「数学分析」浅谈多元函数的高阶全微分

实际上是数分课上老师提的一嘴, 自己补充并完善了一下写出的文章（）

在数学分析 / 高等数学的学习中, 我们已经学会了多元函数的微分, 偏导, 高阶偏导和 \(C^{r}\) 类. 但大多数教材并没有介绍高阶全微分, 即 \(f^{(r)}(\boldsymbol{x})\) (也可记作 \(D^{r}f(\boldsymbol{x})\)). 在本文, 我将介绍多元函数的高阶全微分, 以及为什么大多数教材都不介绍多元函数的高阶全微分.

高阶微分的形式

设 \(E \in \mathbb{R}^{n}\), 映射 \(f \colon E \to \mathbb{R}^{m}\) 可微, 则有导映射 \(f' \colon E \to \mathscr{L}(\mathbb{R}^{n}, \mathbb{R}^{m})\) (其中 \(\mathscr{L}(\mathbb{R}^{n}, \mathbb{R}^{m})\) 表示 \(\mathbb{R}^{n}\) 到 \(\mathbb{R}^{m}\) 的线性映射所成之集, 也可记作 \(\operatorname{Hom}(\mathbb{R}^{n}, \mathbb{R}^{n})\)). 同理, 二阶导映射的形式为:

\[f'' \colon E \to \mathscr{L}(\mathbb{R}^{n}, \mathscr{L}(\mathbb{R}^{n}, \mathbb{R}^{m})).\]

三阶导映射的形式为:

\[f''' \colon E \to \mathscr{L}(\mathbb{R}^{n}, \mathscr{L}(\mathbb{R}^{n}, \mathscr{L}(\mathbb{R}^{n}, \mathbb{R}^{m}))).\]

可见该映射形态较为复杂, 在高等代数的视角下, 可以考虑利用同构进行简化. 下文以二阶微分为例, 说明与 \(\mathscr{L}(\mathbb{R}^{n}, \mathscr{L}(\mathbb{R}^{n}, \mathbb{R}^{m}))\) 同构的空间.

双线性映射

考虑如下定义.

定义 设 \(E, F, G\) 是三个线性空间, 用 \(\mathscr{L}(E, F; G)\) 表示全体定义在 \(E \times F\) 上, 取值于 \(G\) 上的双线性映射所组成的集合.

定义 (双线性映射) 设 \(u \colon E \times F \to G\), 如果

1. 对任意的 \(x \in E, \ y_{1}, y_{2} \in F, \ \alpha, \beta \in \mathbb{R}\), 均有 \[u(x, \alpha y_{1} + \beta y_{2}) = \alpha u(x, y_{1}) + \beta u(x, y_{2});\]
2. 对任意的 \(x_{1}, x_{2} \in E, \ y \in F, \ \alpha, \beta \in \mathbb{R}\), 均有 \[u(\alpha x_{1} + \beta x_{2}, y) = \alpha u(x_{1}, y) + \beta u(x_{2}, y);\] 则称 \(u\) 是双线性的.

双线性映射的一个经典例子是内积.

不难看出 \(\mathscr{L}(\mathbb{R}^{n}, \mathbb{R}^{n}; \mathbb{R}^{m})\) 是线性空间. 于是我们可得如下同构关系.

命题 \[\mathscr{L}(\mathbb{R}^{n}, \mathscr{L}(\mathbb{R}^{n}, \mathbb{R}^{m})) \cong \mathscr{L}(\mathbb{R}^{n}, \mathbb{R}^{n}; \mathbb{R}^{m}).\]

证明. 设 \(\forall u \in \mathscr{L}(\mathbb{R}^{n}, \mathbb{R}^{n}; \mathbb{R}^{m})\), 考虑构造同构映射.

对于 \(\forall \boldsymbol{x} \in \mathbb{R}^{n}\), 令 \(u_{\boldsymbol{x}} \colon \boldsymbol{y} \mapsto u(\boldsymbol{x}, \boldsymbol{y})\), 则 \(u_{\boldsymbol{x}} \in \mathscr{L}(\mathbb{R}^{n}, \mathbb{R}^{m})\).

再令 \(\widetilde{u} \colon \boldsymbol{x} \mapsto u_{\boldsymbol{x}}\), 只需说明 \(\widetilde{u} \in \mathscr{L}(\mathbb{R}^{n}, \mathscr{L}(\mathbb{R}^{n}, \mathbb{R}^{m}))\).

