If every rearrangement of a Schauder basis is also a basis, it is an unconditional one

Let $(e_n)_{n=1}^{\infty}$ be a Schauder basis for the Banach space $X$. Suppose for any bijection $\sigma:\mathbb N \to \mathbb N$, $(e_{\sigma(n)})_{n=1}^{\infty}$ is also a Schauder basis, then is it true that $(e_n)_{n=1}^{\infty}$ is an unconditional Schauder basis?(The question is raised by one of my friends. The converse is part of the definition.)

I have tried very hard to prove that for any $x \in X$, there exists $(a_n)_{n=1}^{\infty}$ such that $$\sum _{n=1}^{\infty}a_n e_n=x=\sum _{n=1}^{\infty}a_{\sigma(n)} e_{\sigma(n)}$$

Since there is no conditions about the absolute convergence, we cannot use the proof for "absolute convergence $\Rightarrow$ unconditional convergence". Can anybody show me a proof or an counterexample for this?

• See this question, specifically the comments to the answer. I believe your question is addressed there. Commented Nov 24, 2015 at 22:53

As David Mitra wrote in comments, the continuity of coordinate functionals is the key here. To each Schauder basis $B=(x_n)$ we associate coordinate functionals $\phi_{B,n}$ such that $\phi_{B,n}(x)=c_n$ where $\sum_{ n} c_n x_n = x$. By construction, $\phi_{B, n}(x_m)=0$ for $m\ne n$.
Since the permuted family $B' = (x_{\sigma(n)})$ is still a Schauder basis, we have $\sum_{ n}\phi_{B',n }(x) x_{\sigma(n)} = x$. The goal is to show that $$\phi_{B',n }(x) = \phi_{B,\sigma(n)}(x)$$
The continuity of coordinate functionals (see below) implies $$\phi_{B',n}(x) = \lim_{m\to\infty}\phi_{B',n} \left( \sum_{k\le m} \phi_{B,k}(x) x_{k} \right) = \lim_{m\to\infty}\phi_{B',n}(x) \left( \phi_{B,\sigma(n)}(x) x_{\sigma(n)} \right) = \phi_{B,\sigma(n)} (x)$$ as claimed. The second equality above holds because $\phi_{B',n}$ vanishes on $x_k$ for $k\ne \sigma(n)$.
For a proof that the coordinate functionals are continuous see, e.g., section 1.1 of Victor Zeh's notes on Schauder bases. The proof is too long to reproduce here in full, but the main idea is to show that the formula $$\|x\|_B=\sup_m \left\|\sum_{n\le m} \phi_{B,n}(x) x_n\right\|$$ defines a norm with respect to which $X$ is complete. Since $\|x\|_B\ge \|x\|$ by construction, the open mapping theorem implies $\|x\|_B\le C\|x\|$ for some $C$. This shows that the partial sums of Schauder basis are continuous with respect to $x$, and the conclusion follows.