Problem
ALG-B3-M09-P012 Too Large a Lower Bound
#12
★★★★☆ Level 4 of 5
Prove that there is no additive \(f:\mathbb R\to\mathbb R\) such that \(f(x)\ge x^2\) for all \(x\).
Substitute \(qx\) and divide by \(q\).
Let \(x e0\). For \(q>0\), \(qf(x)=f(qx)\ge q^2x^2\), so \(f(x)\ge qx^2\). As \(q o\infty\), the right-hand side grows without bound, impossible for fixed \(f(x)\). Contradiction.
The same scaling idea in the opposite direction.