*Note: This question is based on [a previous question](https://mathoverflow.net/q/371972/165539)*


I was continuing my research from last time, and I realized my question was too strict! Instead of the polynomial being strictly increasing, it only has to be only positive with the *maximum* smaller than $p(0)$. So, my new question is below:


Given $b$ and $c$ with $b,c>1$, is it possible to construct a polynomial $p(x)$, whose degree $n$ depends on neither $c$ nor $b$, such that:


* $p$ is non-negative on $[0,c]$
* and $b \cdot \max\_{x \in [1,c] }p(x) < p(0)$? (if it can't be done, I will also accept a polynomial that satisfies the previous condition and this condition where $b=c$)



Is it possible to create a polynomial $p(x)$ with this relation between $p(0)$ and $p(c)$? -- Part 2
Suppose that $p$ is a polynomial with the required properties. Let $n:=\deg(p)$. Normalizing, we can assume that $p(0)=1$. Consequently, $0\le p(x)<1/b$ for any $x\in[1,c]$. As a result, the reciprocal polynomial $P(x):=x^np(1/x)$ is monic and satisfies $0<P(x)<x^n/b<1/b$ for any $x\in[c^{-1},1]$.


We now use the following fact: for any monic polynomial of degree $n$, one has $\max\_{x\in[-1,1]} |P(x)|\ge 2^{1-n}$ (equality holds for the Chebyshev polynomials of the first kind). As an easy corollary, any monic polynomial of degree $n$ deviates from $0$ on any interval of length $1/2$ by $2^{1-3n}$ at least. Assuming $c>2$ for simplicity, we thus conclude that $1/b>2^{-3n}$; that is, $n>\gamma \log(b)$, where $\gamma=3\log(2)$. Therefore, the degree of $p$ cannot be bounded by a quantity independent from $b$ and $c$.


