Problem
ALG-B3-M05-P011 A Square Reveals Linearity
#11
★★★★☆ Level 4 of 5
Let \(f:\mathbb R\to\mathbb R\) be additive and satisfy \(f(x^2)=x f(x)\) for all \(x\). Prove that \(f(x)=cx\) for some constant \(c\).
Expand \(f((x+y)^2)\) in two ways.
By additivity, \(f((x+y)^2)=f(x^2)+2f(xy)+f(y^2)\). On the other hand, by the condition this equals \((x+y)f(x+y)=(x+y)(f(x)+f(y))\). After cancelling \(x f(x)+y f(y)\), we get \(2f(xy)=x f(y)+y f(x)\). Put \(y=1\). Then \(2f(x)=x f(1)+f(x)\), hence \(f(x)=x f(1)\). Take \(c=f(1)\).
A nontrivial proof of linearity without continuity.