x)Γ(1− β)

∫ 1

x

Γ(1− β)λdx

=1

(1− x)Γ(1− β)Γ(1− β)λ(1− x)

= λ.

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Thus, f ′1−β(x) exists for all x ε [0, 1]. Furthermore, f ′1−β(x) = 1Γ(1−β)

g(x) for all

x ε [0, 1]. Therefore, Dβf(x) exists on [0, 1]. �

We now apply Proposition 4.0.2 to the extensions of the examples we defined in

Chapter 1.

4.1. The Generalized van der Waerden–Takagi Function

Proposition 4.1.1. Let f(x) =∑∞

n=0a0(bnx)

cn , where c > 1 and b ε N such that

b ≥ c. Let f ∗(x) =

f(x), 0 < x ≤ 1

f(0), x ≤ 0. Then for all 0 < β < log c

log b, Dβf ∗(x) exists

for x ε [0, 1].

Proof. If b = c, then f is Holder continuous of exponent k for all 0 ≤ k < 1 = log clog b

by Theorem 2.1.2. If b > c, then f is a Holder continuous function of exponent log clog b

by

Theorem 2.1.5. In both cases, f(0) = 0 so f satisfies the hypotheses of Proposition

4.0.2 for all 0 < β < log clog b

. Thus, for all 0 < β < log clog b

, Dβf ∗(x) exists for x ε [0, 1]. �

4.2. Kiesswetter’s Function

Proposition 4.2.1. Let g(t) = limn→∞ gn(t), where g0(t) = t and gn(t) is defined

by (3) for all n ε N. Let g∗(t) =

g(t), 0 < t ≤ 1

g(0), t ≤ 0. Then for all 0 < β < 1

2, Dβg∗(t)

exists for t ε [0, 1].

Proof. Since g is a Holder continuous function of exponent 12

by Theorem 2.2.3

and g(0) = 0, g satisfies the hypotheses of Proposition 4.0.2 for all 0 < β < 12. Thus,

for all 0 < β < 12, Dβg∗(t) exists for all t ε [0, 1]. �

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Chapter 5

Dimensions of Graphs of Nowhere

Differentiable Functions

We have examined continuous nowhere differentiable functions in terms of Holder

continuity and fractional differentiability. Another way to study nowhere differen-

tiable functions is to consider their graphs as subsets of the plane. We want to be

able to compare the size of the graph of a continuous nowhere differentiable function

to the sizes of the graphs of other continuous nowhere differentiable functions and to

the sizes of the graphs of everywhere differentiable functions.

An obvious measure of size to consider is the length of a graph. However, the

graphs of continuous nowhere differentiable functions have infinite length. To see this,

we let f be a real-valued function on [a, b] and let P : a = x0 < x1 < . . . < xn = b be

a partition of [a, b]. Then we denote

L(P) =n∑

i=1

((xi − xi−1)

2 + (f(xi)− f(xi−1))2) 1

2

and define the length of the graph of f to be supP L(P). The function f is of bounded

variation on [a, b], denoted f ε BV [a, b], if and only if supP L(P) < ∞. If f ε BV [a, b],

then, by Jordan’s Theorem [8, page 103, theorem 5], f can be written as f = g − h,

where g and h are monotone functions. Since Lebesgue’s Theorem states that a

monotone function on (a, b) is differentiable almost everywhere on (a, b) [8, page 100,

theorem 3], f ′(x) exists almost everywhere on (a, b) if f ε BV [a, b]. Since nowhere

40

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differentiable functions are not differentiable at any point in their domains, they can-

not be of bounded variation. Thus, the graph of a continuous nowhere differentiable

function must have infinite length.

Since we cannot compare the relative sizes of graphs of infinite length, we need

to refine our notion of size. When we considered the smoothness of functions, we

used fractional derivatives to refine our notion of smoothness into a concept that

can distinguish differences in smoothness between examples of continuous nowhere

differentiable functions. Similarly, we will use the ideas of dimension from fractal

geometry to be able to compare the sizes of graphs that have infinite length.

