TITLE: Is it possible to create a polynomial $p(x)$ with this relation between $p(0)$ and $p(c)$?
QUESTION [8 upvotes]: Given $b$ and $c$ with $b,c>1$, is it possible to construct a polynomial $p(x)$, whose degree is $n$ for all $c$ and $b$, such that:

$|p|$ is strictly increasing on $[1,c]$

and $|b \cdot p(c)| < |p(0)|$?


This might be satisfied by an interpolating polynomial, but how to actually construct it is beyond me.

REPLY [2 votes]: Let me present a more explicit version of Fedor’s argument.
Choose distinct $x_0,\dots,x_n\in[1,c]$.
By Lagrange’s interpolation formula, there exist constants $a_0,\dots,a_n$ such that
$$
  p(0)=\sum_{I=0}^n a_ip(x_i)
$$
for each polynomial $p$ of degree not exceeding $n$. Therefore,
$$
  |p(0)|\leq \sum_{I=0}^n |a_i|\cdot |p(x_i)|\leq 
     |p(c)|\cdot \sum_{I=0}^n |a_i|.
$$