3
$\begingroup$

Prove that in Banach space $X$ the following are equivalent for series $\sum_{i\in\mathbb{N}} x_i$:

1) $\forall\varepsilon>0 \hspace{2mm} \exists $ finite $F\subset\mathbb{N} $ such that for every finite $F'\subset \mathbb{N}$ with $F\subseteq F'$, $\lVert\sum_{i\in F'}x_i-x\rVert\ < \varepsilon$

2) For every permutation $\pi$ of $\mathbb{N}$, $\sum_{i\in\mathbb{N}} x_{\pi(i)}$ converges to $x$.

For 1)$\Rightarrow$2) Let $\pi$ be a permutation of the natural numbers and $\varepsilon>0$. By the assumption in 1), there exists F with the properties above. For some $n_0$ we have that $F\subseteq\{\pi(0),\pi(1),\dots,\pi(n_0)\}$ and thus $\lVert\sum_{i=1}^nx_i-x\rVert\ < \varepsilon$ whenever $n\geq n_0$

For 2)$\Rightarrow 1)$ let's assume that 1) does not hold, that is $\exists\varepsilon_0>0 \hspace{2mm} \forall $ finite $F\subset\mathbb{N} \hspace{2mm} \exists F'\subset \mathbb{N}$ with $F\subseteq F'$ and $\lVert\sum_{i\in F'}x_i-x\rVert\ \geq \varepsilon_0$.

We have that there is $n_0$ such that $\lVert\sum_{i=1}^{n_0}x_i-x\rVert\ < \varepsilon_0$ (because the series converjes for the identity permutation). By the assumption there is a finite set $F_1\supset F$ with $\lVert\sum_{i\in F_1}x_i-x\rVert\ \geq \varepsilon_0$. Let's look at the permutation $\pi=F\cup(F_1\setminus F)\cup (N\setminus F)$. Since 2) holds for $\pi$there is a $j_0>\max\{F\}$ such that $\lVert\sum_{i=0}^{\pi(i)=j_0}x_\pi(i)-x\rVert\ < \varepsilon_0$. Now there is a set $F_2$ such that ... and continue by recursion. I'm not 100% sure if the permutation that I get is such that the series doesn't converge to $x$. And is there a simpler construction to get a contradiction?

$\endgroup$
0
$\begingroup$

I have to admit that I didn't manage to digest your final paragraph, but I think I can see a way of generating a permutation $\pi$ such that $\sum_{i \in \mathbb N}x_{\pi(i)}$ does not converge to $x$.

The key is the look at the statement of the negation of (1). This says that there exists an $\epsilon_0 > 0$ such that for all finite $F \subset \mathbb N$, there exists a finite $F' \subset \mathbb N$ with $F \subseteq F'$ and $|| \sum_{i \in F'} x_i - x || \geq \epsilon_0$.

A consequence is that there exists a strictly increasing sequence $$ F_1 \subset F_2 \subset F_3 \subset\dots \subset \mathbb N $$ where each $F_j$ is finite and $||\sum_{i \in F_j} x_i -x || \geq \epsilon_0$ for each $j$.

[You can generate this sequence recursively. For example, you can start with $F_1 = \emptyset$. And for each $F_j$, you can take $F_{j+1}$ to be the $F'$ provided by applying the above statement with $F = F_j \cup \{ p_j \}$, where $p_j$ is any integer that isn't in $F_j$. (Including the extra element $p_j$ ensures that the inclusion $F_j \subset F_{j+1}$ obtained in this way is a strict inclusion.)]

Then (to use your notation), we can consider a permutation $\pi$ of the form $ F_1 \cup (F_2 \setminus F_1) \cup (F_3 \setminus F_2) \cup \dots$, which clearly violates condition (2).

$\endgroup$
  • $\begingroup$ My idea, basicly, was the same, but I constructed that sets $\endgroup$ – user3701033 Jun 16 '18 at 20:14

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.