# Hahn Banach Theorem, First geometric form

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 $$F(x) for all $$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}