Problem
ALG-B3-M09-P004 Global Upper Bound
#4
★★☆☆☆ Level 2 of 5
Let \(f\) be additive and \(f(x)\le100\) for all \(x\). Prove that \(f\equiv0\).
If \(f(a)>0\), consider \(na\). If \(f(a)<0\), consider negative multiples.
If \(f(a)>0\), then \(f(na)=nf(a)\) exceeds \(100\) for large \(n\), contradiction. If \(f(a)<0\), then \(f((-n)a)=-nf(a)\) exceeds \(100\). Hence \(f(a)=0\) for every \(a\).
Global boundedness is stronger than local boundedness.