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.