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 !

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