On non-negativeness of continuous and polynomial functions

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On non-negativeness of continuous and polynomial functions

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Source: Romania NMO 2025 12.3

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#1 h Apr 2, 2025, 2:51 PM • 1 Y

Y by IvannavI

a) Let $a\in \mathbb{R}$ and $f \colon \mathbb{R} \to \mathbb{R}$ be a continuous function for which there exists an antiderivative $F \colon \mathbb{R} \to \mathbb{R}$ , such that $F(x)+a\cdot f(x) \geq 0$ , for any $x \in \mathbb{R}$ , and $\lim_{|x| \to \infty} \frac{F(x)}{e^{|\alpha \cdot x|}}=0$ holds for any $\alpha \in \mathbb{R}^*$ . Prove that $F(x) \geq 0$ for all $x \in \mathbb{R}$ .
b) Let $n\geq 2$ be a positive integer, $g \in \mathbb{R}[X]$ , $g = X^n + a_1X^{n-1}+ \dots + a_{n-1}X+a_n$ be a polynomial with all of its roots being real, and $f \colon \mathbb{R} \to \mathbb{R}$ a polynomial function such that $f(x)+a_1\cdot f'(x)+a_2\cdot f^{(2)}(x)+\dots+a_n\cdot f^{(n)}(x) \geq 0$ for any $x \in \mathbb{R}$ . Prove that $f(x) \geq 0$ for all $x \in \mathbb{R}$ .

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#2 h Apr 4, 2025, 11:27 AM • 1 Y

Y by soryn

I can only do the first part.
a) We may assume $a\not=0$ . Set $k:=1/a$ and $g(x)=e^{kx}F(x)$ then $g'(x)=e^{kx}(kF+f)=ke^{kx} (F+af)$ .
If $a>0$ then $g'\ge 0$ and $g(-\infty)=\lim_{x\rightarrow -\infty } e^{kx}F(x)=\lim {{F(x)}\over {e^{|kx|}}}=0$ . Hence $g(x)\ge 0$ for all $x$ and the same holds for $F$ . If $a<0$ then $g'\le 0, g(\infty)=0$ and we get the same result.

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#4 h Apr 11, 2025, 1:07 PM

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If you've had an introductory course in differential equations, then you know what to do. If $D:\mathcal{C}(\mathbb{R})\to \mathcal{C}(\mathbb{R})$ denotes the differentiation operator then the statement becomes $g(D)(f)\geq 0$ . If you write $g=(X-x_1)\dots(X-x_n)$ where $x_i\in\mathbb{R}$ then we're done by induction using a).
I don't believe this problem is appropriate for a High-School Olympiad, as it's impossible to do without having seen the trick before.

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