Does weak* convergence together with convergence of norms implies strong convergence in l1 as dual of c0? Let $x^{*}_{n}$ is weak$^*$ convergent to $x^*$ in weak$^*$ topology on $l_{1}$ induced by $c_{0}$ and $\|x^{*}_{n}\|\rightarrow \|x^{*}\|.$ It is true that then we have $\|x^{*}_{n}-x^{*}\|\rightarrow 0$?
 A: The answer is yes.  In order to prove it, let me make a change in your notation, denoting the given sequence  by
$\{x^n\}_n$, the limit by $x$, where each $x^n = (x^n_i)_i$, and  $x = (x_i)_i$.
Given  $\varepsilon >0$, choose an integer $p>0$ such that
$$
  \sum_{i=p+1}^\infty  |x_i| <\varepsilon ,
  $$
and observe that, for every $n$
$$
  \sum_{i=p+1}^\infty  |x^n_i| =   \|x^n\| - \sum_{i=1}^p |x^n_i|
  \ \buildrel n\to \infty  \over {\longrightarrow} \
  \|x\| - \sum_{i=1}^p |x_i| =   \sum_{i=p+1}^\infty  |x_i| <\varepsilon ,
  $$
so there is an $n_0$ such that
$$
  \sum_{i=p+1}^\infty  |x^n_i| <\varepsilon ,
  $$
for all $n\geq n_0$.  By increasing $n_0$ we may assume that also
$$
  \sum_{i=1}^p |x_i-x^n_i|<\varepsilon .
  $$
For $n\geq n_0$,   we than have that,
$$
  \|x-x^n\| =
  \sum_{i=1}^\infty  |x_i-x^n_i| = $$$$ =
  \sum_{i=1}^p |x_i-x^n_i| +   \sum_{i=p+1}^\infty  |x_i-x^n_i| \leq  $$$$ \leq 
  \sum_{i=1}^p |x_i-x^n_i| +   \sum_{i=p+1}^\infty  |x_i| +   \sum_{i=1}^p |x^n_i|  < 3\varepsilon ,
  $$
so   $\|x-x^n\| \to  0$, as desired.
