(Clairaut 定理)设 $E$ 是 $\mathbf{R}^n$ 的开子集合,并设 $f:\mathbf{E}\to \mathbf{R}^{m}$ 是 $E$ 上的二次连续可微函数.那么对于一切$x_0\in E$ 和 $1\leq i,j\leq n$,

  \begin{align*}

    \frac{\partial }{\partial x_j}\frac{\partial f}{\partial

      x_i}(x_0)= \frac{\partial }{\partial x_i}\frac{\partial

      f}{\partial x_j}(x_0)

  \end{align*}

证明:这个定理的本质是二重极限的顺序问题,在题设条件下,交换极限的顺序对结果无影响.我们依照定义来证明.不妨设 $j<i$.设 $x_0$ 在 $\mathbf{R}^n$ 中的坐标

  为$(a_1,a_2,\cdots,a_n)$.则

  \begin{equation}

    \label{eq:8.00}

    \frac{\partial f}{\partial x_i}(x_0)=\lim_{\Delta x_{i}\to 0;\Delta

      x_{i}\neq 0}\frac{f(a_1,\cdots,a_i+\Delta

      x_{i},\cdots,a_n)-f(a_1,\cdots,a_i,\cdots,a_n)}{\Delta x_{i}}.

  \end{equation}

易得

  \begin{align*}

    &\frac{\partial }{\partial x_j}\frac{\partial f}{\partial

      x_i}(x_0)\\&=\lim_{\Delta x_j\to 0;\Delta x_j\neq

      0}\lim_{\Delta x_i\to 0;\Delta x_i\neq

      0}\frac{\frac{f(a_1,\cdots,a_j+\Delta x_j,\cdots,a_i+\Delta

      x_i,\cdots,a_n)-f(a_1,\cdots,a_j+\Delta

      x_j,\cdots,a_i,\cdots,a_n)}{\Delta x_i}-\frac{f(a_1,\cdots,a_j,\cdots,a_i+\Delta

      x_i,\cdots,a_n)-f(a_1,\cdots,a_i,\cdots,a_n)}{\Delta x_i}}{\Delta x_j}.

  \end{align*}

且易得

\begin{align*}

    &\frac{\partial }{\partial x_i}\frac{\partial f}{\partial

      x_j}(x_0)\\&=\lim_{\Delta x_i\to 0;\Delta x_i\neq

      0}\lim_{\Delta x_j\to 0;\Delta x_j\neq

      0}\frac{\frac{f(a_1,\cdots,a_j+\Delta x_j,\cdots,a_i+\Delta

      x_i,\cdots,a_n)-f(a_1,\cdots,a_j+\Delta

      x_j,\cdots,a_i,\cdots,a_n)}{\Delta x_i}-\frac{f(a_1,\cdots,a_j,\cdots,a_i+\Delta

      x_i,\cdots,a_n)-f(a_1,\cdots,a_i,\cdots,a_n)}{\Delta x_i}}{\Delta x_j}.

  \end{align*}

结合微分中值定理,再加上二阶偏导数连续,因此极限可以交换顺序,而结果值不变.得证.