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 !

  • $\begingroup$ 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
    Commented 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. $\endgroup$ Commented 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 $A$. $\endgroup$ Commented 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. $\endgroup$ Commented Jun 9, 2022 at 10:21
  • 1
    $\begingroup$ What are the definitions of $d$ and $\hat{d}$? $\endgroup$ Commented Jun 9, 2022 at 16:29

1 Answer 1


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.


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