Let $f\in C^\infty([0,1])$ and consider the sequence $$f_n=f^{(n)}$$
where $f^{(n)}$ denote the derivative of order $n$ of $f$. My question is: What is a necessary condition to impose on $f$, such that $f_n$ converges uniformly to some $g\in C([0,1])$, i.e. $\|f_n-g\|_\infty\to 0$?
I could find some examples that maybe can help.
I - If $f$ is a polynomious, then $f_n\to 0$,
II - If $f(x)=e^{\lambda x}$, then $f_n\to 0$ if $\lambda\in [0,1)$ and $f_n\to f$ if $\lambda =1$.
Thank you
Assume that $f^{(n)}\to l$ uniformly on $[0,1]$. Then there is a constant $M$ such that for each $n$ and $x\in [0,1]$, $|f_n(x)|\leqslant M$. Conversely, this conditions implies convergence of $(f^{(n_k)},k\geqslant 1)$, where $n_k\uparrow \infty$.
We assume that $f^{(n)}\to l$ uniformly on $[0,1]$. Using $$f^{(n)}(x)-f^{(n)}(0)=\int_0^xf^{(n+1)}(t)\mathrm dt,$$ and an argument of dominated convergence, we get $$l(x)-l(0)=\int_0^xl(t)\mathrm ,dt$$ hence we get a nice expression for the candidate $l$.