Checking a machinery heavy proof for "separability of continuous dual implies separability of original space"

Question

For the following

Proposition. $X$ normed space. If $X^*$ separable, then $X$ separable.

I'm aware of an elementary proof, as presented here. However, I prefer using theorems, instead of using tricks. So I wonder if my proof below is correct. The listed theorems/lemmas are allowed to be used.

Thm. $X$ Banach separable. Then closed unit ball $ball X^*$ of the dual is $wk^*$-metrizable.

Banach-Alaoglu. $X$ normed. Then $ball X^*$ is $wk^*$ compact.

Lemma. $X$ normed. Then the canonical map $\pi: X\to X^{**}$ is a $wk$-$\sigma(X^{**}, X^*)$ embedding, and $\pi(X)$ is $\sigma(X^{**}, X^*)$ dense in $X^{**}$.

Lemma'. $X$ locally convex with top $\tau$. A subset $C\subset X$ nonempty and convex. Then $cl_{wk} C=cl_\tau C$.

Lemma''. A compact metric space $Y$ is totally bounded and separable. If defining $$E_n:=\left\{y_{n}^{k}\in Y: \bigcup_{k=1}^{m_n} B_Y\left(y_n^k, 1/n\right)=Y, m_n< +\infty\right\},$$ then $E:=\bigcup_{n}E_n$ is a countable dense subset of $Y$. Moreover, if $Y'\subset Y$ is a dense subset, then $Y'$ is also separable and contains a countable subset $E'$ that is dense in $Y$.

Proof of Proposition. (We use $wk^{**}$ to denote $\sigma(X^{**}, X^*)$ topology of the double dual $X^{**}$)

Since $X^*$ is Banach and separable. By Thm, $Y:=ball X^{**}$ is $wk^{**}$-metrizable. It is also compact by Banach-Alaoglu. Let $d$ be a metric inducing the relative $wk^{**}$ topology on $Y$. Then, $(Y, d)$ is metric compact. By lemma, $\pi(ball X)$ is $d$-dense in $Y$. By lemma'', there exists a countable set $D\subset ball X$ s.t. $\pi(D)$ is $d$-dense in $Y$. This gives $$cl_{wk}D=ball X.$$

Then, consider the convex hull $co(D)$. We have $D\subset co(D)\subset ball X.$ Thus, lemma' gives $$cl_{\|\cdot\|}co(D)=cl_{wk}co(D)=cl_{wk}D=ball X.$$

However, the convex hull is not countable. To fix this, take $D'$ as the rational convex hull of $D$. That is $$D':=\left\{\sum_{i=1}^m r_ix_i: x_1,\cdots, x_m\in D; r_1,\cdots, r_m\in \mathbb{Q}\cap [0,1]; \sum_{i=1}^mr_i=1\right\}$$ Then $D'\subset ball X$ is countable. Note $D'$ is $\|\cdot\|$-dense in $co(D)$, and so we finally get $$cl_{\|\cdot\|}D'=cl_{\|\cdot\|} co(D)=ball X.$$ $\square$

I have not seen any proof taking this route. But I can't seem to find any error. Thanks in advance.

Answer 1

Here are a few minor observations about your theorems and lemmas.

• You have two results called "Thm". It would be more clear if you made a distinction between them.

• If $X$ is a separable normed space, then the assumption that $X$ is complete is not needed to conclude that the closed unit ball of $X^{*}$ is metrisable with respect to the weak$^{*}$ topology.

• In Lemma' the expression "$cl_{wk} A=cl_\tau A$" should presumably be replaced with "$cl_{wk} C=cl_\tau C$".

• The set in Lemma'' does not appear to be well-defined. I assume you meant that for each $n\in\mathbb{N}$ there is some $m_{n}\in\mathbb{N}$ and a finite set $\{y_{n}^{1}, \ldots , y_{n}^{m_{n}} \} \subseteq Y$ such that $\bigcup_{k=1}^{m_{n}} B_{Y}(y_{n}^{k}, 1/n ) = Y$.

Regarding your main result, the proof is fine. In particular, there is no problem with the argument involving the convex hull and the rational convex hull. However, since you are already using many results that are much deeper than needed such as the Banach-Alaoglu theorem, you can shorten the proof by using another such result, namely that a subset $A\subseteq X$ of the normed space $X$ is separable with respect to the weak topology if and only if it is separable with respect to the norm topology. See here for more details. Then you can argue as follows.

Suppose $X$ is a normed space such that $X^{*}$ is separable. By a metrisability result and the Banach-Alaoglu theorem we have that $({\rm ball} \, X^{**}, {\rm weak}^{*})$ is compact and metrisable. In particular, we have that $({\rm ball}\, X^{**}, {\rm weak}^{*})$ is separable. Let $\pi \colon X \to X^{**}$ denote the canonical embedding. By Goldstine's theorem we have that $({\rm ball}\, X, {\rm weak})$ is homeomorphic to $(\pi ({\rm ball}\, X), {\rm weak}^{*})$. Since a subset of a separable metrisable space is separable, it follows that $({\rm ball}\, X, {\rm weak})$ is separable. Furthermore, as a subset of $X$ is separable with respect to the weak topology if and only if it is separable with respect to the norm topology, it follows that $({\rm ball}\, X, {\rm norm})$ is separable. Therefore $X$ is separable.