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 $b=c$)


2 Answers 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 $c$.

  • $\begingroup$ I'm having trouble with your quantifiers involving $\kappa$. Does the statement mean: $\exists \kappa > 0\,\forall p(x) = x^n + \dotsb\,\forall a \in \mathbb R\,\exists x \in \mathbb R,\quad\text{($\lvert x - a\rvert < 1/4$ and $\lvert p(x)\rvert \ge \kappa^n$)}$? $\endgroup$
    – LSpice
    Sep 26, 2020 at 19:49
  • 1
    $\begingroup$ @LSpice: Yes, this seems a correct restatement to me. I am trying to find a reference. $\endgroup$
    – Seva
    Sep 26, 2020 at 19:55
  • $\begingroup$ (Actually I guess it's $\exists\kappa\,\forall n\,\cdots$.) I asked because I thought at first reading that you meant that the polynomial was bounded away from $0$, which didn't make sense to me. $\endgroup$
    – LSpice
    Sep 26, 2020 at 21:39
  • $\begingroup$ @Seva Thanks! +1 and accept! $\endgroup$
    – DUO Labs
    Sep 27, 2020 at 21:00

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.

  • $\begingroup$ (after writing this answer I saw the preceding question already gave a similar explanation. In fact, both versions of the question are quite explicitly asking for a discontinuous linear form on a finite dimensional normed space) $\endgroup$ Sep 27, 2020 at 12:37

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