函数导数十二：高阶偏导数和Taylor公式

定义1

设函数$f$在开集$D$上的每一点处存在偏导数：

$$D_if(\boldsymbol{x}) = \frac{\partial f}{\partial x_i}(\boldsymbol{x}) \quad (i=1,2,\cdots,n)$$

称它们为$f$的一阶偏导函数，如果对这些偏导函数又可以取偏导数，得出的就是$f$的二阶偏导函数，依次可以定义三阶偏导数以及更高阶的偏导数。对于二阶偏导数，将一阶偏导函数$\displaystyle \frac{\partial f}{\partial x_j}$再对$x_i$求偏导数，即$\displaystyle \frac{\partial f}{\partial x_i}\left(\frac{\partial f}{\partial x_j}\right)$记作$\displaystyle \frac{\partial^2 f}{\partial x_i \partial x_j}$，这里$i,j$独立地从$1$变到$n$，如果$i=j$，那么把$\displaystyle \frac{\partial^2 f}{\partial x_i \partial x_i}$记作$\displaystyle \frac{\partial^2 f}{\partial x_i^2}(i=1,2,\cdots,n)$；如果$i\ne j$，这类二阶偏导数称为混合偏导数。

定理1

设开集$D \subset \mathbb{R}^2$，$f: D \to \mathbb{R}$，如果$\displaystyle \frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial^2 f}{\partial y \partial x}$在$(x_0,y_0)$的某个邻域上存在，且$\displaystyle \frac{\partial^2 f}{\partial y\partial x}$在$(x_0,y_0)$处连续，那么$\displaystyle \frac{\partial^2 f}{\partial x \partial y}$在$(x_0,y_0)$处存在，而且

$$\frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x}$$

证：记

$$\varphi(h, k) = f(x_0 + h, y_0 + k) - f(x_0+h, y_0) - f(x_0, y_0+k) + f(x_0, y_0)$$

令

$$g(x) = f(x, y_0+k) - f(x, y_0)$$

从而由微分中值定理可知

$$\begin{aligned} \varphi(h, k) & = g(x_0 + h) - g(x_0) \\ & = g^\prime(x_0 + \theta_1 h)h \\ & = \left(\frac{\partial f}{\partial x}(x_0 + \theta_1 h, y_0 + k) - \frac{\partial f}{\partial x}(x_0 + \theta_1 h, y_0) \right)h \\ & = \frac{\partial^2 f}{\partial y \partial x}(x_0 + \theta_1h, y_0 + \theta_2k)hk \end{aligned}$$

由于$\displaystyle \frac{\partial^2 f}{\partial y \partial x}$在$(x_0, y_0)$处连续，从而有

$$\lim \limits_{h\to 0, k \to 0} \frac{\varphi(h ,k)}{hk} = \frac{\partial^2 f}{\partial y \partial x}(x_0, y_0)$$

而又有

$$\lim \limits_{k \to 0} \frac{\varphi(h ,k)}{hk} = \lim \limits_{k \to 0} \frac{1}{h} \left( \frac{f(x_0+h, y_0+k) - f(x_0+h, y_0)}{k} - \frac{f(x_0, y_0+k) - f(x_0, y_0)}{k}\right) = \frac{1}{h}\left( \frac{\partial f}{\partial y}(x_0 +h, y_0) - \frac{\partial f}{\partial y}(x_0, y_0)\right)$$

所以

$$\lim \limits_{h \to 0, k \to 0} \frac{\varphi(h ,k)}{hk} = \lim \limits_{h \to 0} \frac{1}{h}\left( \frac{\partial f}{\partial y}(x_0 +h, y_0) - \frac{\partial f}{\partial y}(x_0, y_0)\right) = \frac{\partial^2 f}{\partial x \partial y}(x_0, y_0)$$

所以$\displaystyle \frac{\partial^2 f}{\partial x \partial y}(x_0, y_0)$存在，而且

$$\frac{\partial^2 f}{\partial x \partial y}(x_0, y_0) = \frac{\partial^2 f}{\partial y \partial x}(x_0, y_0)$$

Q.E.D.

