unconditional convergent

Question

I have to decide whether the following statement is true or false : every permutation of a basic sequence is equivalent to the entire sequence ! where a sequence $(x_n)$ in a Banach space $X$ is called basic if it's a basis of $[x_1,x_2,x_3,.........]$ (it's closed span).
I think it's false, and I think that if it's true then $\displaystyle\sum_{1}^{\infty}x_n$ converges unconditionaly . Any ideas. Thank you!

Answer 1

A (Schauder) basis $(x_i)_{i=1}^\infty$ of a Banach space $X$ is unconditional if and only if for every $x\in X$, its expansion $\sum\limits_{i=1}^\infty \alpha_i x_i$ in terms of the basis converges unconditionally. This is equivalent to saying that any permutation of the basis is a basic sequence.

There are, of course, bases that are not unconditional. For instance, it can be shown that the sequence $\{e_1, e_1-e_2, e_1-e_3,\ldots\}$ is a basis of $\ell_1$ that is not unconditional.

The property that you require is stronger. A basis $(x_i)_{i=1}^\infty$ for which every permutation of $(x_i)_{i=1}^\infty$ is equivalent to $(x_i)_{i=1}^\infty$ is called symmetric (recall that two basic sequences $(x_i)$ and $(y_i)$ are equivalent if the convergence of $\sum\alpha_i x_i$ is equivalent to that of $\sum\alpha_i y_i$).

Of note:

• For $1<p<\infty$, the space $L_p(0,1)$ does not have a symmetric basis (it in fact does not possess a subsymmetric basis)$\,^1$. So the Haar system in these spaces gives an example of an unconditional basis that is not symmetric.
• The sequence $(x_i)$ constructed in my answer to this post is an unconditional basic sequence in $\ell_1$. It follows from the results there that this sequence is not equivalent to the standard unit vector basis of $\ell_1$. So from the last point below, it follows that $(x_n)$ is not symmetric.
• If $X$ is one of $c_0$, $\ell_1$, or $\ell_2$, then every unconditional basis of $X$ is symmetric. This follows from the fact that these spaces have unique unconditional bases$\,^2$.
• If $X$ is one of $c_0$, $\ell_p$ $1\le p<\infty$, then every symmetric basis of $X$ is equivalent to the unit vector basis of $X$. In fact, any symmetric basic sequence in $X$ is equivalent to the standard unit vector basis of $X$$\,^3$.

$^1$ c.f., Ivan Singer, Bases in Banach Spaces, Chapter 22.

$^2$ This is a well known fact. See, e.g., Singer or Joeseph Diestel's Sequences and Series in Banach Spaces.

$^3$ c.f. Lindenstrauss and Tzafriri Classical Banach Spaces I, proposition 3.b.5 and the remark following.