0
$\begingroup$

Every weakly null sequence in a Banach space, as a subset, is clearly relatively weakly compact. To quantify the elementary fact, we need the following quantities:

$$\delta_{0}((x_{n})_{n}):=\sup_{x^{*}\in B_{X^{*}}}\limsup_{n}|\langle x^{*},x_{n}\rangle |$$ for a bounded sequence $(x_{n})_{n}$ of a Banach space $X$, where $B_{X^{*}}$ is the closed unit ball of $X^{*}$.

For a bounded subset $A$ of $X$ we set: $$\operatorname{wck}_{X}(A)=\sup\{\textrm{d}(\textrm{clust}_{X^{**}}((x_{n})_{n}),X)\colon (x_{n})_{n}\text{ is a sequence in }A\}, $$ where $\textrm{clust}_{X^{**}}((x_{n})_{n})=\cap_{n=1}^{\infty}\overline{\{x_{m}:m>n\}}^{w^*}$ is the set of all weak$^{*}$-cluster points of $(x_{n})_{n}$ in $X^{**}$. It follows from the Eberlein-Smulyan theorem that $\operatorname{wck}_{X}(A)=0$ if and only if $A$ is relatively weakly compact.

Let $A$ and $B$ be non-empty subsets of a Banach space $X$, we set $$\textrm{d}(A,B)=\inf\{\|a-b\|\colon a\in A,b\in B\},$$ $$\widehat{\textrm{d}}(A,B)=\sup\{\textrm{d}(a,B)\colon a\in A\}.$$ $\textrm{d}(A,B)$ is the ordinary distance between $A$ and $B$, and $\widehat{\textrm{d}}(A,B)$ is the (non-symmetrised) Hausdorff distance from $A$ to $B$. When $A$ is a bounded subset of $X$, we set $$\textrm{wk}_{X}(A)=\widehat{\textrm{d}}\big(\overline{A}^{\sigma(X^{**},X^{*})},X\big). $$

It is a direct consequence of the Banach-Alaoglu theorem that $A$ is relatively weakly compact if and only if $\textrm{wk}_{X}(A)=0$.

Question 1. $\operatorname{wck}_{X}(\{x_{n}:n=1,2,\cdots\})\leq \delta_{0}((x_{n})_{n})$ for every bounded sequence $(x_{n})_{n}$ of a Banach space $X$ ?

Question 2. $\operatorname{wk}_{X}(\{x_{n}:n=1,2,\cdots\})\leq \delta_{0}((x_{n})_{n})$ for every bounded sequence $(x_{n})_{n}$ of a Banach space $X$ ?

Thank you !

$\endgroup$
7
  • $\begingroup$ for a bounded sequence $(x_n)$, let $C\subseteq X^{**}$ be the set of its weak$^*$ cluster points. Isn't $\delta_0 = \sup\{ \|\mu\|_{X^{**}} : \mu\in C \}$ and $\textrm{wck}_X = \sup\{ \|\mu+X\|_{X^{**}/X} : \mu\in C\}$ ? If so, the inequality is evident. $\endgroup$
    – Onur Oktay
    Jun 9, 2022 at 5:41
  • $\begingroup$ I'm a bit confused: I assume by a cluster point of a sequence you mean the limit of a convergent subnet of the sequence? If so, you don't need the Eberlein-Smulyan theorem to see the characterization of weak compactness that you mention. $\endgroup$ Jun 9, 2022 at 6:41
  • $\begingroup$ @OnurOktay In your argument, the second equality seems to be false because in my question, the sequence $(x_{n})_{n}$ is considered as a subset $A$. $\endgroup$ Jun 9, 2022 at 8:07
  • $\begingroup$ @JochenGlueck A cluster point of a sequence does not mean the limit of a convergent subnet of the sequence. I add the definition of the set of all weak*-cluster points of a sequence in my question. $\endgroup$ Jun 9, 2022 at 10:21
  • 1
    $\begingroup$ What are the definitions of $d$ and $\hat{d}$? $\endgroup$ Jun 9, 2022 at 16:29

1 Answer 1

0
$\begingroup$

I answer Question 1 by myself and I am sure my proof is correct.