定理2

设定义在凸区域$D \subset \mathbb{R}^n$上的函数$f$可微，则对任何两点$\boldsymbol{a}, \boldsymbol{b} \in D$，在由$\boldsymbol{a},\boldsymbol{b}$确定的线段上存在一点$\boldsymbol{\xi}$，使得

$$f(\boldsymbol{b}) - f(\boldsymbol{a}) = Jf(\boldsymbol{\xi})(\boldsymbol{b} - \boldsymbol{a})$$

证：由$\boldsymbol{a}$与$\boldsymbol{b}$确定的线段上的点可表示为$\boldsymbol{a} + t(\boldsymbol{b} - \boldsymbol{a})$，这里$t \in [0, 1]$，令

$$\varphi(t) = f(\boldsymbol{a} + t(\boldsymbol{b} - \boldsymbol{a}))$$

那么$\varphi$是$[0, 1]$上的可微函数，由单变量的微分中值定理可知，存在$\theta \in (0,1)$，使得

$$\varphi(1) - \varphi(0) = \varphi^\prime(\theta)$$

即

$$f(\boldsymbol{b}) - f(\boldsymbol{a}) = \boldsymbol{J}f(\boldsymbol{a} + \theta (\boldsymbol{b} - \boldsymbol{a}))(\boldsymbol{b} - \boldsymbol{a})$$

再令$\boldsymbol{\xi} = \boldsymbol{a} + \theta(\boldsymbol{b} - \boldsymbol{a})$即证得结论。

Q.E.D.

定理3

设$D$是$\mathbb{R}^n$中的区域，如果对任意的$\boldsymbol{x} \in D$，有

$$\frac{\partial f}{\partial x_1}(\boldsymbol{x}) = \cdots = \frac{\partial f}{\partial x_n}(\boldsymbol{x}) = 0$$

那么$f$在$D$上为一个常数。

证：如果$D$是凸区域，则由定理1立即得出结论。如果$D$不是凸区域，任取$\boldsymbol{x}_0 \in D$，令

$$\begin{aligned} A = \{ \boldsymbol{x} \in D: f(\boldsymbol{x}) = f(\boldsymbol{x_0})\} \\ B = \{ \boldsymbol{x} \in D: f(\boldsymbol{x}) \ne f(\boldsymbol{x_0}) \} \end{aligned}$$

显然$A$非空，而$D=A \cup B$，由于$D$是连通开集，若能证明$A,B$是开集，则由点列极限六的定理3可知，$B = \varnothing$，从而证得结论。为了证明$A$是开集，任取$\boldsymbol{a} \in A \subset D$，存在$B_{\boldsymbol{r}}(\boldsymbol{a}) \in D$，由于$B_{\boldsymbol{r}}(\boldsymbol{a})$是凸区域，从而$f$在$B_{\boldsymbol{r}}(\boldsymbol{a})$上是常数，且对任意的$\boldsymbol{x} \in B_{\boldsymbol{r}}(\boldsymbol{a})$，有

$$f(\boldsymbol{x}) = f(\boldsymbol{a}) = f(\boldsymbol{x}_0)$$

从而$B_{\boldsymbol{r}}(\boldsymbol{a}) \subset A$，也就说明$A$是开集。同样的方法也可以证明$B$是开集。再由上面分析可知命题成立。

Q.E.D.

定理4

设$k,n$是两个正整数，那么

$$(x_1 + \cdots + x_n)^k = \sum_{\alpha_1 + \cdots + \alpha_n = k} \frac{k!}{\alpha_1!\cdots\alpha_n!}x_1^{\alpha_1} \cdots x_n^{\alpha_n}$$

这里$\alpha_1,\cdots,\alpha_n$是非负整数。如果记$\boldsymbol{\alpha} = (\alpha_1, \cdots, \alpha_n)$，$\boldsymbol{x}=(x_1,\cdots,x_n)$，且

$$\begin{aligned} |\boldsymbol{\alpha}| &= \alpha_1 + \cdots + \alpha_n \\ \boldsymbol{\alpha}! &= \alpha_1!\cdots\alpha_n! \\ \boldsymbol{x}^{\boldsymbol{\alpha}} &= x_1^{\alpha_1} \cdots x_n^{\alpha_n} \end{aligned}$$

