Does weak* convergence together with convergence of norms implies strong convergence in l1 as dual of c0?

Question

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 $\|x^{*}_{n}-x^{*}\|\rightarrow 0$?

Answer 1

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.