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Let us call a series $\sum_n x_n$ in a Banach space "good" if there exists a permutation $\sigma:\mathbb N\to\mathbb N$ such that the rearranged series $\sum_n x_{\sigma(n)}$ converges.

Find a simple proof of the following theorem (which was proved by E.Steinitz in 1913 according to V.Kadets).

Theorem. A series $\sum_n x_n$ in a finite-dimensional Banach space $X$ is "good" if and only if for every linear function $f:X\to\mathbb R$ the series $\sum_n f(x_n)$ is "good".

(This problem was posed 24.09.2017 by Vaja Tarieladze from Tbilisi on page 72 of Volume 1 of the Lviv Scottish Book).

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  • $\begingroup$ Well that is obviously false, take for example the function that maps every element of $X$ to zero. Then the sum over the $f(x_n)$ is always good, so it can't be an equivalence. Do you maybe want to consider only injective $f$? $\endgroup$ – Dirk Sep 25 '17 at 10:41
  • $\begingroup$ Oh, I shouldn't read things so fast, sorry... $\endgroup$ – Dirk Sep 26 '17 at 8:48
  • $\begingroup$ With $X=\mathbb{R}^n$, I wonder whether it's enough to take the $f$'s to be the coordinate projections. This would be too good to be true, so does anyone know a counterexample? $\endgroup$ – Tobias Fritz Sep 26 '17 at 14:08
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    $\begingroup$ @TobiasFritz Let $(x_n)=((1,1)$, $(-1/2,-1/2)$, $(1/3,1/(3\cdot 2))$, $(1/5,1/(3\cdot 2))$, $(-1/4,-1/4)$, $(1/7,1/(5\cdot 4))$, $(1/9,1/(5\cdot 4))$, $(1/11,1/(5\cdot 4))$, $(1/13,1/(5\cdot 4))$, $(-1/6,-1/6)$,$\dots)$. For $k\ge 1$ there's a block of length $2^{k-1}$ where the first coordinate runs through odd fractions and the second coordinate is $1/((2k-1)\cdot 2^{k-1})$, followed by $(-1/(2k),-1/(2k))$. Each coordinate is good, but $f:(a,b)\mapsto a-b$ sends it to a nonnegative series that diverges like $\log(n)-\log\log(n)$. $\endgroup$ – MTyson Sep 26 '17 at 16:18
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    $\begingroup$ @PietroMajer This problem was resolved by Kadets in (mathnet.ru/links/4cc776c55afef1328f3b1c249757914d/ivm4725.pdf) in case of all Frechet spaces: the Steinitz rearrangement theorem is true for a Frechet space $X$ if and only if $X$ is topologically isomorphic to the countable product of real lines. $\endgroup$ – Taras Banakh Nov 26 '17 at 14:17
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A relatively short inductive proof of Steinitz Theorem can be founded in this paper. Here we present a sketch of the proof, which is based on 3 lemmas whose proof is left to the reader. First we recall the formulation of the result.

Theorem. A series $\sum_{n=1}^\infty x_n$ in a finite-dimensional Banach space $X$ is "good" if and only if for any linear functional $f:X\to\mathbb R$ the series $\sum_{n=1}^\infty f(x_n)$ is "good".

Proof. This theorem will be proved by induction on the dimension of $X$. The theorem is trivially true for Banach spaces of dimension $\le 1$. Assume that for some number $d$ the theorem has been proved for all Banach spaces of dimension $<d$.

Fix a Banach space $X$ of dimension $d$. We can assume that $X=\mathbb R^d$ and the norm $\|\cdot\|$ of $X$ is generated by the standard inner product $\langle\cdot,\cdot\rangle$.

Let $\sum_{n=1}^\infty x_n$ be a series in $X$ such that for every linear functional $f:X\to\mathbb R$ the series $\sum_{n=1}^\infty f(x_n)$ is "good". We lose no generality assuming that $x_n\ne 0$ for all $n$.