5.1. Definitions and Facts

One important dimension is the Hausdorff dimension. Let F ⊂ Rn. The diameter of

a set V is defined to be |V | = sup{|x − y| : x, y ε V }. Then a countable collection,

{Vi}, of sets of diameter less than or equal to δ that covers F is a δ-cover of F . Now

for all δ > 0, the quantity Hsδ(F ) is defined as

Hsδ(F ) = inf

{∞∑i=1

|Vi|s : {Vi} is a δ-cover of F

}.

The s-dimensional Hausdorff measure of F, Hs(F ), is defined to be

Hs(F ) = limδ→0+

Hsδ(F ).

The Hausdorff dimension of F is then defined as

dimH F = inf {s : Hs(F ) = 0} = sup {s : Hs(F ) = ∞}

so that

Hs(F ) =

∞, s < dimH F

0, s > dimH F.

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Another commonly used notion of dimension is the box–counting dimension. We

again let F ⊂ Rn. For δ > 0, the collection of cubes

{[M1δ, (M1 + 1)δ]× . . .× [Mnδ, (Mn + 1)δ] : M1, . . . ,Mn ε Z}

is called the δ–mesh of Rn. Let Nδ(F ) be the number of δ–mesh cubes that intersect

F . The lower box–counting dimension of F is defined as

dimBF = lim infδ→0+

log Nδ(F )

− log δ. (10)

The upper box–counting dimension of F is defined by

dimBF = lim supδ→0+

log Nδ(F )

− log δ. (11)

If the limits in (10) and (11) are equal, we call the value of the limit the box–counting

dimension of F and denote it dimB F . It is sufficient to consider the limits in (10)

and (11) as δ → 0+ through any decreasing sequence δk such that δk = ck for some

constant 0 < c < 1.

Before we consider the Hausdorff and box–counting dimensions of the graphs of

specific functions, we note the following two facts from [2, pages 42, 146] relating

these two concepts of dimension.

Fact 5.1.1. For all F ⊂ Rn, dimH F ≤ dimBF ≤ dimBF.

Fact 5.1.2. If f : [0, 1] → R has a continuous derivative and F is the graph of f ,

then dimH F = dimB F = 1.

To bound the Hausdorff and upper box–counting dimensions of the graph of the

Generalized van der Waerden–Takagi function and the graph of Kiesswetter’s func-

tion, we use the following facts from [2, pages 147, 30].

Fact 5.1.3. Let f : [0, 1] → R be a continuous function and let F be the graph of

f . If f is Holder continuous for the exponent 2 − s on [0, 1], where 1 ≤ s ≤ 2, then

Hs(F ) < ∞ and dimH F ≤ dimBF ≤ s.

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Fact 5.1.4. Let F ⊂ Rn. Let g : F → Rm. If |g(x) − g(y)| ≤ M |x − y| for all

x, y ε F, then dimH g(F ) ≤ dimH F.

There are two corollaries of Fact 5.1.4 that are important for obtaining a lower

bound for the Hausdorff dimension for the graph of a function.

Corollary 5.1.5. Let F be the graph of a function, f : R → R. Let T be the

projection of F onto the x–axis. Then dimH T ≤ dimH F.

Proof. Let the function g : R2 → R be defined by g(x, y) = x for all x, y ε R. Let

(x1, y1), (x2, y2) ε F. Then

|g(x1, y1)− g(x2, y2)| = |x1 − x2|

≤((x1 − x2)

2 + (y1 − y2)2) 1

2

= |(x1, y1)− (x2, y2)|

Thus, by Fact 5.1.4, dimH g(F ) ≤ dimH F. Since T = g(F ), dimH T ≤ dimH F. �

Corollary 5.1.6. Let f : [0, 1] → R. Let F be the graph of f. Then dimH F ≥ 1.

Proof. By Corollary 5.1.5, dimH([0, 1]) ≤ dimH F. To determine dimH([0, 1]),

we consider H1([0, 1]).

