📝 题目
例 23 设 $U\left( {{x}_{0},{\delta }_{0}}\right) \subset {\mathbf{R}}^{n},f \in {C}^{2}\left( {U\left( {{x}_{0},{\delta }_{0}}\right) }\right) ,\nabla f\left( {x}_{0}\right) = 0$ . 又设对任意单位向量 $\mathbf{\alpha } \in {\mathbf{R}}^{n},{\left( \mathbf{\alpha } \cdot \nabla \right) }^{2}f\left( {\mathbf{x}}_{0}\right) > 0$ . 求证: $\exists 0 < \delta < {\delta }_{0}$ ,使
$$ \left( {x - {x}_{0}}\right) \cdot \nabla f\left( x\right) > 0.\;\left( {\forall x \in U\left( {{x}_{0},\delta }\right) \smallsetminus \left\{ {x}_{0}\right\} }\right) . $$
💡 答案与解析
证 因为 $g\left( \mathbf{\alpha }\right) \frac{\text{ 定义 }}{}{\left( \mathbf{\alpha } \cdot \nabla \right) }^{2}f\left( {\mathbf{x}}_{0}\right)$ 在闭集
$$ S\overset{\text{ 定义 }}{ = }\left\{ {\left. {a \in {\mathbf{R}}^{n}}\right| \;\left| {\;a}\right| = 1}\right\} $$
上是连续函数,所以 $m\overset{\text{ 定义 }}{ = }\inf \{ g\left( \mathbf{\alpha }\right) \mid \mathbf{\alpha } \in S\} > 0$ . 又对 $\forall \mathbf{\alpha } \in S$ ,有 $\left( {\mathbf{\alpha } = \left( {{\alpha }^{1},{\alpha }^{2},\cdots ,{\alpha }^{n}}\right) }\right)$
$$ \left| {{\left( \mathbf{\alpha } \cdot \nabla \right) }^{2}f\left( \mathbf{x}\right) - {\left( \mathbf{\alpha } \cdot \nabla \right) }^{2}f\left( {\mathbf{x}}_{0}\right) }\right| $$
$$ = \left| {\mathop{\sum }\limits_{{i,j = 1}}^{n}\left( {\frac{{\partial }^{2}f\left( \mathbf{x}\right) }{\partial {x}^{j}\partial {x}^{i}} - \frac{{\partial }^{2}f\left( {\mathbf{x}}_{0}\right) }{\partial {x}^{j}\partial {x}^{i}}}\right) {\alpha }^{i}{\alpha }^{j}}\right| $$
$$ \leq \mathop{\sum }\limits_{{i,j = 1}}^{n}\left| {\frac{{\partial }^{2}f\left( \mathbf{x}\right) }{\partial {x}^{j}\partial {x}^{i}} - \frac{{\partial }^{2}f\left( {\mathbf{x}}_{0}\right) }{\partial {x}^{j}\partial {x}^{i}}}\right| . $$
因此 $\exists 0 < \delta < {\delta }_{0}$ ,使得
$$ \left| {{\left( \mathbf{\alpha } \cdot \nabla \right) }^{2}f\left( x\right) - {\left( \mathbf{\alpha } \cdot \nabla \right) }^{2}f\left( {x}_{0}\right) }\right| < \frac{m}{2}\;\left( {\forall x \in U\left( {{x}_{0},\delta }\right) }\right) . $$
于是 $\forall x \in U\left( {{x}_{0},\delta }\right)$ 时,有
$$ {\left( a \cdot \nabla \right) }^{2}f\left( x\right) \geq {\left( a \cdot \nabla \right) }^{2}f\left( {x}_{0}\right) - \left| {{\left( a \cdot \nabla \right) }^{2}f\left( x\right) - {\left( a \cdot \nabla \right) }^{2}f\left( {x}_{0}\right) }\right| $$
$$ \geq \frac{m}{2} > 0\text{ . } $$
今对 $\forall x \in U\left( {{x}_{0},\delta }\right) \smallsetminus \left\{ {x}_{0}\right\}$ ,考虑辅助函数
$$ F\left( t\right) \overset{\text{ 定义 }}{ = }f\left( {{x}_{0} + t\left( {x - {x}_{0}}\right) }\right) \;\left( {0 \leq t \leq 1}\right) , $$
则有
$$ {F}^{\prime }\left( t\right) = \left( {\mathbf{x} - {\mathbf{x}}_{0}}\right) \cdot \nabla f\left( {{\mathbf{x}}_{0} + t\left( {\mathbf{x} - {\mathbf{x}}_{0}}\right) }\right) , $$
$$ {F}^{\prime \prime }\left( t\right) = {\left( \left( \mathbf{x} - {\mathbf{x}}_{0}\right) \cdot \nabla \right) }^{2}f\left( {{\mathbf{x}}_{0} + t\left( {\mathbf{x} - {\mathbf{x}}_{0}}\right) }\right) $$
$$ = {\left| \mathbf{x} - {\mathbf{x}}_{0}\right| }^{2}{\left( \mathbf{a} \cdot \nabla \right) }^{2}f\left( {{\mathbf{x}}_{0} + t\left( {\mathbf{x} - {\mathbf{x}}_{0}}\right) }\right) $$
$$ \geq \frac{m}{2}{\left| x - {x}_{0}\right| }^{2} > 0\;\left( {0 \leq t \leq 1}\right) , $$
其中
$$ \alpha \frac{\text{ 定义 }\frac{x - {x}_{0}}{\left| x - {x}_{0}\right| }}{\left| x - {x}_{0}\right| } \in S. $$
由此可见, ${F}^{\prime }\left( t\right)$ 在 $\left\lbrack {0,1}\right\rbrack$ 上严格单调增加,于是对于 $\forall x \in U\left( {{x}_{0},\delta }\right) \smallsetminus$ $\left\{ {x}_{0}\right\}$ ,有
$$ \left( {\mathbf{x} - {\mathbf{x}}_{0}}\right) \cdot \nabla f\left( \mathbf{x}\right) = {F}^{\prime }\left( 1\right) > {F}^{\prime }\left( 0\right) = \left( {\mathbf{x} - {\mathbf{x}}_{0}}\right) \cdot \nabla f\left( {\mathbf{x}}_{0}\right) = 0. $$