要证对于 \(\forall \boldsymbol{x}_{1}, \boldsymbol{x}_{2} \in \mathbb{R}^{n}, \ \alpha, \beta \in \mathbb{R}\), 有 \(\widetilde{u}(\alpha \boldsymbol{x}_{1} + \beta \boldsymbol{x}_{2}) = \alpha\widetilde{u}(\boldsymbol{x}_{1}) + \beta\widetilde{u}(\boldsymbol{x}_{2})\), 即 \(u_{\alpha \boldsymbol{x}_{1} + \beta \boldsymbol{x}_{2}} = \alpha u_{\boldsymbol{x}_{1}} + \beta u_{\boldsymbol{x}_{2}}\).

故只需证对于 \(\forall \boldsymbol{y} \in \mathbb{R}^{n}\), 有 \(u_{\alpha \boldsymbol{x}_{1} + \beta \boldsymbol{x}_{2}}(\boldsymbol{y}) = \alpha u_{\boldsymbol{x}_{1}}(\boldsymbol{y}) + \beta u_{\boldsymbol{x}_{2}}(\boldsymbol{y})\), 即 \(u(\alpha \boldsymbol{x}_{1} + \beta \boldsymbol{x}_{2}, \boldsymbol{y}) = \alpha u(\boldsymbol{x}_{1}, \boldsymbol{y}) + \beta u(\boldsymbol{x}_{2}, \boldsymbol{y})\).

由 \(u\) 的双线性性得上述等式成立.

令 \(g \colon u \mapsto \widetilde{u}\), 不难验证 \(g\) 是同构映射, 故有 \(\mathscr{L}(\mathbb{R}^{n}, \mathscr{L}(\mathbb{R}^{n}, \mathbb{R}^{m})) \cong \mathscr{L}(\mathbb{R}^{n}, \mathbb{R}^{n}; \mathbb{R}^{m})\).

Q.E.D.

于是研究 \(\mathscr{L}(\mathbb{R}^{n}, \mathscr{L}(\mathbb{R}^{n}, \mathbb{R}^{m}))\) 可转化为研究 \(\mathscr{L}(\mathbb{R}^{n}, \mathbb{R}^{n}; \mathbb{R}^{m})\). 但由于有连续这个概念, 需要定义合适的距离, 使得该同构是保持距离不变的同构. 在此需要定义范数.

定义 (线性映射的范数) 设 \(L \in \mathscr{L}(\mathbb{R}^{n}, \mathbb{R}^{m})\), 我们称 \[\|L\| = \sup_{|\boldsymbol{h}| = 1} |L\boldsymbol{h}|\] 为 \(L\) 的范数.

定义 (双线性映射的范数) 设 \(u\) 是双线性映射, 我们称 \[\|u\| = \sup_{|\boldsymbol{x}| = 1, |\boldsymbol{y}| = 1} |u(\boldsymbol{x}, \boldsymbol{y})|\] 为 \(u\) 的范数.

利用范数可定义 \(u, v\) 之间的距离为 \(\|u - v\|\). 于是通过下列推导可得该同构保持距离.

命题 对于上述定义的同构映射 \(g \colon u \mapsto \widetilde{u}\), 有 \[\|u\| = \|\widetilde{u}\|.\]

证明. 由上述定义可得

\[ \begin{align*} \|u\| &= \sup_{|\boldsymbol{x}| = 1, |\boldsymbol{y}| = 1} |u(\boldsymbol{x}, \boldsymbol{y})|. \\ \|\widetilde{u}\| &= \sup_{|\boldsymbol{x}| = 1} \|\widetilde{u}(\boldsymbol{x})\| \\ &= \sup_{|\boldsymbol{x}| = 1} \|u_{\boldsymbol{x}}\| \\ &= \sup_{|\boldsymbol{x}| = 1} \sup_{|\boldsymbol{y}| = 1} |u_{\boldsymbol{x}}(\boldsymbol{y})| \\ &= \sup_{|\boldsymbol{x}| = 1} \sup_{|\boldsymbol{y}| = 1} |u(\boldsymbol{x}, \boldsymbol{y})|. \end{align*} \]

不难看出 \(\|u\| \geq \|\widetilde{u}\|\), 下证 \(\|u\| \leq \|\widetilde{u}\|\). 对 \(\forall|\boldsymbol{x}| = 1, \boldsymbol{x} \in \mathbb{R}^{n}\), 有

\[\|\widetilde{u}\| \geq \sup_{|\boldsymbol{y}| = 1} |u(\boldsymbol{x}, \boldsymbol{y})|.\]

则对 \(\forall|\boldsymbol{y}| = 1, \boldsymbol{y} \in \mathbb{R}^{n}\), 有

\[\|\widetilde{u}\| \geq |u(\boldsymbol{x}, \boldsymbol{y})|.\]

即对 \(\forall |\boldsymbol{x}| = 1, |\boldsymbol{y}| = 1, \ \boldsymbol{x}, \boldsymbol{y} \in \mathbb{R}^{n}\), 均有 \(\|\widetilde{u}\| \geq |u(\boldsymbol{x}, \boldsymbol{y})|\).

