切比雪夫单调不等式(Chebyshev's monotonic inequality)(一般分配律)

发布时间 2023-09-30 16:28:10作者: Exotic_sum

前置知识:

一般分配律:

\(\displaystyle\sum_{\substack{j\in J\\k\in K}}a_jb_k\)

\(=\displaystyle\sum_{\substack{j\in J}}\displaystyle\sum_{\substack{k\in K}}a_jb_k\)

\(=(\displaystyle\sum_{\substack{j\in J}}a_j)(\displaystyle\sum_{\substack{k\in K}}b_k)\)

考虑如下二重和式:

\(S=\displaystyle\sum_{1≤j<k≤n}(a_k-a_j)(b_k-b_j)\)

当交换 \(j,k\),有对称性:

\(S=\displaystyle\sum_{1≤k<j≤n}(a_j-a_k)(b_j-b_k)=\displaystyle\sum_{1≤k<j≤n}(a_k-a_j)(b_k-b_j)\)

考虑将S与自己相加,利用恒等式

\([1≤j<k≤n]+[1≤k<j≤n]=[1≤j,k≤n]-[1≤j=k≤n]\)

得到:

\(2S=\displaystyle\sum_{1≤j,k≤n}(a_j-a_k)(b_j-b_k)-\displaystyle\sum_{1≤j=k≤n}(a_j-a_k)(b_j-b_k)\)

注意到 \(\displaystyle\sum_{1≤j=k≤n}(a_j-a_k)(b_j-b_k)\) 等于 \(0\),得

\(2S=\displaystyle\sum_{1≤j,k≤n}(a_j-a_k)(b_j-b_k)\)

考虑将 \((a_j-a_k)(b_j-b_k)\) 拆开:

\(2S=\displaystyle\sum_{1≤j,k≤n}(a_jb_j-a_jb_k-a_kb_j+a_kb_k)\)

\(2S=\displaystyle\sum_{1≤j,k≤n}a_jb_j-\displaystyle\sum_{1≤j,k≤n}a_jb_k-\displaystyle\sum_{1≤j,k≤n}a_kb_j+\displaystyle\sum_{1≤j,k≤n}a_kb_k\)

\(2S=2\displaystyle\sum_{1≤j,k≤n}a_kb_k-2\displaystyle\sum_{1≤j,k≤n}a_jb_k\)

考虑一般分配律:

\(2S=2n\displaystyle\sum_{1≤k≤n}a_kb_k-2(\displaystyle\sum_{k=1}^na_k)(\displaystyle\sum_{k=1}^nb_k)\)

\(S=n\displaystyle\sum_{1≤k≤n}a_kb_k-(\displaystyle\sum_{k=1}^na_k)(\displaystyle\sum_{k=1}^nb_k)\)

又因为 \(S=\displaystyle\sum_{1≤j<k≤n}(a_k-a_j)(b_k-b_j)\)

\(\displaystyle\sum_{1≤j<k≤n}(a_k-a_j)(b_k-b_j)=n\displaystyle\sum_{1≤k≤n}a_kb_k-(\displaystyle\sum_{k=1}^na_k)(\displaystyle\sum_{k=1}^nb_k)\)

移项:

\((\displaystyle\sum_{k=1}^na_k)(\displaystyle\sum_{k=1}^nb_k)=n\displaystyle\sum_{1≤k≤n}a_kb_k-\displaystyle\sum_{1≤j<k≤n}(a_k-a_j)(b_k-b_j)\)

这个恒等式得到称为切比雪夫单调不等式(Chebyshev's monotonic inequality)的一个特例:

\((\displaystyle\sum_{k=1}^na_k)(\displaystyle\sum_{k=1}^nb_k)≤n\displaystyle\sum_{1≤k≤n}a_kb_k\)\(a\) 单调递增,\(b\) 单调递增)

\((\displaystyle\sum_{k=1}^na_k)(\displaystyle\sum_{k=1}^nb_k)≥n\displaystyle\sum_{1≤k≤n}a_kb_k\)\(a\) 单调递增,\(b\) 单调递减)

一般来说,如果 \(a\) 单调递增,且 \(p\)\(\{ 1,..,n\}\) 的一个排列,不难证明,当 \(b_{p(1)}≤...≤b_{p(n)}\) 时,有 \(\displaystyle\sum_{1≤k≤n}a_kb_{p(k)}\) 的最大值出现,而当 \(b_{p(1)}≥...≥b_{p(n)}\) 时有最小值出现。