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函数极限七:向量值函数的连续与一致连续

定义1:向量值函数

\(D \subset \mathbb{R}^n\),定义\(\boldsymbol{f}: D \to \mathbb{R}^m\)为向量值函数,其中\(D\)称为\(\boldsymbol{f}\)定义域,\(f(D)\)称为值域。当用\(\boldsymbol{y} = \boldsymbol{f}(\boldsymbol{x})(\boldsymbol{x} \in D)\)来表示这种映射时,\(\boldsymbol{x} \in \mathbb{R}^n, \boldsymbol{y} \in \mathbb{R}^m\)。按分量形式写出\(\boldsymbol{x} = (x_1,x_2,\cdots,x_n), \boldsymbol{y}=(y_1,y_2,\cdots,y_m)\),那么该映射函数相当于给定了\(m\)\(n\)元函数:
\[ \left\{ \begin{aligned} & y_1 = f_1(x_1,x_2,\cdots,x_n) \\ & y_2 = f_2(x_1,x_2,\cdots,x_n) \quad (x_1,x_2,\cdots,x_n) \in D \subset \mathbb{R}^n \\ & \cdots \\ & y_m = f_m(x_1,x_2,\cdots,x_n) \end{aligned} \right. \]
反过来也成立。也就是说,如果给定了\(m\)个定义在\(D \subset \mathbb{R}^n\)上的函数,就相当于给定了定义在\(D\)上,映射到\(\mathbb{R}^m\)中的一个映射,或者说在\(D\)上定义了一个在\(\mathbb{R}^m\)中取值的向量值函数。这一事实表示为
\[ \boldsymbol{f} = (f_1,f_2,\cdots,f_m) \]
其中\(f_i: D \to \mathbb{R}\)称为\(\boldsymbol{f}\)的第\(i\)个分量函数\((i=1,2,\cdots,m)\)

定义2

\(D \subset \mathbb{R}^n\)\(\boldsymbol{f}: D \to \mathbb{R}^m\),又设\(\boldsymbol{a} \in D^\prime, \boldsymbol{p}\in \mathbb{R}^m\)。如果\(\forall \varepsilon > 0\)\(\exists \delta > 0\),使得当\(\boldsymbol{x} \in D\)\(0 < \Vert \boldsymbol{x} - \boldsymbol{a} \Vert < \delta\)时,有
\[ \Vert \boldsymbol{f}(\boldsymbol{x}) - \boldsymbol{p} \Vert < \varepsilon \]
则称映射\(f\)在点\(\boldsymbol{a}\)处有极限\(p\),记为
\[ \lim \limits_{\boldsymbol{x} \to \boldsymbol{a}} \boldsymbol{f}(\boldsymbol{x}) = \boldsymbol{p} \]
也可以简记为
\[ \boldsymbol{f}(\boldsymbol{x}) \to \boldsymbol{p} \quad (\boldsymbol{x} \to \boldsymbol{a}) \]

定理1

\(D \subset \mathbb{R}^n\)\(\boldsymbol{f}: D \to \mathbb{R}^m,\boldsymbol{a} \in D^\prime, \boldsymbol{p}=(p_1,p_2,\cdots,p_m)\in \mathbb{R}^m, \boldsymbol{f} = (f_1,f_2,\cdots,f_m)\)。那么
\[ \lim \limits_{\boldsymbol{x} \to \boldsymbol{a}} \boldsymbol{f}(\boldsymbol{x}) = \boldsymbol{p} \]
当且仅当
\[ \lim \limits_{\boldsymbol{x} \to \boldsymbol{a}} f_i(\boldsymbol{x}) = p_i \quad (i=1,2,\cdots,m) \]

