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.

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  1. November 9, 2013 at 2:43 pm

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