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?

  • $\begingroup$ 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

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.

  • 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

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