又对 \(\forall \varepsilon > 0\), 均 \(\exists |\boldsymbol{x}'| = 1, |\boldsymbol{y}'| = 1, \ \boldsymbol{x}', \boldsymbol{y}' \in \mathbb{R}^{n}\), 使得

\[\|u\| < |u(\boldsymbol{x}', \boldsymbol{y}')| + \varepsilon \leq \|\widetilde{u}\| + \varepsilon.\]

故由 \(\varepsilon\) 的任意性知 \(\|u\| \leq \|\widetilde{u}\|\).

Q.E.D.

于是该同构是保持距离的同构, 该问题转化为了讨论双线性型的性质.

与 \(C^{2}\) 类等价

\(C^{r}(E)\) 定义如下.

定义 (\(C^{r}\) 类) 如果 \(n\) 元函数 \(f\) 的每个阶数不大于 \(r\) 的偏导均在 \(E\) 上存在且连续, 我们就称 \(f\) 在 \(E\) 上是 \(C^{r}\) 类的, 记作 \(f \in C^{r}(E)\).

我们先讨论 \(f'\) 的情况.

命题 设 \(E\) 是 \(\mathbb{R}^{n}\) 的开集, \(f \colon E \to \mathbb{R}^{m}\) 在 \(E\) 上可微. 则 \(f \in C^{1}(E)\) 的充要条件是: 对 \(\forall \boldsymbol{a} \in E, \ \varepsilon > 0\), 均 \(\exists \delta > 0\), 使得对于满足 \(|\boldsymbol{x} - \boldsymbol{a}| < \delta\) 的任意的 \(\boldsymbol{x} \in E\) 均有 \[\|f'(\boldsymbol{x}) - f'(\boldsymbol{a})\| < \varepsilon.\]

证明.

必要性: 设 \(f = (f_{1}, f_{2}, \cdots, f_{m})^{\text{T}}\), 有

\[ f'(\boldsymbol{a}) = \begin{bmatrix} \frac{ \partial f_{1} }{ \partial x_{1} }(\boldsymbol{a}) & \frac{ \partial f_{1} }{ \partial x_{2} }(\boldsymbol{a}) & \cdots & \frac{ \partial f_{1} }{ \partial x_{n} }(\boldsymbol{a}) \\ \frac{ \partial f_{2} }{ \partial x_{1} }(\boldsymbol{a}) & \frac{ \partial f_{2} }{ \partial x_{2} }(\boldsymbol{a}) & \cdots & \frac{ \partial f_{2} }{ \partial x_{n} }(\boldsymbol{a}) \\ \vdots & \vdots && \vdots \\ \frac{ \partial f_{m} }{ \partial x_{1} }(\boldsymbol{a}) & \frac{ \partial f_{m} }{ \partial x_{2} }(\boldsymbol{a}) & \cdots & \frac{ \partial f_{m} }{ \partial x_{n} }(\boldsymbol{a}) \end{bmatrix}. \]

