I'm trying to prove that if $f\in C^{\infty}(]-1,1[,\mathbb{R})$ and there exists a $p \in \mathbb{N}$ so that for all $n\in\mathbb{N}$, $f^{(n)}$ has at most $p$ zeros in $]-1,1[$ then $f$ is analytic.
I still don't have any concrete idea to move on so any kind of help is fine.
Thanks
References and steps for the proof below:
This is a highly non-trivial result proved by Schaeffer in 1943 in this paper,
It generalizes a fairly famous result of S Bernstein (the case $p=0$) and it is based on an estimate of the type $|f(x)| \le M$ in an interval of length $L$ around some $a$, then if $f^{(n)}$ doesn't change sign more than $n-1$ times on the interval we have $|f'(a)| \le \frac{(10n)^{2n}M}{L}$.
Applying carefully this, one then proves by induction that under the OP hypothesis on $f$ and all its derivatives not changing signs more than $p$ times (which is implied by the hypothesis on the zeroes of course) $|\frac{f^{(n)}(x)}{n!}| \le (\frac{a}{1-|x|})^n, -1 < x < 1, a=a(p)$ some huge but fixed positive integer and that obviously shows the analyticity of $f$