Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 2 / 11F (Topics In Analysis)

The Baire category theorem states that every countable intersection of open dense subsets of a complete metric space is dense. Equivalently, no nonempty open subset of a complete metric space is a countable union of nowhere dense sets.

To prove the first form, let $G_1,G_2,\dots$ be open dense subsets of a complete metric space $X$, and let $V\subseteq X$ be nonempty and open. Choose a closed ballInductively, density and openness of $G_{n+1}$ allow a closed ballThe balls are nested and $d(x_m,x_n)<2^{-n}$ for $m>n$, so $(x_n)$ is Cauchy. Completeness gives $x_n\to x\in X$. For every $n$, all later centres lie in $\overline{B}(x_n,r_n)$; closedness gives $x\in \overline{B}(x_n,r_n)\subseteq G_n$. The first ball also lies in $V$, henceSince every nonempty open $V$ meets the intersection, that intersection is dense.

Every summand is at most $2^{-r}$, and ${\sum }_{r\ge 0}{2}^{-r}=2$, so $d(f,g)$ is finite and well defined. Nonnegativity and symmetry follow from the uniform norm. If $d(f,g)=0$, its $r=0$ summand gives $\Vert f-g{\Vert }_{\infty }=0$, hence $f=g$. Finally,and the triangle inequality for each uniform norm give the triangle inequality for $d$. Thus $d$ is a metric.

Let $(f_n)$ be $d$-Cauchy. For fixed $r$ and $0<\epsilon <1$, sufficiently large $m,n$ satisfywhich forces $\Vert f_m^{(r)}-f_n^{(r)}{\Vert }_{\infty }<\epsilon$. Hence, for every $r$, the continuous functions $f_n^{(r)}$ converge uniformly to some continuous $g_r$. The fundamental theorem of calculus givesUniform convergence permits passage to the limit, yieldingTherefore $g_r^{\prime }=g_{r+1}$ for every $r$, so $g_0\in C^{\infty }([0,1])$ and $g_0^{(r)}=g_r$.

To prove convergence in $d$, first choose $R$ so that ${\sum }_{r\ge R}{2}^{-r}$ is small, then use uniform convergence for the finitely many derivatives $r<R$. Thus $d(f_n,g_0)\to 0$. This proves the completeness of the smooth-function metric.

For rational $q\in (0,1)$ and positive integer $M$, defineEvaluation of the $m$th derivative at $q$ is continuous in the smooth-function metric, so $U_{q,M}$ is open.

It is also dense. Given $f$ and a metric tolerance $\epsilon >0$, choose $R\ge 1$ so that the contribution of all derivatives of order at least $R$ is below $\epsilon /2$, and choose $m\ge \max \{M,R\}$. For large $K$, perturb $f$ byFor every $r<R$, $\Vert h_K^{(r)}{\Vert }_{\infty }\le \epsilon /(4R)$, so $d(f+h_K,f)<\epsilon$. On the other hand,For sufficiently large $K$, $|f^{(m)}(q)+h_K^{(m)}(q)|>m!m^m$, proving density.

The space is complete by part (b). The Baire category theorem therefore makesdense. Put $E=C^{\infty }([0,1])\setminus G$. Each complement $U_{q,M}^c$ is closed and nowhere dense, so $E$ is a meagre set, or a set of first category. Every $f\notin E$ has the required derivative growth. This is the generic superfactorial derivative growth at rational points.

For $f\notin E$ and rational $q\in (0,1)$, every $M$ admits $m\ge M$ withConsequentlyBy the Cauchy-Hadamard theorem, the Taylor series of $f$ at $q$ has radius of convergence zero.

Suppose some Taylor series represented $f$ throughout a neighborhood of a point $x\in [0,1]$. That neighborhood contains a rational $q\in (0,1)$. A function represented locally by a convergent power series is a real analytic function, and re-expanding that series about $q$ gives the Taylor series of $f$ at $q$ a positive radius of convergence. This contradicts the preceding conclusion. Thus $f$ agrees with no Taylor series on a neighborhood of any point; it is a nowhere-analytic generic smooth function.