Is an lsc sublinear function $X^* \rightarrow (-\infty, \infty]$ always a support function for some closed non-empty $C \subset X$? I can't seem to find any resources on this, even though it seems like an obvious question to ask. The separation theorem implies that, if we have an lsc sublinear function $\phi : X^* \rightarrow (-\infty, \infty]$, and it is equal to a support function $\mathrm{supp}_C$, then $C$ must be the set:
$$C = \bigcap\limits_{f \in X^*} f^{-1}(-\infty, \phi(f)].$$
Moreover, we can easily see that, if we define $C$ as above, then $\mathrm{supp}_C \le \phi$.
Certainly $C$ is closed and convex. I can't seem to prove even that $C$ is non-empty, but I have a suspicion that this may be most of the battle. I also suspect that this is conditional on $X$ being reflexive. Any suggestions?
 A: Let $\phi:X^*\rightarrow\mathbb{R}\cup\{\infty\}$ be a sublinear weak* lsc functional. If we assume $\phi$ is proper (meaning $\phi\not\equiv \infty$ or equivalently its essential domain is non-empty) then for some $x^*$ in its essential domain $\phi(x^*)<\infty$. The following two subsets $\{(x^*,\phi (x^*)-1)\}$ (a weak* compact subset) and  $\mathbb{epi}\phi:=\{(f,\beta)\in X^*\times\mathbb{R};\ \phi(f)\leq\beta\}$ (a weak* closed subset) are convex and disjoint so by a separation theorem they can be separated by a weak*-continuous linear functional $(x,\lambda)\in X\times\mathbb{R}$ hence
$$x^*(x)+\lambda\cdot(\phi(x^*)-1)\leq f(x)+\lambda\cdot\beta$$
for any $(f,\beta)\in\mathbb{epi}\phi$. Since $\beta$ in not bounded above $\lambda$  must be non-negative, and for $\beta=\phi(f)$ we get 
$$x^*(x)+\lambda\cdot(\phi(x^*)-1)\leq f(x)+\lambda \phi(f)$$
for any $f\in \mathbb{dom}\phi$. Of course the above inequality holds even for $f\notin \mathbb{dom}\phi$ since in that case the right side of the inequality is $\infty$. Since $x(\cdot)+\lambda \phi(\cdot):X^*\rightarrow \mathbb{R}$ is sublinear and bounded below (by the constant $x^*(x)+\lambda\cdot(\phi(x^*)-1)$) it must be non-negative (exercise) hence
$$f(-\frac{1}{\lambda}x)\leq\phi(f) \ \forall f\in X^*$$
so we may conclude that $-\frac{1}{\lambda}x\in C$ and $C\not=\emptyset$.
A few remarks:


*

*We assumed that $\phi$ is proper. If it isn't then  $C=X$ hence $C\not = \emptyset$.

*Since you mentioned reflexivity, I'll add that in the case that $X$
is reflexive the weak-* and norm topologies coincide so you may
consider the latter instead if the former.

*You also claimed that if $\phi$  is a support function on some set, this set must be $C$ which is false as can be easily seen by the example
$\phi:\mathbb{R}\rightarrow\mathbb{R}$ defined by
$\phi(x):=\max\{x,-x\}$ (try to find $C$ in this case).

*The separation theorem used above states that in a locally convex
space (in particular a dual space), a compact subset and a closed subset which are convex and disjoint can be (strongly) separated by a continuous linear functional.

*One can replace the space $(X^*,\sigma(X^*,X))$ above with any locally convex space.
