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

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.

• 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. – N. S. Sep 11 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.