3
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ What is the metric $d$ (in $\ell_\infty$ and in $\ell_\infty^{**}$)? $\endgroup$ Mar 1, 2021 at 7:44
  • $\begingroup$ The metric $d$ in $l_{\infty}$ and in $l_{\infty}^{**}$ is induced by the norm. $\endgroup$ Mar 1, 2021 at 8:17
  • $\begingroup$ I don't quite understand the notation: what does $\overline{ (s_n)_n }^{\sigma(l_\infty, l_1)}$ mean? From context, this is not the weak$^*$-closure of the set of values of the sequence. Is it means to be the set of weak$^*$-limit points of the sequence? $\endgroup$ Mar 1, 2021 at 8:59
  • 1
    $\begingroup$ @MatthewDaws I edited the question. I am sorry. $\endgroup$ Mar 1, 2021 at 9:15

1 Answer 1

1
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Thanks, Matthew. I'll check your argument. $\endgroup$ Mar 1, 2021 at 9:57
  • $\begingroup$ I check your argument and it is correct.. Many thanks! Matthew. $\endgroup$ Mar 3, 2021 at 2:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.