IMC 2000 · Day 2 · P8

Let $(x - \alpha, x + \alpha) \subset [0,1]$ be an arbitrary non-empty open interval. The function $f$ is not monoton in the intervals $[x - \alpha, x]$ and $[x, x + \alpha]$, thus there exist some real numbers $x - \alpha \le p < q \le x$, $x \le r < s \le x + \alpha$ so that $f(p) > f(q)$ and $f(r) < f(s)$.

By Weierstrass' theorem, $f$ has a global minimum in the interval $[p, s]$. The values $f(p)$ and $f(s)$ are not the minimum, because they are greater than $f(q)$ and $f(s)$, respectively. Thus the minimum is in the interior of the interval, it is a local minimum. So each non-empty interval $(x - \alpha, x + \alpha) \subset [0,1]$ contains at least one local minimum.