Let $A=\{x_{n}:n=1,2,\cdots\}$ and let $0<c<\operatorname{wck}_{X}(A)$ be arbitrary. Then there exists a sequence $(z_{n})_{n}$ in $A$ so that $\textrm{d}(\textrm{clust}_{X^{**}}((z_{n})_{n}),X)>c$. Let $Y=\overline{\textrm{span}}\{x_{n}\colon n=1,2,\ldots\}$ and $i_{Y}\colon Y\rightarrow X$ be the inclusion map. Since $i_{Y}^{**}\colon Y^{**}\rightarrow X^{**}$ is an isometric embedding, we get $$\textrm{d}(\textrm{clust}_{Y^{**}}((z_{n})_{n}),Y)\geqslant \textrm{d}(\textrm{clust}_{X^{**}}((z_{n})_{n}),X)>c.$$ Let $\varepsilon>0$. Take any $y^{**}_{0}\in \textrm{clust}_{Y^{**}}((z_{n})_{n})$ and let $d=\textrm{d}(y^{**}_{0},Y)$. By the Hahn-Banach theorem, there exists $y^{***}_{0}\in S_{Y^{***}}$ so that $\langle y^{***}_{0},y^{**}_{0}\rangle=d$ and $\langle y^{***}_{0},y\rangle=0$ for all $y\in Y$. We let $$C=B_{Y^{*}}\cap \{y^{***}\in Y^{***}\colon |\langle y^{***},y^{**}_{0}\rangle-d|<\varepsilon\}.$$ By Goldstine's theorem, $y^{***}_{0}\in \overline{C}^{\sigma(Y^{***},Y^{**})}$. Since $\langle y^{***}_{0},y\rangle=0$ for all $y\in Y$, we get $0\in \overline{C}^{\sigma(Y^{*},Y)}$. Since $Y$ is separable, there exists a weak$^{*}$ null sequence $(f_{m})_{m}$ in $C$. By passing to a subsequence, we may assume that the limit $\lim\limits_{m}\langle y^{**}_{0},f_{m}\rangle$ exists, which is denoted by $a$. By the definition of $C$, $|a-d|\leqslant \varepsilon$. Since $y^{**}_{0}\in \textrm{clust}_{X^{**}}((z_{n})_{n})$, we get a subsequence $(y_{n})_{n}$ of $(z_{n})_{n}$ so that $|\langle y^{**}_{0}-y_{n},f_{m}\rangle|<\frac{1}{n}$ for $m=1,2,\ldots,n$. This implies that $\lim\limits_{n\to\infty}\langle f_{m},y_{n}\rangle=\langle y^{**}_{0},f_{m}\rangle$ for each $m$ and then $\lim\limits_{m\to\infty}\lim\limits_{n\to\infty}\langle f_{m},y_{n}\rangle=a$.

Claim. $|a|\leq \delta_{0}((x_{n})_{n})$.

Case 1. the set $\{y_{n}:n=1,2,\cdots\}$ is finite.

In this case, there exists a subsequence $(y_{k_{n}})_{n}$ and $y_{0}\in Y$ so that $y_{k_{n}}=y_{0}$ for all $n$. Hence $$a=\lim_{m}\lim_{n}\langle f_{m},y_{k_{n}}\rangle=\lim_{m}\langle f_{m},y_{0}\rangle=0.$$ The claim holds trivially.

Case 2. the set $\{y_{n}:n=1,2,\cdots\}$ is infinite.

In this case, we get two strictly increasing sequences $(k_{i})_{i},(l_{i})_{i}$ of positive integers so that $y_{k_{i}}=x_{l_{i}}(i=1,2,\cdots)$. Hence, for each $m$, we get $$|\langle y^{**}_{0},f_{m}\rangle|=\lim\limits_{i\to\infty}|\langle f_{m},y_{k_{i}}\rangle|\leq \delta_{0}^{Y}((x_{n})_{n}):=\sup_{y^{*}\in B_{Y^{*}}}\limsup_{n}|\langle y^{*},x_{n}\rangle|.$$ By the Hahn-Banach theorem, we get $|a|\leq \delta_{0}((x_{n})_{n})$.

By Claim, we get $d\leq \delta_{0}((x_{n})_{n})+\varepsilon$ and so is $c$. Letting $\varepsilon\rightarrow 0$, we get $c\leq \delta_{0}((x_{n})_{n})$. As $c$ was arbitrary, the proof is complete.

$\endgroup$

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.