Alphabeta Math
RemarkRemark: Literature-sourcedProof: Not applicablePipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-12
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Carleson hunt does not include the lone endpoint

Statement

Assume AC as in the Kolmogorov construction. Its L1(T) witness excludes a bound for the Fourier partial-sum maximal operator on all of L1. The proposed Carleson–Hunt theorem concerns only 1<p<∞; this endpoint observation does not establish any of those positive bounds.

Facts & Assumptions

Given: Normalized Haar measure on the circle and the Kolmogorov witness f.

[F1]

The locally authored conclusion of Kolmogorov lone fourier series diverges almost everywhere gives f∈L1 with unbounded symmetric partial sums almost everywhere. The unavailable original backing was separately resolved by the owner using the complete local proof; no original-source retrieval is claimed.

[F2]

The Axiom of Choice is inherited from the countable construction of that witness.

Proof

1.1F1F2givenalgebra

Put Mf=sup⁡N≥0∣SNf∣. Each finite partial sum is measurable, hence this countable supremum is measurable. By F1 it equals infinity outside a null set on a space of measure one. For every positive integer k, the nonnegative simple function k on that conull set is bounded above by Mf, so the definition of the nonnegative integral gives ∫Mf≥k. Consequently ∥Mf∥1=∞.

2.1F1step 1.1algebra∎

Since ∥f∥1<∞ by F1, step 1.1 contradicts every proposed finite-constant inequality ∥Mh∥1≤C∥h∥1 for all h∈L1. It also contradicts every finite weak-(1,1) constant: m{Mf>t}=1 for all t, whereas C∥f∥1/t<1 for sufficiently large t. Thus neither assertion follows by adjoining the endpoint to the proposed positive-p range.

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