Problem
ALG-B3-M11-P016 Product and Sign
#16
★★★★★ Level 5 of 5
Let \(f\) be additive and \(f(x)f(y)\le xy\) for all \(x,y\). Find \(f\).
First \(y=x\), then different signs of \(xy\).
With \(y=x\), \(f(x)^2\le x^2\), so \(f\) is bounded on \([-1,1]\), and \(f(x)=cx\). Then \(c^2xy\le xy\). If \(xy>0\), \(c^2\le1\); if \(xy<0\), \(c^2\ge1\). Hence \(c^2=1\). The answer is \(f(x)=x\), \(f(x)=-x\).
Inequality as regularity.