Romanian Mathematical Olympiad

We shall prove that the answer is $f:\mathbb{R}\to \mathbb{R}$, $f(x)=mx+n$, with $m,n\in \mathbb{R}$. It is trivial to see that this family of functions satisfy the property.

For the converse, let $f$ with the given property. Consider $a,b\in \mathbb{R}$, $a<b$, and define $$m=\frac{f(b)-f(a)}{b-a}\text{and}n=\frac{bf(a)-af(b)}{b-a}.$$ We shall show that $f(x)=mx+n$, for all $x\in [a,b]$. By way of contradiction, let $x_0\in [a,b]$, such that $f(x_0)\ne m{x}_0+n$. Consider the sets $$A=\{x\in [a,x_0]\mid f(x)=mx+n\}\text{and}B=\{x\in [x_0,b]\mid f(x)=mx+n\}.$$ $A$ and $B$ are nonempty because $a\in A$ and $b\in B$. Consider the numbers $\alpha =\sup (A)$ and $\beta =\inf (B)$. By the continuity of $f$ we get $\alpha \in A$, $\beta \in B$ and $\alpha <x_0<\beta$. By the hypothesis, there is $t\in (0,1)$ such that $$f((1-t)\alpha +t\beta )=(1-t)f(\alpha )+tf(\beta )=m((1-t)\alpha +t\beta )+n.$$ Then $\gamma =(1-t)\alpha +t\beta \in (\alpha ,\beta )$ and $f(\gamma )=m\gamma +n$, in contradiction with the definition of $\alpha$ and $\beta$. So, $f(x)=mx+n$, for all $x\in [a,b]$.

To extend the answer to $\mathbb{R}$, observe that if $f:[a,c]\to \mathbb{R}$, $c>b$ is a function obtained from the preceding considerations, say $f(x)=m^{\prime }x+n^{\prime }$, we must have by continuity $m=m^{\prime }$, $n=n^{\prime }$. The same remark can be made on intervals of the form $[c,b]$ with $c<a$.