Which convex functions are characterized as the supremum of linear functions? A convex function $f$ can be represented as the supremum of all the affine functions that are dominated by $f$. Is there a simple characterization of those convex functions $g$ that can be represented as the supremum of all the linear functions that are dominated by $g$?
 A: Assuming we're talking about functions from $\Bbb R$ to $\Bbb R$, the function $g$ has this property if and only if there exist $a,b$ with $a\ge b$ and $g=g_{a,b}$, where $$g_{a,b}(t)=\begin{cases}at&(t\ge0),
\\bt&(t<0).\end{cases}$$
It's clear that every such $g_{a,b}$ is the sup of a family of linear functions. Conversely, say $g$ is convex and $g$ is the sup of a family $F$ of linear functions. Let $L_a(t)=at$. There exists $S\subset\Bbb R$ such that $$F=\{L_a:a\in S\}.$$Hence $$g(t)=\sup_{f\in F}f(t)=g_{\alpha,\beta}(t),$$where $\alpha=\sup S$ and $\beta=\inf S$. 
(Note we never used the assumption that $g$ is convex. This makes sense: It's clear that the sup of any family of linear functions is convex...)
Edit: In fact $g:\Bbb R^n\to\Bbb R$ is the supremum of  a family of linear functions if and only if $g$ is convex and $$g(tx)=tg(x)\quad(t\ge 0).$$
Again it's clear that the sup of a family of linear functions has this form. Suppose conversely that $g$ is convex and has that homogeneity property. We know that $g$ is the sup of a family of affine functions. So the following shows that $g$ is the sup of a family of linear functions:


Suppose $A$ is affine; write $A=\alpha+L$, where $\alpha\in\Bbb R$ and $L$ is linear. Then $A\le g$ if and only if $\alpha\le0$ and $L\le g$.


Proof: Suppose first that $\alpha\le 0$ and $L\le g$. Since $A\le L$ it is clear that $A\le g$.
Conversely, suppose $A\le g$. Then $\alpha=A(0)\le g(0)=0$, and for $t>0$ we have
$$\alpha+tL(x)\le g(tx)=tg(x).$$So $\alpha/t+L\le g$; let $t\to\infty$ and it follows that $L\le g$
Edit 2: This is in reply to a comment, wouldn't fit nicely in the comment box. It's been said that a convex function on $\Bbb R^n$ need not be continuous. Seems to me that yes it is continuous:
Note I'm taking convex function to be real-valued, as I thought was standard. If we allow convex functions to take values in $(-\infty,\infty]$ then yes, there are certainly discontinuous convex functions on $\Bbb R^n$.
Heh, proof that it is standard to take "real-valued" as part of the definition: A standard result says that any convex function on $\Bbb R$ is continuous..


Lemma If $f$ is convex on $\Bbb R^n$ then $f$ is bounded above on compact sets.


Proof: Given $K$ compact there exist $x_1,\dots,x_{n+1}$ such that $K$ is contained in the convex hull of the $x_j$: hence $f\le\max(f(x_j))$ on $K$.
Now suppose that $f$ is convex on $\Bbb R^n$. Choose $c$ so that $f(x)\le c$ for $|x|\le 10$.
Suppose $t=|y|< 1$. There exists $z$ with $|z|=1$ such that $y=tz=(1-t)0+tz$.Hence $$f(y)\le tf(z)+(1-t)f(0)\le ct+(1-t)f(0).$$Now $t\to0$ as $y\to0$, so $$f(0)\ge\limsup_{y\to0}f(y).$$For the other direction, say $\alpha=\limsup_{y\to0}f(y)$. Say $y_j\to0$, $f(y_j)\to\alpha$. Let $z_j=-y_j/|y_j|$. Then $0$ is a convex combination of $y_j$ and $z_j$: $$0=t_jz_j+(1-t_j)y_j.$$In particular, $t_j=|(1-t_j)y_j|\to0$. So
$$f(0)\le ct_j+(1-t_j)f(y_j)\to\alpha.$$So $$\limsup_{y\to0}f(y)\le f(0)\le\liminf_{y\to0}f(y),$$hence $\lim_{y\to0}f(y)=f(0).$
A: By the Fenchel-Moreau (aka Biconjugate) Theorem, all lower-semicontinuous convex funnctions on a Hausdorff locally convex space can be represented as a supremum of affine functions
Indeed, for such functions, one has $f^{**} = f$, i.e
$$
f(x) \equiv \sup_{y}x^Ty - f^*(y),
$$
a supremum of affine functions.
