函数导数九：映射微分和复合求导

设开集$D \subset \mathbb{R}^n$，$\boldsymbol{f}: D \to \mathbb{R}^m$。记$\boldsymbol{f}$的分量依次为$f_1,f_2,\cdots,f_m$，可以把$\boldsymbol{f}(\boldsymbol{x})$写成

$$\boldsymbol{f(x)} = \left( \begin{matrix} f_1(\boldsymbol{x}) \\ f_2(\boldsymbol{x}) \\ \vdots \\ f_m(\boldsymbol{x}) \end{matrix} \right) \quad (\boldsymbol{x} \in D)$$

设点$\boldsymbol{x}_0 \in D, \boldsymbol{h} \in \mathbb{R}^n$。由于$\boldsymbol{x}_0$是$D$的内点，所以总可以找到充分小的$\Vert \boldsymbol{h} \Vert$使得$\boldsymbol{x}_0 + \boldsymbol{h} \in D$。

定义1

如果映射$\boldsymbol{f}$满足

$$\boldsymbol{f}(\boldsymbol{x}_0 + \boldsymbol{h}) - \boldsymbol{f}(\boldsymbol{x}_0) = \boldsymbol{Ah} + \boldsymbol{r}(\boldsymbol{h})$$

式中$\boldsymbol{A}$是一个$m \times n$矩阵，它的元素不依赖于$\boldsymbol{h}$，且

$$\lim \limits_{\boldsymbol{h} \to \boldsymbol{0}} \frac{\Vert \boldsymbol{r}(\boldsymbol{h}) \Vert}{\Vert \boldsymbol{h} \Vert} = 0$$

则称映射$\boldsymbol{f}$在点$\boldsymbol{x}_0$处可微，并称$\boldsymbol{Ah}$是$\boldsymbol{f}$在点$\boldsymbol{x}_0$处的微分，记作

$$\mathrm{d}\boldsymbol{f}(\boldsymbol{x}_0) = \boldsymbol{Ah}$$

定理1

若映射$\boldsymbol{f}$在点$\boldsymbol{x}_0$处可微，则有

$$\mathrm{d}\boldsymbol{f}(\boldsymbol{x}_0) = \boldsymbol{Jf}(\boldsymbol{x}_0)\boldsymbol{h}$$

其中

$$\boldsymbol{Jf}(\boldsymbol{x}_0) = \left[ \begin{matrix} \frac{\partial f_1(\boldsymbol{x}_0)}{\partial x_1} & \cdots & \frac{\partial f_1(\boldsymbol{x}_0)}{\partial x_n} \\ \vdots & & \vdots \\ \frac{\partial f_m(\boldsymbol{x}_0)}{\partial x_1} & \cdots & \frac{\partial f_m(\boldsymbol{x}_0)}{\partial x_n} \end{matrix} \right]$$

称之为映射$\boldsymbol{f}$在点$\boldsymbol{x}_0$处的Jacobi矩阵，也称为导数。

证：有定义可知，

$$\boldsymbol{f}(\boldsymbol{x}_0 + \boldsymbol{h}) - \boldsymbol{f}(\boldsymbol{x}_0) = \boldsymbol{Ah} + \boldsymbol{r}(\boldsymbol{h})$$

设

$$\boldsymbol{A} = \left[\begin{matrix} a_{11} & \cdots & a_{1n} \\ \vdots & & \vdots \\ a_{m1} & \cdots & a_{mn} \end{matrix}\right]$$

将其代入上式中得

$$f_i(\boldsymbol{x}_0 + \boldsymbol{h}) - f_i(\boldsymbol{x}_0) = \sum_{j=1}^n a_{ij} h_j + r_i(\boldsymbol{h})$$

其中$r_i(\boldsymbol{h})$表示$\boldsymbol{r}(\boldsymbol{h})$的第$i$个分量，由$\boldsymbol{r}(\boldsymbol{h})$的性质可知

$$r_i(\boldsymbol{h}) = o(\Vert \boldsymbol{h} \Vert) \quad (\boldsymbol{h} \to 0, i=1,2,\cdots,m)$$

从而继续由函数导数八定理1可知

$$a_{ij} = \frac{\partial f_i(\boldsymbol{x}_0)}{\partial x_j} (i=1,2,\cdots,m;j=1,2,\cdots,n)$$

从而得证。

Q.E.D.

