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 $\mathrm{[}\mathrm{0}\mathrm{,}\mathrm{1}\mathrm{]}$ 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 $P(I)\subseteq X$ consisting of all non-decreasing functions, the subset $Q(I)\subseteq X$ 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 $I\subseteq J$. 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, $M\subset X$ 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

$$f(x)=\sum _{k=0}^{\mathrm{\infty }}{f}_k(x)$$

where the functions $f_k$ are defined by

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 is dense in $[0,1]$. Given any interval $I\subset [0,1]$ we can then find a point of the form $\frac{L}{2^k}$ in its interior.

It is easily seen that $f_j(\frac{L}{2^k})=0$ for $j\ge k$.

For any integer $j>k$, consider the points $a_j:=\frac{2^{j-k}L-1}{2^j}$ and $b_j:=\frac{2^{j-k}L+1}{2^j}$. The points $a_j$ (resp. $b_j$) are just the points on the left (resp. on the right) of $\frac{L}{2^k}$ when we divide the unit interval in segments of size $\frac{1}{2^j}$.

A direct calculation would show that

and similarly for $b_j$.

Evaluating $f$ in the points $a_j$ and $b_j$ we obtain

$f(a_j)$ $=$ ${\displaystyle \sum _{s=0}^{\mathrm{\infty }}}{f}_s(a_j)$

$=$ ${\displaystyle \sum _{s=0}^{j-1}}{f}_s(a_j)$

$=$ ${\displaystyle \sum _{s=0}^{k-1}}{f}_s(a_j)+{\displaystyle \sum _{s=k}^{j-1}}{f}_s(a_j)$

$=$ ${\displaystyle \sum _{s=0}^{k-1}}{f}_s({\displaystyle \frac{L}{2^k}})+{\displaystyle \sum _{s=0}^{k-1}}(\pm {\displaystyle \frac{1}{2^j}})+{\displaystyle \frac{j-k}{2^j}}$

$=$ $f({\displaystyle \frac{L}{2^k}})+{\displaystyle \sum _{s=0}^{k-1}}(\pm {\displaystyle \frac{1}{2^j}})+{\displaystyle \frac{j-k}{2^j}}$

and similarly for $f(b_j)$.

The least value we can obtain is $f(a_j)=f({\displaystyle \frac{L}{2^k}})-{\displaystyle \frac{k}{2^j}}+{\displaystyle \frac{j-k}{2^j}}=f({\displaystyle \frac{L}{2^k}})+{\displaystyle \frac{j-2k}{2^j}}$, and even in this extreme case we can still choose $j$ large enough so that $a_j\in I$ and $j>2k$.

For this appropriate $j$ we see that $f(a_j)>f(\frac{L}{2^k})$, and similarly $f(b_j)>f(\frac{L}{2^k})$.

Recall that . 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