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By James's Theorem, A. Ulger (Weak compactness in $L^{1}(\mu.X)$, Proc. Amer. Math. Soc. 113(1991),143-149.) proved that a bounded subset $A$ of a Banach space $X$ is relatively weakly compact if and only if given any sequence $(x_{n})_{n}$ in $A$, there exists a sequence $(z_{n})_{n}$ with $z_{n}\in conv(x_{i}:i\geq n)$ that converges weakly. J. Diestel, W. M. Ruess and W. Schachermayer (Weak compactness $L^{1}(\mu,X)$, Proc. Amer. Math. Soc. 118(1993),447-453) proved that a bounded subset $A$ of a Banach space $X$ is relatively weakly compact if and only if given any sequence $(x_{n})_{n}$ in $A$, there exists a sequence $(z_{n})_{n}$ with $z_{n}\in conv(x_{i}:i\geq n)$ that is norm convergent.

Let $(x_{n})_{n}$ be a sequence in a Banach space $X$. We say that a sequence $(z_{n})_{n}$ in $X$ is a convex block subsequence of $(x_{n})_{n}$ if there exists $0=k_{0}<k_{1}<k_{2}<\cdots <k_{n}<\cdots$ so that $z_{n}\in conv(x_{i})_{i=k_{n-1}+1}^{k_{n}}$ for all $n$. The collection of all convex block subsequences of $(x_{n})_{n}$ is denoted by $cbs((x_{n})_{n})$. By Mazur's theorem, we get the following result:

Theorem. The following statements are equivalent for a bounded subset $A$ of a Banach space $X$

(1)$A$ is relatively weakly compact.

(2)Every sequence in $A$ admits a convex block subsequence that is norm convergent.

(3)Every sequence in $A$ admits a convex block subsequence that is weakly convergent.

For a bounded sequence $(x_{n})_{n}$ in a Banach space $X$. We set $$ca((x_{n})_{n})=\inf_{n}\sup_{k,l\geq n}\|x_{k}-x_{l}\|.$$ Then $ca((x_{n})_{n})=0$ if and only if $(x_{n})_{n}$ is norm convergent.

For a Banach space $X$, we set $$R(X)=\sup_{(x_{n})_{n}\subseteq B_{X}}\inf_{(z_{n})_{n}\in cbs((x_{n})_{n})}ca((z_{n})_{n}).$$ It follows from Theorem that $R(X)=0$ if $X$ is reflexive. But I do not know whether the converse is true.

Thank you!

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Yes, I believe the converse does hold.

Beanland/Freeman proved that an operator $T\in\mathcal{L}(X,Y)$ is weakly compact if and only if for every normalized basic sequence $(x_n)\in\mathcal{NB}_X$, the image sequence $(Tx_n)$ fails to dominate the summing basis $(s_n)$ for $c_0$. Consequently, by considering the identity operator, $X$ is reflexive if and only if none of its normalized basic sequences dominates $(s_n)$.

Now select $(x_n)\in\mathcal{NB}_X$ and $\varepsilon>0$. Due to $R(X)=0$, there is $(z_n)\in\text{cbs}(x_n)$ such that $\text{ca}(z_n)<\varepsilon/2$. Hence, there are $k,l\in\mathbb{N}$ such that $\|z_k-z_l\|<\varepsilon$. On the other hand, if $u_k$ and $u_l$ are the corresponding convex blocks of $(s_n)$ then $\|u_k-u_l\|\geqslant 1$.

Did I miss anything?

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  • $\begingroup$ You are right, Ben. Thank you. $\endgroup$ – Dongyang Chen Sep 27 '20 at 13:41

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