Alphabeta Math
CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-04
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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.
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Local Holder regularity implies Fourier convergence at a point

Statement

Assume the Axiom of Countable Choice.

Let f be a one-period integrable function and let x∈R. Suppose there are constants C>0, α>0, and δ∈(0,1/2) such that

∣f(x+t)−f(x)∣≤Ctαand∣f(x−t)−f(x)∣≤Ctα

for every 0<t<δ. Then

SNf(x)→f(x)as N→∞.

Facts & Assumptions

Given: The Axiom of Countable Choice, a one-period integrable function f, a real x, constants C>0 and α>0, and a real δ with 0<δ<1/2 such that ∣f(x+t)−f(x)∣≤Ctαand∣f(x−t)−f(x)∣≤Ctα for every 0<t<δ.

[L1]

Assuming the Axiom of Countable Choice, if ∫0δ∣f(x+t)+f(x−t)−2s∣t dt<∞, then SNf(x)→s (Dini pointwise convergence criterion for Fourier series).

Proof

technique · direct
1.1L1givenalgebra

For 0<t<δ, the triangle inequality and the hypotheses give ∣f(x+t)+f(x−t)−2f(x)∣≤∣f(x+t)−f(x)∣+∣f(x−t)−f(x)∣≤2Ctα.

2.1step 1.1algebra

Therefore ∫0δ∣f(x+t)+f(x−t)−2f(x)∣t dt≤2C∫0δtα−1 dt=2Cαδα<∞.

3.1L1step 2.1∎

Applying [L1] with s=f(x) and using step 2.1 gives SNf(x)→f(x).

Depends on

Used by

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Dependency tree · two levels

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Sources