下册 7.2 多元函数的可微性 第30题

\section*{解题过程：} （1）令 $\varphi(t)=f(t \Delta x, t \Delta y),(\Delta x, \Delta y) \neq(0,0)$ ．由题设条件，$\varphi(t)$ 可导，$t \neq 0$ ． $$ f(\Delta x, \Delta y)-f(0,0)=\varphi(1)-\varphi(0)=\varphi^{\prime}(\xi)=\frac{\partial f(\xi \Delta x, \xi \Delta y)}{\partial x} \Delta x+\frac{\partial f(\xi \Delta x, \xi \Delta y)}{\partial y} \Delta y $$ 再由条件得 $\displaystyle \lim _{\substack{\Delta x \rightarrow 0 \\ \Delta y \rightarrow 0}} \frac{f(\Delta x, \Delta y)-f(0,0)-0 \Delta x-0 \Delta y}{\sqrt{\Delta x^{2}+\Delta y^{2}}}=0$ ．故 $f(x, y)$ 在点 $(0,0)$ 处可微． （2）设 $P\left(x_{1}, x_{2}, \cdots, x_{n}\right) \in \mathbf{R}^{n}-\{O\}$ ，由于 $$ \begin{aligned} |f(P)-f(O)| \leqslant & f\left(x_{1}, x_{2}, \cdots, x_{n}\right)-f\left(0, x_{2}, \cdots, x_{n}\right)\left|+\left|f\left(0, x_{2}, \cdots, x_{n}\right)-f\left(0,0, x_{3}, \cdots, x_{n}\right)\right|+\right. \\ & \left|f\left(0,0, x_{3}, \cdots, x_{n}\right)-f\left(0,0,0, x_{4}, \cdots, x_{n}\right)\right|+\cdots+\left|f\left(0,0, \cdots, 0, x_{n}\right)-f(0,0,0, \cdots, 0)\right| \\ \leqslant & \left|f_{x_{1}}^{\prime}\left(\xi_{1}, x_{2}, \cdots, x_{n}\right)\right| \cdot\left|x_{1}\right|+\left|f_{x_{2}}^{\prime}\left(x_{1}, \xi_{2}, \cdots, x_{n}\right)\right| \cdot\left|x_{2}\right|+\cdots+\left|f_{x_{n}}^{\prime}\left(x_{1}, x_{2}, \cdots, \xi_{n}\right)\right| \cdot\left|x_{n}\right|, \end{aligned} $$ 其中 $\xi_{i}$ 在 0 与 $x_{i}$ 之间， 由 $\displaystyle \lim _{P \rightarrow o} \frac{\partial f}{\partial x_{i}}=0$ 得 $\displaystyle \lim _{P \rightarrow o} \frac{|f(P)-f(O)|}{|O P|}=0$ ，即 $f(P)$ 在 $(O)$ 处可微． （3）因为 $$ \begin{aligned} & f\left(x_{0}+\Delta x, y_{0}+\Delta y\right)-f\left(x_{0}, y_{0}\right) \\ & =f\left(x_{0}+\Delta x, y_{0}+\Delta y\right)-f\left(x_{0}, y_{0}+\Delta y\right)+f\left(x_{0}, y_{0}+\Delta y\right)-f\left(x_{0}, y_{0}\right) \\ & =f_{x}\left(x_{0}+\theta_{1} \Delta x, y_{0}+\Delta y\right) \Delta x+f_{y}\left(x_{0}, y_{0}\right) \Delta y+\beta \Delta y=f_{x}\left(x_{0}, y_{0}\right) \Delta x+\alpha \Delta x+f_{y}\left(x_{0}, y_{0}\right) \Delta y+\beta \Delta y \\ & =f_{x}\left(x_{0}, y_{0}\right) \Delta x+f_{y}\left(x_{0}, y_{0}\right) \Delta y+\alpha \Delta x+\beta \Delta y, \text { 其中, 当 }(\Delta x, \Delta y) \rightarrow 0 \text { 时, } \alpha \rightarrow 0, \beta \rightarrow 0 . \end{aligned} $$ 所以 $f(x, y)$ 在 $\left(x_{0}, y_{0}\right)$ 处可微。 （4）转证 $\displaystyle f_{x y}\left(x_{0}, y_{0}\right)=\lim _{\Delta y \rightarrow 0} \frac{f_{x}\left(x_{0}, y_{0}+\Delta y\right)-f_{x}\left(x_{0}, y_{0}\right)}{\Delta y}=f_{y x}\left(x_{0}, y_{0}\right)$ ． 记 $F(\Delta x, \Delta y)=\left(f\left(x_{0}+\Delta x, y_{0}+\Delta y\right)-f\left(x_{0}, y_{0}+\Delta y\right)\right)-\left(f\left(x_{0}+\Delta x, y_{0}\right)-f\left(x_{0}, y_{0}\right)\right)$ ．