$f:\Omega\subseteq\mathbf{R}^n\to \mathbf{R}^m$ is differentiable at $a$ if and only if there is a function $\epsilon(x)$ so that for $x\in\Omega$ we have $$ f(x)=f(a)+T(x-a)+\epsilon(x)||x-a|| $$with $\epsilon(x)\to 0$ as $x\to a$.

 

 

Proof:

 

$\Leftarrow$:

\begin{equation}

||f(x)-f(a)-T(x-a)||=||x-a||||\epsilon(x)||

\end{equation}

When $x\to a$,$\epsilon(x)\to 0$,so $||\epsilon(x)||\to 0$.Thus

\begin{equation}

\lim_{x\to a;x\in\Omega\backslash\{a\}}\frac{||f(x)-f(a)-T(x-a)||}{||x-a||}=0

\end{equation}

So $f$ is differentiable at $a$.

 

 

$\Rightarrow$:

\begin{equation}

\lim_{x\to a;x\in\Omega\backslash\{a\}}\frac{||f(x)-f(a)-T(x-a)||}{||x-a||}=0

\end{equation}

means $\forall \varepsilon>0$,there exists $\delta>0$,such that for all $x'\in\{x\in\Omega\backslash\{a\}:||x-a||\leq \delta\}$,we have

\begin{equation}\label{eq:4}

\frac{||f(x')-f(a)-T(x'-a)||}{||x'-a||}\leq \varepsilon

\end{equation}

equation \ref{eq:4} is equivalent to

\begin{equation}\label{eq:5}

||f(x')-f(a)-T(x'-a)||\leq \varepsilon||x'-a||

\end{equation}

Let $p(x)=f(x)-f(a)-T(x-a)$.Then equation \ref{eq:5} becomes

\begin{equation}

||p(x')||\leq\varepsilon||x'-a||

\end{equation}

Now let $\epsilon(x)=\frac{p(x)}{||x-a||}$,then $||\epsilon(x')||\leq\varepsilon$.And it is easy to verify that $f(x)=f(a)+T(x-a)+\epsilon(x)||x-a||$.Done.$\Box$

 

注:T(x-a) 叫做全微分.它的意义是自变量进行微小变化后因变量随之而发生的改变.