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
Contents
Summary
iv
Abstract
vii
Introduction
Chapter 1
1
Nowhere Differentiable Functions
1.1. The Generalized van der Waerden–Takagi Function
1.2. Kiesswetter’s Function
Chapter 2
3
3
11
Hölder 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
Hölder 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
References
54
iii
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 properties of continuous nowhere differentiable functions. One quantity that we want to
consider is Hölder continuity, which relates the difference in the values of a function
iv
at two points to the distance between the two points. If there exists a positive constant, 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 Hölder continuous of exponent α. We
determine the exponents for which the Generalized van der Waerden–Takagi function
and Kiesswetter’s function are Hölder 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 Hölder 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 continuous 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 Hölder 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
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 properties of Hölder continuity, fractional derivatives, and dimension allow us to obtain a
better understanding of continuous nowhere differentiable functions.
vi
Abstract
In this thesis we investigate two classes of continuous nowhere differentiable funcP
a0 (bn x)
tions on [0, 1] : the Generalized van der Waerden–Takagi function, f (x) = ∞
n=0
cn
where a0 (x) = dist(x, Z), c > 1, b N, and b ≥ c, and Kiesswetter’s function. We
investigate the Hölder 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
Introduction
In this thesis, we consider continuous functions on [0, 1] that are nowhere differentiable 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 reintroduced 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 differentiable 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 Hölder continuity. We will study the Hölder continuity of
both the Generalized van der Waerden–Takagi function and Kiesswetter’s function.
Although continuous nowhere differentiable functions do not have first-order derivatives, 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 exponents and not of the value of the fractional derivative, we will extend a result of
1
Hardy and Littlewood [3] to relate Hölder continuity and fractional differentiability.
We can then apply our results about Hölder continuity to determine for which exponents 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
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) =
∞
X
a0 (bn x)
n=0
cn
.
where c > 1 and b 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
P
such that |fn (x)| ≤ Mn for all x D and n N∪{0}. If the series
Mn is convergent,
P
then
fn is uniformly convergent on D.
Proposition 1.1.2. If b N and c > 1 such that b < c, then f (x) =
P∞
n=0
is Lipschitz on [0, 1] and is differentiable at all points x [0, 1) such that x 6=
all k, m N ∪ {0}.
3
a0 (bn x)
cn
k
2bm
for
n
Proof. Let b N and c > 1 such that b < c. We let fn (x) = a0 (bcn x) on [0, 1].
P
Then f (x) = ∞
n=0 fn (x). As we will show in Proposition 2.1.1, |fn (x) − fn (y)| ≤
b n
|x − y| for all x, y [0, 1] and n N ∪ {0}. Thus, for all x, y [0, 1],
c
|f (x) − f (y)| ≤
∞
X
|fn (x) − fn (y)| ≤
n=0
since
P∞
n=0
b n
c
∞ n
X
b
n=0
c
|x − y| =
c
|x − y|
b−c
is 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 N ∪ {0}, we denote the right-hand
P
n (x)
+
derivative of fn by D+ fn (x) = limt→x+ fn (t)−f
. Let g(x) = ∞
n=0 D fn (x). For all
t−x
n
n
P
+
n N ∪ {0} and x [0, 1), D+ fn (x) is either cb or − cb . Thus ∞
n=0 D fn (x) =
P∞
b n
±
is uniformly convergent on [0, 1) by the Weierstrass M-Test since b < c.
n=0
c
Since D+ fn is continuous at all points in [0, 1) except on the set {{ 2bkn } : k Z ∩
[0, 2bn − 1]}, D+ fn is Riemann integrable on [0, 1). Then
x
Z
Z
g(t)dt =
0
0
∞
xX
+
D fn (t)dt =
∞ Z
X
n=0
n=0
0
x
D+ fn (t)dt =
∞
X
fn (x) = f (x)
n=0
P∞
D+ fn (x) is uniformly convergent on [0, 1). Then, by the Fundamental
P∞
P
+
+
Theorem of Calculus, f 0 (x) = ∞
n=0 D fn (x) for all x [0, 1) such that
n=0 D fn (x)
since
n=0
is continuous at x. Therefore, f (x) is differentiable at all points x [0, 1) such that
x 6=
k
2bm
for all k, m 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 =
k
2bm
for some k, m N ∪ {0}. The differentiability
of the function at these points is dependent on the specific values of b and c. In
P
a0 (bn x)
fact, for certain values of b and c such that b < c, the function f (x) = ∞
n=0
cn
is everywhere differentiable on its domain. To see this, we consider the following
proposition suggested by [12, pages 33–38].
4
Proposition 1.1.3. Let f : [0, 1] → R be defined by f (x) =
P∞
n=0
a0 (2n x)
4n
and
g : [0, 1] → R be defined by g(x) = 2x − 2x2 . Then f = g.
n
x)
Proof. For all i, k N ∪ {0} such that i ≤ 2k − 1, the function a0 (2
is linear on
4n
i i+1 ,
for all n N ∪ {0} such that n ≤ k − 1. Then since 22i+1
k+1 is the midpoint of
2k 2k
i i+1 ,
,
2k 2k
!
a0 2n 22i+1
a0 2n i+1
1 a0 2n 2ik
k+1
2k
=
+
4n
2
4n
4n
for all n N ∪ {0} such that n ≤ k − 1. Using these facts, we observe that f and g
satisfy the same difference equation for i, k N ∪ {0} such that i ≤ 2k − 1
2i + 1
2k+1
f
1
−
2
k
2i+1
X
a0 2k+1−n
i
i+1
f
+f
=
2k
2k
4n
n=0
=
=
a0
1
−
2
2i+1
k−1
X
a0
n=0
i
2k−n
+ a0
4n
i+1
2k−n
!
2
k
4
1 1
2
= k+1
k
24
4
and
g
2i + 1
2k+1
1
−
2
i
i+1
2i + 1 4i2 + 4i + 1
−
g
+
g
=
2k
2k
2k
22k+1
2i + 1 2i2 + 2i + 1
−
−
2k
22k
4i2 + 4i + 2 4i2 + 4i + 1
−
22k+1
22k+1
1
2
= 2k+1 = k+1 .
2
4
=
Thus, f and g agree on the set 2ik : i, k N ∪ {0}, i ≤ 2k − 1 . Since f and g are
continuous on [0, 1], 2ik : i, k 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 [0, 1].
5
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
P∞
n=0
a0 (bn x)
,
cn
where c > 1 and b N such that
b ≥ c, is uniformly convergent on [0, 1].
for all x R, then a0 (bn x) ≤ 21 for all x [0, 1] , n N∪{0}.
n Let Mn = 2c1n for all n N ∪ {0}. Then a0 (bcn x) ≤ Mn for all n N ∪ {0}. The series
P∞ 1 n
P∞
1
is a geometric series and is convergent since c > 1. Thus,
n=0 c
n=0 Mn = 2
P
a0 (bn x)
by the Weierstrass M-test, ∞
is uniformly convergent on [0, 1] .
n=0
cn
Proof. Since a0 (x) ≤
1
2
Proposition 1.1.5. The function a0 (x) is continuous on R.
Proof. Since a0 (x) is linear on z, z + 21 ∪ z + 12 , z + 1 : z Z , then a0 (x)
is continuous on R \ Z ∪ z2 : z Z . It is obvious that a0 (x) is continuous at t for
all t Z ∪ z2 : z Z . Therefore, a0 (x) is continuous on R.
Proposition 1.1.6. For all n N ∪ {0}, the function fn (x) =
a0 (bn x)
,
cn
where c > 1
and b N such that b ≥ c, is continuous on [0, 1].
Proof. Since a continuous function multiplied by a constant is a continuous
function, bn x is continuous on [0, 1] for all n 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 (bn x) is a continuous function on [0, 1] for all n N ∪ {0}.
Thus, for all n N ∪ {0},
a0 (bn x)
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) =
P∞
n=0
a0 (bn x)
cn
6
is continuous on [0, 1].
n
Proof. By Proposition 1.1.6, we know that a0 (bcn x) is continuous on [0, 1] for all
P
a0 (bn x)
n N∪{0}. By Proposition 1.1.4, ∞
is uniformly convergent on [0, 1]. Thus,
n=0
cn
P
P
m
a0 (bn x)
a0 (bn x)
the sequence
is uniformly convergent on [0, 1], and ∞
is
n=0
n=0
cn
cn
mN
continuous on [0, 1] by Fact 1.1.7.
To show that the Generalized van der Waerden–Takagi function is nowhere differentiable on [0, 1], we need the following lemma.
Lemma 1.1.9. If f is differentiable at x = c, then if (un )n N is an increasing
sequence and (vn )n N is a decreasing sequence such that un 6= vn and un ≤ c ≤ vn for
all n N and such that limn→∞ vn − un = 0, then limn→∞
f (vn )−f (un )
vn −un
= f 0 (c).
Proof. Fix ε > 0. Since f is differentiable at x = c, there exists δ > 0 such that
ε
(c)
0
if 0 < |x − c| < δ, then f (x)−f
−
f
(c)
< 2.
x−c
Case 1. There exists M N such that un = c < vn for all n > M .
Let (un )n N be an increasing sequence and (vn )n N be a decreasing sequence such
that there exists M N such that un = c < vn for all n N such that n > M . Then
since limn→∞ vn − un = 0, there exists N N such that if n N and n > N, then
|vn − un | < δ. Let n N such that n > max{N, M }. Since un = c, then |vn − c| < δ.
Thus
f (vn ) − f (c)
ε
f (vn ) − f (un )
0
0
< .
−
f
(c)
=
−
f
(c)
v n − un
vn − c
2
Therefore, limn→∞
f (vn )−f (un )
vn −un
= f 0 (c).
Case 2. There exists M 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 −un
is f 0 (c).
Case 3. un < c < vn for all n N.
Let (un )n N and (vn )n N be sequences such that un < c < vn for all n N. Then
since limn→∞ vn − un = 0, there exists N N such that if n N and n > N, then
|vn − un | < δ. Let n N such that n > N . Since |vn − un | < δ and un < c < vn ,
7
|vn − c| < δ and |un − c| < δ. Then
f (vn ) − f (un ) − (vn − un )f 0 (c) f (vn ) − f (un )
0
− f (c) = vn − un
v n − un
f (vn ) − f (un ) − f (c) + f (c) − (vn + c − c − un )f 0 (c) = v n − un
f (vn ) − f (c) − (vn − c)f 0 (c) ≤ vn − un
f (un ) − f (c) − (un − c)f 0 (c) + un − vn
f (vn ) − f (c) − (vn − c)f 0 (c) ≤ vn − c
f (un ) − f (c) − (un − c)f 0 (c) + un − c
f (vn ) − f (c)
f (un ) − f (c)
0
0
= − f (c) + − f (c)
vn − c
un − c
ε ε
<
+
2 2
= ε.
Therefore, limn→∞
f (vn )−f (un )
vn −un
= f 0 (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) =
P∞
n=0
a0 (bn x)
,
cn
where c > 1 and b N
such that b ≥ c, is nowhere differentiable on [0, 1].
Proof. Let t [0, 1]. We want to show that f (x) is not differentiable at x = t.
Case 1. t = 1.
Let (um )m N and (vm )m N be sequences such that um = 1 −
1
bm
for all m N and
vm = 1 for all m N. Clearly, limm→∞ vm − um = 0. Then
f (vm ) − f (um )
=
v m − um
P∞
n=0
a0 (bn )
cn
−
P∞
n=0
1
bm
8
1
a0 ((1− bm
)bn )
n
c
= bm −
m−1
X
n=0
= −
m−1
X
n=0
= −
a0 b n −
cn
bn
bm
!
bm+n
bm c n
m−1
X
n=0
b
c
n
.
f (vm )−f (um )
=
6
0.
Thus
the
sequence
, which is
c
vm −um
mN
Pm−1
b n
equivalent to the sequence
, does not converge as m → ∞. By
n=0 − c
mN
Since b ≥ c > 0, limn→∞
b n
Lemma 1.1.9, f (x) is not differentiable at x = 1.
Case 2. t [0, 1).
Since t [0, 1), then for all m N there exists km Z ∩ [0, 2bm − 1] such that
t<
km +1
.
2bm
Let um =
km
2bm
for all m N and let vm =
km +1
2bm
km
2bm
≤
for all m N. Then (vm )m N is
a decreasing sequence and (um )m N is an increasing sequence such that limm→∞ vm −
um = 0.
Let m N. Since um =
km
2bm
and vm =
km +1
,
2bm
then for all n such that 0 ≤ n ≤ m − 1
are either both contained in z, z + 12 or
there exists z Z such that bn vm and bn um
n
are both contained in z + 21 , z + 1 . Thus for all n such that 0 ≤ n ≤ m − 1, a0 (bcn x)
n
n
is linear on [um , vm ] and has a slope of either cb or − cb . Then for all n such that
0 ≤ n ≤ m − 1,
a0 (bn vm )
cn
n
− a0 (bcnum )
=±
v m − um
n
b
.
c
For n ≥ m, we must consider 2 subcases.
Subcase 1. b is even.
Let n = m. If km is even, then a0
a0 (bn vm )
cn
bn km
2bm
n
− a0 (bcnum )
=
vm − um
= 0 and a0
1
2cm
−0
1
2bm
bn (km +1)
2bm
m
b
=
.
c
By a similar argument, if km is odd, then
a0 (bn vm )
cn
n
− a0 (bcnum )
=−
vm − um
9
m
b
.
c
= 12 . Thus,
Now let n > m. Then bn−m km and bn−m (km + 1) are both even since b is even.
n
b (km +1)
bn km
Thus, a0 2bm = 0 and a0
= 0. Then, for all n N such that n > m,
2bm
a0 (bn vm )
cn
n
− a0 (bcnum )
= 0.
v m − um
Therefore if b is even, then
m
f (vm ) − f (um ) X
=
±
v m − um
n=0
b n
c
Since b ≥ c > 0, limn→∞ ±
verge. Then limm→∞
f (vm )−f (um )
vm −um
n
b
.
c
6= 0. Thus limm→∞
(1)
Pm
n=0
±
b n
c
does not con-
does 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
bn km
2bm
= 0 and a0
bn (km +1)
2bm
= 12 . Thus if km is
even, then
a0 (bn vm )
cn
n
− a0 (bcnum )
=
v m − um
1
2cn
−0
1
2bm
=
bm
.
cn
for all n N such that n ≥ m. If km is odd, then by a similar argument
a0 (bn vm )
cn
n
− a0 (bcnum )
bm
=− n
v m − um
c
for all n N such that n ≥ m. Thus,
n
m−1
∞ n
X
f (vm ) − f (um ) X
b
1
m
=
±
±b
.
v m − um
c
c
n=0
n=m
Since c > 1,
P∞
n=m
1 n
c
=
c
c−1
1 m
c
. Therefore if b is odd, then
n
m
m−1
f (vm ) − f (um ) X
b
c
b
=
±
±
.
vm − um
c
c−1 c
n=0
(2)
n
c
b m
Since b ≥ c > 1, then limn→∞ ± cb
6= 0 and limm→∞ ± c−1
6= 0. Thus
c
Pm−1
n
(um )
c
b m
limm→∞ n=0 ± cb ± c−1
does not converge. Then limm→∞ f (vvmm)−f
does
c
−um
not exist. By Lemma 1.1.9, f (x) is not differentiable at x = t.
10
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
 
