= 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 $y+y'\in Y$ and the vector $x+x'$:
$$
S(y)+S(y')\leq p(x+x'+y+y')-q(x+x').
$$
<Subadditivity> of $p$, with the inserted vectors $z$ and $-z$, gives
$$
p(x+x'+y+y')\leq p(x+y+z)+p(x'+y'-z).
$$
Superadditivity of $q$ gives $q(x+x')\geq q(x)+q(x')$. Combining these inequalities and moving terms yields exactly
$$
\boxed{S(y')-p(x'+y'-z)+q(x')
\leq-S(y)+p(x+y+z)-q(x).}
$$
Every vector appearing in the comparison is permitted by the assumed domination, so no membership of $x,x',z$ in $Y$ is needed.
Back to article page
