方企勤 第五章 多元函数微分学 第10题
📝 题目
例 10 设 $f\left( {x,y}\right)$ 定义在开集 $\Omega$ 内,若 $f\left( {x,y}\right)$ 对 $x$ 连续,对 $y$ 满足李普希兹条件,即 $\forall \left( {x,{y}^{\prime }}\right) ,\left( {x,{y}^{\prime \prime }}\right) \in \Omega$ ,有
$$ \left| {f\left( {x,{y}^{\prime }}\right) - f\left( {x,{y}^{\prime \prime }}\right) }\right| \leq L\left| {{y}^{\prime } - {y}^{\prime \prime }}\right| \;\left( {L\text{ 为常数 }}\right) . $$
求证: $f\left( {x,y}\right)$ 在 $\Omega$ 上连续.
💡 答案解析
证 $\forall \left( {{x}_{0},{y}_{0}}\right) \in \Omega$ ,由于 $f\left( {x,{y}_{0}}\right)$ 在 ${x}_{0}$ 点连续,所以 $\forall \varepsilon > 0$ , $\exists {\delta }_{1}\left( {{x}_{0},{y}_{0}}\right) > 0$ ,当 $\left| {x - {x}_{0}}\right| < {\delta }_{1}$ 时,有
$$ \left| {f\left( {x,{y}_{0}}\right) - f\left( {{x}_{0},{y}_{0}}\right) }\right| < \varepsilon /2. \tag{1.5} $$
取 ${\delta }_{2} = \varepsilon /\left( {2L}\right) > 0$ ,当 $\left| {y - {y}_{0}}\right| < {\delta }_{2}$ 时,由条件可得
$$ \left| {f\left( {x,y}\right) - f\left( {x,{y}_{0}}\right) }\right| \leq L\left| {y - {y}_{0}}\right| < L \cdot \varepsilon /\left( {2L}\right) = \varepsilon /2. $$
(1.6)
只要取 $\delta = \min \left( {{\delta }_{1},{\delta }_{2}}\right) > 0$ ,当 $\left| {x - {x}_{0}}\right| < \delta ,\left| {y - {y}_{0}}\right| < \delta$ ,且邻域 $U\left( {\left( {{x}_{0},{y}_{0}}\right) ;\delta }\right) \subset \Omega$ ,则有
$$ \left| {f\left( {x,y}\right) - f\left( {{x}_{0},{y}_{0}}\right) }\right| $$
$$ \leq \left| {f\left( {x,y}\right) - f\left( {x,{y}_{0}}\right) }\right| + \left| {f\left( {x,{y}_{0}}\right) - f\left( {{x}_{0},{y}_{0}}\right) }\right| $$
$$ \leq \varepsilon /2 + \varepsilon /2 = \varepsilon \text{ (由式子(1.5) 与(1.6)), } $$
即 $f\left( {x,y}\right)$ 在 $\left( {{x}_{0},{y}_{0}}\right)$ 点连续. 由 $\left( {{x}_{0},{y}_{0}}\right)$ 的任意性,知 $f\left( {x,y}\right) \in {C\Omega }$ .