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 !



Weakly null sequences in Banach spaces
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.