x
f1  
y
 
x
f2  
y
 
x
f3  
y
 
x
f4  
y

=
1
4
0

1
= 4
0

1
= 4
0

1
= 4
0
 
x
 
−1
y
2
   
1
x
0
  +  4 
1
−1
y
2
2
   
1
0
x
   + 2
1
y
0
2
   
3
0
x
   + 4 .
1
1
y
2
2
0
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 )nN 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 [0, 1],
the point (t, gn (t)) 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 N, we develop
the following recurrence relation.
11
Proposition 1.2.1. For all t [0, 1] and all n N,


−1

g (4t),
0 ≤ t ≤ 14

2 n−1




 1 gn−1 (4t − 1) − 1 , 1 < t ≤ 1
2
2
4
2
gn (t) =

1
1

g (4t − 2),
< t ≤ 43

2 n−1
2




 1 gn−1 (4t − 3) + 1 , 3 < t ≤ 1.
2
2
4
(3)
Proof. Let t [0, 1] and n N. Then (t, gn−1 (t)) Ln−1 .
By the definition of the Kiesswetter’s curve,



1
4
t
=
f1 
gn−1 (t)
0
0




1
t
4
t

 Ln .
=
−1
−1
g (t)
gn−1 (t)
2
2 n−1
g (4t) Ln .
Thus if 0 ≤ t ≤ 41 , then t, −1
2 n−1
Similarly, since


t

1
4
=
f2 
gn−1 (t)
0
then, for
1
4
0
1
2


t

1
4


1
t
4
+

1
4

+ =
 Ln ,
1
−1
1
gn−1 (t)
g (t) − 2
2
2 n−1
< t ≤ 12 , t, 12 gn−1 (4t − 1) −
1
2
Ln .
Applying f3 , we show that


t

1
4
0
=
f3 
0
gn−1 (t)
Thus if
1
2
1
2


t
 

1
2
1
t
4
+
1
2


+
 Ln .
=
1
gn−1 (t)
0
g (t)
2 n−1
< t ≤ 34 , then t, 21 gn−1 (4t − 2) Ln .
Similarly, since the definition of Kiesswetter’s curve states that

