Let $X$ be a Banach space and $\{e_n\}$ be a unconditional Schauder basis for $X$, i.e., if for every $x \in X$, its expansion $$x = \sum_i a_i e_i$$ converges unconditionally.

(Recall that a series $\sum_i x_i$ in a Banach space $X$ is unconditionally convergent if $\sum_i \epsilon_i x_i$ converges for all choices of signs $\epsilon = \pm 1$.)


For given $A \subseteq \mathbb{N}$, $\sum_{i \in A} a_i e_i$ converges whenever $\sum_{i=1}^{\infty} a_i e_i$ converges in $X$.

It looks simple, but I couldn't check this.

Any help will be appreciated.


1 Answer 1


Let $\epsilon_i = 1$ when $i\in A$ and $\epsilon_1=-1$ otherwise. By assumption, $\sum \epsilon_i a_i x_i$ converges. We also know $\sum a_i x_i$ converges. The sum of two convergent series converges, and this sum is $2\sum_{i\in A} a_i e_i$.

Note on the order of summation: in the above argument I suppose it is fixed somehow, probably by having $i$ running through natural numbers. Otherwise, the word "unconditional" would not be needed at all.

Indeed, consider a Schauder basis an with arbitrary index sets $J$. In this case, with no preferred order of summation on $J$, convergence $\sum_{j\in J} a_ix_i = x$ has to be understood as: for every $\epsilon>0$ there exists a finite set $E \subset J$ such that for every finite set $F$ containing $E$ we have $$\left\|\sum_{i\in F}a_i x_i - x\right\| < \epsilon$$ In this form, convergence is automatically unconditional. One does not need any assumption with $\epsilon$ to conclude that the sum $\sum_{i\in A} a_i x_i$ converges: just use the definition. Given $\epsilon$, there exists $E$ as above... then let $E'=E\cap A$ and this set works for summation over $A$.

(Aside: to me, unconditional convergence really means what's written in the preceding paragraph, not the $\epsilon_i$ thing. They are equivalent, though.)

  • $\begingroup$ Thanks! But still, as I used this argument, still worried about the order of summation. I conclude that for fixed $n \in \mathbb{N}$, $\sum_{i=1}^{n} (\epsilon_i a_i x_i + a_i x_i) = 2 \sum_{i \in A \cap \{1,\dots,n\} } a_i x_i.$ By letting $n \rightarrow \infty$, LHS converges, and RHS goes to $2 \sum_{i \in A \cap \mathbb{N}} a_i x_i = \sum_{i \in A} a_i x_i$. The last line looks vaild. However, I feel like I'm missing the order of summation here. $\endgroup$
    – cdamle
    Commented Sep 20, 2017 at 5:19
  • 1
    $\begingroup$ If you are worried about the order of summation, you should have gotten worried earlier, when reading the statement of the problem. What does $\sum_i a_ix_i$ mean, what does $\sum_{i\in A} a_i x_i$ mean, etc. Anyway, I expanded the answer. $\endgroup$
    – user357151
    Commented Sep 20, 2017 at 11:37
  • $\begingroup$ Thanks a lot. I think I'm not familiar with the definition of conditional convergence. I need to check the equivalent definitions of this concept. $\endgroup$
    – cdamle
    Commented Sep 27, 2017 at 8:27

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