If $X$ is a separable normed linear space, then we know that every bounded sequence in $X^*$ has a weak-* convergent subsequence . Can we drop the separability condition , i.e. if we don't assume $X$ is separable, then are there counterexamples ?
Please help. Thanks in advance
Take $X=l^\infty$, which is not separable. Define the sequence $f_n$ in $(l^\infty)^*$ by $$ f_n (x) = x_n. $$ Then $(f_n)$ is a bounded sequence, in fact, $\|f_n\|_{(l^\infty)^*}=1$.
However, it does not have a weak-star converging subsequence. Let $(f_{n_k})$ denote a subsequence. Then define $x\in l^\infty$ by $$ x_{n_k}=(-1)^k, $$ set all other entries $x_i=0$. Then $$ f_{n_k}(x) = (-1)^k, $$ which is not convergent.
