 Solution

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.