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The summing basis of c0 is conditional

Example

For $n \geq 1$ put $s_{n} = (1, \dots, 1, 0, \dots)$ , with $n$ initial ones. Then $(s_{n})$ is a conditional Schauder basis of real or complex $c_{0}$ .

Facts & Assumptions

[L1]

A basis is conditional when some basis expansion is not unconditionally convergent (Unconditional and conditional Schauder bases).

[L2]

Unconditional convergence implies convergence of every subseries (Equivalent forms of unconditional convergence).

Verification

technique · counterexample

Given: The objects and hypotheses in the Statement.

1.1algebra

For $x = (x_{k})_{k \geq 0} \in c_{0}$ set $a_{n} = x_{n - 1} - x_{n}$ for $n \geq 1$ . [given] The $k$ th coordinate of $\sum_{n = 1}^{N} a_{n} s_{n}$ is $x_{k} - x_{N}$ when $0 \leq k < N$ and zero otherwise. Hence the error has supremum at most $\max {∣ x_{N} ∣, \sup_{k \geq N} ∣ x_{k} ∣} \to 0$ . Conversely the coordinate identities force $a_{n} = x_{n - 1} - x_{n}$ , so $(s_{n})$ is a Schauder basis.

2.1givenL2L1step 1.1

Take $x_{k} = (- 1)^{k} / (k + 1)$ for $k \geq 0$ . Then $a_{n} = (- 1)^{n - 1} (1 / n + 1 / (n + 1))$ . The subseries over the odd indices has first coordinate

$\sum_{n o d d} (\frac{1}{n} + \frac{1}{n + 1}) = + \infty .$

It therefore does not converge in $c_{0}$ . By [L2] the basis expansion of this $x$ is not unconditional, and [L1] makes the basis conditional. [L1, L2, explicit witness] ∎

Depends on

Used by

Reordering a conditional basis expansion can destroy convergence Counterexample

Dependency tree · two levels

5 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

Thomas Schlumprecht, Course Notes in Functional Analysis, Math 655 (standard reference, not scraped)