Sandwich form of the Hahn-Banach theorem ($q\le U\le p$)

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 $U(y)=U(x+y)-U(x)$.

Suppose $S$ satisfies the mixed domination on a subspace $Y$, and $z\notin Y$. Apply that domination to $y+y^{\prime }$ and $x+x^{\prime }$, split the $p$ argument as $(x+y+z)+(x^{\prime }+y^{\prime }-z)$, and use superadditivity of $q$. This proves every lower candidate in the displayed interval is at most every upper candidate. The zero-vector candidates bound the lower supremum below by $-p(-z)$ and the upper infimum above by $p(z)$, so both endpoints are finite. Choosing any intervening value $a$ permits $U(y+tz)=S(y)+ta$. For $t>0$, scale the upper comparison by $t$; for $t<0$, scale the lower comparison by $-t$. Both give the required mixed domination, proving the interval characterizes the permitted extension values.