 Solution

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.