t


1
4
=
f4 
gn−1 (t)
0
then, for
3
4
0
1
2


t

+
gn−1 (t)
< t ≤ 1, t, 12 gn−1 (4t − 3) +
1
2
 
3
4
Ln .
12
1
2

=
1
t
4
+
1
g (t)
2 n−1

3
4
+
1
2
 Ln ,
Thus,
gn (t) =









−1
g (4t),
2 n−1
0≤t≤
1
4
1
g (4t
2 n−1
− 1) − 21 ,
1
4
<t≤
1
2

1

g (4t − 2),

2 n−1




 1 gn−1 (4t − 3) + 1 ,
2
2
1
2
<t≤
3
4
3
4
< t ≤ 1.
We now prove some properties of the functions, gn for n 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 N. Then, for all n N ∪ {0}, gn (0) = 0.
Proof. Since g0 (t) = t for all t [0, 1], g0 (0) = 0. Let k N ∪ {0} and assume
gk (0) = 0. Then
−1
gk (4(0)) =
gk+1 (0) =
2
−1
2
(0) = 0.
Therefore, gn (0) = 0 for all n 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 N. Then, for all n N ∪ {0}, gn (1) = 1.
Proof. Since g0 (t) = t for all t [0, 1], g0 (1) = 1. Let k N ∪ {0} and assume
gk (1) = 1. Then
1
1
1
1 1
1
gk+1 (1) = gk (4(1) − 3) + = gk (1) + = + = 1.
2
2
2
2
2 2
Thus, gn (1) = 1 for all n 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 N. If j, k N ∪ {0} such that j ≥ k, then gj 4mk = gk 4mk for all
m Z ∩ [0, 4k ].
13
g0
g0
gj
gk
Proof. Let k = 0. Then m = 0 or m = 1. By Proposition 1.2.2, gj 400 =
0
1
=
0
for
all
j
N∪{0}
such
that
j
≥
0.
Similarly
by
Proposition
1.2.3,
g
=
j
0
0
4
4
1
= 1 for all j N∪{0} such that j ≥ 0. Thus, for all m Z such that 0 ≤ m ≤ 40 ,
40
m
= g0 4m0 for all j N ∪ {0} such that j ≥ 0.
40
Now let k N such that k ≥ 0 and assume that for all m Z ∩ [0, 4k ], gj 4mk =
m
for all j N such that j ≥ k.
4k
We now consider k + 1. Fix m Z such that 0 ≤ m ≤ 4k+1 and let j N such that
j ≥ k + 1.
Case 1. 0 ≤
m
4k+1
≤ 41 .
Using (3), we show that
m −1
m
4m
−1
gj k+1 =
gj−1
g
.
=
j−1
4
2
4k+1
2
4k
≤ 14 , then 0 ≤ m ≤ 4k . Also since j ≥ k + 1, then j − 1 ≥ k. Thus, by
the inductive hypothesis, gj−1 4mk = gk 4mk . So
Since 0 ≤
m
4k+1
m m −1 m gk k = gk+1 k+1 .
gj k+1 =
4
2
4
4
Case 2.
1
4
1
4
<
Since
<
m
4k+1
m
4k+1
≤ 21 .
≤ 12 , then 0 < m − 4k ≤ 4k . Using (3) and the inductive hypothesis
as in case 1, we show that
m m 1
m − 4k
1
1
m − 4k
1
− = gk
− = gk+1 k+1 .
gj k+1 = gj−1
4
2
4k
2
2
4k
2
4
Case 3.
Since
1
2
1
2
<
<
m
4k+1
m
4k+1
≤ 43 .
≤ 43 , then 0 < m − 2(4k ) ≤ 4k . Then by (3) and the inductive
hypothesis,
m 1
m m − 2(4k )
1
m − 2(4k )
gj k+1 = gj−1
= gk
= gk+1 k+1 .
4
2
4k
2
4k
4
Case 4.
3
4
<
m
4k+1
≤ 1.
14
Since
3
4
<
m
4k+1
≤ 1, then 0 < m − 3(4k ) ≤ 4k . Thus using (3) and the inductive
hypothesis, we show that
m m 1
m − 3(4k )
1
1
m − 3(4k )
1
+
=
g
+
=
g
.
gj k+1 = gj−1
k
k+1
4
2
4k
2
2
4k
2
4k+1
m
m
Thus, for all m Z ∩ [0, 4k+1 ], gj 4k+1
= gk+1 4k+1
for all j N such that j ≥
k + 1. Therefore for all j, k N ∪ {0} such that j ≥ k, gj 4mk = gk 4mk for all
m 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 N. If k N ∪ {0}, then, for all n, m Z such that n ≥ k and 0 ≤ m ≤ 4k − 1,
= 1k
gn mk − gn m+1
k
4
4
2
Proof. By Proposition 1.2.4, if k, n N ∪ {0} such that n ≥ k, then gn 4mk =
gk 4mk for all m Z ∩ [0, 4k ]. Thus, we need only show that if k N ∪ {0}, then
gk mk − gk m+1
= 1k for all m Z such that 0 ≤ m ≤ 4k − 1.
4
4k
2
First let k = 0. Let m [0, 40 − 1]. Thus, m = 0. Then
0
1 1
g0
= |0 − 1| = 0 .
−
g
0
0
0
4
4
2
= 1j for all m Z∩[0, 4j −1]. Let
Now let k = j N and assume gj 4mj − gj m+1
4j
2
s
=
− gj+1 4s+1
s Z such that 0 ≤ s ≤ 4j+1 − 1. We want to show that gj+1 4j+1
j+1
1
.
2j+1
Case 1. 0 ≤
s
4j+1
≤
1
4
−
1
.
4j+1
s
1
Since 0 ≤ 4j+1
≤ 14 − 4j+1
, then 0 ≤ s ≤ 4j − 1. Thus, by the inductive hypothesis,
gj sj − gj s+1
= 1j . Then using (3), we show that
j
4
4
2
s −1 s −1
s
+
1
s
+
1
gj+1
= gj
−
g
−
g
j+1
j
4j+1
4j+1
2
4j
2
4j
1 s s + 1 =
gj j − gj
2
4
4j
1 1
1
=
= j+1 .
j
2 2
2
15
Case 2.
Since
1
4
1
4
−
−
1
4j+1
1
4j+1
<
<
s
4j+1
s
4j+1
≤
1
2
≤
1
2
−
1
.
4j+1
−
1
4j+1
and s Z, then
1
4
≤
s
4j+1
≤
1
2
−
1
.
4j+1
Thus
0 ≤ s − 4j ≤ 4j − 1. Then by the inductive hypothesis and (3),
1 1 s + 1
s 1 s
s
+
1
= gj
gj+1
− gj+1
− 1 − − gj
+1 +
2
4j+1
4j+1 4j
2 2
4j
1 s − 4j
s − 4j + 1 =
gj
− gj
2
4j
4j
1 1
1
=
= j+1 .
j
2 2
2
1 2
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 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 N. Then the function gn (t) is continuous on [0, 1] for all n N ∪ {0}.
Proof. By the definition of Kiesswetter’s curve, g0 (t) is continuous on [0, 1]. Let
k N ∪ {0} and assume that gk (t) is continuous on [0, 1]. Then, by (3), gk+1 (t) is
continuous on 0, 41 ∪ 14 , 12 ∪ 12 , 43 ∪ 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 = 21 , and t = 34 . 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) =
1
g (4t)
2 k
for 0 ≤ t ≤
1
,
4
thus
limt→ 1 − gk+1 (t) = gk+1 ( 14 ). So we only need to show that the limit limt→ 1 − gk+1 (t) =
4
4
limt→ 1 + gk+1 (t). Since
4
lim− gk+1 (t) = lim−
t→ 14
t→ 14
−1
−1
1
gk (4t) = lim−
gk (t) = −
t→1
2
2
2
16
and
1
1
1
1
1
lim+ gk+1 (t) = lim+ gk (4t − 1) − = lim+ gk (t) − = − ,
2 t→0 2
2
2
t→ 14
t→ 14 2
then limt→ 1 − gk+1 (t) = limt→ 1 + gk+1 (t). Therefore, gk+1 (t) is continuous at t = 14 . 4
4
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 N. Then the sequence {gn }n N∪{0} converges uniformly on [0, 1].
Proof. Fix ε > 0. Pick N N such that 2N +1 > 1ε . Let j, n N such that j > n ≥
N. Then if x [0, 1], there exists m Z ∩ [0, 4n − 1] such that
j > n, then gj 4mn = gn 4mn by Proposition 1.2.4. Then
m
4n
m+1
.
4n
≤x≤
Since
m
m
|gj (x) − gn (x)| = gj (x) − gj n + gn n − gn (x)
4
4
m m ≤ gj (x) − gj n + gn (x) − gn n 4
4
j−n m 4 m = gj (x) − gj
g
(x)
−
g
+
.
n
n
j
n
4
4
Since
m
4n
≤x≤
Since
4j−n m+s
4j
4j−n m
4j
≤x≤
j−n
m+1
4n
and gn is linear on [ 4mn , m+1
], then, by Proposition 1.2.5,
4n
m m 1
m + 1 ≤ n.
gn (x) − gn n ≤ gn n − gn
n
4
4
4
2
≤ x ≤
4j−n m+4j−n
,
4j
4j−n m+s+1
.
4j
then there exists s Z ∩ [0, 4j−n − 1] so that
Since gj (x) is linear on the interval [ 4
j−n m+s
4j
,4
j−n m+s+1
4j
]
j−n
and x [ 4 4jm+s , 4 4jm+s ], then, by Proposition 1.2.5,
j−n j−n
j−n
gj (x) − gj 4 m ≤ gj 4 m + s − gj 4 m + s + 1 ≤ 1 .
2j
j
j
j
4
4
4
Thus,
j−n m 4 m |gj (x) − gn (x)| ≤ gj (x) − gj
+
g
(x)
−
g
n
n
4j
4n
≤
1
1
1
+ n < n−1 < ε.
j
2
2
2
17
We recall that for all n 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 N∪{0} ,
we can define Kiesswetter’s function, g : [0, 1] → R, by g(t) = limn→∞ gn (t) since
