Recall that a basis $(x\_{n})\_{n}$ for a Banach space $X$ is called boundedly complete if for every scalar sequence $(a\_{n})\_{n}$ with $\sup\_{n}\|\sum\_{i=1}^{n}a\_{i}x\_{i}\|<\infty$, the series $\sum\_{n=1}^{\infty}a\_{n}x\_{n}$ converges. It is well-known that $c\_{0}$ has no boundedly complete basis. My question is to give a quantitative version of the fact. More precisely, for a bounded sequence $(x\_{n})\_{n}$, 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 $(x\_{n})\_{n}$ be a basis for a Banach space $X$. We set
$$\textrm{bc}((x\_{n})\_{n})=\sup\Big\{\textrm{ca}((\sum\_{i=1}^{n}a\_{i}x\_{i})\_{n})\colon
(\sum\_{i=1}^{n}a\_{i}x\_{i})\_{n}\subseteq B\_{X}\Big\},$$ where $B\_{X}$ is the closed unit ball of $X$. Clearly, $(x\_{n})\_{n}$ is boundedly complete if and only if $\textrm{bc}((x\_{n})\_{n})=0$.


I have proved that the $\textrm{bc}$-values of the unit vector basis and the summing basis in $c\_{0}$ are both equal to $1$. Therefore, I have the following question.


Question. $\textrm{bc}((x\_{n})\_{n})\geq 1$ for every basis $(x\_{n})\_{n}$ in $c\_{0}$ ?


Thank you!



$c_{0}$ has no boundedly complete basis
Proposition.
Let $(x\_n)$ be a basis for a Banach space $(X,\|\cdot\|\_X)$ that is not boundedly complete. Then $bc(x\_n) \ge 1$.


Proof: Let $(Z,\|\cdot\|\_Z)$ be the Banach space of all sequences $A=(a\_n)$ of scalars for which
$$
\|A\|\_Z = \sup\_N \|\sum\_{n=1}^N a\_n x\_n\|\_X < \infty.
$$
For $N =1,2,...$, define the operator $R\_N$ on $Z$ by
letting the first $N-1$ coordinates of $R\_N(A)$ be zero and the other coordinates to agree with the coordinates of $A$ (so $R\_1(A)= A$). Let $A$ be any element of $Z$ s.t.
$$
!A! := \limsup\_N \|R\_N(A)\|\_Z >0.
$$
That such an $A$ exists is equivalent to the statement that $(x\_n)$ is not boundedly complete.


Now if $\|A\|\_Z$ were less than or equal to $ !A!$, then we would be done. That need not be true, but to complete the proof it is enough to show that
$$
\liminf\_N \|R\_N(A)\|\_Z -!R\_N(A)! \le 0,
$$
which is obvious from the definition of $!\cdot!$ and the observation that for every $N$, $!R\_N(A)! = !A!$.


Added 4/15/22.


Look at it this way. Since $!A! > 0$, divide by $!A!$ to see that WLOG $!A! = 1$. Now maybe
$\|R\_1(A)\|\_Z$ is much larger than one, but there is $N\_0$ so that for all $N>N\_0$,
$\|R\_N(A)\|\_Z$ is less than $1+\epsilon$. By replacing $A$ with $R\_{N\_0}(A)$, we can assume WLOG that $\|R\_N(A)\|\_Z$ is less than $1+\epsilon$ for all $N$ ("WLOG" because $!A!=!R\_N(A)!$ for all $N$). This means that ALL partial sums $\sum\_{n=k}^j a\_n x\_n$ in $X$ with any starting point have norm at most $1+\epsilon$, but arbitrarily far out you have partial sums $\sum\_{n=k}^j a\_n x\_n$ that have norm in $X$ at least $1-\epsilon$.


