# minimal value polynomials with integer coefficients

Let $D$ be the set of polynomials of integer coefficients $f\in\mathbb{Z}[x]$ such that $f(x)\ge 0$ at $x\in[-2,2]$, where the zero polynomial $f=0$ is excluded. Can I find a finite "minimal" set $F\subseteq D$ on $[-2,2]$? That is, $F$ is finite and for each $f\in D$, there exists $g\in F$ such that $f(x)\ge g(x)$ for every $x\in[-2,2]$. If not, at least can I find a finite minimal set of each degree? I don't know even how I approach this problem, except for the degree 1 case..

• Why not $F:=\{0\}$? – Guest May 17 '15 at 5:35
• Ah, it's my mistake. We must trivially exclude the zero. – D. Lee May 17 '15 at 5:37

Let $n$ be a natural number greater than the degrees of every polynomial in $F$, and consider $x^n$. We must have $0 \le g(x) \le x^n$ for all $x\in [-2,2]$. Note that in particular we have $g(0) = 0$, and since $g(x) \ge 0$ in a neighborhood, the nonzero coefficient of lowest degree must be positive. However, $x^n < |g(x)|$ for $x$ suitably close to $0$ for such polynomials $g$, since that lowest coefficient must necessarily be in degree less than $n$. This contradiction shows that no such $F$ can exist.
If you want counterexamples in a specified degree, note that each such polynomial intersects the $x$-axis finitely often, and therefore the set of those $x$ coordinates such that $g(x) = 0$ is finite. You can choose a rational $p/q$ that is not in this set, and then set $f(x) = (qx-p)^2h(x)$, where $h(x)$ is any polynomial that is positive on $[-2,2]$. Then $f(p/q) = 0 < g(p/q)$ for all $g$ in your finite set $F$. This covers every case except the case of linear functions.
In that case choose a point $(r_1,r_2)$ lying below the graphs of every element of $F$ (such a point exists if you take $r_1 = p/q$ as above, and then set $r_2$ less than all values of $g(p/q)$). Then the line between the points $(r_1, r_2)$ and $(-2, 0)$ will provide the counterexample.