故有

\[ f'(\boldsymbol{x}) - f'(\boldsymbol{a}) = \begin{bmatrix} \frac{ \partial f_{1} }{ \partial x_{1} }(\boldsymbol{x}) - \frac{ \partial f_{1} }{ \partial x_{1} }(\boldsymbol{a}) & \frac{ \partial f_{1} }{ \partial x_{2} }(\boldsymbol{x}) - \frac{ \partial f_{1} }{ \partial x_{2} }(\boldsymbol{a}) & \cdots & \frac{ \partial f_{1} }{ \partial x_{n} }(\boldsymbol{x}) - \frac{ \partial f_{1} }{ \partial x_{n} }(\boldsymbol{a}) \\ \frac{ \partial f_{2} }{ \partial x_{1} }(\boldsymbol{x}) - \frac{ \partial f_{2} }{ \partial x_{1} }(\boldsymbol{a}) & \frac{ \partial f_{2} }{ \partial x_{2} }(\boldsymbol{x}) - \frac{ \partial f_{2} }{ \partial x_{2} }(\boldsymbol{a}) & \cdots & \frac{ \partial f_{2} }{ \partial x_{n} }(\boldsymbol{x}) - \frac{ \partial f_{2} }{ \partial x_{n} }(\boldsymbol{a}) \\ \vdots & \vdots && \vdots \\ \frac{ \partial f_{m} }{ \partial x_{1} }(\boldsymbol{x}) - \frac{ \partial f_{m} }{ \partial x_{1} }(\boldsymbol{a}) & \frac{ \partial f_{m} }{ \partial x_{2} }(\boldsymbol{x}) - \frac{ \partial f_{m} }{ \partial x_{2} }(\boldsymbol{a}) & \cdots & \frac{ \partial f_{m} }{ \partial x_{n} }(\boldsymbol{x}) - \frac{ \partial f_{m} }{ \partial x_{n} }(\boldsymbol{a}) \end{bmatrix}. \]

又有

\[ \|f'(\boldsymbol{x}) - f'(\boldsymbol{a})\| = \sup_{|\boldsymbol{y}| = 1} |(f'(\boldsymbol{x}) - f'(\boldsymbol{a}))\boldsymbol{y}|. \]

且 \(|(f'(\boldsymbol{x}) - f'(\boldsymbol{a}))\boldsymbol{y}| \leq M\sqrt{ mn } \cdot |\boldsymbol{y}| = M\sqrt{ mn }\), 其中

\[ M = \max\left\{ \left| \frac{ \partial f_{i} }{ \partial x_{j} }(\boldsymbol{x}) - \frac{ \partial f_{i} }{ \partial x_{j} }(\boldsymbol{a}) \right| : 1 \leq i \leq m, 1 \leq j \leq n \right\}. \]

又 \(f \in C^{1}(E)\), 则对 \(\forall \varepsilon > 0, \ 1 \leq i \leq m, 1 \leq j \leq n\), 均 \(\exists \delta > 0\), 使得对于满足 \(|\boldsymbol{x} - \boldsymbol{a}| < \delta\) 的任意的 \(\boldsymbol{x} \in E\) 均有

\[\left| \frac{ \partial f_{i} }{ \partial x_{j} }(\boldsymbol{x}) - \frac{ \partial f_{i} }{ \partial x_{j} }(\boldsymbol{a}) \right| < \frac{\varepsilon}{\sqrt{ mn }}.\]

故有

\[\|f'(\boldsymbol{x}) - f'(\boldsymbol{a})\| = \sup_{|\boldsymbol{y}| = 1} |(f'(\boldsymbol{x}) - f'(\boldsymbol{a}))\boldsymbol{y}| \leq M\sqrt{ mn } < \frac{\varepsilon}{\sqrt{ mn }} \cdot \sqrt{ mn } = \varepsilon.\]

充分性: 若对 \(\forall \boldsymbol{a} \in E, \ \varepsilon > 0\), 均 \(\exists \delta > 0\), 使得对于满足 \(|\boldsymbol{x} - \boldsymbol{a}| < \delta\) 的任意的 \(\boldsymbol{x} \in E\) 均有 \(\|f'(\boldsymbol{x}) - f'(\boldsymbol{a})\| < \varepsilon\), 则对任意的 \(1 \leq i \leq m, 1 \leq j \leq n\), 有

\[ \begin{align*} \left| \frac{ \partial f_{i} }{ \partial x_{j} }(\boldsymbol{x}) - \frac{ \partial f_{i} }{ \partial x_{j} }(\boldsymbol{a}) \right| &\leq \sqrt{ \sum_{k=1}^{m} \left( \frac{ \partial f_{k} }{ \partial x_{j} }(\boldsymbol{x}) - \frac{ \partial f_{k} }{ \partial x_{j} }(\boldsymbol{a}) \right)^2 } \\ &= |(f'(\boldsymbol{x}) - f'(\boldsymbol{a}))\boldsymbol{e}_{j}| \\ &\leq \|f'(\boldsymbol{x}) - f'(\boldsymbol{a})\| \\ &< \varepsilon. \end{align*} \]

其中 \(\boldsymbol{e}_i\) 为坐标向量. 则对 \(\forall \boldsymbol{a} \in E, \ 1 \leq i \leq m, 1 \leq j \leq n\), 均有 \(\frac{ \partial f_{i} }{ \partial x_{j} }(\boldsymbol{a})\) 连续, 即 \(f \in C^{1}(E)\).

