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I've come across a proof online of the H.B theorem using the goemetric version.

There is a step in the proof which im not sure why is true and it is as follows:

Let $X$ be a linear space and $M \subset X \times \Bbb R$ be a maximal subspace.

Then $M = G(F)$ for $F :X \to \Bbb R$ linear.

Someone can show me why this is true?

Maybe I misunderstood the proof, its the last paragraph of Matrin's answer here -https://mathoverflow.net/questions/134508/direct-proof-of-the-separation-theorem-of-hahn-banach.

Thanks a lot for helping!

Bernard
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    Please do not delete questions once they got an answer, especially not without explantion. – quid Dec 28 '18 at 23:22
  • @quid you are right, i realized that i need to emphasize the fact i dont have topology, that $X $ is not a normed space just a linear space. –  Dec 29 '18 at 09:31

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To me Martin's answer is rather clear. I'll change all the notations for clarity.

Let $B$ be a normed real vector space, $A$ a sub-vector space, $f$ a linear map $A \to \mathbb{R}$ satisfying $|f(x)| \le c \|x\|$.

Let $E = B \times \mathbb{R}$ and $graph(f) = \{ (x,f(x)) \in E | x\in B\}$ and $U = \{ (x,t) \in E | x \in B, t > c\|x\|\}$. Then $U$ is open and convex and $graph(f)$ is a closed sub-vector space disjoint from $U$ so there is an hyperplane $H$ (that is a sub-vector space such that for some $v \not \in H$ then $E = \{ h+av | h \in H, a \in \mathbb{R}\}$) containing $graph(f)$ and disjoint from $U$. Those conditions ensure that for any $x \in E$ there is a unique $t \in \mathbb{R}$ such that $(x,t) \in H$ and $\tilde{f}(x) = t$ extends $f$ and satisfies $|\tilde{f}(x)| \le c \|x\|$.

$\tilde{f}$ is linear because $\tilde{f}(x_1)= t_1,\tilde{f}(x_2) =t_2 \implies (x_1+x_2,t_1+t_2) \in H \implies \tilde{f}(x_1+x_2) = t_1+t_2$

reuns
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  • $f$ is defined in subspace $A$, when you wrote the graph definition, you made $x$ vary in $B$. Why can you do that? – Ilovemath Mar 30 '21 at 21:07