We first determine a lower bound for H1([0, 1]). Let δ > 0. Let {Vi}i ε N be a δ–

cover of [0, 1]. Since [0, 1] ⊂⋃∞

i=1 Vi,∑∞

i=1 |Vi| ≥ 1. Thus, H1δ([0, 1]) ≥ 1 for all δ > 0.

Then

H1([0, 1]) = limδ→0+

H1δ([0, 1]) ≥ 1.

Now we determine an upper bound for H1([0, 1]). Let δ > 0. Let m = d1δe. Then

[0,1] can be covered by m sets of diameter δ. Thus,

H1δ([0, 1]) = inf

{∞∑i=1

|Vi| : {Vi} is a δ-cover of [0,1]

}

≤ mδ ≤(

1 +1

δ

)δ = δ + 1.

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Thus,

H1([0, 1]) = limδ→0+

H1δ([0, 1]) ≤ lim

δ→0+δ + 1 = 1.

So H1([0, 1]) = 1. Since 0 < H1([0, 1]) < ∞, dimH [0, 1] = 1. Therefore, dimH F ≥

1. �

5.2. The Generalized van der Waerden–Takagi Function

We now apply the above general statements regarding the Hausdorff and box–counting

dimensions to the specific example of the graph of the Generalized van der Waerden–

Takagi function with b = c.

Proposition 5.2.1. Let F be the graph of the function f : [0, 1] → R defined by

f(x) =∑∞

n=0a0(cnx)

cn , where c ε N and c ≥ 2. Then dimH F ≤ dimBF ≤ 1.

Proof. By Theorem 2.1.2, f is a Holder continuous function of exponent α for

all α < 1. So f satisfies the hypothesis of Fact 5.1.3 for 1 < s ≤ 2. Then, for all

s ε (1, 2] ,

dimH F ≤ dimBF ≤ s.

Thus,

dimH F ≤ dimBF ≤ 1.

Theorem 5.2.2. Let F be the graph of the function f : [0, 1] → R defined by

f(x) =∑∞

n=0a0(cnx)

cn , where c ε N and c ≥ 2. Then dimH F = 1.

Proof. By Proposition 5.2.1, dimH F ≤ 1. Since f : [0, 1] → R, dimH F ≥ 1 by

Corollary 5.1.6. Therefore, dimH F = 1. �

Theorem 5.2.3. Let F be the graph of the function f : [0, 1] → R defined by

f(x) =∑∞

n=0a0(cnx)

cn , where c ε N and c ≥ 2. Then dimB F exists and equals 1.

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Proof. By Fact 5.1.1 and Propositions 5.2.1 and 5.2.2,

1 = dimH F ≤ dimBF ≤ dimBF ≤ 1.

Thus, dimBF = dimBF = 1. Therefore, dimB F = 1. �

We now consider the more general version of the Generalized van der Waerden-

Takagi function with b > c.

Proposition 5.2.4. Let F be the graph of the function f : [0, 1] → R defined by

f(x) =∑∞

n=0a0(bnx)

cn , where c > 1 and b ε N such that b > c. Then dimH F ≤ dimBF ≤

2− log clog b

.

Proof. By Theorem 2.1.5, f is a Holder continuous function of exponent α for

all α ≤ log clog b

. So f satisfies the hypothesis of Fact 5.1.3 for 2− log clog b

≤ s ≤ 2. Thus,

dimH F ≤ dimBF ≤ 2− log c

log b.

Using Fact 5.1.1 and Corollary 5.1.6, we can obtain 1 as a lower bound for the

Hausdorff and lower box–counting dimensions of the graph of the Generalized van der

Waerden–Takagi function with c > 1 and b ε N such that b > c. Although obtaining

a tighter bound for the Hausdorff dimension is beyond the scope of this work, we do

want to obtain a tighter bound for the lower box–counting dimension of the graph.

The following fact from [2, pages 146–147] will be important for obtaining this bound.

