TITLE: Prove that there exists $n\in\mathbb{N}$ such that $f^{(n)}$ has at least n+1 zeros on $(-1,1)$
QUESTION [28 upvotes]: 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.

REPLY [2 votes]: $\def\cH{\mathcal{H}}\def\RR{\mathbb{R}}$A bit more than the question asks for is true. Let $\cH$ be the class of function $f: \RR \to \RR$ for which $f^{(n)}(-1) = f^{(n)}(1) = 0$ for all $n$. Note that, for  $f \in \cH$, the sequence $Z(f^{(n)})$ is weakly increasing. I will show that $Z(f^{(n)})-n \to \infty$. Let's define $L(f) = \lim_{n \to \infty} Z(f^{(n)})-n$.
We recall the lemma from my other answer:
Lemma Let $a \in (-\infty, -1] \cup [1, \infty)$ and $p \in \cH$. Put $q(x) = (x-a) p(x)$. Then $Z(q^{(n+1)}) \geq Z(p^{(n)})+1$.
We can rewrite this as $Z(q^{n+1}) - (n+1) \geq Z(p^{(n)}) - n$ and take the limit to deduce.
Corollary 1 With notation as in the lemma, $L(q) \geq L(p)$.
Using this repeatedly
Corollary 2 For any positive integers $A$ and $B$, we have $L(f) \geq L(f (1+z)^{-A} (1-z)^{-B})$.
Now, suppose for the sake of contradiction that $L(f)<\infty$. Replacing $f$ by one of its derivatives, we may assume that $Z(f^{(n)})  = n +c$ for all $n$. Let the zeroes of $f$ be $-1< r_1 < r_2 < \cdots < r_c< 1$ and without loss of generality suppose $f$ is positive on $(-1,r_1)$. But then if we choose $A$ and $B$ very large with $(A-B)/(A+B) \in (-1,r_1)$, then $g(z) : = f (1+z)^{-A} (1-z)^{-B})$ will have positive second derivative at $(A-B)/(A+B)$, so $g'$ has at least $3$ zeroes in $(-1, r_1)$, as well as at least one zero in each of $(r_1,r_2)$, $(r_2, r_3)$, ..., $(r_{c-1},r_c)$, $(r_c, 1)$, so $Z(g') \geq c+3$ and $L(g) \geq c+2$. 
This contradicts corollary $2$.

In a different question, the OP asks about functions with $f^{(n)}(-1) = f^{(n)}(1) = 0$ for $n \geq 1$, $f(-1)=0$ and $f(1)=1$. Clearly, if $f$ is such a function then $f' \in \cH$, so the bound also applies to such functions. In the comments to that question, there were a bunch of off by $1$ errors, so I thought I'd write up an answer to show that $L(f)$ is infinite anyway, and the off by $1$'s don't matter.