CounterexampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck pass

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.

FALSE: every $L^{2}$ field on $T^{3}$ has absolutely convergent Fourier coefficients. An explicit $H^{0}$ family with $\sum_{k} ∣ \hat{u} (k) ∣ = \infty$

Statement refuted

False statement. Every field $u \in H^{0} (T^{3}; R^{3}) = L^{2} (T^{3}; R^{3})$ has absolutely convergent Fourier coefficients: $\sum_{k \in Z^{3}} ∣ \hat{u} (k) ∣_{2} < + \infty .$ (By Sobolev embedding on the torus: $H^{σ} \subseteq C^{j}$ for $σ > j + 3 / 2$ , with continuous representatives the corresponding statement for $H^{σ}$ with $σ > 3 / 2$ is true; the false statement drops the hypothesis, and that is exactly what fails.)

Facts & Assumptions

Given: The weights $w_{k}$ , the shell decomposition ${∣ k ∣_{\infty} = L}$ for $L \in N$ , and the coefficient family $\hat{u}$ .

[F1]

If $f : [0, \infty) \to R$ is nonincreasing and nonnegative, then $\sum_{k \geq 0} f (k)$ converges if and only if the integrals $\int_{0}^{N} f$ are bounded (The integral test: for $f \geq 0$ nonincreasing on $[0, \infty)$ , $\sum_{k} f (k)$ converges if and only if the sequence $(\int_{0}^{N} f)_{N}$ is bounded, with $\int_{0}^{N} f \leq \sum_{k < N} f (k) \leq f (0) + \int_{0}^{N} f$ ); in particular $\sum_{L \geq 2} L^{- 1} (\log L)^{- 2} < \infty$ and $\sum_{L \geq 2} L^{1 / 2} (\log L)^{- 1} = + \infty$ .

[F2]

A nonnegative extended sum over an index set is the supremum of its finite partial sums, and it is finite exactly when those partial sums are bounded (Series in the nonnegative extended real line).

[F3]

The Fourier coefficient map is a bijection from the real classes of $L^{2} (T^{3}; R^{3})$ onto the real-symmetric families in $ℓ^{2} (Z^{3}; C^{3})$ , and $∥ u ∥_{L^{2}}^{2} = \sum_{k} ∣ \hat{u} (k) ∣_{2}^{2}$ (The Fourier basis and Parseval's identity on the finite torus, Periodic vector Sobolev spaces on the three-torus, the weight family, and the Leray projection).

[F4]

The statement refuted would, if true, force $\sum_{k} ∣ \hat{u} (k) ∣_{2} < \infty$ for every real-symmetric square-summable family (Sobolev embedding on the torus: $H^{σ} \subseteq C^{j}$ for $σ > j + 3 / 2$ , with continuous representatives records the correct positive result under $σ > 3 / 2$ ).

Counterexample

Use the conventions of Periodic vector Sobolev spaces on the three-torus, the weight family, and the Leray projection: $w_{k} = (1 + 4 π^{2} ∣ k ∣_{2}^{2})^{1 / 2}$ and $∥ u ∥_{H^{0}}^{2} = \sum_{k} ∣ \hat{u} (k) ∣_{2}^{2}$ . Let $v : = \frac{1}{\sqrt{3}} (1, 1, 1) \in R^{3}$ and define the real vector field $u$ by the coefficient family $\hat{u} (k) : = {\begin{cases} 0, & ∣ k ∣_{\infty} \leq 1, \\ \frac{w_{k}^{- 3 / 2}}{\log w_{k}} v, & ∣ k ∣_{\infty} \geq 2. \end{cases}$ The family is real-symmetric ( $\hat{u} (- k) = \hat{u} (k) = \overline{\hat{u} (k)}$ ), so it defines a real field $u \in H^{0}$ . Then $∥ u ∥_{H^{0}}^{2} < + \infty$ while $\sum_{k} ∣ \hat{u} (k) ∣_{2} = + \infty$ : the coefficient series of this $L^{2}$ field is not absolutely convergent.

