75th Romanian Mathematical Olympiad

a) For $a=0$ the result is obvious. For $a\ne 0$, consider the differentiable function $g:\mathbb{R}\to \mathbb{R}$ defined by $g(x)=F(x)\cdot e^{\frac{x}{a}}$. We have $$g^{\prime }(x)=f(x)\cdot e^{\frac{x}{a}}+\frac{1}{a}\cdot F(x)\cdot e^{\frac{x}{a}}=\frac{1}{a}\cdot e^{\frac{x}{a}}\cdot \left(F(x)+a\cdot f(x)\right).$$ For $a>0$ it is obvious that $g^{\prime }(x)\ge 0$, for $x\in \mathbb{R}$, implying that $g$ is non-decreasing. As ${\lim }_{x\to -\infty }g(x)={\lim }_{x\to -\infty }F(x)\cdot e^{\frac{x}{a}}=0$, we get $g(x)\ge 0$, for any $x\in \mathbb{R}$, thus $F(x)=g(x)\cdot e^{-\frac{x}{a}}\ge 0$, for $x\in \mathbb{R}$. If $a<0$, $g^{\prime }(x)\le 0$, for $x\in \mathbb{R}$, implying that $g$ is non-increasing. As ${\lim }_{x\to \infty }g(x)={\lim }_{x\to -\infty }F(x)\cdot e^{\frac{x}{a}}=0$, we obtain $g(x)\ge 0$, for $x\in \mathbb{R}$. It follows that $F(x)=g(x)\cdot e^{-\frac{x}{a}}\ge 0$, for any $x\in \mathbb{R}$.

b) Let $\mathcal{P}=\{f\mid f:\mathbb{R}\to \mathbb{R},f\text{ is a polynomial function}\}$. For any real $a$ consider the function $T_a:\mathcal{P}\to \mathcal{P}$, defined by $T_a(f)=f+a\cdot f^{\prime }$, i.e., $T_a(f)(x)=f(x)+a\cdot f^{\prime }(x)$, for any $x\in \mathbb{R}$. As for any $f\in \mathcal{P}$ and any $\alpha \in \mathbb{R}$, we have ${\lim }_{|x|\to \infty }\frac{f(x)}{e^{|\alpha \cdot x|}}=0$, by a), for real $a$, we have the implication $$f\in \mathcal{P}:T_a(f)(x)\ge 0,\forall x\in \mathbb{R}⟹f(x)\ge 0,\forall x\in \mathbb{R}.$$ Consider the roots $r_1,r_2,\dots ,r_n$ of $g$, and $s_k=-r_k$ for $k\in \{1,2,\dots ,n\}$ their opposites. Then $g=(X-r_1)(X-r_2)\dots (X-r_n)=(X+s_1)(X+s_2)\dots (X+s_n)$. By Vieta, we obtain the coefficients of $g$: $$a_k=\sum _{1\le i_1<i_2<\cdots <i_k\le n}{s}_{i_1}{s}_{i_2}\dots s_{i_k},\text{for any }k\in \{1,2,\dots ,n\}.$$ For $m\in \{1,2,\dots ,n\}$ and any $f\in \mathcal{P}$, we get $$\begin{gathered} (T_{s_m}\circ \cdots \circ T_{s_2}\circ T_{s_1})(f)=f+\left(\sum _{i=1}^m{s}_i\right)f^{\prime }+\left(\sum _{1\le i_1<i_2\le m}{s}_{i_1}{s}_{i_2}\right)f^{\prime \prime }+\dots \\ \cdots +\left(\sum _{1\le i_1<i_2<\cdots <i_k\le m}{s}_{i_1}{s}_{i_2}\dots s_{i_k}\right)f^{(k)}+\cdots +(s_1{s}_2\dots s_m)f^{(m)}.(Q(m)) \end{gathered}$$ If for $m\in \{1,2,\dots ,n-1\}$, $Q(m)$ is supposed to be true, we have $$\begin{array}{rl}(T_{s_{m+1}}\circ T_m\circ \dots T_{s_2}\circ T_{s_1})(f)& =T_{s_{m+1}}(T_m\circ \cdots \circ T_{s_1})(f)=\\ & =T_{s_{m+1}}\left(\sum _{J\subseteq \{1,2,\dots ,m\}}\left(\prod _{j\in J}{s}_j\right)f^{|J|}\right)=\\ & =\left(\sum _{J\subseteq \{1,2,\dots ,m\}}\left(\prod _{j\in J}{s}_j\right)f^{|J|}\right)+\\ & +s_{m+1}\cdot {\left(\sum _{J\subseteq \{1,2,\dots ,m\}}\left(\prod _{j\in J}{s}_j\right)f^{|J|}\right)}^{\prime }\\ & =\sum _{J_1\subseteq \{1,2,\dots ,m,m+1\}}\left(\prod _{j\in J_1}{s}_j\right)f^{|J_1|}\end{array}$$ so $Q(m+1)$ is also true.

From $Q(n)$, we have $$(T_{s_n}\circ \cdots \circ T_{s_2}\circ T_{s_1})(f)=f+a_1\cdot f^{\prime }+a_2\cdot f^{\prime \prime }+\cdots +a_n\cdot f^{(n)}.$$ By the hypothesis, we have $$(T_{s_n}\circ \cdots \circ T_{s_2}\circ T_{s_1})(f)(x)=f(x)+a_1\cdot f^{\prime }(x)+a_2\cdot f^{\prime \prime }(x)+\cdots +a_n\cdot f^{(n)}(x)\ge 0,$$ for all $x\in \mathbb{R}$. Successively, applying a), for $m\in \{1,2,\dots ,n\}$, we get $$(T_{s_m}\circ \cdots \circ T_{s_2}\circ T_{s_1})(f)(x)\ge 0,\text{for any }x\in \mathbb{R}.$$ In particular $f(x)\ge 0$, for any $x\in \mathbb{R}$.