Theorem 1 (Hahn-Banach)

Let $p$ be a sub-linear function on a vector space $E$. Let $V$ be a vector subspace of $E$ and $f$ a linear form on $V$ such that, for all $y \in V$, we have $f(y) \leq p(y)$. Then there exists a linear form $h$ on $E$ that is an extension of $f$ and such that $h(x) \leq p(x)$ for $x \in E$.

The set of pairs $(x, a)$ such that $p(x) \leq a$ is a convex subset $P$ of the vector space $E_1 = E \times \mathbf{R}$ (II, p. 17, prop. 19), and it is clearly a pointed cone. Let $V_1$ be the subspace $V \times \mathbf{R}$ of $E_1$ and $g(y, a) = -f(y) + a$ for each point $(y, a) \in V_1$. Then $g$ is a positive linear form for the preorder structure on $V_1$ defined by $P \cap V_1$; for if $(y, a) \in P \cap V_1$, then $a \geq p(y) \geq f(y)$, therefore $g(y, a) \geq 0$. Next let $(x, a) \in E_1$; we show that $(x, a)$ is less than a point of $V_1$ for the preorder defined by $P$. If $(x', a') \in V_1$ then $(x, a) \leq (x', a')$ if, and only if, $p(x' - x) \leq a' - a$, taking $a' \geq p(-x) + a$, we see that $(0, a')$ of $V_1$ satisfies the requirements. Thus we can apply prop. I of II, p. 21; there is a linear form $u$ on $E_1$ extending $g$ and positive for the preorder defined by $P$. Therefore $u(0, 1) = g(0, 1) = 1$ and $u$ is of the form $u(x, a) = -h(x) + a$, where $h$ is a linear form on $E$ that extends $f$; further, for all $x \in E$ and all $a \geq p(x)$, we have $h(x) \leq a$, therefore $h(x) \leq p(x)$. Q.E.D.