characterization of weak sequential convergence in $l_1$ space with sliding hump method Show that weak sequential convergence in $l_1$ implies norm convergence and inverse, but the norm topology on $l_1$ is not equivalent to $\sigma(l_1,l_\infty)$
Hint: Use "sliding hump" method
Any response will be appreciated (even if not using the sliding hump method), thanks in advance.
 A: It is enough to show that if $(x_n)$ is a sequence in $\ell_1$ that converges weakly to 0, then $\Vert x_n\Vert\rightarrow 0$.  Suppose that $(x_n)$ is weakly convergent to $0$, but $\Vert x_n\Vert $ does not converge to 0. Without loss of generality, we may assume that there is a positive number $\alpha$ such that  $\Vert x_n\Vert>\alpha$ for all $n$.
Essentially, we will find a subsequence of $(x_n)$ such that for $i\ne j$, the coordinates of $x_i$ which give it most of its mass are disjoint from the coordinates of $x_j$ which give it most of its mass. Such a sequence cannot be weakly null (and in fact, upon normilazation,  would be equivalent to the unit vector (Schauder) basis of $\ell_1$). This would of course, be a contradiction.
Choose $\epsilon$ such that $\alpha/2-2\epsilon>0$.
We will use the notation $x_n\bigl|_{j\in A}$ to denote the vector $x_n$ restricted to the set $A$.
Set $y_1=x_1$.  Choose an integer $m_1$ so that $\Vert y_1\bigl|_{ j\le m_1 }\Vert\ge \alpha/2 $ and $\Vert y_1\bigl|_{ j> m_1 }\Vert<\epsilon$.
Since $(x_n)$ converges weakly to $0$, it follows that $(x_n)$ converges coordinatewise to $0$.  That is, for each $j$ we have $\lim\limits_{n\rightarrow\infty} x_n(j)=0$. Using this, we may find a positive integer $n_1>m_1$ such that for any $n\ge n_1$, we have  $\Vert x_n\bigl|_{j\le m_1}\Vert<\epsilon$.
Set $y_2=x_{n_1}$.  Choose an integer $m_2>m_1$ so that $\Vert y_2\bigl|_{m_1< j\le m_2 }\Vert\ge \alpha/2 $ and $\Vert y_2\bigl|_{ j> m_2 }\Vert<\epsilon$.
Next, choose $n_2>n_1$ so that for all $n\ge n_2$, we have  $\Vert x_n\bigl|_{j\le m_2}\Vert< \epsilon$.
Set $y_3=x_{n_2}$.  Choose an integer $m_3>m_2$ so that $\Vert y_3\bigl|_{m_2< j\le m_3 }\Vert\ge \alpha/2 $  and $\Vert y_3\bigl|_{ j> m_3 }\Vert<\epsilon$.
Continuing in this manner, we obtain a subsequence $(y_n)$ of $(x_n)$ which is not weakly null.
Indeed, define a sequence of signs $z$ so that for $m_{i-1}< j\le m_i$ the coordinate $z(j)$ agrees with the sign of $y_i(j)$.   Then $z\in\ell_\infty$ and, for each $i$, we have $|z(y_i) | \ge \alpha/2-2\epsilon$.
