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How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Divergence and Almost Everywhere Convergence of Fourier Series

1 · Prerequisites

2 · Summary

The symmetric Fourier partial sums on T=R/Z can diverge at a prescribed point even for a continuous real function. The local proofs connect this failure to the exact Lebesgue-constant operator norm and show how a weak maximal estimate would close almost-everywhere convergence from Fejer polynomial approximation.

Kolmogorov's almost-everywhere divergence theorem and the Carleson–Hunt maximal estimate are explicitly recorded literature results. Their deep proofs are not supplied. The convergence corollary retains the maximal estimate as a hypothesis, and the endpoint discussion keeps that literature boundary visible. All integrals use Haar measure of total mass one.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

Carleson maximal partial-sum operator

Definition

Use T=R/Z with Haar measure m(T)=1 and the negative-sign Fourier coefficients and symmetric partial sums of Period-one Fourier coefficients, partial sums, and convolution on the torus. For fL1(T) define the Carleson maximal partial-sum operator by

Cf(x):=supNZ0SNf(x)[0,].

Each coefficient is a finite integral independent of the representative of f, and each SNf is a continuous trigonometric polynomial. Thus C depends only on the L1 class, at every point. For every real a, the set {Cf>a}=N0{SNf>a} is measurable.

Linearity of finite Fourier sums gives C(f+g)Cf+Cg and C(af)=aCf for nonzero scalars a. Also C0=0; with 0:=0 the homogeneity identity holds for a=0 as well. This is sublinearity with extended nonnegative values. Including N=0 retains the constant Fourier coefficient.

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

Fourier partial-sum operator norm equals the Lebesgue constant

Statement

On T=R/Z with Haar mass one, for each integer N0 and each prescribed x0T, let TN,x0f=SNf(x0). Over either R or C, on the continuous periodic functions with supremum norm,

TN,x0=SN:C(T)C(T)=TDN(t)dm(t).

For N1 these norms are at least 13πlog(N+1), and hence are unbounded as N.

Facts & Assumptions

Given: An integer N0, a point x0T, and either scalar field, with normalized Haar measure.

[F1]

For every one-period integrable f, every N0 and every x, SNf(x)=01f(xt)DN(t)dt (Fourier partial sums are Dirichlet convolutions).

[F2]

DN(t)=1+2k=1Ncos(2πkt) is real, even, continuous and has integral one (Dirichlet and Fejer kernels).

[F3]

The norm of a bounded linear map is the supremum of its output norms over the closed unit ball (The operator norm as the least bound and as the unit-sphere or unit-ball supremum).

[F4]

For tZ, DN(t)=sin((2N+1)πt)/sin(πt); at integers it equals 2N+1 (Closed form and size bounds for the Dirichlet kernel).

Proof

technique · direct estimates and continuous approximation
1.1

Put LN=01DN(t)dt. The convolution formula gives SNf(x)LNf for every x. Finite Fourier sums are linear and continuous as functions of x, so both maps in the statement are bounded linear maps and TN,x0SNLN.

F1F3
1.2

For N1 and 0jN1, set aj=(j+1/6)/(2N+1) and bj=(j+5/6)/(2N+1). These disjoint intervals lie in (0,1/2). On them sin((2N+1)πt)1/2 and 0<sin(πt)πt, so LN12πj=0N1ajbjdt/t. Each integral is at least (bjaj)/bj=2/3j+5/623(j+1).

F4algebra
2.1

For δ>0 define fδ(u)=DN(x0u)/max{δ,DN(x0u)}. The denominator is positive, so this is a real continuous periodic function with norm at most one, also admissible in the complex space. Writing a=DN(x0u), we have 0aa2/max{δ,a}δ. Changing variables in the periodic integral yields 0LNTN,x0fδδ. Thus TN,x0LNδ for every δ>0, proving both norm identities. Zeros of the kernel cause no discontinuity in this test.

F1F2F3step 1.1
3.1

Consequently LN13πj=1N1/j13π1N+1dt/t=13πlog(N+1). For N=0, D0=1 and both norms equal one by the already proved identities (the test f=1 attains the value). This covers the initial index and proves the asserted unboundedness.

