Sublinear functional as supremum of linear functionals Given a sublinear functional on a Vector space $V$, is it possible to write it as supremum of family of linear functionals?
 A: Yes, every sublinear functional $p$ is the supremum of all linear $\varphi$ with $\varphi \leqslant p$. And conversely, the supremum of any family of linear $\varphi \colon V \to \mathbb{R}$ is a sublinear functional (if we don't allow the value $+\infty$ for sublinear functionals, that holds only for pointwise bounded families).
It's an easy consequence of one of the Hahn-Banach theorems, Theorem 3.2 in Rudin (FA):

Suppose
(a) $M$ is a subspace of a real vector space $X$,
  (b) $p \colon X \to \mathbb{R}$ satisfies
  $$p(x+y) \leqslant p(x) + p(y)\quad \text{and}\quad p(tx) = tp(x)$$
  $\quad$  if $x \in X,\, y \in X,\, t \geqslant 0$,
  (c) $f \colon M \to \mathbb{R}$ is linear and $f(x) \leqslant p(x)$ on $M$.
Then there exists a linear $\Lambda \colon X \to \mathbb{R}$ such that
  $$\Lambda x = f(x) \qquad (x \in M)$$
  and
  $$-p(-x) \leqslant \Lambda x \leqslant p(x) \qquad (x \in X).$$

For any $x \in V$, define $f \colon \mathbb{R}\cdot x \to \mathbb{R}$ by $f(\lambda x) = \lambda p(x)$. For $\lambda \geqslant 0$ we have $f(\lambda x) = p(\lambda x)$, and for $\lambda < 0$ we have $f(\lambda x) \leqslant p(\lambda x)$ by the sublinearity of $p$ (that gives $0 = p(0) \leqslant p(\lambda x) + p(\lvert\lambda\rvert x)$, whence $p(\lambda x) \geqslant -p(\lvert\lambda\rvert x) = -f(\lvert\lambda\rvert x) = f(\lambda x)$ for $\lambda < 0$). Extending $f$ to all of $V$ as guaranteed by the theorem yields
$$p(x) = \sup \{ \Lambda x : \Lambda \leqslant p\}.$$
