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PreprintCC BY 4.0arXiv

An explicit counterexample to the Hinrichs--Vybiral conjecture

Abstract

Conjecture~2 of Hinrichs and Vybíral asserts that every continuous, nonnegative, positive definite function f on Rd, normalized by f(0)=1, satisfies [f(xj−xk)]j,k=1n⪰ oneoneT/n. We give an explicit trigonometric polynomial on R7 and eight points for which this inequality fails. All hypotheses are verified directly: positive definiteness follows from nonnegative Fourier coefficients, and pointwise nonnegativity follows from interpolation on a cube. For an explicit vector of signs, the quadratic form is 22/5, whereas the proposed lower bound is 9/2. The resulting gap is exactly −1/10. The example is compatible with the established bound for functions of the form |g|2 with g positive definite.\\ The solution was found in a single prompt try by a colleague using his private licence of ChatGPT 6. Our aim is to make the solution public, as well as to collect some ideas that emerged during the analysis of the solution.

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BibTeX
@misc{vybiral2026explicit,
  title = {{An explicit counterexample to the Hinrichs--Vybiral conjecture}},
  author = {Jan Vybiral},
  howpublished = {arXiv},
  year = {2026},
  month = sep,
  doi = {10.48550/arxiv.2609.21733},
  eprint = {2609.21733},
  archivePrefix = {arXiv},
  url = {https://arxiv.org/abs/2609.21733},
  note = {CC BY 4.0},
}
RIS
TY  - UNPB
TI  - An explicit counterexample to the Hinrichs--Vybiral conjecture
AU  - Jan Vybiral
PY  - 2026
DA  - 2026/09/18
PB  - arXiv
DO  - 10.48550/arxiv.2609.21733
UR  - https://arxiv.org/abs/2609.21733
AB  - Conjecture~2 of Hinrichs and Vybíral asserts that every continuous, nonnegative, positive definite function f on R^d, normalized by f(0)=1, satisfies [f(x_j−x_k)]_j,k=1^n⪰ oneone^T/n. We give an explicit trigonometric polynomial on R^7 and eight points for which this inequality fails. All hypotheses are verified directly: positive definiteness follows from nonnegative Fourier coefficients, and pointwise nonnegativity follows from interpolation on a cube. For an explicit vector of signs, the quadratic form is 22/5, whereas the proposed lower bound is 9/2. The resulting gap is exactly −1/10. The example is compatible with the established bound for functions of the form |g|^2 with g positive definite.\\ The solution was found in a single prompt try by a colleague using his private licence of ChatGPT 6. Our aim is to make the solution public, as well as to collect some ideas that emerged during the analysis of the solution.
ER  -