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Here $X$ is some normed space.

I know the converse is true, but I don't know a proof for the other direction.

That is, if $F\subset X^*$ is a subspace that is weak*-dense how would one show that $F$ separates points (i.e. if $f(x)=0$ for all $f\in F$ then $x=0$)?

Belgi
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Han Qin
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  • Do not "destroy" your own question. – Belgi Oct 01 '15 at 21:42
  • What is the annihilator of $F$ in $X$? Call it $A$. Then what is the annihilator of $A$ in $X^\ast$? – Daniel Fischer Oct 01 '15 at 21:52
  • Not sure what your definition of separating points is. Ignore me if this is not what you need. Let $x,y\in X.$ then $\exists s\in X, c\in\mathbb{R}, e>{0}$ such that $s\left(x\right)-e\leq c \leq s\left(y\right)+e$. Choose $s_n\rightharpoonup s$. Then $\exists n_x:\forall n\geq n_x \left|s_n\left(x\right)-s\left(x\right)\right|\leq\frac{e}{2}$ and $\exists n_y:\forall n\geq n_y \left|s_n\left(y\right)-s\left(y\right)\right|\leq\frac{e}{2}$ and thus for $n\geq n_0:=max\left(n_x,n_y\right) : s_n\left(x\right)-\frac{e}{2}\leq c \leq s_n\left(y\right)+\frac{e}{2}$ – Max Oct 01 '15 at 21:58

1 Answers1

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Suppose that $f(x)=0$ for all $f\in F$.

Let $g\in X^*$. Then there exists a net $\{f_j\}\subset F$ with $f_j\to g$ weak$^*$. So $$ g(x)=\lim f_j(x)=\lim 0=0. $$ Thus $g(x)=0$ for all $g\in X^*$, and then $x=0$.

Martin Argerami
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