Zeroes of derivatives of high order The problem is following.
Let $f:(-1,1)\to [-1,1]$ has $n$ derivatives. Prove that there exists a number $\alpha_n$ (independent from $f$) such that condition $|f'(0)|\geq \alpha_n$ implies that equation $f^{(n)}(t)=0$ has at least $n-1$ distinct zeroes on $(-1,1).$
I can prove this for $n=2.$ In this case it is enough to take $\alpha_2>2.$ Unfortunatley, I can't proceed further and prove the general case.
 A: By Darboux's theorem $f^{(n)}$, being a derivative, has the intermediate value property, so it suffices to show that $f^{(n)}$ has $n$ values with alternating signs. This we can prove by induction.
I will first prove a lemma so as not to interrupt the flow of the proof later: For every non-negative integer $n$ and every length $L$, there is $\beta_n(L)$ such that if $f:(-1,1)\to[-1,1]$ has $n$ derivatives, then there is no interval of length $L$ over which $|f^{(n)}|\gt \beta_n(L)$. The proof proceeds by induction. For the base case $n=0$, we can choose $b_0(L)=1$ since $|f|\le 1$. Now assume for the sake of contradiction that there is an interval of length $L$ over which $|f^{(n+1)}|\gt \beta_{n+1}(L)$, with $\beta_{n+1}(L)$ to be determined. Then by Darboux's theorem $f^{(n+1)}$ doesn't change sign on the interval, and by the mean value theorem in each quarter of the interval $f^{(n)}$ must change by at least $\beta_{n+1}(L)/(L/4)$ in the same direction. It can vanish in at most one of these quarters, and if so there is at least one other quarter which is not adjacent to that quarter; thus there is a quarter in which $|f^{(n)}|\gt \beta_{n+1}(L)/(L/4)$, and we obtain a contradiction by choosing $\beta_{n+1}(L)=(L/4)\beta_n(L/4)$.
We can use this to prove a strengthening of the claim: For every $\epsilon,M\gt0$ there is $\alpha_n(\epsilon,M)$ such that if $f:(-1,1)\to[-1,1]$ has $n$ derivatives and $|f'(0)|\ge\alpha_n(\epsilon,M)$, then $f^{(n)}$ has $n$ values with alternating signs in $[-\epsilon,\epsilon]$ and the outermost of these values have absolute value $\ge M$.
The base case $n=1$ is again trivial: Choose $\alpha_1(\epsilon,M)=M$; then $f'(0)$ is the required value.
Now assume that $f^{(n)}$ has $n$ values in $[-\epsilon,\epsilon]$ with alternating signs and outermost absolute values $\ge M$ if $|f'(0)|\ge\alpha_n(\epsilon,M)$. By the mean value theorem, that implies that $f^{(n+1)}$ has $n-1$ values with alternating signs in between, so we only have to find two more such values, one on each side.
Let $\epsilon,M\gt0$ be given. If $|f'(0)|\ge\alpha_{n}(\epsilon/2,\kappa)$ with $\kappa$ to be determined, then $f^{(n)}$ has $n$ values in $[-\epsilon/2,\epsilon/2]$ with alternating signs and outermost absolute values $\ge\kappa$. Let the rightmost of these values be at $a$. By the lemma, there is $\xi\in[a,a+\epsilon/2]$ with $|f^{(n)}(\xi)|\le\beta_n(\epsilon/2)$, and then by the mean value theorem there is $\zeta\in[a,\xi]$ with
$$|f^{(n+1)}(\zeta)|=|f^{(n)}(\xi)-f^{(n)}(a)|/(\xi-a)\ge\left(\kappa-\beta_n(\epsilon/2)\right)/(\epsilon/2)\;,$$
and the sign of $f^{(n+1)}(\zeta)$ is the one required for the alternation. A leftmost value is found similarly. Thus, if we choose $\kappa$ such that $\left(\kappa-\beta_n(\epsilon/2)\right)/(\epsilon/2)=M$, i.e. $\kappa=M\epsilon/2+\beta_n(\epsilon/2)$, then $\alpha_{n+1}(\epsilon,M)=\alpha_n(\epsilon/2,\kappa)$ suffices.