Q.E.D.

接下来讨论 \(f''\) 的情况.

命题 设 \(E\) 是 \(\mathbb{R}^{n}\) 的开集, \(f \colon E \to \mathbb{R}^{m}\) 在 \(E\) 上二次可微. 则 \(f \in C^{2}(E)\) 的充要条件是: 对 \(\forall \boldsymbol{a} \in E, \ \varepsilon > 0\), 均 \(\exists \delta > 0\), 使得对于满足 \(|\boldsymbol{x} - \boldsymbol{a}| < \delta\) 的任意的 \(\boldsymbol{x} \in E\) 均有 \[\|f''(\boldsymbol{x}) - f''(\boldsymbol{a})\| < \varepsilon.\]

证明.

必要性: 有

\[\|f''(\boldsymbol{x}) - f''(\boldsymbol{a})\| = \sup_{|\boldsymbol{h}| = 1, |\boldsymbol{k}| = 1} |(f''(\boldsymbol{x}) - f''(\boldsymbol{a}))(\boldsymbol{h}, \boldsymbol{k})|.\]

又有

\[f''(\boldsymbol{a})\boldsymbol{e}_{i} = \lim_{ t \to 0 } \frac{f'(\boldsymbol{a} + t\boldsymbol{e}_{i}) - f'(\boldsymbol{a})}{t}.\]

故

\[ \begin{align*} f''(\boldsymbol{a})(\boldsymbol{e}_{i}, \boldsymbol{e}_{j}) &= (f''(\boldsymbol{a})\boldsymbol{e}_{i})\boldsymbol{e}_{j} \\ &= \lim_{ t \to 0 } \frac{f'(\boldsymbol{a} + t\boldsymbol{e}_{i})\boldsymbol{e}_{j} - f'(\boldsymbol{a})\boldsymbol{e}_{j}}{t} \\ &= \lim_{ t \to 0 } \frac{\frac{ \partial f }{ \partial x_{j} }(\boldsymbol{a} + t\boldsymbol{e}_{i}) - \frac{ \partial f }{ \partial x_{j} }(\boldsymbol{a})}{t} \\ &= \frac{ \partial f }{ \partial x_{i} \partial x_{j} }(\boldsymbol{a}). \end{align*} \]

则对 \(\forall \boldsymbol{h} = (h_{1}, \cdots, h_{n})^{\text{T}}, \boldsymbol{k} = (k_{1}, \cdots, k_{n})^{\text{T}}\), 由 \(f''(\boldsymbol{a})\) 的双线性性得

\[f''(\boldsymbol{a})(\boldsymbol{h}, \boldsymbol{k}) = \sum_{i=1}^{n} \sum_{j=1}^{n} h_{i}k_{j}f''(\boldsymbol{a})(\boldsymbol{e}_{i}, \boldsymbol{e}_{j}) = \sum_{i=1}^{n} \sum_{j=1}^{n} h_{i}k_{j}\frac{ \partial f }{ \partial x_{i} \partial x_{j} }(\boldsymbol{a}).\]

故有

\[(f''(\boldsymbol{x}) - f''(\boldsymbol{a}))(\boldsymbol{h}, \boldsymbol{k}) = \sum_{i=1}^{n} \sum_{j=1}^{n} h_{i}k_{j}\left( \frac{ \partial f }{ \partial x_{i} \partial x_{j} }(\boldsymbol{x}) - \frac{ \partial f }{ \partial x_{i} \partial x_{j} }(\boldsymbol{a}) \right).\]

则当 \(|\boldsymbol{h}| = |\boldsymbol{k}| = 1\) 时, 有 \(|h_{i}|, |k_{j}| \leq 1, \ 1 \leq i, j \leq n\), 故 \(|h_{i}k_{j}| \leq 1\), 有

\[ \begin{align*} (f''(\boldsymbol{x}) - f''(\boldsymbol{a}))(\boldsymbol{h}, \boldsymbol{k}) &\leq \sum_{i=1}^{n} \sum_{j=1}^{n} \left| h_{i}k_{j}\left( \frac{ \partial f }{ \partial x_{i} \partial x_{j} }(\boldsymbol{x}) - \frac{ \partial f }{ \partial x_{i} \partial x_{j} }(\boldsymbol{a}) \right) \right| \\ &\leq \sum_{i=1}^{n} \sum_{j=1}^{n} \left| \frac{ \partial f }{ \partial x_{i} \partial x_{j} }(\boldsymbol{x}) - \frac{ \partial f }{ \partial x_{i} \partial x_{j} }(\boldsymbol{a}) \right|. \end{align*} \]

