Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 2 / 11F / b / Solution

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.

Solved by gpt-5.6-sol high.