TITLE: Separability of Banach Spaces
QUESTION [15 upvotes]: A homework problem from Folland Chapter 5, problem 5.25.
If $\mathcal{X}$ is a Banach space and $\mathcal{X}^{\star}$ is separable, then $\mathcal{X}$ is separable.
I tried the following approach: For every $\epsilon >0$ I wanted to show the existence of a linear map from $x_{1},\ldots,x_{n}$ such that for any $x\in\mathcal{X}$ $\| x-L(x_{1},\ldots,x_{n})\|\leq \epsilon$.

REPLY [3 votes]: This is a rather indirect proof. 
We know from Alaoglu's theorem that $B(H^{*})$ is weak-star compact. If we can prove $H^{*}$ is weak star metrizable, then we can show $H$ is separable. If I am not mistaken, proving $B(H^{*})$ is metrizable is enough to show $H^{*}$ is metrizable. By Urysohn's metrization theorem, a compact space is metrizable if and only if it is Hausdauff and second countable. But I think if $H$ is a Banach space over $\mathbb{C}$ or $\mathbb{R}$, then $H^{*}$ must be Hausdauff as functionals separate points. 
We still need to show $H^{*}$ is second countable. Given any open set $U$ in $H^{*}$, since $H^{*}$ is separable there is a countable subset $f_{n}$ dense in $U$. Let $x_{n}\not \in ker(f_{n})$ such that $f_{n}(x_{n})=1$, $f_{n}(x_{m})=0$. This is possible since $\bigcup \ker(f_{n})\not=H$ by Baire Category Theorem (each one of them is nowhere dense). Now the set $$U_{n,m}=\{f:\langle f,x_{n}\rangle<\frac{1}{m}\}$$must somehow cover $U$. So $H$ must be separable.