Recall that $(y\_{n})\_{n}$ is a convex block subsequence of a sequence $(x\_{n})\_{n}$ in a Banach space $X$ provided that there exists a strictly increasing sequence of positive integers $(k\_{n})\_{n}$ so that $y\_{n}\in \textrm{co}(x\_{i})\_{i=k\_{n-1}+1}^{k\_{n}}$ for every $n$ ($k\_{0}=0$).
The collection of all convex block subsequences of $(x\_{n})\_{n}$ will be denoted by $\textrm{cbs}((x\_{n})\_{n})$.


Let $(x\_{n})\_{n}$ be a bounded sequence in a Banach space $X$. We set $$\textrm{ca}((x\_{n})\_{n})=\inf\_{n}\sup\_{k,l\geq n}\|x\_{k}-x\_{l}\|.$$
Then $(x\_{n})\_{n}$ is norm-Cauchy if and only if $\textrm{ca}((x\_{n})\_{n})=0$.


Let $X$ be a Banach space. We set $$R(X)=\sup\_{(x\_{n})\_{n}\subseteq B\_{X}}\inf\_{(z\_{n})\_{n}\in \textrm{cbs}((x\_{n})\_{n})}\textrm{ca}((z\_{n})\_{n}).$$


We have proved that a Banach space $X$ is reflexive if and only if $R(X)=0$.


Furthermore, we have shown that $R(X)=2$ if $X$ contains a subspace isomorphic to $l\_{1}$ (not contains $l\_{1}$ as a subset), and $R(c)=2$, where $c$ denotes the space of all convergent scalar sequences equipped with the supremum norm. My concern is $R(c\_{0})$.


Question. Is $R(c\_{0})\leq 1$ ?


Thank you.



A quantity measuring the reflexivity of Banach spaces
Here is a sketch of a proof that $R(c\_0)\le 4/3$.


Suppose that $(x\_n)$ is a sequence in the unit ball of $c\_0$. By passing to a subsequence we can assume that $x\_n$ converges coordinate wise to a vector $x$ in the unit ball of $\ell\_\infty$. By passing to another subsequence and making a small perturbation we can assume that there are $k\_1<k\_2<\dots$ so that $x\_n$ is supported on $1,\dots,k\_n$ and $1\_{[1,k\_n]} x\_{n+1} = 1\_{[1,k\_n]} x$. Let $z\_n:= (2/3)x\_{2n} + (1/3)x\_{2n+1}$. A simple computation shows that if $z$ is in the convex hull of
$\{z\_k : k>n \}$, then $\|z\_n-z\|\le 4/3$. Thus $ca((z\_n))\_n \le 4/3$.


EDIT 27.12.2020: On any interval $(k\_{n-1}, k\_n]$, if $k\not= n$, then $1\_{(k\_{n-1}, k\_n]}x\_k $ is either $0$ or $1\_{(k\_{n-1},k\_n]}x$. Consequently, if $z$ and $w$ are both averages of $N$ different $x\_n$'s, then $\|z-w\| \le 1 + 1/N$. From this it follows that $R(c\_0)\le 1$.