the sequence {gn }n 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 N, is continuous on [0, 1].
Proof. By Proposition 1.2.6, gn (x) is continuous on [0, 1] for all n N ∩ {0}.
Then by Proposition 1.2.7, the sequence {gn }n 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 N, is nowhere differentiable on [0, 1].
Proof. Let c [0, 1]. We want to show that g(x) is not differentiable at c.
Case 1. c = 1.
Let (um )m N and (vm )m N be sequences such that um =
vm =
4m
4m
4m −1
4m
for all m N and
= 1 for all m N. Clearly, limm→∞ vm − um = 0. Then by Propositions 1.2.4
and 1.2.5,
lim
4m
4m −1 g(vm ) − g(um ) n→∞ gn 4m − limn→∞ gn
m
4
1
vm − um = 4m
m
m
4
4 − 1 m
− gm
= 4 gm
4m
4m
1
= 4m m = 2m .
2
g(vm )−g(um )
m) Then g(vvmm)−g(u
→
∞
as
m
→
∞
so
the
sequence
−um
vm −um
mN
converge. Thus, by Lemma 1.1.9, g(x) is not differentiable at x = c.
Case 2. c [0, 1).
18
does not
Since c [0, 1), then for all m N there exists km Z∩[0, 4m −1] such that
km +1
.
4m
Let um =
km
4m
for all m N and let vm =
km +1
4m
km
4m
≤c<
for all m N. Then (um )m N is an
increasing sequence and (vm )m N is a decreasing sequence. Clearly, limm→∞ vm −um =
limm→∞
1
4m
= 0.
Then by Propositions 1.2.4 and 1.2.5,
lim
km +1
g(vm ) − g(um ) − limn→∞ gn
n→∞ gn
m
4
1
vm − um = 4m
km + 1
km m
= 4 gm
− gm
m
4
4m 1
= 4m m = 2m .
2
km
4m
g(vm )−g(um )
m) m
Since limm→∞ g(vvmm)−g(u
=
lim
2
=
∞,
the
sequence
m→∞
−um
vm −um
mN
does not converge. Therefore, g(x) is not differentiable at x = c by Lemma 1.1.9. 19
Chapter 2
Hölder Continuity
A function f : [0, 1] → R is said to be a Hölder continuous function of exponent
α if, for some constant M ,
|f (x) − f (y)| ≤ M |x − y|α
for all x, y [0, 1]. We note that if f is Hölder continuous of exponent 1, then f is
Lipschitz. The maximal value of α such that f is a Hölder continuous function
of exponent α is related to the existence of the derivative of f on [0, 1]. We will
investigate the Hölder 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 Hölder continuous
of exponent α for α ≥ 1. Lipschitz functions are differentiable almost everywhere
[8, pages 108–112], and functions that are Hölder 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 Hölder 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
P
P∞ a0 (bn y) a0 (bn x)
−
for x, y [0, 1], we need to obtain a
be able to consider ∞
n=0
n=0
cn
cn n
n
bound for a0 (bcn x) − a0 (bcn y) for x, y [0, 1] and n N ∪ {0}.
20
Proposition 2.1.1. For all x, y [0, 1] and n N ∪ {0},
n
a0 (bn x) a0 (bn y) ≤ b |x − y|.
−
cn
n
c
c
Proof. Let n N ∪ {0} and x, y [0, 1]. Let k Z such that k is the closest integer
to bn y. Then
a0 (bn x) ≤ |bn x − k| ≤ |bn x − bn y| + |bn y − k| = |bn x − bn y| + a0 (bn y) .
Thus,
a0 (bn x) − a0 (bn y) ≤ |bn x − bn y| .
By a similar argument, it can be shown that
a0 (bn y) − a0 (bn x) ≤ |bn y − bn x| .
Thus,
|a0 (bn x) − a0 (bn y)| ≤ bn |x − y| .
Therefore,
n
a0 (bn x) a0 (bn y) ≤ b |x − y| .
−
cn
n
c
c
We first consider the Hölder 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 N such that c ≥ 2, then f (x) =
P∞
n=0
a0 (cn x)
cn
is Hölder
continuous of exponent α on [0, 1] if and only if α < 1.
Proof. Let α < 1. Let x and y [0, 1] such that x 6= y. Thus 0 < |x − y| ≤ 1.
Then there exists k Z such that k ≥ 0 and
1
ck+1
< |x − y| ≤
1
.
ck
Then by Proposition 2.1.1,
k−1 k−1 X
a0 (cn x) a0 (cn y) X
c n
1
|x − y| = k|x − y| ≤ k k(1−α) |x − y|α .
cn − cn ≤
c
c
n=0
n=0
21
(4)
Since 0 ≤
a0 (cn t)
cn
≤
1
cn
for all n N ∪ {0} and t [0, 1], then
a0 (cn x) a0 (cn y) 1
cn − cn ≤ cn
for all n N ∪ {0}. Thus
∞ ∞
X
a0 (cn x) a0 (cn y) X
1
1
≤ k−1 ≤ c2 |x − y| ≤ c2 |x − y|α
cn − cn ≤
n
c
c
n=k
n=k
since
1
ck+1
(5)
< |x − y| ≤ 1.
Then combining (4) and (5), we obtain
∞ X
a0 (cn x) a0 (cn y) 1
k
α
2
α
2
α
cn − cn ≤ k ck(1−α) |x − y| + c |x − y| = c + ck(1−α) |x − y| .
n=0
k( 1−α )
ck
1
(1−α)
k
for all k N ∪ {0} and c > 1, limk→∞ ck(1−α)
= 0.
k
k
is bounded,
Thus, limk→∞ c2 + ck(1−α)
= c2 . Then the sequence c2 + ck(1−α)
k N∪{0}
k
≤ Mα for all k N ∪ {0}.
and there exists Mα > 0 such that c2 + ck(1−α)
Since
k
ck(1−α)
=
Then
∞ X
a0 (cn x) a0 (cn y) α
cn − cn ≤ Mα |x − y| .
n=0
Thus, f is Hölder 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 Hölder continuous of order α for all α ≥ 1.
We now consider the Hölder continuity of the Generalized van der Waerden-Takagi
function with b > c.
Proposition 2.1.3. If c > 1 and b N such that b > c, then f (x) =
is Hölder continuous of exponent α on [0, 1] for all α ≤
Proof. Let α ≤
log c
log b
P∞
n=0
a0 (bn x)
cn
log c
.
log b
< 1. Let x and y [0, 1] such that x 6= y. Thus 0 < |x−y| ≤ 1.
Then there exists k Z such that k ≥ 0 and
22
1
bk+1
< |x − y| ≤
1
.
bk
Thus by Proposition
2.1.1,
k−1 k−1 n
X
X
a0 (bn x) a0 (bn y) b
|x − y|
cn − cn ≤
c
n=0
n=0
"
#
k−1 n
1 X b
≤
|x − y|α
bk(1−α) n=0 c
"
!
#
b k
−
1
1
c =
|x − y|α
b
k(1−α)
b
−
1
c
"
!#
α k
c
b
1
=
− k(1−α)
|x − y|α .
b−c
c
b
Since 0 ≤
a0 (bn t)
cn
≤
1
cn
for all n N ∪ {0} and t [0, 1], then
a0 (bn x) a0 (bn y) 1
cn − cn ≤ cn .
for all n N ∪ {0}. Thus, by the same argument as in (5) in Proposition 2.1.2,
∞ X
a0 (bn x) a0 (bn y) 2
α
cn − cn ≤ c |x − y| .
n=k
Then
"
∞ X
a0 (bn x) a0 (bn y) c
cn − cn ≤ b − c
n=0
bα
c
k
−
!
1
bk(1−α)
#
+ c2 |x − y|α .
α
k
b
1
c
1
Since b > 1, limk→∞ bk(1−α)
= 0. Thus, the limit limk→∞ b−c
+ c2
−
c
bk(1−α)
α k
c
bα k
c
+ c2 exists if and only if limk→∞ bc exists. Since α ≤ log
, then
= limk→∞ b−c
c
log b
α k
bα
c
bα k
1
≤ 1. Thus limk→∞ bc is finite. The sequence b−c
− bk(1−α)
+ c2
c
c
k N∪{0}
α
k
c
b
1
2
is then bounded, and there exists Mα > 0 such that b−c
− bk(1−α) + c ≤ Mα
c
for all k N ∪ {0}.
Then
∞ X
a0 (bn x) a0 (bn y) α
cn − cn ≤ Mα |x − y| .
n=0
Proposition 2.1.4. If c > 1 and b N such that b > c, then f (x) =
is not Hölder continuous of exponent α on [0, 1] for α >
23
log c
.
log b
P∞
n=0
a0 (bn x)
cn
Proof. Since b is an integer, c > 1, and b > c, f is a continuous nowhere differentiable function by Propositions 1.1.8 and 1.1.10. Thus, f is not Hölder continuous
of exponent greater than or equal to 1.
To determine the Hölder continuity of f of exponent α for α < 1, let
Let k N. Since b ≥ 2,
