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!



A characterization of reflexivity of Banach spaces via convex block sequences
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?