F2step 2.1step 1.2algebra

Context

The harmonic lower estimate is included here to support the functional and operator norm assertion. It reuses the classical Lebesgue-constant calculation; it is not a separate growth theorem.

CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

A continuous function with divergent Fourier series at a prescribed point

Statement refuted

Continuity of a one-periodic real function guarantees convergence of its Fourier series at a prescribed point.

More precisely, assume DC. For every x0T there exists fC(T,R) such that

supN0SNf(x0)=.

Facts & Assumptions

Given: DC and a prescribed point x0T=R/Z.

[F1]

On real or complex C(T) the functional fSNf(x0) is bounded, has norm DN1, and these norms are unbounded (Fourier partial-sum operator norm equals the Lebesgue constant).

[F2]

Assuming DC, a pointwise bounded family of bounded linear maps from a Banach space to a normed space has uniformly bounded operator norms (Uniform boundedness principle).

[F3]

For a nonempty compact metric space K, C(K,R) is complete in the supremum metric (C(K,R) is complete in the supremum metric for every nonempty compact metric space K).

[F4]

Assuming countable choice, if a one-period integrable function h satisfies 0δh(x+t)+h(xt)2sdt/t< for some δ(0,1/2), then SNh(x)s (Dini pointwise convergence criterion for Fourier series).

Counterexample

technique · direct application of uniform boundedness
1.1

Let X={fC([0,1],R):f(0)=f(1)} with the supremum norm. The interval is nonempty and compact, so a Cauchy sequence in X has a continuous uniform limit by the completeness theorem. Its endpoint values remain equal, since f(0)f(1)2ffj for every approximating member fj. Thus X is a real Banach space, identified isometrically with the real continuous periodic functions.

F3algebra
2.1

Define TN:XR by TNf=SNf(x0). These maps are real-valued bounded linear functionals, and supNTN=. If all fX had supNTNf<, uniform boundedness on this Banach space would make the operator norms uniformly bounded. Hence there exists a real fX with supNTNf=. Every individual value is finite, so this sequence cannot converge.

F1F2step 1.1
3.1

For this witness, vanishing on any neighborhood of x0 is impossible: if it vanished there, choose 0<δ<1/2 within that neighborhood. The Dini integral with s=0 would be zero, giving SNf(x0)0. DC supplies the countable choice assumed by that criterion. This contradicts the unboundedness in step 2.1 and proves the stated localization observation.

F4step 2.1given
LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

A weak maximal bound implies almost-everywhere Fourier convergence

Statement

Assume countable choice and fix 1p<. On the period-one torus with normalized Haar measure, suppose a finite constant A0 satisfies

m{Cg>λ}Aλpgpp(gLp(T), λ>0).

Then for every fLp(T), SNf(x)f(x) almost everywhere. The conclusion holds for any measurable representative of f.

Facts & Assumptions

Given: Countable choice, 1p<, a finite weak-bound constant A0 as in the statement, and fLp(T).

[F1]

For gL1(T), Cg=supN0SNg is a measurable extended nonnegative function, defined from the continuous finite Fourier sums and independent of the representative (Carleson maximal partial-sum operator).

[F2]

Assuming countable choice, for each one-periodic complex fLp([0,1]), 1p<, the Fejer means satisfy σjffp0 (Fejer means converge in L^p for 1 <= p < infinity).

[F3]

For a measurable nonnegative extended function h and t>0, m{ht}t1hdm (Chebyshev-Markov inequality for the integral).

Proof

technique · polynomial approximation and level-set estimates
1.1

Use a finite-valued measurable representative of f, changing it on a null set if needed. It is integrable since f1+fp and the torus has mass one. Set Qj=σjf=(j+1)1r=0jSrf. These are explicit polynomials and fQjp0. Integrating e2πi(k)x over [0,1] gives one for k= and zero otherwise, so SNQj=Qj whenever Nj, including constants and the zero polynomial.

F1F2given
2.1

The extended function H(x)=lim supNSNf(x)f(x) is measurable: a limsup is a countable infimum of countable suprema of measurable functions. For each j and Nj, linearity and the triangle inequality give SNffSN(fQj)+fQj. Hence HC(fQj)+fQj.

