Bump functions whose derivatives $f^{(n)}(x)$ converge to $0$ uniformly as $n\to\infty$

Solution 1:

Suppose $f$ was such a function. From Taylor's theorem, for an arbitrary $a\in\mathbb R$, $$ f(x)=T_nf(x;a)+R_n(x;a)$$ where $T_nf(x;a)$ is the $n$th Taylor polynomial around $a$, and $R_n$ is the remainder term, which can be written e.g. in mean value form, for some $\xi$ between $x$ and $a$: $$ R_n(x;a)=\frac{f^{(n)}(\xi)(x-a)^n}{n!} $$ Note that for our function $f$, we easily have $$ \sup_{x:|x-a|<1/2}|R_n(x;a)| \to 0$$ Hence, $f$ is real analytic at each $a$. But a (real) analytic function that is compactly supported must be the zero function, by the identity theorem. So no such $f$ exists. In fact the same proof works if $\|f^{(n)}\|_\infty = O(n!)$.