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I am across two definitions for unconditional convergence for which it is not immediately obvious to me that they are equivalent. Here are the definitions. Throughout, $\frak{X}$ will denote a Banach space.

Definition 1. Given a series $\sum x_n$ in $\frak{X}$, we say that this series converges unconditionally to $x$ if for every $\varepsilon>0$, there is a finite subset $J\subseteq \mathbb{N}$ such that for every finite subset $I$ such that $J\subseteq I\subseteq\mathbb{N}$ one has $\|x - \sum_{i\in I}x_i\|<\varepsilon$.

Definition 2. A series $\sum x_n$ in $\frak{X}$ is said to converge unconditionally to $x$ if for any permutation $\sigma:\mathbb{N}\to \mathbb{N}$, the series $\sum x_{\sigma(n)}$ converges to $x$.

I was wondering how one might go about proving that these two definitions are equivalent.

Integrand
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CWcx
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2 Answers2

12

This is proved in detail in:

T.H. Hildebrandt, On unconditional convergence in normed vector spaces, Bull. Amer. Math. Soc. 46 (12) (1940), 959–962, MR0003448.

In fact, Hildebrandt proves the equivalence of five properties A–E. Your definition 1 is his condition E and your definition 2 is his condition A.

Since Hildebrandt proves the equivalence of A and E (both directions) on page 960 and the paper is freely accessible it makes little sense to reproduce the clean and clear argument here.

Let me mention that condition A goes back to

W. Orlicz, Über unbedingte Konvergenz in Funktionenräumen (I), Studia Math. 4 (1933), 33–37.

while (according to Hildebrandt) condition E was studied by Moore in

E.H. Moore, General analysis, Memoirs of the American Philosophical Society, vol. 1, part 2, 1939, p. 63.

t.b.
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10

1 implies 2

Let consider that $\sum x_n$ converge unconditionally to $x$ according to definition 1. Let $\sigma$ be an arbitrary permutation of $\mathbb{N}$, now by definition 1 given $\varepsilon >0$, there is a finite subset $J_\varepsilon\subset\mathbb{N}$ such that for every finite subset $I$ such that $J_\varepsilon \subset I$,we have $\|x - \sum_{i\in I}x_i\|<\varepsilon$.

Now, given $\varepsilon > 0$ take $N_\varepsilon\in\mathbb{N}$ such that $$J_\varepsilon\subset \{\sigma(n)\,|\,n\leq N_\varepsilon\}$$ which can be easily calculated by $N_\varepsilon=\max \sigma^{-1}(J_\varepsilon)$. Now, we have that for every $m\geq N_\varepsilon$, $$\{\sigma(n)\,|\,n\leq m\}$$ contains $J_\varepsilon$ and it is finite. So $$\|x - \sum_{i\in \{\sigma(n)\,|\,n\leq m\}}x_i\|<\varepsilon$$ i.e. $$\|x-\sum_{n\leq m}x_{\sigma(n)}\|<\varepsilon$$ Hence, $$\sum x_{\sigma(n)}$$ converge to $x$ by definition, because for every $\varepsilon>0$ there is a $N_\varepsilon$ such that for every $m\geq N_\varepsilon$, $\|x-\sum_{n\leq m}x_{\sigma(n)}\|<\varepsilon$.

2 implies 1

For the other sense, we will use counterimplication. Suppose that $\sum x_n$ does not converges unconditionally to $x$ according definition 1, then we have that there is a $\varepsilon>0$ such that for every finite set $J\subset\mathbb{N}$ there exist a bigger finite subset $I_J$ containing it such that $$\|x - \sum_{i\in I_J}x_i\|\geq\varepsilon$$ By the axiom of choice let $g$ be the function that for each finite sets $J$ selects the finite set $J_I$ with the special property. And define a sequence of sets as follows by the recursion theorem $$A_0=\{0\}$$ $$A_{2n+1}=g(A_{2n})$$ $$A_{2n+2}=\{n\in\mathbb{N}\,|\,n\leq (\max A_{2n+1}+1)\}$$ We can see that each $A_i$ is finite and contained in the next, and that they cover all $\mathbb{N}$. We only have to apply induction.

Now, we can define a permutation $\sigma:\mathbb{N}\rightarrow\mathbb{N}$ such that covers the $A_i$ one by one in order (i.e. we have an increasing sequence $\{n_i\}$ such that $\sigma\{n\,|\,n\leq n_i\}=A_i$), which makes that $\sigma$ has as images from some initial segments of $\mathbb{N}$ sets of the form $g(A_{2n})$. $\sum x_{\sigma(n)}$ does not converge to $x$. Since for the existenting $\varepsilon$ we have that given any $N\in\mathbb{N}$ there will be an $M_N>N$ such that $$\sigma(\{n\leq M_N\})=g(A_{2i_N})$$ for some $i_N$. So, $$\|\sum_{n\leq M_n}x_{\sigma(n)}-x\|\geq \varepsilon$$ as desired.

Josué Tonelli-Cueto
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