Is there a polynomial f with integer coefficients that satisfies the following criteria:
- f is not constant;
- for all $x\in[0,1]$, $1-\frac{1}{x}\leq f(x)\leq \frac{1}{x}$;
- For all $x\in [1,4]$, $0\leq f(x)\leq 1$.
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$\begingroup$ We have $f(1), f(2), f(3), f(4) \in \{0, 1\}$, so there are at most 16 possibilities to $f(x) \mod (x-1)(x-2)(x-3)(x-4)$. This modulus must also have integer coefficients (because $(x-1)(x-2)(x-3)(x-4)$ is monic), and checking all the possibilities, it must be either $0$ or $1$. $\endgroup$– Daniel WeberCommented Feb 13 at 17:25
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$\begingroup$ Because $f(1), f(2), f(3), f(4)$ must actually be extrema, we can get the value $\mod (x-1)^2 (x-2)^2 (x-3)^2 (x-4)^2$ $\endgroup$– Daniel WeberCommented Feb 13 at 17:32
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$\begingroup$ @Yanlong Hao Doesn't the technique in David Speyer's answer of your very similar previous question mathoverflow.net/questions/458834 apply here? $\endgroup$– Peter MuellerCommented Feb 13 at 22:57
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$\begingroup$ @PeterMueller Thanks for asking. The $L_\infty$ bounds here is $\infty$. Even considering the function g(x)=xf(x), the possible $L_\infty$ bounds after shift down is possible close to 2.5. Hence it is related but does not answer the question.. $\endgroup$– Yanlong HaoCommented Feb 14 at 0:02
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1 Answer
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Set $g(x)=x\cdot(f(x)-\frac{1}{2})$. Then the maximum of $\lvert g(x)\rvert$ on $[0,4]$ is at most $2$. From David Speyer's answer of a similar question and the reference given there, it should follow that $g(x)=2T_n(\frac{x}{2}-1)$, where $T_n$ is the Chebyshev function of degree $n$, and $n$ is the degree of $g$. However, $g(0)=0$, while $-1$ is not a root of $T_n$. So there is no such $f$.
answered Feb 14 at 15:54