Prove that there exists $n\in\mathbb{N}$ such that $f^{(n)}$ has at least n+1 zeros on $(-1,1)$ 
Let $f\in C^{\infty}(\mathbb{R},\mathbb{R})$ such that $f(x)=0$ on $\mathbb{R}\setminus (-1,1)$. Prove that there exists $n\in\mathbb{N}$ such that $f^{(n)}$ has at least $n+1$ zeros on $(-1,1)$

I have already post this exercice in Mathematics Stack Exchange and I have succeed to prove it with the following lemma :

Lemma: Let $f$ be a smooth function vanishing outside $(-1,1)$ and $P\in \mathbb{R}[X]$ a polynomial of degree $\deg(P)=d$ without zeros in $(-1,1)$. Then: 
  $$\forall n\in \mathbb{N}, Z((fP)^{(n+d)}) \geq Z(f^{(n)})+d,$$
  where $Z(g)$ the number of zeros of $g$ on $(-1,1)$


In fact, we don't understand why, after reading, this mysterious zero will appear. Plus, it is not really highlighted the role of the compactness of the support which is essential here.
It would be very interesting to have an another proof explaining the appearance of this additional zero.
References


*

*The exercise in Mathematics Stack Exchange  

*Oral exercice of ENS Paris (2011)


NB: I'm still a student but I am convinced that this is an interesting question that I hope deserves its place here.
Thank you in advance for your time,
Julien.
 A: Unfortunately I haven't been able to find an intuitive explanation for this behavior or even an alternative proof. But I've played a bit with your proof and I'm posting my version here in the hope that it might help someone else to find an explanation.
Definition


*

*Let $\mathcal{S} = \{f \in C^\infty([-1, 1])\colon \forall n\in \mathbb{N}\colon f^{(n)}(-1) = f^{(n)}(1) = 0 \}$.

*For $n\in\mathbb{N}$ and $\alpha\in\mathbb{R}\setminus[-1, 1]$ let $D_\alpha^n = \frac{1}{(x + \alpha)^{n-1}}\frac{d}{dx}(x+\alpha)^n$.

*For $f\in\mathcal{S}$ let $Z(f)\in \mathbb{N}\cup \{\infty\}$ be the number of zeros of $f$ in $(-1, 1)$.


Lemma


*

*$Z(D^n_\alpha f) \ge 1 + Z(f)$.

*$\left((x+\alpha)f\right)^{(n)} = D_\alpha^n f^{(n-1)}$.


Proof


*

*$f$ (and thus $(x+\alpha)^n f$) has $Z(f)+2$ zeros on $[-1, 1]$. By Rolle's theorem, $D_\alpha^n f$ has at least $Z(f)+1$ zeros in $(-1, 1)$.

*$$((x+\alpha)f)^{(n)} = nf^{(n-1)} + (x + \alpha)f^{(n)} = \frac{1}{(x+\alpha)^{n-1}}\frac{d}{dx}\left((x+\alpha)^n f^{(n-1)}\right) = D_\alpha^n f^{(n-1)}$$ $\square$


Theorem
For all $f\in\mathcal{S}$ there is an $n\in\mathbb{N}$ such that $Z(f^{(n)})> n + 1$.
Proof
If $f$ has a zero in $(-1, 1)$, we are done. Otherwise assume that $f>0$ in $(-1, 1)$ and let $\alpha = \frac{6}{5}$ and $\beta = \frac{4}{5}$. Note that $|\alpha^2 - \beta^2|<1$. Let $g_k = \frac{f}{(x^2 - \alpha^2)^{2k}}$. Then  


*

*$g_k(-1) = 0$,

*$g_k(-\beta) \to \infty$,

*$g_k(0) \to 0$,

*$g_k(\beta) \to \infty$, and

*$g_k(1) = 0$


for $k\to\infty$. Hence, for $k$ large enough, $g_k$ has at least two local maxima and one local minimum in $(-1, 1)$. Thus, $Z(g_k') > 2$.
By factoring $x^2 - \alpha^2 = (x-\alpha)(x+\alpha)$ and applying the lemma $4k$ times, we see $$f^{(4k+1)} = ((x^2 - \alpha^2)^{2k}g_k)^{(4k+1)} = D_{-\alpha}^{4k+1}D_\alpha^{4k}\cdots D_{-\alpha}^{3}D_\alpha^{2}g_k'.$$ Hence, $Z(f^{(4k+1)}) \ge 4k + Z(g_k') > 4k + 2$. $\square$
