Problem
ALG-B3-M03-P012 Nonnegativity on a ray
#12
★★★★★ Level 5 of 5
Let \(f:\mathbb R\to\mathbb R\) be additive and \(f(x)\ge0\) for all \(x\ge0\). Prove that \(f(x)=cx\) and \(c\ge0\).
Hint. Nonnegativity gives monotonicity.
If \(x\le y\), then \(y-x\ge0\), so \(f(y)-f(x)=f(y-x)\ge0\). Hence \(f\) is nondecreasing. By the previous result, \(f(x)=cx\), with \(c=f(1)\ge0\).
Goal: show which conditions actually force a function to be linear or affine.