CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-06

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.

The Bernstein Hölder-one-half endpoint can fail

Statement refuted

Every $C^{1 / 2} (T)$ function belongs to $A (T)$ .

Facts & Assumptions

Given: For $k \geq 2$ , put $u_{k} (x) : = e^{i k \log k} e^{2 π i k x}$ and $A_{N} (x) : = \sum_{k = 2}^{N} u_{k} (x)$ , and set $g (x) : = \sum_{k = 2}^{\infty} u_{k} (x) / k$ .

[F1]

Grafakos's Exercise 3.3.8 supplies the estimate $\sup_{x} ∣ A_{N} (x) ∣ \leq C \sqrt{N}$ for every $N \geq 2$ and gives summation by parts as the hint for this construction.

Counterexample

1.1

Summation by parts and [F1] give, uniformly in $x$ and $L \geq M \geq 2$ , $∣ \sum_{k = M}^{L} u_{k} (x) / k ∣ \leq C_{1} M^{- 1 / 2}$ . Hence the series defining $g$ converges uniformly and $g$ is continuous.

F1algebra

1.2

Let $0 < H \leq 1 / 2$ be the circular size of a displacement $h$ , choose its representative with $∣ h ∣ = H$ , put $N = ⌊ H^{- 1} ⌋$ , and set $b_{k} = (e^{2 π i k h} - 1) / k$ . Then $∣ b_{N} ∣ \leq 2 π H$ and $∣ b_{k} - b_{k + 1} ∣ \leq 4 π H / k$ for $2 \leq k < N$ . Summation by parts and [F1] now bound $\sum_{k = 2}^{N} u_{k} (x) b_{k}$ by $C_{2} H \sqrt{N} \leq C_{2} \sqrt{H}$ .

F1givenalgebra

2.1

Uniform convergence permits termwise integration, so $\hat{g} (k) = e^{i k \log k} / k$ for $k \geq 2$ and $\hat{g} (k)$ vanishes otherwise. Therefore $\sum_{k} ∣ \hat{g} (k) ∣ = \sum_{k = 2}^{\infty} 1 / k = \infty$ , and $g \notin A (T)$ .

step 1.1algebra

2.2

The terms with $k > N$ equal the difference of the two series tails at $x + h$ and $x$ . Step 1.1 bounds their sum by $2 C_{1} N^{- 1 / 2} \leq 2 \sqrt{2} C_{1} \sqrt{H}$ .

step 1.1algebra

3.1

For $H = 0$ the difference vanishes; for $H > 0$ , steps 1.2 and 2.2 give $∣ g (x + h) - g (x) ∣ \leq C_{3} H^{1 / 2}$ , so $g \in C^{1 / 2} (T)$ but $g \notin A (T)$ .

step 2.1step 1.2step 2.2∎

Used by

Nothing in the library uses this result yet.

Dependency tree · 0 levels

Nothing. This result depends on no other item in the library.

Sources

Loukas Grafakos, Classical Fourier Analysis, Exercise 3.3.8 (standard reference, not scraped)