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 $X$ be a Banach space with a basis $(x_{n})_{n=1}^\infty$. Recall that $(x_{n})_{n=1}^\infty$ 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.
Let $(x_{n})_{n=1}^\infty$ be a basis for a Banach space $X$. We set
$$\textrm{bc}((x_{n})_{n=1}^\infty)=\sup\Big\{\textrm{ca}((\sum_{i=1}^{n}a_{i}x_{i})_{n=1}^\infty)\colon \|\sum_{i=1}^{n}a_{i}x_{i}\|\leq 1, \forall n\Big\}.$$
Clearly, $(x_{n})_{n=1}^\infty$ is boundedly complete if and only if $\textrm{bc}((x_{n})_{n=1}^\infty)=0$.
We consider some classical non-boundedly complete bases as follows:
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$\operatorname{bc}((e_{n})_{n})=1$, where $(e_{n})_{n}$ is the unit vector basis of $c_{0}$.
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$\operatorname{bc}((s_{n})_{n})=1$, where $(s_{n})_{n}$ is the summing basis of $c_{0}$.
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$\operatorname{bc}((e_{n})_{n=0}^{\infty})=2$, where $(e_{n})_{n=0}^{\infty}$ is the unit vector basis of $c$ ($e_{0}=(1,1,1,\ldots)$).
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$\operatorname{bc}((e_{n})_{n})=1$, where $(e_{n})_{n}$ is the unit vector basis of the James space $\mathcal{J}$.
Question 1. $\textrm{bc}((x_{n})_{n=1}^\infty)\in \{0,1,2\}$ for every basis $(x_{n})_{n}$ ?
Since $c_{0}, C[0,1]$ or $L_{1}[0,1]$ has no boundedly complete basis, we have the following natural questions:
Question 2. $\textrm{bc}((x_{n})_{n=0}^\infty)=$ ?, where $(x_{n})_{n=0}^\infty$ is the Faber-Schauder basis for $C[0,1]$ (See I. Singer, Bases in Banach spaces I, pp. 11 for the Faber-Schauder basis).
Question 3. $\textrm{bc}((h_{n})_{n=1}^\infty)=$ ?, where $(h_{n})_{n=1}^\infty$ is the Haar basis for $L_{1}[0,1]$.
PS: We have proved that if $(x_{n})_{n}$ is an $1$-unconditional basis, then $\textrm{bc}((x_{n})_{n=1}^\infty)=0$ or $1$.
Best Answer
This is an answer to Dongyang's query in the comments. I show $\text{bc}(x_n)\ge 1$ if $(x_n)$ is a monotone basis for $X$ that is not boundedly complete. Let $(x_n^*)$ be the functionals biorthogonal to $(x_n)$ and let $Y$ be the closed linear span in $X^*$ of $(x_n^*)$, so that $(x_n^*)$ is a monotone basis for $Y$. By monotonicity of $(x_n)$, the natural evaluation map from $X$ into $Y^*$ is an isometric isomorphism and, since $(x_n)$ is not boundedly complete, this map is not surjective. Therefore we can regard $X$ as being a proper subspace of $Y^*$. Thus there is a unit vector $y^*$ in $Y^*$ whose distance to $X$ is arbitrarily close to one (say, larger than $1-\epsilon$). You can write $y^*=\sum a_n x_n$ with the understanding that the convergence of the sum is in the weak$^*$ topology on $Y^*$. By monotonicity, for every $N$, $\|\sum_{n=1}^N a_n x_n\| \le 1 = \|y^*\|$. On the other hand, for every $N$, $\|y^* - \sum_{n=1}^N a_n x_n\|$ is at least as large as the distance from $y^*$ to $X$, which implies that for every $N$, $\|\sum_{n=N}^\infty a_n x_n\| > 1-\epsilon$. Now for any fixed $N$, $\sum_{n=N}^M a_n x_n$ converges weak$^*$ to $\sum_{n=N}^\infty a_n x_n$ and hence for large $M$, $\|\sum_{n=N}^M a_n x_n\| > 1-\epsilon$. From this it follows that $\text{bc}(x_n)\ge 1$.