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.