Let us recall that a basis $(x\_{n})\_{n}$ for a Banach space $X$ is boundedly complete if for every scalar sequence $(a\_{n})\_{n}$ with $\sup\limits\_{n}\|\sum\limits\_{i=1}^{n}a\_{i}x\_{i}\|<\infty$, the series $\sum\limits\_{n=1}^{\infty}a\_{n}x\_{n}$ converges in norm.


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 $(x\_{n})\_{n}$ be a basis for a Banach space $X$. We introduce a quantity measuring non-bounded completeness as follows:
$$\textrm{bc}((x\_{n})\_{n})=\sup\Bigg\{\textrm{ca}\Big(\big(\sum\_{i=1}^{n}a\_{i}x\_{i}\big)\_{n}\Big)\colon
\big(\sum\_{i=1}^{n}a\_{i}x\_{i}\big)\_{n}\subseteq B\_{X}\Bigg\},$$ where $B\_{X}$ is the closed unit ball of $X$. Clearly, $(x\_{n})\_{n}$ is boundedly complete if and only if $\textrm{bc}((x\_{n})\_{n})=0$.


We think about the $\textrm{bc}$-values of some known bases and obtain the following:


1. $\operatorname{bc}((e\_{n})\_{n})=1$, where $(e\_{n})\_{n}$ is the unit vector basis of $c\_{0}$.
2. $\operatorname{bc}((s\_{n})\_{n})=1$, where $(s\_{n})\_{n}$ is the summing basis of $c\_{0}$.
3. $\operatorname{bc}((e\_{n})\_{n=0}^{\infty})=2$, where $(e\_{n})\_{n=0}^{\infty}$ is the unit vector basis of $c$ ($e\_{0}=(1,1,1,\ldots)$).
4. $\operatorname{bc}((e\_{n})\_{n})=1$, where $(e\_{n})\_{n}$ is the unit vector basis of the James space $\mathcal{J}$.
5. Let $x\_{1}=e\_{1}$ and $x\_{n}=-x\_{n-1}+(n-1)e\_{n}$ for $n\geq 2$, where $(e\_{n})\_{n}$ is the unit vector basis of $c\_{0}$. Then $\operatorname{bc}((x\_{n})\_{n})=1$.
6. $\operatorname{bc}((f\_{n})\_{n})=2$, where $(f\_{n})\_{n}$ is the Haar basis of $L\_{1}[0,1]$.
7. $\operatorname{bc}((f\_{n})\_{n})=2$, where $(f\_{n})\_{n}$ is the Faber-Schauder basis of $C[0,1]$.


The examples above yield naturally the following question:


Question. $\textrm{bc}((x\_{n})\_{n})=1$ or $2$ for every basis $(x\_{n})\_{n}$ that is not boundedly complete ?


Prof. William B. Johnson has the following guess with respect to Question:


Guess. Let $(x\_{n})\_{n}$ be a basis for a Banach space $X$ that is not boundedly complete. If $(x\_{n})\_{n}$ is monotone and shrinking, then $\textrm{bc}((x\_{n})\_{n})=1$ or $2$.


I can not prove Guess. Thank you !



Boundedly complete bases
As to the question, the answer is "no". The simplest counterexample is the closure of finite support sequences in the norm $\sup\_{j\ge 1}|a\_j|+\sum\_{j\ge 1}{|a\_j-a\_{j+1}|}$ (decaying to $0$ sequences of finite total variation) with the standard basis $x\_n(k)=\delta\_{nk}$. Then for any partial sum sequence in the unit ball, the total variation component of the norm gets exhausted eventually and the differences become essentially $(0,0,\dots,0,a\dots, a,0,0,\dots)$ with norm $3|a|\le 3/2$ (every entry of a sequence in the unit ball should be at most $1/2$ in absolute value). On the other hand, the coefficient sequence $1/2,1/2,1/2,\dots$ realizes this bound giving $\mathbf{bc}((x\_n)\_n)=3/2$.


I cannot tell anything about the Guess because I don't know what the adjectives "monotone" and "shrinking" mean applied to a basis in a Banach space and you didn't bother to explain ;-)


*Edit:* OK, since Bill explained the words, I'll attempt to refute the guess as well. It will be essentially the same construction but using the quadratic variance instead of the usual one.


Define $X$ to be the space of decaying to $0$ sequences $a=(a\_j)\_{j\ge 1}$ with
$$
\|a\|\_0=\sup\_{j\ge 1}|a\_j|+\left[\sum\_{j\ge 1}|a\_j-a\_{j+1}|^2\right]^{\frac 12}<+\infty\,.
$$
To make the standard basis monotone, we'll just change the norm to the equivalent one given by
$\|a\|=\sup\_m\|a^m\|\_0$ where $a^m=(a\_1,\dots,a\_m,0,0,\dots)$.


Now, if $\|a\|\le 1$, then we still have all $|a\_j|\le\frac 12$, so, since the variation component of the norm will eventually exhaust itself except for the endpoint jump, we still can say that $\mathbb{bc}\le \frac12(1+\sqrt 2)$ with the sequence of all $1/2$ giving exactly that value.


The only thing that remains to show is that our standard basis is shrinking. One can, probably, compute the dual space directly, but we'll take a shortcut.


One key property of our space is that if we have finitely many elements $a\_q\in X$ of norm $1$ with *separated* supports and numbers $b\_q$ with $\sum\_q|b\_q|^2\le 1$, then $\|\sum\_q b\_qa\_q\|\le 1$.


Now let $\psi$ be a linear functional on $X$ corresponding to the sequence $(\psi\_1,\psi\_2,\dots)$ with the usual pairing
$$
\psi(a)=\sum\_{j\ge 1}\psi\_ja\_j\,.
$$
If an arbitrarily faraway tail of $\psi$ has norm at least $\varepsilon$, then for every $Q$, we can inductively construct $Q$ sequences $a\_q\in X$ of norm $1$ with finite separated supports such that $\psi(a\_q)\ge\varepsilon/4$. But then $\psi(Q^{-1/2}\sum\_qa\_q)\ge \varepsilon\sqrt Q/4$, which contradicts the boundedness of $\psi$. Thus the tails must tend to $0$ in the norm and we are done.


If you look at this construction closely, you'll realize that it can yield any $\mathbb bc\in(1,2)$ but not one in $(0,1)$. So, the interesting question really is


**Question:** Can we construct a basis (any or monotone and shrinking) in a Banach space with $\mathbb{bc}\in (0,1)$?


