I am studying the proof of Hahn Banach Theorem (First geometric form). And the following is a part of the proof.

Let $U\subseteq E$ be open, convex and nonempty and let $x_0\in E\backslash U$. Then, there exists $F\in E^{*}$ such that $F(x)<F(x_0)$ for all $x\in U.$

Proof (in part)

Consider the subspace generated by $x_0\in E\backslash U$, which is given by

\begin{align} M=\{x:\,x=\lambda x_0,\,\lambda\in \Bbb{R} \} \end{align} Define \begin{align} f:&M\to \Bbb{R}\\&x\mapsto f(x)\equiv f(\lambda x_0)=\lambda \end{align} Let $\gamma, \eta\in \Bbb{R} $ and $x,y\in E$, then there exists $\lambda,\beta\in \Bbb{R}$ such that $x=\lambda x_0$ and $y=\beta x_0$ and \begin{align} f(\gamma x+\eta y)&=f\left(\gamma (\lambda x_0)+\eta(\beta x_0) \right) \\&=f\left((\gamma \lambda+\eta\beta)x_0 \right)\\&=\gamma \lambda+\eta\beta \\&= \gamma f( x)+\eta f( y).\end{align} This implies that $f$ is linear on $M.$ Introducing the gauge $p$ on $U,$ we have that \begin{align} f( x)\leq p(x),\;\forall\;x\in M.\end{align} Then, we can apply the Hanh-Banach Theorem to find a linear functional $f:E\to \Bbb{R},$ extending $f$ such that \begin{align} f( x)\leq p(x),\;\forall\;x\in E.\end{align}

The following is definition of gauge that I know:

Definition: Given that $E$ is a normed linear space. Let $M\subseteq E$ be an open, convex set with $0\in M.$ For all $x\in E$, define

\begin{align} p(x)=\inf\{\alpha>0:\,\alpha^{-1}x\in M \} \end{align} Then, $p$ is called a gauge of $M$.

Question: What kind of gauge $p,$ was defined on $U$ such that \begin{align} f( x)\leq p(x),\;\forall\;x\in M?\end{align}


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