Completeness of the smooth-function metric
= Completeness of the smooth-function metric
If $(f_n)$ is Cauchy in the smooth-function metric, each derivative sequence $(f_n^{(r)})$ converges uniformly to a continuous function $g_r$. Passing to the limit in
$$
f_n^{(r)}(x)-f_n^{(r)}(0)=\int_0^x f_n^{(r+1)}(t)\,dt
$$
shows that $g_r'=g_{r+1}$. Hence $g_0$ is smooth and $f_n\to g_0$ in the metric.