Prove that there exists $n\in\mathbb{N}$ such that $f^{(n)}$ has at least n+1 zeros on $(-1,1)$

I believe Julien's proof is right and is also elegant, although there are a lot of typos in his write up. I'll rewrite it here (as Community Wiki), since there seems to be some doubt about that. Of course, this doesn't answer Julien's request for a different proof, but I see nothing wrong with his proof

I introduce the notation $Z(h)$ for the number of zeroes of $h$ on $(-1,1)$.

Let $f$ be as stated. If $f$ vanishes on $(-1,1)$ we are done so, without loss of generality, let $f(x)$ be positive on $(-1,1)$.

For any positive integer $k$, define $$ g:=x\to \frac{f(x)}{(1-x^2)^k} \mbox{ for } |x|<1, \mbox{ and } 0 \mbox{ otherwise} $$ L'Hospital's rule shows that $g$ is in $C^\infty(\mathbb{R},\mathbb{R})$, and $g$ is identically zero outside $(-1,1)$.

Choose $k$ such that $$ 0 = f(-1) \ < \ f\left(-\frac{1}{2}\right) (1-\frac14)^{-k} \ >\ f(0)\ < \ f\left(\frac{1}{2}\right) (1-\frac14)^{-k} \ > \ f(1) = 0. $$ So $$ 0 = g(-1) \ < \ g(-1/2) \ >\ g(0)\ <\ g(1/2)\ >\ g(1) = 0.$$

Then $g'$ must be positive somewhere on $(-1,-1/2)$, negative somewhere on $(-1/2,0)$, positive somewhere on $(0,1/2)$ and negative somewhere on $(1/2,1)$. Thus $Z(g^{(1)}) \geq 3$, and the result holds for $g$ (instead of $f$), with $n=1$.

We will show that $Z(f^{(2k+1)}) \geq 2k+3$.

Lemma Let $p(x)$ be identically $0$ outside $(-1,1)$ and let $a$ be a real number not in $(-1,1)$. Set $q(x) = p(x) (x-a)$. Then $Z(q^{(n+1)}(x)) \geq Z(p^{(n)}(x))+1$.

Proof Let the zeroes of $p^{(n)}$ be $-1 < r_1 < r_2 < \cdots < r_m < 1$ and set $r_0=-1$ and $r_{m+1}=1$. Then $(x-a)^{n+1} p^{(n)}(x)$ vanishes at all the $r_i$, so by Rolle's theorem, $Z(\frac{d}{dx} ( (x-a)^{n+1} p^{(n)}(x) )) \geq m+1$. But $\frac{d}{dx} ( (x-a)^{n+1} p^{(n)}(x) ) = (x-a)^n q^{(n+1)}$, and $(x-a)^{n}$ doesn't vanish on $[-1,1]$, so $Z(q^{(n+1)}) \geq m+1$ as desired. $\square$

Basically, the lemma says that throwing in a non vanishing linear factor doesn't make the Rolle's theorem bound worse.

Now set $h_0 = g$, $h_1 = (1-x) h_0$, $h_2 = (1+x) h_1$, $h_3 = (1-x) h_2$, $h_4 = (1+x) h_3$ etcetera, until $h_{2k} = (1-x^2)^k h_0 = f$. Applying the lemma repeatedly gives $$Z(f^{(2k+1)}) \geq Z(h_{2k-1}^{(2k)} )+1 \geq Z(h_{2k-2}^{2k-1})+2 \geq \cdots \geq Z(h_0^{(1)}) + 2k \geq 3+2k$$ as desired.

• Lemma: Let f a such function and $P\in \mathbb{R}[X]$ a polynomial and $deg(P)=d$ without zeros on $(-1,1)$. Then: $$\forall n\in \mathbb{N}, Z((fP)^{(n+d)}) \geq Z(f^{(n)})+d $$

Proof. By induction it's sufficient to prove $\forall n\in \mathbb{N}$, $\forall \alpha \in \mathbb{R}$\ $(-1,1)$ : $$Z((g)^{(n+1)}) \geq Z(f^{(n)})+1 $$ where $g(x)=(x-\alpha)f(x)$ ,

Moreover, $g^{(n+1)}(x)=(x-\alpha)^{-(1-n)}\frac{d}{dx}\Bigl((x-\alpha)^{n}f^{(n)}(x)\Bigr)$. Then $g^{(n+1)}$ vanishes at least once again on (-1,1). QED

• Let p an integer such that
  $$ \Bigl(\frac{3}{4}\Bigr)^p < \min\left(\frac{f(\frac12)}{f(0)},\frac{f(-\frac12)}{f(0)}\right) $$ and $$ g(x)=(1-x^2)^{-p}f(x),\qquad g(-1)=g(1)=0 $$ Then $$ g(-1)<g(-1/2),\qquad g(-1/2)>g(0),\qquad g(0)<g(1/2),\qquad g(1/2)>g(1). $$ Thus, $g'$ vanishes at least three times on (-1,1), by the lemma we get $$ Z(f^{(n+2p)}) \geq Z(g^{(n)})+2p $$ and we have (with $n=1$) $$ Z(f^{(1+2p)}) \geq 3+2p $$ Therefore, $f^{(m)}$ for $m\geq 2p+1$ has at least $m+2$ zeros on $(-1,1)$