I am studying the proof of the Eberlein-Smulian Theorem via basic sequences in the book "Topics in Banach Space Theory" from F.Albiac and N. Kalton. Although, I found myself stuck in the following criterion (Theorem 1.5.6 from the book)
Let $S$ be a bounded subset of a Banach space $X$ such that $0\notin\overline{S}^{\Vert \cdot\Vert}$. Then the following are equivalent:
- $S$ fails to contain a basic sequence;
- $\overline{S}^{weak}$ is weakly compact and fails to contain $0$.
$(2)\implies (1)$ is done
$(1) \implies (2)$: I've already proved that $0\notin \overline{S}^{weak}$. To show that $\overline{S}^{weak}$ is weakly compact, we need to show that any weak* cluster point of $i(S)\subset X^{**}$ is already in $i(X)$ where $i:X\longrightarrow X^{**}$ is the canonical embedding.
Once we prove it, we will have that $\overline{S}^{weak}=i^{-1}(\overline{i(S)}^{weak^*})$ is the pre-image of a w*-compact.
Suppose $x^{**}$ is a weak* cluster point of $i(S)$ and $x^{**} \in X^{**}\setminus i(X)$. Using the theorem 1.5.2, the set $i(S) - x^{**} = \{i(s)-x^{**}: s\in S \}$ there is a sequence $(x_n)\subset S$ such that $(i(x_n) - x^{**})$ is basic.
I couldn't prove that: $x^{**}\notin \overline{[i(x_n) - x^{**}: n>N]}^{\Vert\cdot\Vert}$ for some choice of $N \in \mathbb{N}$
I tried to write $x^{**}=\sum\limits_{n=m}^\infty \lambda_n [i(x_n) - x^{**}]$ to get a contradiction with the fact that $0\in \overline{i(S) - x^{**}}^{weak*}\setminus\overline{i(S) - x^{**}}^{\Vert\cdot\Vert}$ but I couldn't conclude anything from it.
Find $N\in\mathbb{N}$ so that $x_n\neq 0$ for $n\geq N$, and assume towards a contradiction that $x^{**}\in[x_n-x^{**}]_{n=N}^\infty$. Write \begin{equation}x^{**}=\sum_{n=N}^\infty\lambda_n(x_n-x^{**}).\end{equation} Set $a_k:=1+\sum_{n=N}^k\lambda_n$, and find a subsequence $(a_{k_j})$ s.t. for all $j$, $|\frac{a_{k_j}}{|a_{k_j}|}-\alpha|<\frac{\epsilon}{\|x^{**}\|}$ for some nonzero $\alpha\in\mathbb{C}$, and either $|a_{k_j}|\geq 1$ or else $0<\frac{1}{|a_{k_j}|}<\frac{2}{\beta}$ for some $\beta\in(0,1]$. Note that $\forall\epsilon>0\exists M\geq N$ s.t. if $j\geq M$, \begin{equation}\|a_{k_j}x^{**}-\sum_{n=N}^{k_j}\lambda_nx_n\|=\|x^{**}-\sum_{n=N}^{k_j}\lambda_n(x_n-x^{**})\|<\epsilon.\end{equation} Then \begin{equation}\|\alpha x^{**}-\frac{1}{|a_{k_j}|}\sum_{n=N}^{k_j}\lambda_nx_n\|\leq\frac{a_{k_j}}{|a_{k_j}|}x^{**}-\frac{1}{|a_{k_j}|}\sum_{n=N}^{k_j}\lambda_nx_n\|+\|\alpha x^{**}-\frac{a_{k_j}}{|a_{k_j}|}x^{**}\|\end{equation} \begin{equation}<\frac{\epsilon}{|a_{k_j}|}+\epsilon<\frac{2\epsilon}{\beta}+\epsilon\to 0\end{equation} as $\epsilon\to 0$. It follows that $\alpha x^{**}\in X$ and hence $x^{**}\in X$, which is impossible.