Show that $\mu_n=\frac1n\sum_1^n\epsilon(j)$ has infinitely many weak$^*$ limit point

Question

Let $\mu \in (\ell^\infty)^*$ defined by $$\mu_n=\frac1n\sum_1^n\epsilon(j)$$ Where $\epsilon(j)(x_n)=x_j$ for $(x_n)\in\ell^\infty$.

Show that $\mu_n=\frac1n\sum_{j=1}^n\epsilon(j)$ has infinitely many weak$^*$ limit point.

I think since $\mu_n\in B_{\ell^\infty}^*$ (the closed unit ball of $(\ell^\infty)^*$) so by Banach-Alaoglu $\mu_n$ has at least one limit point. But how to show the existence of infinitely many weak$^*$ limit point?

Thanks.

Answer 1

Major hint: Let $$E=\{\mu_n:n=3^{2k}\}$$and $$O=\{\mu_n:n=3^{2k+1}\}.$$You can explicitly write down an $x\in\ell^\infty$ such that $\mu_nx\ge1$ for every $\mu_n\in E$ while $\mu_nx\le-1$ for every $\mu_n\in O$. So $E$ and $O$ are contained in disjoint closed sets. That shows there are at least two limit points; modify this to show that for every $n\in\Bbb N$ there are at least $n$.