0
$\begingroup$

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!

$\endgroup$

1 Answer 1

1
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ You are right, Ben. Thank you. $\endgroup$ Sep 27, 2020 at 13:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.