又 \(f \in C^{2}(E)\), 则对 \(\forall \varepsilon > 0, \ 1 \leq k \leq m, \ 1 \leq i, j \leq n\), 均 \(\exists \delta > 0\), 使得对于满足 \(|\boldsymbol{x} - \boldsymbol{a}| < \delta\) 的任意的 \(\boldsymbol{x} \in E\) 均有

\[\left| \frac{ \partial f_{k} }{ \partial x_{i} \partial x_{j} }(\boldsymbol{x}) - \frac{ \partial f_{k} }{ \partial x_{i} \partial x_{j} }(\boldsymbol{a}) \right| < \frac{\varepsilon}{n^{2}\sqrt{ m }}.\]

故有

\[\left| \frac{ \partial f }{ \partial x_{i} \partial x_{j} }(\boldsymbol{x}) - \frac{ \partial f }{ \partial x_{i} \partial x_{j} }(\boldsymbol{a}) \right| = \sqrt{ \sum_{k=1}^{m} \left( \frac{ \partial f_{k} }{ \partial x_{i} \partial x_{j} }(\boldsymbol{x}) - \frac{ \partial f_{k} }{ \partial x_{i} \partial x_{j} }(\boldsymbol{a}) \right)^2 } < \frac{\varepsilon}{n^{2}}.\]

故有

\[ \begin{align*} \|f''(\boldsymbol{x}) - f''(\boldsymbol{a})\| &= \sup_{|\boldsymbol{h}| = 1, |\boldsymbol{k}| = 1} |(f''(\boldsymbol{x}) - f''(\boldsymbol{a}))(\boldsymbol{h}, \boldsymbol{k})| \\ &\leq \sum_{i=1}^{n} \sum_{j=1}^{n} \left| \frac{ \partial f }{ \partial x_{i} \partial x_{j} }(\boldsymbol{x}) - \frac{ \partial f }{ \partial x_{i} \partial x_{j} }(\boldsymbol{a}) \right| \\ &< \varepsilon. \end{align*} \]

充分性: 若对 \(\forall \boldsymbol{a} \in E, \ \varepsilon > 0\), 均 \(\exists \delta > 0\), 使得对于满足 \(|\boldsymbol{x} - \boldsymbol{a}| < \delta\) 的任意的 \(\boldsymbol{x} \in E\) 均有 \(\|f''(\boldsymbol{x}) - f''(\boldsymbol{a})\| < \varepsilon\), 则对任意的 \(1 \leq k \leq m, \ 1 \leq i, j \leq n\), 有

\[ \begin{align*} \left| \frac{ \partial f_{k} }{ \partial x_{i} \partial x_{j} }(\boldsymbol{x}) - \frac{ \partial f_{k} }{ \partial x_{i} \partial x_{j} }(\boldsymbol{a}) \right| &\leq \sqrt{ \sum_{t=1}^{m} \left( \frac{ \partial f_{t} }{ \partial x_{i} \partial x_{j} }(\boldsymbol{x}) - \frac{ \partial f_{t} }{ \partial x_{i} \partial x_{j} }(\boldsymbol{a}) \right)^2 } \\ &= |(f''(\boldsymbol{x}) - f''(\boldsymbol{a}))(\boldsymbol{e}_{i}, \boldsymbol{e}_{j})| \\ &\leq \|f''(\boldsymbol{x}) - f''(\boldsymbol{a})\| \\ &< \varepsilon. \end{align*} \]

则对 \(\forall \boldsymbol{a} \in E, \ 1 \leq k \leq m, \ 1 \leq i, j \leq n\), 均有 \(\frac{ \partial f_{k} }{ \partial x_{i} \partial x_{j} }(\boldsymbol{a})\) 连续, 即 \(f \in C^{2}(E)\).

Q.E.D.

于是我们证明了二阶微分连续等价于 \(C^{2}\) 类. 由于高阶微分过于复杂且使用繁琐, 我们可用 \(C^{2}\) 类替代这种复杂定义. 这也是大多数教材不介绍高阶微分的原因.

这种做法可拓展至 \(r\) 阶微分, 替换成 \(r\) 重线性映射即可 (多重线性映射的一个经典例子是行列式, 其按行 / 列展开体现了其多重线性性).