I'm reading Royden Chap 15. The last remark in section 1 reads:
For $X = l^{\infty}$, $B^*$, the closed unity ball of $X^*$, is not weak-* sequentially compact. Indeed the sequence $\{\psi_n\}\subset B^*$ defined for each $n$ by $$\psi_n(\{x_k\})=x_n \qquad \forall \{x_k\}\in l^{\infty}$$ fails to have a weak-* convergent subsequence.
I am having trouble seeing how $\{\psi_n\}$ has no weak-* convergent subsequence. I'm having trouble even stating what I want to show. I think what I want to show is:
$$\exists (c>0 \text{ and }\hat{x}\in X^{**}) \, \forall(n,m\in\mathbb{N})\text{ s.t. }c<|\hat{x}(\psi_n)-\hat{x}(\psi_m)|$$
Best Answer
Let's start by writing down what it would mean to have a weak$^*$-convergent subsequence of $\{\psi_n\}_{n \geq 1}$.
We would then have an increasing sequence $n_k$ and $\psi \in X^*$ such that for every $x \in \ell^\infty$, $\psi_{n_k}(x) \to \psi(x)$ as $k \to \infty$. Note that $\psi_{n_k}(x) \to \psi(x)$ if and only if $x_{n_k} \to \psi(x)$.
It is then not too hard to see that this cannot happen, since you can take $x$ to be the sequence $$x_n = \begin{cases} (-1)^k \qquad n = n_k \\ 0 \qquad \text{otherwise} \end{cases}$$ and then $\psi_{n_k}(x) = (-1)^k$ does not even converge.