Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 4 / 12I (Topics in Analysis)

A subset $A$ of a metric space $X$ is a nowhere dense set whenEquivalently, every nonempty open set contains a nonempty open subset disjoint from $A$.

The Baire category theorem says that if $X$ is a complete metric space and $G_1,G_2,\dots$ are open dense subsets of $X$, then ${\bigcap }_{n\ge 1}{G}_n$ is dense in $X$. Equivalently, no nonempty open subset of $X$ is a countable union of nowhere-dense sets.

To prove it, take a nonempty open set $V$. Since $G_1$ is open and dense, there is a closed ballwith $0<r_1<2^{-1}$. Inductively, openness and density of $G_{n+1}$ allow us to chooseThe balls are nested, and for $m>n$ their centres satisfyThus $(x_n)$ is Cauchy and has a limit $x\in X$. Every closed ball contains the tail of the sequence, so it contains $x$. HenceAs $V$ was arbitrary, the intersection is dense.

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.

For $n\ge 0$, putEach $S_n$ is closed, the hypothesis gives $\mathbb{R}={\bigcup }_n{S}_n$, and $E_n\subset E_{n+1}$ because a function that vanishes on an open set has derivative zero there.

The open setis dense. Indeed, inside any nonempty open interval choose a nondegenerate closed subinterval $J$. The complete metric space $J$ is covered by the closed sets $J\cap S_n$, so the Baire's theorem makes one of them contain a relative open interval, which lies in some $E_n$.

On each connected component $I$ of $\Omega$, the function $F$ is one polynomial. To see this, any compact subinterval $K\subset I$ is covered by the increasing family $(E_n)$. A finite subcover therefore gives $K\subset E_N$ for some $N$, and $F^{(N)}=0$ on $K$. Hence $F$ is polynomial on $K$; overlapping compact intervals force these polynomials to agree throughout $I$.

Let $D=\mathbb{R}\setminus \Omega$. This closed set has no isolated points. Otherwise, on the two sides of an isolated point $x$, $F$ would equal polynomials. Smoothness makes all their one-sided derivatives agree at $x$, so the two polynomials are identical and extend across $x$. Some derivative would then vanish on a neighborhood of $x$, contrary to $x\in D$.

Suppose that $D$ is nonempty. It is complete andis a countable closed cover. A second application of Baire's theorem gives an index $k$ and an open interval $U_0$ such thatChoose $x_0\in D\cap U_0$ and a bounded open interval $U$ centred at $x_0$ whose closure lies in $U_0$.

Because $D$ has no isolated points, every $x\in D\cap U$ is approached by distinct points of $D\cap U$. Difference quotients first give $F^{(k+1)}(x)=0$, and induction gives

No component of $\Omega$ meeting $U$ can cross $x_0$, so it has an endpoint in $D\cap U$. If its polynomial had degree $d\ge k$, then its $d$th derivative would approach a nonzero constant at that endpoint, contradicting the preceding display. Its degree is therefore less than $k$, so $F^{(k)}$ also vanishes on the part of that component in $U$. We conclude that $F^{(k)}=0$ throughout $U$, which says $U\subset E_k\subset \Omega$ and contradicts $D\cap U\ne \varnothing$.