Problem
ALG-B3-M04-P013 Square Root of Two from Injectivity
#13
★★★★☆ Level 4 of 5
Find all increasing functions \(f:\mathbb R\to\mathbb R\) such that \(f(x+f(y))=f(x)+2y\) for all \(x,y\).
Prove \(f(f(y))=2y\), then use monotone additivity.
If \(f(a)=f(b)\), comparing the equalities for \(y=a\) and \(y=b\) gives \(2a=2b\), so \(a=b\). The function is injective. With \(y=0\), \(f(x+f(0))=f(x)\), hence \(f(0)=0\). With \(x=0\), \(f(f(y))=2y\), so \(f\) is surjective.
Let \(t=f(y)\). Then \(f(t)=2y\), hence \(f(x+t)=f(x)+f(t)\). The function is additive and increasing, so \(f(x)=cx\). From \(f(f(y))=2y\), we get \(c^2=2\), and increasing gives \(c>0\). The answer is \(f(x)=\sqrt{2}x\). Direct checking works.
Good after rational-domain tasks: on \(\mathbb R\), a regularity condition is needed, here monotonicity.