I want to prove that the geometric Hahn Banach theorem implies the analytic one.
Edit:
To avoid confusion I will state the vesion of H.B theorems im familiar with:
Analytic H.B:
Let $X$ be a linear space(over $\Bbb R$) and $Y\subset X$ a subspace. Let $p:X \to \Bbb R$ be a sub-additive function. Suppose $f:Y\to \Bbb R$ is a linear map s.t $f(y) \le p(y)$ for all $y\in Y$ then there exists an extention $g:X\to \Bbb R$ of $f$ s.t. $g(x)\le p(x) $ for all $x\in X$.
Geometric H.B:
Let $X$ be a linear space. $K \subset X$ convex s.t. each point in $K$ is an internal point. Let $D$ be a plane disjoint from $K$ then there exists hyperplane that contains $D$ and disjoint from $K$.
In my problem $X$ is NOT a normed space so there are no open sets.
Given $X$ a linear space and $Y\subset X$ a subspace $p:X\to \Bbb R$ sub-additive, and $f:Y\to \Bbb R$ linear s.t $f(y)\le p(y)$ for $y\in Y$ we need to extend $f$ to $g:X\to \Bbb R $ and $g(x)\le p(x)$ fo r all $x\in X$
So, we look at $X \times \Bbb R $and define $K = \{(x,t) : t>p(x)\}$.
The fact that $K$ is convex is easy. How can I show directly that every point in $K$ is internal? (can't say that $K$ is open).
Now after showing that, we can look at $Graph(f)$ and observe that $Graph(f)\cap K = \emptyset$.
So by the geometric H.B theorem we have a hyperplane (which is a maximal subspace in this case because $(0,0)\in Graph(f)$ ) M containing $Graph(f) $ and disjoint from $K$.
Now my problem is to show that each $x\in X$ has a unique $t\in \Bbb R$ s.t. $(x,t) \in M$. (I need it in order to extend $f$).
I know that if we take $v_0\notin M$ then each $v \in X \times \Bbb R$ has a unique representation as $v = \alpha v_0 + m$ for $m \in M$ , this is because $M$ is a maximal subspace. not sure if that helps.
Thanks for helping!
Best Answer
Here's another approach, according to guidance in the book Functional Analysis by Prof.s Weiss, Lindenstrauss and Pazi (there's no English translation as far as I'm aware of, sorry) in page 217, Q.13:
Using the notation of the analytical H.B. you've written above, define $K:=\{x\in X:p(x)<1\}$, $\alpha:=\sup_{y\in Y\cap K}\{f(y)\}$ and $D:=\{y\in Y:f(y)=\alpha\}$.
Note that $K$ is convex, and that every point of $K$ is an internal point (take $k\in K$ and you can write explicit $\epsilon$ for any $y\in X$ to show the definition in the link you referred to holds).
If $f\equiv 0$ it's easy to extend $f$ to be identically $0$ on all of $X$. Otherwise, there is a $y\in Y$ for which $f(y)>0$, and as $Y$ is a vector space and $f$ is linear, we can scale this point to get $f(y_0)=\alpha$. Note that $D=\ker(f)+y_0$ so it is a plane (an affine subspace).
Lastly note that $D\cap K=\emptyset$ (this is true thanks to the definition of $\alpha$ as a supremum and $K$ as a preimage of an open interval).
Apply Geometric H.B. and achieve a hyperplane $\hat D$ that contains $D$ and does not intersect with $K$ (specifically, $0\notin\hat D$). So $\hat D-y_0$ is a maximal subspace which contains $\ker(f)$ and does not contain $y_0$, so we may define the extension of $f$ as $0$ on $\hat D-y_0$ and extend linearly. (Note that $Y=\ker(f)\oplus span\{y_0\}$ so this definition actually extends $f$).