Suppose a continuous function $f:\mathbb{R} \rightarrow \mathbb{R}$ is nowhere monotone. Show that there exists a local minimum for each interval.
This question is from Moscow institute. First of all, I can't even construct a nowhere monotone function. What I can think of, linear functions, clearly do not satisfy this, as we can always zoom in to get monotonicity.
Remark: A function $f$ is nowhere monotone if for every interval $[a,b]$, $f$ is not increasing or decreasing.
You want to show that the set of local minima is dense in $[a,b]$, any closed and bounded interval of $R$:
Consider the interval $[a,b]$. Take $(x-\epsilon ,x+\epsilon )$ any non-empty open subinterval. Then $f$ is not monotone on the intervals $[x-\epsilon ,x]$ and $[x, x+\epsilon ]$ so we can find numbers $p,q,r,s$ such that $x-\epsilon \leq p< q\leq x$ and $x\leq r< s\leq x+\epsilon $ such that $f(p)> f(q)$ and $f(r)< f(s)$. Now $f$ has a global minimum on $[p,s]$ but neither $f(p)$ nor $f(s)$ can be this minimum. So the minimum must be interior to $[p,s]$.
Also, OP can't figure out a nowhere monotone function. I have one beautifull example >:)
$$ \mbox{Let be }f\colon\mathbb{R}\to\mathbb{R}\mbox{ such that }f(x)=\begin{cases} x, & \mbox{ if } x\in\mathbb{Q} \\ -x, & \mbox{ if } x\notin\mathbb{Q} \end{cases} $$
such beautifull funtion, is isnt?
