67th Romanian Mathematical Olympiad

$$a=\inf \{t\in (-\infty ,x_0)\mid f^2(x)>0,\forall x\in [t,x_0]\},$$

$$b=\sup \{t\in (x_0,\infty )\mid f^2(x)>0,\forall x\in [x_0,t]\}.$$ We have $-\infty \le a<x_0<b\le \infty$ and $f(x)\ne 0,\forall x\in (a,b)$. Since $f=(f^2{)}^{\prime }$ has the intermediate value property (following from the theorem of Darboux), we deduce that $f$ has a constant sign on the interval $(a,b)$. If $f(x)>0,\forall x\in (a,b)$, then $f=\sqrt{f^2}$ on $(a,b)$, hence $f$ is differentiable on $(a,b)$. From the hypothesis we obtain $2f(x)f^{\prime }(x)=f(x)$, $\forall x\in (a,b)$, therefore $f^{\prime }(x)=\frac{1}{2},\forall x\in (a,b)$. We deduce that for some $c\in \mathbb{R}$, we have $f(x)=\frac{x}{2}+c$, $x\in (a,b)$. Since $f(x)>0,\forall x\in (a,b)$, we obtain $a\ge -2c$ (so $a$ is finite). Because $f^2$ is continuous, it results that $f^2(a)=0$, and then $c=-\frac{a}{2}$, hence $f(x)=\frac{x-a}{2}$, $x\in [a,b)$. If $b$ were finite, we would obtain $0=f^2(b)={\lim }_{x\uparrow b}{\left(\frac{x-a}{2}\right)}^2=\frac{(b-a{)}^2}{4}$; contradiction. It follows that $b=\infty$ and that $f(x)=\frac{x-a}{2},x\in [a,\infty )$.

Moreover, $f(x)\le 0,\forall x\in (-\infty ,a)$. Indeed, if there were some $x_1<a$ such that $f(x_1)>0$, then $f(a)>0$ which is a contradiction. If $f(x)<0,\forall x\in (a,b)$, we obtain, in a similar way that $b$ is finite, $f(x)=\frac{x-b}{2}$, $x\in (-\infty ,b]$ and $f(x)\ge 0,\forall x\in (b,\infty )$.

We conclude that, along the null function, the following types of functions $f:\mathbb{R}\to \mathbb{R}$ are solutions: $$1.f(x)=\{\begin{array}{ll}0,& x<a\\ \frac{x-a}{2},& x\ge a\end{array},a\in \mathbb{R};$$ $$2.f(x)=\{\begin{array}{ll}\frac{x-b}{2},& x\le b\\ 0,& x>b\end{array},b\in \mathbb{R};$$ $$3.f(x)=\{\begin{array}{ll}\frac{x-b}{2},& x\le b\\ 0,& x\in (b,a)\\ \frac{x-a}{2},& x\ge a\end{array},a,b\in \mathbb{R},b<a;$$ $$4.f(x)=\frac{x}{2}+c,x\in \mathbb{R},c\in \mathbb{R}.$$