The question is out of Rudin Functional analysis Chapter 3 problem 1.
Call a set $H \subset \mathbb{R}$ a hyperplane if there exists real numbers $a_1,\ldots, a_n, c$ (with $a_i \neq 0$ for at least one $i$) such that $H$ consists of all points $x=(x_1,\ldots, x_n)$ that satisfy $\sum a_i x_i = c$.
Suppose $E$ is a convex set in $\mathbb{R}^n$, with nonempty interior, and $y$ is a boundary point of $E$. Prove that there is a hyperplane $H$ such that $y \in H$ and $E$ lies entirely on one side of $H$ (State the conclusion more precisely.) Suggestion: Suppose $0$ is an interior point of $E$, let $M$ be the one-dimensional subspace that contains $y$, apply Theorem 3.2. (The Hahn Banach theorem.)
I think it correct to say the following for a restatement of the conclusion.
We wish to find a linear functional $\Lambda : \mathbb{R}^n \rightarrow \mathbb{R}$ such that if $y\not \in int E$ then $$ \Lambda (x) \leq \Lambda (y) \;\;\; for \;\;\; all \;\;\; x \in E, $$
I am struggling to define a semi-norm, to satisfy the Hahn Banach requirements. I am pretty sure that my semi-norm needs to involve the equation of the hyper-plane. If anyone could please help me get the semi-norm for this one figured out I would be over joyed.
I have an attempt at the proof but I don't think its correct at all. I will post what I have.
Let $E$ be a convex set in $\mathbb{R}^n$, with non empty interior. Suppose $0 \in E$ is interior and $y$ is a boundary point of $E$. Let $M$ be the one-dimensional subspace that cointains $y$. Let $p(x)$ be the seminorm satisfying the hyper plane requirements, and let $p(0) = 0$. Then if $x \neq 0 \neq y$ $$ p(x + y) = \sum a_i (x_i + y_i) = c $$ $$ p(x) + p(y) = \sum a_i x_i + \sum b_i y_i = c + c = 2c $$
So, $p(x+y) \leq p(x) + p(y)$ in addition $p(tx) = \sum a_i (t x_i) = c = \sum a_i x_i = p(x)$.
Define $f(x)$ on $M$ by $f(x) = c$. Then $f(x) \leq p(x)$ on $M$. Then by the Hahn Banach theorem there exists a linear functional $\Lambda x = f(x)$ for $x \in M$. and $$ \Lambda x \leq p(x) \mbox{ for } x \in \mathbb{R}^n $$ Since $\Lambda y = f(y) = p(y)$ on $M$ we have the desired result that $$ \Lambda x \leq \Lambda y $$
Let $p$ be the Minkowski functional of $E$ (section 1.33, page 24 of Rudin). Clearly it satisfies the condition (b) of section 3.2, p. 56 even though $E$ need not be balanced. $E$ is absorbing, as $0\in\operatorname{int}E$.
Set $f(ty)=t$ $(t\in\mathbb R)$. Apply Theorem 3.2.
Clearly $p\le1$ on $E$ and $p=1$ on the boundary of $E$, hence $\Lambda\le 1$ on $E$ and $\Lambda y=1$.