定理2

若映射$\boldsymbol{f}$在点$\boldsymbol{x}_0$的某一邻域内存在Jacobi矩阵$\boldsymbol{Jf}$，且$\boldsymbol{Jf}$的各元素在$\boldsymbol{x}_0$处都连续，则映射$\boldsymbol{f}$在点$\boldsymbol{x}_0$处可微。

证：由定理1与函数导数八的定理2易证。

Q.E.D.

定义2

设开集$D \subset \mathbb{R}^n$，$\boldsymbol{f}: D \to \mathbb{R}^m$。如果$\boldsymbol{f}$在$D$上的每一点处都连续，则记$\boldsymbol{f} \in C(D)$；如果$\boldsymbol{Jf}$在$\boldsymbol{D}$上的每一点处都连续，则记$\boldsymbol{f} \in C^1(D)$。

定理3

设开集$D \subset \mathbb{R}^n$，$\boldsymbol{g}: D \to \mathbb{R}^m$，$\boldsymbol{g}$在点$\boldsymbol{x}_0 \in D$处可微，又设$\boldsymbol{f}$把包含$\boldsymbol{g}(D)$的一个开集映射至$\mathbb{R}^l$，并且$\boldsymbol{f}$在$\boldsymbol{g}(\boldsymbol{x}_0)$处可微，那么复合映射$\boldsymbol{f} \circ \boldsymbol{g}$在点$\boldsymbol{x}_0$处可微，并且

$$\boldsymbol{J}(\boldsymbol{f}\circ \boldsymbol{g})(\boldsymbol{x}_0) = \boldsymbol{Jf}(\boldsymbol{g}(\boldsymbol{x}_0)) \boldsymbol{Jg}(\boldsymbol{x}_0)$$

证：令$\boldsymbol{y}_0 = \boldsymbol{g}(\boldsymbol{x}_0)$，$\boldsymbol{A} = \boldsymbol{Jf}(\boldsymbol{y}_0)，\boldsymbol{B} = \boldsymbol{Jg}(\boldsymbol{x}_0)$，易知$\boldsymbol{A}$是$l\times m$矩阵，$\boldsymbol{B}$是$m \times n$矩阵，从而若能证明

$$\lim \limits_{\Vert \boldsymbol{h} \Vert \to 0} \frac{\Vert (\boldsymbol{f} \circ \boldsymbol{g})(\boldsymbol{x}_0 + \boldsymbol{h}) - (\boldsymbol{f} \circ \boldsymbol{g})(\boldsymbol{x}_0) - \boldsymbol{ABh} \Vert}{\Vert \boldsymbol{h} \Vert} = 0 \tag{1}$$

则由定义1可知，$\boldsymbol{J}(\boldsymbol{f}\circ \boldsymbol{g})(\boldsymbol{x}_0) = \boldsymbol{AB}$。
由于$\boldsymbol{g},\boldsymbol{f}$分别在$\boldsymbol{x}_0,\boldsymbol{y}_0$处可微，从而有

$$\begin{aligned} \boldsymbol{g}(\boldsymbol{x}_0 + \boldsymbol{h}) - \boldsymbol{g}(\boldsymbol{x}_0) = \boldsymbol{Bh} + \boldsymbol{u}(\boldsymbol{h}) \\ \boldsymbol{f}(\boldsymbol{y}_0 + \boldsymbol{k}) - \boldsymbol{f}(\boldsymbol{y}_0) = \boldsymbol{Ak} + \boldsymbol{v}(\boldsymbol{k}) \end{aligned} \tag{2}$$

其中$\boldsymbol{h}，\boldsymbol{k}$分别为$n \times 1$矩阵和$m \times 1$矩阵，且

$$\begin{aligned} \frac{\Vert \boldsymbol{u}(\boldsymbol{h}) \Vert}{\Vert \boldsymbol{h} \Vert} \to 0 \quad (\Vert \boldsymbol{h} \Vert \to 0) \\ \frac{\Vert \boldsymbol{v}(\boldsymbol{k}) \Vert}{\Vert \boldsymbol{k} \Vert} \to 0 \quad (\Vert \boldsymbol{k} \Vert \to 0) \end{aligned}$$

