Functional Analysis – Bases in c?

banach-spacesfa.functional-analysis

$c_{0}$, the space of the scalar sequence that converges to $0$ endowed with the sup norm, has two well-known bases: the unit vector basis $(e_{n})_{n}$, where $e_{n}(k)=1$ if $k=n$ and $0$ otherwise, and the summing basis $(s_{n})_{n}$, where $s_{n}=\sum_{k=1}^{n}e_{k}$. It is known that $c_{0}$ has no boundedly complete basis. Recall that a basis $(x_{n})_{n}$ for a Banach space $X$ is boundedly complete if for every scalar sequence $(a_{n})_{n=1}^{\infty}$ with $\sup_{n}\|\sum_{i=1}^{n}a_{i}x_{i}\|<\infty$, the series $\sum_{n=1}^{\infty}a_{n}x_{n}$ converges.

We introduce a quantity measuring (non-)bounded completeness of a basis as follows:

Let $(x_{n})_{n=1}^\infty$ be a bounded sequence in a Banach space $X$. We set $$\textrm{ca}((x_{n})_{n=1}^\infty)=\inf_{n}\sup_{k,l\geq n}\|x_{k}-x_{l}\|.$$ Then $(x_{n})_{n=1}^\infty$ is norm-Cauchy if and only if $\textrm{ca}((x_{n})_{n=1}^\infty)=0$.

Let $(x_{n})_{n=1}^\infty$ be a basis for a Banach space $X$. We set $$\textrm{bc}((x_{n})_{n=1}^\infty)=\sup\Big\{\textrm{ca}((\sum_{i=1}^{n}a_{i}x_{i})_{n=1}^\infty)\colon
(\sum_{i=1}^{n}a_{i}x_{i})_{n=1}^\infty\subseteq B_{X}\Big\},$$
where $B_{X}$ is the closed unit ball of $X$.

Since we have proved that $\operatorname{bc}((e_{n})_{n})=\operatorname{bc}((s_{n})_{n})=1$, we have the following natural question:

Question. $\textrm{bc}((x_{n})_{n=1}^\infty)=1$ for every basis $(x_{n})_{n}$ in $c_{0}$ ?

To answer the question, I want to know if there are other bases in $c_{0}$ besides these two important classes of bases. Are there more references about bases in $c_{0}$? Thank you.

Best Answer

You have answered your own question modulo a lemma that is basically obvious:

Lemma. Let $(x_n)$ be a basis for a Banach space $X$ and let $Y=(\sum_{n=0}^\infty E_n)_0$ be the $c_0$-sum of $E_n := \text{span} \{ x_0,\dots,x_n\}$. Let $(y_k)$ be the obvious basis for $Y$ induced by concatenating the given bases for $E_n$; $n=1,2,\dots$. Then $\text{bc}((y_n) \ge \text{bc} (x_n)$.

In your other post about defining a quantitative version of non bounded completeness you pointed out that there is a basis $(x_n)$ for $c$ for which $\text{bc} (x_n)=2$. For this basis, $E_n := \text{span} \{ x_0,\dots,x_n\}$ is obviously isometrically isomorphic to $\ell_\infty^{ n+1}$ and hence $Y:= (\sum_{n=0}^\infty E_n)_0$ is isometrically isomorphic to $c_0$.