则上式可简写为

$$(x_1 + \cdots + x_n)^k = \sum_{|\boldsymbol{\alpha}|=k}\frac{k!}{\boldsymbol{\alpha}!}\boldsymbol{x}^{\boldsymbol{\alpha}}$$

证：对加项的个数$n$作归纳。当$n=2$时，该定理就是二项式定理，固然成立。先设$n-1$时命题成立，那么当加项的个数为$n$时，有

$$\begin{aligned} (x_1 + \cdots + x_n)^k &= ((x_1 + \cdots + x_{n-1}) + x_n)^k \\ &= \sum_{\alpha_n=0}^k \frac{k!}{\alpha_n!(k-\alpha_n)!}(x_1+\cdots+x_{n-1})^{k-\alpha_n}x_n^{\alpha_n} \\ &= \sum_{\alpha_n=0}^k \frac{k!}{\alpha_n!(k-\alpha_n)!} \sum_{\alpha_1 + \alpha_{n-1}=k-\alpha_n}\frac{(k-\alpha_n)!}{\alpha_1!\cdots\alpha_{n-1}!} x_1^{\alpha_1} \cdots x_{n-1}^{\alpha_{n-1}} x_n^{\alpha_n} \\ & = \sum_{\alpha_1 + \cdots + \alpha_n = k} \frac{k!}{\alpha_1!\cdots\alpha_n!}x_1^{\alpha_1} \cdots x_n^{\alpha_n} \end{aligned}$$

Q.E.D.

定理5：Taylor公式

设$D \subset \mathbb{R}^n$是一个凸区域，$f \in C^{m+1}(D)$，$\boldsymbol{a}=(a_1,\cdots,a_n)$，$\boldsymbol{a}+\boldsymbol{h} = (a_1+h_1,\cdots,a_n+h_n)$是$D$中的两个点，则必存在$\theta \in (0, 1)$，使得

$$f(\boldsymbol{a} + \boldsymbol{h}) = \sum_{k=0}^m \sum_{|\boldsymbol{a}|=k} \frac{D^{\boldsymbol{\alpha}}f(\boldsymbol{a})}{\boldsymbol{\alpha}!} \boldsymbol{h}^{\boldsymbol{\alpha}} + \boldsymbol{R}_m$$

其中

$$D^{\boldsymbol{\alpha}}f(\boldsymbol{a}) = \frac{\partial^{\alpha_1+\cdots+\alpha_n}f}{\partial x_1^{\alpha_1} \cdots \partial x_n^{\alpha_n}}(\boldsymbol{a})$$

且

$$\boldsymbol{R}_m = \sum_{|\boldsymbol{\alpha}|=m+1} \frac{D^{\boldsymbol{\alpha}}f(\boldsymbol{a} + \theta \boldsymbol{h})}{\boldsymbol{\alpha}!} \boldsymbol{h}^{\boldsymbol{\alpha}}$$

称为Lagrange余项。

证：固定$\boldsymbol{a},\boldsymbol{h}$，设$t \in [0,1]$，考虑$[0,1]$上的函数$\varphi(t) = f(\boldsymbol{a} + t \boldsymbol{h})$，显然$\varphi$在$[0,1]$上有$m+1$阶的连续导数，对$\varphi$用单变量函数的Taylor公式，得

$$\varphi(1) = \varphi(0) + \varphi^\prime(0) + \frac{1}{2!}\varphi^{\prime\prime}(0) + \cdots + \frac{1}{m!}\varphi^{(m)}(0) + \frac{1}{(m+1)!}\varphi^{(m+1)}(\theta) \tag{1}$$

