75th Romanian Mathematical Olympiad

(i) $\Rightarrow$ (ii)

Suppose $f:\mathbb{R}\to \mathbb{R}$ is continuously differentiable on $\mathbb{R}$. Let $a\in \mathbb{R}$, and two sequences $(x_n{)}_{n\ge 1}$ and $(y_n{)}_{n\ge 1}$, convergent to $a$, such that $x_n\ne y_n$, for $n\in {\mathbb{N}}^{\ast }$. By Lagrange theorem, there are points $a_n\in (\min \{x_n,y_n\},\max \{x_n,y_n\})$ such that $$\frac{f(x_n)-f(y_n)}{x_n-y_n}=f^{\prime }(a_n).$$ We deduce that the sequence $(a_n{)}_{n\ge 1}$ converges also to $a$. By the continuity of $f^{\prime }$, we obtain $$\underset{n\to \infty }{\lim }\frac{f(x_n)-f(y_n)}{x_n-y_n}=\underset{n\to \infty }{\lim }{f}^{\prime }(a_n)=f^{\prime }(a),$$ implying that the sequence ${\left(\frac{f(x_n)-f(y_n)}{x_n-y_n}\right)}_{n\ge 1}$ is convergent.

(ii) $\Rightarrow$ (i)

Let $f:\mathbb{R}\to \mathbb{R}$ be a function having property (ii). For $a\in \mathbb{R}$, there exists ${\ell }_a={\lim }_{n\to \infty }\frac{f(a+1/n)-f(a)}{1/n}\in \mathbb{R}$.

Consider two sequences $(x_n{)}_{n\ge 1}$ and $(y_n{)}_{n\ge 1}$ with limit $a$, such that $x_n\ne y_n$, for any $n\in {\mathbb{N}}^{\ast }$. Define sequences $(z_n{)}_{n\ge 1}$ and $(t_n{)}_{n\ge 1}$ by $z_{2n-1}=x_n$, $z_{2n}=a+1/n$, $t_{2n-1}=y_n$ and $t_{2n}=a$, for $n\in {\mathbb{N}}^{\ast }$. We have ${\lim }_{n\to \infty }{z}_n={\lim }_{n\to \infty }{t}_n=a$ and $z_n\ne t_n$, for any $n\in {\mathbb{N}}^{\ast }$. By hypothesis ${\left(\frac{f(z_n)-f(t_n)}{z_n-t_n}\right)}_{n\ge 1}$ is convergent. We get $$\underset{n\to \infty }{\lim }\frac{f(x_n)-f(y_n)}{x_n-y_n}=\underset{n\to \infty }{\lim }\frac{f(a+1/n)-f(a)}{1/n}={\ell }_a.$$ In particular, for any sequence $(x_n{)}_{n\ge 1}$ satisfying ${\lim }_{n\to \infty }{x}_n=a$ and $x_n\ne a$, $\forall n\in {\mathbb{N}}^{\ast }$, we have $$\underset{n\to \infty }{\lim }\frac{f(x_n)-f(a)}{x_n-a}={\ell }_a.$$ So, ${\lim }_{x\to a}\frac{f(x)-f(a)}{x-a}={\ell }_a$. As a conclusion $f$ is differentiable on $\mathbb{R}$, and $f^{\prime }(a)={\ell }_a$, $\forall a\in \mathbb{R}$.

Suppose, by contraposition, that $f^{\prime }$ is discontinuous at a point $a\in \mathbb{R}$. Then, there exists $\epsilon >0$ and a sequence $(a_n{)}_{n\ge 1}$ with limit $a$, and $a_n\ne a$, for all $n\in {\mathbb{N}}^{\ast }$, such that $|f^{\prime }(a)-f^{\prime }(a_n)|>\epsilon$, $\forall n\in {\mathbb{N}}^{\ast }$. As $f$ is differentiable at $a_n$, we can choose $x_n\in (a_n,a_n+1/n)$ such that $$|f^{\prime }(a_n)-\frac{f(x_n)-f(a_n)}{x_n-a_n}|<\frac{\epsilon }{2}.$$ The sequence $(x_n{)}_{n\ge 1}$ converges to $a$, $x_n\ne a_n$, for $n\in {\mathbb{N}}^{\ast }$, but $$|f^{\prime }(a)-\frac{f(x_n)-f(a_n)}{x_n-a_n}|\ge |f^{\prime }(a)-f^{\prime }(a_n)|-|f^{\prime }(a_n)-\frac{f(x_n)-f(a_n)}{x_n-a_n}|>\frac{\epsilon }{2},$$ for all $n\in {\mathbb{N}}^{\ast }$, in contradiction with ${\lim }_{n\to \infty }\frac{f(x_n)-f(a_n)}{x_n-a_n}=f^{\prime }(a)$. In conclusion, $f^{\prime }$ is continuous on $\mathbb{R}$.