Polynomials $f$ with integer coefficients such that $f(x) \geq 0$ on $[-2,2]$ and $f(x) \leq \frac{1}{1+x}$ on $(-1,2]$

Find polynomials with integer coefficients $f\in\mathbb{Z}[x]$ such that $f(x)\ge 0$ on $x\in[-2,2]$ and $\frac{1}{1+x}\ge f(x)$ on $x\in(-1,2]$.

I guess only such polynomial is just $0$, but it seems hard to prove. I found an example $f(x)=x^2(x+1)(x-1)^2(x-2)^2$ satisfying $0\le f(x)\le \frac{1}{1+x}$ on $x\in(-1,2]$, but it satisfies $f(x)<0$ on $x\in [-2,-1)$. Are there nontrivial (i.e., other than the zero function) examples?

Solution 1:

This is not true, there are many polynomials satisfying much tighter inequalities. I will show:

Key Claim For any $\epsilon > 0$, there is a polynomial $f$ with integer coefficients obeying $|f(x)| < \epsilon$ for $x \in [-1,2]$.

Therefore, $f(x)^2 \geq 0$ everywhere and, if we choose $\epsilon < 1/\sqrt{3}$, then $f(x)^2 < 1/3 < 1/(1+x)$ on $(-1,2]$. Of course, using smaller values of $\epsilon$, we could instead look at $f_1^2+f_2^2$, where $|f_1|$, $|f_2|<\epsilon$, or $f^2+f^4$, or many more complicated things. I think there is no help of classifying such polynomials.

Before getting into the general theory, let me prove that at least one such polynomial exists by exhibiting one: Set $$g(x) = x^{10}-5 x^9+6 x^8+6 x^7-14 x^6+9 x^4-x^3-2 x^2=(x-2) (x-1)^2 x^2 (x+1) (x^2-x-1)^2.$$ Then the minimum value of $|g|$ on $[-1,2]$ is $\approx 0.431058 < 1/\sqrt{3} \approx 0.57735$. So $g^2$ meets the criteria.

I knew the Key Claim held due to a theorem of Fekete: Let $a<b<a+4$ be real numbers. Then there is a polynomial $p$ with integer coefficients such that $$|p(x)| \leq 2^{1-2^{-n-1}} (n-1) \left(\frac{b-a}{4} \right)^{n/2} \ \mbox{for} \ x \in [a,b].$$ In particular, taking $[a,b]=[-1,2]$, the right hand side goes to $0$ as $n \to \infty$, proving the claim.

Here is a constructive proof of the Key Claim.

Lemma For every $n$, there is a degree $n$ polynomial $p_n(x)$ with integer coefficients such that, if we expand $p_n(3/2 \cos \theta + 1/2)$ as a (finite)Fourier series, the coefficient of $\cos (m \theta)$ is at most $(3/4)^m$.

Proof Sketch We choose the coefficients of $p_n$ starting with $x^n$ and working down. We take the coefficient of $x^n$ to be $1$. For all lower terms, we choose the coefficient of $x^m$ which makes the coefficient of $\cos (m \theta)$ as small as possible. Since $\left( (3/2) \cos \theta + 1/2 \right)^m = 2 (3/4)^m \cos (m \theta) + \cdots$, we can always make the coefficient of $\cos (m \theta)$ be at most $(3/4)^m$. $\square$

The polynomial $g$ above was chosen by the above method with $n=10$. We have $$g((3/2) \cos \theta + 1/2) = -0.136745 - 0.0875816 \cos(2 \theta) - 0.0482025 \cos (4 \theta) + 0.0848179 \cos(6 \theta) + 0.0750847 \cos(8 \theta) + 0.112627 \cos(10 \theta).$$

In the above example, we got lucky that a single polynomial was small enough. In general, choose $N$ large enough that $\sum_{m=N}^{\infty} (3/4)^m < \epsilon/3$. Then, by the pigeonhole principle, there must be two indices $i$ and $j$ such that the first $N$ Fourier terms of $p_i(3/2\cos \theta+1/2) - p_j(3/2 \cos \theta+1/2)$ are all $< \epsilon/(3 N)$. Then $p_i-p_j$ satisfies the Key Claim, since $(3/2) \cos \theta+1/2$ parametrizes $[-1,2]$ and $|\cos (m \theta)| \leq 1$.

References: Fekete's original paper (German), an English exposition and generalization of Fekete's result which inspired the above proof, a very readable paper on small integer polynomials.