发表机构
Czech Technical University in Prague（布拉格捷克理工大学）
机构由 AI 辅助整理，请以论文原文为准。AI 中文总结
本文构造了 $\R^7$ 上的显式三角多项式和八个点，给出 Hinrichs--Vybíral 猜想的一个反例，其中二次型与下界差距为 $-1/10$，并验证了所有假设。
AI 中文摘要
Hinrichs 和 Vybíral 的猜想 2 断言：在 $\R^d$ 上，任意连续、非负、正定函数 $f$，若满足归一化条件 $f(0)=1$，则不等式 $[f(x_j-x_k)]_{j,k=1}^n\succeq \one\one^T/n$ 成立。我们给出 $\R^7$ 上的一个显式三角多项式以及八个点，使得该不等式不成立。所有假设均被直接验证：正定性由非负傅里叶系数保证，逐点非负性通过在立方体上的插值得到。对于显式符号向量，二次型值为 $22/5$，而所提出的下界为 $9/2$。由此产生的差距恰好为 $-1/10$。该例子与对形如 $|g|^2$（其中 $g$ 正定）的函数所建立的已知界相容。该解是由一位同事使用其私人 ChatGPT 6 许可证在一次提示尝试中发现的。我们的目的是公开该解，并收集在分析该解过程中涌现的一些想法。
英文摘要
Conjecture~2 of Hinrichs and Vybíral asserts that every continuous, nonnegative, positive definite function $f$ on $\R^d$, normalized by $f(0)=1$, satisfies $[f(x_j-x_k)]_{j,k=1}^n\succeq \one\one^T/n$. 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.
