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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 $x\in\mathcal{X}$ $\| x-L(x_{1},\ldots,x_{n})\|\leq \epsilon$.

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the sentence in which you describe your approach does not seem to be complete. – Mariano Suárez-Alvarez Feb 8 '12 at 6:16
The book provides a hint for this problem: Let $\{f_n\}_1^\infty$ be a countable dense subset of $\mathcal{X}^*$. For each $n$ choose $x_n\in\mathcal{X}$ with $\|x_n\|=1$ and $|f_n(x_n)|\geq \frac{1}{2}\|f_n\|$. Then the linear combinations of $\{x_n\}_1^\infty$ are dense in $\mathcal{X}$. – john w. Feb 8 '12 at 6:24
sorry I forgot to mention the hint. – user24367 Feb 8 '12 at 6:34
Same question:… – user39158 May 7 '14 at 10:01
up vote 10 down vote accepted

Use the Hahn-Banach Theorem:

Taking $f_n$ and $x_n$ as in your hint.

Let $Y$ be the set of all linear combinations of the $x_i$ with rational coefficients.

Suppose $Y$ were not dense in $X$. Then the closure of $Y$ is a proper subspace of $X$, and thus, there is an $f\in X^*$ of norm 1 with $f(Y)=\{0\}$. Then $$ {1\over 2}\Vert f_n\Vert\le|f_n(x_n)| =|f_n(x_n) - f(x_n)| \le \Vert f_n-f\Vert \Vert x_n\Vert =\Vert f_n-f\Vert $$

Take $\Vert f_{n_i}-f\Vert\rightarrow 0$. Then from the above, $\Vert f\Vert=0$, a contradiction.

You could also use Riesz' lemma:

Let $Y$ be a proper closed subspace of the normed space $X$ and $0<\theta<1$. Then there is an $x_\theta$ of norm 1 for which $\Vert x_\theta-y\Vert>\theta$ for all $y\in Y$.

If $X$ were not seperable, you could use Hahn Banach to construct uncountably many functionals $f_\alpha\in X^*$ with $\Vert f_\alpha-f_\beta\Vert\ge \theta$ whenever $\alpha\ne\beta$.

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And, of course, Hahn-Banach (or some other consequence of the Axiom of Choice) is essential. The non-separable Banach space $l_\infty / c_0$ has dual $\{0\}$, as far as we can tell without AC... – GEdgar Feb 8 '12 at 15:26
Why is $f(Y) = 0$ in your fourth sentence? – Tyler Hilton Dec 9 '13 at 1:45
Could you please expand a bit on your Riesz lemma argument? I don't see how to construct those functionals. – Potato Aug 14 '15 at 23:01

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.

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