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.



Is it possible to create a polynomial $p(x)$ with this relation between $p(0)$ and $p(c)$?
No, it's not possible to construct such a $p$ whose degree $n$ is bounded independently of $b$ and $c$. In fact, it's not possible even if we fix the value of $c$. I'll prove this for $c=2$ below, but the same argument works in general.


Suppose to the contrary that it were possible. Then for every $b>1$ we could choose a polynomial $p\_b$ such that


* $p\_b$ has degree $n$;
* $p\_b$ takes non-negative values and is strictly increasing on $[1,2]$;
* $b\cdot|p\_b(2)|<|p\_b(0)|=1$.


(We take a polynomial satisfying the desired conditions and multiply by an appropriate scalar.)


*Key claim:* The coefficients of a polynomial $p\_b$ satisfying these three conditions are bounded independently of $b$. That is, there is a constant $B$ such that the absolute value of every coefficient of every polynomial $p\_b$ is at most $B$.


*Proof of claim:* Fix any $n+1$ distinct real numbers $x\_0,x\_1,\dots,x\_n$ between $1$ and $2$ inclusive, and for $i=0,1,\dots,n$ define the polynomial $f\_i(x)$ by$$f\_i(x)=\prod\_{j\neq i}\frac{x-x\_j}{x\_i-x\_j} \,,$$where the product is taken over all indices $i=0,1,\dots,n$ except $i=j$. Thus, $f\_i$ is the unique degree $n$ polynomial such that $f\_i(x\_i)=1$ and $f\_i(x\_j)=0$ for $j\neq i$.


Now the theory of Lagrange interpolation says that for any polynomial $p$ of degree at most $n$, we have
$$p(x) = \sum\_{i=0}^np(x\_i)f\_i(x) \,.$$
But in our case, we know that we have
$$|p\_b(x\_i)|\leq|p\_b(2)|\leq b^{-1}\cdot|p(0)|<1$$
for every $b$, since $p\_b$ is strictly increasing on $[1,2]$. Thus, the absolute value of every coefficient of every polynomial $p\_b=\sum\_{i=0}^np(x\_i)f\_i$ is at most $(n+1)B'$, where $B'$ is the largest absolute value of any coefficient of any of the $f\_i$. This gives our bound independently of $b$, and proves the key claim.


Now by a compactness argument (a.k.a. the Bolzano--Weierstraß Theorem), our key claim implies that we may choose an increasing sequence of integers $b\_1<b\_2<\dots$ such that the polynomials $p\_{b\_i}$ converge coefficientwise to a polynomial $p$. What can we say about this limiting polynomial $p$? Well, by taking an appropriate limit of the above properties of the $p\_{b\_i}$, we find:


* $p$ has degree $n$;
* $p$ takes non-negative values and is weakly increasing on $[1,2]$;
* $|p(0)|=1$; and
* $|p(2)|\leq b\_i^{-1}$ for every $i$.


Since the integers $b\_i$ increase without bound, this final condition implies that actually $|p(2)|=0$. Since $p$ is non-negatively valued and weakly increasing on $[1,2]$, we find that $p$ actually has to be equal to $0$ on all of $[1,2]$. This implies that $p$ must be the zero polynomial. But this contradicts the assumption that $|p(0)|=1$.


