If f\in \mathbb{Z}[X]f\in \mathbb{Z}[X] has the property that |f(x)|<1, \forall x\in (-2, 2)|f(x)|<1, \forall x\in (-2, 2), then prove that f=0f=0.

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.

Answer

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.

Attribution
Source : Link , Question Author : ChemistryGeek , Answer Author : user159517

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