Problem
ALG-B1-M08-P022 Equation with invertibility
#22
★★★★★ Level 5 of 5
Let \(f:\mathbb R\to\mathbb R\) be increasing and satisfy \(f(x+f(y))=f(x)+y\) for all \(x,y\). Find \(f\).
First substitute \(y=0\), then \(x=0\), and then replace \(y\) by \(f(t)\).
Let \(a=f(0)\). At \(y=0\), we have \(f(x+a)=f(x)\). Since \(f\) is increasing, it is injective, so \(a=0\). At \(x=0\), we get \(f(f(y))=y\).
Now put \(y=f(t)\) in the original equation. Then \(f(x+t)=f(x)+f(t)\), so \(f\) is additive. An increasing additive function has the form \(f(x)=cx\), where \(c>0\). From \(f(f(y))=y\), we get \(c^2y=y\) for all \(y\), hence \(c=1\). Answer: \(f(x)=x\).
A strong problem for the module: several substitutions, injectivity, and Cauchy regularity work together.