I'm wondering if the space $$S=\left\{f\in C^{\infty}([0,1])\ \ :\ \ f(0)=0\bigwedge f(1)=1\bigwedge\forall n\in\mathbb{N},\ f^{(n)}(0)=f^{(n)}(1)=0\right\}$$ contains an element $f_0$ such that $f_0^{(n)}$ has at most $n+1$ roots for all $n\in\mathbb{N}_0$.  Intuitively I'm somewhat inclined to think it's false. Do anyone know if this holds, or how I should go on about finding out?  If $f_0^{(n)}$ has $n+1$ roots, then by the mean value theorem $f_0^{(n+1)}$ has $n$ distinct roots in the interior of the support in addition to the boundary points being roots. Hence by induction $f_0^{(n)}$ must have $n+1$ roots for all $n\in\mathbb{N}$.  As to whether the lower bound is non-strict, I think maybe the way to go is to assume that $(a_i)_{i\in\mathbb{N}}$ is a sequence with $a_i \in S$ and $a_i^{(i)}$ having at most $i+1$ roots. Because if those properties imply that not all derivatives of $a_i$ remain bounded as $i\to\infty$, then no such sequence can have an element of $S$ as limit, and so the lower bound must be strict.   I made some numerically experiments with linear programs which indicated the existence of such a sequence where also $\sup_x a_i^{(1)}(x)&lt;2$. So several derivatives must be considered simultaneously.
Let $f\in C^{\infty}(\mathbb{R},\mathbb{R})$ such that $f(x)=0$ on $\mathbb{R}$\ $(-1,1)$. Prove that there exists $n\in\mathbb{N}$ such that $f^{(n)}$ has at least n+1 zeros on $(-1,1)$  My attempt :   Assume $f$ is strictly positive  (if not n=0 works). Let $Z(f^{(n)})$ the number of zeros of $f^{(n)}$ on $(-1,1)$. Then we have: $$ \forall n\in\mathbb{N}:\;f^{(n)}(-1)=f^{(n)}(1)=0 $$ so that iterating the Rolle theorem to the successive derivatives of f we have  $$  \;Z(f^{(k+1)}) \geq Z(f^{(k)}) +1 $$ thus, by induction : $$ \forall n\in\mathbb{N}: \;Z(f^{(n+1)}) \geq Z(f^{(n)}) +1 $$ Therefore, $$ Z(f^{(n)}) \geq n $$ How can I find an another zero ?   I'm still looking for an another proof which does not use the lemma ..,  Thank you in advance.
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