Problem
ALG-B3-M04-P017 Midpoints from the Image
#17
★★★★★ Level 5 of 5
Let \(f:\mathbb R\to\mathbb R\) be continuous and surjective. Find all \(f\) such that \(f(x+f(y))+f(x-f(y))=2f(x)\) for all \(x,y\).
By surjectivity replace \(f(y)\) with arbitrary \(t\).
Since \(f\) is surjective, \(t=f(y)\) is arbitrary. Therefore \(f(x+t)+f(x-t)=2f(x)\) for all \(x,t\). Taking \(u=x+t\), \(v=x-t\), we get \(f\left(\frac{u+v}{2}\right)=\frac{f(u)+f(v)}{2}\). A continuous function preserving midpoints is affine: \(f(x)=ax+b\).
Surjectivity requires \(a\ne0\). Checking shows that every function \(f(x)=ax+b\), \(a\ne0\), satisfies the original equation. The answer is \(f(x)=ax+b\), where \(a\ne0\).
A bridge to Jensen-type equations: surjectivity turns a special increment into an arbitrary one.