Let $(X, \| \cdot \|)$ be a Banach space and $(x_n) \subset X$ such that $\sum_{n} \|x_n\| = +\infty$. Is it true that for each $x\in X$ there is a bijection $\phi :\mathbb{N}\to\mathbb{N}$ such that $\sum_{n} x_{\phi(n)} = x$ ? I believe that inductively this can be proven if $X$ is of finite dimension. But what about higher dimensions?

  • $\begingroup$ You (at least) also have to assume $\displaystyle\sum x_n$ to be conditionnally convergent. But still, I don't think it holds : pick $X=\Bbb{R}^2$, pick your favourite conditionnaly convergent not absolutely convergent series $\displaystyle\sum x_n$, and your favourite absolutely convergent series $\displaystyle\sum y_n$. Then the sequence $(x_n, y_n)$ would be a counterexample because no matter how you rearrabge the terms, the sum of the second coordinate will always be $\displaystyle\sum_{n=0}^\infty y_n$ (if the limit makes sense) $\endgroup$ – Max Jun 25 '17 at 10:40
  • 1
    $\begingroup$ Some references to literature related to some results on rearrangements of series in Banach spaces are mentioned in this answer. $\endgroup$ – Martin Sleziak Jun 25 '17 at 12:02

Your conjecture fails for trivial reasons.

Suppose $X$ is just $\mathbb R^2$ with basis $\{x,y\}$.Take the standard conditionally convergent sequence of real numbers $a_n = (-1)^nn^{-1}$ and consider the sequence $v_n=(a_n,0)$ of vectors. By Riemann's theorem we can for any $r \in \mathbb R$ rearrange the sequence to make any $(r,0)$ the limit. But we cannot for example rearrange to make $(0,1)$ the limit.

Exercise: What condition can be placed on the series of vectors to make the rearrangement theorem hold?

  • $\begingroup$ No coordinate-sequence is absolutely convergent would be my first guess $\endgroup$ – JustDroppedIn Jun 25 '17 at 10:55
  • $\begingroup$ @Kps Look at $\frac{(-1)^n}n\, (1,1)$. If you rearrange you will never get anything that does not lie on the diagonal of $\Bbb R\times \Bbb R$. $\endgroup$ – s.harp Jun 25 '17 at 11:05
  • $\begingroup$ Let $S\subset X^*$ countable, such that the span$(S)$ is dense in $X^*$. If $\sum x^*(x_n)$ converges, while $\sum |x^*(x_n)|=\infty$, for all $x^*\in X^*$, then for every $x\in X$, there is a rearrangement $x_{j_n}$ of $x_n$, such that $\sum x_{j_n}=x$. $\endgroup$ – Yiorgos S. Smyrlis Jun 25 '17 at 11:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.