Past exam of the analysis course of the University of Oxford / 2007 / B / 5 / b / ii / Solution

Repeated additivity gives $P(mu,v)=mP(u,v)$ for every positive integer $m$. Since $u$ is the sum of $n$ copies of $u/n$, $P(u/n,v)=P(u,v)/n$. Combining these provesThe case $m=0$ also holds by additivity at zero.