AI 中文总结
解决Frankl和Tokushige关于r - 交叉相交族的乘积猜想，证明相关尖锐不等式及有偏测度定理。通过有序划分耦合、星校准上影子不等式和解析不等式三步克服不等参数问题，耦合捕捉交叉相交组合信息简化证明。
AI 中文摘要
我们解决了Frankl和Tokushige关于r - 交叉相交族的均匀和有偏乘积猜想。设r≥2，0≤ki≤(r - 1)n/r，且Fi⊆([n]/ki)为r - 交叉相交的。我们证明了尖锐不等式∏i = 1r(|Fi|/(n/ki)) ≤ ∏i = 1r(ki/n)，等式由一个公共1 - 星的相应层达到。作为结果，我们得到了0≤pi≤(r - 1)/r时类似的pi - 有偏测度定理，∏i = 1rμpi(Fi) ≤ ∏i = 1rpi。主要困难在于不等参数不能确定单个公共目标层；相反，目标层ℓ1，…，ℓr必须满足∑i = 1rℓi=(r - 1)n。我们分三步克服这种不对称性。有序划分耦合为每个这样的目标层选择给出了一个尖锐的加法不等式。星校准上影子不等式将一个族在其原始层上的密度与其在适当选择的目标层上的上影子的密度联系起来；通过对n进行归纳证明，归纳步骤简化为一个两点不等式。最后，一个解析不等式表明得到的不对称加法估计意味着所需的乘积界。令人惊讶的是，耦合捕捉了交叉相交的所有组合信息，将证明的其余部分简化为一个解析论证。
英文摘要
Let $\mathcal{F}\subseteq\binom{[n]}k$, and let $\partial_{k\to\ell}\mathcal{F}$ be the family of all $\ell$-sets containing a member of $\mathcal{F}$. Writing $μ_k(\mathcal{F})=|\mathcal{F}|/\binom nk$, $p=k/n$, and $q=\ell/n$, we prove $$ μ_\ell(\partial_{k\to\ell}\mathcal{F})\geq \begin{cases}μ_k(\mathcal{F})^{\frac{\log q}{\log p}}, &0\leq μ_k(\mathcal{F})\leq p^2,\\ q\left[1-\left(1-\frac{μ_k(\mathcal{F})}{p}\right)^{ \frac{\log(1-q)}{\log(1-p)}} \ \ \right], &p^2\leq μ_k(\mathcal{F})\leq p,\\ 1-(1-μ_k(\mathcal{F}))^{\frac{\log(1-q)}{\log(1-p)}}, &p\leq μ_k(\mathcal{F})\leq1. \end{cases} $$ This yields a dimension-free closed-form lower bound for the finite Kruskal--Katona profile and each branch is asymptotically sharp. As the main application, we settle both the uniform and biased product conjectures of Frankl and Tokushige. If $r\geq2$, $0\leq k_i\leq(r-1)n/r$, and $\mathcal{F}_i\subseteq\binom{[n]}{k_i}$ are $r$-cross-intersecting, then $$ \prod_{i=1}^rμ_{k_i}(\mathcal{F}_i)\leq\prod_{i=1}^r\frac{k_i}{n}. $$ In biased product setting, we prove that for $r$-cross-intersecting families $\mathcal{A}_i\subseteq 2^{[n]}$ and $0\leq p_i\leq(r-1)/r$, $\prod_{i=1}^rμ_{p_i}(\mathcal{A}_i)\leq\prod_{i=1}^rp_i.$ We further determine all equality cases in both settings.
Comments32 pages, a new version gives a clearer formulation of the upper-shadow comparisons on the slice and a complete characterization of all equality cases
