Question: Given $f\in C^{\infty}$ which is not analytic on a bounded domain $\Omega \subseteq \mathbb{R}$. What can we say about the sequence $\lbrace f^{(m)} \rbrace _{m=1}^{\infty} $? Specifically – what can we say about the convergence and/or decay of its $L^2$ and $L^{\infty}$ norms? If nothing, why? Of special importance is to bound globally Taylor like terms of the form $f^{(m)}(\xi)/m! $ for some $\xi \in \Omega$.
Motivation, not Neccessary: My question is motivated by the residue terms of numerical integration formulas. We can usually have nice convergence rates for analytical functions due to the exponential decay in their taylor series coefficients $f^{(m)}/m! $.
I have read some of the math.stackexchange posts about it, and while this one does contain some possible leads, I couldn't find the answers to my questions there yet.
Thanks
Best Answer
For the $L^\infty$ norms, nothing at all apart from Alexander Eremenko's answer, because of Borel's theorem: any sequence of real (resp. complex) numbers can be realized as the sequence of Taylor coefficients of some real (resp. complex) valued smooth function $f$ at any given point. By the Sobolev embedding theorem and the fact that $f$ may be chosen compactly supported in the interior of $\Omega$, one can bound the $L^\infty$ norm of $f$ from above by a positive constant times the sum of the $L^1$ norms of the $n$-th order derivatives of $f$, where $n$ is the dimension of the domain $\Omega$. Therefore, if the sequence of $L^\infty$ norms of the derivatives of $f$ blows up, so does the corresponding sequence of $L^1$ norms. Since you assumed $\Omega$ to be bounded, this implies the same property for the $L^p$ norms for all $1\leq p\leq\infty$ by Hölder's inequality.
(side remark: the case of Sobolev's embedding theorem which is relevant to the present context is an easy consequence of the fundamental theorem of Calculus applied once to each variable of $f$, which also shows that the constant above may be taken equal to 1)