continuous nowhere monotonic function

continuous nowhere monotonic function

Let

f

be a real-valued

continuous function

defined on the

unit interval

[

0

,

1

]

. It seems intuitively clear that

f

should be

monotonic

on some

subinterval

I

of

[

0

,

1

]

. Most of the concrete examples seem to

support

this. A

counterexample

is termed

nowhere monotonic

, meaning that the

function

is not monotonic in any subinterval of

[

0

,

1

]

. Surprisingly, nowhere

monotonic functions

do exist:

Proposition 1

.

There exists a real-valued continuous function defined on

[

0

,

1

]

that is nowhere monotonic.

A sketch of the proof goes as follows:

1.

Let

X

be the set of all

continuous

real-valued functions on [0,1]. Then

X

is a

complete metric space

given the sup norm. Clearly

X

is non-empty.

2.

Given any subinterval

I

of

[

0

,

1

]

, the subset

P

⁢

(

I

)

⊆

X

consisting of all non-decreasing functions, the subset

Q

⁢

(

I

)

⊆

X

consisting of all non-increasing functions, and hence their union

M

⁢

(

I

)

, are closed.

3.

Furthermore,

M

⁢

(

I

)

is nowhere dense.

4.

Let

S

be the set of all

rational

intervals

in

[

0

,

1

]

(a rational interval is an interval whose

endpoints

are rational numbers). Then

S

is

countably infinite

. Take the union

M

of all

M

⁢

(

I

)

, where

I

ranges over

S

.

5.

If

f

is

monotone

on some interval

J

in

[

0

,

1

]

, then

f

is monotone on some rational interval

I

⊆

J

. If the

theorem

is false, then every continuous function is monotone on some rational interval, which means

M

=

X

.

6.

However,

M

is a

countable

union of nowhere

dense sets

and

X

is a non-empty complete metric space. By

Baire Category Theorem

, this can not happen. Therefore,

M

⊂

X

strictly and there exists a continuous nowhere monotone real-valued function defined on

[

0

,

1

]

.

Example : van der Waerden function

The above shows the

existence

of such a function. Here is an actual example of a nowhere monotonic continuous function, called the

van der Waerden function

. This function, which we designate by

f

, is given by a series

f

⁢

(

x

)

=

∑

k

=

0

∞

f

k

⁢

(

x

)

where the functions

f

k

are defined by

f

0

⁢

(

x

)

=

{

x

,

if

⁢

0

≤

x

≤

1

2

-

x

+

1

,

if

⁢

1

2

≤

x

≤

1

and

f

k

⁢

(

x

)

=

1

2

k

⁢

f

0

⁢

(

2

k

⁢

x

)

where each

f

k

⁢

(

x

)

is defined on

[

0

,

2

-

k

]

. Since

f

k

agrees on the endpoints, we can extend the its domain to the entire unit interval by periodic extension (so that the graph of

f

k

⁢

(

x

)

has the shape of a sawtooth).

Figure 1:

The graphics of

f

0

(left),

f

1

(middle) and

f

2

(right).

Figure 2:

The graphic of the van der Waerden function (the dashed lines are the graphics of each

f

k

).

- It is easy to check that each

f

k

is continuous. Using the

Weierstrass M-test

we can also see that the series

converges uniformly

, and therefore conclude that

f

itself is a continuous function (it is the uniform limit of continuous functions).

- We now prove that

f

is nowhere monotonic:

The set

{

L

2

k

:

k

∈

ℕ

,

0

<

L

<

2

k

}

is dense in

[

0

,

1

]

. Given any interval

I

⊂

[

0

,

1

]

we can then find a point of the form

L

2

k

in its

interior

.

It is easily seen that

f

j

⁢

(

L

2

k

)

=

0

for

j

≥

k

.

For any integer

j

>

k

, consider the points

a

j

:=

2

j

-

k

⁢

L

-

1

2

j

and

b

j

:=

2

j

-

k

⁢

L

+

1

2

j

. The points

a

j

(resp.

b

j

) are just the points on the left (resp. on the right) of

L

2

k

when we divide the unit interval in segments of size

1

2

j

.

A direct calculation would show that

{

f

s

⁢

(

a

j

)

=

0

,

s

≥

j

f

s

⁢

(

a

j

)

=

1

2

j

,

j

>

s

≥

k

f

s

⁢

(

a

j

)

=

f

s

⁢

(

L

2

k

)

±

1

2

j

,

k

>

s

≥

0

and similarly for

b

j

.

Evaluating

f

in the points

a

j

and

b

j

we obtain

f

⁢

(

a

j

)

=

∑

s

=

0

∞

f

s

⁢

(

a

j

)

=

∑

s

=

0

j

-

1

f

s

⁢

(

a

j

)

=

∑

s

=

0

k

-

1

f

s

⁢

(

a

j

)

+

∑

s

=

k

j

-

1

f

s

⁢

(

a

j

)

=

∑

s

=

0

k

-

1

f

s

⁢

(

L

2

k

)

+

∑

s

=

0

k

-

1

(

±

1

2

j

)

+

j

-

k

2

j

=

f

⁢

(

L

2

k

)

+

∑

s

=

0

k

-

1

(

±

1

2

j

)

+

j

-

k

2

j

and similarly for

f

⁢

(

b

j

)

.

The least value we can obtain is

f

⁢

(

a

j

)

=

f

⁢

(

L

2

k

)

-

k

2

j

+

j

-

k

2

j

=

f

⁢

(

L

2

k

)

+

j

-

2

⁢

k

2

j

, and even in this extreme case we can still choose

j

large enough so that

a

j

∈

I

and

j

>

2

⁢

k

.

For this appropriate

j

we see that

f

⁢

(

a

j

)

>

f

⁢

(

L

2

k

)

, and similarly

f

⁢

(

b

j

)

>

f

⁢

(

L

2

k

)

.

Recall that

a

j

<

L

2

k

<

b

j

. We conclude that

f

is not monotonic in

I

, and hence it is nowhere monotonic.

Remark

. The van der Waerden function turns out to be nowhere

differentiable

as well.

Title

continuous nowhere monotonic function

Canonical name

ContinuousNowhereMonotonicFunction

Date of creation

2013-03-22 14:59:08

Last modified on

2013-03-22 14:59:08

Owner

asteroid (17536)

Last modified by

asteroid (17536)

Numerical id

19

Author

asteroid (17536)

Entry type

Result

Classification

msc 54E52

Defines

van der Waerden function

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