Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)
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Kolmogorov gliding hump series converges in lone

Statement

Assume AC. For j1 one can choose analytic polynomials Pj of degree dj with Pj1=1 and m(Ej)<2j, where Ej={APjj2j}. Set aj=2j, m1=1, mj+1=mj+dj+1 and Qj=ajemjPj. Their positive frequency intervals are pairwise disjoint and increasing. The series jQj converges in complex L1(T) to f, with f11 and fjJQj12J. Moreover jQj(x)< almost everywhere, and its pointwise sum represents f.

Facts & Assumptions

[F1]

Analytic norm-one blocks can have arbitrarily large partial-sum maxima outside a set of prescribed small measure Kolmogorov block polynomial with large partial sums.

[F2]

Modulation translates frequency support and preserves pointwise magnitude up to the scalar amplitude Separated frequency blocks do not disturb earlier partial sum maxima.

[F3]

Complex L1 is complete under countable choice Complex Lp completeness and almost-everywhere subsequences.

[F4]

Nonnegative increasing measurable functions satisfy monotone convergence Monotone convergence for the integral.

[F5]

Proof

Given: AC and the block supplier F1.

1.1

Apply F1 for each j1 with H=j2j and η=2j. AC selects one polynomial from each nonempty set of possible polynomials. Its degree can be taken as its greatest nonzero coefficient index, which exists because its norm is one. The stated exceptional sets are measurable because the maxima are finite continuous maxima. The recursion for mj is explicit; it gives mj+1>mj+dj and mjj1. F2 therefore places each Qj in the claimed disjoint positive interval and gives Qj1=aj=2j.

F1F2F5
2.1

For partial sums fJ=jJQj and K>J, the triangle inequality gives fKfJ1j=J+1K2j2J. They are Cauchy, so F3 supplies an L1 limit f. Passing K to infinity in the norm inequality gives ffJ12J, and f1j12j=1. AC includes the countable choice required by F3.

F3F5step 1.1
3.1

Apply F4 to GJ=jJQj. Its increasing limit G is measurable and has integral G=limJjJ2j=1. For every positive integer M, Mm{G=}G=1, so G< off a null set. There the numerical series converges absolutely; set its sum h to zero on that measurable exceptional set. The resulting function is measurable, hG a.e., and hfJj>JQj a.e. A further application of F4 to the nonnegative tail gives hfJ12J. Consequently hf1hfJ1+fJf10, so h represents f. No smallness of completed blocks is attributed to their frequency shifts.

F4step 1.1step 2.1

Depends on

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Sources