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.

  • $\begingroup$ It doesn't make sense to say that $n$ depends on neither $c$ nor $b$. If you were to word it as "Does there exist an $n$ such that for all $b,c$, there exist a polynomial of degree $n$ such that ..." Also, your first condition can almost be simplified to "the absolute value is strictly increasing". $\endgroup$ Dec 17, 2020 at 1:43
  • $\begingroup$ @Acccumulation Noted and changed accordingly. $\endgroup$
    – DUO Labs
    Dec 17, 2020 at 2:22

3 Answers 3


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 $|p(0)|=1$.


Your first condition yields $$|p(c)|=|p(1)|+\int_{1}^c |p'(x)|dx:=\|p(x)\|.$$ All linear functionals on a finite-dimensional space are bounded, so if $\deg p\leqslant n$, we get $|p(0)|\leqslant C_n \|p(x)\|$ for certain $C_n$. Thus, if $b>C_n$, the second condition is not achievable.

  • 3
    $\begingroup$ I think this answer really gets to the point of what's going on here. I think it might be helpful to remark that both Fedor's solution and the one I gave revolve around similar ideas: using the first condition to bound the polynomial $p$, and deriving a contradiction by playing off this bound against the value of $p(0)$ using the second condition. However, in the proof I gave, these two steps are bound up together in the choosing to normalise everything such that $|p(0)|=1$, so I think Fedor's proof makes this underlying structure clearer. $\endgroup$ Sep 18, 2020 at 10:45
  • 2
    $\begingroup$ @AlexanderBetts Fedor is a genius. Your answer is good too. $\endgroup$
    – Nik Weaver
    Sep 18, 2020 at 15:35

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, $$ |p(0)|\leq \sum_{I=0}^n |a_i|\cdot |p(x_i)|\leq |p(c)|\cdot \sum_{I=0}^n |a_i|. $$


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