Problem
ALG-B3-M11-P007 Quadratic Bound
#7
★★★★☆ Level 4 of 5
Let \(f\) be additive and \(f(x)\le x^2\) for all \(x\). Prove that \(f\equiv0\).
Substitute \(qx\) and let \(q\to0+\).
For \(q>0\), \(qf(x)=f(qx)\le q^2x^2\), so \(f(x)\le qx^2\). Letting \(q o0+\), \(f(x)\le0\). For \(-x\), we get \(-f(x)\le0\), hence \(f(x)=0\).
Scaling idea.