Prove that E is reflexive. QUESTION: Let $E$ be a Banach space where $E'$ is the dual space of $E$. Assume the $E'$ is separable and suppose that for all neighborhood $V$ of $\sigma(E', E'')$ is there exists a neighborhood of $W\in \sigma(E', E)$ such that $W\subset V$. Show that $E$ is reflexive.
MY ATTEMPT:
We have that $E$ is also separable, because $E$ is a Banach space, and $E'$ is separable. Therefore, to show that $E$ is reflexive, we only need to show that $E'$ is reflexive (by another theorem). Now, by Kakutani's theorem, it is enough show that the closure of $B_{E'}$ (the unit ball in $E'$) is compact in $\sigma(E', E'')$. Let $\{V_{\lambda}\}_{\lambda\in \Gamma}$ be an open cover of the ball $\overline{B_{E'}}$.
On the other hand, by Banach-Alaogly-Bourbaki's theorem, $\overline{B_{E'}}$ is compact in $\sigma(E', E)$. Now, from the statement we can infer that $\{W_{\lambda}\}_{\lambda \in \Gamma}$ is another an open cover of $\overline{B_{E'}}$, then is there exists $k\in \mathbb{N}$ such that $\overline{B_{E'}}\subset \displaystyle\bigcup_{k=1}^{n}W_{\lambda_k}$. And because $W_{\lambda_k}\subset V_{\lambda_k}$ (I'm not very sure in how to explain this). Thus, $\overline{B_{E'}}\subset \displaystyle\bigcup_{k=1}^{n}V_{\lambda_k}$. Therefore $\overline{B_{E'}}$ is compact in $\sigma(E, E'')$, then by Kakutani's theorem this implies that $E'$ is reflexive, which implies (by other result) that $E$ is itself reflexive.
DOUBT: Well, I do not feel that every step above is right. I mean I would like to give more explanations, to provide another proof more step by step.
Would you help me with this, please?
Thanks in advance.
P.S.: I thought maybe it is possible to use Eberlein-Smulian theorem's too.
Other idea, using @David Mitra
MY SECOND ATTEMPT:
Because $E'$ is separable then is there exists a sequence $(x_n)\in E$ such that $\|x_n\|=1$, $\forall \; n$ and holds that $x_n\rightharpoonup 0$ weakly. (I've proved this using that $E'$ is separable (by definition is there exist a subset $D$ which is contable and dense in $E'$.)). At the end I've used this result to prove that $\|f_n-f\|\leq \epsilon$ for every $f, f_n\in D$,$\forall \epsilon>0$. Thus forall $k>n$ we have that $\|f(x_k)-0\|\longrightarrow 0$ as $m\rightarrow +\infty$. Therefore, $x_k\rightharpoonup$ weakly. Then, there is $M>0$, such that $\|x_k\|\leq M$, i.e., $(x_k)$ is bounded. Hence, by the Eberlein-Smulian theorem this implies that $E$ is reflexive.
 A: To me the first attempt looks good. I'm posting my idea in reply form as it's too long for a comment, but in case I'm wrong you can let me know that I'll fix or delete the post. When you get the open coverage $\{V_{\lambda}\}_{\lambda\in \Gamma}$ in the topology $\sigma(E', E'')$, by hypothesis for each open $V_{\lambda}$ exists $W_{\lambda}$ open in $\sigma(E', E)$ such that $W_{\lambda} \subset V_{\lambda}.$ To use the Banach-Alaogly-Bourbaki's theorem, you need to ensure that the family $\{W_{\lambda}\}_{\lambda\in \Gamma}$ is coverage of $\overline{B_{E'}}$. This at first does not seem to be true for this family $\{W_{\lambda}\}_{\lambda\in \Gamma}$ even because the open $W_{\lambda}$ are inside $V_{\lambda}$. Hence I agree with @Berci argument: each $V_{\lambda}$ is the union $W_{\lambda}$ with the weak star neighborhoods of the points of $V_{\lambda}\setminus W_{\lambda}$, which are obtained considering for each $x \in V_{\lambda}\setminus W_{\lambda}$ the weak star neighborhood of the origin $0$, since $V_{\lambda}\setminus\{x\}$ contains $0$. If we denote these weak star neighborhoods by $W_x$, we have to $V_{\lambda} = W_{\lambda} \cup \left( \bigcup_{x \in V_{\lambda}\setminus W_{\lambda}} W_x \right):=W_{\lambda,x}.$ With a little more detail to find the neighborhoods $W_x$ so that the equality is true: for every $x \in V_{\lambda}$ there is a basic neighborhood in the weak topology $V(x, \varepsilon)$ containing $x$. So when you translate $V_{\lambda}$ to the origin, $V(x, \varepsilon)$ becomes a basic neighborhood of the origin in the weak topology. Since $\overline{B_{E'}}$ is metrizable, I can choose a weak star neighborhood contained in the translated base neighborhood. Translating this by $x$, we arrive at $W_x$.
Using the Banach-Alaogly-Bourbaki's theorem for the cover $\{W_{\lambda,x}\}_{\lambda \in \Gamma}$ of $\overline{B_{E'}}$, exists a finite subcover say $\{ W_{\lambda_1,x}, \cdots , W_{\lambda_n,x} \}$. Therefore $\{ V_{\lambda_1}, \cdots , V_{\lambda_n} \}$ is a finite subcover of $\overline{B_{E'}}$.
A: The argument you mention at the beginning of your proof shows that it is enough if you show that $E'$ is reflexive. Let's do that.
Let $\mathcal{W}$ be the neighborhood filter of $0$ in $\sigma(E', E)$, and $\mathcal{V}$ be the neighborhood filter of $0$ in $\sigma(E', E'')$.
All topologies in concern are "translation invariant". Therefore, it is enough if you show that $\mathcal{V} = \mathcal{W}$. You already know that $\mathcal{W} \subset \mathcal{V}$, because $E \subset E''$.
Take $V \in \mathcal{V}$. Then, the hypothesis say that there is $W \in \mathcal{W}$ such that $W \subset V$. And this implies that $V \in \mathcal{W}$. That is,
$$\mathcal{V} \subset \mathcal{W}.$$
Now, I don't really understand what you mean by neighborhood of $\sigma(E', E)$... in the above, I assume you mean that both are neighborhood of the same point. If this is not the case, you can take $W \subset \mathrm{int}(V)$ and "translate $V$ and $W$" so that they are now both neighborhoods of $0$.
