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I have a question about a step in proving $(M^\perp)^\perp=\overline{M}$ where $M$ is a linear subspace of a normed vector space $E$. And $M^\perp=\{f\in E^*|\langle f,x\rangle =0\}$

This is the proof from Brezis book:

Proof: Assuming the other direction, show $\subset$ direction. Suppose there exists $x_0\in (M^\perp)^{\perp}$ such that $x_0\notin \overline{M}$. Then, Hahn-Banach says there exists a hyperplane that strictly separates $\{x_0\}$ and $\overline{M}$. Thus there are some $f\in E^*$ and $\alpha\in\mathbb{R}$ such that $$ \langle f,x\rangle <\alpha<\langle f, x_0\rangle, \forall x\in M$$

Since $M$ is a linear space it follows that $\langle f,x\rangle =0$(WHY??), for all $x$ and $\langle f, x_0\rangle >0$. Therefore $f\in M^\perp$ and consequently $\langle f, x_0\rangle =0$, a contradiction.

Can someone tell me why the place (WHY??) is true? I couldn't find out the line follows from the fact that $M$ is a linear space.

Thanks

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  • $\begingroup$ Why $x_0\in E$? Is it necessary to add $E$ is a reflexive space? $\endgroup$ – C.Ding Apr 17 '16 at 13:32
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Pick $\beta \in \mathbb R$ then $$\beta \langle f, x\rangle=\langle f, \beta x\rangle < \alpha $$

Since this happens for all $\beta \in \mathbb R$, it follows that $\langle f,x\rangle=0$.

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    $\begingroup$ @soundwave If $x \in \overline{M}$ then you can find some $x_n \in M$ such that $x_n \to x$... Then, for each $f \in M^\perp$ you have $$<f,x> = \lim_n <f, x_n> = \lim_n 0=0$$ $\endgroup$ – N. S. Jan 28 '19 at 15:36
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    $\begingroup$ @soundwave If $<f,x> \neq 0$ then it is either negative or positive. If $< f,x> >0$ then pick $\beta > \frac{\alpha}{<f,x> }$. If $< f,x> < 0$ then pick $\beta < \frac{\alpha}{<f,x> }$. $\endgroup$ – N. S. Jan 29 '19 at 18:23
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    $\begingroup$ @soundwave To make this a bot more clear, if $<f,x> \neq 0$ then $\beta <f,x>$ goes to $+ \infty$ at one of $\pm \infty$, depending on the sign of $<f,x>$. Therefore it cannot be bounded. $\endgroup$ – N. S. Jan 29 '19 at 18:25
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    $\begingroup$ @soundwave And your argument explains why we must also have $\alpha >0$ in this case... Note that the hyperplane separating these points means that the inequalities could actually be opposite way, so this part should be WLOG :) $\endgroup$ – N. S. Jan 29 '19 at 18:26
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    $\begingroup$ @soundwave Exactly. $<f,x> >0$ implies some restriction on $\beta$... But you can pick $\beta$ anything you want.... Therefore, picking $\beta$ to cpntradict the restriction you get a ..... contradiction. $\endgroup$ – N. S. Jan 30 '19 at 14:20

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