其中$\theta \in (0, 1)$。显然$\varphi(1) = f(\boldsymbol{a} + \boldsymbol{h})$，$\varphi(0) = f(\boldsymbol{a})$，根据复合函数的求导公式得

$$\varphi^\prime(t) = \frac{\partial f}{\partial x_1}(\boldsymbol{a} + t \boldsymbol{h})h_1 + \cdots + \frac{\partial f}{\partial x_n}(\boldsymbol{a} + t \boldsymbol{h})h_n = \left(h_1\frac{\partial}{\partial x_1} + \cdots + h_n\frac{\partial}{\partial x_n}\right)f(\boldsymbol{a} + t \boldsymbol{h})$$

从而可得

$$\begin{aligned} \varphi^{\prime\prime}(t) &= \left(h_1\frac{\partial}{\partial x_1} + \cdots + h_n\frac{\partial}{\partial x_n}\right)^2f(\boldsymbol{a} + t\boldsymbol{h}) \\ \cdots, \\ \varphi^{(m)}(t) &= \left(h_1\frac{\partial}{\partial x_1} + \cdots + h_n\frac{\partial}{\partial x_n}\right)^mf(\boldsymbol{a} + t\boldsymbol{h}) \end{aligned}$$

根据定理4可知，

$$\varphi^{(k)}(t) = \sum_{|\boldsymbol{\alpha}|=k}\frac{k!}{\boldsymbol{\alpha}!}\frac{\partial^{\alpha_1}}{\partial x_1^{\alpha_1}}\cdots \frac{\partial ^{\alpha_n}}{\partial x_n^{\alpha_n}}f(\boldsymbol{a} + t\boldsymbol{h})\boldsymbol{h}^{\boldsymbol{\alpha}} = \sum_{|\boldsymbol{\alpha}|=k} \frac{k!}{\boldsymbol{\alpha}!}D^{\boldsymbol{\alpha}}f(\boldsymbol{a} + t\boldsymbol{h}) \boldsymbol{h}^{\boldsymbol{\alpha}}$$

所以

$$\varphi^{(k)}(0) = \sum_{|\boldsymbol{\alpha}|=k} \frac{k!}{\boldsymbol{\alpha}!}D^{\boldsymbol{\alpha}}f(\boldsymbol{a}) \boldsymbol{h}^{\boldsymbol{\alpha}}$$

将其代入(1)式，即得证明的结论。

Q.E.D.

特别地

Taylor公式的前三项写出来就是

$$f(\boldsymbol{a} + \boldsymbol{h}) = f(\boldsymbol{a}) + \frac{\partial f}{\partial x_1}(\boldsymbol{a})h_1 + \cdots + \frac{\partial f}{\partial x_n}(\boldsymbol{a})h_n + \frac{1}{2} \sum_{i,j=1}^n\frac{\partial^2 f}{\partial x_i \partial x_j}(\boldsymbol{a})h_ih_j + \cdots$$

如果记

$$Hf(\boldsymbol{a}) = \begin{bmatrix} \frac{\partial^2 f}{\partial x_1^2}(\boldsymbol{a}) & \cdots & \frac{\partial^2 f}{\partial x_1 \partial x_n}(\boldsymbol{a}) \\ \vdots & & \vdots \\ \frac{\partial^2 f}{\partial x_n \partial x_1}(\boldsymbol{a}) & \cdots & \frac{\partial^2 f}{\partial x_n^2}(\boldsymbol{a}) \end{bmatrix}$$

那么上式可写成

$$f(\boldsymbol{a} + \boldsymbol{h}) = f(\boldsymbol{a}) + Jf(\boldsymbol{a})\boldsymbol{h} + \frac{1}{2}\boldsymbol{h}^T Hf(\boldsymbol{a}) \boldsymbol{h} + \cdots$$

