Is there a non zero bump function $f:\mathbb{R}\to\mathbb{R}$ such that the sequence $f^{(n)}(x)$ given by the $n^\text{th}$ derivatives of $f$ converges uniformly to $0$?
An equivalent question was asked but it was marked as a duplicate for some reason.
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 $\|f^{(n)}\|_\infty = O(n!)$.