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Recall that a bounded subset $A$ of a Banach space $X$ is said to be weakly precompact if every sequence in $A$ admits a weakly Cauchy subsequence. Rosenthal's $l_{1}$-theorem states that a bounded subset $A$ is weakly precompact if and only if it contains no $l_{1}$-sequence. For a bounded subset $A$ of $X$, we let $$\operatorname{wpc}_{X}(A)=\inf\{\epsilon>0:A\subseteq K_{\epsilon}+\epsilon B_{X}\},$$ where $K_{\epsilon}$ is weakly precompact.

It is easy to see that $\operatorname{wpc}_{X}(A)=0$ if and only if $A$ is weakly precompact. Clearly, every weakly Cauchy sequence is weakly precompact. Next I want to know how to quantify this simple fact.

Let $(x_{n})_{n}$ be a bounded sequence of $X$. We let $$\delta((x_{n})_{n})=\sup_{x^{*}\in B_{X^{*}}}\inf_{n}\sup_{k,l\geq n}|\langle x^{*},x_{k}-x_{l}\rangle|.$$ Clearly, $\delta((x_{n})_{n})=0$ if and only if $(x_{n})_{n}$ is weakly Cauchy. It is easy to see that $\delta((x_{n})_{n})$ is the diameter of the set of all $weak^{*}$-cluster points of $(x_{n})_{n}$ in $X^{**}$.

Question. $\operatorname{wpc}_{X}((x_{n})_{n})\leq \delta((x_{n})_{n})$

Thank you !

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  • $\begingroup$ "It is easy to see that $\operatorname{wpc}_{X}(A)=0$ if and only if $A$ is weakly precompact." Hmm, I doesn't seem easy enough for me to prove it. Could you please elaborate or provide a reference? $\endgroup$ Sep 12, 2021 at 17:06
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    $\begingroup$ @JochenGlueck Suppose that $\operatorname{wpc}_{X}(A)=0$ and $A$ is not weakly precompact. Then $A$ contains an $l_{1}$-sequence $(x_{n})_{n}$ with constant $c$. For $0<\epsilon<c$, there exist a weakly precompact subset $K_{\epsilon}$ that contains an $l_{1}$-sequence $(z_{n})_{n}$ with constant $c-\epsilon$. This is a contradiction. $\endgroup$ Sep 13, 2021 at 1:33
  • $\begingroup$ Thanks for your reply, that's a nice argument! (I had not been aware of Rosenthal's characterization of weakly sequentially precompact sets.) $\endgroup$ Sep 13, 2021 at 12:20

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