证:必要性。因为\(\Vert f_i(\boldsymbol{x}) - p_i \Vert \le \Vert \boldsymbol{f}(\boldsymbol{x}) - \boldsymbol{p} \Vert \quad (i=1,2,\cdots,m)\),所以\(\forall \varepsilon > 0\)\(\exists \delta > 0\),使得当\(\boldsymbol{x} \in D\)\(0 < \Vert \boldsymbol{x} - \boldsymbol{a} \Vert < \delta\)时,有
\[ | f_i(\boldsymbol{x}) - p_i | \le \Vert \boldsymbol{f}(\boldsymbol{x}) - \boldsymbol{p} \Vert < \varepsilon \]
这就表明
\[ \lim \limits_{\boldsymbol{x} \to \boldsymbol{a}} f_i(\boldsymbol{x}) = p_i \quad (i=1,2,\cdots,m) \]

充分性。因为
\[ \Vert \boldsymbol{f}(\boldsymbol{x}) - \boldsymbol{p} \Vert \le \sum_{i=1}^m |f_i(\boldsymbol{x}) - p_i| \]
\(\varepsilon > 0\)\(\exists \delta > 0\),使得当\(0 < \Vert \boldsymbol{x} - \boldsymbol{a} \Vert < \delta\)时,有\(|f(\boldsymbol{x}_i) - p_i| < \varepsilon / m \quad(i=1,2,\cdots,m)\),所以
\[ \Vert \boldsymbol{f}(\boldsymbol{x}) - \boldsymbol{p} \Vert \le \sum_{i=1}^m |f_i(\boldsymbol{x}) - p_i| < m \cdot \frac{\varepsilon}{m} = \varepsilon \]

Q.E.D.

定理2

\(D \subset \mathbb{R}^n\)\(\boldsymbol{f},\boldsymbol{g}: D \to \mathbb{R}^m,\boldsymbol{a} \in D^\prime\),并且
\[ \lim \limits_{\boldsymbol{x} \to \boldsymbol{a}} \boldsymbol{f}(\boldsymbol{x}) = \boldsymbol{p},\lim \limits_{\boldsymbol{x} \to \boldsymbol{a}} \boldsymbol{g}(\boldsymbol{x}) = \boldsymbol{q} \]
于是我们有:
(1)对任何\(\lambda \in \mathbb{R}\),可以得出\(\lim \limits_{\boldsymbol{x} \to \boldsymbol{a}}(\lambda \boldsymbol{f}(\boldsymbol{a})) = \lambda \boldsymbol{p}\)
(2)\(\lim \limits_{\boldsymbol{x} \to \boldsymbol{a}} (\boldsymbol{f}(\boldsymbol{x}) + \boldsymbol{g}(\boldsymbol{x})) = \boldsymbol{p} + \boldsymbol{q}\)

证:利用不等式很容易证明。

Q.E.D.

定义3:连续映射

\(D \subset \mathbb{R}^n\)\(\boldsymbol{f}: D \to \mathbb{R}^m,\boldsymbol{a} \in D\)。如果\(\forall \varepsilon > 0\)\(\exists \delta > 0\),使得当\(\boldsymbol{x} \in D \cap B_\delta(\boldsymbol{a})\)时,有\(\boldsymbol{f}(\boldsymbol{x}) \in B_\varepsilon(\boldsymbol{f}(\boldsymbol{a}))\),则称映射\(\boldsymbol{f}\)在点\(\boldsymbol{a}\)处连续。当\(\boldsymbol{f}\)\(D\)中的每一点都连续,称映射\(f\)\(D\)上连续。

定理3

\(D\)\(\mathbb{R}^n\)中的开集,\(\boldsymbol{f}:D \to \mathbb{R}^m\)\(\boldsymbol{f}\)\(D\)上连续的充分必要条件是,对任意的\(\mathbb{R}^m\)上的开集\(G\)\(\boldsymbol{f}^{-1}(G)\)\(\mathbb{R}^n\)中的开集。这里
\[ \boldsymbol{f}^{-1}(G) = \{ \boldsymbol{x} \in D : \boldsymbol{f}(\boldsymbol{x}) \in G \} \]

