# 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.

• Chebyshev polynomials certainly give a minimum possible supremum norm, and I believe I recall that these give rise to nontrivial and optimal lower bounds on integrals of polynomials. I'll search for those. – Robert Wolfe Nov 16 '18 at 15:33
• I've found this: ac.els-cdn.com/0022247X85900654/… but I recall a somewhat more recent paper. Can't recall if they were lecture notes or an actual article. – Robert Wolfe Nov 16 '18 at 15:50

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 $$k\geq 1$$.