Construct a sequence in a Banach Space 
The following two are equivalent.

*

*For all $x \in B_E = \{y \in E:\|y\| \leq 1 \}$ there is a sequence $(x_n) \subseteq E$ such that $\|x_n\|=1$ and $x_n \rightarrow_w x$ (weak
convergence).

*There is a sequence $(x_n) \subseteq E$ such that $x_n \rightarrow_w 0$ and $x_n \nrightarrow 0$ (convergence in norm).


$1 \Rightarrow 2$ is trivial. I need some help or hint about how to construct the sequence for $2 \Rightarrow 1$.
 A: I will make the following assumption:
(2'): $\exists (x_n)\subset E$ and $c>0$ such that $x_n$ converges weakly to $0$ and $\|x_n\|>c$ for all $n$.
This follows from (2), since non-convergence in norm implies the existence of a subsequence bounded away from $0$. (As a side note, this also implies (2).)
Let $(x_n)$, $c$ satisfy (2'), and let $x\in B_E$. Let $y_n = x + \alpha_n x_n$, where $\alpha_n\ge 0$ is chosen such that $\|y_n\| = 1$. To see that this is possible, let $f(\alpha) = \|x+\alpha x_n\|$, and note that
$$f(\alpha) = \|\alpha x_n + x\|\ge |\alpha|\|x_n\|-\|x\| \ge c|\alpha|-\|x\|. $$
Since $f$ is continuous, $f(0) = \|x\|\le 1$ and $f(\alpha)\rightarrow\infty$ as $\alpha\rightarrow\infty$, it follows from IVT that there exists $\alpha_n$ such that $f(\alpha_n) = 1$. Moreover, we have
$$ 1 = f(\alpha_n)\ge c\alpha_n - \|x\|\implies \alpha_n\le \frac{1+\|x\|}{c}\le \frac{2}{c}. $$
Since $\alpha_n$ is uniformly bounded, the sequence $\alpha_nx_n$ converges weakly to $0$, and so $y_n$ converges weakly to $x$, while by construction $\|y_n\| = 1$.
