$f\in C^{n}(-1,1)$ and $\left|f\right|\leq1.$Show $a_{n}$ exists.Here $a_{n}$ satisfy if $|f'(0)|\geq a_{n}$ then $f^{(n)}(x)$ has at least $n-1$ different zeros.
A friend of mine told me this problem.The only thing I can know is that we can take $a_{2}=3$.If $f''(x)$ has no zeros then $f$ is convex (assume f''>0).Then $f(x)\geq f'(0)x+f(0).$ If $f'(0)\geq 3$ then $f(1)\geq 3-1=2>1.$ If $f'(0)\leq -3$ then $f(-1)\geq 3-1=2>1.$
But I cannot use the same way to solve $n=3$ and the general case.
I think the general case is related to polynominal.Any help or hint will be thankful.