# A quantity measuring the reflexivity of Banach spaces

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.

• 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 $B_X$? Dec 16, 2020 at 15:04
• @DieterKadelka $B_{X}$ denotes the closed unit ball of $X$. Dec 16, 2020 at 15:47
• 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? Dec 16, 2020 at 20:00
• Please note: "contains an isomorphic copy of $\ell_1$'' is not the same as "contains $\ell_1$ as a subset''. Dec 16, 2020 at 21:18
• @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, Dec 17, 2020 at 0:04

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 $$k_1 so that $$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 $$R(c_0)\le 1$$.

• 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 $R(c_{0})\leq 1$. Dec 18, 2020 at 14:29
• @DongyangChen Just take $z_n:=\frac{p-1}{p}x_{2n}+\frac{1}{p}x_{2n+1}$. Dec 18, 2020 at 14:52
• @BenW What's the $p$? You means the $p$ is bigger than 1 ? Dec 18, 2020 at 14:53
• $p$ is meant to be an arbitrary integer $>1$, yes. Then $\|z_n-z\|\leq 1+\frac{1}{p}$ which yields $R(c_0)=1$. Dec 18, 2020 at 15:01
• Nice! Thanks, Bill and Ben. 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 $$n, we get $$\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$$.