Problem
ALG-B3-M10-P011 Increasing Square Root of Two
#11
★★★★★ Level 5 of 5
Find all increasing \(f:\mathbb R\to\mathbb R\) such that \(f(x+f(y))=f(x)+2y\).
Prove injectivity, then additivity.
If \(f(a)=f(b)\), comparing the equations for \(y=a,b\) gives \(2a=2b\), so \(a=b\). With \(y=0\), \(f(x+f(0))=f(x)\), hence \(f(0)=0\). With \(x=0\), \(f(f(y))=2y\). Let \(t=f(y)\). Then \(f(t)=2y\), and the original equation gives \(f(x+t)=f(x)+f(t)\). The function is additive and increasing, so \(f(x)=cx\). From \(c^2=2\) and increasingness, \(c>0\). The answer is \(f(x)=\sqrt{2}x\).
Mixes Module 4 with the inequality module.