Let $X=\ell^\infty$, $f_m \in X^*$ be defined as $f_m(\langle x_n\rangle)=x_m$. Does $(f_n)_{n\in \bf N}$ has a weak* convergent subsequence? 
Let $X=\ell^\infty$, $f_m \in X^*$ be defined as $f_m(\langle x_n\rangle)=x_m$.Does $(f_n)_{n\in \bf N}$ has a weak* convergent subsequence?

I think $(f_n)$ does not have convergent subsequence but unable to prove this properly. Please help!  

The sequence $(f_n)$ does not have a weak-* convergent subsequence. To see this, suppose $f_{n_j}$ is a subsequence converging weak-* to some $f \in (\ell^\infty)^*$ as $j \to \infty$. Thus
$$x_{n_j} = f_{n_j}(x) \to f(x)$$
as $j \to \infty$ for every $x \in X = \ell^\infty$.
Define $x \in \ell^\infty$ by $x_{n_j} = (-1)^j$ and $x_n = 0$ if $n$ is not equal to any $n_j$.  Then the sequence $(x_{n_j})$ does not converge to anything, which contradicts the displayed statement above.
Although $(f_n)$ does not have a weak-* convergent subsequence, it does have a weak-* convergent subnet.
