Exercise: Let $f\in C^{(n)}(]-1,1[)$ and $\sup_{-1<x<1}|f(x)|\leq1$. Let $m_k(I)=\operatorname*{inf}_{x\in I}|f^{(k)}(x)|$, where $I$ is an interval contained in $]-1, I[$. Show that
a) if $I$ is partitioned into three successive intervals $I_1,I_2$ and $I_3$ and $\mu$ is the length of $I_2$, then$m_{k}(I)\leq\frac{1}{\mu}\big(m_{k-1}(I_{1})+m_{k-1}(I_{3})\big)$;
b) if $I$ has length $\lambda$, then $m_{k}(I)\leq\frac{2^{k(k+1)/2}k^{k}}{\lambda^{k}}$;
My puzzle: The bound of (b) is not strict. For example, when $\lambda = 2$, $m_1(I)$ is the minimum of first order derivative, and its upper bound is $\frac{2}{\lambda} = 1$ which is strictly bounded by (b), the function can be $f(x) = x$. But for $m_2(I)$, the minimum of second order derivative can only reach $4$ by $f(x) = 2x^2-1$, but $m_{k}(I)\leq\frac{2^{k(k+1)/2}k^{k}}{\lambda^{k}}$ claims it is $\frac{2^3\ 2^2}{2^2} = 8$. When $k = 3$ or more, the upper bound of minimum of $k$ order derivative has more gap. Then I give my proof of (a) and (b), then maybe there is a strict upper bound of it.
Proof of (a): It is not hard to see for all $x_1 \in I_1, x_3 \in I_3$ we have $\frac{f(x_1)-f(x_3)}{x_1-x_3} = f'(\xi)$. $x_3-x_1 \leq \mu$, and so we have $f'(\xi) \leq \frac1\mu(|f(x_1)|+|f(x_3)|)$, then by the arbitrariness of $x_1, x_3$, $m_1(I) \leq f'(\xi) \leq \frac1\mu(m_0(I_1)+m_0(I_3))$. When $k$ is larger than $1$, similar to this one.
(b): When $k = 1$, $m_k(I) \leq \frac{2}{\mu} \leq \frac{2}{\lambda}$. When $k = 2$, $m_2(I) \leq \frac{1}{\mu}(m_1(I_1)+m_1(I_3)) \leq \frac{1}{\mu}(\frac{2}{\mu_1}+\frac{2}{\mu_3})$, where the last inequality comes from inductive hypothesis. Then by Lagrange multiplier we get when $\mu_1 = \mu_3 = \frac14\lambda, \mu = \frac12\lambda$, $m_2(I) \leq \frac{32}{\lambda}$. When $k$ is larger, the only thing to do is solving the extremum of $\frac{1}{\mu}(\frac{1}{\mu_1^k}+\frac{1}{\mu_3^k})$.
(My conjecture: The upper bound of $m_k(I)$ is too loose, and it should be $m_{k}(I)\leq\frac{2^{k(k+1)/2}k^{k}}{\lambda^{k}}\frac{1}{2^{k-1}}$ because the inductive hypothesis neglect the continuous of derivatives.)
Hope someone can give me the strict bound of it or a nicer approach for this exercise. If I am wrong, please correct me. I am stuck on this for a long time. Thanks in advance.