Lemma 1. $\lim_{n\to\infty}x_n=0$.

Definition. A point $s$ on the unit sphere $S=\{x\in X:\|x\|=1\}$ is called a divergence direction of the series $\sum_{n=1}^\infty x_n$ if for any neighborhood $U\subset S$ of $x$ the set $\mathbb N_U=\{n\in\mathbb N:\frac{x_n}{\|x_n\|}\in U\}$ is infinite and $\sum_{n\in\mathbb N_U}\|x_n\|=\infty$.

Let $D$ be the (closed) set of all divergence directions of $(x_n)_{n\in\mathbb N}$.

If $D=\emptyset$, then by the compactness of the sphere $S$, $\sum_{n=0}^\infty \|x_n\|<\infty$ and the series $\sum_{n=1}^\infty x_n$ is convergent, so is "good".

So, we assume that $D$ is not empty. Using Hahn-Banach Theorem it can be shown that the convex hull $conv(D)$ of $D$ contains zero.

Let $D_0$ be a subset of smallest cardinality in $D$ such that $0\in conv(D_0)$.

It can be shown that the set $D_0$ is affinely independent and zero is contained in the interior of the simplex $conv(D_0)$.

Let $X_0$ be the linear hull of the set $D_0$ and $X_1$ be the orhogonal complement of $X_0$ in the Euclidean space $X$. So, $X=X_0\oplus X_1$.

For $i\in\{0,1\}$ let $pr_i:X\to X_i$ be the orthogonal projections of $X$ onto $X_i$.

Lemma 2. There exists a subset $\Omega\subset \mathbb N$ such that

1) $\sum_{n\in\Omega}\|pr_1(x_n)\|<+\infty$;

2) for any $s\in D$ and a neighborhood $U\subset S$ of $s$ the set $\Omega_U=\{n\in\Omega:\frac{x_n}{\|x_n\|}\in U\}$ is infinite and $\sum\limits_{n\in\Omega_U}\|pr_0(x_n)\|=\infty$.

Lemma 3. There exists a constant $C$ (dependent on $D_0$) such that for any $x\in X_0$, $\varepsilon>0$, and a finite set $\Omega_0\subset \Omega$ there exists a finite set $F\subset \Omega\setminus \Omega_0$ such that

1) $\|x-\sum_{n\in F}pr_0(x_n)\|<\varepsilon$;

2) $\sum_{n\in F}\|x_n\|\le C\max\{\|x\|,\varepsilon\}$.

Since $X_1$ has dimension $<d$, we can apply the inductive assumption and conclude that the series $\sum_{n\in\mathbb N\setminus\Omega}pr_1(x_n)$ is "good", which means that for some permutation $\sigma_1$ of $\mathbb N\setminus\Omega$ the series $\sum_{n\in\mathbb N\setminus\Omega}pr_1(x_{\sigma_1(n)})$ is convergent in the Banach space $X_1$.

Then using Lemma 3, we can extend the permutation $\sigma_1$ of $\mathbb N\setminus\Omega$ to a permutation $\sigma$ of $\mathbb N$ such that $\sum_{n=1}^\infty pr_0(x_{\sigma(n)})=0$.

Since $\sum_{n\in\Omega}\|pr_1(x_n)\|<+\infty$, the series $\sum_{n\in\Omega}pr_1(x_{\sigma(n)})$ converges and then the series $\sum_{n\in\mathbb N}pr_1(x_{\sigma(n)})=\sum_{n\in\mathbb N\setminus\Omega}pr_1(x_{\sigma(n)})+\sum_{n\in\Omega}pr_1(x_{\sigma(n)})$ converges, too.

Then the series $\sum_{n\in\mathbb N}x_{\sigma(n)}=\sum_{n\in\mathbb N}pr_0(x_{\sigma(n)})+pr_1(x_{\sigma(n)})$ also converges and hence the series $\sum_{n\in\mathbb N}x_n$ is "good".

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