The weak*-convergence of the summing basis of $c_{0}$

Question

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.

Answer 1

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.