74th Romanian Mathematical Olympiad

a) Let $\epsilon >0$, be arbitrary. By the hypothesis there are ${\delta }_1,M>0$ such that $|f(x)|<M$, for all $x\in (-{\delta }_1,{\delta }_1)$. As $g$ is continuous at $0$, we get a ${\delta }_2>0$, depending on $\epsilon$, such that $|g(x)-g(0)|<\frac{\epsilon }{2}$ for any $x\in (-{\delta }_2,{\delta }_2)$. Let $\delta =\min \{{\delta }_1,{\delta }_2,1\}$. From $0<\delta \le 1$, we get $a^2<\delta ,\forall a\in (-\delta ,\delta )$.

Let $x\in (-\delta ,\delta )$. As $|x|<\delta \le {\delta }_2$, we have $$\begin{array}{rl}|f(x)-f(0)|& =\frac{|2f(x)-2f(0)|}{2}\le \frac{|2f(x)+f(x^2)-3f(0)|+|f(x^2)-f(0)|}{2}\\ & =\frac{|g(x)-g(0)|}{2}+\frac{|f(|x{|}^2)-f(0)|}{2}<\frac{\epsilon }{4}+\frac{|f(|x{|}^2)-f(0)|}{2}.\end{array}$$ Inductively, we have $$|f(x)-f(0)|<\epsilon \left(\frac{1}{2^2}+\frac{1}{2^3}+\cdots +\frac{1}{2^{n+1}}\right)+\frac{|f(|x{|}^{2^n})-f(0)|}{2^n},n\in {\mathbb{N}}^{\ast }$$

b) For $a\in (0,1)$ and $A=\{\pm a^{2^n},n\in \mathbb{Z}\}$, define $$f:\mathbb{R}\to \mathbb{R},f(x)=\{\begin{array}{ll}(-1{)}^n{2}^n,& |x|=a^{2^n},n\in \mathbb{Z}\\ 0,& x\in \mathbb{R}\setminus A\end{array}.$$ Because ${\lim }_{k\to \infty }{a}^{2^{2k}}=0$ and ${\lim }_{k\to \infty }f(a^{2^{2k}})={\lim }_{k\to \infty }{2}^{2k}=\infty$, $f$ is discontinuous at the origin. For $x\in \mathbb{R}\setminus A$, $x^2\in \mathbb{R}\setminus A$, so $g(x)=0$. For $x\in A$, there is $n\in \mathbb{Z}$ such that $|x|=a^{2^n}$, so $$g(x)=2f(a^{2^n})+f(a^{2^{n+1}})=2^{n+1}[(-1{)}^n+(-1{)}^{n+1}]=0.$$ Therefore, $g$ is the null function.