Must a function squeezed by two polynomials be necessarily a polynomial?

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?

• Well it depends on the context. I don't see any context for the problem ... But I guess the answer is no. Dec 17 '19 at 13:13
• For example, $x-1\le x-\sin x\le x+1$. Dec 17 '19 at 13:16

If you want something given by a formula, consider for instance $$f(x)=|x|^{3/2}.$$ It satisfies $$\tfrac x2 for all $$x\in\mathbb R$$.

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.

• 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 $f$.
– lhf
Dec 17 '19 at 14:22
• @fhl: nope, alphabetic order has advantages.
– user65203
Dec 17 '19 at 14:24

This function works, is $$C^1$$, but is not a polynomial: $$f(x)= \begin{cases} 4 & x \le 1 \\x^2 -2x +5 & x \ge 1 \end{cases}$$