方企勤 第五章 多元函数微分学 第1题
📝 题目
例 1 设 ${x}_{n},{y}_{n} \in {\mathbf{R}}^{m}\left( {n = 1,2,\cdots }\right) ,\mathop{\lim }\limits_{{n \rightarrow \infty }}{x}_{n} = \mathbf{a},\mathop{\lim }\limits_{{n \rightarrow \infty }}{y}_{n} = \mathbf{b}$ ,证明:
$$ \mathop{\lim }\limits_{{n \rightarrow \infty }}\left( {{x}_{n} \cdot {y}_{n}}\right) = a \cdot b. $$
💡 答案解析
证法 $1\forall \varepsilon > 0$ ,因有极限点列必为有界点列,故存在 ${M}_{1} > 0$ , 使 $\left| {\mathbf{y}}_{n}\right| \leq {M}_{1}$ . 令 $M = \max \left( {{M}_{1},\left| \mathbf{a}\right| }\right)$ . 由 $\displaystyle{\mathop{\lim }\limits_{{n \rightarrow \infty }}{\mathbf{x}}_{n} = \mathbf{a},\mathop{\lim }\limits_{{n \rightarrow \infty }}{\mathbf{y}}_{n} = \mathbf{b},\exists N}$ ,当 $n > N$ 时,有
$$ \left| {{x}_{n} - a}\right| < \varepsilon /{2M},\;\left| {{y}_{n} - b}\right| < \varepsilon /{2M}. $$
于是当 $n > N$ 时,有
$$ \left| {{x}_{n} \cdot {y}_{n} - a \cdot b}\right| = \left| {{x}_{n} \cdot {y}_{n} - a \cdot {y}_{n} + a \cdot {y}_{n} - a \cdot b}\right| $$
$$ = \left| {\left( {{\mathbf{x}}_{n} - \mathbf{a}}\right) \cdot {\mathbf{y}}_{n} + \mathbf{a} \cdot \left( {{\mathbf{y}}_{n} - \mathbf{b}}\right) }\right| $$
$$ \leq \left| {\left( {{\mathbf{x}}_{n} - \mathbf{a}}\right) \cdot {\mathbf{y}}_{n}}\right| + \left| {\mathbf{a} \cdot \left( {{\mathbf{y}}_{n} - \mathbf{b}}\right) }\right| $$
$$ \leq \left| {{x}_{n} - a}\right| \cdot \left| {y}_{n}\right| + \left| a\right| \cdot \left| {{y}_{n} - b}\right| $$
$$ \leq M\left( {\left| {{x}_{n} - a}\right| + \left| {{y}_{n} - b}\right| }\right) < \varepsilon , $$
即
$$ \mathop{\lim }\limits_{{n \rightarrow \infty }}\left( {{x}_{n} \cdot {y}_{n}}\right) = a \cdot b. $$
证法 2 设
$$ {x}_{n} = \left( {{x}_{n}^{1},\cdots ,{x}_{n}^{m}}\right) ,\;{y}_{n} = \left( {{y}_{n}^{1},\cdots ,{y}_{n}^{m}}\right) \;\left( {n = 1,2,\cdots }\right) , $$
$$ \mathbf{a} = \left( {{a}^{1},\cdots ,{a}^{m}}\right) ,\;\mathbf{b} = \left( {{b}^{1},\cdots ,{b}^{m}}\right) . $$
由 $\displaystyle{\mathop{\lim }\limits_{{n \rightarrow \infty }}{\mathbf{x}}_{n} = \mathbf{a},\mathop{\lim }\limits_{{n \rightarrow \infty }}{\mathbf{y}}_{n} = \mathbf{b}}$ ,可得
$$ \mathop{\lim }\limits_{{n \rightarrow \infty }}{x}_{n}^{i} = {a}^{i},\;\mathop{\lim }\limits_{{n \rightarrow \infty }}{y}_{n}^{i} = {b}^{i}\;\left( {i = 1,\cdots ,m}\right) , $$
所以 $\mathop{\lim }\limits_{{n \rightarrow \infty }}\left( {{\mathbf{x}}_{n} \cdot {\mathbf{y}}_{n}}\right) = \mathop{\lim }\limits_{{n \rightarrow \infty }}\mathop{\sum }\limits_{{i = 1}}^{m}\left( {{x}_{n}^{i} \cdot {y}_{n}^{i}}\right) = \mathop{\sum }\limits_{{i = 1}}^{m}{a}^{i} \cdot {b}^{i} = \mathbf{a} \cdot \mathbf{b}$ .
评注 证法 2 用到空间是有限维这一性质, 而证法 1 没有用到空间是有限维这一性质, 所以它对任一具有内积的线性空间都适用.