1
bk−n
≤
1
2
log c
log b
< α < 1.
for all n N ∪ {0} such that n < k. Thus,
∞
n 1
X
a
1
1
0 b bk
f
− f (0) = f
=
bk
bk
cn
n=0
k−1 b n
X
c
=
k
b
n=0
Pk−1 b n α
1
n=0 c
− 0
=
bk(1−α) bk
α
b k
− 1 c 1
c
− 0
= k(1−α)
b
(b − c) bk
c
=
b−c
Thus, limk→∞
c
b−c
bα k
c
−
1
bα
c
k
!
α
1
− k(1−α) k − 0 .
b
b
1
must exist in order for f to be Hölder continu
c
bα k
α
→ ∞ as k → ∞. Thus,
,
b
>
c
and
ous of exponent α. However, since α > log
log b
c
α
k
c
b
1
1
is unbounded as k → ∞ since limk→∞ bk(1−α)
− bk(1−α)
= 0. Therefore, f
b−c
c
bk(1−α)
is not Hölder continuous of exponent α on [0, 1].
Theorem 2.1.5. Let c > 1 and b N such that b > c. Then f (x) =
is Hölder continuous of exponent α on [0, 1] if and only if α ≤
P∞
n=0
a0 (bn x)
cn
log c
.
log b
Proof. This theorem is the immediate result of Propositions 2.1.3 and 2.1.4. Corollary 2.1.6. For all m N such that m ≥ 2, the function,
f (x) =
∞
X
a0 ((cm )n x)
n=0
cn
,
with c > 1 such that cm N, is Hölder continuous on [0, 1] if and only if α ≤
24
1
.
m
Proof. Fix m N \ {1}. Let c > 1 such that cm N. Since m ≥ 2, cm > c. Thus,
f satisfies the hypotheses for Theorem 2.1.5. Therefore, f is Hölder continuous of
exponent α on [0, 1] for α ≤
α on [0, 1] for α >
log c
log cm
=
log c
log cm
=
1
,
m
and f is not Hölder continuous of exponent
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 N, is Hölder continuous of exponent α on [0, 1] for all
α ≤ 21 .
Proof. Let α ≤ 21 . Let x, y [0, 1] so that x 6= y. Then there exists k N ∪ {0}
such that
1
4k+1
< |x − y| ≤
1
.
4k
Then by Proposition 1.2.7,
|gj (x) − gk (x)| ≤
1
1
+ j
k
2
2
for all j N such that j > k. Thus,
1
1
1
+ j = k.
k
j→∞ 2
2
2
|g(x) − gk (x)| = lim |gj (x) − gk (x)| ≤ lim
j→∞
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
1
+ |gk (x) − gk (y)| + k .
k
2
2
25
In order to bound |gk (x) − gk (y)|, we assume, without loss of generality, that
.
y > x. Then there exists m Z ∩ 0, 4k − 1 such that 4mk ≤ x < m+1
4k
Case 1. x < y ≤
m+1
.
4k
Then by Proposition 1.2.5,
m
m + 1 1
|gk (x) − gk (y)| ≤ gk k − gk
= k.
k
4
4
2
Case 2.
m+1
4k
<y<
m+2
.
4k
Then by Proposition 1.2.5,
m
+
1
m
+
1
+ gk
|gk (x) − gk (y)| ≤ gk (x) − gk
− gk (y)
k
k
4
4
m
m + 1 m+1
m + 2 ≤ gk k − gk
− gk
+ gk
4
4k
4k
4k
1
= 2 k .
2
So in both cases |gk (x) − gk (y)| ≤ 2 21k . Thus
1
1
1
|g(x) − g(y)| ≤ 4 k = 8 k+1 < 8 |x − y| 2 ≤ 8|x − y|α .
2
2
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 N, is not Hölder continuous of exponent α on [0, 1] for all
α > 21 .
Proof. Let α > 12 . Let xk =
m
4k
and yk =
[0, 4k − 1]. Then xk , yk [0, 1] and |xk − yk | =
1
.
4k
|g(xk ) − g(yk )| = |gk (xk ) − gk (yk )| =
m+1
4k
where k N ∪ {0} and m Z ∩
By Propositions 1.2.4 and 1.2.5,
1
1
= 4k(α− 2 ) |xk − yk |α .
k
2
1
However 4k(α− 2 ) is not bounded, so there does not exist a constant M such that
|g(xk ) − g(yk )| ≤ M |xk − yk |α for all k N ∪ {0}.
26
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 N, is a Hölder continuous function of exponent α on [0, 1] if
and only if α ≤ 21 .
Proof. This theorem is an immediate consequence of Propositions 2.2.1 and
2.2.2.
27
Chapter 3
Fractional Derivatives
Although continuous nowhere differentiable functions do not have first-order derivatives 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 different 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−n f , then
Z t
−1
f (ξ)dξ.
D f (t) =
(6)
0
Proposition 3.0.1. For all n N,
D
−n
1
f (t) =
(n − 1)!
Z
t
(t − ξ)n−1 f (ξ)dξ.
(7)
0
Proof. Let n = 1. Then by (6),
Z t
Z t
−1
f (ξ)dξ =
(t − ξ)1−1 f (ξ)dξ.
D f (t) =
0
0
Thus, (7) holds for n = 1. Now let n = k N and assume (7) holds. Then let n = k + 1
and consider
D
−(k+1)
Z
t
D−k f (x)dx
f (t) =
0
Z x
1
=
(x − ξ)k−1 f (ξ)dξdx
(k
−
1)!
0
0
Z t
Z t
1
=
f (ξ) (x − ξ)k−1 dxdξ
(k − 1)! 0
ξ
Z
t
28
x=t
Z t
1
(x − ξ)k
dξ
=
f (ξ)
(k − 1)! 0
k
x=ξ
Z
1 t
=
f (ξ)(t − ξ)k dξ.
k! 0
So (7) holds for n = k + 1. Thus, by induction, (7) holds for all n N.
The definition of the n-fold integral can then be generalized to define fractional
integrals of order υ > 0 by
D
−υ
1
f (t) =
Γ(υ)
Z
t
(t − ξ)υ−1 f (ξ)dξ,
0
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
Dµ f (t) = Dm D−(m−µ) f (t)
where m N such that m ≥ dµe, where dxe is the smallest integer greater than or equal
to x. To show that the value of Dµ f (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 [t1 , t2 ], x [a(t), b(t)]}. Then for t1 < t < t2 ,
d
dt
Z
b(t)
0
0
Z
b(t)
f (x, t)dx = f (b(t), t)b (t) − f (a(t), t)a (t) +
a(t)
a(t)
∂
f (x, t) dx.
∂t
Fact 3.0.3. Let Γ(x) be the Gamma function. Then Γ(x + 1) = xΓ(x) for all
x R such that x is not a negative integer.
Proposition 3.0.4. Let µ > 0. Then for all m N such that m ≥ dµe,
Dm D−(m−µ) f (t) = Ddµe D−(dµe−µ) f (t) .
29
(8)
Proof. Let m = dµe. Clearly, (8) holds for m = dµe. Now let k 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,
Z t
−(k+1−µ)
1
k−µ
k+1
k+1
(t − ξ) f (ξ)dξ
D
D
f (t) = D
Γ(k + 1 − µ) 0
Z
d t
1
k
k−µ
= D
(t − ξ) f (ξ)dξ
Γ(k + 1 − µ) dt 0
Z t
1
k−µ−1
k
(k − µ)(t − ξ)
f (ξ)dξ
= D
Γ(k + 1 − µ) 0
Z t
1
k
k−µ−1
(t − ξ)
f (ξ)dξ
= D
Γ(k − µ) 0
= 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 N such that
m ≥ dµe.
To illustrate that fractional derivatives are indeed different from classical derivatives, we now calculate the fractional derivative of order
1
2
of f (t) = c, where c is a
constant.
Since d 21 e = 1, we write the fractional derivative as
1
2
h
D
−1
2
(t − ξ)
−1
2
D c=D
1
and show that
D
−1
2
1
c= 1
Γ( 2 )
Z
t
0
c
i
√
2c t
cdξ = √ .
π
Then the fractional derivative of f of order 12 for t 6= 0 is
1
2c √
c
1
D2c = D √
t =√ ,
π
πt
30
(9)
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
Chapter 4
Hölder 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 Hölder continuity and fractional derivatives. We
want to show that a function, f ∗ (x), that is Hölder 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 Hölder continuous of exponent k for all k ≥ 0, but, as we showed
in (9), g does not have a fractional derivative of order
1
2
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 Hölder function for the exponent α on
[0, 1], then the extension of f ∗ to (−∞, 1] defined by