F1step 1.1
3.1

For every λ>0, the set {H>2λ} is contained in {C(fQj)>λ}{fQj>λ}. Apply the assumed weak estimate to the first set and the integral inequality to h=fQjp, t=λp, for the second. The latter strict superlevel set is contained in {hλp}. Subadditivity yields m{H>2λ}(A+1)λpfQjpp.

F3step 2.1given
4.1

Let j with λ fixed. The right side tends to zero, so m{H>2λ}=0. Because H0, {H>0}=r=1{H>2/r} has measure at most the sum of these zero measures, hence zero. Outside it the nonnegative errors have limsup zero and therefore tend to zero. Changing the representative alters the conclusion on only a null set. This proves convergence almost everywhere, including p=1 whenever its hypothesized weak estimate holds.

step 1.1step 3.1algebra
RemarkRemark: Literature-sourcedProof: Not supplied sources checked 2026-09-07 not proved hereOpen item page →
Recorded, not proved here. This statement is included so the library can refer to it honestly, with a citation to the literature. It is not proved anywhere in this library: the track that would prove it has not been developed here yet.

Kolmogorov polynomial blocks — recorded construction lemma

Recorded construction lemma

For every integer M1 there exist a nonnegative trigonometric polynomial gM on T=R/Z and a measurable set AMT such that, for normalized Haar measure,

gM1=1,m(AM)>12M,infxAMsupN1SNgM(x)>2M.

The supremum is of the same partial sums as Carleson maximal partial-sum operator. This is the integer version of Grafakos, Lemma 4.2.4, including nonnegativity from its construction. The proof is not supplied here. “Block” does not mean disjoint frequency support.

Construction cost in the source

Grafakos's Lemma 4.2.2 aligns phases: if 1,x1,,xn are linearly independent over Q, then for any unimodular zj and ε>0 some integer L satisfies e2πiLxjzj<ε for all j. Its Fourier-averaging argument is not established locally.

Lemma 4.2.3 constructs atomic probability measures μn with supL1DLμnclogn almost everywhere, for an absolute c>0. Rational independence supplies simultaneous alignment of the kernel terms. Lemma 4.2.4 then selects a finite maximal truncation and smooths μn with a Fejer kernel. Positivity and mass one survive this smoothing, while the finitely many partial sums remain close enough to preserve the required height. These are descriptions of unproved source machinery, not local facts available as dependencies.

RemarkRemark: Literature-sourcedProof: Not supplied sources checked 2026-09-07 not proved hereOpen item page →
Recorded, not proved here. This statement is included so the library can refer to it honestly, with a citation to the literature. It is not proved anywhere in this library: the track that would prove it has not been developed here yet.

Kolmogorov almost-everywhere divergence — recorded theorem

Recorded theorem

There exists fL1(T), where T=R/Z has Haar mass one, such that

supN0SNf(x)=for almost every xT.

Equivalently, the function in Carleson maximal partial-sum operator is infinite almost everywhere for this f. Every individual partial sum is finite. The Fourier series therefore diverges almost everywhere.

This records Grafakos, Theorem 4.2.1 and the explicit conclusion (4.2.13). No local proof is supplied, and the claim is not strengthened to divergence at every point.

Source architecture

The source forms a summable weighted series of the polynomials in Kolmogorov polynomial blocks — recorded construction lemma . It chooses weights and polynomial degrees recursively. The large contribution of the current polynomial must exceed both the contribution of earlier polynomials and the tail; those two errors require separate estimates. A full-measure limsup of the good sets supplies arbitrarily large partial sums of the final integrable function. The polynomial construction and this summation argument remain external, so the linked record is a bibliographic mention rather than a logical prerequisite.

RemarkRemark: Literature-sourcedProof: Not supplied sources checked 2026-09-07 not proved hereOpen item page →
Recorded, not proved here. This statement is included so the library can refer to it honestly, with a citation to the literature. It is not proved anywhere in this library: the track that would prove it has not been developed here yet.

Carleson–Hunt maximal bound and almost-everywhere convergence — recorded theorem

Recorded theorem

On the period-one torus with normalized Haar measure, for every 1<p< there is a finite constant Cp such that

CfpCpfp(fLp(T)),

where C is Carleson maximal partial-sum operator. Consequently,

SNf(x)f(x)for almost every xT.

