derivative

Prerequisites

• 向量空间

Multivariate complex vector-valued functions

• Complex vector space: $<\mathbb{C},{\mathbb{C}}^n>$

Complex vector space can define inner product, which can induce norm, which can induce metric, which can induce topology.

• Multivariate complex-valued function $f:{\mathbb{C}}^n\to \mathbb{C}$

• Multivariate complex vector-valued function $f:{\mathbb{C}}^n\to {\mathbb{C}}^m$

adherent point Let $X$ be a topological space, $E$ be a subset of $X$, $x$ be a point of $X$. If every neighborhood $N_x$ of $x$ has $N_x\cap E\ne \varnothing$, then $x$ is called an adherent point of $E$. Equivalently, $x\in \overline{E}$ ($\overline{E}$ is the closure of $E$).

limit point Let $X$ be a topological space, $E$ be a subset of $X$, $x$ be a point of $X$. If every neighborhood $N_x$ of $x$ has $N_x\cap E\backslash \{x\}\ne \varnothing$, then $x$ is called a limit point of $E$.

limit point is also an adherent point.

Derivative

Differentiability at a point Let $X$ be a subset of ${\mathbb{C}}^n$, and let $x_0\in X$ be an element of $X$ which is also a limit point of $X$. Let $f:X\to {\mathbb{C}}^m$ be a function. Let $L$ be a linear transformation from ${\mathbb{C}}^n$ to ${\mathbb{C}}^m$. We say that $f$ is differentiable at $x_0$ on $X$ with derivative $L$ and write $f^{\prime }(x_0):=L$ if

$$\underset{x\to x_0,x\in X\setminus \{x_0\}}{\lim }\frac{d(f(x),f(x_0)+L(x-x_0))}{d(x,x_0)}=0$$

Uniqueness of derivatives Let $E$ be a subset of ${\mathbb{C}}^n$, $f:E\to {\mathbb{C}}^m$ be a function, $x_0\in E$ be an interior point of $E$, and let $L_1:{\mathbb{C}}^n\to {\mathbb{C}}^m$ and $L_2:{\mathbb{C}}^n\to {\mathbb{C}}^m$ be linear transformations. Suppose that $f$ is differentiable at $x_0$ with derivative $L_1$, and also differentiable at $x_0$ with derivative $L_2$. Then $L_1=L_2$.

proof:
because $L_1\ne L_2$, so there exists $v\in {\mathbb{C}}^n\wedge v\ne 0$, s.t. $L_1(v)\ne L_2(v)$.
$R_1(t):=f(x_0+tv)-f(x_0)-L_1(tv)$.
$R_2(t):=f(x_0+tv)-f(x_0)-L_2(tv)$.
$t(L_1(v)+L_2(v))=L_1(tv)-L_2(tv)=R_2(t)-R_1(t)$.
$\frac{||(L_1(v)-L_2(v))||}{||v||}=\frac{||t(L_1(v)-L_2(v))||}{||tv||}=\frac{||R_2(t)-R_1(t)||}{||tv||}\le \frac{||R_2(t)||}{||tv||}+\frac{||R_1(t)||}{||tv||}$.
so $L_1(v)-L_2(v)=0$, contradiction.