Problem
ALG-B3-M09-P019 Product with One Sign
#19
★★★★★ Level 5 of 5
Let \(f\) be additive and \(f(x)f(y)\ge xy\) for all \(x,y\). Find \(f\).
First put \(y=x\), then consider \(xy<0\).
With \(y=x\), \(f(x)^2\ge x^2\), which does not give boundedness. But with \(y=-x\), \(f(x)f(-x)\ge -x^2\), so \(-f(x)^2\ge -x^2\), hence \(f(x)^2\le x^2\). Thus \(f\) is bounded on \([-1,1]\), so \(f(x)=cx\). The condition becomes \(c^2xy\ge xy\) for all \(x,y\). If \(xy>0\), then \(c^2\ge1\); if \(xy<0\), then \(c^2\le1\). Hence \(c^2=1\). The answer is \(f(x)=x\), \(f(x)=-x\).
Harder than the \(\le\) version because boundedness must be extracted through \(y=-x\).