Carleson's original result concerns p=2; Hunt obtained the full open range. This is recorded literature, with no local proof of the maximal estimate. Laugesen's Theorem 8.7 and its omitted-proof discussion give the torus convergence statement and the required strong maximal estimate; the source's period 2π is rescaled by t=2πx with normalized measure.

The norm here is supNSNfp. Uniformly bounding the separate numbers SNfp does not supply this estimate. The endpoint p=1 is excluded.

CorollaryStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07 rests on unproved materialOpen item page →

Almost-everywhere convergence from the Carleson–Hunt estimate

Statement

Assume countable choice and let 1<p<. Suppose explicitly that a finite Cp0 satisfies CgpCpgp for every gLp(T), with symmetric sums and normalized Haar measure. Then every fLp(T) satisfies SNff almost everywhere.

Facts & Assumptions

Given: Countable choice, 1<p<, and the strong maximal estimate in the statement for all gLp(T).

[F1]

Assuming countable choice, for fixed 1p<, a bound m{Cg>λ}Aλpgpp for all gLp and all λ>0, with finite A0, implies SNff almost everywhere for every fLp (A weak maximal bound implies almost-everywhere Fourier convergence).

[F2]

If h is nonnegative and measurable and t>0, then m{ht}t1hdm (Chebyshev-Markov inequality for the integral).

Proof

technique · strong-to-weak estimate and the maximal convergence principle
1.1

For gLp and λ>0, apply the integral inequality to the nonnegative measurable function (Cg)p at t=λp. The assumed norm bound makes its integral finite and yields m{Cg>λ}λp(Cg)pdmCppλpgpp.

F2given
2.1

Thus the weak estimate holds for every g and every positive threshold with finite A=Cpp0. The fixed exponent is within 1p< and countable choice is given, so the maximal convergence principle proves the assertion for every f.

F1step 1.1given

Literature boundary

Carleson–Hunt maximal bound and almost-everywhere convergence — recorded theorem records that the strong estimate is true. This proof establishes the implication from that estimate as an explicit hypothesis; it does not prove or discharge the estimate itself.

RemarkRemark: Literature-sourcedProof: Not applicableaudited 2026-09-07 rests on unproved materialOpen item page →

What the Carleson–Hunt proof requires

One proof route

The Lacey–Thiele route to Carleson–Hunt maximal bound and almost-everywhere convergence — recorded theorem uses a decomposition in time and frequency. Lacey's survey works with a real-line model; it is not a local transference argument to the torus.

In §3, tiles are organized into trees. Lemma 3.6 reduces residual density and controls the total length of selected tree tops by inverse density. Lemma 3.9 reduces residual size with an inverse-square size bound on that total length. Lemma 3.11 bounds a tree contribution by its top length times its size and density. Matching density with squared size balances these estimates, and (3.13)–(3.16) leave scale contributions bounded by multiples of min{2n,2n}, summable over nZ.

The Lp extension in §7 uses distributional estimates and interpolation. For the large testing-set case, its opening removes an exceptional set defined by a maximal function and separately treats tiles inside and outside that set. This is a sourced roadmap of one method. The density, size, tree and exceptional-set estimates have not been proved on this page.

RemarkRemark: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-09-07 rests on unproved materialOpen item page →

The L1 endpoint is excluded

Endpoint interpretation

Assume countable choice. The assertion in Kolmogorov almost-everywhere divergence — recorded theorem prevents extending Carleson–Hunt maximal bound and almost-everywhere convergence — recorded theorem to all of L1(T). This comparison relies on the recorded Kolmogorov existence theorem, whose proof is not supplied here.

Indeed, a weak (1,1) estimate m{Cg>λ}Ag1/λ for all gL1 and λ>0 would imply almost-everywhere convergence for every such g by A weak maximal bound implies almost-everywhere Fourier convergence. That implication is incompatible with the recorded witness. A strong (1,1) estimate would imply this weak estimate by Chebyshev-Markov inequality for the integral applied to Cg.

This is an interpretation of external literature, not an independently proved weak-endpoint theorem. Failure at one prescribed point for a continuous function is a different phenomenon from failure on a full-measure set for an integrable function.

5 · Examples, counterexamples and false statements

None yet.

Sources