existence of a weakly cauchy sequence if the dual space is separable

Question

Let X be a normed space such that $X^*$ is separable. Given any sequence $(x_n)\in X$ then there exist a subsequence weakly cauchy

Answer 1

We will moreover suppose that $(x_n)$ is bounded, as the statement is wrong otherwise. Let $(x_n^*)$ be a dense sequence in $X^*$. As $\mathbb K$ is complete, there is a subsequence $(x_{n,1})_n$ of $(x_n)$ such that $\bigl(x_1^*(x_{n,1})\bigr)$ converges, by induction, we find subsequences $(x_{n,k})_n$ of $(x_{n,k-1})_n$ such that $\bigl(x_{i}^*(x_{n,k})\bigr)$ converges for $i \le k$. Now let $(y_n) := (x_{n,n})$, by construction $(y_n)$ is a subsequence of $(x_n)$ and $\bigl(x_i^*(y_n)\bigr)_n$ converges for every $i$. Now let $x^* \in X^*$ and $\epsilon > 0$. There is an $i \in \mathbb N$ such that $\|x^* - x_i^*\| \le \frac \epsilon {3\sup\|x_n\|}$, and an $N \in \mathbb N$ such that $\|x^*_i(y_n) - x_i^*(y_m)\| \le \frac \epsilon 3$ for $n,m \ge N$. Then $\|x^*(y_n) - x^*(y_m)\| \le \epsilon$ for $n, m \ge N$ and $(y_n)$ is weakly Cauchy.