Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 6 / 2

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.

We prove the extension by Zorn's lemma, retaining the full domination condition. Consider all pairs $(Y,S)$ extending $(Y_0,T_0)$, with $Y$ a subspace and $S$ linear, satisfying $S(y)\le p(x+y)-q(x)$ for every $x\in X,y\in Y$. Order these pairs by extension. A chain has an upper bound: take the union of its subspaces and maps. The union is a subspace, and its map is well defined and linear because any finitely many vectors lie together in one member of the chain. Its domination is inherited there. Thus a maximal extension exists.

Suppose its domain $Y$ is proper, and choose $z\notin Y$. A linear extension to $Y+\mathbb{R}z$ has the form $S_a(y+tz)=S(y)+ta$. Define the admissible values for a sandwich extension bywith $x,x^{\prime }\in X$ and $y,y^{\prime }\in Y$. Part (i) shows every quantity in the supremum is at most every quantity in the infimum, hence $A\le B$. Both bounds are real and finite: choosing zero vectors provides $A\ge -p(-z)$ and $B\le p(z)$, and the pairwise comparison then bounds them from the other sides. Choose $a\in [A,B]$.

For $t>0$, positive homogeneity and the upper bound on $a$, used with $x/t,y/t$, giveFor $t=-s<0$, the lower bound on $a$ applied to $x/s,y/s$ giveswhich rearranges to $S(y)-sa\le p(x+y-sz)-q(x)$. For $t=0$ this is the existing domination. Thus $S_a$ is a legitimate larger extension, contradicting maximality. The maximal domain is therefore all of $X$; call its map $T$.

Setting the first vector in the domination to zero yields $T(v)\le p(v)$. Setting it to $v$ and the domain vector to $-v$ yields $-T(v)\le -q(v)$. ConsequentlyThe one-dimensional extension has checked both signs of its coefficient; positive homogeneity alone does not permit treating negative coefficients as positive ones.

Take $Y_0=\{0\}$ and $T_0(0)=0$. The hypothesis $p(x)\ge q(x)$ is exactly the required initial domination $0\le p(x+0)-q(x)$. Applying the extension proved in part (ii) giveswith $U$ linear. There is no reason that $U$ must be nonzero; for instance $p=q=0$ forces $U=0$.

This is the sandwich form of the Hahn-Banach theorem. Conversely, any linear map between these two functionals satisfies the stronger mixed domination, since $U(y)=U(x+y)-U(x)\le p(x+y)-q(x)$. Thus the mixed inequality used during extension is the appropriate invariant, not an extra final restriction.