这里$Hf$称为$f$的Hesse方阵，它是一个$n$阶对称方阵。

定理6

设$D \subset \mathbb{R}^n$是一个凸区域，$f \in C^m(D)$，$\boldsymbol{a}$和$\boldsymbol{a}+\boldsymbol{h}$是$D$中的两个点，那么

$$f(\boldsymbol{a} + \boldsymbol{h}) = \sum_{k=0}^m \sum_{|\boldsymbol{\alpha}|=k} \frac{D^{\boldsymbol{\alpha}}f(\boldsymbol{a})}{\boldsymbol{\alpha}!} \boldsymbol{h}^{\boldsymbol{\alpha}} + o(\Vert \boldsymbol{h} \Vert^m) \quad (\boldsymbol{h} \to \boldsymbol{0})$$

证：由定理5可知

$$f(\boldsymbol{a} + \boldsymbol{h}) = \sum_{k=0}^{m-1} \sum_{|\boldsymbol{\alpha}|=k} \frac{D^{\boldsymbol{\alpha}}f(\boldsymbol{a})}{\boldsymbol{\alpha}!} \boldsymbol{h}^{\boldsymbol{\alpha}} + \sum_{|\boldsymbol{\alpha}|=m}\frac{D^{\boldsymbol{\alpha}}f(\boldsymbol{a} + \theta \boldsymbol{h})}{\boldsymbol{\alpha}!} \boldsymbol{h}^{\boldsymbol{\alpha}} \tag{2}$$

其中$\theta \in (0, 1)$，因为$f$的$m$阶偏导数连续，所以

$$\lim \limits_{\boldsymbol{h} \to \boldsymbol{0}} D^{\boldsymbol{\alpha}}f(\boldsymbol{a} + \theta \boldsymbol{h}) = D^{\boldsymbol{\alpha}}f(\boldsymbol{a}) \quad (|\boldsymbol{\alpha}|=m)$$

从而有

$$D^{\boldsymbol{\alpha}}f(\boldsymbol{a} + \theta \boldsymbol{h}) = D^{\boldsymbol{\alpha}}f(\boldsymbol{a}) + o(1) \quad (\boldsymbol{h} \to 0)$$

所以

$$\frac{D^{\boldsymbol{\alpha}}f(\boldsymbol{a} + \theta \boldsymbol{h})}{\boldsymbol{\alpha}!} \boldsymbol{h}^{\boldsymbol{\alpha}} = \frac{D^{\boldsymbol{\alpha}}f(\boldsymbol{a})}{\boldsymbol{\alpha}!} \boldsymbol{h}^{\boldsymbol{\alpha}} + o(\boldsymbol{h}^{\boldsymbol{\alpha}}) \quad (\boldsymbol{h} \to \boldsymbol{0})$$

当$|\boldsymbol{\alpha}|=m$时，有

$$|\boldsymbol{h}^{\boldsymbol{\alpha}}| = |h_1^{\alpha_1} \cdots h_n^{\alpha_n}| = |h_1|^{\alpha_1} \cdots |h_n|^{\alpha_n} \le \Vert \boldsymbol{h} \Vert^{m}$$

从而

$$\sum_{|\boldsymbol{\alpha}| = m}\frac{D^{\boldsymbol{\alpha}}f(\boldsymbol{a} + \theta \boldsymbol{h})}{\boldsymbol{\alpha}!} \boldsymbol{h}^{\boldsymbol{\alpha}} = \sum_{|\boldsymbol{\alpha}| = m}\frac{D^{\boldsymbol{\alpha}}f(\boldsymbol{a})}{\boldsymbol{\alpha}!} \boldsymbol{h}^{\boldsymbol{\alpha}} + o(\Vert \boldsymbol{h} \Vert^{m}) \quad (\boldsymbol{h} \to \boldsymbol{0})$$

将上式代入(2)式中，即证得命题成立。

Q.E.D.

