TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)

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Kolmogorov lone fourier series diverges almost everywhere

Statement

Assume AC. There exists a complex-valued $f \in L^{1} (T)$ such that $\sup_{N \geq 0} ∣ S_{N} f (x) ∣ = \infty$ for almost every x. The torus has normalized measure and $e_{k} (x) = e^{2 π i k x}$ .

Facts & Assumptions

[F1]

The explicit gliding-hump sequence has an $L^{1}$ limit with norm at most one Kolmogorov gliding hump series converges in lone.

[F2]

Its symmetric Fourier partial sums are unbounded off a null set Kolmogorov block maxima diverge off a null limsup set.

[F3]

Assume AC The Axiom of Choice.

Proof

Given: AC and normalized torus conventions.

1.1

Take f to be the $L^{1}$ limit supplied by F1, represented by the absolutely convergent block sum where defined and zero on its measurable null exceptional set. This is a complex integrable function with norm at most one, so all its Fourier coefficients are well defined.

F1F3

2.1

F2 applies to exactly this sequence and limit. It gives $\sup_{N} ∣ S_{N} f (x) ∣ = \infty$ outside a measurable null set. A convergent complex sequence is bounded: beyond some index it lies within one of its limit, and its finite initial segment has a finite maximum. Therefore these partial sums in particular fail to converge almost everywhere. The choice assumption is the one used for the countable norm-one block selection and the complex completeness supplier in F1.

F2F3step 1.1∎

Depends on

Used by

Carleson hunt does not include the lone endpoint Remark

Dependency tree · two levels

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Sources

Grafakos, Classical Fourier Analysis, third edition (standard reference, not scraped)