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Let $\mathbb K$ be the field $\mathbb R$ or $\mathbb C$. If $E$ is a vector space on $\mathbb C$, let the real space subjacent to $E$ be $E_0$, i.e. the space obtained by restricting the product on $\mathbb C\times E\:\:$ to $\:\:\mathbb R\times E$.

Let $E^*$ be the algebraic dual of $E$ and let $E'$ be the topological dual of $E$.


Let $E$ be a vector space over $\mathbb K$. Let $S$ be a closed subspace of $E$. Then there exists $f\in (E_0)'\backslash\{0\}$, and $\alpha\in\mathbb R$ such that either $S=\{x\in E\mid f(x)\geq \alpha\}$ or $S=\{x\in E\mid f(x)\leq \alpha\}$.

Conversely, if $f\in(E_0)'\backslash\{0\}$ and $\alpha\in\mathbb R$, then the sets $\{x\in E\mid f(x)\geq \alpha\}$ and $S=\{x\in E\mid f(x)\leq \alpha\}$ are closed.

Finally, if $f\in(E_0)^*\backslash\{0\}$ and $\alpha\in\mathbb R$, then the sets $S=\{x\in E\mid f(x)\geq \alpha\}$ and $S=\{x\in E\mid f(x)\leq \alpha\}$ are closed if and only if $f$ is continuous.

Attempt: I am sure this descends from Hahn Banach, however I'm not convinced at how $S$ must be equal to one of that sets pointed above in the first question, and then I am not able to write down a neat solution of the remaining two question, so I'm asking you your help. Thanks in advance.

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What do you mean by closed "subspace"? A closed subset, a closed convex subset, a closed affine subspace or a closed vector subspace? I take it you probably mean a closed vector subspace, but either way, the first assertion is false. – Olivier Bégassat Nov 2 '11 at 0:15
that was my first thought.. indeed as it is written the set is a semi space while $S$ could be just a line hence there is no hope for the two to be equal. So what should i assume? that the text is wrong? – uforoboa Nov 2 '11 at 7:12
What is the name of the section and subsection this is in? What theorem precede this? Give us some context to decide what's going on. – Olivier Bégassat Nov 2 '11 at 7:39
The section is about topological vector space, precisely about locally convex vector space. The text is copied from the book. – uforoboa Nov 2 '11 at 12:32

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