TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-14

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Dvoretzky--Rogers theorem

Statement

Assume Countable Choice. Every infinite-dimensional real or complex Banach space contains an unconditionally convergent series that is not absolutely convergent.

Facts & Assumptions

[A1]

Countable Choice holds (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[L1]

Each sufficiently high-dimensional finite block admits vectors with prescribed squared norms and the uniform subset-sum estimate (The Dvoretzky--Rogers finite-block estimate).

[L2]

Uniform smallness of all finite tails is equivalent to unconditional convergence in a Banach space (Equivalent forms of unconditional convergence).

Proof

technique · construction

Given: The objects and hypotheses in the Statement.

1.1

Put $c_{n} = (8 n^{2})^{- 1}$ for $n \geq 1$ . Since $\sum_{n \geq 1} n^{- 2} < 2$ , we have $\sum_{n} c_{n} < 1 / 4$ , while $\sum_{n} \sqrt{c_{n}} = 8^{- 1 / 2} \sum_{n} 1 / n = \infty$ . Put $N_{1} = 1$ and, for $m \geq 2$ , recursively take $N_{m}$ to be the least integer greater than $N_{m - 1}$ such that $\sum_{n \geq N_{m}} c_{n} < 4^{- m}$ . Thus

given

$\sum_{m} {(\sum_{N_{m} \leq n < N_{m + 1}} c_{n})}^{1 / 2} < \infty .$

[explicit least-index recursion, scalar series]

2.1

Let $r_{m} = N_{m + 1} - N_{m}$ . Infinite-dimensionality supplies a subspace of dimension at least $r_{m} (r_{m} - 1)$ (the cases $r_{m} \leq 1$ are chosen directly). Use [A1] exactly here to select, for all $m$ , one family $(x_{n})_{N_{m} \leq n < N_{m + 1}}$ given by [L1] with $d_{n} = c_{n}$ . Then $∥ x_{n} ∥ = 1 / (\sqrt{8} n)$ and every subset $F$ of the $m$ th block satisfies

givenA1L1step 1.1

$∥ \sum_{n \in F} x_{n} ∥ \leq \sqrt{3} {(\sum_{n \in F} c_{n})}^{1 / 2} .$

[A1, L1, step 1.1]

3.1

For any finite set $F$ contained in the tail beginning at $N_{M}$ , split it by blocks and use the triangle inequality and step 2.1:

givenL2step 2.1step 1.1

$∥ \sum_{n \in F} x_{n} ∥ \leq \sqrt{3} \sum_{m \geq M} {(\sum_{N_{m} \leq n < N_{m + 1}} c_{n})}^{1 / 2} .$

The right side tends to zero by step 1.1. Condition (3) of [L2] therefore holds, so $\sum_{n} x_{n}$ converges unconditionally. [L2, steps 1.1, 2.1]

4.1

On the other hand, [given, step 2.1, step 3.1] $\sum_{n} ∥ x_{n} ∥ = 8^{- 1 / 2} \sum_{n} 1 / n = \infty$ , so the same series is not absolutely convergent.

step 2.1algebra∎

Depends on

Used by

Universal agreement of absolute and unconditional convergence Corollary

Dependency tree · two levels

11 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

A. Dvoretzky and C. A. Rogers, Absolute and Unconditional Convergence in Normed Linear Spaces (standard reference, not scraped)