 Solution

The hypotheses make $p$ a sublinear functional and $q$ a superlinear functional. In particular $p(0)=q(0)=0$. Apply the domination property to $y+y^{\prime }\in Y$ and the vector $x+x^{\prime }$:Subadditivity of $p$, with the inserted vectors $z$ and $-z$, givesSuperadditivity of $q$ gives $q(x+x^{\prime })\ge q(x)+q(x^{\prime })$. Combining these inequalities and moving terms yields exactlyEvery vector appearing in the comparison is permitted by the assumed domination, so no membership of $x,x^{\prime },z$ in $Y$ is needed.