定理7：拟微分平均值定理

设$\boldsymbol{f}: [a,b] \to \mathbb{R}^m$是$[a,b]$上的连续映射，在开区间$(a,b)$上可微，那么存在一点$\xi \in (a,b)$使得

$$\Vert \boldsymbol{f}(b) - \boldsymbol{f}(a) \Vert \le \Vert J\boldsymbol{f}(\xi) \Vert (b-a)$$

证：设$\boldsymbol{u} = \boldsymbol{f}(b) - \boldsymbol{f}(a)$，利用$\mathbb{R}^m$中的内积来定义函数

$$\varphi(t) = \left<\boldsymbol{u}, \boldsymbol{f}(t)\right> \quad (a \le t \le b)$$

易知$\varphi$是$[a,b]$上的连续函数，并在开区间$(a,b)$上可微，对$\varphi$使用微分中值定理，可知存在一点$\xi \in (a,b)$使得

$$\varphi(b) - \varphi(a) = (b-a)\varphi^\prime(\xi) = (b-a)\left<\boldsymbol{u}, J\boldsymbol{f}(\xi) \right>$$

而

$$\varphi(b) - \varphi(a) = \left< \boldsymbol{u}, \boldsymbol{f}(b) \right> - \left< \boldsymbol{u}, \boldsymbol{f}(a) \right> = \left< \boldsymbol{u}, \boldsymbol{f}(b) - \boldsymbol{f}(a)\right> = \left< \boldsymbol{u}, \boldsymbol{u} \right> = \Vert \boldsymbol{u} \Vert^2$$

由Cauchy-Schwarz不等式，可得

$$\Vert \boldsymbol{u} \Vert^2 = (b-a)\left< \boldsymbol{u}, J\boldsymbol{f}(\xi) \right> \le (b-a)\Vert \boldsymbol{u} \Vert \Vert J\boldsymbol{f}(\xi) \Vert$$

当$\boldsymbol{u} \ne \boldsymbol{0}$时，式子两边消去$\Vert \boldsymbol{u} \Vert$即得证原命题；若$\boldsymbol{u} = \boldsymbol{0}$，命题自然成立。

Q.E.D.

定理8

设凸区域$D \subset \mathbb{R}^n$，且映射$\boldsymbol{f}: D \to \mathbb{R}^m$在$D$上可微，则对任何$\boldsymbol{a},\boldsymbol{b} \in D$，在由$\boldsymbol{a}, \boldsymbol{b}$所决定的线段上必有一点$\boldsymbol{\xi}$，使得

$$\boldsymbol{f}(\boldsymbol{b}) - \boldsymbol{f}(\boldsymbol{a}) \le \Vert J\boldsymbol{f}(\boldsymbol{\xi}) \Vert \Vert \boldsymbol{b} - \boldsymbol{a}\Vert$$

证：由$\boldsymbol{a}$与$\boldsymbol{b}$所决定的线段可表示为

$$\boldsymbol{r}(t) = \boldsymbol{a} + t (\boldsymbol{b} - \boldsymbol{a}) \quad (0 \le t \le 1)$$

令

$$\boldsymbol{g}(t) = \boldsymbol{f} \circ \boldsymbol{r}(t)$$

映射$g$在$[0,1]$上连续，在$(0,1)$内可微，从而

$$J\boldsymbol{g}(t) = J\boldsymbol{f}(\boldsymbol{r}(t))(\boldsymbol{b} - \boldsymbol{a})$$

由定理7可知存在$\tau \in (0, 1)$，使得

$$\Vert \boldsymbol{g}(1) - \boldsymbol{g}(0)\Vert = \Vert J\boldsymbol{g}(\tau)\Vert$$

即

$$\Vert \boldsymbol{f}(\boldsymbol{b}) - \boldsymbol{f}(\boldsymbol{a}) \Vert \le \Vert J\boldsymbol{f}(\boldsymbol{r}(\tau))(\boldsymbol{b} - \boldsymbol{a}) \Vert$$

令$\boldsymbol{\xi} = \boldsymbol{r}(\tau)$，可得

$$\Vert \boldsymbol{f}(\boldsymbol{b}) - \boldsymbol{f}(\boldsymbol{a}) \Vert \le \Vert J\boldsymbol{f}(\boldsymbol{\xi})(\boldsymbol{b} - \boldsymbol{a}) \Vert \le \Vert J\boldsymbol{f}(\boldsymbol{\xi}) \Vert \Vert \boldsymbol{b} - \boldsymbol{a} \Vert$$

Q.E.D.