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.

  • $\begingroup$ Looks interesting but is difficult to read. Is it possible to introduce $Y := X^{\mathbb{N}}$ and replace $(x_n)_n$ by $x$? And what is $B_X$? $\endgroup$ Dec 16, 2020 at 15:04
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    $\begingroup$ @DieterKadelka $B_{X}$ denotes the closed unit ball of $X$. $\endgroup$ Dec 16, 2020 at 15:47
  • $\begingroup$ If I understand your claim correctly that: if $X$ contains $\ell_1$, then $R(X)=2$. Thus $R(c_0)=2$, because $\ell_1\subset c_0$. Or am I missing something? $\endgroup$ Dec 16, 2020 at 20:00
  • $\begingroup$ Please note: "contains an isomorphic copy of $\ell_1$'' is not the same as "contains $\ell_1$ as a subset''. $\endgroup$ Dec 16, 2020 at 21:18
  • $\begingroup$ @JohannLangemets $X$ contains an isomorphic copy of $l_{1}$ means that $X$ contains a subspace isomorphic to $l_{1}$, not just contains $l_{1}$ as a subset, $\endgroup$ Dec 17, 2020 at 0:04

2 Answers 2


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 $R(c_0)\le 1$.

  • $\begingroup$ Thanks, Bill. We can prove that $R(X)\geq 1$ if $X$ is non-reflexive. Therefore $R(c_{0})\geq 1$. I hope to prove that $R(c_{0})\leq 1$ and then $R(c_{0})=1$. This is optimal. I do not know if your proof yields $R(c_{0})\leq 1$. $\endgroup$ Dec 18, 2020 at 14:29
  • $\begingroup$ @DongyangChen Just take $z_n:=\frac{p-1}{p}x_{2n}+\frac{1}{p}x_{2n+1}$. $\endgroup$
    – Ben W
    Dec 18, 2020 at 14:52
  • $\begingroup$ @BenW What's the $p$? You means the $p$ is bigger than 1 ? $\endgroup$ Dec 18, 2020 at 14:53
  • $\begingroup$ $p$ is meant to be an arbitrary integer $>1$, yes. Then $\|z_n-z\|\leq 1+\frac{1}{p}$ which yields $R(c_0)=1$. $\endgroup$
    – Ben W
    Dec 18, 2020 at 15:01
  • $\begingroup$ Nice! Thanks, Bill and Ben. $\endgroup$ Dec 18, 2020 at 15:05

For each $n$, we define $x_{n}(i)=1, i\leq n$ and $x_{n}(i)=-1,i>n$. Given any $(z_{n})_{n}\in \textrm{cbs}((x_{n})_{n})$, $z_{n}=\sum\limits_{i=k_{n-1}+1}^{k_{n}}\lambda_{i}x_{i}$. Then, for $n<m$, we get $$\sum\limits_{i=k_{n-1}+1}^{k_{n}}\lambda_{i}x_{i}(k_{m-1}+1)=-1, \quad \sum\limits_{i=k_{m-1}+1}^{k_{m}}\lambda_{i}x_{i}(k_{m-1}+1)=1.$$ This implies that $\|z_{n}-z_{m}\|=2$ and so $\textrm{ca}((z_{n})_{n})=2$. Thus, we obtain $R(c)=2$.


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