Problem
ALG-B3-M04-P014 Continuous Invertibility
#14
★★★★☆ Level 4 of 5
Find all continuous functions \(f:\mathbb R\to\mathbb R\) satisfying \(f(f(x)+y)=x+f(y)\) for all \(x,y\).
Start with \(y=0\) and \(x=0\).
With \(y=0\), \(f(f(x))=x+f(0)\). This implies injectivity: if \(f(a)=f(b)\), then \(a+f(0)=b+f(0)\). It also implies surjectivity. With \(x=0\), \(f(f(0)+y)=f(y)\), so injectivity gives \(f(0)=0\). Hence \(f(f(x))=x\).
Put \(y=f(t)\): \(f(f(x)+f(t))=x+t=f(f(x+t))\). By injectivity, \(f(x)+f(t)=f(x+t)\). The function is continuous and additive, so \(f(x)=cx\). The condition \(f(f(x))=x\) gives \(c^2=1\). The answer is \(f(x)=x\) and \(f(x)=-x\).
A key example for moving from bijectivity to additivity.