Young's theorem杨氏定理

杨氏定理

定理叙述

参考百度百科

Young's Theorem: Let \(f\) be a differentiable function of \(n\) variables. If each of the cross-partials \(f_{ij}^{\prime \prime}\) and \(f_{ji}^{\prime \prime}\) exists and is continuous at all points in some open set \(S\) of values of \((x_1, \cdots, x_n)\) then

\(\forall (x_1, \cdots, x_n) \in S, f_{ij}^{\prime \prime}(x_1, \cdots, x_n)=f_{ji}^{\prime \prime}(x_1, \cdots, x_n)\)

定理的证明

Proof for Young's Theorem:

\[\forall (x_1^0, \cdots, x_n^0) \in S \]

\[f_{ij}^{\prime \prime}(x_1^0, \cdots, x_n^0)=\lim\limits_{x_j\rightarrow x_j^0} \dfrac{ f_{i}^{\prime}(x_1^0, \cdots x_i \cdots x_j \cdots x_n^0)- f_{i}^{\prime}(x_1^0, \cdots x_i \cdots x_j^0 \cdots x_n^0)}{x_j-x_j^0} \\=\lim\limits_{x_j\rightarrow x_j^0} \dfrac{\lim\limits_{x_i\rightarrow x_i^0} \frac{ f(x_1^0, \cdots x_i \cdots x_j \cdots x_n^0)- f(x_1^0, \cdots x_i^0 \cdots x_j \cdots x_n^0)}{x_i-x_i^0} - \lim\limits_{x_i\rightarrow x_i^0} \frac{ f(x_1^0, \cdots x_i \cdots x_j^0 \cdots x_n^0)- f(x_1^0, \cdots x_i^0 \cdots x_j^0 \cdots x_n^0)}{x_i-x_i^0}}{x_j-x_j^0}\\ = \lim\limits_{x_j\rightarrow x_j^0}\lim\limits_{x_i\rightarrow x_i^0} \dfrac{f(x_1^0, \cdots x_i \cdots x_j \cdots x_n^0) + f(x_1^0, \cdots x_i^0 \cdots x_j^0 \cdots x_n^0) - f(x_1^0, \cdots x_i^0 \cdots x_j \cdots x_n^0) - f(x_1^0, \cdots x_i \cdots x_j^0 \cdots x_n^0)}{(x_i-x_i^0)(x_j-x_j^0)}. \]

similarly,

\[f_{ji}^{\prime \prime}(x_1^0, \cdots, x_n^0)=\lim\limits_{x_j\rightarrow x_j^0}\lim\limits_{x_i\rightarrow x_i^0} \dfrac{f(x_1^0, \cdots x_i \cdots x_j \cdots x_n^0) + f(x_1^0, \cdots x_i^0 \cdots x_j^0 \cdots x_n^0) - f(x_1^0, \cdots x_i^0 \cdots x_j \cdots x_n^0) - f(x_1^0, \cdots x_i \cdots x_j^0 \cdots x_n^0)}{(x_i-x_i^0)(x_j-x_j^0)}. \]

\[\therefore \forall (x_1, \cdots, x_n) \in S, f_{ij}^{\prime \prime}(x_1, \cdots, x_n)=f_{ji}^{\prime \prime}(x_1, \cdots, x_n). \]

posted @ 2022-08-14 09:30  TaOQC  阅读(948)  评论(0)    收藏  举报