Proof technique: direct; count the shells $∣ k ∣_{\infty} = L$ , compare the two series with the integral test, and invoke the Fourier coefficient isometry.

Given: The coefficient family $\hat{u}$ and the facts [F1]-[F4].

1.1givenalgebra

For $∣ k ∣_{\infty} = L$ the weight satisfies $2 π L \leq w_{k} \leq c L$ with $c : = \sqrt{1 + 12 π^{2}}$ , and the shell has exactly $(2 L + 1)^{3} - (2 L - 1)^{3} = 24 L^{2} + 2$ points. Hence for $L \geq 2$ , with $ℓ_{L} : = \log (2 π L)$ one has two-sided bounds $c^{- 3 / 2} L^{- 3 / 2} / \log (c L) \leq ∣ \hat{u} (k) ∣_{2} \leq (2 π L)^{- 3 / 2} / ℓ_{L}$ on that shell, so shell summation (multiplicity $24 L^{2} + 2$ ) gives $c_{1} \sum_{L \geq 2} L^{1 / 2} / \log L \leq \sum_{k} ∣ \hat{u} (k) ∣_{2} \leq c_{2} \sum_{L \geq 2} L^{1 / 2} / \log L$ and $c_{3} \sum_{L \geq 2} 1 / (L \log^{2} L) \leq \sum_{k} ∣ \hat{u} (k) ∣_{2}^{2} \leq c_{4} \sum_{L \geq 2} 1 / (L \log^{2} L)$ for positive constants $c_{i}$ .

2.1step 1.1F1F2

By [F1] the series $\sum_{L \geq 2} 1 / (L \log^{2} L)$ converges (its integral test function $t^{- 1} (\log t)^{- 2}$ has finite integral from $2$ to $\infty$ , the antiderivative being $- 1 / \log t$ ) while $\sum_{L \geq 2} L^{1 / 2} / \log L$ diverges (its summands do not tend to $0$ ); by [F2] this means $∥ u ∥_{H^{0}}^{2} = \sum_{k} ∣ \hat{u} (k) ∣_{2}^{2} < + \infty$ and $\sum_{k} ∣ \hat{u} (k) ∣_{2} = + \infty$ .

3.1step 2.1F3F4∎

The family is real-symmetric by construction, so by [F3] it is the coefficient family of a real field $u \in H^{0}$ ; by step 2.1 the hypothesis of the false statement is satisfied but its conclusion fails. Hence the false statement is refuted.

Remarks

The counterexample isolates the exact role of the hypothesis $σ > 3 / 2$ in Sobolev embedding on the torus: $H^{σ} \subseteq C^{j}$ for $σ > j + 3 / 2$ , with continuous representatives: the proof there bounds $\sum ∣ \hat{u} (k) ∣$ by the Cauchy-Schwarz factor $(\sum_{k} w_{k}^{- 2 σ})^{1 / 2}$ , which is finite precisely when $2 σ > 3$ . At $σ = 0$ this mechanism is unavailable and, as shown, the conclusion is false.
Nothing is asserted here about the continuity of this particular field; the refuted statement concerns absolute convergence of its Fourier coefficients only, and the witness is a coefficient family in $H^{0}$ .

FALSE: every $L^{2}$ field on $T^{3}$ has absolutely convergent Fourier coefficients. An explicit $H^{0}$ family with $\sum_{k} ∣ \hat{u} (k) ∣ = \infty$

Statement refuted

Facts & Assumptions

Counterexample

Remarks

Depends on

Dependency tree · 2 levels

Sources

FALSE: every L2L2 field on T3T3 has absolutely convergent Fourier coefficients. An explicit H0H0 family with ∑k∣u^(k)∣=∞∑k​∣u(k)∣=∞

Statement refuted

Facts & Assumptions

Counterexample

Remarks

Depends on

Dependency tree · 2 levels

Sources

FALSE: every $L^{2}$ field on $T^{3}$ has absolutely convergent Fourier coefficients. An explicit $H^{0}$ family with $\sum_{k} ∣ \hat{u} (k) ∣ = \infty$