 f ∗ (x), 0 < x ≤ 1
f (x) =
 f ∗ (0), x ≤ 0
is Hölder continuous for α on (−∞, 1].
32
Proof. Since f ∗ is Hölder continuous for α, there exists M > 0 such that for all
x, y [0, 1]
|f ∗ (x) − f ∗ (y)| ≤ M |x − y|α .
Let x, y (−∞, 1].
Case 1. x, y (−∞, 0].
Then |f (x) − f (y)| = |f ∗ (0) − f ∗ (0)| = 0 ≤ M |x − y|α .
Case 2. x, y (0, 1].
Then |f (x) − f (y)| = |f ∗ (x) − f ∗ (y)| ≤ M |x − y|α .
Case 3. x (0, 1], y (−∞, 0].
Then |f (x) − f (y)| = |f ∗ (x) − f ∗ (0)| ≤ M |x − 0|α ≤ M |x − y|α .
Case 4. x (−∞, 0], y (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 (−∞, 1].
Using Proposition 4.0.1 and the ideas of Hardy and Littlewood [3], we are now able
to establish the connection between Hölder 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 Hölder 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),
0<x≤1
f (x) =
,
 f ∗ (0) = 0, x ≤ 0
then Dβ f (x) exists for all x [0, 1].
Proof. For ε > 0 and x [0, 1], we define
Z x−ε
1
f (t)(x − t)−β dt.
f1−β,ε (x) =
Γ(1 − β) 0
33
Claim 4.0.3. For x [0, 1], the limit limε→0+
R x−ε
0
f (t)(x − t)−β dt 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 (−∞, 1]. Let
(εn )n N be a sequence of positive numbers that converges to 0. Then there exists N N
1
λ(1−β) 1−β
.
such that if n N so that n > N, then εn <
2R
Let n, m N such that n > m > N and let x [0, 1]. Then since β − 1 < 0
Z
x−εm
−β
f (t)(x − t)
x−εn
Z x−εm
−β
|f (t)|(x − t) dt
dt ≤ x−εn
Z x−εm
≤ R(x − t)−β dt
x−εn
R(x − t)−β+1 t=x−εm
= β−1
t=x−εn
R(εm )−β+1 R(εn )−β+1 = −
β−1
β−1 R(εm )−β+1 R(εn )−β+1
+
1−β
1−β
λ λ
<
+ = λ.
2 2
≤
Thus,
R
x−εn
0
f (t)(x − t)−β dt
is a Cauchy sequence and is uniformly converR x−ε
gent. Therefore, the uniform limit limε→0+ 0 f (t)(x − t)−β dt exists for x [0, 1] and
Rx
is denoted 0 f (t)(x − t)−β dt.
nN
R x−ε
1
Thus, f1−β can be defined on [0, 1] as the limit of Γ(1−β)
f (t)(x − t)−β dt as
0
Rx
1
f (t)(x − t)−β dt. We note that,
ε → 0+ which can be denoted as f1−β (x) = Γ(1−β)
0
0
by the definition of the fractional derivative of order β, f1−β
= Dβ f.
Then for ε > 0
Γ(1 −
0
β)f1−β,ε
(x)
= f (x − ε)ε
−β
Z
−β
0
34
x−ε
f (t)(x − t)−β−1 dt
for x [0, 1] by Fact 3.0.2. If x (0, 1], then
Z x−ε
0
(f (x)−f (t))(x−t)−β−1 dt−ε−β (f (x)−f (x−ε))+f (x)x−β .
Γ(1−β)f1−β,ε (x) = β
0
Since f is a Hölder 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 (−∞, 1].
Claim 4.0.4. As ε → 0+ , ε−β (f (x) − f (x − ε)) converges uniformly to 0 on [0, 1].
1
λ k−β
Proof. Fix λ > 0. Let δ < M
. Let x [0, 1]. If 0 < ε < δ, then
f (x) − f (x − ε)
M |ε|k
≤
−
0
= M εk−β < M δ k−β = λ.
β
β
ε
ε
Claim 4.0.5. The limit limε→0+
R x−ε
0
(f (x) − f (t))(x − t)−β−1 dt exists for x [0, 1],
and the convergence is uniform.
Proof. Fix λ > 0. Let (εn )n N be a sequence of positive numbers that converges
1
λ(k−β) k−β
to 0. Then there exists N N such that if n N so that n > N , then εn <
.
2M
Let n, m N such that n > m > N . Let x [0, 1]. Then
Z x−εm
Z x−εm
−β−1
−β−1
≤ (f
(x)
−
f
(t))(x
−
t)
dt
|f
(x)
−
f
(t)||x
−
t|
dt
x−εn
x−εn
Z x−εm
≤ M |x − t|k−β−1 dt
x−εn
t=x−εm
−M
k−β (x − t) = k−β
t=x−εn
M
k−β
M ≤
(εm )k−β + (εn )k−β k−β
M
λ(k − β) λ(k − β)
<
+
= λ.
k−β
2M
2M
=
35
(εm )k−β − (εn )k−β So
R
x−εn
(f (x)
0
−β−1
− f (t))(x − t)
dt
is a Cauchy sequence and is uniformly
R x−ε
convergent. Thus, the uniform limit limε→0+ 0 (f (x) − f (t))(x − t)−β−1 dt exists
Rx
and is denoted 0 (f (x) − f (t))(x − t)−β−1 dt.
nN
Thus, for x (0, 1], we define
Z x
(f (x) − f (t))(x − t)−β−1 dt + f (x)x−β
g(x) = β
0
0
on (0, 1]. Since for ε > 0,
and note that g is the uniform limit of Γ(1 − β)f1−β,ε
Z −ε
−β−1
0
−β
f (t)(−t)
dt
Γ(1 − β)f1−β,ε (0) = Γ(1 − β) f (−ε)ε − β
0
Z
= Γ(1 − β) 0 − β
−ε
0dt = 0,
0
0
we define g(0) = 0. Since Γ(1 − β)f1−β,ε
converges uniformly to g(x) on (0, 1] as
0
0
(x) converges uniformly
(0) = g(0) for all ε > 0, then f1−β,ε
ε → 0+ and Γ(1 − β)f1−β,ε
to
1
g(x)
Γ(1−β)
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.
1
λ(k−β) k−β
. If x [0, 1] such that |x−0| < δ,
Fix λ > 0. Then there exists 0 < δ <
Mk
then
Z x
−β−1
−β |g(x) − 0| = β
(f (x) − f (t))(x − t)
dt + f (x)x 0
Z x
≤ β
|f (x) − f (t)||x − t|−β−1 dt + |f (x) − f (0)|x−β
0
Z x
≤ β
M (x − t)k−β−1 dt + M xk−β
0
≤
=
−M β
(x − t)k−β
k−β
t=x
t=0
M β k−β
x
+ M xk−β
k−β
36
+ M xk−β
Mk
k−β
Mk
k−β
=
<
xk−β
δ k−β = λ.
Thus, g is continuous at x = 0 and is therefore continuous on [0, 1].
Then for x [0, 1],
f1−β (x) − f1−β (0) =
lim (f1−β,ε (x) − f1−β,ε (0))
Z x
0
= lim+
f1−β,ε
(t)dt
ε→0
0
Z x
0
lim+ f1−β,ε
=
(t)dt
0 ε→0
Z x
1
g(t)dt.
=
Γ(1 − β) 0
ε→0+
So
1
f1−β (x) =
Γ(1 − β)
Z
x
g(t)dt − f1−β (0).
0
Then by the Fundamental Theorem of Calculus,
0
f1−β
(x) =
1
g(x)
Γ(1 − β)
0
for 0 < x < 1. Thus, f1−β
(x) = Dβ f (x) exists for all x (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 convergence of
f1−β (x)−f1−β (0)
x
as 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 [0, 1] and |x − 0| < δ, then |g(x) − g(0)| < Γ(1 − β)λ. Let x [0, 1] such that
37
|x − 0| < δ, then
Z x
f1−β (x) − f1−β (0)
1
g(0)
= −
g(x)dx
x
Γ(1 − β)
xΓ(1 − β) 0
Z x
1
=
|g(x) − g(0)|dx
xΓ(1 − β) 0
Z x
1
Γ(1 − β)λdx
<
xΓ(1 − β) 0
1
=
(Γ(1 − β)λx)
xΓ(1 − β)
= λ.
Similarly to show that f1−β has a left–hand derivative at x = 1, we consider the
convergence of
f1−β (1)−f1−β (x)
1−x
Claim 4.0.8. limx→1−
as x → 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 [0, 1] and |x − 1| < δ, then |g(x) − g(1)| < Γ(1 − β)λ. Let x [0, 1] such that
|x − 1| < δ, then
Z 1
f1−β (1) − f1−β (x)
1
1
g(1) −
=
g(x)dx
−
g(1)
1−x
Γ(1 − β) Γ(1 − β) 1 − x x
R 1
1
x g(x) − g(1)dx
=
+ g(1) − g(1)
Γ(1 − β) 1−x
Z 1
1
≤
|g(x) − g(1)|dx
(1 − x)Γ(1 − β) x
Z 1
1
<
Γ(1 − β)λdx
(1 − x)Γ(1 − β) x
1
=
Γ(1 − β)λ(1 − x)
(1 − x)Γ(1 − β)
= λ.
38
0
0
(x) =
(x) exists for all x [0, 1]. Furthermore, f1−β
Thus, f1−β
1
g(x)
Γ(1−β)
for all
x [0, 1]. Therefore, Dβ f (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
P
a0 (bn x)
, where c > 1 and b N such that
Proposition 4.1.1. Let f (x) = ∞
n=0
cn

