Lemma 9.6.1. Let $E$ be a vector space over $\real$, $\rho: E \to \real$ be a sublinear functional, $F \subset E$ be a subspace, $\phi \in \hom(F; \real)$ with $\phi(x) \le \rho(x)$ for all $x \in F$.
Let $x_{0} \in E \setminus F$, $\lambda \in \real$, and define
\[\phi_{x_0, \lambda}: F + \real x_{0} \quad (y + tx_{0}) \mapsto \phi(y) + \lambda t\]
then there exists $\lambda \in \real$ such that $\phi_{x_0, \lambda}(x) \le \rho(x)$ for all $x \in F + \real x_{0}$.
Proof. Let $x, y \in F$, then
\begin{align*}\phi(x) + \phi(y) = \phi(x + y)&\le \rho(x + y) \le \rho(x - x_{0}) + \rho(y + x_{0}) \\ \phi(x) - \rho(x - x_{0})&\le \rho(y + x_{0}) - \phi(y) \\ \sup_{x \in F}[\phi(x) - \rho(x - x_{0})]&\le \inf_{x \in F}[\rho(x + x_{0}) - \phi(x)]\end{align*}
Let
\[\lambda \in \braks{\sup_{x \in F}[\phi(x) - \rho(x - x_0)], \inf_{x \in F}[\rho(x + x_0) - \phi(x)]}\]
then for any $x \in F$ and $t > 0$,
\begin{align*}\phi(x + tx_{0})&= \phi(x) + t\lambda = t\braks{\phi(t^{-1}x) + \lambda}\\&\le t\braks{\rho(t^{-1}x + x_0) - \phi(t^{-1}x) + \phi(t^{-1}x)}\\&= t\rho(t^{-1}x + x_{0}) = \rho(x + tx_{0}) \\ \phi(x - tx_{0})&= \phi(x) - t\lambda = t\braks{\phi(t^{-1}x) - \lambda}\\&\ge t\braks{\rho(t^{-1}x - x_0) + \phi(t^{-1}x) - \phi(t^{-1}x)}\\&= t\rho(t^{-1}x - x_{0}) = \rho(x - tx_{0})\end{align*}
$\square$