Lemma 4.4.1.label Let $E$ be a vector space over $\R$, $\rho:E\to\R$ be a sublinear functional, $F\suf E$ be a subspace, $\phi\in \text{Hom}(F;\R)$ with $\phi(x)\leq \rho(x)$ for all $x\in F$. For each $x_{0}\in E\del F,\lam\in \R$ define the extension
\begin{align*}\phi_{x_0,\lam}:F+\R x_{0}\to \R&&y+tx_{0}\mto \phi(y)+\lam t\end{align*}
then there exists $\lam\in \R$ depending on $x_{0}$ such that $\phi_{x_0,\lam(x)}\leq\rho(x)$ for all $x\in F+\R x_{0}$.
Proof. Let $x,y\in F$, then
\begin{align*}\phi(x)+\phi(y)=\phi(x+y)&\leq\rho(x+y)\leq \rho(x-x_{0})+\rho(y+x_{0})\\ \phi(x)-\rho(x-x_{0})&\leq \rho(y+x_{0})-\phi(y)\\ \SUP{x\in F}\braks{\phi(x)-\rho(x-x_0)}&\leq\INF{x\in F}\braks{\rho(x+x_0)-\phi(x)}\end{align*}
Let
\begin{align*}\lam\in \braks{\SUP{x\in F}\braks{\phi(x)-\rho(x-x_0)},\INF{x\in F}\braks{\rho(x+x_0)-\phi(x)}}\end{align*}
then for any $x\in F,t>0$
\begin{align*}\phi_{x_0,\lam}(x+tx_{0})&=\phi(x)+\lam t=t \braks{\phi(t\inv x)+\lam}\\&\leq t \braks{\rho(t\inv x+x_0)-\phi(t\inv x)+\phi(t\inv x)}\\&=t\rho(t\inv x+x_{0})=\rho(x+tx_{0})\\ \phi(x-tx_{0})&=\phi(x)-t\lam=t \braks{\phi(t\inv x)-\lam}\\&\geq t \braks{\rho(t\inv x-x_0)+\phi(t\inv x)-\phi(t\inv x)}\\&=t \rho(t\inv x-x_{0})=\rho(x-tx_{0})\end{align*}
$\square$
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