Lower bound for an integral of a polynomial over an interval of finite length Assume that $Q$ is a degree $k$ polynomial and $I$ an interval of finite length $c$. Can you please give me some hints on how to show that
$$ \left( \sum_{m=0}^k \frac{c^{m}}{m!} |Q^{(m)} (t_0) |  \right)^2  \leq \frac{C(k)^{2c}}{c} \int_{I} Q^{2}(t) dt, $$
for some positive constant $C(k)$ depending only on the degree $k$ and any $t_0 \in I$. The statement is very puzzling and I have had a hard time establishing it. All hints are therefore greatly appreciated.
EDIT: If this is wrong, as it looks like, can it be repaired under some additional assumptions? Thank you.
 A: I think the statement is false.
For $I \subset \mathbb R$, $k\in \mathbb N$ and $C>0$ let  $P(I,C,k)$ be the proposition : $$\forall Q\in \mathbb R_k[x],\quad\quad \max_{t_0\in I}\left(\sum_{m=0}^{k} \frac{l(I)^m}{m!}|Q^{(m)}(t_0)|\right)^2 \leq \frac{C^{2l(I)}}{l(I)}\int_I Q^2(t)\mathrm d t$$ 
where $l(I)$  is the length of $I$. 
You can notice that for all $a\in \mathbb R, P(I,k,C)\Leftrightarrow P(I+a,k,C)$. 
Then, notice that the change of variable $t= \lambda s$ transforms $Q(X)$ into $Q(\lambda X)$ which is of same degree and $(Q(\lambda X))^{(m)}=\lambda^m Q^{(m)}(\lambda X)$. Furthermore, this change of variable leaves invariant the integral over $l(I)$ on the right hand side. 
Therefore, 
$$\forall \lambda,C>0, \forall k\in \mathbb N, \quad\quad P(I,C,k) \Leftrightarrow P(\lambda I, C^{1/\lambda},k).$$
If the statement is true for some constant $k$ and some constant $C(k)$ then $C$ can be chosen equal to $1$ : 
$$\forall k\in \mathbb N, \exists C>0, \forall I\subset \mathbb R, P(I,C,k) $$
$$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \Rightarrow \forall k\in \mathbb N, \forall I\subset \mathbb R, P(I,1,k).$$ 
We thus showed : 
$$\forall k\in \mathbb N, \exists C>0, \forall I\subset \mathbb R, P(I,C,k) \quad \quad\Leftrightarrow \quad\quad \forall k\in \mathbb N,  P([0,1],1,k) $$
Finally, lets test this on $Q=X^k$, we have : 
$$ \max_{t_0\in [0,1]} \left(\sum_{m=0}^k \frac{k!}{(k-m)!m!} t_0^{k-m}\right)^2 
= 2^k$$
and $$ \int_{0}^1 X^k = \frac{1}{k+1}$$
thus contradicting $P([0,1],1,k)$  for $k\geq 1$.
