= Solution
Every summand is at most $2^{-r}$, and $\sum_{r\geq0}2^{-r}=2$, 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 $\|f-g\|_\infty=0$, hence $f=g$. Finally,
$$
\min\{1,a+b\}\leq\min\{1,a\}+\min\{1,b\}
$$
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 $0<\varepsilon<1$, sufficiently large $m,n$ satisfy
$$
d(f_m,f_n)<2^{-r}\varepsilon,
$$
which forces $\|f_m^{(r)}-f_n^{(r)}\|_\infty<\varepsilon$. Hence, for every $r$, the continuous functions $f_n^{(r)}$ converge uniformly to some continuous $g_r$. The fundamental theorem of calculus gives
$$
f_n^{(r)}(x)-f_n^{(r)}(0)
=\int_0^x f_n^{(r+1)}(t)\,dt.
$$
Uniform convergence permits passage to the limit, yielding
$$
g_r(x)-g_r(0)=\int_0^xg_{r+1}(t)\,dt.
$$
Therefore $g_r'=g_{r+1}$ 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 $\sum_{r\geq R}2^{-r}$ is small, then use uniform convergence for the finitely many derivatives $r<R$. Thus $d(f_n,g_0)\to0$. This proves the <completeness of the smooth-function metric>.
Back to article page