Fact 5.2.5. Let f : [0, 1] → R be continuous. Let 0 < δ < 1 and let m = d1δe. If Nδ

is the number of squares of the δ–mesh that intersect the graph of f and Rf [iδ, (i+1)δ]

is the maximum range of f over the interval [iδ, (i + 1)δ], then

δ−1

m−1∑i=0

Rf [iδ, (i + 1)δ] ≤ Nδ ≤ 2m + δ−1

m−1∑i=0

Rf [iδ, (i + 1)δ].

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By using Fact 5.2.5, the only difficulty in determining a lower bound for the box–

counting dimension is the calculation of a lower bound for the range of the function

over an interval of a given length δ. However, if we use δk = 12bk , then the horizontal

endpoints of each square in the δk–mesh are points of the form i2bk and i+1

2bk where

i ε Z ∩ [0, 2bk − 1]. The value of f(

i+12bk

)− f

(i

2bk

)has already been calculated in the

proof of Proposition 1.1.10. If b is even, then (1) implies that∣∣∣∣f (i + 1

2bk

)− f

(i

2bk

)∣∣∣∣ =

∣∣∣∣∣k∑

n=0

±(

b

c

)n∣∣∣∣∣∣∣∣∣ 1

2bk

∣∣∣∣ . (12)

If b is odd, then (2) implies that∣∣∣∣f (i + 1

2bk

)− f

(i

2bk

)∣∣∣∣ =

∣∣∣∣∣k−1∑n=0

±(

b

c

)n

± c

c− 1

(b

c

)k∣∣∣∣∣∣∣∣∣ 1

2bk

∣∣∣∣ . (13)

Using Fact 5.2.5, we can now obtain a bound for the lower box–counting dimension

of the graph of the Generalized van der Waerden-Takagi function for all values of b

and c such that we can obtain a positive lower bound for the quantities given in (12)

and (13).

Proposition 5.2.6. Let F be the graph of the function f : [0, 1] → R defined by

f(x) =∑∞

n=0a0(bnx)

cn , where c > 1 and b ε N such that b is even and b ≥ 2c. Then

2− log clog b

≤ dimBF.

Proof. For all k ε N, let δk = 12bk and let mk = 1

δk= 2bk. Then for all k ε N and

i ε Z ∩ [0, mk − 1],

Rf [iδk, (i + 1)δk] ≥∣∣∣∣f (i + 1

2bk

)− f

(i

2bk

)∣∣∣∣=

∣∣∣∣∣k∑

n=0

±(

b

c

)n∣∣∣∣∣∣∣∣∣ 1

2bk

∣∣∣∣≥

(bc

)k −∑k−1n=0

(bc

)n2bk

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=

(bc

)k [( bc

)− 2]+ 1

2bk((

bc

)− 1)

≥bk−1

ck (b− 2c)

2bk≥ 0

by (12) since b ≥ 2c.

Then by Fact 5.2.5,

Nδk≥ δ−1

k

mk−1∑i=0

Rf [iδk, (i + 1)δk]

≥ 2bk

2bk−1∑i=0

bk−1

ck (b− 2c)

2bk

=2b2k−1

ck(b− 2c) .

Then

dimBF = lim infδ→0+

log Nδ(F )

− log δ

≥ lim infk→∞

log 2b2k−1

ck (b− 2c)

− log 12bk

= lim infk→∞

log(b− 2c) + log 2 + (2k − 1) log b− k log c

log 2 + k log b

= 2− log c

log b.

Proposition 5.2.7. Let F be the graph of the function f : [0, 1] → R defined by

f(x) =∑∞

n=0a0(bnx)

cn , where c > 1 and b ε N such that b is odd and b ≥ 2c − 1. Then

2− log clog b

≤ dimBF.