由 $$ f_{x}\left(x_{0}, y_{0}+\Delta y\right)=\lim _{\Delta x \rightarrow 0} \frac{f\left(x_{0}+\Delta x, y_{0}+\Delta y\right)-f\left(x_{0}, y_{0}+\Delta y\right)}{\Delta x}, f_{x}\left(x_{0}, y_{0}\right)=\lim _{\Delta x \rightarrow 0} \frac{f\left(x_{0}+\Delta x, y_{0}\right)-f\left(x_{0}, y_{0}\right)}{\Delta x} $$ 得 $$ \begin{aligned} f_{x y}\left(x_{0}, y_{0}\right) & =\lim _{\Delta y \rightarrow 0} \frac{\lim _{\Delta x \rightarrow 0} \frac{f\left(x_{0}+\Delta x, y_{0}+\Delta y\right)-f\left(x_{0}, y_{0}+\Delta y\right)}{\Delta x}-\lim _{\Delta x \rightarrow 0} \frac{f\left(x_{0}+\Delta x, y_{0}\right)-f\left(x_{0}, y_{0}\right)}{\Delta x}}{\Delta y} \\ & =\lim _{\Delta y \rightarrow 0} \lim _{\Delta x \rightarrow 0} \frac{1}{\Delta y \Delta x} F(\Delta x, \Delta y) \end{aligned} $$ 由于 $f_{x}$ 存在，故当 $\Delta y$ 充分小时， $\displaystyle \lim _{\Delta x \rightarrow 0} \frac{1}{\Delta x} F(\Delta x, \Delta y)$ 存在． 同理 $\displaystyle f_{y x}\left(x_{0}, y_{0}\right)=\lim _{\Delta x \rightarrow 0} \lim _{\Delta y \rightarrow 0} \frac{1}{\Delta x \Delta y} F(\Delta x, \Delta y)$ ． 记 $\varphi(y)=f\left(x_{0}+\Delta x, y\right)-f\left(x_{0}, y\right)$ ，则 $$ \begin{aligned} \frac{1}{\Delta x \Delta y} F(\Delta x, \Delta y) & =\frac{1}{\Delta x \Delta y}\left(\varphi\left(y_{0}+\Delta y\right)-\varphi\left(y_{0}\right)\right)=\frac{1}{\Delta x} \varphi_{y}\left(y_{0}+\theta_{1} \Delta y\right) \\ & =\frac{1}{\Delta x}\left(f_{y}\left(x_{0}+\Delta x, y_{0}+\theta_{1} \Delta y\right)-f_{y}\left(x_{0}, y_{0}+\theta_{1} \Delta y\right)\right)=f_{y x}\left(x_{0}+\theta_{2} \Delta x, y_{0}+\theta_{1} \Delta y\right) \end{aligned} $$ 因 $f_{y x}$ 在 $\left(x_{0}, y_{0}\right)$ 连续，故 $$ \lim _{\substack{\Delta x \rightarrow 0 \\ \Delta y \rightarrow 0}} \frac{1}{\Delta x \Delta y} F(\Delta x, \Delta y)=\lim _{\substack{\Delta x \rightarrow 0 \\ \Delta y \rightarrow 0}} f_{y x}\left(x_{0}+\theta_{2} \Delta x, y_{0}+\theta_{1} \Delta y\right)=f_{y x}\left(x_{0}, y_{0}\right) $$ 这说明重极限等于累次极限． 因 $f_{x}, f_{y}$ 在点 $\left(x_{0}, y_{0}\right)$ 的某邻域内存在，故由累次极限定理可知 $\displaystyle \lim _{\Delta y \rightarrow 0} \lim _{\Delta x \rightarrow 0} \frac{1}{\Delta y \Delta x} F(\Delta x, \Delta y)$ ， $\displaystyle \lim _{\Delta x \rightarrow 0} \lim _{\Delta y \rightarrow 0} \frac{1}{\Delta x \Delta y} F(\Delta x, \Delta y)$ 存在．故 $$ \lim _{\Delta y \rightarrow 0} \lim _{\Delta x \rightarrow 0} \frac{1}{\Delta y \Delta x} F(\Delta x, \Delta y)=\lim _{\Delta x \rightarrow 0} \lim _{\Delta y \rightarrow 0} \frac{1}{\Delta x \Delta y} F(\Delta x, \Delta y)=\lim _{\substack{\Delta x \rightarrow 0 \\ \Delta y \rightarrow 0}} \frac{1}{\Delta x \Delta y} F(\Delta x, \Delta y) . $$ 所以 $f_{x y}\left(x_{0}, y_{0}\right)=f_{y x}\left(x_{0}, y_{0}\right)$ 。