= Solution
Here $\phi$ is a finite real <convex function> on the whole real <vector space>; no topology or continuity is required. We prove the <convex domination form of the Hahn-Banach theorem> by extending across one vector. Suppose a <linear functional> $f$ is already dominated on a <vector subspace> $W$ and choose $v\notin W$. For $w\in W$ and positive $t,s$, the proposed extension $g(w+\tau v)=f(w)+\tau c$ must satisfy the bounds
$$
L=\sup_{w\in W,\ t>0}\frac{f(w)-\phi(w-tv)}{t}\le c\le
U=\inf_{w\in W,\ s>0}\frac{\phi(w+sv)-f(w)}{s}.
$$
To prove compatibility, take $w_1,w_2\in W$ and $t,s>0$. The <convex combination> of $w_1-tv$ and $w_2+sv$ with weights $s/(s+t)$ and $t/(s+t)$ lies in $W$. Domination there and <convexity> give
$$
sf(w_1)+tf(w_2)\le s\phi(w_1-tv)+t\phi(w_2+sv).
$$
Rearranging proves every lower candidate is at most every upper candidate. Also the candidates $w=0,t=s=1$ show
$$
-\phi(-v)\le L\le U\le\phi(v).
$$
Thus both endpoints are finite, and some $c\in[L,U]$ exists. For $\tau>0$ the upper bound gives domination of $f(w)+\tau c$; for $\tau<0$ the lower bound gives it; and for $\tau=0$ domination was assumed. This defines a dominated <linear functional> on $W+\mathbb Rv$ extending $f$.
Order all dominated extensions by inclusion of their domains and agreement of values. A chain has the union extension, still linear and dominated. By <Zorn's lemma> there is a maximal extension. The one-vector construction would enlarge any proper domain, so its domain is $V$. Hence \b[there is a full linear extension with $\boxed{g|_W=f,\quad g\le\phi}$]. A <convex function> here need not be positively homogeneous; this is why the bounds included all positive $s,t$.
Now fix $y\in V$. The translated <convex function> $\varphi_y(z)=\phi(y+z)-\phi(y)$ vanishes at zero. Apply the just-proved extension theorem to the zero <linear functional> on $\{0\}$ to obtain a <linear functional> $\ell_y$ with $\ell_y(z)\le\varphi_y(z)$ for every $z$. Set
$$
a_y(x)=\phi(y)+\ell_y(x-y).
$$
This is an <affine function>, satisfies $a_y(x)\le\phi(x)$ for every $x$, and has $a_y(y)=\phi(y)$. Let $A=\{a_y:y\in V\}$. Every member is an affine minorant, while the member indexed by $x$ attains $\phi(x)$ at $x$. Thus \b[the exact representation is]
$$
\boxed{\phi(x)=\sup_{a\in A}a(x)\quad(x\in V).}
$$
This is the <finite convex function as supremum of affine minorants> property. The <linear functionals> are algebraic; without a topology no assertion of continuous affine minorants is intended.
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