Space of smooth functions on a compact interval ($C^{\infty }([0,1])$)

The space $C^{\infty }([0,1])$ becomes a complete metric space underConvergence in this metric is exactly uniform convergence of every derivative separately.

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 inshows that $g_r^{\prime }=g_{r+1}$. Hence $g_0$ is smooth and $f_n\to g_0$ in the metric.

Outside a meagre set in $C^{\infty }([0,1])$, every rational $q\in (0,1)$ and every positive integer $M$ admit $m\ge M$ such thatFor fixed $q,M$, the union over $m\ge M$ of these strict-inequality sets is open. It is dense because a perturbation $\delta \cos (K(x-q)-m\pi /2)$ can make the $m$th derivative arbitrarily large while keeping any prescribed finite collection of lower derivatives arbitrarily small. The Baire category theorem applied over the countable pairs $(q,M)$ gives the claim.

The generic derivative bound implies that at every rational $q\in (0,1)$,The Cauchy-Hadamard theorem therefore gives radius zero to the Taylor series at every such $q$. If a Taylor series represented $f$ on a neighborhood of any point, that neighborhood would contain a rational point at which $f$ was real analytic, a contradiction.