Quantifications of unconditional bases

Question

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

Recall that a series $\sum\limits_{i=1}^{\infty}x_{i}$ in a Banach space $X$ is called unconditionally convergent if $\sum\limits_{i=1}^{\infty}\varepsilon_{i}x_{i}$ is (norm) convergent for every $(\varepsilon_{n})_{n=1}^{\infty}\subseteq \{1,-1\}$.

Let $(x_{n})_{n=1}^{\infty}$ be a basis for a Banach space $X$ with biorthogonal functionals $(x^{*}_{n})_{n=1}^{\infty}$. Then $(x_{n})_{n=1}^{\infty}$ is called unconditional if the series $\sum\limits_{i=1}^{\infty}\langle x^{*}_{i},x\rangle x_{i}$ is unconditionally convergent for every $x\in X$.

Let $(x_{n})_{n=1}^{\infty}$ be a basis for a Banach space $X$ with biorthogonal functionals $(x^{*}_{n})_{n=1}^{\infty}$. We set $$\textrm{ucb}((x_{n})_{n=1}^{\infty})=\sup_{x\in B_{X}}\sup_{(\varepsilon_{n})_{n=1}^{\infty}\subseteq\{1,-1\}}\textrm{ca}((\sum_{i=1}^{n}\varepsilon_{i}\langle x^{*}_{i},x\rangle x_{i})_{n=1}^{\infty}).$$ Clearly, $(x_{n})_{n=1}^{\infty}$ is unconditional if and only if $\textrm{ucb}((x_{n})_{n=1}^{\infty})=0$. We have obtained the following results:

1. $\textrm{ucb}((s_{n})_{n=1}^{\infty})=\infty$, where $(s_{n})_{n=1}^{\infty}$ is the summing basis of $c_{0}$ which is defined as $s_{n}=\sum\limits_{i=1}^{n}e_{i}$ ($n\in \mathbb{N}$)($(e_{n})_{n=1}^{\infty}$ is the unit vector basis of $c_{0}$).
2. $\textrm{ucb}((e_{n})_{n=1}^{\infty})=\infty$, where $(e_{n})_{n=1}^{\infty}$ is the unit vector basis of the James space $\mathcal{J}$.
3. Let $(x_{n})_{n=1}^{\infty}$ be the basis of $c_{0}$ defined by $x_{1}=e_{1}, x_{n}=((-1)^{n+1},(-1)^{n+2},\cdots, (-1)^{2n},0,0,\cdots)$ for $n\geq 2$. Then $\textrm{ucb}((x_{n})_{n=1}^{\infty})=\infty.$
4. Let $(x_{n})_{n=1}^{\infty}$ be the basis of $c_{0}$ defined by $x_{1}=e_{1}, x_{n}=-x_{n-1}+(n-1)e_{n}$ for $n\geq 2$. Then $\textrm{ucb}((x_{n})_{n=1}^{\infty})=\infty.$

The above examples yield naturally the following question:

Question. Is $\textrm{ucb}((x_{n})_{n=1}^{\infty})=\infty$ for every conditional basis $(x_{n})_{n=1}^{\infty}$?

Answer 1

The answer is yes! Let $(x_{n})_{n=1}^{\infty}$ be a conditional basis for a Banach space $X$ with biorthogonal functionals $(x^{*}_{n})_{n=1}^{\infty}$. By Theorem 16.1 in I. Singer's ``Bases in Banach spaces I'', there exist $x\in X$ and $x^{*}\in B_{X^{*}}$ so that $\sum_{i=1}^{\infty}|\langle x^{*}_{i},x\rangle||\langle x^{*},x_{i}\rangle|=\infty$. For each $i$, we choose $\varepsilon_{i}\in \{1,-1\}$ with $\varepsilon_{i}\langle x^{*}_{i},x\rangle \langle x^{*},x_{i}\rangle=|\langle x^{*}_{i},x\rangle||\langle x^{*},x_{i}\rangle|$. Then, for each $n$, we get $$\|\sum\limits_{i=1}^{n}\varepsilon_{i}\langle x^{*}_{i},x\rangle x_{i}\|\geq \sum_{i=1}^{n}\varepsilon_{i}\langle x^{*}_{i},x\rangle \langle x^{*},x_{i}\rangle=\sum_{i=1}^{n}|\langle x^{*}_{i},x\rangle||\langle x^{*},x_{i}\rangle|\rightarrow \infty$$ as $n\rightarrow \infty$.