 f (x), 0 < x ≤ 1
c
∗
, Dβ f ∗ (x) exists
b ≥ c. Let f (x) =
. Then for all 0 < β < log
log b
 f (0), x ≤ 0
for x [0, 1].
Proof. If b = c, then f is Hölder continuous of exponent k for all 0 ≤ k < 1 =
by Theorem 2.1.2. If b > c, then f is a Hölder continuous function of exponent
log c
log b
log c
log 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 c
.
log b
Thus, for all 0 < β <
log c
,
log b
Dβ f ∗ (x) exists for x [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
 g(t), 0 < t ≤ 1
by (3) for all n N. Let g ∗ (t) =
. Then for all 0 < β < 21 , Dβ g ∗ (t)
 g(0), t ≤ 0
exists for t [0, 1].
Proof. Since g is a Hölder continuous function of exponent
1
2
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 , Dβ g ∗ (t) exists for all t [0, 1].
39
Chapter 5
Dimensions of Graphs of Nowhere
Differentiable Functions
We have examined continuous nowhere differentiable functions in terms of Hölder
continuity and fractional differentiability. Another way to study nowhere differentiable 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
X
(xi − xi−1 )2 + (f (xi ) − f (xi−1 ))2
12
i=1
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 BV [a, b], if and only if supP L(P) < ∞. If f 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 0 (x) exists almost everywhere on (a, b) if f BV [a, b]. Since nowhere
40
differentiable functions are not differentiable at any point in their domains, they cannot 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 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 Hδs (F ) is defined as
Hδs (F ) = inf
(∞
X
)
|Vi |s : {Vi } is a δ-cover of F .
i=1
The s-dimensional Hausdorff measure of F, Hs (F ), is defined to be
Hs (F ) = lim+ Hδs (F ).
δ→0
The Hausdorff dimension of F is then defined as
dimH F = inf {s : Hs (F ) = 0} = sup {s : Hs (F ) = ∞}
so that