Proof. For all k ε N, let δk = 12bk and let mk = 1

δk= 2bk. Then for all k ε N and

i ε Z ∩ [0, mk − 1],

Rf [iδk, (i + 1)δk] ≥∣∣∣∣f (i + 1

2bk

)− f

(i

2bk

)∣∣∣∣47

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=

∣∣∣∣∣k−1∑n=0

±(

b

c

)n

± c

c− 1

(b

c

)k∣∣∣∣∣∣∣∣∣ 1

2bk

∣∣∣∣≥ 1

2bk

(c

c− 1

(b

c

)k

−k−1∑n=0

(b

c

)n)

=1

2bk

(bc

)k ( b−2c+1c−1

)+ 1(

bc

)− 1

≥bk−1

ck (b− 2c + 1)

2bk≥ 0

by (13) since b ≥ 2c− 1.

Then by Fact 5.2.5,

Nδk≥ δ−1

k

mk−1∑i=0

Rf [iδk, (i + 1)δk]

≥ 2bk

2bk−1∑i=0

bk−1

ck (b− 2c + 1)

2bk

=2b2k−1

ck(b− 2c + 1) .

Then

dimBF = lim infδ→0+

log Nδ(F )

− log δ

≥ lim infk→∞

log 2b2k−1

ck (b− 2c + 1)

− log 12bk

= lim infk→∞

log(b− 2c + 1) + log 2 + (2k − 1) log b− k log c

log 2 + k log b

= 2− log c

log b.

We know summarize the above discussion and propositions in the following theo-

rem.

48

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Theorem 5.2.8. Let F be the graph of the function f : [0, 1] → R defined by

f(x) =∑∞

n=0a0(bnx)

cn , where c > 1 and b ε N such that b > c. Then 1 ≤ dimH F ≤

dimBF ≤ dimBF ≤ 2 − log clog b

. Moreover, if b is even and b ≥ 2c or b is odd and

b ≥ 2c− 1, then dimB F exists and equals 2− log clog b

.

5.3. Kiesswetter’s Function

We now consider the Hausdorff and box–counting dimensions of the graph of Kiess-

wetter’s function. The calculations are similar to those made for the graph of the

Generalized van der Waerden–Takagi function.

Proposition 5.3.1. Let g : [0, 1] → R be defined by g(t) = limn→∞ gn(t), where

g0(t) = t and gn(t) is defined by (3) for all n ε N. Let G be the graph of g. Then

dimH G ≤ dimBG ≤ 32.

Proof. By Theorem 2.2.3, g is a Holder continuous function of exponent α for

all α ≤ 12. So g satisfies the hypothesis of Fact 5.1.3 for 3

2≤ s ≤ 2. Thus,

dimH G ≤ dimBG ≤ 3

2.

Proposition 5.3.2. Let g : [0, 1] → R be defined by g(t) = limn→∞ gn(t), where

g0(t) = t and gn(t) is defined by (3) for all n ε N. Let G be the graph of g. Then

32≤ dimBG.

Proof. For all k ε N, let δk = 14k and let mk = 1

δk= 4k. By Proposition 1.2.5,∣∣g ( i+1

4k

)− g

(i

4k

)∣∣ = 12k for all k ε N and i ε Z ∩ [0, mk − 1]. Thus, for all k ε N and

i ε Z ∩ [0, mk − 1],

Rg[iδk, (i + 1)δk] ≥1

2k= δ

12k .

49

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Then by Fact 5.2.5,

Nδk(G) ≥ δ−1

k

mk−1∑i=0

Rg[iδk, (i + 1)δk]

≥ δ−1k

mk−1∑i=0

δ12k

= δ−1k δ−1

k δ12k = δ

−32

k .

Then

dimBG = lim infδ→0+

log Nδ(G)

− log δ

≥ lim infk→∞

log( 14k )

−32

− log 14k

= lim infk→∞

−32log 1

4k

− log 14k

=3

2.

Theorem 5.3.3. Let g : [0, 1] → R be defined by g(t) = limn→∞ gn(t), where

g0(t) = t and gn(t) is defined by (3) for all n ε N. Let G be the graph of g. Then

dimB G exists and equals 32.

