4
$\begingroup$

I am currently reading through the proof of the following result: If the dual of a Banach space $X$ is separable, then $X$ is separable.

Proof: Let $\{ f_n \}_{n=1}^{\infty}$ be a dense subset of the unit ball $\mathcal{B}$ in $X^{\ast}$. For each $n \in \mathbb{N}$, pick $x_n \in X$ such that $f_n(x_n) > \frac{1}{2}$. Let $Y = \overline{\text{span}\{x_n\}}$ and observe that $Y$ is separable, since finite rational combinations of $\{ x_n \}$ are dense in $Y$. It is now sufficient to show that $X=Y$. We proceed by contradiction. Suppose that $X \neq Y$. Then there is an $f \in X^{\ast}$, with $\| f \| =1$ such that $f(x) =0$ for all $x \in Y$. Now choose $n$ such that $\| f_n - f \| < \frac{1}{4}$. Then \begin{eqnarray*} \left| f(x_n) \right| & = & \left| f_n(x_n) - f_n(x_n) + f(x_n) \right| \\ & \geq & \left| f_n(x_n) \right| - \left| f_n(x_n) - f(x_n) \right| \\ & \geq & \left| f_n(x_n) \right| - \| f_n - f \| \cdot \| x_n \| \\ & > & \frac{1}{2} - \frac{1}{4} = \frac{1}{4}. \end{eqnarray*}

I understand the mechanics of this proof, but want to understand the rationale behind the structure of the proof.

So, we want to start with a countably dense subset of the dual, and show that this necessitates the existence of a countably dense subset of $X$.

 1. Why choose the dense subset to be on the unit ball? 
 2. Is there any foresight that one may grasp on thinking of a Banach space as the
 closure of the span of a sequence on the unit ball of the dual? 
 3. I do not see where the assumption of f(x)=0 is used in the inequality. 

Does anyone have a more intuitive proof of this result?

$\endgroup$
  • $\begingroup$ My edit was to fix a typo. In the proof it should say "pick $x_n\in X$", not "pick $x_n\in \mathcal B$". $\endgroup$ – DanielWainfleet Aug 10 '17 at 5:09
  • $\begingroup$ There is a flaw in the proof. You should say that $\{f_n\}$ is dense in the unit $sphere,$ so each $\|f_n\|=1.$ And you should say that each $\|x_n\|=1$ for each $n$. So in the second-last line we have $\|f_n-f\|\cdot \|x_n\|<1/4\|x_n\|=1/4.$.... The idea is that if $\|f\|=1$ then $f(x_n)$ can't be $ 0$ for all $n,$ else $\|f-f_n\|\geq |f(x_n)-f_n(x_n)|>1/2$ for all $n$, contradicting the denseness of $\{f_n\}_n$ on the unit sphere. $\endgroup$ – DanielWainfleet Aug 10 '17 at 5:38
6
$\begingroup$

A normed space has the property (that all metric spaces have) that $X$ is separable then all subsets are separable too in their subspace topology. So $X^\ast$ (norm)-separable implies that $B_{X^\ast}$ is separable. The reverse also holds in all normed spaces $Y$ by "scaling": if $x \neq 0$ then choose $\alpha = \frac{1}{\|x_n\|}$ so that $\|\alpha x\| =1$. If we then (by separability of the unit ball) find a sequence $d_n$ on the ball $B_Y$ that converges to $\alpha x$, and also a sequence of rationals $q_n \to \frac{1}{\alpha}$. Then $q_n d_n \to \frac{1}{\alpha} (\alpha x) = x$. This essentially shows that if $D$ is countable and dense in $B_Y$ then $\{qd: q \in \mathbb{Q}, d \in D\}$ is (countable and) dense in $Y$. So $B_Y$ separable iff $Y$ separable. So we lose nothing by using the unit ball (here the sphere really). And the Hahn-Banach theorem which links things in $X$ to $X^\ast$, gives us functionals on $B_{X^\ast}$ anyway.

Using the unit sphere makes things easier, because you know the norm of all dense elements, namely 1, which allows for the choice of the $x_n$ (otherwise we'd need to scale there too which makes for a more messy proof). We have to prove something on $X$, so we can find points $x_n$ on which $f_n$ is relatively large: we know that $\|f_n\| = \sup \{|f_n(x)|: x \in B_X \}$, so we can find $x_n \in B_X$ such that $|f(x_n)|$ is as close to $1$ as we like. Here more than $\frac{1}{2}$ is sufficient.

As above a countable dense set in $B_X$ is enough to get one on $X$, using the span with rational coefficients. So try that for the $D = \{x_n: n \in \mathbb{N}\}$ we now have: taking finite sums from $D$ with rational coefficients we can approximate all members of the linear span of $D$ (just approximate real coefficients in $\mathbb{R}$ by members of $\mathbb{Q}$; we use that the field is separable). This $\operatorname{span}_{\mathbb{Q}}(D)$ is still countable (standard set theory argument: finite products of countable sets are countable and a union of countably many countable sets is countable). So $Y = \overline{\operatorname{span}(D)}$ has a countable dense set $\operatorname{span}_{\mathbb{Q}}(D)$. So we'd be done if $Y =X$. So assume it's not.

Then, Hahn-Banach allows us to find a functional $f$ (back to $B_{X^\ast}$ where we know something about the $f_n$) that has norm $1$ and is $0$ on $Y$. (In particular $f$ is such, that it is $0$ on the set $D$ where the $f_n$ are chosen to be large, and so the $f$ is very "different" from the $f_n$ and so we cannot approximate it in norm by the $f_n$.)

So we know $f(x_n) = 0$ (as $x_n \in Y$!) and the final part of the proof shows that if $f_n$ is chosen to be close to $f$, the point $x_n$ where $f_n$ is large shows that $f$ should also be at least $\frac{1}{4}$ in that point as well. This gives the needed contradiction.

$\endgroup$
  • 1
    $\begingroup$ For those familiar with ordinals we can show that If $X$ is not separable then $X^*$ isn't separable, as follows: For $S\subset X$ let $B(S)$ be the closed linear subspace generated by $S$ (with $B(\phi)=\{0\}$). For $a< \omega_1$ let $f_a \in X^*$ with $\|f_a\|=1$ and $ f(B(\{x_b: b<a\})=\{0\}$ and let $x_a\in X$ with $\|x_a\|=1$ and $f_a(x_a)>1/2$. Then $\{f_a: a<\omega_1\}$ is an uncountable closed discrete subset of $X^*,$ because $b<a<\omega_1$ implies $\|f_b-f_a\|\geq |f_b(x_b)-f_a(x_b)|/\|x_b\|>1/2.$ $\endgroup$ – DanielWainfleet Aug 10 '17 at 6:06
  • $\begingroup$ $ \mathbb R $ is separable normed space. Is the set of irrational numbers separable in the subspace topology? $\endgroup$ – izimath May 30 at 14:18
  • $\begingroup$ @izimath it is. All rational linear combinations of $\sqrt{2}$ are dense, e.g. $\endgroup$ – Henno Brandsma May 30 at 14:20
  • $\begingroup$ Then how can I prove the general result? Can you give some guidelines? $\endgroup$ – izimath May 30 at 14:22
  • $\begingroup$ @izimath here I show that $X$ separable (metric) implies that $X$ has a countable base (topologically!) and this implies that all subspaces have a countable base and thus all subspaces are also separable. $\endgroup$ – Henno Brandsma May 30 at 14:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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