Problem
ALG-B3-M09-P018 Two Different Bounds
#18
★★★★★ Level 5 of 5
Let \(f\) be additive and suppose \(-x^2\le f(x)\le x^2\) for all \(x\). Prove that \(f\equiv0\).
It is enough to use the upper bound and apply it to \(-x\).
From the upper bound, as before, for \(q>0\): \(qf(x)=f(qx)\le q^2x^2\), so \(f(x)\le qx^2\). Letting \(q o0+\), we get \(f(x)\le0\). Apply the same to \(-x\): \(f(-x)\le0\), i.e. \(-f(x)\le0\), so \(f(x)\ge0\). Therefore \(f(x)=0\).
Shows that the lower bound may be redundant because additivity has symmetry.