 Solution

Suppose $a^k\rightharpoonup a$ in ${\ell }^1$ but $a^k$ does not converge to $a$ in the norm topology. There are $\epsilon >0$ and a subsequence $(a^{m_k})$ such thatSet $x^k=a^{m_k}-a$. Thenwhich proves the first assertion.

We now use a gliding hump argument. Put $b_0=0$. Having chosen $b_{n-1}$ and $k_{n-1}$, coordinatewise convergence $x_i^k\to 0$ lets us choose $k_n>k_{n-1}$ so thatFor this fixed element of ${\ell }^1$, choose $b_n>b_{n-1}$ so far out thatDefine one sequence $y$ byThe blocks partition the positive integers and $|y_i|\le 1$, so $y\in {\ell }^{\infty }$. On the $n$th assigned block, the signs agree; outside it, use $|y_i|\le 1$. The duality of l1 and l infinity givesBut $x^{k_n}\rightharpoonup 0$ requires $⟨x^{k_n},y⟩\to 0$ for this fixed $y\in ({\ell }^1{)}^{\ast }={\ell }^{\infty }$, a contradiction. Therefore every weakly convergent sequence in ${\ell }^1$ converges in norm. This is the Schur property of l1.