证:必要性。如果\(\boldsymbol{f}^{-1}(G)\)是空集,自然是开集;如果非空,任取\(\boldsymbol{x}_0 \in \boldsymbol{f}^{-1}(G)\),则\(\boldsymbol{f}(\boldsymbol{x_0}) \in G\),由于\(G\)是开集,从而存在\(\varepsilon > 0\),使得\(B_\varepsilon(\boldsymbol{f}(\boldsymbol{x}_0)) \subset G\)。又因为\(\boldsymbol{f}\)\(\boldsymbol{x}_0\)处连续,从而存在\(\delta > 0\),使得当\(\boldsymbol{x} \in B_\delta(\boldsymbol{x}_0)\)时,有\(\boldsymbol{f}(\boldsymbol{x}) \in B_\varepsilon(\boldsymbol{f}(\boldsymbol{x}_0)) \subset G\),即\(\boldsymbol{x} \in \boldsymbol{f}^{-1}(G)\),从而\(B_\delta(\boldsymbol{x}_0) \subset \boldsymbol{f}^{-1}(G)\)。这就说明\(\boldsymbol{f}^{-1}(G)\)是开集。

充分性。任取\(\boldsymbol{x}_0 \in D\),对任意小的\(\varepsilon > 0\),取\(G=B_\varepsilon(\boldsymbol{f}(\boldsymbol{x}_0))\),可知\(G\)是开集,利用条件知\(\boldsymbol{f}^{-1}(G)\)也是开集。因为\(\boldsymbol{x}_0 \in \boldsymbol{f}^{-1}(G)\),从而存在\(\delta > 0\),使得\(B_\delta(\boldsymbol{x}_0) \subset \boldsymbol{f}^{-1}(G)\)。即对任意的\(\boldsymbol{x} \in B_\delta(\boldsymbol{x}_0)\),有\(\boldsymbol{f}(\boldsymbol{x}) \in G = B_\varepsilon(\boldsymbol{f}(\boldsymbol{x}_0))\)。从而\(\boldsymbol{f}\)\(\boldsymbol{x}_0\)处连续。由于\(\boldsymbol{x}_0\)是任取的,所以\(\boldsymbol{f}\)\(D\)上连续。

定义4:一致连续映射

\(D \subset \mathbb{R}^n\)\(\boldsymbol{f}: D \to \mathbb{R}^m\),如果\(\forall \varepsilon > 0\)\(\exists \delta> 0\),使得当\(\boldsymbol{x},\boldsymbol{y} \in D\)\(\Vert \boldsymbol{x} - \boldsymbol{y} \Vert < \delta\)时,有\(\Vert \boldsymbol{f}(\boldsymbol{x}) - \boldsymbol{f}(\boldsymbol{y}) \Vert < \varepsilon\),则称\(\boldsymbol{f}\)\(D\)上一致连续。

定理4

\(D \subset \mathbb{R}^n\)\(\boldsymbol{f}: D \to \mathbb{R}^m\)\(D\)上一致连续当且仅当\(\boldsymbol{f}\)的每一个分量函数在\(D\)上一致连续。

证:利用\(|f_i(\boldsymbol{x}) - f_i(\boldsymbol{y})| \le \Vert \boldsymbol{f}(\boldsymbol{x}) - \boldsymbol{f}(\boldsymbol{y}) \Vert\),很容易证明。

Q.E.D.

定理5

\(D \subset \mathbb{R}^n\)\(\boldsymbol{f}: D \to \mathbb{R}^m\)\(D\)上的连续映射。如果\(D\)是紧致集,那么\(\boldsymbol{f}\)\(D\)上是一致连续的。

证:与函数极限六的定理1证法一样。

Q.E.D.

定理6

\(D \subset \mathbb{R}^n\)\(\boldsymbol{f}: D \to \mathbb{R}^m\)\(D\)上的连续映射。如果\(D\)\(\mathbb{R}^n\)中的连通集,那么\(\boldsymbol{f}(D)\)\(\mathbb{R}^m\)中的连通集。

证:与函数极限六的定理4证法一样。

定理7

\(D \subset \mathbb{R}^n\)\(\boldsymbol{f}: D \to \mathbb{R}^m\)\(D\)上的连续映射。如果\(D\)\(\mathbb{R}^n\)中紧致集,那么\(\boldsymbol{f}(D)\)\(\mathbb{R}^m\)中的紧致集。

证:与函数极限六的定理2证法一样。