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For a finite real convex function on an entire real vector space, apply the convex domination form of the Hahn-Banach theorem to $\varphi (y+z)-\varphi (y)$ and the zero functional on $\{0\}$. It supplies a linear functional ${\ell }_y$ dominated by this translated function. The affine function $a_y(x)=\varphi (y)+{\ell }_y(x-y)$ lies below $\varphi$ everywhere and equals it at $y$. Taking the supremum proves the formula. These are algebraic affine minorants; continuity requires additional topological hypotheses.

Here $\varphi$ 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 boundsTo 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 giveRearranging proves every lower candidate is at most every upper candidate. Also the candidates $w=0,t=s=1$ showThus 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 $W+\mathbb{R}v$ 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 there is a full linear extension with $\boxed{g{|}_W=f,g\le \varphi }$. 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 ${\phi }_y(z)=\varphi (y+z)-\varphi (y)$ vanishes at zero. Apply the just-proved extension theorem to the zero linear functional on $\{0\}$ to obtain a linear functional ${\ell }_y$ with ${\ell }_y(z)\le {\phi }_y(z)$ for every $z$. SetThis is an affine function, satisfies $a_y(x)\le \varphi (x)$ for every $x$, and has $a_y(y)=\varphi (y)$. Let $A=\{a_y:y\in V\}$. Every member is an affine minorant, while the member indexed by $x$ attains $\varphi (x)$ at $x$. Thus the exact representation isThis 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.