The question is easy to state: Is there a non-constant $f\in\mathbb{Z}[x]$ such that for all $x\in [0,4]$, we have $|f(x)|\leq 1$? I do not know where to find a useful reference for it. I did a few tries, and the best I can get is some variation of $(x-1)(x-2)(x-3)(x-4)$, which could take a small value when $0.72<x\leq 4$. I can also show that there does not exist such a $f$ for $[-\frac{\sqrt{2}}{2},4]$, but it is probably a bad estimate.
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5$\begingroup$ I would be interested to know the idea for the proof of the $[-1/\sqrt 2,4]$ result! $\endgroup$– Ethan DlugieCommented Nov 21, 2023 at 16:27
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No. Integer coefficients are a red herring; the point is that, if $f(x) = f_n x^n + \cdots + f_0$ is a real polynomial with $f_n \neq 0$, then the $L_{\infty}$ norm of $f(x)$ on an interval of the form $[a, a+4]$ is at least $2 |f_n|$. So, if $f_n$ is a nonzero integer, this is $\geq 2$.
See Wikipedia for the analogous result for $L_{\infty}$ norm on $[-1, 1]$. To deduce the corresponding formula on $[0,4]$, make the change of variable $y = 2x+2$ and keep track of the powers of $2$.
answered Nov 21, 2023 at 15:55
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1$\begingroup$ The red herring does suggest the follow-up question: What is $M=\inf\|p\|_{L^{\infty}(0,4)}$ when we minimize over non-constant $p\in\mathbb Z[x]$? Your answer shows that $M\ge 2$, but presumably $M>2$. $\endgroup$ Commented Nov 21, 2023 at 17:53
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4$\begingroup$ In fact, it's still $M=2$. To see this, define the polynomial $T_n$, monic of degree $n$, by $T_n(z+1/z)=z^n+1/z^n$. Note that $T_n$ has integer coefficients. Then put $z:=e^{i\theta}$, giving $T_n(2\cos\theta)=2\cos(n\theta)$. Thus $|T_n(x)|$ is bounded by $2$ on $[-2,2]$. Hence $|T_n(x-2)|$ is bounded by $2$ on $[0,4]$. (Note that $T_n$ is closely related to the standard Chebychev polynomial $t_n(x)$, which is defined by $t_n(\cos\theta)=\cos(n\theta)$.) $\endgroup$– Chris76Commented Nov 22, 2023 at 10:50
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$\begingroup$ @Chris76 How can I extend your result for arbitrary intervals $[-B, B]$? Is it still the case, for larger intervals, that the real-coefficients and integer-coefficients case have the same lower bound? $\endgroup$ Commented Feb 18 at 21:14