Home > Algebra, IMO, Olympiad, Problem Solving > Miklos Schweitzer 2013 Problem 8

## Miklos Schweitzer 2013 Problem 8

Problem 8. Let ${f : \Bbb{R} \rightarrow \Bbb{R}}$ be a continuous and strictly increasing function for which

$\displaystyle f^{-1}\left(\frac{f(x)+f(y)}{2}\right)(f(x)+f(y)) =(x+y)f\left(\frac{x+y}{2}\right)$

for all ${x,y \in \Bbb{R}}$ (${f^{-1}}$ denotes the inverse of ${f}$). Prove that there exist real constants ${a \neq 0}$ and ${b}$ such that ${f(x)=ax+b}$ for all ${x \in \Bbb{R}}$.

Miklos Schweitzer 2013

Solution: Pick ${y=k-x}$. Then we have

$\displaystyle f^{-1}(\frac{f(x)+f(k-x)}{2}) (f(x)+f(k-x))=kf(k).$

Using the hypothesis we find that ${r \mapsto f^{-1}(r/2)r}$ is strictly monotone on a a finite number of subintervals (depending on the sign of ${f^{-1}}$ and ${r}$) so as a consequence ${f(x)+f(k-x)}$ can take only a finite number of values. Since ${x \mapsto f(x)+f(k-x)}$ is continuous we conclude that for every ${k}$ the ${f(x)+f(k-x)=c_k}$, a real constant. As a consequence we have

$\displaystyle f(x+k)-f(x)=f(x+k)+f(k/2-x)-f(k/2-x)-f(x)=Const(k)$

for every ${x \in \Bbb{R}}$.

Let ${a = f(1)-f(0),\ b=f(0)}$ and consider a positive integer ${n}$. Denote ${r=f(x+1/n)-f(x)}$. Then

$\displaystyle a= f(1)-f(\frac{n-1}{n})+...+f(1/n)-f(0)=nr,$

so ${r=a/n}$ and ${f(1/n)=a \cdot \frac{1}{n}+b}$. Pick now ${m}$ a positive integer and notice that

$\displaystyle f(m/n)-f(0)= f(m/n)-f((m-1)/n)+...+f(1/n)-f(0)=m a/n,$

so we have ${f(m/n)=a\cdot \frac{m}{n}+b}$. This proves by continuity that for ${x\geq 0}$ we have ${f(x)=ax+b}$.

Using the identity ${f(x)-f(x-1)=f(1)-f(0)=a}$ we can extend the result for negative ${x}$ and we are done.