Proof.

1. (1)

   Let $x_{0}\in E\del F$, then by Lemma 4.4.1, there exists $\lam\in\R$ such that if

   \begin{align*}\phi_{x_0,\lam}:F+\R x_{0}\to \R&&(x+tx_{0})\mto x+\lam t\end{align*}

   
   

   then $\phi_{x_0,\lam}(x)\leq \rho(x)$ for all $x\in F+\R x_{0}$. Thus for any $F\sub E$, $\phi$ admits an extension to a subspace that strictly contains $F$. Let

   \begin{align*}\textbf{F}=\curl{\Phi\in\hom{}{F'}{\R}:F'\supf,\Phi|_F=\phi,\Phi\leq \rho|_{F'}}\end{align*}

   
   

   For any $\Phi,\Phi'\in \textbf{F}$, define $\Phi\leq\Phi'$ if $\Phi'$ is an extension of $\Phi$. For any $\Phi\in\textbf{F}$, let $\cali{D}(\Phi)$ be its domain. Let $\cali{F}\suf \textbf{F}$ be a chain then $\cups{}{\Phi\in\cali{F}}\cali{D}(\Phi)$ is a subspace, hence there exists $\Phi\in\hom{}{\cups{}{\Phi'\in \cali{F}}\cali{D}(\Phi')}{\R}$ such that $\Phi|_{\cali{D}(\Phi')}=\Phi'$ for all $\Phi'\in\cali{F}$. Thus $\Phi\in \textbf{F}$ is an upper bound of $\cali{F}$. By Zorn’s lemma, $\textbf{F}$ admits a maximal element $\Phi$. If $\cali{D}(\Phi)\sub E$, then $\Phi$ is not maximal by Lemma 4.4.1. Therefore $\cali{D}(\Phi)=E$ and $\Phi$ is a desired extension.

2. (2)

   By absolute homogeneity of $\rho$, for any $u\in \hom{}{E}{\R}$, $u\leq \rho$ if and only if $\abs{u}\leq\rho$. By (1) we assume WLOG $K=\com$.

   By Proposition 4.4.2 $\Re(\phi)\in \hom{}{E}{\R}$. By (1), there exists $u\in\hom{}{E}{\R}$ such that $\abs{u}\leq\rho$ and $u|_{F}=\Re(\phi)$ so by Proposition 4.4.2 we are done.

$\square$