 ∞, s < dimH F
s
H (F ) =
.
 0, s > dim F
H
41
Another commonly used notion of dimension is the box–counting dimension. We
again let F ⊂ Rn . For δ > 0, the collection of cubes
{[M1 δ, (M1 + 1)δ] × . . . × [Mn δ, (Mn + 1)δ] : M1 , . . . , Mn 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
dimB F = lim inf
+
δ→0
log Nδ (F )
.
− log δ
(10)
The upper box–counting dimension of F is defined by
dimB F = lim sup
δ→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 ≤ dimB F ≤ dimB F.
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 function, 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 Hölder continuous for the exponent 2 − s on [0, 1], where 1 ≤ s ≤ 2, then
Hs (F ) < ∞ and dimH F ≤ dimB F ≤ s.
42
Fact 5.1.4. Let F ⊂ Rn . Let g : F → Rm . If |g(x) − g(y)| ≤ M |x − y| for all
x, y 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 R. Let
(x1 , y1 ), (x2 , y2 ) F. Then
|g(x1 , y1 ) − g(x2 , y2 )| = |x1 − x2 |
(x1 − x2 )2 + (y1 − y2 )2
≤
21
= |(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 N be a δ–
S
P∞
1
cover of [0, 1]. Since [0, 1] ⊂ ∞
i=1 Vi ,
i=1 |Vi | ≥ 1. Thus, Hδ ([0, 1]) ≥ 1 for all δ > 0.
Then
H1 ([0, 1]) = lim+ Hδ1 ([0, 1]) ≥ 1.
δ→0
Now we determine an upper bound for H1 ([0, 1]). Let δ > 0. Let m = d 1δ e. Then
[0,1] can be covered by m sets of diameter δ. Thus,
(∞
)
X
Hδ1 ([0, 1]) = inf
|Vi | : {Vi } is a δ-cover of [0,1]
i=1
≤ mδ ≤
1
1+
δ
43
δ = δ + 1.
Thus,
H1 ([0, 1]) = lim+ Hδ1 ([0, 1]) ≤ lim+ δ + 1 = 1.
δ→0
δ→0
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
P
a0 (cn x)
f (x) = ∞
, where c N and c ≥ 2. Then dimH F ≤ dimB F ≤ 1.
n=0
cn
Proof. By Theorem 2.1.2, f is a Hölder 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 (1, 2] ,
dimH F ≤ dimB F ≤ s.
Thus,
dimH F ≤ dimB F ≤ 1.
Theorem 5.2.2. Let F be the graph of the function f : [0, 1] → R defined by
P
a0 (cn x)
f (x) = ∞
, where c N and c ≥ 2. Then dimH F = 1.
n=0
cn
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
P
a0 (cn x)
f (x) = ∞
, where c N and c ≥ 2. Then dimB F exists and equals 1.
n=0
cn
44
Proof. By Fact 5.1.1 and Propositions 5.2.1 and 5.2.2,
1 = dimH F ≤ dimB F ≤ dimB F ≤ 1.
Thus, dimB F = dimB F = 1. Therefore, dimB F = 1.
We now consider the more general version of the Generalized van der WaerdenTakagi function with b > c.
Proposition 5.2.4. Let F be the graph of the function f : [0, 1] → R defined by
P
a0 (bn x)
, where c > 1 and b N such that b > c. Then dimH F ≤ dimB F ≤
f (x) = ∞
n=0
cn
2−
log c
.
log b
Proof. By Theorem 2.1.5, f is a Hölder continuous function of exponent α for
all α ≤
log c
.
log b
So f satisfies the hypothesis of Fact 5.1.3 for 2 −
dimH F ≤ dimB F ≤ 2 −
log c
log b
≤ s ≤ 2. Thus,
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 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 = d 1δ e. 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
X
Rf [iδ, (i + 1)δ] ≤ Nδ ≤ 2m + δ
i=0
−1
m−1
X
i=0
45
Rf [iδ, (i + 1)δ].
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 =
1
,
2bk
then the horizontal
where
endpoints of each square in the δk –mesh are points of the form 2bik and i+1
2bk
i Z ∩ [0, 2bk − 1]. The value of f i+1
− f 2bik has already been calculated in the
2bk
proof of Proposition 1.1.10. If b is even, then (1) implies that
X
n k
i
+
1
i
b 1 f
−f
±
=
.
2bk
2bk c 2bk (12)
n=0
If b is odd, then (2) implies that
X
n
k k−1
1 i
b
c
b
i
+
1
=
.
f
−f
±
±
2bk
2bk n=0
c
c − 1 c 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
P
a0 (bn x)
, where c > 1 and b N such that b is even and b ≥ 2c. Then
f (x) = ∞
n=0
cn
2−
log c
log b
≤ dimB F.
Proof. For all k N, let δk =
1
2bk
and let mk =
1
δk
= 2bk . Then for all k N and
i Z ∩ [0, mk − 1],
i+1
i Rf [iδk , (i + 1)δk ] ≥ f
−f
2bk
2bk n k
X
b 1 = ±
c 2bk n=0
≥
b k
c
−
Pk−1
n=0
2bk
46
b n
c
b k
c
=
b
−
2
+1
c b
−1
c
2bk
bk−1
ck
≥
(b − 2c)
≥0
2bk
by (12) since b ≥ 2c.
Then by Fact 5.2.5,
Nδ k ≥
δk−1
m
k −1
X
Rf [iδk , (i + 1)δk ]
i=0
≥ 2b
k
k −1 k−1
2b
b
X
ck
i=0
(b − 2c)
2bk
2k−1
=
2b
ck
(b − 2c) .
Then
dimB F = lim inf
+
δ→0
log Nδ (F )
− log δ
2k−1
log 2b ck (b − 2c)
≥ lim inf
k→∞
− log 2b1k
log(b − 2c) + log 2 + (2k − 1) log b − k log c
k→∞
log 2 + k log b
log c
.
= 2−
log b
= lim inf
Proposition 5.2.7. Let F be the graph of the function f : [0, 1] → R defined by
P
a0 (bn x)
f (x) = ∞
, where c > 1 and b N such that b is odd and b ≥ 2c − 1. Then
n=0
cn
2−
log c
log b
≤ dimB F.
Proof. For all k N, let δk =
1
2bk
and let mk =
1
δk
= 2bk . Then for all k N and
i Z ∩ [0, mk − 1],
i+1
i Rf [iδk , (i + 1)δk ] ≥ f
−f
2bk
2bk 47
k−1 k n
X
b
b 1 c
±
= ±
c
c − 1 c 2bk n=0
!
k X
k−1 n
1
c
b
b
≥
−
2bk c − 1 c
c
n=0
b k b−2c+1
+1
1 c
c−1
=
b
2bk
−1
c
bk−1
ck
≥
(b − 2c + 1)
≥0
2bk
by (13) since b ≥ 2c − 1.
Then by Fact 5.2.5,
Nδk ≥ δk−1
m
k −1
X
Rf [iδk , (i + 1)δk ]
i=0
≥ 2b
k
k −1 k−1
2b
b
X
ck
i=0
(b − 2c + 1)
2bk
2k−1
=
2b
ck
(b − 2c + 1) .
Then
dimB F = lim inf
+
δ→0
log Nδ (F )
− log δ
2k−1
log 2b ck (b − 2c + 1)
≥ lim inf
k→∞
− log 2b1k
log(b − 2c + 1) + log 2 + (2k − 1) log b − k log c
k→∞
log 2 + k log b
log c
= 2−
.
log b
= lim inf
We know summarize the above discussion and propositions in the following theorem.
48
Theorem 5.2.8. Let F be the graph of the function f : [0, 1] → R defined by
P
a0 (bn x)
f (x) = ∞
, where c > 1 and b N such that b > c. Then 1 ≤ dimH F ≤
n=0
cn
dimB F ≤ dimB F ≤ 2 −
log c
.
log 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 c
.
log b
5.3. Kiesswetter’s Function
We now consider the Hausdorff and box–counting dimensions of the graph of Kiesswetter’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 N. Let G be the graph of g. Then
dimH G ≤ dimB G ≤ 32 .
Proof. By Theorem 2.2.3, g is a Hölder continuous function of exponent α for
all α ≤ 21 . So g satisfies the hypothesis of Fact 5.1.3 for
3
2
≤ s ≤ 2. Thus,
3
dimH G ≤ dimB G ≤ .
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 N. Let G be the graph of g. Then
3
2
≤ dimB G.
Proof. For all k N, let δk = 41k and let mk = δ1k = 4k . By Proposition 1.2.5,
i+1 g k − g ik = 1k for all k N and i Z ∩ [0, mk − 1]. Thus, for all k N and
4
4
2
i Z ∩ [0, mk − 1],
Rg [iδk , (i + 1)δk ] ≥
49
1
1
2
=
δ
k.
2k
Then by Fact 5.2.5,
Nδk (G) ≥
δk−1
m
k −1
X
Rg [iδk , (i + 1)δk ]
i=0
≥ δk−1
m
k −1
X
1
δk2
i=0
1
−3
= δk−1 δk−1 δk2 = δk2 .
Then
dimB G = lim inf
+
δ→0
log Nδ (G)
− log δ
−3
log( 41k ) 2
≥ lim inf
k→∞
− log 41k
− 23 log 41k
= lim inf
k→∞ − log 1k
4
=
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 N. Let G be the graph of g. Then
dimB G exists and equals 23 .
Proof. By Fact 5.1.1 and Propositions 5.3.1 and 5.3.2,
3
3
≤ dimB G ≤ dimB G ≤ .
2
2
Thus, dimB G = dimB G = 32 . Therefore, dimB G = 32 .
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.
50
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 N. Let G be the graph of g. Then
1 ≤ dimH G ≤ 23 .
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 ≤ 32 .
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Conclusion
In this thesis we investigated two special classes of continuous nowhere differentiable 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 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 Hölder continuity. We found that all of the
functions in both classes were Hölder continuous of exponent α on [0, 1] for some
α > 0. In fact, the Generalized van der Waerden–Takagi function with b = c is Hölder
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 Hölder 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 differentiable 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
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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 Generalized 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 Hölder 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 differentiable 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 Hölder continuity of the other functions 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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References
[1] Gerald Edgar, Measure, Topology, and Fractal Geometry, Springer–Verlag, New York, 1990.
[2] Kenneth Falconer, Fractal geometry, John Wiley & Sons Ltd., Chichester, 1990.
[3] G. H. Hardy and J. E. Littlewood, Some Properties of Fractional Integrals I, Math. Z. 27 (1928),
565–606.
[4] Wilfred Kaplan, Advanced calculus, Addison-Wesley Press, Inc., Cambridge, Mass., 1952.
[5] Karl Kiesswetter, Ein einfaches Beispiel für eine Funktion, welche überall stetig und nicht differenzierbar ist, Math.-Phys. Semesterber. 13 (1966), 216–221 (German).
[6] Konrad Knopp, Ein einfaches Verfahren zur Bildung stetiger nirgends differenzierbarer Funktionen, Math. Z. 2 (1918), 1–26 (German).
[7] Keith B. Oldham and Jerome Spanier, The Fractional Calculus, Academic Press, New York,
1974.
[8] H. L. Royden, Real analysis, 3rd ed., Macmillan Publishing Company, New York, 1988.
[9] Abdullah Shidfar and Kazem Sabetfakhri, On the Continuity of Van Der Waerden’s Function
in the Hölder Sense, Amer. Math. Monthly 93 (1986), 375–376.
[10] T. Takagi, A simple example of the continuous function without derivative, Proc. Phys.-Math.
Soc. Tokyo Ser. II 1 (1903), 176–177.
[11] B. L. van der Waerden, Ein einfaches Beispiel einer nichtdifferenzierbaren stetigen Funktion,
Math. Z. 32 (1930), 474–475 (German).
[12] Masaya Yamaguti, Masayoshi Hata, and Jun Kigami, Mathematics of Fractals, Translations of
Mathematical Monographs, vol. 167, American Mathematical Society, Providence, RI, 1997.
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