Can someone tell me if I got the following right:
Assume $X$ to be a normed vector space over $\mathbb{R}$. Prove that if the dual space $X^\ast$ is separable then $X$ is separable as well.
I'm supposed to use the following hint: First show that for each $x_n^\ast$ we may choose a unit vector $x_n \in X$ such that $x_n^\ast (x_n) \geq \frac{\| x_n^\ast \|}{2}$. Then show that \overline{Y} = \overline{span_\mathbb{Q} \{ x_n\}} = X.
My answer:
(i) (by contradiction)
Assume for all $x_n \in X$ with $\| x_n \|_X = 1$ that
x_n^\ast (x_n) < \frac{\| x_n^\ast \|}{2} \iff 2 x_n^\ast (x_n) < \sup_{x_n; \| x_n \|_X = 1 } \{ | x_n^\ast(x_n) | \}
Claim: Then $\exists r \in \mathbb{R}: x_n^\ast(x_n) < r < \sup_{\dots}\{\dots \}$:
(i) $x_n^\ast (x_n) \neq 0$: because $\| x_n \| = 1$ and $x_n^\ast$ linear
(ii) if $x_n^\ast (x_n) < 0$ then $2 x_n^\ast (x_n) < x_n^\ast (x_n) < x_n^\ast(- x_n) < 2 x_n^\ast(-x_n) < \sup_{\dots} \{ \dots \}$
(iii) $x_n^\ast(x_n) > 0$ then $\forall x_n: x_n^\ast (x_n) < 2 x_n^\ast (x_n) < \sup$
$\implies $ the $\sup$ is not the l.u.b., contradiction, so the claim $x_n^\ast (x_n) \geq \frac{\| x_n^\ast \|}{2}$ is true.
(ii) Using:
$ x \in \overline{Y} \iff \lnot \exists $ bounded linear functional $f^\ast$ such that $f^\ast (y ) = 0 \forall y \in Y$ and $f^\ast (x) \neq 0$
Let $f^\ast \in X^\ast$. Then \{ x_n^\ast \} dense in $X^\ast \implies$
\forall \varepsilon > 0 \exists x_n^\ast : \| x_n^\ast - f^\ast \| = \sup_{x \in X: \| x \| \leq 1} \{ |x_n^\ast (x) - f^\ast(x)| \} < \varepsilon
But for $x_n \in Y \subset X (\| x_n \| = 1)$ we know $| x_n^\ast (x_n)| \geq x_n^\ast (x_n) \geq \frac{\| x_n \|}{2} > 0$
so if $f^\ast (y) = 0 \forall y \in Y$ then
$ 0 < \frac{\| x_n \|}{2} \leq |x_n^\ast (x_n)| = |x_n^\ast (x_n) - f^\ast(x_n)|$
$ \implies \exists \varepsilon: \sup |x_n^\ast (x_n) - f^\ast(x_n)| > \varepsilon$
$ \implies \lnot \exists f^\ast \in X^\ast : \forall y \in Y: f^\ast (y) = 0$
$ x \in \overline{Y}$
Thanks for your help.