75th Romanian Mathematical Olympiad

a) If $f([a,b])\subseteq [0,\infty )$ or $f([a,b])\subseteq (-\infty ,0]$, then $m=|f(\frac{a+b}{2})|>0$, and on one of the intervals $(a,\frac{a+b}{2})$ or $(\frac{a+b}{2},b)$ the inequality $|f(x)|>m$ holds for any $x$ in that interval. Then $$\begin{array}{rl}0& =\left|{\int }_a^{b}f(x)dx\right|={\int }_a^b|f(x)|dx={\int }_a^{\frac{a+b}{2}}|f(x)|dx+{\int }_{\frac{a+b}{2}}^b|f(x)|dx\\ & \ge m\cdot \frac{b-a}{2}>0,\end{array}$$ which is impossible. It follows that $f(a)\cdot f(b)<0$.

b) We will show that the only sequences of real numbers $(a_n{)}_{n\ge 1}$ which satisfy the condition in the statement are the constant sequences and the sequences assuming exactly two distinct real values, which become stationary beginning with a certain rank. Obviously, if $(a_n{)}_{n\ge 1}$ is a constant sequence, with $a_n=a$, $a\in \mathbb{R}$, then the sequence is convergent, with ${\lim }_{n\to \infty }{a}_n=a$ and ${\int }_{a_k}^{a_{k+1}}f(x)dx=0$, for any $k\ge 1$, and any function $f:\mathbb{R}\to \mathbb{R}$. If $\{a_n|n\ge 1\}=\{a,b\}$ and there is a rank $n_0\in {\mathbb{N}}^{\ast }$ with $a_n=a$, for any $n\ge n_0$, then $(a_n{)}_{n\ge 1}$ is convergent, with ${\lim }_{n\to \infty }{a}_n=a$, and there is the function $f:\mathbb{R}\to \mathbb{R}$ defined by $f(x)=2x-a-b$, which is strict monotone and satisfies the equalities ${\int }_{a_k}^{a_{k+1}}f(x)dx=0$, for any $k\ge 1$.

Let $(a_n{)}_{n\ge 1}$ be a convergent sequence of real numbers for which there is a strict monotone function $f:\mathbb{R}\to \mathbb{R}$ such that $${\int }_{a_1}^{a_2}f(x)dx={\int }_{a_2}^{a_3}f(x)dx=\cdots ={\int }_{a_n}^{a_{n+1}}f(x)dx=\dots$$

Let $I={\int }_{a_k}^{a_{k+1}}f(x)dx$, for every $k\ge 1$, and $a={\lim }_{n\to \infty }{a}_n$. For a fixed $r>0$ there is a rank $n_r\ge 1$, such that $a_n\in (a-r,a+r)$, for any $n\ge n_r$. Since $f$ is strict monotone, if $M=\max (|f(a-r)|,|f(a+r)|)$, then $|f(x)|<M$, for every $x\in (a-r,a+r)$. Then, for any $p\in {\mathbb{N}}^{\ast }$ we have: $$\begin{array}{rl}p\cdot |I|& =|p\cdot I|=\left|\sum _{k=1}^p{\int }_{a_{n_r+k-1}}^{a_{n_r+k}}f(x)dx\right|=\left|{\int }_{a_{n_r}}^{a_{n_r+p}}f(x)dx\right|\\ & \le \left|{\int }_{a_{n_r}}^{a_{n_r+p}}|f(x)|dx\right|<{\int }_{a-r}^{a+r}|f(x)|dx\le 2r\cdot M.\end{array}$$ It follows that $0\le |I|<\frac{2Mr}{p}$, for any $p\in {\mathbb{N}}^{\ast }$, so that $I=0$. We show now that $\text{card}(\{a_n|n\in {\mathbb{N}}^{\ast }\})\le 2$. Assuming the opposite, there exist $i,j,k\in {\mathbb{N}}^{\ast }$, with $i<j<k$ such that $a_i\ne a_j\ne a_k\ne a_i$. Then $${\int }_{a_i}^{a_j}f(x)dx=\sum _{l=i}^{j-1}{\int }_{a_l}^{a_{l+1}}f(x)dx=(j-i)\cdot I=0$$ and, also, ${\int }_{a_j}^{a_k}f(x)dx=(k-j)\cdot I=0$, respectively ${\int }_{a_i}^{a_k}f(x)dx=0$. The function $f$ being strict monotone, it follows that $f(a_i)\cdot f(a_j)<0$, $f(a_i)\cdot f(a_k)<0$, and $f(a_j)\cdot f(a_k)<0$. But then $$(f(a_i)\cdot f(a_j)\cdot f(a_k){)}^2=(f(a_i)\cdot f(a_j))\cdot (f(a_i)\cdot f(a_k))\cdot (f(a_j)\cdot f(a_k))<0,$$ which is absurd.