Problem
ALG-B3-M09-P014 Lower Bound
#14
★★★★☆ Level 4 of 5
Let \(f\) be additive and \(f(x)>-1\) for all \(x\). Prove that \(f\equiv0\).
If \(f(a) e0\), take suitable integer multiples of \(a\).
If \(f(a)>0\), then for large negative \(n\), \(nf(a)<-1\), i.e. \(f(na)<-1\), contradiction. If \(f(a)<0\), take large positive \(n\). Therefore \(f(a)=0\) for every \(a\).
Global lower bound.