A function that is nowhere monotonic

Give an example of a function that is continuous but is nowhere monotonic. In continuity give an example of a function that is nowhere monotonic and has no extrema values.

Solution

We begin with the first question. We will construct a function that is continuous but nowhere monotonic. Let $f_1(x)  \equiv |x| , \; |x| \leq 1/2$ and let $f_1(x)$ be defined for other values of $x$ by periodic continuation with period $1$, i.e $f_1(x+n)=f_1(x)$  for every real number $x$ and integer $n$. For $n>1$ define:

$$f_n(x)= 4^{-n+1} f_1(4^{n-1}x)$$

so that for every positive integer $n$ , $f_n$ is a periodic function of period $4^{-n+1}$ and maximum value $\frac{1}{2} \cdot 4^{-n+1}$. Finally define $f$ with domain $\mathbb{R}$

$$f(x)=\sum_{n=1}^{\infty} f_n(x) = \sum_{n=1}^{\infty} \frac{f_1(4^{n-1}x)}{4^{n-1}}$$

The series converges uniformly (this is immediate from the M-test) on $\mathbb{R}$ and is everywhere continuous.

For any point $a$ of the form $a=k4^{-m}$ where $k$ is an integer and $m$ is a positive integer , $f_n(a)=0$ for $n>m$ and hence:

$$f(a)=f_1(a)+\cdots +f_m(a)$$

For any positive integer $m$ let $h_m$ be the positive number $4^{-2m-1}$. Then we see that $f_n(a+h_m)=0$ for $n>2m+1$ and hence:

$$\begin{align*}
f\left ( a+h_m \right )-f(a) &= \left [ f_1\left ( a+h_m \right )-f_1(a) \right ]+\cdots + \left [ f_m\left ( a+h_m \right ) -f_m(a) \right ]\\
 & \quad \quad \quad \quad \qquad \qquad \quad \quad \; + f_{m+1}\left (a+h_m  \right )+\cdots + f_{2m+1}\left ( a+h_m \right ) \\
 &\geq  -mh_m + (m+1)h_m = h_m >0
\end{align*}$$

Similarly,.

$$f\left ( a-h_m \right )-f(a) \geq -m h_m +(m+1)h_m= h_m >0$$

Since members of the form $a=k4^{-m}$ are dense , it follows that in no open interval is $f$ monotonic and this completes the first question.


For the second question, the function:

$$f(x)= \left\{\begin{matrix}
x &,  &x \in \mathbb{Q} \\
 -x& , &  x \notin \mathbb{Q}
\end{matrix}\right.$$

qualifies. The details are left to the reader.