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I could not make sense of a step when proving two definitions for unconditional Schauder basis are equivalent:

  1. For any scalar sequence $(\alpha_{n})$, if $\sum{\alpha_{n}x_{n}} $ converge, then it converge unconditionally.
  2. There exist $C > 0$, finite, such that: \begin{align} \Bigg\|\sum_{i=0}^n\epsilon_i\alpha_ix_i\Bigg\| \le C\Bigg\|\sum_{i=0}^n\alpha_ix_i\Bigg\| \end{align} uniformly over $n$ and all sequences $\epsilon_n$ with $|\epsilon_n| \le 1$, and all scalar coefficients ${\alpha_n}$.

To prove 2 $\Rightarrow$ 1, supposing $\sum \alpha_ix_i$ converges, i.e. $\forall \ \ \epsilon > 0, \exists\ \ n = n(\epsilon)$, s.t. \begin{align} \Bigg\|\sum_{i>n}\alpha_ix_i\Bigg\| < \frac{\epsilon}{2C}. \end{align}

Given any finite $A \subset (n, \infty)\cap\mathbb{Z}$, by choosing the sign sequence of $\{\epsilon_n\}$ s.t. equals $1$ on $A$, equals $0$ on $B = \{1, 2, ..., n\}$, and equals $\pm 1$ constantly on $\mathbb{N}\backslash (A\cup B)$. Using 2, we obtain: \begin{align} \Bigg\|\sum_{i\in A}\alpha_ix_i\Bigg\| \le 2C\Bigg\|\sum_{i>n}\alpha_ix_i\Bigg\| < \epsilon \ \ \ \ (\textbf{I could NOT understand this step}) \end{align} I could understand the rest steps, i.e. $\|\sum_{i\in A}x_i\| \le \epsilon \Leftrightarrow \sum x_n$ converges unconditionally.

Could anyone help me to understand the step above I put in bold?

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  • $\begingroup$ Sorry about the formatting, I'll edit it so that the math format works $\endgroup$
    – Jay Zha
    Commented Oct 18, 2016 at 2:57
  • $\begingroup$ Thank you Jacky. I meant for $\mathbb N$$\setminus$A$\setminus$B - I also corrected the format, but I could not submit it due to non-significant changes $\endgroup$
    – Jay Zha
    Commented Oct 18, 2016 at 3:19
  • $\begingroup$ See if there is anything else that you need to correct. $\endgroup$ Commented Oct 18, 2016 at 3:21
  • $\begingroup$ That's perfect. Thanks a lot Jacky! $\endgroup$
    – Jay Zha
    Commented Oct 18, 2016 at 3:25

1 Answer 1

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It should say that for any $\epsilon >0$ there exists n such that that $$V(n)=sup_{n'>n} \left\{ \|\sum_{i=n+1}^{n'}a_ix_i\|\right\}<\epsilon /C.$$ (This is directly from the def'n of a Cauchy sequence.)

So take any $m>n.$ Let $e_{i,m}=0$ for $i\leq n$ and for $i>m.$ And for $n<i\leq m, $ let $|e_{i,m}|=1.$ Let $b_{i,m}=a_i$ for $n<i\leq m.$ And let $b_{i,m}=0$ for $i\leq n$ and for $i>m. $ Then for all $m>n$ we have $$\epsilon> C\cdot V(n) \geq C\|\sum_{i=n+1}^ma_ix_i\|= C\|\sum_{i=0}^mb_{i,m}x_i\|\geq \|\sum_{i=0}^me_{i,m}b_{i,m}x_i\|=\|\sum_{i=n+1}^me_{i,m}a_ix_i\|.$$

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  • $\begingroup$ Thanks a lot! This is it. And each A could be decomposed into finitely many unions of [k, l], and each ||Sum over [k, l]|| is bounded by that of [n+1, l] + that of [n+1, k] $\endgroup$
    – Jay Zha
    Commented Oct 18, 2016 at 21:15
  • $\begingroup$ Good. I was wondering about the "$2$" in "$\epsilon /2C$" which seems to suggest that there was another step or another term involved. $\endgroup$ Commented Oct 18, 2016 at 21:35

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