Is it possible to create a polynomial $p(x)$ with this relation between $p(0)$ and $p(c)$? -- Part 2 Note: This question is based on a previous question
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 $b=c$)
 A: 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 $c$.
A: I think it is not true, not even if we allow $n$ depend on $c$, for the simple reason that all linear forms are continuous on a finite dimensional normed space.
Given $n\in\mathbb{N}$, and $c>1$, consider  the $n+1$ dimensional linear space $V$ of polynomials of degree less than or equal to $n$  with the norm $\|\cdot\|_{\infty,[1,c]}$. The evaluation at $0$ is a continuous linear form on this space, therefore there exists $b>0$ such that for all $p\in V$ there holds $|p(0)|\le b\|p\|_{\infty,[1,c]}$. Therefore,  for the couple $(b,c)$, there is no $p$ as wanted.
