Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 4 / 12I / i / Solution

Let $C=C([0,1])$ with the uniform norm. For rational numbers $0\le a<b\le 1$, let $M_{a,b}^{+}$ and $M_{a,b}^{-}$ be the sets of functions that are respectively nondecreasing and nonincreasing on $[a,b]$.

Both sets are closed. For example, if $f_j\in M_{a,b}^{+}$ and $f_j\to f$ uniformly, then for $x<y$ in $[a,b]$,They also have empty interior. Given $f\in M_{a,b}^{+}$ and $\epsilon >0$, continuity lets us choose $a<x<y<b$ sufficiently close thatAdd a continuous triangular bump $h$ supported near $x$, with $h(x)=3\epsilon /4$, $h(y)=0$, and $\Vert h{\Vert }_{\infty }<\epsilon$. Then $(f+h)(x)>(f+h)(y)$, so the $\epsilon$-ball about $f$ is not contained in $M_{a,b}^{+}$. The analogous upward bump at $y$ deals with $M_{a,b}^{-}$. Thus both sets are nowhere dense.

There are only countably many rational pairs $(a,b)$, sois meagre. Completeness of $C$ and the Baire category theorem show that its complement is nonempty. Choose $f$ in that complement. If $f$ were monotone on an interval of positive length, that interval would contain a closed interval with rational endpoints, putting $f$ in one of the displayed sets. Hence $f$ is monotone on no interval of positive length, as described by the generic nowhere-monotone continuous function result.