Problem
ALG-B3-M09-P020 Two-Sided Product Sign
#20
★★★★★ Level 5 of 5
Let \(f\) be additive and \(f(x)f(y)\le xy\) for all \(x,y\). Find all such functions.
First get boundedness from \(y=x\), then use both signs of \(xy\).
With \(y=x\), \(f(x)^2\le x^2\). Hence \(|f(x)|\le1\) on \([-1,1]\), so the additive function is bounded on an interval. Therefore \(f(x)=cx\). Substitute: \(c^2xy\le xy\) for all \(x,y\). If \(xy>0\), then \(c^2\le1\). If \(xy<0\), division reverses the sign and gives \(c^2\ge1\). Thus \(c^2=1\). The answer is \(f(x)=x\) and \(f(x)=-x\). The check is immediate.
Final task of the module: boundedness plus sign analysis of a product.