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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.

If $p$ is a sublinear functional, $q$ a superlinear functional and $q\le p$ on a real vector space, there exists a linear functional $U$ between them. More generally, a linear functional $S$ on a subspace extends with this domination whenever $S(y)\le p(x+y)-q(x)$ for all ambient $x$ and subspace $y$. In a one-dimensional extension the admissible values for a sandwich extension form a nonempty real interval. Positive homogeneity checks the extension on both signs of the new coordinate. Chain unions preserve linearity and the mixed domination, so Zorn's lemma gives a maximal extension, whose domain must be the whole space. Finally $x=0$ in the mixed inequality gives the upper bound; using the pair $(x,-x)$ gives the lower bound. Conversely, any linear $U$ between $q,p$ obeys the mixed inequality by $U(y)=U(x+y)-U(x)$.