Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
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: math.stackexchange.com/questions/82385/… –  user39158 May 7 at 10:01

2 Answers 2

up vote 5 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 $\alpha\ne\beta$.

share|improve this answer
    
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

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.