# 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 $$b=c$$)

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 $$0 for any $$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 $$c$$.

• 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$)}$? Sep 26, 2020 at 19:49
• @LSpice: Yes, this seems a correct restatement to me. I am trying to find a reference.
– Seva
Sep 26, 2020 at 19:55
• (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. Sep 26, 2020 at 21:39
• @Seva Thanks! +1 and accept! 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.

• (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) Sep 27, 2020 at 12:37