Suppose that $(x\_{n})\_{n}$ is a sequence in a Banach space $X$. We let $\textrm{clust}\_{X^{\*\*}}((x\_{n})\_{n})$ be collection of all the weak\*-limit points of $(x\_{n})\_{n}$ in $X^{\*\*}$.


Let $(e\_{n})\_{n}$ be the unit vector basis of $c\_{0}$. Let $s\_{n}=\sum\limits\_{i=1}^{n}e\_{i}(n=1,2,\cdots)$. It is easy to see that $s\_{n}\rightarrow e\_{0}=(1,1,\cdots)$ in $\sigma(l\_{\infty},l\_{1})$. Hence $\textrm{d}(\textrm{clust}\_{c\_{0}^{\*\*}}((s\_{n})\_{n}),c\_{0})=1$.
If we consider $(s\_{n})\_{n}$ as a sequence of $l\_{\infty}$, what is
$\textrm{d}(\textrm{clust}\_{l\_{\infty}^{\*\*}}((s\_{n})\_{n}),l\_{\infty})$ ?


Thank you.



The weak*-convergence of the summing basis of $c_{0}$
For a Banach space $E$, let $\kappa = \kappa\_E:E\rightarrow E^{\*\*}$ be the canonical inclusion. Consider
$$ E^\perp = \{ M\in E^{\*\*\*} : M(\kappa\_E(x))=0 \ (x\in E) \} = \ker\kappa\_E^\*. $$
A simple calculation shows that $\kappa\_E^\* \circ \kappa\_{E^\*} = 1\_{E^\*}$ and so $\kappa\_{E^\*}\circ\kappa\_E^\*$ is a projection of $E^{\*\*\*}$ onto (the image of) $E^\*$ with complementary subspace $E^\perp$. In nice situations (like when $E=c\_0$) you even get an $\ell\_1$-direct sum.


Let $(x\_n)$ be a sequence in $E$, and consider the sequence $(\kappa\_E(x\_n))$ in $E^{\*\*}$. Consider the sequence $(\kappa\_{E^{\*\*}}\kappa\_E(x\_n))$ in $E^{(4)}$, and let a subnet converge to $\mu$ in $\sigma(E^{(4)}, E^{\*\*\*})$.
By moving to a sub-subnet if necessary, we may suppose that our subnet, in $E^{\*\*}$, converges to $F$ in $\sigma(E^{\*\*}, E^\*)$. For $M + \kappa\_{E^\*}(f) \in E^\perp \oplus E^\* \cong E^{\*\*\*}$, we see that
$$ \mu(M+\kappa(f)) = F(f), $$
because $M(\kappa\_E(x\_n))=0$ for all $n$. Thus
$$ \mu(N) = F(\kappa\_{E}^\*(N)) = \kappa\_E^{\*\*}(F) (N) \qquad (N\in E^{\*\*\*}). $$
So $\mu = \kappa\_E^{\*\*}(F)$. (**Note:** This is not the same as $\kappa\_{E^{\*\*}}(F)$. It is a common and easy mistake to think this!)


We want to compute $d(\kappa\_E^{\*\*}(F), \kappa\_{E^{\*\*}}(E^{\*\*}))$. Given $N\in E^{\*\*\*}$ and $G\in E^{\*\*}$,
$$ \kappa\_E^{\*\*}(F)(N) - \kappa\_{E^{\*\*}}(G)(N)
= F(\kappa\_E^\*(N)) - N(G). $$
Let $N = \kappa\_{E^\*}(f) + M$ for some $f\in E^\*$ and $M\in E^\perp$, so we get
$$ F(f) - G(f) - M(G) = (F-G)(f) - M(G). $$
If $E^{\*\*\*} = E^\* \oplus\_1 E^\perp$ then taking the supremum over $\|N\|=1$ is the same as taking the supremum over $\|f\| + \|M\|=1$, which yields
$$ \max\big( \|F-G\|, \|G\|\big). $$
We conclude that
$$ d(\kappa\_E^{\*\*}(F), \kappa\_{E^{\*\*}}(E^{\*\*})) = \inf\_{G\in E^{\*\*}}
\max\big( \|F-G\|, \|G\|\big). $$


In your case, $E=c\_0$ and $x\_n=s\_n \rightarrow 1$ in $\sigma(E^{\*\*},E^\*)$. We have the $\ell^1$-sum property, and so the value you want is
$$ \inf\_{G\in \ell^\infty} \max\big( \|1-G\|\_\infty, \|G\|\_\infty\big) = \frac12. $$
I must say that I find this result counter-intuitive, but I believe the argument is correct.


