Problem
ALG-B3-M09-P011 Quadratic Upper Bound
#11
★★★★☆ Level 4 of 5
Let \(f\) be additive and \(f(x)\le x^2\) for all \(x\). Prove that \(f\equiv0\).
Replace \(x\) by \(qx\) and let \(q\to0+\).
For \(q>0\), \(qf(x)=f(qx)\le q^2x^2\). Divide by \(q\): \(f(x)\le qx^2\). Letting \(q o0+\), we get \(f(x)\le0\). Applying this to \(-x\), we get \(-f(x)\le0\), i.e. \(f(x)\ge0\). Hence \(f(x)=0\).
Key scaling trick.