记

$$\frac{\Vert \boldsymbol{u}(\boldsymbol{h}) \Vert}{\Vert \boldsymbol{h} \Vert} = \varepsilon(\boldsymbol{h})， \quad \frac{\Vert \boldsymbol{v}(\boldsymbol{h}) \Vert}{\Vert \boldsymbol{h} \Vert} = \eta(\boldsymbol{h})$$

则

$$\Vert \boldsymbol{u}(\boldsymbol{h}) \Vert = \varepsilon(\boldsymbol{h}) \Vert \boldsymbol{h} \Vert， \quad \Vert \boldsymbol{v}(\boldsymbol{k}) \Vert = \eta(\boldsymbol{k}) \Vert \boldsymbol{k} \Vert$$

且

$$\lim \limits_{\Vert \boldsymbol{h} \Vert \to 0} \varepsilon(\boldsymbol{h}) = 0， \quad \lim \limits_{\Vert \boldsymbol{k} \Vert \to 0} \eta(\boldsymbol{k}) = 0 \tag{3}$$

对给定的$\boldsymbol{h}$，令$\boldsymbol{k} = \boldsymbol{g}(\boldsymbol{x}_0 + \boldsymbol{h}) - \boldsymbol{g}(\boldsymbol{x}_0)$，由式(2)可得

$$\Vert \boldsymbol{k} \Vert \le \Vert \boldsymbol{Bh} \Vert + \Vert \boldsymbol{u}(\boldsymbol{h})\Vert \le (\Vert \boldsymbol{B} \Vert + \varepsilon(\boldsymbol{h})) \Vert \boldsymbol{h} \Vert$$

从而有

$$\begin{aligned} & \Vert \boldsymbol{f} \circ \boldsymbol{g} (\boldsymbol{x}_0 + \boldsymbol{h}) - \boldsymbol{f} \circ \boldsymbol{g} (\boldsymbol{x}_0) - \boldsymbol{ABh} \Vert \\ & = \Vert \boldsymbol{f}(\boldsymbol{g} (\boldsymbol{x}_0 + \boldsymbol{h})) - \boldsymbol{f}(\boldsymbol{g} (\boldsymbol{x}_0)) - \boldsymbol{ABh} \Vert \\ & = \Vert \boldsymbol{f}(\boldsymbol{y}_0 + \boldsymbol{k}) - \boldsymbol{f}(\boldsymbol{y}_0) - \boldsymbol{ABh} \Vert \\ & = \Vert \boldsymbol{Ak} + \boldsymbol{v}(\boldsymbol{k}) - \boldsymbol{ABh} \Vert \\ & = \Vert \boldsymbol{A}(\boldsymbol{k} - \boldsymbol{Bh}) + \boldsymbol{v}{\boldsymbol{k}} \Vert \\ & \le \Vert \boldsymbol{A} \Vert u(\boldsymbol{h}) + \eta(\boldsymbol{k})\Vert \boldsymbol{k} \Vert \\ & \le \Vert \boldsymbol{A} \Vert \varepsilon(\boldsymbol{h}) \Vert \boldsymbol{h} \Vert + \eta(\boldsymbol{k})(\Vert \boldsymbol{B} \Vert + \varepsilon(\boldsymbol{h})) \Vert \boldsymbol{h} \Vert \end{aligned}$$

所以

$$\frac{\Vert (\boldsymbol{f} \circ \boldsymbol{g})(\boldsymbol{x}_0 + \boldsymbol{h}) - (\boldsymbol{f} \circ \boldsymbol{g})(\boldsymbol{x}_0) - \boldsymbol{ABh} \Vert}{\Vert \boldsymbol{h} \Vert} \le \Vert \boldsymbol{A} \Vert \varepsilon(\boldsymbol{h}) + \eta(\boldsymbol{k})(\Vert \boldsymbol{B} \Vert + \varepsilon(\boldsymbol{h}))$$

再由式(3)可知式(1)成立。

Q.E.D.