Let $f\in \mathbb{Z}[X]$ such that $|f(x)|<1, \forall x\in (-2, 2)$. Prove that $f=0$.
I couldn't make too much progress on this problem. I tried considering $f=a_nX^n+a_{n-1}X^{n-1}+...+a_1X+a_0$ and by setting $x=0$ in the hypothesis I got that $|a_0|<1$ and this doesn't look useful.
Then I thought about looking at $f$'s degree, but I couldn't make any observations on this.
I believe that the key of the problem should be that the polynomial's coefficients are integers, but I don't know how to use that. Apart from the Rational Root Theorem (which doesn't seem useful here) I don't have in mind other results regarding polynomials with integer coefficients.

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    $\begingroup$ You can show that there exists some $Q(x) \in \mathbb Z[X]$ such that $$f(x)=x(x-1)(x+2)(x-2)(x+2)Q(x)$$ but I do not see how this helps. $\endgroup$ – N. S. Sep 11 '20 at 18:21

I can provide you with an answer based on Chebyshev polynomials. Suppose that $f$ satisfies the stated assumptions, let $k$ be the degree of $f$, such that we have $a_k \neq 0$ and define $g(x) := f(x)/a_k$. As $a_k \in \mathbb{Z}$ we must have $|a_k|\geq 1$ and therefore $|g(x)| = |f(x)/a_k| < 1$ for $x \in (-2,2)$. Define the $n$-th Chebyshev polynomial $$ T_n(x) := \cos(n\arccos(x)) $$ on $[-1,1]$. Note that $2^{1-n}T_n(x)$ has minimal supremum norm among all monic polynomials of degree $n$ on $[-1,1]$ for $n \geq 1$ (see here). By a rescaling argument, it follows that $2T_n(x/2)$ has minimal supremum norm among all monic polynomials of degree $n$ on $[-2,2]$ for $n \geq 1$. It is easy to see from the definition that for $n \geq 1$ it holds that $$\sup_{x \in [-2,2]} 2|T_n(x/2)| = 2.$$ By continuity of $g$, we have $$\sup_{x \in [-2,2]} |g(x)| \leq 1,$$ and therefore we must have $k < 1$, as otherwise we would have found a monic polynomial with smaller supremum norm than the Chebyshev polynomial of the corresponding degree, a contradiction. It follows that $g$ is constant and therefore equal to $\frac{a_0}{a_k}$. Because of $|f(0)| < 1$ we must have $|a_0| < 1$ and as $a_0 \in \mathbb{Z}$ it follows that $a_0 = 0$, therefore $g \equiv 0 $ and hence $f \equiv 0$, which contradicts the assumption that $a_k \neq 0$. Therefore, the only polynomial satisfying the hypotheses is the zero polynomial.

  • $\begingroup$ Ah, the beautiful Chebyshev polynomials! +1 $\endgroup$ – Teresa Lisbon Sep 12 '20 at 3:50

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