Bourgain的$\Lambda(p)$选择定理：具有多项式失败界的贪心证明
Bourgain's Lambda(p) selection theorem: a greedy proof with polynomial failure bounds
AI总结：
本文给出Bourgain有限$\Lambda(p)$选择定理的自包含贪心证明，通过$L^r$中的势能递减和加权并界，获得显式的多项式失败概率界。
AI中文摘要：
我们给出了Bourgain有限$\Lambda(p)$选择定理的一个自包含证明，并带有显式的概率界。对于每个$p>2$和$N\ge2$，给定概率空间上$N$个两两正交且以1为界的函数，一个均匀选择的基数为$\lceil N^{\frac{2}{p}}\rceil$的子系统以至少$1-\frac{1}{N}$的概率满足$\Lambda(p)$不等式，其中常数仅依赖于$p$。该不等式对所有复系数向量同时成立。更一般地，对于每个固定的$A>0$，成功概率至少为$1-\frac{1}{N^A}$，常数仅依赖于$p$和$A$，且与$N$无关。证明使用了$L^r$（其中$r>2$）中的贪心逼近，其势能在所有逼近尺度上递减。一个关于累积更新次数的加权界控制了离散更新历史的并界，增加分配给每个历史的权重可得到指定的失败指数，而不改变基数指数。
英文摘要：
We give a self-contained proof of Bourgain's finite $Λ(p)$ selection theorem with an explicit probability bound. For every $p>2$ and $N\ge2$, given $N$ pairwise orthogonal functions bounded by one on a probability space, a uniformly chosen subsystem of cardinality $\lceil N^{\frac{2}{p}}\rceil$ satisfies the $Λ(p)$ inequality with probability at least $1-\frac{1}{N}$, with a constant depending only on $p$. The inequality holds simultaneously for all complex coefficient vectors. More generally, for every fixed $A>0$, the success probability is at least $1-\frac{1}{N^A}$ with a constant depending only on $p$ and $A$, and independent of $N$. The proof uses a greedy approximation in $L^r$, with $r>2$, whose potential decreases throughout all approximation scales. A single weighted bound on the cumulative number of updates controls a union bound over discrete update histories, and increasing the weight assigned to each history gives the prescribed failure exponent without changing the cardinality exponent.
