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Let $f(x)$ be squeezed by $\frac{x}{2} \le f(x) \le x^2 -2x +5$. Must the function $f(x) $ squeezed between these two polynomials be necessarily a polynomial?

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    $\begingroup$ Well it depends on the context. I don't see any context for the problem ... But I guess the answer is no. $\endgroup$
    – Matti P.
    Dec 17 '19 at 13:13
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    $\begingroup$ For example, $x-1\le x-\sin x\le x+1$. $\endgroup$
    – Berci
    Dec 17 '19 at 13:16
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If you want something given by a formula, consider for instance $$ f(x)=|x|^{3/2}. $$ It satisfies $$ \tfrac x2<f(x)<x^2-2x+5 $$ for all $x\in\mathbb R$.

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Consider

$$f(x)\le g(x)\le h(x)$$ and two arbitrary functions $\phi(x), \psi(x)>0$. Then

$$g(x)=\dfrac{\phi(x)f(x)+\psi(x)h(x)}{\phi(x)+\psi(x)}$$ satisfies the inequalities.

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  • $\begingroup$ I'd be clearer if you wrote $g(x)\le f(x)\le h(x)$. The function in the middle in OP's formulation is $f$. $\endgroup$
    – lhf
    Dec 17 '19 at 14:22
  • $\begingroup$ @fhl: nope, alphabetic order has advantages. $\endgroup$
    – user65203
    Dec 17 '19 at 14:24
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This function works, is $C^1$, but is not a polynomial: $$ f(x)= \begin{cases} 4 & x \le 1 \\x^2 -2x +5 & x \ge 1 \end{cases} $$

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