Proof. By Fact 5.1.1 and Propositions 5.3.1 and 5.3.2,

3

2≤ dimBG ≤ dimBG ≤ 3

2.

Thus, dimBG = dimBG = 32. Therefore, dimB G = 3

2. �

Although we are not able to obtain an exact value for the Hausdorff dimension of

the graph of Kiesswetter’s function, the preceding propositions and corollaries provide

bounds for the value. These bounds are stated explicitly in the following proposition

for completeness.

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Proposition 5.3.4. Let g : [0, 1] → R be defined by g(t) = limn→∞ gn(t), where

g0(t) = t and gn(t) is defined by (3) for all n ε N. Let G be the graph of g. Then

1 ≤ dimH G ≤ 32.

Proof. Since g : [0, 1] → R, dimH G ≥ 1 by Corollary 5.1.6. By Fact 5.1.1 and

Proposition 5.3.3, dimH G ≤ dimB G ≤ 32. Therefore, 1 ≤ dimH G ≤ 3

2. �

51

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Conclusion

In this thesis we investigated two special classes of continuous nowhere differ-

entiable functions. One class consisted of the Generalized van der Waerden–Takagi

function. Our methods allowed us to remove the restriction that b ≥ 4c imposed

by Knopp and to show that the Generalized van der Waerden–Takagi function is

nowhere differentiable on [0, 1] for c > 1 and b ε N such that b ≥ c. The other class we

considered was Kiesswetter’s function, which we showed was nowhere differentiable

on [0, 1].

Although these functions do not have first–order derivatives, they do share some

smoothness properties. In order to study these properties, we had to refine our

methods to reveal the finer structure that differentiability cannot see. One way we

measured smoothness was by using Holder continuity. We found that all of the

functions in both classes were Holder continuous of exponent α on [0, 1] for some

α > 0. In fact, the Generalized van der Waerden–Takagi function with b = c is Holder

continuous of exponent α on [0, 1] for all α < 1.

Another way of measuring the smoothness of the continuous nowhere differentiable

functions was the existence of fractional derivatives. We used Holder continuity to

determine that the functions in both classes have fractional derivatives of order β for

some β > 0. We found that the Generalized van der Waerden–Takagi function with

b = c has fractional derivatives of order β for all 0 ≤ β < 1.

Our final approach for measuring the smoothness of continuous nowhere differen-

tiable functions utilized the ideas of dimension from fractal geometry to measure the

size of the graphs of the functions as subsets of the plane. The potential range for the

52

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Hausdorff and box–counting dimensions of the graphs of the functions is [1, 2] since

they are continuous functions on [0, 1], but for all the functions in the two classes,

we were able to show that these values are less than 2. For the graph of the General-

ized van der Waerden–Takagi function with b = c, we found that the Hausdorff and

box–counting dimensions equal 1.

Our measures of smoothness allow us to compare these classes of functions to

continuous everywhere differentiable functions. The fact that the Generalized van

der Waerden–Takagi function with b = c is Holder continuous of exponent α for all

α < 1, has fractional derivatives everywhere of order β for all 0 ≤ β < 1, and has

a graph of dimension 1 reveals the strong similarities between this special case of

the Generalized van der Waerden–Takagi function and continuous everywhere differ-

entiable functions. With respect to these three measures, the Generalized van der

Waerden–Takagi function with b = c is as close to being everywhere differentiable as

a nowhere differentiable function can be. The Holder continuity of the other func-

tions we studied and the dimensions of their graphs indicate that these functions

differ much more from everywhere differentiable functions than the Generalized van

der Waerden–Takagi function with b = c does.

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[2] Kenneth Falconer, Fractal geometry, John Wiley & Sons Ltd., Chichester, 1990.

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[4] Wilfred Kaplan, Advanced calculus, Addison-Wesley Press, Inc., Cambridge, Mass., 1952.

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[9] Abdullah Shidfar and Kazem Sabetfakhri, On the Continuity of Van Der Waerden’s Function

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54